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|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
931
|
B
|
World Cup
|
PROGRAMMING
| 1,200
|
[
"constructive algorithms",
"implementation"
] | null | null |
The last stage of Football World Cup is played using the play-off system.
There are *n* teams left in this stage, they are enumerated from 1 to *n*. Several rounds are held, in each round the remaining teams are sorted in the order of their ids, then the first in this order plays with the second, the third — with the fourth, the fifth — with the sixth, and so on. It is guaranteed that in each round there is even number of teams. The winner of each game advances to the next round, the loser is eliminated from the tournament, there are no draws. In the last round there is the only game with two remaining teams: the round is called the Final, the winner is called the champion, and the tournament is over.
Arkady wants his two favorite teams to play in the Final. Unfortunately, the team ids are already determined, and it may happen that it is impossible for teams to meet in the Final, because they are to meet in some earlier stage, if they are strong enough. Determine, in which round the teams with ids *a* and *b* can meet.
|
The only line contains three integers *n*, *a* and *b* (2<=≤<=*n*<=≤<=256, 1<=≤<=*a*,<=*b*<=≤<=*n*) — the total number of teams, and the ids of the teams that Arkady is interested in.
It is guaranteed that *n* is such that in each round an even number of team advance, and that *a* and *b* are not equal.
|
In the only line print "Final!" (without quotes), if teams *a* and *b* can meet in the Final.
Otherwise, print a single integer — the number of the round in which teams *a* and *b* can meet. The round are enumerated from 1.
|
[
"4 1 2\n",
"8 2 6\n",
"8 7 5\n"
] |
[
"1\n",
"Final!\n",
"2\n"
] |
In the first example teams 1 and 2 meet in the first round.
In the second example teams 2 and 6 can only meet in the third round, which is the Final, if they win all their opponents in earlier rounds.
In the third example the teams with ids 7 and 5 can meet in the second round, if they win their opponents in the first round.
| 1,000
|
[
{
"input": "4 1 2",
"output": "1"
},
{
"input": "8 2 6",
"output": "Final!"
},
{
"input": "8 7 5",
"output": "2"
},
{
"input": "128 30 98",
"output": "Final!"
},
{
"input": "256 128 256",
"output": "Final!"
},
{
"input": "256 2 127",
"output": "7"
},
{
"input": "2 1 2",
"output": "Final!"
},
{
"input": "2 2 1",
"output": "Final!"
},
{
"input": "4 1 3",
"output": "Final!"
},
{
"input": "4 1 4",
"output": "Final!"
},
{
"input": "4 2 1",
"output": "1"
},
{
"input": "4 2 3",
"output": "Final!"
},
{
"input": "4 2 4",
"output": "Final!"
},
{
"input": "4 3 1",
"output": "Final!"
},
{
"input": "4 3 2",
"output": "Final!"
},
{
"input": "4 3 4",
"output": "1"
},
{
"input": "4 4 1",
"output": "Final!"
},
{
"input": "4 4 2",
"output": "Final!"
},
{
"input": "4 4 3",
"output": "1"
},
{
"input": "8 8 7",
"output": "1"
},
{
"input": "8 8 5",
"output": "2"
},
{
"input": "8 8 1",
"output": "Final!"
},
{
"input": "16 4 3",
"output": "1"
},
{
"input": "16 2 4",
"output": "2"
},
{
"input": "16 14 11",
"output": "3"
},
{
"input": "16 3 11",
"output": "Final!"
},
{
"input": "32 10 9",
"output": "1"
},
{
"input": "32 25 28",
"output": "2"
},
{
"input": "32 22 18",
"output": "3"
},
{
"input": "32 17 25",
"output": "4"
},
{
"input": "32 18 3",
"output": "Final!"
},
{
"input": "64 40 39",
"output": "1"
},
{
"input": "64 60 58",
"output": "2"
},
{
"input": "64 34 37",
"output": "3"
},
{
"input": "64 26 24",
"output": "4"
},
{
"input": "64 50 43",
"output": "5"
},
{
"input": "64 17 42",
"output": "Final!"
},
{
"input": "128 116 115",
"output": "1"
},
{
"input": "128 35 33",
"output": "2"
},
{
"input": "128 61 59",
"output": "3"
},
{
"input": "128 116 123",
"output": "4"
},
{
"input": "128 17 15",
"output": "5"
},
{
"input": "128 124 77",
"output": "6"
},
{
"input": "128 4 80",
"output": "Final!"
},
{
"input": "256 224 223",
"output": "1"
},
{
"input": "256 24 22",
"output": "2"
},
{
"input": "256 199 196",
"output": "3"
},
{
"input": "256 148 159",
"output": "4"
},
{
"input": "256 178 166",
"output": "5"
},
{
"input": "256 75 97",
"output": "6"
},
{
"input": "256 185 200",
"output": "7"
},
{
"input": "256 3 238",
"output": "Final!"
},
{
"input": "256 128 129",
"output": "Final!"
},
{
"input": "256 255 129",
"output": "7"
},
{
"input": "256 255 128",
"output": "Final!"
},
{
"input": "256 129 256",
"output": "7"
},
{
"input": "128 98 69",
"output": "6"
},
{
"input": "128 47 83",
"output": "Final!"
},
{
"input": "16 2 3",
"output": "2"
},
{
"input": "64 32 30",
"output": "2"
},
{
"input": "8 4 5",
"output": "Final!"
},
{
"input": "8 7 8",
"output": "1"
},
{
"input": "8 2 3",
"output": "2"
},
{
"input": "8 2 5",
"output": "Final!"
}
] | 1,520,179,288
| 1,588
|
Python 3
|
OK
|
TESTS
| 64
| 77
| 5,632,000
|
arr = input().split(' ')
n = int(arr[0])
a = int(arr[1])
b = int(arr[2])
a -= 1
b -= 1
ra = a//2
rb = b//2
c = 1
while ra!=rb:
ra = ra//2
rb = rb//2
c += 1
if(pow(2,c)==n): print("Final!")
else: print(c)
|
Title: World Cup
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The last stage of Football World Cup is played using the play-off system.
There are *n* teams left in this stage, they are enumerated from 1 to *n*. Several rounds are held, in each round the remaining teams are sorted in the order of their ids, then the first in this order plays with the second, the third — with the fourth, the fifth — with the sixth, and so on. It is guaranteed that in each round there is even number of teams. The winner of each game advances to the next round, the loser is eliminated from the tournament, there are no draws. In the last round there is the only game with two remaining teams: the round is called the Final, the winner is called the champion, and the tournament is over.
Arkady wants his two favorite teams to play in the Final. Unfortunately, the team ids are already determined, and it may happen that it is impossible for teams to meet in the Final, because they are to meet in some earlier stage, if they are strong enough. Determine, in which round the teams with ids *a* and *b* can meet.
Input Specification:
The only line contains three integers *n*, *a* and *b* (2<=≤<=*n*<=≤<=256, 1<=≤<=*a*,<=*b*<=≤<=*n*) — the total number of teams, and the ids of the teams that Arkady is interested in.
It is guaranteed that *n* is such that in each round an even number of team advance, and that *a* and *b* are not equal.
Output Specification:
In the only line print "Final!" (without quotes), if teams *a* and *b* can meet in the Final.
Otherwise, print a single integer — the number of the round in which teams *a* and *b* can meet. The round are enumerated from 1.
Demo Input:
['4 1 2\n', '8 2 6\n', '8 7 5\n']
Demo Output:
['1\n', 'Final!\n', '2\n']
Note:
In the first example teams 1 and 2 meet in the first round.
In the second example teams 2 and 6 can only meet in the third round, which is the Final, if they win all their opponents in earlier rounds.
In the third example the teams with ids 7 and 5 can meet in the second round, if they win their opponents in the first round.
|
```python
arr = input().split(' ')
n = int(arr[0])
a = int(arr[1])
b = int(arr[2])
a -= 1
b -= 1
ra = a//2
rb = b//2
c = 1
while ra!=rb:
ra = ra//2
rb = rb//2
c += 1
if(pow(2,c)==n): print("Final!")
else: print(c)
```
| 3
|
|
69
|
A
|
Young Physicist
|
PROGRAMMING
| 1,000
|
[
"implementation",
"math"
] |
A. Young Physicist
|
2
|
256
|
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
|
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100).
|
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
|
[
"3\n4 1 7\n-2 4 -1\n1 -5 -3\n",
"3\n3 -1 7\n-5 2 -4\n2 -1 -3\n"
] |
[
"NO",
"YES"
] |
none
| 500
|
[
{
"input": "3\n4 1 7\n-2 4 -1\n1 -5 -3",
"output": "NO"
},
{
"input": "3\n3 -1 7\n-5 2 -4\n2 -1 -3",
"output": "YES"
},
{
"input": "10\n21 32 -46\n43 -35 21\n42 2 -50\n22 40 20\n-27 -9 38\n-4 1 1\n-40 6 -31\n-13 -2 34\n-21 34 -12\n-32 -29 41",
"output": "NO"
},
{
"input": "10\n25 -33 43\n-27 -42 28\n-35 -20 19\n41 -42 -1\n49 -39 -4\n-49 -22 7\n-19 29 41\n8 -27 -43\n8 34 9\n-11 -3 33",
"output": "NO"
},
{
"input": "10\n-6 21 18\n20 -11 -8\n37 -11 41\n-5 8 33\n29 23 32\n30 -33 -11\n39 -49 -36\n28 34 -49\n22 29 -34\n-18 -6 7",
"output": "NO"
},
{
"input": "10\n47 -2 -27\n0 26 -14\n5 -12 33\n2 18 3\n45 -30 -49\n4 -18 8\n-46 -44 -41\n-22 -10 -40\n-35 -21 26\n33 20 38",
"output": "NO"
},
{
"input": "13\n-3 -36 -46\n-11 -50 37\n42 -11 -15\n9 42 44\n-29 -12 24\n3 9 -40\n-35 13 50\n14 43 18\n-13 8 24\n-48 -15 10\n50 9 -50\n21 0 -50\n0 0 -6",
"output": "YES"
},
{
"input": "14\n43 23 17\n4 17 44\n5 -5 -16\n-43 -7 -6\n47 -48 12\n50 47 -45\n2 14 43\n37 -30 15\n4 -17 -11\n17 9 -45\n-50 -3 -8\n-50 0 0\n-50 0 0\n-16 0 0",
"output": "YES"
},
{
"input": "13\n29 49 -11\n38 -11 -20\n25 1 -40\n-11 28 11\n23 -19 1\n45 -41 -17\n-3 0 -19\n-13 -33 49\n-30 0 28\n34 17 45\n-50 9 -27\n-50 0 0\n-37 0 0",
"output": "YES"
},
{
"input": "12\n3 28 -35\n-32 -44 -17\n9 -25 -6\n-42 -22 20\n-19 15 38\n-21 38 48\n-1 -37 -28\n-10 -13 -50\n-5 21 29\n34 28 50\n50 11 -49\n34 0 0",
"output": "YES"
},
{
"input": "37\n-64 -79 26\n-22 59 93\n-5 39 -12\n77 -9 76\n55 -86 57\n83 100 -97\n-70 94 84\n-14 46 -94\n26 72 35\n14 78 -62\n17 82 92\n-57 11 91\n23 15 92\n-80 -1 1\n12 39 18\n-23 -99 -75\n-34 50 19\n-39 84 -7\n45 -30 -39\n-60 49 37\n45 -16 -72\n33 -51 -56\n-48 28 5\n97 91 88\n45 -82 -11\n-21 -15 -90\n-53 73 -26\n-74 85 -90\n-40 23 38\n100 -13 49\n32 -100 -100\n0 -100 -70\n0 -100 0\n0 -100 0\n0 -100 0\n0 -100 0\n0 -37 0",
"output": "YES"
},
{
"input": "4\n68 3 100\n68 21 -100\n-100 -24 0\n-36 0 0",
"output": "YES"
},
{
"input": "33\n-1 -46 -12\n45 -16 -21\n-11 45 -21\n-60 -42 -93\n-22 -45 93\n37 96 85\n-76 26 83\n-4 9 55\n7 -52 -9\n66 8 -85\n-100 -54 11\n-29 59 74\n-24 12 2\n-56 81 85\n-92 69 -52\n-26 -97 91\n54 59 -51\n58 21 -57\n7 68 56\n-47 -20 -51\n-59 77 -13\n-85 27 91\n79 60 -56\n66 -80 5\n21 -99 42\n-31 -29 98\n66 93 76\n-49 45 61\n100 -100 -100\n100 -100 -100\n66 -75 -100\n0 0 -100\n0 0 -87",
"output": "YES"
},
{
"input": "3\n1 2 3\n3 2 1\n0 0 0",
"output": "NO"
},
{
"input": "2\n5 -23 12\n0 0 0",
"output": "NO"
},
{
"input": "1\n0 0 0",
"output": "YES"
},
{
"input": "1\n1 -2 0",
"output": "NO"
},
{
"input": "2\n-23 77 -86\n23 -77 86",
"output": "YES"
},
{
"input": "26\n86 7 20\n-57 -64 39\n-45 6 -93\n-44 -21 100\n-11 -49 21\n73 -71 -80\n-2 -89 56\n-65 -2 7\n5 14 84\n57 41 13\n-12 69 54\n40 -25 27\n-17 -59 0\n64 -91 -30\n-53 9 42\n-54 -8 14\n-35 82 27\n-48 -59 -80\n88 70 79\n94 57 97\n44 63 25\n84 -90 -40\n-100 100 -100\n-92 100 -100\n0 10 -100\n0 0 -82",
"output": "YES"
},
{
"input": "42\n11 27 92\n-18 -56 -57\n1 71 81\n33 -92 30\n82 83 49\n-87 -61 -1\n-49 45 49\n73 26 15\n-22 22 -77\n29 -93 87\n-68 44 -90\n-4 -84 20\n85 67 -6\n-39 26 77\n-28 -64 20\n65 -97 24\n-72 -39 51\n35 -75 -91\n39 -44 -8\n-25 -27 -57\n91 8 -46\n-98 -94 56\n94 -60 59\n-9 -95 18\n-53 -37 98\n-8 -94 -84\n-52 55 60\n15 -14 37\n65 -43 -25\n94 12 66\n-8 -19 -83\n29 81 -78\n-58 57 33\n24 86 -84\n-53 32 -88\n-14 7 3\n89 97 -53\n-5 -28 -91\n-100 100 -6\n-84 100 0\n0 100 0\n0 70 0",
"output": "YES"
},
{
"input": "3\n96 49 -12\n2 -66 28\n-98 17 -16",
"output": "YES"
},
{
"input": "5\n70 -46 86\n-100 94 24\n-27 63 -63\n57 -100 -47\n0 -11 0",
"output": "YES"
},
{
"input": "18\n-86 -28 70\n-31 -89 42\n31 -48 -55\n95 -17 -43\n24 -95 -85\n-21 -14 31\n68 -18 81\n13 31 60\n-15 28 99\n-42 15 9\n28 -61 -62\n-16 71 29\n-28 75 -48\n-77 -67 36\n-100 83 89\n100 100 -100\n57 34 -100\n0 0 -53",
"output": "YES"
},
{
"input": "44\n52 -54 -29\n-82 -5 -94\n-54 43 43\n91 16 71\n7 80 -91\n3 15 29\n-99 -6 -77\n-3 -77 -64\n73 67 34\n25 -10 -18\n-29 91 63\n-72 86 -16\n-68 85 -81\n-3 36 44\n-74 -14 -80\n34 -96 -97\n-76 -78 -33\n-24 44 -58\n98 12 77\n95 -63 -6\n-51 3 -90\n-92 -10 72\n7 3 -68\n57 -53 71\n29 57 -48\n35 -60 10\n79 -70 -61\n-20 77 55\n-86 -15 -35\n84 -88 -18\n100 -42 77\n-20 46 8\n-41 -43 -65\n38 -98 -23\n-100 65 45\n-7 -91 -63\n46 88 -85\n48 59 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 1",
"output": "YES"
},
{
"input": "18\n-14 -64 -91\n-8 -66 -86\n-23 92 -40\n6 -3 -53\n57 41 78\n-79 42 -22\n-88 -17 45\n4 -45 44\n83 -18 -25\n34 86 -92\n75 -30 12\n44 99 11\n-67 -13 72\n22 83 -56\n-37 71 72\n-9 -100 100\n0 -100 31\n0 -58 0",
"output": "YES"
},
{
"input": "23\n-70 37 78\n42 84 6\n28 -94 -24\n-49 76 95\n-67 18 84\n-53 78 -5\n65 -63 -64\n-66 -64 -69\n81 -93 95\n10 52 -79\n-89 -61 -64\n-64 -47 43\n-81 -35 55\n80 82 73\n-60 -81 -18\n49 -10 -19\n-58 70 12\n-24 -15 -93\n98 -93 -54\n-28 -75 11\n100 100 -63\n100 100 0\n56 34 0",
"output": "YES"
},
{
"input": "3\n0 2 -2\n1 -1 3\n-3 0 0",
"output": "NO"
}
] | 1,528,311,496
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 81
| 186
| 0
|
n = int(input())
x = y = z = 0
for k in range(n):
m = list(map(int, input().split()))
x += m[0]
y += m[1]
z += m[2]
if x == y == z == 0:
print('YES')
else:
print('NO')
|
Title: Young Physicist
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100).
Output Specification:
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
Demo Input:
['3\n4 1 7\n-2 4 -1\n1 -5 -3\n', '3\n3 -1 7\n-5 2 -4\n2 -1 -3\n']
Demo Output:
['NO', 'YES']
Note:
none
|
```python
n = int(input())
x = y = z = 0
for k in range(n):
m = list(map(int, input().split()))
x += m[0]
y += m[1]
z += m[2]
if x == y == z == 0:
print('YES')
else:
print('NO')
```
| 3.9535
|
748
|
A
|
Santa Claus and a Place in a Class
|
PROGRAMMING
| 800
|
[
"implementation",
"math"
] | null | null |
Santa Claus is the first who came to the Christmas Olympiad, and he is going to be the first to take his place at a desk! In the classroom there are *n* lanes of *m* desks each, and there are two working places at each of the desks. The lanes are numbered from 1 to *n* from the left to the right, the desks in a lane are numbered from 1 to *m* starting from the blackboard. Note that the lanes go perpendicularly to the blackboard, not along it (see picture).
The organizers numbered all the working places from 1 to 2*nm*. The places are numbered by lanes (i. e. all the places of the first lane go first, then all the places of the second lane, and so on), in a lane the places are numbered starting from the nearest to the blackboard (i. e. from the first desk in the lane), at each desk, the place on the left is numbered before the place on the right.
Santa Clause knows that his place has number *k*. Help him to determine at which lane at which desk he should sit, and whether his place is on the left or on the right!
|
The only line contains three integers *n*, *m* and *k* (1<=≤<=*n*,<=*m*<=≤<=10<=000, 1<=≤<=*k*<=≤<=2*nm*) — the number of lanes, the number of desks in each lane and the number of Santa Claus' place.
|
Print two integers: the number of lane *r*, the number of desk *d*, and a character *s*, which stands for the side of the desk Santa Claus. The character *s* should be "L", if Santa Clause should sit on the left, and "R" if his place is on the right.
|
[
"4 3 9\n",
"4 3 24\n",
"2 4 4\n"
] |
[
"2 2 L\n",
"4 3 R\n",
"1 2 R\n"
] |
The first and the second samples are shown on the picture. The green place corresponds to Santa Claus' place in the first example, the blue place corresponds to Santa Claus' place in the second example.
In the third sample there are two lanes with four desks in each, and Santa Claus has the fourth place. Thus, his place is in the first lane at the second desk on the right.
| 500
|
[
{
"input": "4 3 9",
"output": "2 2 L"
},
{
"input": "4 3 24",
"output": "4 3 R"
},
{
"input": "2 4 4",
"output": "1 2 R"
},
{
"input": "3 10 24",
"output": "2 2 R"
},
{
"input": "10 3 59",
"output": "10 3 L"
},
{
"input": "10000 10000 160845880",
"output": "8043 2940 R"
},
{
"input": "1 1 1",
"output": "1 1 L"
},
{
"input": "1 1 2",
"output": "1 1 R"
},
{
"input": "1 10000 1",
"output": "1 1 L"
},
{
"input": "1 10000 20000",
"output": "1 10000 R"
},
{
"input": "10000 1 1",
"output": "1 1 L"
},
{
"input": "10000 1 10000",
"output": "5000 1 R"
},
{
"input": "10000 1 20000",
"output": "10000 1 R"
},
{
"input": "3 2 1",
"output": "1 1 L"
},
{
"input": "3 2 2",
"output": "1 1 R"
},
{
"input": "3 2 3",
"output": "1 2 L"
},
{
"input": "3 2 4",
"output": "1 2 R"
},
{
"input": "3 2 5",
"output": "2 1 L"
},
{
"input": "3 2 6",
"output": "2 1 R"
},
{
"input": "3 2 7",
"output": "2 2 L"
},
{
"input": "3 2 8",
"output": "2 2 R"
},
{
"input": "3 2 9",
"output": "3 1 L"
},
{
"input": "3 2 10",
"output": "3 1 R"
},
{
"input": "3 2 11",
"output": "3 2 L"
},
{
"input": "3 2 12",
"output": "3 2 R"
},
{
"input": "300 2000 1068628",
"output": "268 314 R"
},
{
"input": "300 2000 584756",
"output": "147 378 R"
},
{
"input": "300 2000 268181",
"output": "68 91 L"
},
{
"input": "10000 9999 186450844",
"output": "9324 4745 R"
},
{
"input": "10000 9999 197114268",
"output": "9857 6990 R"
},
{
"input": "10000 9999 112390396",
"output": "5621 818 R"
},
{
"input": "10000 10000 1",
"output": "1 1 L"
},
{
"input": "10000 10000 2",
"output": "1 1 R"
},
{
"input": "10000 10000 100000001",
"output": "5001 1 L"
},
{
"input": "10000 10000 199999999",
"output": "10000 10000 L"
},
{
"input": "10000 10000 200000000",
"output": "10000 10000 R"
},
{
"input": "1 2 1",
"output": "1 1 L"
},
{
"input": "1 2 2",
"output": "1 1 R"
},
{
"input": "1 2 3",
"output": "1 2 L"
},
{
"input": "1 2 4",
"output": "1 2 R"
},
{
"input": "2 1 1",
"output": "1 1 L"
},
{
"input": "2 1 2",
"output": "1 1 R"
},
{
"input": "2 1 3",
"output": "2 1 L"
},
{
"input": "2 1 4",
"output": "2 1 R"
},
{
"input": "4 3 7",
"output": "2 1 L"
},
{
"input": "1 1 1",
"output": "1 1 L"
}
] | 1,482,657,367
| 667
|
Python 3
|
OK
|
TESTS
| 46
| 77
| 4,608,000
|
n,m,k=[int(i) for i in input().split()]
ryad=k//(2*m)+1
if k%(2*m)==0:
ryad-=1
parta=(k-2*m*(ryad-1))//2+k%2
if k%2==0:
mesto='R'
else:
mesto='L'
print(ryad,parta,mesto)
|
Title: Santa Claus and a Place in a Class
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Santa Claus is the first who came to the Christmas Olympiad, and he is going to be the first to take his place at a desk! In the classroom there are *n* lanes of *m* desks each, and there are two working places at each of the desks. The lanes are numbered from 1 to *n* from the left to the right, the desks in a lane are numbered from 1 to *m* starting from the blackboard. Note that the lanes go perpendicularly to the blackboard, not along it (see picture).
The organizers numbered all the working places from 1 to 2*nm*. The places are numbered by lanes (i. e. all the places of the first lane go first, then all the places of the second lane, and so on), in a lane the places are numbered starting from the nearest to the blackboard (i. e. from the first desk in the lane), at each desk, the place on the left is numbered before the place on the right.
Santa Clause knows that his place has number *k*. Help him to determine at which lane at which desk he should sit, and whether his place is on the left or on the right!
Input Specification:
The only line contains three integers *n*, *m* and *k* (1<=≤<=*n*,<=*m*<=≤<=10<=000, 1<=≤<=*k*<=≤<=2*nm*) — the number of lanes, the number of desks in each lane and the number of Santa Claus' place.
Output Specification:
Print two integers: the number of lane *r*, the number of desk *d*, and a character *s*, which stands for the side of the desk Santa Claus. The character *s* should be "L", if Santa Clause should sit on the left, and "R" if his place is on the right.
Demo Input:
['4 3 9\n', '4 3 24\n', '2 4 4\n']
Demo Output:
['2 2 L\n', '4 3 R\n', '1 2 R\n']
Note:
The first and the second samples are shown on the picture. The green place corresponds to Santa Claus' place in the first example, the blue place corresponds to Santa Claus' place in the second example.
In the third sample there are two lanes with four desks in each, and Santa Claus has the fourth place. Thus, his place is in the first lane at the second desk on the right.
|
```python
n,m,k=[int(i) for i in input().split()]
ryad=k//(2*m)+1
if k%(2*m)==0:
ryad-=1
parta=(k-2*m*(ryad-1))//2+k%2
if k%2==0:
mesto='R'
else:
mesto='L'
print(ryad,parta,mesto)
```
| 3
|
|
910
|
A
|
The Way to Home
|
PROGRAMMING
| 800
|
[
"dfs and similar",
"dp",
"greedy",
"implementation"
] | null | null |
A frog lives on the axis *Ox* and needs to reach home which is in the point *n*. She starts from the point 1. The frog can jump to the right at a distance not more than *d*. So, after she jumped from the point *x* she can reach the point *x*<=+<=*a*, where *a* is an integer from 1 to *d*.
For each point from 1 to *n* is known if there is a lily flower in it. The frog can jump only in points with a lilies. Guaranteed that there are lilies in the points 1 and *n*.
Determine the minimal number of jumps that the frog needs to reach home which is in the point *n* from the point 1. Consider that initially the frog is in the point 1. If the frog can not reach home, print -1.
|
The first line contains two integers *n* and *d* (2<=≤<=*n*<=≤<=100, 1<=≤<=*d*<=≤<=*n*<=-<=1) — the point, which the frog wants to reach, and the maximal length of the frog jump.
The second line contains a string *s* of length *n*, consisting of zeros and ones. If a character of the string *s* equals to zero, then in the corresponding point there is no lily flower. In the other case, in the corresponding point there is a lily flower. Guaranteed that the first and the last characters of the string *s* equal to one.
|
If the frog can not reach the home, print -1.
In the other case, print the minimal number of jumps that the frog needs to reach the home which is in the point *n* from the point 1.
|
[
"8 4\n10010101\n",
"4 2\n1001\n",
"8 4\n11100101\n",
"12 3\n101111100101\n"
] |
[
"2\n",
"-1\n",
"3\n",
"4\n"
] |
In the first example the from can reach home in two jumps: the first jump from the point 1 to the point 4 (the length of the jump is three), and the second jump from the point 4 to the point 8 (the length of the jump is four).
In the second example the frog can not reach home, because to make it she need to jump on a distance three, but the maximum length of her jump equals to two.
| 500
|
[
{
"input": "8 4\n10010101",
"output": "2"
},
{
"input": "4 2\n1001",
"output": "-1"
},
{
"input": "8 4\n11100101",
"output": "3"
},
{
"input": "12 3\n101111100101",
"output": "4"
},
{
"input": "5 4\n11011",
"output": "1"
},
{
"input": "5 4\n10001",
"output": "1"
},
{
"input": "10 7\n1101111011",
"output": "2"
},
{
"input": "10 9\n1110000101",
"output": "1"
},
{
"input": "10 9\n1100000001",
"output": "1"
},
{
"input": "20 5\n11111111110111101001",
"output": "4"
},
{
"input": "20 11\n11100000111000011011",
"output": "2"
},
{
"input": "20 19\n10100000000000000001",
"output": "1"
},
{
"input": "50 13\n10011010100010100111010000010000000000010100000101",
"output": "5"
},
{
"input": "50 8\n11010100000011001100001100010001110000101100110011",
"output": "8"
},
{
"input": "99 4\n111111111111111111111111111111111111111111111111111111111011111111111111111111111111111111111111111",
"output": "25"
},
{
"input": "99 98\n100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001",
"output": "1"
},
{
"input": "100 5\n1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111",
"output": "20"
},
{
"input": "100 4\n1111111111111111111111111111111111111111111111111111111111111111111111111111110111111111111111111111",
"output": "25"
},
{
"input": "100 4\n1111111111111111111111111111111111111111111111111111111111111101111111011111111111111111111111111111",
"output": "25"
},
{
"input": "100 3\n1111110111111111111111111111111111111111101111111111111111111111111101111111111111111111111111111111",
"output": "34"
},
{
"input": "100 8\n1111111111101110111111111111111111111111111111111111111111111111111111110011111111111111011111111111",
"output": "13"
},
{
"input": "100 7\n1011111111111111111011101111111011111101111111111101111011110111111111111111111111110111111011111111",
"output": "15"
},
{
"input": "100 9\n1101111110111110101111111111111111011001110111011101011111111111010101111111100011011111111010111111",
"output": "12"
},
{
"input": "100 6\n1011111011111111111011010110011001010101111110111111000111011011111110101101110110101111110000100111",
"output": "18"
},
{
"input": "100 7\n1110001111101001110011111111111101111101101001010001101000101100000101101101011111111101101000100001",
"output": "16"
},
{
"input": "100 11\n1000010100011100011011100000010011001111011110100100001011010100011011111001101101110110010110001101",
"output": "10"
},
{
"input": "100 9\n1001001110000011100100000001000110111101101010101001000101001010011001101100110011011110110011011111",
"output": "13"
},
{
"input": "100 7\n1010100001110101111011000111000001110100100110110001110110011010100001100100001110111100110000101001",
"output": "18"
},
{
"input": "100 10\n1110110000000110000000101110100000111000001011100000100110010001110111001010101000011000000001011011",
"output": "12"
},
{
"input": "100 13\n1000000100000000100011000010010000101010011110000000001000011000110100001000010001100000011001011001",
"output": "9"
},
{
"input": "100 11\n1000000000100000010000100001000100000000010000100100000000100100001000000001011000110001000000000101",
"output": "12"
},
{
"input": "100 22\n1000100000001010000000000000000001000000100000000000000000010000000000001000000000000000000100000001",
"output": "7"
},
{
"input": "100 48\n1000000000000000011000000000000000000000000000000001100000000000000000000000000000000000000000000001",
"output": "3"
},
{
"input": "100 48\n1000000000000000000000100000000000000000000000000000000000000000000001000000000000000000100000000001",
"output": "3"
},
{
"input": "100 75\n1000000100000000000000000000000000000000000000000000000000000000000000000000000001000000000000000001",
"output": "3"
},
{
"input": "100 73\n1000000000000000000000000000000100000000000000000000000000000000000000000000000000000000000000000001",
"output": "2"
},
{
"input": "100 99\n1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001",
"output": "1"
},
{
"input": "100 1\n1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111",
"output": "99"
},
{
"input": "100 2\n1111111111111111111111111111111110111111111111111111111111111111111111111111111111111111111111111111",
"output": "50"
},
{
"input": "100 1\n1111111111111111011111111111111111111111111111111111111111111111111101111111111111111111111111111111",
"output": "-1"
},
{
"input": "100 3\n1111111111111111111111111101111111111111111111111011111111111111111111111111111011111111111111111111",
"output": "33"
},
{
"input": "100 1\n1101111111111111111111101111111111111111111111111111111111111011111111101111101111111111111111111111",
"output": "-1"
},
{
"input": "100 6\n1111111111111111111111101111111101011110001111111111111111110111111111111111111111111110010111111111",
"output": "17"
},
{
"input": "100 2\n1111111101111010110111011011110111101111111011111101010101011111011111111111111011111001101111101111",
"output": "-1"
},
{
"input": "100 8\n1100110101111001101001111000111100110100011110111011001011111110000110101000001110111011100111011011",
"output": "14"
},
{
"input": "100 10\n1000111110100000001001101100000010011100010101001100010011111001001101111110110111101111001010001101",
"output": "11"
},
{
"input": "100 7\n1110000011010001110101011010000011110001000000011101110111010110001000011101111010010001101111110001",
"output": "-1"
},
{
"input": "100 3\n1111010001000001011011000011001111000100101000101101000010111101111000010000011110110011001101010111",
"output": "-1"
},
{
"input": "100 9\n1101010101101100010111011000010100001010000101010011001001100010110110000000010000101000000001101101",
"output": "13"
},
{
"input": "100 14\n1010100000000000010101000010001100000000000011100010000001000001011010001110001010100000100001101101",
"output": "9"
},
{
"input": "100 13\n1000000001101001110000010000011001000000000000001010000000100001001010000000000000000100010000000001",
"output": "-1"
},
{
"input": "100 18\n1000000000000000110000000000000000010000000001000001000001000000000100000000000010000000000000000001",
"output": "-1"
},
{
"input": "100 32\n1000000000000000000000000001000000000000000000000101000000000000000000000000000000000001000000000001",
"output": "-1"
},
{
"input": "100 79\n1000000001000000000101000000000000000000000000000000000000000000000000000000000000000000000000000001",
"output": "2"
},
{
"input": "100 41\n1000000000000000000000000000000000010000000000000000000000000000000000000000100000000000000000000001",
"output": "3"
},
{
"input": "100 82\n1000000000000000000100000000000000000000000000000000000000000000000000000000000000000000000000000001",
"output": "2"
},
{
"input": "100 96\n1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001",
"output": "-1"
},
{
"input": "43 30\n1001000001111111010100100100110101011101101",
"output": "2"
},
{
"input": "7 1\n1111111",
"output": "6"
},
{
"input": "9 3\n101000001",
"output": "-1"
},
{
"input": "10 3\n1100000001",
"output": "-1"
},
{
"input": "8 2\n10000101",
"output": "-1"
},
{
"input": "2 1\n11",
"output": "1"
}
] | 1,603,292,179
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 63
| 109
| 307,200
|
n, d = list(map(int, input().split(" ")))
s = input()
location = 0
jump_counter = 0
i = 0
while i < len(s):
i += d
if i >= len(s) - 1:
jump_counter += 1
print(jump_counter)
exit()
while not int(s[i]) and i > location:
i -= 1
if i == location:
print("-1")
exit()
else:
location = i
jump_counter += 1
print(jump_counter)
|
Title: The Way to Home
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A frog lives on the axis *Ox* and needs to reach home which is in the point *n*. She starts from the point 1. The frog can jump to the right at a distance not more than *d*. So, after she jumped from the point *x* she can reach the point *x*<=+<=*a*, where *a* is an integer from 1 to *d*.
For each point from 1 to *n* is known if there is a lily flower in it. The frog can jump only in points with a lilies. Guaranteed that there are lilies in the points 1 and *n*.
Determine the minimal number of jumps that the frog needs to reach home which is in the point *n* from the point 1. Consider that initially the frog is in the point 1. If the frog can not reach home, print -1.
Input Specification:
The first line contains two integers *n* and *d* (2<=≤<=*n*<=≤<=100, 1<=≤<=*d*<=≤<=*n*<=-<=1) — the point, which the frog wants to reach, and the maximal length of the frog jump.
The second line contains a string *s* of length *n*, consisting of zeros and ones. If a character of the string *s* equals to zero, then in the corresponding point there is no lily flower. In the other case, in the corresponding point there is a lily flower. Guaranteed that the first and the last characters of the string *s* equal to one.
Output Specification:
If the frog can not reach the home, print -1.
In the other case, print the minimal number of jumps that the frog needs to reach the home which is in the point *n* from the point 1.
Demo Input:
['8 4\n10010101\n', '4 2\n1001\n', '8 4\n11100101\n', '12 3\n101111100101\n']
Demo Output:
['2\n', '-1\n', '3\n', '4\n']
Note:
In the first example the from can reach home in two jumps: the first jump from the point 1 to the point 4 (the length of the jump is three), and the second jump from the point 4 to the point 8 (the length of the jump is four).
In the second example the frog can not reach home, because to make it she need to jump on a distance three, but the maximum length of her jump equals to two.
|
```python
n, d = list(map(int, input().split(" ")))
s = input()
location = 0
jump_counter = 0
i = 0
while i < len(s):
i += d
if i >= len(s) - 1:
jump_counter += 1
print(jump_counter)
exit()
while not int(s[i]) and i > location:
i -= 1
if i == location:
print("-1")
exit()
else:
location = i
jump_counter += 1
print(jump_counter)
```
| 3
|
|
698
|
A
|
Vacations
|
PROGRAMMING
| 1,400
|
[
"dp"
] | null | null |
Vasya has *n* days of vacations! So he decided to improve his IT skills and do sport. Vasya knows the following information about each of this *n* days: whether that gym opened and whether a contest was carried out in the Internet on that day. For the *i*-th day there are four options:
1. on this day the gym is closed and the contest is not carried out; 1. on this day the gym is closed and the contest is carried out; 1. on this day the gym is open and the contest is not carried out; 1. on this day the gym is open and the contest is carried out.
On each of days Vasya can either have a rest or write the contest (if it is carried out on this day), or do sport (if the gym is open on this day).
Find the minimum number of days on which Vasya will have a rest (it means, he will not do sport and write the contest at the same time). The only limitation that Vasya has — he does not want to do the same activity on two consecutive days: it means, he will not do sport on two consecutive days, and write the contest on two consecutive days.
|
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100) — the number of days of Vasya's vacations.
The second line contains the sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=3) separated by space, where:
- *a**i* equals 0, if on the *i*-th day of vacations the gym is closed and the contest is not carried out; - *a**i* equals 1, if on the *i*-th day of vacations the gym is closed, but the contest is carried out; - *a**i* equals 2, if on the *i*-th day of vacations the gym is open and the contest is not carried out; - *a**i* equals 3, if on the *i*-th day of vacations the gym is open and the contest is carried out.
|
Print the minimum possible number of days on which Vasya will have a rest. Remember that Vasya refuses:
- to do sport on any two consecutive days, - to write the contest on any two consecutive days.
|
[
"4\n1 3 2 0\n",
"7\n1 3 3 2 1 2 3\n",
"2\n2 2\n"
] |
[
"2\n",
"0\n",
"1\n"
] |
In the first test Vasya can write the contest on the day number 1 and do sport on the day number 3. Thus, he will have a rest for only 2 days.
In the second test Vasya should write contests on days number 1, 3, 5 and 7, in other days do sport. Thus, he will not have a rest for a single day.
In the third test Vasya can do sport either on a day number 1 or number 2. He can not do sport in two days, because it will be contrary to the his limitation. Thus, he will have a rest for only one day.
| 500
|
[
{
"input": "4\n1 3 2 0",
"output": "2"
},
{
"input": "7\n1 3 3 2 1 2 3",
"output": "0"
},
{
"input": "2\n2 2",
"output": "1"
},
{
"input": "1\n0",
"output": "1"
},
{
"input": "10\n0 0 1 1 0 0 0 0 1 0",
"output": "8"
},
{
"input": "100\n3 2 3 3 3 2 3 1 3 2 2 3 2 3 3 3 3 3 3 1 2 2 3 1 3 3 2 2 2 3 1 0 3 3 3 2 3 3 1 1 3 1 3 3 3 1 3 1 3 0 1 3 2 3 2 1 1 3 2 3 3 3 2 3 1 3 3 3 3 2 2 2 1 3 1 3 3 3 3 1 3 2 3 3 0 3 3 3 3 3 1 0 2 1 3 3 0 2 3 3",
"output": "16"
},
{
"input": "10\n2 3 0 1 3 1 2 2 1 0",
"output": "3"
},
{
"input": "45\n3 3 2 3 2 3 3 3 0 3 3 3 3 3 3 3 1 3 2 3 2 3 2 2 2 3 2 3 3 3 3 3 1 2 3 3 2 2 2 3 3 3 3 1 3",
"output": "6"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "1\n2",
"output": "0"
},
{
"input": "1\n3",
"output": "0"
},
{
"input": "2\n1 1",
"output": "1"
},
{
"input": "2\n1 3",
"output": "0"
},
{
"input": "2\n0 1",
"output": "1"
},
{
"input": "2\n0 0",
"output": "2"
},
{
"input": "2\n3 3",
"output": "0"
},
{
"input": "3\n3 3 3",
"output": "0"
},
{
"input": "2\n3 2",
"output": "0"
},
{
"input": "2\n0 2",
"output": "1"
},
{
"input": "10\n2 2 3 3 3 3 2 1 3 2",
"output": "2"
},
{
"input": "15\n0 1 0 0 0 2 0 1 0 0 0 2 0 0 0",
"output": "11"
},
{
"input": "15\n1 3 2 2 2 3 3 3 3 2 3 2 2 1 1",
"output": "4"
},
{
"input": "15\n3 1 3 2 3 2 2 2 3 3 3 3 2 3 2",
"output": "3"
},
{
"input": "20\n0 2 0 1 0 0 0 1 2 0 1 1 1 0 1 1 0 1 1 0",
"output": "12"
},
{
"input": "20\n2 3 2 3 3 3 3 2 0 3 1 1 2 3 0 3 2 3 0 3",
"output": "5"
},
{
"input": "20\n3 3 3 3 2 3 3 2 1 3 3 2 2 2 3 2 2 2 2 2",
"output": "4"
},
{
"input": "25\n0 0 1 0 0 1 0 0 1 0 0 1 0 2 0 0 2 0 0 1 0 2 0 1 1",
"output": "16"
},
{
"input": "25\n1 3 3 2 2 3 3 3 3 3 1 2 2 3 2 0 2 1 0 1 3 2 2 3 3",
"output": "5"
},
{
"input": "25\n2 3 1 3 3 2 1 3 3 3 1 3 3 1 3 2 3 3 1 3 3 3 2 3 3",
"output": "3"
},
{
"input": "30\n0 0 1 0 1 0 1 1 0 0 0 0 0 0 1 0 0 1 1 0 0 2 0 0 1 1 2 0 0 0",
"output": "22"
},
{
"input": "30\n1 1 3 2 2 0 3 2 3 3 1 2 0 1 1 2 3 3 2 3 1 3 2 3 0 2 0 3 3 2",
"output": "9"
},
{
"input": "30\n1 2 3 2 2 3 3 3 3 3 3 3 3 3 3 1 2 2 3 2 3 3 3 2 1 3 3 3 1 3",
"output": "2"
},
{
"input": "35\n0 1 1 0 0 2 0 0 1 0 0 0 1 0 1 0 1 0 0 0 1 2 1 0 2 2 1 0 1 0 1 1 1 0 0",
"output": "21"
},
{
"input": "35\n2 2 0 3 2 2 0 3 3 1 1 3 3 1 2 2 0 2 2 2 2 3 1 0 2 1 3 2 2 3 2 3 3 1 2",
"output": "11"
},
{
"input": "35\n1 2 2 3 3 3 3 3 2 2 3 3 2 3 3 2 3 2 3 3 2 2 2 3 3 2 3 3 3 1 3 3 2 2 2",
"output": "7"
},
{
"input": "40\n2 0 1 1 0 0 0 0 2 0 1 1 1 0 0 1 0 0 0 0 0 2 0 0 0 2 1 1 1 3 0 0 0 0 0 0 0 1 1 0",
"output": "28"
},
{
"input": "40\n2 2 3 2 0 2 3 2 1 2 3 0 2 3 2 1 1 3 1 1 0 2 3 1 3 3 1 1 3 3 2 2 1 3 3 3 2 3 3 1",
"output": "10"
},
{
"input": "40\n1 3 2 3 3 2 3 3 2 2 3 1 2 1 2 2 3 1 2 2 1 2 2 2 1 2 2 3 2 3 2 3 2 3 3 3 1 3 2 3",
"output": "8"
},
{
"input": "45\n2 1 0 0 0 2 1 0 1 0 0 2 2 1 1 0 0 2 0 0 0 0 0 0 1 0 0 2 0 0 1 1 0 0 1 0 0 1 1 2 0 0 2 0 2",
"output": "29"
},
{
"input": "45\n3 3 2 3 3 3 2 2 3 2 3 1 3 2 3 2 2 1 1 3 2 3 2 1 3 1 2 3 2 2 0 3 3 2 3 2 3 2 3 2 0 3 1 1 3",
"output": "8"
},
{
"input": "50\n3 0 0 0 2 0 0 0 0 0 0 0 2 1 0 2 0 1 0 1 3 0 2 1 1 0 0 1 1 0 0 1 2 1 1 2 1 1 0 0 0 0 0 0 0 1 2 2 0 0",
"output": "32"
},
{
"input": "50\n3 3 3 3 1 0 3 3 0 2 3 1 1 1 3 2 3 3 3 3 3 1 0 1 2 2 3 3 2 3 0 0 0 2 1 0 1 2 2 2 2 0 2 2 2 1 2 3 3 2",
"output": "16"
},
{
"input": "50\n3 2 3 1 2 1 2 3 3 2 3 3 2 1 3 3 3 3 3 3 2 3 2 3 2 2 3 3 3 2 3 3 3 3 2 3 1 2 3 3 2 3 3 1 2 2 1 1 3 3",
"output": "7"
},
{
"input": "55\n0 0 1 1 0 1 0 0 1 0 1 0 0 0 2 0 0 1 0 0 0 1 0 0 0 0 3 1 0 0 0 1 0 0 0 0 2 0 0 0 2 0 2 1 0 0 0 0 0 0 0 0 2 0 0",
"output": "40"
},
{
"input": "55\n3 0 3 3 3 2 0 2 3 0 3 2 3 3 0 3 3 1 3 3 1 2 3 2 0 3 3 2 1 2 3 2 3 0 3 2 2 1 2 3 2 2 1 3 2 2 3 1 3 2 2 3 3 2 2",
"output": "13"
},
{
"input": "55\n3 3 1 3 2 3 2 3 2 2 3 3 3 3 3 1 1 3 3 2 3 2 3 2 0 1 3 3 3 3 2 3 2 3 1 1 2 2 2 3 3 3 3 3 2 2 2 3 2 3 3 3 3 1 3",
"output": "7"
},
{
"input": "60\n0 1 0 0 0 0 0 0 0 2 1 1 3 0 0 0 0 0 1 0 1 1 0 0 0 3 0 1 0 1 0 2 0 0 0 0 0 1 0 0 0 0 1 1 0 1 0 0 0 0 0 1 0 0 1 0 1 0 0 0",
"output": "44"
},
{
"input": "60\n3 2 1 3 2 2 3 3 3 1 1 3 2 2 3 3 1 3 2 2 3 3 2 2 2 2 0 2 2 3 2 3 0 3 3 3 2 3 3 0 1 3 2 1 3 1 1 2 1 3 1 1 2 2 1 3 3 3 2 2",
"output": "15"
},
{
"input": "60\n3 2 2 3 2 3 2 3 3 2 3 2 3 3 2 3 3 3 3 3 3 2 3 3 1 2 3 3 3 2 1 3 3 1 3 1 3 0 3 3 3 2 3 2 3 2 3 3 1 1 2 3 3 3 3 2 1 3 2 3",
"output": "8"
},
{
"input": "65\n1 0 2 1 1 0 1 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 0 1 2 0 2 1 0 2 1 0 1 0 1 1 0 1 1 1 2 1 0 1 0 0 0 0 1 2 2 1 0 0 1 2 1 2 0 2 0 0 0 1 1",
"output": "35"
},
{
"input": "65\n2 2 2 3 0 2 1 2 3 3 1 3 1 2 1 3 2 3 2 2 2 1 2 0 3 1 3 1 1 3 1 3 3 3 3 3 1 3 0 3 1 3 1 2 2 3 2 0 3 1 3 2 1 2 2 2 3 3 2 3 3 3 2 2 3",
"output": "13"
},
{
"input": "65\n3 2 3 3 3 2 3 2 3 3 3 3 3 3 3 3 3 2 3 2 3 2 2 3 3 3 3 3 2 2 2 3 3 2 3 3 2 3 3 3 3 2 3 3 3 2 2 3 3 3 3 3 3 2 2 3 3 2 3 3 1 3 3 3 3",
"output": "6"
},
{
"input": "70\n1 0 0 0 1 0 1 0 0 0 1 1 0 1 0 0 1 1 1 0 1 1 0 0 1 1 1 3 1 1 0 1 2 0 2 1 0 0 0 1 1 1 1 1 0 0 1 0 0 0 1 1 1 3 0 0 1 0 0 0 1 0 0 0 0 0 1 0 1 1",
"output": "43"
},
{
"input": "70\n2 3 3 3 1 3 3 1 2 1 1 2 2 3 0 2 3 3 1 3 3 2 2 3 3 3 2 2 2 2 1 3 3 0 2 1 1 3 2 3 3 2 2 3 1 3 1 2 3 2 3 3 2 2 2 3 1 1 2 1 3 3 2 2 3 3 3 1 1 1",
"output": "16"
},
{
"input": "70\n3 3 2 2 1 2 1 2 2 2 2 2 3 3 2 3 3 3 3 2 2 2 2 3 3 3 1 3 3 3 2 3 3 3 3 2 3 3 1 3 1 3 2 3 3 2 3 3 3 2 3 2 3 3 1 2 3 3 2 2 2 3 2 3 3 3 3 3 3 1",
"output": "10"
},
{
"input": "75\n1 0 0 1 1 0 0 1 0 1 2 0 0 2 1 1 0 0 0 0 0 0 2 1 1 0 0 0 0 1 0 1 0 1 1 1 0 1 0 0 1 0 0 0 0 0 0 1 1 0 0 1 2 1 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 1 1 1 0 1 0",
"output": "51"
},
{
"input": "75\n1 3 3 3 1 1 3 2 3 3 1 3 3 3 2 1 3 2 2 3 1 1 1 1 1 1 2 3 3 3 3 3 3 2 3 3 3 3 3 2 3 3 2 2 2 1 2 3 3 2 2 3 0 1 1 3 3 0 0 1 1 3 2 3 3 3 3 1 2 2 3 3 3 3 1",
"output": "16"
},
{
"input": "75\n3 3 3 3 2 2 3 2 2 3 2 2 1 2 3 3 2 2 3 3 1 2 2 2 1 3 3 3 1 2 2 3 3 3 2 3 2 2 2 3 3 1 3 2 2 3 3 3 0 3 2 1 3 3 2 3 3 3 3 1 2 3 3 3 2 2 3 3 3 3 2 2 3 3 1",
"output": "11"
},
{
"input": "80\n0 0 0 0 2 0 1 1 1 1 1 0 0 0 0 2 0 0 1 0 0 0 0 1 1 0 2 2 1 1 0 1 0 1 0 1 1 1 0 1 2 1 1 0 0 0 1 1 0 1 1 0 1 0 0 1 0 0 1 0 0 0 0 0 0 0 2 2 0 1 1 0 0 0 0 0 0 0 0 1",
"output": "56"
},
{
"input": "80\n2 2 3 3 2 1 0 1 0 3 2 2 3 2 1 3 1 3 3 2 3 3 3 2 3 3 3 2 1 3 3 1 3 3 3 3 3 3 2 2 2 1 3 2 1 3 2 1 1 0 1 1 2 1 3 0 1 2 3 2 2 3 2 3 1 3 3 2 1 1 0 3 3 3 3 1 2 1 2 0",
"output": "17"
},
{
"input": "80\n2 3 3 2 2 2 3 3 2 3 3 3 3 3 2 3 2 3 2 3 3 3 3 3 3 3 3 3 2 3 1 3 2 3 3 0 3 1 2 3 3 1 2 3 2 3 3 2 3 3 3 3 3 2 2 3 0 3 3 3 3 3 2 2 3 2 3 3 3 3 3 2 3 2 3 3 3 3 2 3",
"output": "9"
},
{
"input": "85\n0 1 1 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 2 0 1 0 0 2 0 1 1 0 0 0 0 2 2 0 0 0 1 0 0 0 1 2 0 1 0 0 0 2 1 1 2 0 3 1 0 2 2 1 0 0 1 1 0 0 0 0 1 0 2 1 1 2 1 0 0 1 2 1 2 0 0 1 0 1 0",
"output": "54"
},
{
"input": "85\n2 3 1 3 2 3 1 3 3 2 1 2 1 2 2 3 2 2 3 2 0 3 3 2 1 2 2 2 3 3 2 3 3 3 2 1 1 3 1 3 2 2 2 3 3 2 3 2 3 1 1 3 2 3 1 3 3 2 3 3 2 2 3 0 1 1 2 2 2 2 1 2 3 1 3 3 1 3 2 2 3 2 3 3 3",
"output": "19"
},
{
"input": "85\n1 2 1 2 3 2 3 3 3 3 3 3 3 2 1 3 2 3 3 3 3 2 3 3 3 1 3 3 3 3 2 3 3 3 3 3 3 2 2 1 3 3 3 3 2 2 3 1 1 2 3 3 3 2 3 3 3 3 3 2 3 3 3 2 2 3 3 1 1 1 3 3 3 3 1 3 3 3 1 3 3 1 3 2 3",
"output": "9"
},
{
"input": "90\n2 0 1 0 0 0 0 0 0 1 1 2 0 0 0 0 0 0 0 2 2 0 2 0 0 2 1 0 2 0 1 0 1 0 0 1 2 2 0 0 1 0 0 1 0 1 0 2 0 1 1 1 0 1 1 0 1 0 2 0 1 0 1 0 0 0 1 0 0 1 2 0 0 0 1 0 0 2 2 0 0 0 0 0 1 3 1 1 0 1",
"output": "57"
},
{
"input": "90\n2 3 3 3 2 3 2 1 3 0 3 2 3 3 2 1 3 3 2 3 2 3 3 2 1 3 1 3 3 1 2 2 3 3 2 1 2 3 2 3 0 3 3 2 2 3 1 0 3 3 1 3 3 3 3 2 1 2 2 1 3 2 1 3 3 1 2 0 2 2 3 2 2 3 3 3 1 3 2 1 2 3 3 2 3 2 3 3 2 1",
"output": "17"
},
{
"input": "90\n2 3 2 3 2 2 3 3 2 3 2 1 2 3 3 3 2 3 2 3 3 2 3 3 3 1 3 3 1 3 2 3 2 2 1 3 3 3 3 3 3 3 3 3 3 2 3 2 3 2 1 3 3 3 3 2 2 3 3 3 3 3 3 3 3 3 3 3 3 2 2 3 3 3 3 1 3 2 3 3 3 2 2 3 2 3 2 1 3 2",
"output": "9"
},
{
"input": "95\n0 0 3 0 2 0 1 0 0 2 0 0 0 0 0 0 0 1 0 0 0 2 0 0 0 0 0 1 0 0 2 1 0 0 1 0 0 0 1 0 0 0 0 1 0 1 0 0 1 0 1 2 0 1 2 2 0 0 1 0 2 0 0 0 1 0 2 1 2 1 0 1 0 0 0 1 0 0 1 1 2 1 1 1 1 2 0 0 0 0 0 1 1 0 1",
"output": "61"
},
{
"input": "95\n2 3 3 2 1 1 3 3 3 2 3 3 3 2 3 2 3 3 3 2 3 2 2 3 3 2 1 2 3 3 3 1 3 0 3 3 1 3 3 1 0 1 3 3 3 0 2 1 3 3 3 3 0 1 3 2 3 3 2 1 3 1 2 1 1 2 3 0 3 3 2 1 3 2 1 3 3 3 2 2 3 2 3 3 3 2 1 3 3 3 2 3 3 1 2",
"output": "15"
},
{
"input": "95\n2 3 3 2 3 2 2 1 3 1 2 1 2 3 1 2 3 3 1 3 3 3 1 2 3 2 2 2 2 3 3 3 2 2 3 3 3 3 3 1 2 2 3 3 3 3 2 3 2 2 2 3 3 2 3 3 3 3 3 3 3 0 3 2 0 3 3 1 3 3 3 2 3 2 3 2 3 3 3 3 2 2 1 1 3 3 3 3 3 1 3 3 3 3 2",
"output": "14"
},
{
"input": "100\n1 0 2 0 0 0 0 2 0 0 0 1 0 1 0 0 1 0 1 2 0 1 1 0 0 1 0 1 1 0 0 0 2 0 1 0 0 2 0 0 0 0 0 1 1 1 0 0 1 0 2 0 0 0 0 1 0 1 0 1 0 1 0 1 2 2 0 0 2 0 1 0 1 0 1 0 0 0 1 0 0 2 1 1 1 0 0 1 0 0 0 2 0 0 2 1 1 0 0 2",
"output": "63"
},
{
"input": "100\n3 2 1 3 2 3 2 3 2 2 3 1 3 3 3 3 3 2 2 3 2 2 3 2 3 3 3 2 3 1 2 1 3 3 3 3 1 3 3 3 3 3 2 3 2 1 3 3 1 2 2 3 1 3 3 1 2 2 1 3 1 3 2 2 3 3 1 3 2 3 1 2 1 2 3 3 2 2 1 2 3 3 3 3 3 1 3 3 3 3 2 1 3 0 3 3 3 2 3 3",
"output": "15"
},
{
"input": "100\n1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2",
"output": "0"
},
{
"input": "100\n3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3",
"output": "0"
},
{
"input": "100\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "50"
},
{
"input": "100\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2",
"output": "50"
},
{
"input": "99\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "49"
},
{
"input": "100\n2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1",
"output": "0"
},
{
"input": "100\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0",
"output": "100"
},
{
"input": "2\n0 3",
"output": "1"
},
{
"input": "2\n1 0",
"output": "1"
},
{
"input": "2\n1 2",
"output": "0"
},
{
"input": "2\n2 0",
"output": "1"
},
{
"input": "2\n2 1",
"output": "0"
},
{
"input": "2\n2 3",
"output": "0"
},
{
"input": "2\n3 0",
"output": "1"
},
{
"input": "2\n3 1",
"output": "0"
},
{
"input": "100\n3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0",
"output": "50"
}
] | 1,569,516,487
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 88
| 139
| 307,200
|
n = int(input())
days = list(map(int, input().split()))
dp = [[999,999,999] for x in range(100)]
dp[0][0] = 1
if (days[0] == 1):
dp[0][1] = 0
elif (days[0] == 2):
dp[0][2] = 0
elif days[0] == 3:
dp[0][1] = 0
dp[0][2] = 0
for i in range(1, n):
dp[i][0] = 1 + min(dp[i-1][0], dp[i-1][1], dp[i-1][2])
if (days[i] == 1):
dp[i][1] = min(dp[i-1][0], dp[i-1][2])
elif days[i] == 2:
dp[i][2] = min(dp[i-1][0], dp[i-1][1])
elif days[i] == 3:
dp[i][1] = min(dp[i-1][0], dp[i-1][2])
dp[i][2] = min(dp[i-1][0], dp[i-1][1])
print(min(dp[n-1][0], dp[n-1][1], dp[n-1][2]))
|
Title: Vacations
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya has *n* days of vacations! So he decided to improve his IT skills and do sport. Vasya knows the following information about each of this *n* days: whether that gym opened and whether a contest was carried out in the Internet on that day. For the *i*-th day there are four options:
1. on this day the gym is closed and the contest is not carried out; 1. on this day the gym is closed and the contest is carried out; 1. on this day the gym is open and the contest is not carried out; 1. on this day the gym is open and the contest is carried out.
On each of days Vasya can either have a rest or write the contest (if it is carried out on this day), or do sport (if the gym is open on this day).
Find the minimum number of days on which Vasya will have a rest (it means, he will not do sport and write the contest at the same time). The only limitation that Vasya has — he does not want to do the same activity on two consecutive days: it means, he will not do sport on two consecutive days, and write the contest on two consecutive days.
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100) — the number of days of Vasya's vacations.
The second line contains the sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=3) separated by space, where:
- *a**i* equals 0, if on the *i*-th day of vacations the gym is closed and the contest is not carried out; - *a**i* equals 1, if on the *i*-th day of vacations the gym is closed, but the contest is carried out; - *a**i* equals 2, if on the *i*-th day of vacations the gym is open and the contest is not carried out; - *a**i* equals 3, if on the *i*-th day of vacations the gym is open and the contest is carried out.
Output Specification:
Print the minimum possible number of days on which Vasya will have a rest. Remember that Vasya refuses:
- to do sport on any two consecutive days, - to write the contest on any two consecutive days.
Demo Input:
['4\n1 3 2 0\n', '7\n1 3 3 2 1 2 3\n', '2\n2 2\n']
Demo Output:
['2\n', '0\n', '1\n']
Note:
In the first test Vasya can write the contest on the day number 1 and do sport on the day number 3. Thus, he will have a rest for only 2 days.
In the second test Vasya should write contests on days number 1, 3, 5 and 7, in other days do sport. Thus, he will not have a rest for a single day.
In the third test Vasya can do sport either on a day number 1 or number 2. He can not do sport in two days, because it will be contrary to the his limitation. Thus, he will have a rest for only one day.
|
```python
n = int(input())
days = list(map(int, input().split()))
dp = [[999,999,999] for x in range(100)]
dp[0][0] = 1
if (days[0] == 1):
dp[0][1] = 0
elif (days[0] == 2):
dp[0][2] = 0
elif days[0] == 3:
dp[0][1] = 0
dp[0][2] = 0
for i in range(1, n):
dp[i][0] = 1 + min(dp[i-1][0], dp[i-1][1], dp[i-1][2])
if (days[i] == 1):
dp[i][1] = min(dp[i-1][0], dp[i-1][2])
elif days[i] == 2:
dp[i][2] = min(dp[i-1][0], dp[i-1][1])
elif days[i] == 3:
dp[i][1] = min(dp[i-1][0], dp[i-1][2])
dp[i][2] = min(dp[i-1][0], dp[i-1][1])
print(min(dp[n-1][0], dp[n-1][1], dp[n-1][2]))
```
| 3
|
|
225
|
A
|
Dice Tower
|
PROGRAMMING
| 1,100
|
[
"constructive algorithms",
"greedy"
] | null | null |
A dice is a cube, its faces contain distinct integers from 1 to 6 as black points. The sum of numbers at the opposite dice faces always equals 7. Please note that there are only two dice (these dices are mirror of each other) that satisfy the given constraints (both of them are shown on the picture on the left).
Alice and Bob play dice. Alice has built a tower from *n* dice. We know that in this tower the adjacent dice contact with faces with distinct numbers. Bob wants to uniquely identify the numbers written on the faces of all dice, from which the tower is built. Unfortunately, Bob is looking at the tower from the face, and so he does not see all the numbers on the faces. Bob sees the number on the top of the tower and the numbers on the two adjacent sides (on the right side of the picture shown what Bob sees).
Help Bob, tell whether it is possible to uniquely identify the numbers on the faces of all the dice in the tower, or not.
|
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of dice in the tower.
The second line contains an integer *x* (1<=≤<=*x*<=≤<=6) — the number Bob sees at the top of the tower. Next *n* lines contain two space-separated integers each: the *i*-th line contains numbers *a**i*,<=*b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=6; *a**i*<=≠<=*b**i*) — the numbers Bob sees on the two sidelong faces of the *i*-th dice in the tower.
Consider the dice in the tower indexed from top to bottom from 1 to *n*. That is, the topmost dice has index 1 (the dice whose top face Bob can see). It is guaranteed that it is possible to make a dice tower that will look as described in the input.
|
Print "YES" (without the quotes), if it is possible to to uniquely identify the numbers on the faces of all the dice in the tower. If it is impossible, print "NO" (without the quotes).
|
[
"3\n6\n3 2\n5 4\n2 4\n",
"3\n3\n2 6\n4 1\n5 3\n"
] |
[
"YES",
"NO"
] |
none
| 500
|
[
{
"input": "3\n6\n3 2\n5 4\n2 4",
"output": "YES"
},
{
"input": "3\n3\n2 6\n4 1\n5 3",
"output": "NO"
},
{
"input": "1\n3\n2 1",
"output": "YES"
},
{
"input": "2\n2\n3 1\n1 5",
"output": "NO"
},
{
"input": "3\n2\n1 4\n5 3\n6 4",
"output": "NO"
},
{
"input": "4\n3\n5 6\n1 3\n1 5\n4 1",
"output": "NO"
},
{
"input": "2\n2\n3 1\n1 3",
"output": "YES"
},
{
"input": "3\n2\n1 4\n3 1\n4 6",
"output": "YES"
},
{
"input": "4\n3\n5 6\n1 5\n5 1\n1 5",
"output": "YES"
},
{
"input": "5\n1\n2 3\n5 3\n5 4\n5 1\n3 5",
"output": "NO"
},
{
"input": "10\n5\n1 3\n2 3\n6 5\n6 5\n4 5\n1 3\n1 2\n3 2\n4 2\n1 2",
"output": "NO"
},
{
"input": "15\n4\n2 1\n2 4\n6 4\n5 3\n4 1\n4 2\n6 3\n4 5\n3 5\n2 6\n5 6\n1 5\n3 5\n6 4\n3 2",
"output": "NO"
},
{
"input": "20\n6\n3 2\n4 6\n3 6\n6 4\n5 1\n1 5\n2 6\n1 2\n1 4\n5 3\n2 3\n6 2\n5 4\n2 6\n1 3\n4 6\n4 5\n6 3\n3 1\n6 2",
"output": "NO"
},
{
"input": "25\n4\n1 2\n4 1\n3 5\n2 1\n3 5\n6 5\n3 5\n5 6\n1 2\n2 4\n6 2\n2 3\n2 4\n6 5\n2 3\n6 3\n2 3\n1 3\n2 1\n3 1\n5 6\n3 1\n6 4\n3 6\n2 3",
"output": "NO"
},
{
"input": "100\n3\n6 5\n5 1\n3 2\n1 5\n3 6\n5 4\n2 6\n4 1\n6 3\n4 5\n1 5\n1 4\n4 2\n2 6\n5 4\n4 1\n1 3\n6 5\n5 1\n2 1\n2 4\n2 1\n3 6\n4 1\n6 3\n2 3\n5 1\n2 6\n6 4\n3 5\n4 1\n6 5\n1 5\n1 5\n2 3\n4 1\n5 3\n6 4\n1 3\n5 3\n4 1\n1 4\n2 1\n6 2\n1 5\n6 2\n6 2\n4 5\n4 2\n5 6\n6 3\n1 3\n2 3\n5 4\n6 5\n3 1\n1 2\n4 1\n1 3\n1 3\n6 5\n4 6\n3 1\n2 1\n2 3\n3 2\n4 1\n1 5\n4 1\n6 3\n1 5\n4 5\n4 2\n4 5\n2 6\n2 1\n3 5\n4 6\n4 2\n4 5\n2 4\n3 1\n6 4\n5 6\n3 1\n1 4\n4 5\n6 3\n6 3\n2 1\n5 1\n3 6\n3 5\n2 1\n4 6\n4 2\n5 6\n3 1\n3 5\n3 6",
"output": "NO"
},
{
"input": "99\n3\n2 1\n6 2\n3 6\n1 3\n5 1\n2 6\n4 6\n6 4\n6 4\n6 5\n3 6\n2 6\n1 5\n2 3\n4 6\n1 4\n4 1\n2 3\n4 5\n4 1\n5 1\n1 2\n6 5\n4 6\n6 5\n6 2\n3 6\n6 4\n2 1\n3 1\n2 1\n6 2\n3 5\n4 1\n5 3\n3 1\n1 5\n3 6\n6 2\n1 5\n2 1\n5 1\n4 1\n2 6\n5 4\n4 2\n2 1\n1 5\n1 3\n4 6\n4 6\n4 5\n2 3\n6 2\n3 2\n2 1\n4 6\n6 2\n3 5\n3 6\n3 1\n2 3\n2 1\n3 6\n6 5\n6 3\n1 2\n5 1\n1 4\n6 2\n5 3\n1 3\n5 4\n2 3\n6 3\n1 5\n1 2\n2 6\n5 6\n5 6\n3 5\n3 1\n4 6\n3 1\n4 5\n4 2\n3 5\n6 2\n2 4\n4 6\n6 2\n4 2\n2 3\n2 4\n1 5\n1 4\n3 5\n1 2\n4 5",
"output": "NO"
},
{
"input": "98\n6\n4 2\n1 2\n3 2\n2 1\n2 1\n3 2\n2 3\n6 5\n4 6\n1 5\n4 5\n5 1\n6 5\n1 4\n1 2\n2 4\n6 5\n4 5\n4 6\n3 1\n2 3\n4 1\n4 2\n6 5\n3 2\n4 2\n5 1\n2 4\n1 3\n4 5\n3 2\n1 2\n3 1\n3 2\n3 6\n6 4\n3 6\n3 5\n4 6\n6 5\n3 5\n3 2\n4 2\n6 4\n1 3\n2 4\n5 3\n2 3\n1 3\n5 6\n5 3\n5 3\n4 6\n4 6\n3 6\n4 1\n6 5\n6 2\n1 5\n2 1\n6 2\n5 4\n6 3\n1 5\n2 3\n2 6\n5 6\n2 6\n5 1\n3 2\n6 2\n6 2\n1 2\n2 1\n3 5\n2 1\n4 6\n1 4\n4 5\n3 2\n3 2\n5 4\n1 3\n5 1\n2 3\n6 2\n2 6\n1 5\n5 1\n5 4\n5 1\n5 4\n2 1\n6 5\n1 4\n6 5\n1 2\n3 5",
"output": "NO"
},
{
"input": "97\n3\n2 1\n6 5\n4 1\n6 5\n3 2\n1 2\n6 3\n6 4\n6 3\n1 3\n1 3\n3 1\n3 6\n3 2\n5 6\n4 2\n3 6\n1 5\n2 6\n3 2\n6 2\n2 1\n2 4\n1 3\n3 1\n2 6\n3 6\n4 6\n6 2\n5 1\n6 3\n2 6\n3 6\n2 4\n4 5\n6 5\n4 1\n5 6\n6 2\n5 4\n5 1\n6 5\n1 4\n2 1\n4 5\n4 5\n4 1\n5 4\n1 4\n2 6\n2 6\n1 5\n5 6\n3 2\n2 3\n1 4\n4 1\n3 6\n6 2\n5 3\n6 2\n4 5\n6 2\n2 6\n6 5\n1 4\n2 6\n3 5\n2 6\n4 1\n4 5\n1 3\n4 2\n3 2\n1 2\n5 6\n1 5\n3 5\n2 1\n1 2\n1 2\n6 4\n5 1\n1 2\n2 4\n6 3\n4 5\n1 5\n4 2\n5 1\n3 1\n6 4\n4 2\n1 5\n4 6\n2 1\n2 6",
"output": "NO"
},
{
"input": "96\n4\n1 5\n1 5\n4 6\n1 2\n4 2\n3 2\n4 6\n6 4\n6 3\n6 2\n4 1\n6 4\n5 1\n2 4\n5 6\n6 5\n3 2\n6 2\n3 1\n1 4\n3 2\n6 2\n2 4\n1 3\n5 4\n1 3\n6 2\n6 2\n5 6\n1 4\n4 2\n6 2\n3 1\n6 5\n3 1\n4 2\n6 3\n3 2\n3 6\n1 3\n5 6\n6 4\n1 4\n5 4\n2 6\n3 5\n5 4\n5 1\n2 4\n1 5\n1 3\n1 2\n1 3\n6 4\n6 3\n4 5\n4 1\n3 6\n1 2\n6 4\n1 2\n2 3\n2 1\n4 6\n1 3\n5 1\n4 5\n5 4\n6 3\n2 6\n5 1\n6 2\n3 1\n3 1\n5 4\n3 1\n5 6\n2 6\n5 6\n4 2\n6 5\n3 2\n6 5\n2 3\n6 4\n6 2\n1 2\n4 1\n1 2\n6 3\n2 1\n5 1\n6 5\n5 4\n4 5\n1 2",
"output": "NO"
},
{
"input": "5\n1\n2 3\n3 5\n4 5\n5 4\n5 3",
"output": "YES"
},
{
"input": "10\n5\n1 3\n3 1\n6 3\n6 3\n4 6\n3 1\n1 4\n3 1\n4 6\n1 3",
"output": "YES"
},
{
"input": "15\n4\n2 1\n2 6\n6 5\n5 1\n1 5\n2 1\n6 5\n5 1\n5 1\n6 2\n6 5\n5 1\n5 1\n6 5\n2 6",
"output": "YES"
},
{
"input": "20\n6\n3 2\n4 2\n3 5\n4 2\n5 3\n5 4\n2 3\n2 3\n4 5\n3 5\n3 2\n2 4\n4 5\n2 4\n3 2\n4 2\n5 4\n3 2\n3 5\n2 4",
"output": "YES"
},
{
"input": "25\n4\n1 2\n1 5\n5 6\n1 2\n5 1\n5 6\n5 1\n6 5\n2 1\n2 6\n2 6\n2 6\n2 6\n5 6\n2 6\n6 5\n2 1\n1 5\n1 2\n1 2\n6 5\n1 2\n6 5\n6 2\n2 6",
"output": "YES"
},
{
"input": "100\n3\n6 5\n1 5\n2 1\n5 1\n6 5\n5 1\n6 2\n1 2\n6 5\n5 1\n5 1\n1 5\n2 6\n6 2\n5 6\n1 2\n1 5\n5 6\n1 5\n1 2\n2 6\n1 2\n6 2\n1 5\n6 2\n2 6\n1 5\n6 2\n6 5\n5 6\n1 5\n5 6\n5 1\n5 1\n2 1\n1 2\n5 6\n6 5\n1 5\n5 1\n1 2\n1 5\n1 2\n2 6\n5 1\n2 6\n2 6\n5 6\n2 6\n6 5\n6 5\n1 5\n2 1\n5 6\n5 6\n1 2\n2 1\n1 2\n1 2\n1 2\n5 6\n6 2\n1 5\n1 2\n2 1\n2 6\n1 2\n5 1\n1 5\n6 5\n5 1\n5 1\n2 6\n5 6\n6 2\n1 2\n5 1\n6 2\n2 1\n5 6\n2 1\n1 5\n6 5\n6 5\n1 2\n1 2\n5 1\n6 2\n6 2\n1 2\n1 5\n6 5\n5 6\n1 2\n6 5\n2 1\n6 5\n1 5\n5 6\n6 5",
"output": "YES"
},
{
"input": "99\n3\n2 1\n2 6\n6 2\n1 5\n1 5\n6 2\n6 5\n6 5\n6 2\n5 6\n6 5\n6 2\n5 1\n2 6\n6 5\n1 5\n1 5\n2 6\n5 1\n1 5\n1 5\n2 1\n5 6\n6 5\n5 6\n2 6\n6 2\n6 5\n1 2\n1 2\n1 2\n2 6\n5 6\n1 2\n5 6\n1 2\n5 1\n6 5\n2 6\n5 1\n1 2\n1 5\n1 5\n6 2\n5 1\n2 6\n1 2\n5 1\n1 5\n6 5\n6 5\n5 6\n2 1\n2 6\n2 6\n1 2\n6 2\n2 6\n5 6\n6 5\n1 5\n2 1\n1 2\n6 2\n5 6\n6 5\n2 1\n1 5\n1 5\n2 6\n5 1\n1 2\n5 6\n2 1\n6 5\n5 1\n2 1\n6 2\n6 5\n6 5\n5 6\n1 2\n6 5\n1 2\n5 1\n2 1\n5 1\n2 6\n2 1\n6 2\n2 6\n2 6\n2 1\n2 1\n5 1\n1 5\n5 6\n2 1\n5 6",
"output": "YES"
},
{
"input": "98\n6\n4 2\n2 3\n2 3\n2 3\n2 3\n2 3\n3 2\n5 4\n4 2\n5 4\n5 4\n5 4\n5 3\n4 5\n2 3\n4 2\n5 3\n5 4\n4 5\n3 5\n3 2\n4 2\n2 4\n5 4\n2 3\n2 4\n5 4\n4 2\n3 5\n5 4\n2 3\n2 4\n3 5\n2 3\n3 5\n4 2\n3 5\n5 3\n4 2\n5 3\n5 3\n2 3\n2 4\n4 5\n3 2\n4 2\n3 5\n3 2\n3 5\n5 4\n3 5\n3 5\n4 2\n4 2\n3 2\n4 5\n5 4\n2 3\n5 4\n2 4\n2 3\n4 5\n3 5\n5 4\n3 2\n2 3\n5 3\n2 3\n5 3\n2 3\n2 3\n2 4\n2 3\n2 3\n5 3\n2 3\n4 2\n4 2\n5 4\n2 3\n2 3\n4 5\n3 2\n5 3\n3 2\n2 4\n2 4\n5 3\n5 4\n4 5\n5 3\n4 5\n2 4\n5 3\n4 2\n5 4\n2 4\n5 3",
"output": "YES"
},
{
"input": "97\n3\n2 1\n5 6\n1 2\n5 6\n2 6\n2 1\n6 2\n6 5\n6 2\n1 5\n1 2\n1 2\n6 2\n2 6\n6 5\n2 6\n6 5\n5 1\n6 2\n2 6\n2 6\n1 2\n2 6\n1 2\n1 5\n6 2\n6 5\n6 5\n2 6\n1 5\n6 5\n6 2\n6 2\n2 6\n5 6\n5 6\n1 5\n6 5\n2 6\n5 6\n1 5\n5 6\n1 5\n1 2\n5 1\n5 1\n1 5\n5 1\n1 5\n6 2\n6 2\n5 1\n6 5\n2 1\n2 6\n1 5\n1 5\n6 2\n2 6\n5 6\n2 6\n5 6\n2 6\n6 2\n5 6\n1 2\n6 2\n5 6\n6 2\n1 5\n5 6\n1 5\n2 6\n2 6\n2 1\n6 5\n5 1\n5 1\n1 2\n2 1\n2 1\n6 2\n1 5\n2 1\n2 1\n6 2\n5 1\n5 1\n2 6\n1 5\n1 2\n6 2\n2 6\n5 1\n6 5\n1 2\n6 2",
"output": "YES"
},
{
"input": "96\n4\n1 5\n5 1\n6 5\n2 1\n2 1\n2 6\n6 5\n6 5\n6 2\n2 6\n1 5\n6 5\n1 5\n2 6\n6 5\n5 6\n2 1\n2 6\n1 2\n1 5\n2 6\n2 6\n2 1\n1 5\n5 1\n1 2\n2 6\n2 6\n6 5\n1 5\n2 1\n2 6\n1 2\n5 6\n1 5\n2 6\n6 2\n2 6\n6 5\n1 5\n6 5\n6 5\n1 5\n5 1\n6 2\n5 1\n5 1\n1 5\n2 6\n5 1\n1 5\n2 1\n1 2\n6 2\n6 2\n5 6\n1 5\n6 5\n2 1\n6 5\n2 1\n2 1\n1 2\n6 2\n1 2\n1 5\n5 1\n5 6\n6 5\n6 2\n1 5\n2 6\n1 2\n1 2\n5 1\n1 5\n6 5\n6 2\n6 5\n2 6\n5 6\n2 1\n5 6\n2 1\n6 5\n2 6\n2 1\n1 5\n2 1\n6 2\n1 2\n1 5\n5 6\n5 1\n5 6\n2 1",
"output": "YES"
},
{
"input": "3\n6\n3 2\n5 4\n2 6",
"output": "NO"
},
{
"input": "4\n1\n2 3\n2 3\n2 3\n1 3",
"output": "NO"
},
{
"input": "2\n6\n3 2\n6 4",
"output": "NO"
},
{
"input": "3\n6\n3 2\n5 6\n2 4",
"output": "NO"
},
{
"input": "2\n5\n6 3\n4 5",
"output": "NO"
},
{
"input": "2\n6\n3 2\n6 5",
"output": "NO"
},
{
"input": "2\n1\n3 2\n1 2",
"output": "NO"
},
{
"input": "2\n3\n5 1\n3 5",
"output": "NO"
},
{
"input": "2\n1\n2 3\n1 2",
"output": "NO"
},
{
"input": "2\n1\n2 3\n2 1",
"output": "NO"
},
{
"input": "3\n1\n4 5\n4 1\n4 5",
"output": "NO"
},
{
"input": "2\n4\n2 6\n5 4",
"output": "NO"
},
{
"input": "2\n6\n3 2\n6 2",
"output": "NO"
},
{
"input": "2\n3\n2 1\n3 5",
"output": "NO"
},
{
"input": "2\n3\n1 2\n3 1",
"output": "NO"
},
{
"input": "2\n3\n2 6\n5 3",
"output": "NO"
},
{
"input": "3\n3\n1 2\n3 2\n3 1",
"output": "NO"
},
{
"input": "3\n5\n3 1\n1 3\n2 3",
"output": "NO"
},
{
"input": "2\n6\n2 4\n6 5",
"output": "NO"
},
{
"input": "2\n6\n4 5\n6 5",
"output": "NO"
},
{
"input": "2\n6\n3 5\n3 6",
"output": "NO"
},
{
"input": "2\n4\n1 2\n4 5",
"output": "NO"
},
{
"input": "2\n3\n2 6\n3 1",
"output": "NO"
}
] | 1,580,852,949
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 52
| 218
| 307,200
|
n = int(input())
x = int(input())
s = {x, 7-x}
print(('YES','NO')[any(s&set(map(int, input().split())) for _ in range(n))])
|
Title: Dice Tower
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A dice is a cube, its faces contain distinct integers from 1 to 6 as black points. The sum of numbers at the opposite dice faces always equals 7. Please note that there are only two dice (these dices are mirror of each other) that satisfy the given constraints (both of them are shown on the picture on the left).
Alice and Bob play dice. Alice has built a tower from *n* dice. We know that in this tower the adjacent dice contact with faces with distinct numbers. Bob wants to uniquely identify the numbers written on the faces of all dice, from which the tower is built. Unfortunately, Bob is looking at the tower from the face, and so he does not see all the numbers on the faces. Bob sees the number on the top of the tower and the numbers on the two adjacent sides (on the right side of the picture shown what Bob sees).
Help Bob, tell whether it is possible to uniquely identify the numbers on the faces of all the dice in the tower, or not.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of dice in the tower.
The second line contains an integer *x* (1<=≤<=*x*<=≤<=6) — the number Bob sees at the top of the tower. Next *n* lines contain two space-separated integers each: the *i*-th line contains numbers *a**i*,<=*b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=6; *a**i*<=≠<=*b**i*) — the numbers Bob sees on the two sidelong faces of the *i*-th dice in the tower.
Consider the dice in the tower indexed from top to bottom from 1 to *n*. That is, the topmost dice has index 1 (the dice whose top face Bob can see). It is guaranteed that it is possible to make a dice tower that will look as described in the input.
Output Specification:
Print "YES" (without the quotes), if it is possible to to uniquely identify the numbers on the faces of all the dice in the tower. If it is impossible, print "NO" (without the quotes).
Demo Input:
['3\n6\n3 2\n5 4\n2 4\n', '3\n3\n2 6\n4 1\n5 3\n']
Demo Output:
['YES', 'NO']
Note:
none
|
```python
n = int(input())
x = int(input())
s = {x, 7-x}
print(('YES','NO')[any(s&set(map(int, input().split())) for _ in range(n))])
```
| 3
|
|
120
|
A
|
Elevator
|
PROGRAMMING
| 1,000
|
[
"brute force",
"implementation",
"math"
] | null | null |
A sky scraper with 1000 floors has been built in the city of N. It has modern superfast elevators to help to travel from one floor to another. Each elevator has two doors, the front one and the back one. If one goes in through the front door, he goes out through the back one and vice versa. The elevator has two rails numbered with numbers 1 and 2. Rail 1 is located to the left of the entrance to the front door (or correspondingly, to the right of the entrance to the back door). Rail 2 is located opposite it, to the right of the entrance to the front door and to the left of the entrance to the back door. We know that each person in the city of N holds at a rail with the strongest hand.
One day a VIP person visited the city and of course, he took a look at the skyscraper and took a ride in the elevator. We know the door through which he entered and the rail he was holding at. Now we need to determine as soon as possible whether he is left-handed or right-handed.
|
The first line indicates the door through which the very important person entered the elevator. It contains "front" if the person enters the elevator through the front door and "back" if he entered the elevator through the back door. The second line contains integer *a* (1<=≤<=*a*<=≤<=2) which denotes the number of the rail at which the person was holding.
|
Print character "R" if the VIP is right-handed or "L" if he is left-handed.
|
[
"front\n1\n"
] |
[
"L\n"
] |
none
| 0
|
[
{
"input": "front\n1",
"output": "L"
},
{
"input": "back\n1",
"output": "R"
},
{
"input": "front\n2",
"output": "R"
},
{
"input": "back\n2",
"output": "L"
}
] | 1,588,716,597
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 4
| 218
| 0
|
#ROUNIAAUDI
files=open("input.txt","r")
t,t1=files.read().split()
print(t,t1)
files.close()
files2=open("output.txt","w")
if t=="front" and t1=="1":
files2.write("L")
elif t == "front" and t1 == "2":
files2.write("R")
elif t == "back" and t1 == "2":
files2.write("L")
elif t == "back" and t1 == "1":
files2.write("R")
files2.close()
|
Title: Elevator
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A sky scraper with 1000 floors has been built in the city of N. It has modern superfast elevators to help to travel from one floor to another. Each elevator has two doors, the front one and the back one. If one goes in through the front door, he goes out through the back one and vice versa. The elevator has two rails numbered with numbers 1 and 2. Rail 1 is located to the left of the entrance to the front door (or correspondingly, to the right of the entrance to the back door). Rail 2 is located opposite it, to the right of the entrance to the front door and to the left of the entrance to the back door. We know that each person in the city of N holds at a rail with the strongest hand.
One day a VIP person visited the city and of course, he took a look at the skyscraper and took a ride in the elevator. We know the door through which he entered and the rail he was holding at. Now we need to determine as soon as possible whether he is left-handed or right-handed.
Input Specification:
The first line indicates the door through which the very important person entered the elevator. It contains "front" if the person enters the elevator through the front door and "back" if he entered the elevator through the back door. The second line contains integer *a* (1<=≤<=*a*<=≤<=2) which denotes the number of the rail at which the person was holding.
Output Specification:
Print character "R" if the VIP is right-handed or "L" if he is left-handed.
Demo Input:
['front\n1\n']
Demo Output:
['L\n']
Note:
none
|
```python
#ROUNIAAUDI
files=open("input.txt","r")
t,t1=files.read().split()
print(t,t1)
files.close()
files2=open("output.txt","w")
if t=="front" and t1=="1":
files2.write("L")
elif t == "front" and t1 == "2":
files2.write("R")
elif t == "back" and t1 == "2":
files2.write("L")
elif t == "back" and t1 == "1":
files2.write("R")
files2.close()
```
| 3
|
|
106
|
A
|
Card Game
|
PROGRAMMING
| 1,000
|
[
"implementation"
] |
A. Card Game
|
2
|
256
|
There is a card game called "Durak", which means "Fool" in Russian. The game is quite popular in the countries that used to form USSR. The problem does not state all the game's rules explicitly — you can find them later yourselves if you want.
To play durak you need a pack of 36 cards. Each card has a suit ("S", "H", "D" and "C") and a rank (in the increasing order "6", "7", "8", "9", "T", "J", "Q", "K" and "A"). At the beginning of the game one suit is arbitrarily chosen as trump.
The players move like that: one player puts one or several of his cards on the table and the other one should beat each of them with his cards.
A card beats another one if both cards have similar suits and the first card has a higher rank then the second one. Besides, a trump card can beat any non-trump card whatever the cards’ ranks are. In all other cases you can not beat the second card with the first one.
You are given the trump suit and two different cards. Determine whether the first one beats the second one or not.
|
The first line contains the tramp suit. It is "S", "H", "D" or "C".
The second line contains the description of the two different cards. Each card is described by one word consisting of two symbols. The first symbol stands for the rank ("6", "7", "8", "9", "T", "J", "Q", "K" and "A"), and the second one stands for the suit ("S", "H", "D" and "C").
|
Print "YES" (without the quotes) if the first cards beats the second one. Otherwise, print "NO" (also without the quotes).
|
[
"H\nQH 9S\n",
"S\n8D 6D\n",
"C\n7H AS\n"
] |
[
"YES\n",
"YES",
"NO"
] |
none
| 500
|
[
{
"input": "H\nQH 9S",
"output": "YES"
},
{
"input": "S\n8D 6D",
"output": "YES"
},
{
"input": "C\n7H AS",
"output": "NO"
},
{
"input": "C\nKC 9C",
"output": "YES"
},
{
"input": "D\n7D KD",
"output": "NO"
},
{
"input": "H\n7H KD",
"output": "YES"
},
{
"input": "D\nAS AH",
"output": "NO"
},
{
"input": "H\nKH KS",
"output": "YES"
},
{
"input": "C\n9H 6C",
"output": "NO"
},
{
"input": "C\n9H JC",
"output": "NO"
},
{
"input": "D\nTD JD",
"output": "NO"
},
{
"input": "H\n6S 7S",
"output": "NO"
},
{
"input": "D\n7S 8S",
"output": "NO"
},
{
"input": "S\n8H 9H",
"output": "NO"
},
{
"input": "C\n9D TD",
"output": "NO"
},
{
"input": "H\nTC JC",
"output": "NO"
},
{
"input": "C\nJH QH",
"output": "NO"
},
{
"input": "H\nQD KD",
"output": "NO"
},
{
"input": "D\nKS AS",
"output": "NO"
},
{
"input": "S\nAH 6H",
"output": "YES"
},
{
"input": "H\n7D 6D",
"output": "YES"
},
{
"input": "S\n8H 7H",
"output": "YES"
},
{
"input": "D\n9S 8S",
"output": "YES"
},
{
"input": "S\nTC 9C",
"output": "YES"
},
{
"input": "H\nJS TS",
"output": "YES"
},
{
"input": "S\nQD JD",
"output": "YES"
},
{
"input": "D\nKH QH",
"output": "YES"
},
{
"input": "H\nAD KD",
"output": "YES"
},
{
"input": "H\nQS QD",
"output": "NO"
},
{
"input": "C\nTS TH",
"output": "NO"
},
{
"input": "C\n6C 6D",
"output": "YES"
},
{
"input": "H\n8H 8D",
"output": "YES"
},
{
"input": "S\n7D 7S",
"output": "NO"
},
{
"input": "H\nJC JH",
"output": "NO"
},
{
"input": "H\n8H 9C",
"output": "YES"
},
{
"input": "D\n9D 6S",
"output": "YES"
},
{
"input": "C\nJC AH",
"output": "YES"
},
{
"input": "S\nAS KD",
"output": "YES"
},
{
"input": "S\n7S JS",
"output": "NO"
},
{
"input": "H\nTH 8H",
"output": "YES"
},
{
"input": "S\n7S QS",
"output": "NO"
},
{
"input": "C\nKC QC",
"output": "YES"
},
{
"input": "S\nAD 9S",
"output": "NO"
},
{
"input": "D\n7H 8D",
"output": "NO"
},
{
"input": "H\nJC 9H",
"output": "NO"
},
{
"input": "C\n7S AC",
"output": "NO"
},
{
"input": "C\n8C 7C",
"output": "YES"
},
{
"input": "H\n9D 8S",
"output": "NO"
},
{
"input": "D\nAC KS",
"output": "NO"
},
{
"input": "H\n8C QH",
"output": "NO"
},
{
"input": "S\n7S TS",
"output": "NO"
},
{
"input": "C\nAH 6S",
"output": "NO"
},
{
"input": "S\nKS QS",
"output": "YES"
},
{
"input": "H\nAC QC",
"output": "YES"
},
{
"input": "S\n9H 8D",
"output": "NO"
},
{
"input": "S\nTS JS",
"output": "NO"
},
{
"input": "S\n8H 7C",
"output": "NO"
},
{
"input": "C\nAH 6S",
"output": "NO"
},
{
"input": "S\n7S QS",
"output": "NO"
},
{
"input": "C\nAH 6S",
"output": "NO"
},
{
"input": "S\nTS KS",
"output": "NO"
},
{
"input": "C\nTH KH",
"output": "NO"
},
{
"input": "H\n9C 6D",
"output": "NO"
},
{
"input": "H\n9C 8D",
"output": "NO"
},
{
"input": "H\nTH AH",
"output": "NO"
},
{
"input": "H\nTH JH",
"output": "NO"
},
{
"input": "H\nQS 9C",
"output": "NO"
},
{
"input": "H\nKC AC",
"output": "NO"
},
{
"input": "H\nAH KH",
"output": "YES"
},
{
"input": "H\nKS QS",
"output": "YES"
},
{
"input": "C\nAD KS",
"output": "NO"
},
{
"input": "H\nQS 9C",
"output": "NO"
},
{
"input": "H\n9D 7S",
"output": "NO"
},
{
"input": "D\n6D 9S",
"output": "YES"
},
{
"input": "H\nAH KH",
"output": "YES"
},
{
"input": "H\nKC AC",
"output": "NO"
},
{
"input": "D\n8S 6C",
"output": "NO"
},
{
"input": "S\nAC KC",
"output": "YES"
}
] | 1,609,044,113
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 78
| 218
| 0
|
t = str(input())
s1, s2 = map(str,input().split())
temp = "6789TJQKA"
if s1[1] == s2[1]:
print("YES" if temp.find(s1[0]) > temp.find(s2[0]) else "NO")
else:
print("YES" if s1[1] == t else "NO")
|
Title: Card Game
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
There is a card game called "Durak", which means "Fool" in Russian. The game is quite popular in the countries that used to form USSR. The problem does not state all the game's rules explicitly — you can find them later yourselves if you want.
To play durak you need a pack of 36 cards. Each card has a suit ("S", "H", "D" and "C") and a rank (in the increasing order "6", "7", "8", "9", "T", "J", "Q", "K" and "A"). At the beginning of the game one suit is arbitrarily chosen as trump.
The players move like that: one player puts one or several of his cards on the table and the other one should beat each of them with his cards.
A card beats another one if both cards have similar suits and the first card has a higher rank then the second one. Besides, a trump card can beat any non-trump card whatever the cards’ ranks are. In all other cases you can not beat the second card with the first one.
You are given the trump suit and two different cards. Determine whether the first one beats the second one or not.
Input Specification:
The first line contains the tramp suit. It is "S", "H", "D" or "C".
The second line contains the description of the two different cards. Each card is described by one word consisting of two symbols. The first symbol stands for the rank ("6", "7", "8", "9", "T", "J", "Q", "K" and "A"), and the second one stands for the suit ("S", "H", "D" and "C").
Output Specification:
Print "YES" (without the quotes) if the first cards beats the second one. Otherwise, print "NO" (also without the quotes).
Demo Input:
['H\nQH 9S\n', 'S\n8D 6D\n', 'C\n7H AS\n']
Demo Output:
['YES\n', 'YES', 'NO']
Note:
none
|
```python
t = str(input())
s1, s2 = map(str,input().split())
temp = "6789TJQKA"
if s1[1] == s2[1]:
print("YES" if temp.find(s1[0]) > temp.find(s2[0]) else "NO")
else:
print("YES" if s1[1] == t else "NO")
```
| 3.9455
|
664
|
A
|
Complicated GCD
|
PROGRAMMING
| 800
|
[
"math",
"number theory"
] | null | null |
Greatest common divisor *GCD*(*a*,<=*b*) of two positive integers *a* and *b* is equal to the biggest integer *d* such that both integers *a* and *b* are divisible by *d*. There are many efficient algorithms to find greatest common divisor *GCD*(*a*,<=*b*), for example, Euclid algorithm.
Formally, find the biggest integer *d*, such that all integers *a*,<=*a*<=+<=1,<=*a*<=+<=2,<=...,<=*b* are divisible by *d*. To make the problem even more complicated we allow *a* and *b* to be up to googol, 10100 — such number do not fit even in 64-bit integer type!
|
The only line of the input contains two integers *a* and *b* (1<=≤<=*a*<=≤<=*b*<=≤<=10100).
|
Output one integer — greatest common divisor of all integers from *a* to *b* inclusive.
|
[
"1 2\n",
"61803398874989484820458683436563811772030917980576 61803398874989484820458683436563811772030917980576\n"
] |
[
"1\n",
"61803398874989484820458683436563811772030917980576\n"
] |
none
| 500
|
[
{
"input": "1 2",
"output": "1"
},
{
"input": "61803398874989484820458683436563811772030917980576 61803398874989484820458683436563811772030917980576",
"output": "61803398874989484820458683436563811772030917980576"
},
{
"input": "1 100",
"output": "1"
},
{
"input": "100 100000",
"output": "1"
},
{
"input": "12345 67890123456789123457",
"output": "1"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "2 2",
"output": "2"
},
{
"input": "8392739158839273915883927391588392739158839273915883927391588392739158839273915883927391588392739158 8392739158839273915883927391588392739158839273915883927391588392739158839273915883927391588392739158",
"output": "8392739158839273915883927391588392739158839273915883927391588392739158839273915883927391588392739158"
},
{
"input": "1 10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "1"
},
{
"input": "8328748239473982794239847237438782379810988324751 9328748239473982794239847237438782379810988324751",
"output": "1"
},
{
"input": "1029398958432734901284327523909481928483573793 1029398958432734901284327523909481928483573794",
"output": "1"
},
{
"input": "10000 1000000000",
"output": "1"
},
{
"input": "10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "11210171722243 65715435710585778347",
"output": "1"
},
{
"input": "2921881079263974825226940825843 767693191032295360887755303860323261471",
"output": "1"
},
{
"input": "8025352957265704896940312528736939363590612908210603 96027920417708260814607687034511406492969694925539085",
"output": "1"
},
{
"input": "23510978780782786207241069904470895053213996267165977112058175452757132930 210352653280909370107314249722987050753257161175393375412301228883856435481424",
"output": "1"
},
{
"input": "8150070767079366215626260746398623663859344142817267779361251788637547414925170226504788118262 49924902262298336032630839998470954964895251605110946547855439236151401194070172107435992986913614",
"output": "1"
},
{
"input": "15943150466658398903 15943150466658398903",
"output": "15943150466658398903"
},
{
"input": "410470228200245407491525399055972 410470228200245407491525399055972",
"output": "410470228200245407491525399055972"
},
{
"input": "51894705655711504622197349350106792045098781545973899451307 51894705655711504622197349350106792045098781545973899451307",
"output": "51894705655711504622197349350106792045098781545973899451307"
},
{
"input": "60353594589897438036015726222485085035927634677598681595162804007836722215668410 60353594589897438036015726222485085035927634677598681595162804007836722215668410",
"output": "60353594589897438036015726222485085035927634677598681595162804007836722215668410"
},
{
"input": "761733780145118977868180796896376577405349682060892737466239031663476251177476275459280340045369535 761733780145118977868180796896376577405349682060892737466239031663476251177476275459280340045369535",
"output": "761733780145118977868180796896376577405349682060892737466239031663476251177476275459280340045369535"
},
{
"input": "1 2000000000",
"output": "1"
},
{
"input": "13 1928834874",
"output": "1"
},
{
"input": "87 2938984237482934238",
"output": "1"
},
{
"input": "213 413",
"output": "1"
},
{
"input": "3 4",
"output": "1"
}
] | 1,631,319,928
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 28
| 77
| 6,758,400
|
if __name__ == '__main__':
numA, numB = input().split()
if numA == numB:
print(numA)
else:
print("1")
|
Title: Complicated GCD
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Greatest common divisor *GCD*(*a*,<=*b*) of two positive integers *a* and *b* is equal to the biggest integer *d* such that both integers *a* and *b* are divisible by *d*. There are many efficient algorithms to find greatest common divisor *GCD*(*a*,<=*b*), for example, Euclid algorithm.
Formally, find the biggest integer *d*, such that all integers *a*,<=*a*<=+<=1,<=*a*<=+<=2,<=...,<=*b* are divisible by *d*. To make the problem even more complicated we allow *a* and *b* to be up to googol, 10100 — such number do not fit even in 64-bit integer type!
Input Specification:
The only line of the input contains two integers *a* and *b* (1<=≤<=*a*<=≤<=*b*<=≤<=10100).
Output Specification:
Output one integer — greatest common divisor of all integers from *a* to *b* inclusive.
Demo Input:
['1 2\n', '61803398874989484820458683436563811772030917980576 61803398874989484820458683436563811772030917980576\n']
Demo Output:
['1\n', '61803398874989484820458683436563811772030917980576\n']
Note:
none
|
```python
if __name__ == '__main__':
numA, numB = input().split()
if numA == numB:
print(numA)
else:
print("1")
```
| 3
|
|
71
|
A
|
Way Too Long Words
|
PROGRAMMING
| 800
|
[
"strings"
] |
A. Way Too Long Words
|
1
|
256
|
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
|
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
|
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
|
[
"4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n"
] |
[
"word\nl10n\ni18n\np43s\n"
] |
none
| 500
|
[
{
"input": "4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis",
"output": "word\nl10n\ni18n\np43s"
},
{
"input": "5\nabcdefgh\nabcdefghi\nabcdefghij\nabcdefghijk\nabcdefghijklm",
"output": "abcdefgh\nabcdefghi\nabcdefghij\na9k\na11m"
},
{
"input": "3\nnjfngnrurunrgunrunvurn\njfvnjfdnvjdbfvsbdubruvbubvkdb\nksdnvidnviudbvibd",
"output": "n20n\nj27b\nk15d"
},
{
"input": "1\ntcyctkktcctrcyvbyiuhihhhgyvyvyvyvjvytchjckt",
"output": "t41t"
},
{
"input": "24\nyou\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nunofficially\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings",
"output": "you\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nu10y\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings"
},
{
"input": "1\na",
"output": "a"
},
{
"input": "26\na\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz",
"output": "a\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz"
},
{
"input": "1\nabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghij",
"output": "a98j"
},
{
"input": "10\ngyartjdxxlcl\nfzsck\nuidwu\nxbymclornemdmtj\nilppyoapitawgje\ncibzc\ndrgbeu\nhezplmsdekhhbo\nfeuzlrimbqbytdu\nkgdco",
"output": "g10l\nfzsck\nuidwu\nx13j\ni13e\ncibzc\ndrgbeu\nh12o\nf13u\nkgdco"
},
{
"input": "20\nlkpmx\nkovxmxorlgwaomlswjxlpnbvltfv\nhykasjxqyjrmybejnmeumzha\ntuevlumpqbbhbww\nqgqsphvrmupxxc\ntrissbaf\nqfgrlinkzvzqdryckaizutd\nzzqtoaxkvwoscyx\noswytrlnhpjvvnwookx\nlpuzqgec\ngyzqfwxggtvpjhzmzmdw\nrlxjgmvdftvrmvbdwudra\nvsntnjpepnvdaxiporggmglhagv\nxlvcqkqgcrbgtgglj\nlyxwxbiszyhlsrgzeedzprbmcpduvq\nyrmqqvrkqskqukzqrwukpsifgtdc\nxpuohcsjhhuhvr\nvvlfrlxpvqejngwrbfbpmqeirxlw\nsvmasocxdvadmaxtrpakysmeaympy\nyuflqboqfdt",
"output": "lkpmx\nk26v\nh22a\nt13w\nq12c\ntrissbaf\nq21d\nz13x\no17x\nlpuzqgec\ng18w\nr19a\nv25v\nx15j\nl28q\ny26c\nx12r\nv26w\ns27y\ny9t"
},
{
"input": "100\nm\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr",
"output": "m\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr"
}
] | 1,673,522,951
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 20
| 46
| 0
|
count = int(input())
toprint = ""
for x in range(count):
word = input()
if len(word)>10:
num = len(word)-2
ans = word[0] + str(num) + word[len(word)-1]
else:
ans = word
if x==0:
toprint += ans
else:
toprint += "\n" + ans
print(toprint)
|
Title: Way Too Long Words
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
Output Specification:
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
Demo Input:
['4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n']
Demo Output:
['word\nl10n\ni18n\np43s\n']
Note:
none
|
```python
count = int(input())
toprint = ""
for x in range(count):
word = input()
if len(word)>10:
num = len(word)-2
ans = word[0] + str(num) + word[len(word)-1]
else:
ans = word
if x==0:
toprint += ans
else:
toprint += "\n" + ans
print(toprint)
```
| 3.977
|
69
|
A
|
Young Physicist
|
PROGRAMMING
| 1,000
|
[
"implementation",
"math"
] |
A. Young Physicist
|
2
|
256
|
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
|
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100).
|
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
|
[
"3\n4 1 7\n-2 4 -1\n1 -5 -3\n",
"3\n3 -1 7\n-5 2 -4\n2 -1 -3\n"
] |
[
"NO",
"YES"
] |
none
| 500
|
[
{
"input": "3\n4 1 7\n-2 4 -1\n1 -5 -3",
"output": "NO"
},
{
"input": "3\n3 -1 7\n-5 2 -4\n2 -1 -3",
"output": "YES"
},
{
"input": "10\n21 32 -46\n43 -35 21\n42 2 -50\n22 40 20\n-27 -9 38\n-4 1 1\n-40 6 -31\n-13 -2 34\n-21 34 -12\n-32 -29 41",
"output": "NO"
},
{
"input": "10\n25 -33 43\n-27 -42 28\n-35 -20 19\n41 -42 -1\n49 -39 -4\n-49 -22 7\n-19 29 41\n8 -27 -43\n8 34 9\n-11 -3 33",
"output": "NO"
},
{
"input": "10\n-6 21 18\n20 -11 -8\n37 -11 41\n-5 8 33\n29 23 32\n30 -33 -11\n39 -49 -36\n28 34 -49\n22 29 -34\n-18 -6 7",
"output": "NO"
},
{
"input": "10\n47 -2 -27\n0 26 -14\n5 -12 33\n2 18 3\n45 -30 -49\n4 -18 8\n-46 -44 -41\n-22 -10 -40\n-35 -21 26\n33 20 38",
"output": "NO"
},
{
"input": "13\n-3 -36 -46\n-11 -50 37\n42 -11 -15\n9 42 44\n-29 -12 24\n3 9 -40\n-35 13 50\n14 43 18\n-13 8 24\n-48 -15 10\n50 9 -50\n21 0 -50\n0 0 -6",
"output": "YES"
},
{
"input": "14\n43 23 17\n4 17 44\n5 -5 -16\n-43 -7 -6\n47 -48 12\n50 47 -45\n2 14 43\n37 -30 15\n4 -17 -11\n17 9 -45\n-50 -3 -8\n-50 0 0\n-50 0 0\n-16 0 0",
"output": "YES"
},
{
"input": "13\n29 49 -11\n38 -11 -20\n25 1 -40\n-11 28 11\n23 -19 1\n45 -41 -17\n-3 0 -19\n-13 -33 49\n-30 0 28\n34 17 45\n-50 9 -27\n-50 0 0\n-37 0 0",
"output": "YES"
},
{
"input": "12\n3 28 -35\n-32 -44 -17\n9 -25 -6\n-42 -22 20\n-19 15 38\n-21 38 48\n-1 -37 -28\n-10 -13 -50\n-5 21 29\n34 28 50\n50 11 -49\n34 0 0",
"output": "YES"
},
{
"input": "37\n-64 -79 26\n-22 59 93\n-5 39 -12\n77 -9 76\n55 -86 57\n83 100 -97\n-70 94 84\n-14 46 -94\n26 72 35\n14 78 -62\n17 82 92\n-57 11 91\n23 15 92\n-80 -1 1\n12 39 18\n-23 -99 -75\n-34 50 19\n-39 84 -7\n45 -30 -39\n-60 49 37\n45 -16 -72\n33 -51 -56\n-48 28 5\n97 91 88\n45 -82 -11\n-21 -15 -90\n-53 73 -26\n-74 85 -90\n-40 23 38\n100 -13 49\n32 -100 -100\n0 -100 -70\n0 -100 0\n0 -100 0\n0 -100 0\n0 -100 0\n0 -37 0",
"output": "YES"
},
{
"input": "4\n68 3 100\n68 21 -100\n-100 -24 0\n-36 0 0",
"output": "YES"
},
{
"input": "33\n-1 -46 -12\n45 -16 -21\n-11 45 -21\n-60 -42 -93\n-22 -45 93\n37 96 85\n-76 26 83\n-4 9 55\n7 -52 -9\n66 8 -85\n-100 -54 11\n-29 59 74\n-24 12 2\n-56 81 85\n-92 69 -52\n-26 -97 91\n54 59 -51\n58 21 -57\n7 68 56\n-47 -20 -51\n-59 77 -13\n-85 27 91\n79 60 -56\n66 -80 5\n21 -99 42\n-31 -29 98\n66 93 76\n-49 45 61\n100 -100 -100\n100 -100 -100\n66 -75 -100\n0 0 -100\n0 0 -87",
"output": "YES"
},
{
"input": "3\n1 2 3\n3 2 1\n0 0 0",
"output": "NO"
},
{
"input": "2\n5 -23 12\n0 0 0",
"output": "NO"
},
{
"input": "1\n0 0 0",
"output": "YES"
},
{
"input": "1\n1 -2 0",
"output": "NO"
},
{
"input": "2\n-23 77 -86\n23 -77 86",
"output": "YES"
},
{
"input": "26\n86 7 20\n-57 -64 39\n-45 6 -93\n-44 -21 100\n-11 -49 21\n73 -71 -80\n-2 -89 56\n-65 -2 7\n5 14 84\n57 41 13\n-12 69 54\n40 -25 27\n-17 -59 0\n64 -91 -30\n-53 9 42\n-54 -8 14\n-35 82 27\n-48 -59 -80\n88 70 79\n94 57 97\n44 63 25\n84 -90 -40\n-100 100 -100\n-92 100 -100\n0 10 -100\n0 0 -82",
"output": "YES"
},
{
"input": "42\n11 27 92\n-18 -56 -57\n1 71 81\n33 -92 30\n82 83 49\n-87 -61 -1\n-49 45 49\n73 26 15\n-22 22 -77\n29 -93 87\n-68 44 -90\n-4 -84 20\n85 67 -6\n-39 26 77\n-28 -64 20\n65 -97 24\n-72 -39 51\n35 -75 -91\n39 -44 -8\n-25 -27 -57\n91 8 -46\n-98 -94 56\n94 -60 59\n-9 -95 18\n-53 -37 98\n-8 -94 -84\n-52 55 60\n15 -14 37\n65 -43 -25\n94 12 66\n-8 -19 -83\n29 81 -78\n-58 57 33\n24 86 -84\n-53 32 -88\n-14 7 3\n89 97 -53\n-5 -28 -91\n-100 100 -6\n-84 100 0\n0 100 0\n0 70 0",
"output": "YES"
},
{
"input": "3\n96 49 -12\n2 -66 28\n-98 17 -16",
"output": "YES"
},
{
"input": "5\n70 -46 86\n-100 94 24\n-27 63 -63\n57 -100 -47\n0 -11 0",
"output": "YES"
},
{
"input": "18\n-86 -28 70\n-31 -89 42\n31 -48 -55\n95 -17 -43\n24 -95 -85\n-21 -14 31\n68 -18 81\n13 31 60\n-15 28 99\n-42 15 9\n28 -61 -62\n-16 71 29\n-28 75 -48\n-77 -67 36\n-100 83 89\n100 100 -100\n57 34 -100\n0 0 -53",
"output": "YES"
},
{
"input": "44\n52 -54 -29\n-82 -5 -94\n-54 43 43\n91 16 71\n7 80 -91\n3 15 29\n-99 -6 -77\n-3 -77 -64\n73 67 34\n25 -10 -18\n-29 91 63\n-72 86 -16\n-68 85 -81\n-3 36 44\n-74 -14 -80\n34 -96 -97\n-76 -78 -33\n-24 44 -58\n98 12 77\n95 -63 -6\n-51 3 -90\n-92 -10 72\n7 3 -68\n57 -53 71\n29 57 -48\n35 -60 10\n79 -70 -61\n-20 77 55\n-86 -15 -35\n84 -88 -18\n100 -42 77\n-20 46 8\n-41 -43 -65\n38 -98 -23\n-100 65 45\n-7 -91 -63\n46 88 -85\n48 59 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 1",
"output": "YES"
},
{
"input": "18\n-14 -64 -91\n-8 -66 -86\n-23 92 -40\n6 -3 -53\n57 41 78\n-79 42 -22\n-88 -17 45\n4 -45 44\n83 -18 -25\n34 86 -92\n75 -30 12\n44 99 11\n-67 -13 72\n22 83 -56\n-37 71 72\n-9 -100 100\n0 -100 31\n0 -58 0",
"output": "YES"
},
{
"input": "23\n-70 37 78\n42 84 6\n28 -94 -24\n-49 76 95\n-67 18 84\n-53 78 -5\n65 -63 -64\n-66 -64 -69\n81 -93 95\n10 52 -79\n-89 -61 -64\n-64 -47 43\n-81 -35 55\n80 82 73\n-60 -81 -18\n49 -10 -19\n-58 70 12\n-24 -15 -93\n98 -93 -54\n-28 -75 11\n100 100 -63\n100 100 0\n56 34 0",
"output": "YES"
},
{
"input": "3\n0 2 -2\n1 -1 3\n-3 0 0",
"output": "NO"
}
] | 1,665,044,943
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 81
| 92
| 0
|
n = int(input())
a = [0,0,0]
for i in range(n):
x,y,z = map(int,input().split())
a[0] += x
a[1] += y
a[2] += z
if any(a):
print("NO")
else:
print("YES")
|
Title: Young Physicist
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100).
Output Specification:
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
Demo Input:
['3\n4 1 7\n-2 4 -1\n1 -5 -3\n', '3\n3 -1 7\n-5 2 -4\n2 -1 -3\n']
Demo Output:
['NO', 'YES']
Note:
none
|
```python
n = int(input())
a = [0,0,0]
for i in range(n):
x,y,z = map(int,input().split())
a[0] += x
a[1] += y
a[2] += z
if any(a):
print("NO")
else:
print("YES")
```
| 3.977
|
129
|
A
|
Cookies
|
PROGRAMMING
| 900
|
[
"implementation"
] | null | null |
Olga came to visit the twins Anna and Maria and saw that they have many cookies. The cookies are distributed into bags. As there are many cookies, Olga decided that it's no big deal if she steals a bag. However, she doesn't want the sisters to quarrel because of nothing when they divide the cookies. That's why Olga wants to steal a bag with cookies so that the number of cookies in the remaining bags was even, that is, so that Anna and Maria could evenly divide it into two (even 0 remaining cookies will do, just as any other even number). How many ways there are to steal exactly one cookie bag so that the total number of cookies in the remaining bags was even?
|
The first line contains the only integer *n* (1<=≤<=*n*<=≤<=100) — the number of cookie bags Anna and Maria have. The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=100) — the number of cookies in the *i*-th bag.
|
Print in the only line the only number — the sought number of ways. If there are no such ways print 0.
|
[
"1\n1\n",
"10\n1 2 2 3 4 4 4 2 2 2\n",
"11\n2 2 2 2 2 2 2 2 2 2 99\n"
] |
[
"1\n",
"8\n",
"1\n"
] |
In the first sample Olga should take the only bag so that the twins ended up with the even number of cookies.
In the second sample Olga can take any of five bags with two cookies or any of three bags with four cookies — 5 + 3 = 8 ways in total.
In the third sample, no matter which bag with two cookies Olga chooses, the twins are left with 2 * 9 + 99 = 117 cookies. Thus, Olga has only one option: to take the bag with 99 cookies.
| 500
|
[
{
"input": "1\n1",
"output": "1"
},
{
"input": "10\n1 2 2 3 4 4 4 2 2 2",
"output": "8"
},
{
"input": "11\n2 2 2 2 2 2 2 2 2 2 99",
"output": "1"
},
{
"input": "2\n1 1",
"output": "0"
},
{
"input": "2\n2 2",
"output": "2"
},
{
"input": "2\n1 2",
"output": "1"
},
{
"input": "7\n7 7 7 7 7 7 7",
"output": "7"
},
{
"input": "8\n1 2 3 4 5 6 7 8",
"output": "4"
},
{
"input": "100\n1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2",
"output": "50"
},
{
"input": "99\n99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99",
"output": "49"
},
{
"input": "82\n43 44 96 33 23 42 33 66 53 87 8 90 43 91 40 88 51 18 48 62 59 10 22 20 54 6 13 63 2 56 31 52 98 42 54 32 26 77 9 24 33 91 16 30 39 34 78 82 73 90 12 15 67 76 30 18 44 86 84 98 65 54 100 79 28 34 40 56 11 43 72 35 86 59 89 40 30 33 7 19 44 15",
"output": "50"
},
{
"input": "17\n50 14 17 77 74 74 38 76 41 27 45 29 66 98 38 73 38",
"output": "7"
},
{
"input": "94\n81 19 90 99 26 11 86 44 78 36 80 59 99 90 78 72 71 20 94 56 42 40 71 84 10 85 10 70 52 27 39 55 90 16 48 25 7 79 99 100 38 10 99 56 3 4 78 9 16 57 14 40 52 54 57 70 30 86 56 84 97 60 59 69 49 66 23 92 90 46 86 73 53 47 1 83 14 20 24 66 13 45 41 14 86 75 55 88 48 95 82 24 47 87",
"output": "39"
},
{
"input": "88\n64 95 12 90 40 65 98 45 52 54 79 7 81 25 98 19 68 82 41 53 35 50 5 22 32 21 8 39 8 6 72 27 81 30 12 79 21 42 60 2 66 87 46 93 62 78 52 71 76 32 78 94 86 85 55 15 34 76 41 20 32 26 94 81 89 45 74 49 11 40 40 39 49 46 80 85 90 23 80 40 86 58 70 26 48 93 23 53",
"output": "37"
},
{
"input": "84\n95 9 43 43 13 84 60 90 1 8 97 99 54 34 59 83 33 15 51 26 40 12 66 65 19 30 29 78 92 60 25 13 19 84 71 73 12 24 54 49 16 41 11 40 57 59 34 40 39 9 71 83 1 77 79 53 94 47 78 55 77 85 29 52 80 90 53 77 97 97 27 79 28 23 83 25 26 22 49 86 63 56 3 32",
"output": "51"
},
{
"input": "47\n61 97 76 94 91 22 2 68 62 73 90 47 16 79 44 71 98 68 43 6 53 52 40 27 68 67 43 96 14 91 60 61 96 24 97 13 32 65 85 96 81 77 34 18 23 14 80",
"output": "21"
},
{
"input": "69\n71 1 78 74 58 89 30 6 100 90 22 61 11 59 14 74 27 25 78 61 45 19 25 33 37 4 52 43 53 38 9 100 56 67 69 38 76 91 63 60 93 52 28 61 9 98 8 14 57 63 89 64 98 51 36 66 36 86 13 82 50 91 52 64 86 78 78 83 81",
"output": "37"
},
{
"input": "52\n38 78 36 75 19 3 56 1 39 97 24 79 84 16 93 55 96 64 12 24 1 86 80 29 12 32 36 36 73 39 76 65 53 98 30 20 28 8 86 43 70 22 75 69 62 65 81 25 53 40 71 59",
"output": "28"
},
{
"input": "74\n81 31 67 97 26 75 69 81 11 13 13 74 77 88 52 20 52 64 66 75 72 28 41 54 26 75 41 91 75 15 18 36 13 83 63 61 14 48 53 63 19 67 35 48 23 65 73 100 44 55 92 88 99 17 73 25 83 7 31 89 12 80 98 39 42 75 14 29 81 35 77 87 33 94",
"output": "47"
},
{
"input": "44\n46 56 31 31 37 71 94 2 14 100 45 72 36 72 80 3 38 54 42 98 50 32 31 42 62 31 45 50 95 100 18 17 64 22 18 25 52 56 70 57 43 40 81 28",
"output": "15"
},
{
"input": "22\n28 57 40 74 51 4 45 84 99 12 95 14 92 60 47 81 84 51 31 91 59 42",
"output": "11"
},
{
"input": "59\n73 45 94 76 41 49 65 13 74 66 36 25 47 75 40 23 92 72 11 32 32 8 81 26 68 56 41 8 76 47 96 55 70 11 84 14 83 18 70 22 30 39 28 100 48 11 92 45 78 69 86 1 54 90 98 91 13 17 35",
"output": "33"
},
{
"input": "63\n20 18 44 94 68 57 16 43 74 55 68 24 21 95 76 84 50 50 47 86 86 12 58 55 28 72 86 18 34 45 81 88 3 72 41 9 60 90 81 93 12 6 9 6 2 41 1 7 9 29 81 14 64 80 20 36 67 54 7 5 35 81 22",
"output": "37"
},
{
"input": "28\n49 84 48 19 44 91 11 82 96 95 88 90 71 82 87 25 31 23 18 13 98 45 26 65 35 12 31 14",
"output": "15"
},
{
"input": "61\n34 18 28 64 28 45 9 77 77 20 63 92 79 16 16 100 86 2 91 91 57 15 31 95 10 88 84 5 82 83 53 98 59 17 97 80 76 80 81 3 91 81 87 93 61 46 10 49 6 22 21 75 63 89 21 81 30 19 67 38 77",
"output": "35"
},
{
"input": "90\n41 90 43 1 28 75 90 50 3 70 76 64 81 63 25 69 83 82 29 91 59 66 21 61 7 55 72 49 38 69 72 20 64 58 30 81 61 29 96 14 39 5 100 20 29 98 75 29 44 78 97 45 26 77 73 59 22 99 41 6 3 96 71 20 9 18 96 18 90 62 34 78 54 5 41 6 73 33 2 54 26 21 18 6 45 57 43 73 95 75",
"output": "42"
},
{
"input": "45\n93 69 4 27 20 14 71 48 79 3 32 26 49 30 57 88 13 56 49 61 37 32 47 41 41 70 45 68 82 18 8 6 25 20 15 13 71 99 28 6 52 34 19 59 26",
"output": "23"
},
{
"input": "33\n29 95 48 49 91 10 83 71 47 25 66 36 51 12 34 10 54 74 41 96 89 26 89 1 42 33 1 62 9 32 49 65 78",
"output": "15"
},
{
"input": "34\n98 24 42 36 41 82 28 58 89 34 77 70 76 44 74 54 66 100 13 79 4 88 21 1 11 45 91 29 87 100 29 54 82 78",
"output": "13"
},
{
"input": "29\n91 84 26 84 9 63 52 9 65 56 90 2 36 7 67 33 91 14 65 38 53 36 81 83 85 14 33 95 51",
"output": "17"
},
{
"input": "100\n2 88 92 82 87 100 78 28 84 43 78 32 43 33 97 19 15 52 29 84 57 72 54 13 99 28 82 79 40 70 34 92 91 53 9 88 27 43 14 92 72 37 26 37 20 95 19 34 49 64 33 37 34 27 80 79 9 54 99 68 25 4 68 73 46 66 24 78 3 87 26 52 50 84 4 95 23 83 39 58 86 36 33 16 98 2 84 19 53 12 69 60 10 11 78 17 79 92 77 59",
"output": "45"
},
{
"input": "100\n2 95 45 73 9 54 20 97 57 82 88 26 18 71 25 27 75 54 31 11 58 85 69 75 72 91 76 5 25 80 45 49 4 73 8 81 81 38 5 12 53 77 7 96 90 35 28 80 73 94 19 69 96 17 94 49 69 9 32 19 5 12 46 29 26 40 59 59 6 95 82 50 72 2 45 69 12 5 72 29 39 72 23 96 81 28 28 56 68 58 37 41 30 1 90 84 15 24 96 43",
"output": "53"
},
{
"input": "100\n27 72 35 91 13 10 35 45 24 55 83 84 63 96 29 79 34 67 63 92 48 83 18 77 28 27 49 66 29 88 55 15 6 58 14 67 94 36 77 7 7 64 61 52 71 18 36 99 76 6 50 67 16 13 41 7 89 73 61 51 78 22 78 32 76 100 3 31 89 71 63 53 15 85 77 54 89 33 68 74 3 23 57 5 43 89 75 35 9 86 90 11 31 46 48 37 74 17 77 8",
"output": "40"
},
{
"input": "100\n69 98 69 88 11 49 55 8 25 91 17 81 47 26 15 73 96 71 18 42 42 61 48 14 92 78 35 72 4 27 62 75 83 79 17 16 46 80 96 90 82 54 37 69 85 21 67 70 96 10 46 63 21 59 56 92 54 88 77 30 75 45 44 29 86 100 51 11 65 69 66 56 82 63 27 1 51 51 13 10 3 55 26 85 34 16 87 72 13 100 81 71 90 95 86 50 83 55 55 54",
"output": "53"
},
{
"input": "100\n34 35 99 64 2 66 78 93 20 48 12 79 19 10 87 7 42 92 60 79 5 2 24 89 57 48 63 92 74 4 16 51 7 12 90 48 87 17 18 73 51 58 97 97 25 38 15 97 96 73 67 91 6 75 14 13 87 79 75 3 15 55 35 95 71 45 10 13 20 37 82 26 2 22 13 83 97 84 39 79 43 100 54 59 98 8 61 34 7 65 75 44 24 77 73 88 34 95 44 77",
"output": "55"
},
{
"input": "100\n15 86 3 1 51 26 74 85 37 87 64 58 10 6 57 26 30 47 85 65 24 72 50 40 12 35 91 47 91 60 47 87 95 34 80 91 26 3 36 39 14 86 28 70 51 44 28 21 72 79 57 61 16 71 100 94 57 67 36 74 24 21 89 85 25 2 97 67 76 53 76 80 97 64 35 13 8 32 21 52 62 61 67 14 74 73 66 44 55 76 24 3 43 42 99 61 36 80 38 66",
"output": "52"
},
{
"input": "100\n45 16 54 54 80 94 74 93 75 85 58 95 79 30 81 2 84 4 57 23 92 64 78 1 50 36 13 27 56 54 10 77 87 1 5 38 85 74 94 82 30 45 72 83 82 30 81 82 82 3 69 82 7 92 39 60 94 42 41 5 3 17 67 21 79 44 79 96 28 3 53 68 79 89 63 83 1 44 4 31 84 15 73 77 19 66 54 6 73 1 67 24 91 11 86 45 96 82 20 89",
"output": "51"
},
{
"input": "100\n84 23 50 32 90 71 92 43 58 70 6 82 7 55 85 19 70 89 12 26 29 56 74 30 2 27 4 39 63 67 91 81 11 33 75 10 82 88 39 43 43 80 68 35 55 67 53 62 73 65 86 74 43 51 14 48 42 92 83 57 22 33 24 99 5 27 78 96 7 28 11 15 8 38 85 67 5 92 24 96 57 59 14 95 91 4 9 18 45 33 74 83 64 85 14 51 51 94 29 2",
"output": "53"
},
{
"input": "100\n77 56 56 45 73 55 32 37 39 50 30 95 79 21 44 34 51 43 86 91 39 30 85 15 35 93 100 14 57 31 80 79 38 40 88 4 91 54 7 95 76 26 62 84 17 33 67 47 6 82 69 51 17 2 59 24 11 12 31 90 12 11 55 38 72 49 30 50 42 46 5 97 9 9 30 45 86 23 19 82 40 42 5 40 35 98 35 32 60 60 5 28 84 35 21 49 68 53 68 23",
"output": "48"
},
{
"input": "100\n78 38 79 61 45 86 83 83 86 90 74 69 2 84 73 39 2 5 20 71 24 80 54 89 58 34 77 40 39 62 2 47 28 53 97 75 88 98 94 96 33 71 44 90 47 36 19 89 87 98 90 87 5 85 34 79 82 3 42 88 89 63 35 7 89 30 40 48 12 41 56 76 83 60 80 80 39 56 77 4 72 96 30 55 57 51 7 19 11 1 66 1 91 87 11 62 95 85 79 25",
"output": "48"
},
{
"input": "100\n5 34 23 20 76 75 19 51 17 82 60 13 83 6 65 16 20 43 66 54 87 10 87 73 50 24 16 98 33 28 80 52 54 82 26 92 14 13 84 92 94 29 61 21 60 20 48 94 24 20 75 70 58 27 68 45 86 89 29 8 67 38 83 48 18 100 11 22 46 84 52 97 70 19 50 75 3 7 52 53 72 41 18 31 1 38 49 53 11 64 99 76 9 87 48 12 100 32 44 71",
"output": "58"
},
{
"input": "100\n76 89 68 78 24 72 73 95 98 72 58 15 2 5 56 32 9 65 50 70 94 31 29 54 89 52 31 93 43 56 26 35 72 95 51 55 78 70 11 92 17 5 54 94 81 31 78 95 73 91 95 37 59 9 53 48 65 55 84 8 45 97 64 37 96 34 36 53 66 17 72 48 99 23 27 18 92 84 44 73 60 78 53 29 68 99 19 39 61 40 69 6 77 12 47 29 15 4 8 45",
"output": "53"
},
{
"input": "100\n82 40 31 53 8 50 85 93 3 84 54 17 96 59 51 42 18 19 35 84 79 31 17 46 54 82 72 49 35 73 26 89 61 73 3 50 12 29 25 77 88 21 58 24 22 89 96 54 82 29 96 56 77 16 1 68 90 93 20 23 57 22 31 18 92 90 51 14 50 72 31 54 12 50 66 62 2 34 17 45 68 50 87 97 23 71 1 72 17 82 42 15 20 78 4 49 66 59 10 17",
"output": "54"
},
{
"input": "100\n32 82 82 24 39 53 48 5 29 24 9 37 91 37 91 95 1 97 84 52 12 56 93 47 22 20 14 17 40 22 79 34 24 2 69 30 69 29 3 89 21 46 60 92 39 29 18 24 49 18 40 22 60 13 77 50 39 64 50 70 99 8 66 31 90 38 20 54 7 21 5 56 41 68 69 20 54 89 69 62 9 53 43 89 81 97 15 2 52 78 89 65 16 61 59 42 56 25 32 52",
"output": "49"
},
{
"input": "100\n72 54 23 24 97 14 99 87 15 25 7 23 17 87 72 31 71 87 34 82 51 77 74 85 62 38 24 7 84 48 98 21 29 71 70 84 25 58 67 92 18 44 32 9 81 15 53 29 63 18 86 16 7 31 38 99 70 32 89 16 23 11 66 96 69 82 97 59 6 9 49 80 85 19 6 9 52 51 85 74 53 46 73 55 31 63 78 61 34 80 77 65 87 77 92 52 89 8 52 31",
"output": "44"
},
{
"input": "100\n56 88 8 19 7 15 11 54 35 50 19 57 63 72 51 43 50 19 57 90 40 100 8 92 11 96 30 32 59 65 93 47 62 3 50 41 30 50 72 83 61 46 83 60 20 46 33 1 5 18 83 22 34 16 41 95 63 63 7 59 55 95 91 29 64 60 64 81 45 45 10 9 88 37 69 85 21 82 41 76 42 34 47 78 51 83 65 100 13 22 59 76 63 1 26 86 36 94 99 74",
"output": "46"
},
{
"input": "100\n27 89 67 60 62 80 43 50 28 88 72 5 94 11 63 91 18 78 99 3 71 26 12 97 74 62 23 24 22 3 100 72 98 7 94 32 12 75 61 88 42 48 10 14 45 9 48 56 73 76 70 70 79 90 35 39 96 37 81 11 19 65 99 39 23 79 34 61 35 74 90 37 73 23 46 21 94 84 73 58 11 89 13 9 10 85 42 78 73 32 53 39 49 90 43 5 28 31 97 75",
"output": "53"
},
{
"input": "100\n33 24 97 96 1 14 99 51 13 65 67 20 46 88 42 44 20 49 5 89 98 83 15 40 74 83 58 3 10 79 34 2 69 28 37 100 55 52 14 8 44 94 97 89 6 42 11 28 30 33 55 56 20 57 52 25 75 1 87 42 62 41 37 12 54 85 95 80 42 36 94 96 28 76 54 36 4 17 26 24 62 15 17 79 84 36 92 78 74 91 96 77 54 92 81 91 62 98 37 37",
"output": "43"
},
{
"input": "100\n86 24 61 15 11 85 1 31 47 36 23 36 59 34 3 27 16 29 82 28 58 52 52 66 71 61 98 39 60 20 67 41 67 90 73 29 92 17 70 95 58 98 58 32 21 73 46 56 87 72 80 75 40 27 94 31 59 92 93 37 14 99 96 21 97 23 81 91 52 52 96 94 92 28 38 29 52 16 57 27 17 24 91 21 79 55 96 98 95 94 23 78 79 12 77 35 32 75 100 82",
"output": "51"
},
{
"input": "100\n88 85 41 37 69 21 7 69 36 5 92 26 64 75 22 46 67 20 70 22 62 66 38 24 47 49 68 30 90 31 67 86 86 82 9 51 43 45 48 42 73 44 31 94 45 60 54 66 20 87 11 94 34 32 87 66 56 28 75 39 37 90 72 93 55 72 31 42 30 71 87 61 4 12 12 81 23 61 56 98 71 32 30 33 96 63 92 16 8 78 47 91 47 54 49 3 81 82 41 5",
"output": "47"
},
{
"input": "1\n2",
"output": "1"
},
{
"input": "5\n1 1 3 2 2",
"output": "3"
}
] | 1,622,058,182
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 52
| 124
| 0
|
n = int(input())
l = list(map(int,input().split()))
odd = even = 0
for i in range(len(l)):
if(l[i]%2!=0):
odd+=1
else:
even+=1
if(odd%2==0):
print(even)
else:
print(odd)
|
Title: Cookies
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Olga came to visit the twins Anna and Maria and saw that they have many cookies. The cookies are distributed into bags. As there are many cookies, Olga decided that it's no big deal if she steals a bag. However, she doesn't want the sisters to quarrel because of nothing when they divide the cookies. That's why Olga wants to steal a bag with cookies so that the number of cookies in the remaining bags was even, that is, so that Anna and Maria could evenly divide it into two (even 0 remaining cookies will do, just as any other even number). How many ways there are to steal exactly one cookie bag so that the total number of cookies in the remaining bags was even?
Input Specification:
The first line contains the only integer *n* (1<=≤<=*n*<=≤<=100) — the number of cookie bags Anna and Maria have. The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=100) — the number of cookies in the *i*-th bag.
Output Specification:
Print in the only line the only number — the sought number of ways. If there are no such ways print 0.
Demo Input:
['1\n1\n', '10\n1 2 2 3 4 4 4 2 2 2\n', '11\n2 2 2 2 2 2 2 2 2 2 99\n']
Demo Output:
['1\n', '8\n', '1\n']
Note:
In the first sample Olga should take the only bag so that the twins ended up with the even number of cookies.
In the second sample Olga can take any of five bags with two cookies or any of three bags with four cookies — 5 + 3 = 8 ways in total.
In the third sample, no matter which bag with two cookies Olga chooses, the twins are left with 2 * 9 + 99 = 117 cookies. Thus, Olga has only one option: to take the bag with 99 cookies.
|
```python
n = int(input())
l = list(map(int,input().split()))
odd = even = 0
for i in range(len(l)):
if(l[i]%2!=0):
odd+=1
else:
even+=1
if(odd%2==0):
print(even)
else:
print(odd)
```
| 3
|
|
312
|
B
|
Archer
|
PROGRAMMING
| 1,300
|
[
"math",
"probabilities"
] | null | null |
SmallR is an archer. SmallR is taking a match of archer with Zanoes. They try to shoot in the target in turns, and SmallR shoots first. The probability of shooting the target each time is for SmallR while for Zanoes. The one who shoots in the target first should be the winner.
Output the probability that SmallR will win the match.
|
A single line contains four integers .
|
Print a single real number, the probability that SmallR will win the match.
The answer will be considered correct if the absolute or relative error doesn't exceed 10<=-<=6.
|
[
"1 2 1 2\n"
] |
[
"0.666666666667"
] |
none
| 1,000
|
[
{
"input": "1 2 1 2",
"output": "0.666666666667"
},
{
"input": "1 3 1 3",
"output": "0.600000000000"
},
{
"input": "1 3 2 3",
"output": "0.428571428571"
},
{
"input": "3 4 3 4",
"output": "0.800000000000"
},
{
"input": "1 2 10 11",
"output": "0.523809523810"
},
{
"input": "4 5 4 5",
"output": "0.833333333333"
},
{
"input": "466 701 95 721",
"output": "0.937693791148"
},
{
"input": "268 470 444 885",
"output": "0.725614009325"
},
{
"input": "632 916 713 821",
"output": "0.719292895126"
},
{
"input": "269 656 918 992",
"output": "0.428937461623"
},
{
"input": "71 657 187 695",
"output": "0.310488463257"
},
{
"input": "435 852 973 978",
"output": "0.511844133157"
},
{
"input": "518 816 243 359",
"output": "0.719734031025"
},
{
"input": "882 962 311 811",
"output": "0.966386645447"
},
{
"input": "684 774 580 736",
"output": "0.906051574446"
},
{
"input": "486 868 929 999",
"output": "0.577723252958"
},
{
"input": "132 359 996 998",
"output": "0.368154532345"
},
{
"input": "933 977 266 450",
"output": "0.972879407907"
},
{
"input": "298 833 615 872",
"output": "0.441270817024"
},
{
"input": "34 554 14 958",
"output": "0.817324099167"
},
{
"input": "836 934 800 905",
"output": "0.906105535462"
},
{
"input": "482 815 69 509",
"output": "0.914365577772"
},
{
"input": "284 423 137 521",
"output": "0.885974839378"
},
{
"input": "648 881 486 703",
"output": "0.800911421248"
},
{
"input": "450 885 755 836",
"output": "0.533901011176"
},
{
"input": "533 773 823 998",
"output": "0.729222130525"
},
{
"input": "897 957 92 898",
"output": "0.993193806364"
},
{
"input": "699 925 441 928",
"output": "0.866816866175"
},
{
"input": "64 704 148 603",
"output": "0.289486317811"
},
{
"input": "719 735 626 990",
"output": "0.986124079764"
},
{
"input": "1 1000 1 1000",
"output": "0.500250125063"
}
] | 1,609,133,642
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 31
| 109
| 0
|
a, b, c, d = map(int, input().split())
m = a / b
n = 1 - (1 - a / b) * (1 - c / d)
print(round(m / n, 6))
|
Title: Archer
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
SmallR is an archer. SmallR is taking a match of archer with Zanoes. They try to shoot in the target in turns, and SmallR shoots first. The probability of shooting the target each time is for SmallR while for Zanoes. The one who shoots in the target first should be the winner.
Output the probability that SmallR will win the match.
Input Specification:
A single line contains four integers .
Output Specification:
Print a single real number, the probability that SmallR will win the match.
The answer will be considered correct if the absolute or relative error doesn't exceed 10<=-<=6.
Demo Input:
['1 2 1 2\n']
Demo Output:
['0.666666666667']
Note:
none
|
```python
a, b, c, d = map(int, input().split())
m = a / b
n = 1 - (1 - a / b) * (1 - c / d)
print(round(m / n, 6))
```
| 3
|
|
339
|
A
|
Helpful Maths
|
PROGRAMMING
| 800
|
[
"greedy",
"implementation",
"sortings",
"strings"
] | null | null |
Xenia the beginner mathematician is a third year student at elementary school. She is now learning the addition operation.
The teacher has written down the sum of multiple numbers. Pupils should calculate the sum. To make the calculation easier, the sum only contains numbers 1, 2 and 3. Still, that isn't enough for Xenia. She is only beginning to count, so she can calculate a sum only if the summands follow in non-decreasing order. For example, she can't calculate sum 1+3+2+1 but she can calculate sums 1+1+2 and 3+3.
You've got the sum that was written on the board. Rearrange the summans and print the sum in such a way that Xenia can calculate the sum.
|
The first line contains a non-empty string *s* — the sum Xenia needs to count. String *s* contains no spaces. It only contains digits and characters "+". Besides, string *s* is a correct sum of numbers 1, 2 and 3. String *s* is at most 100 characters long.
|
Print the new sum that Xenia can count.
|
[
"3+2+1\n",
"1+1+3+1+3\n",
"2\n"
] |
[
"1+2+3\n",
"1+1+1+3+3\n",
"2\n"
] |
none
| 500
|
[
{
"input": "3+2+1",
"output": "1+2+3"
},
{
"input": "1+1+3+1+3",
"output": "1+1+1+3+3"
},
{
"input": "2",
"output": "2"
},
{
"input": "2+2+1+1+3",
"output": "1+1+2+2+3"
},
{
"input": "2+1+2+2+2+3+1+3+1+2",
"output": "1+1+1+2+2+2+2+2+3+3"
},
{
"input": "1+2+1+2+2+2+2+1+3+3",
"output": "1+1+1+2+2+2+2+2+3+3"
},
{
"input": "2+3+3+1+2+2+2+1+1+2+1+3+2+2+3+3+2+2+3+3+3+1+1+1+3+3+3+2+1+3+2+3+2+1+1+3+3+3+1+2+2+1+2+2+1+2+1+3+1+1",
"output": "1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+3+3+3+3+3+3+3+3+3+3+3+3+3+3+3+3+3"
},
{
"input": "1",
"output": "1"
},
{
"input": "2+1+2+2+1+3+2+3+1+1+2+1+2+2+3+1+1+3+3+3+2+2+3+2+2+2+1+2+1+2+3+2+2+2+1+3+1+3+3+3+1+2+1+2+2+2+2+3+1+1",
"output": "1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+3+3+3+3+3+3+3+3+3+3+3+3+3"
},
{
"input": "2+2+1+1+1+3+1+1+3+3+2+3+1+3+1+1+3+1+1+2+2+2+2+1+2+1+2+1+1+1+3+1+3+2+3+2+3+3+1+1+1+2+3+2+1+3+1+3+2+2",
"output": "1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+3+3+3+3+3+3+3+3+3+3+3+3+3+3"
},
{
"input": "3+2+3+3+2+2+1+2+1+2+3+1+2+3+2+3+2+1+2+2+1+1+2+2+3+2+1+3+1+1+3+2+2+2+2+3+3+2+2+3+3+1+1+2+3+3+2+3+3+3",
"output": "1+1+1+1+1+1+1+1+1+1+1+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+3+3+3+3+3+3+3+3+3+3+3+3+3+3+3+3+3+3"
},
{
"input": "3",
"output": "3"
},
{
"input": "1+1",
"output": "1+1"
},
{
"input": "1+2",
"output": "1+2"
},
{
"input": "1+3",
"output": "1+3"
},
{
"input": "2+1",
"output": "1+2"
},
{
"input": "2+2",
"output": "2+2"
},
{
"input": "2+3",
"output": "2+3"
},
{
"input": "3+1",
"output": "1+3"
},
{
"input": "3+2",
"output": "2+3"
},
{
"input": "3+3",
"output": "3+3"
}
] | 1,697,856,774
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 21
| 154
| 0
|
a = input().split("+")
a.sort()
c = ""
for i in a:
c =c+"+"+ i
print(c[1:])
|
Title: Helpful Maths
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Xenia the beginner mathematician is a third year student at elementary school. She is now learning the addition operation.
The teacher has written down the sum of multiple numbers. Pupils should calculate the sum. To make the calculation easier, the sum only contains numbers 1, 2 and 3. Still, that isn't enough for Xenia. She is only beginning to count, so she can calculate a sum only if the summands follow in non-decreasing order. For example, she can't calculate sum 1+3+2+1 but she can calculate sums 1+1+2 and 3+3.
You've got the sum that was written on the board. Rearrange the summans and print the sum in such a way that Xenia can calculate the sum.
Input Specification:
The first line contains a non-empty string *s* — the sum Xenia needs to count. String *s* contains no spaces. It only contains digits and characters "+". Besides, string *s* is a correct sum of numbers 1, 2 and 3. String *s* is at most 100 characters long.
Output Specification:
Print the new sum that Xenia can count.
Demo Input:
['3+2+1\n', '1+1+3+1+3\n', '2\n']
Demo Output:
['1+2+3\n', '1+1+1+3+3\n', '2\n']
Note:
none
|
```python
a = input().split("+")
a.sort()
c = ""
for i in a:
c =c+"+"+ i
print(c[1:])
```
| 3
|
|
139
|
A
|
Petr and Book
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
One Sunday Petr went to a bookshop and bought a new book on sports programming. The book had exactly *n* pages.
Petr decided to start reading it starting from the next day, that is, from Monday. Petr's got a very tight schedule and for each day of the week he knows how many pages he will be able to read on that day. Some days are so busy that Petr will have no time to read whatsoever. However, we know that he will be able to read at least one page a week.
Assuming that Petr will not skip days and will read as much as he can every day, determine on which day of the week he will read the last page of the book.
|
The first input line contains the single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of pages in the book.
The second line contains seven non-negative space-separated integers that do not exceed 1000 — those integers represent how many pages Petr can read on Monday, Tuesday, Wednesday, Thursday, Friday, Saturday and Sunday correspondingly. It is guaranteed that at least one of those numbers is larger than zero.
|
Print a single number — the number of the day of the week, when Petr will finish reading the book. The days of the week are numbered starting with one in the natural order: Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday.
|
[
"100\n15 20 20 15 10 30 45\n",
"2\n1 0 0 0 0 0 0\n"
] |
[
"6\n",
"1\n"
] |
Note to the first sample:
By the end of Monday and therefore, by the beginning of Tuesday Petr has 85 pages left. He has 65 pages left by Wednesday, 45 by Thursday, 30 by Friday, 20 by Saturday and on Saturday Petr finishes reading the book (and he also has time to read 10 pages of something else).
Note to the second sample:
On Monday of the first week Petr will read the first page. On Monday of the second week Petr will read the second page and will finish reading the book.
| 500
|
[
{
"input": "100\n15 20 20 15 10 30 45",
"output": "6"
},
{
"input": "2\n1 0 0 0 0 0 0",
"output": "1"
},
{
"input": "100\n100 200 100 200 300 400 500",
"output": "1"
},
{
"input": "3\n1 1 1 1 1 1 1",
"output": "3"
},
{
"input": "1\n1 1 1 1 1 1 1",
"output": "1"
},
{
"input": "20\n5 3 7 2 1 6 4",
"output": "6"
},
{
"input": "10\n5 1 1 1 1 1 5",
"output": "6"
},
{
"input": "50\n10 1 10 1 10 1 10",
"output": "1"
},
{
"input": "77\n11 11 11 11 11 11 10",
"output": "1"
},
{
"input": "1\n1000 1000 1000 1000 1000 1000 1000",
"output": "1"
},
{
"input": "1000\n100 100 100 100 100 100 100",
"output": "3"
},
{
"input": "999\n10 20 10 20 30 20 10",
"output": "3"
},
{
"input": "433\n109 58 77 10 39 125 15",
"output": "7"
},
{
"input": "1\n0 0 0 0 0 0 1",
"output": "7"
},
{
"input": "5\n1 0 1 0 1 0 1",
"output": "1"
},
{
"input": "997\n1 1 0 0 1 0 1",
"output": "1"
},
{
"input": "1000\n1 1 1 1 1 1 1",
"output": "6"
},
{
"input": "1000\n1000 1000 1000 1000 1000 1000 1000",
"output": "1"
},
{
"input": "1000\n1 0 0 0 0 0 0",
"output": "1"
},
{
"input": "1000\n0 0 0 0 0 0 1",
"output": "7"
},
{
"input": "1000\n1 0 0 1 0 0 1",
"output": "1"
},
{
"input": "509\n105 23 98 0 7 0 155",
"output": "2"
},
{
"input": "7\n1 1 1 1 1 1 1",
"output": "7"
},
{
"input": "2\n1 1 0 0 0 0 0",
"output": "2"
},
{
"input": "1\n0 0 0 0 0 1 0",
"output": "6"
},
{
"input": "10\n0 0 0 0 0 0 1",
"output": "7"
},
{
"input": "5\n0 0 0 0 0 6 0",
"output": "6"
},
{
"input": "3\n0 1 0 0 0 0 0",
"output": "2"
},
{
"input": "10\n0 0 0 0 0 0 10",
"output": "7"
},
{
"input": "28\n1 2 3 4 5 6 7",
"output": "7"
},
{
"input": "100\n5 5 5 5 5 5 5",
"output": "6"
},
{
"input": "4\n1 0 0 0 0 0 1",
"output": "7"
},
{
"input": "2\n0 0 0 0 0 0 1",
"output": "7"
},
{
"input": "7\n0 0 0 0 0 0 7",
"output": "7"
},
{
"input": "7\n2 1 1 1 1 1 0",
"output": "6"
},
{
"input": "2\n0 0 1 1 0 0 0",
"output": "4"
},
{
"input": "6\n1 1 1 1 1 1 0",
"output": "6"
},
{
"input": "5\n1 1 1 0 0 1 1",
"output": "7"
},
{
"input": "100\n10 20 30 10 10 10 10",
"output": "7"
},
{
"input": "1\n0 0 0 1 0 0 0",
"output": "4"
},
{
"input": "70\n10 10 10 10 10 10 10",
"output": "7"
},
{
"input": "22\n1 2 3 4 5 6 10",
"output": "7"
},
{
"input": "5\n0 0 0 1 0 0 0",
"output": "4"
},
{
"input": "2\n0 0 0 1 0 0 0",
"output": "4"
},
{
"input": "6\n1 0 0 0 0 0 2",
"output": "7"
},
{
"input": "10\n1 2 2 1 2 1 1",
"output": "7"
},
{
"input": "5\n0 0 0 0 0 0 10",
"output": "7"
},
{
"input": "4\n0 1 1 0 0 0 0",
"output": "3"
},
{
"input": "100\n0 0 0 0 0 1 0",
"output": "6"
}
] | 1,664,876,683
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 49
| 92
| 0
|
t=int(input())
l=list(map(int,input().split()))
check=0
x=0
while check<t:
for i in range(len(l)):
if check<t:
check+=l[i]
x=i
print(x+1)
|
Title: Petr and Book
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One Sunday Petr went to a bookshop and bought a new book on sports programming. The book had exactly *n* pages.
Petr decided to start reading it starting from the next day, that is, from Monday. Petr's got a very tight schedule and for each day of the week he knows how many pages he will be able to read on that day. Some days are so busy that Petr will have no time to read whatsoever. However, we know that he will be able to read at least one page a week.
Assuming that Petr will not skip days and will read as much as he can every day, determine on which day of the week he will read the last page of the book.
Input Specification:
The first input line contains the single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of pages in the book.
The second line contains seven non-negative space-separated integers that do not exceed 1000 — those integers represent how many pages Petr can read on Monday, Tuesday, Wednesday, Thursday, Friday, Saturday and Sunday correspondingly. It is guaranteed that at least one of those numbers is larger than zero.
Output Specification:
Print a single number — the number of the day of the week, when Petr will finish reading the book. The days of the week are numbered starting with one in the natural order: Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday.
Demo Input:
['100\n15 20 20 15 10 30 45\n', '2\n1 0 0 0 0 0 0\n']
Demo Output:
['6\n', '1\n']
Note:
Note to the first sample:
By the end of Monday and therefore, by the beginning of Tuesday Petr has 85 pages left. He has 65 pages left by Wednesday, 45 by Thursday, 30 by Friday, 20 by Saturday and on Saturday Petr finishes reading the book (and he also has time to read 10 pages of something else).
Note to the second sample:
On Monday of the first week Petr will read the first page. On Monday of the second week Petr will read the second page and will finish reading the book.
|
```python
t=int(input())
l=list(map(int,input().split()))
check=0
x=0
while check<t:
for i in range(len(l)):
if check<t:
check+=l[i]
x=i
print(x+1)
```
| 3
|
|
69
|
A
|
Young Physicist
|
PROGRAMMING
| 1,000
|
[
"implementation",
"math"
] |
A. Young Physicist
|
2
|
256
|
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
|
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100).
|
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
|
[
"3\n4 1 7\n-2 4 -1\n1 -5 -3\n",
"3\n3 -1 7\n-5 2 -4\n2 -1 -3\n"
] |
[
"NO",
"YES"
] |
none
| 500
|
[
{
"input": "3\n4 1 7\n-2 4 -1\n1 -5 -3",
"output": "NO"
},
{
"input": "3\n3 -1 7\n-5 2 -4\n2 -1 -3",
"output": "YES"
},
{
"input": "10\n21 32 -46\n43 -35 21\n42 2 -50\n22 40 20\n-27 -9 38\n-4 1 1\n-40 6 -31\n-13 -2 34\n-21 34 -12\n-32 -29 41",
"output": "NO"
},
{
"input": "10\n25 -33 43\n-27 -42 28\n-35 -20 19\n41 -42 -1\n49 -39 -4\n-49 -22 7\n-19 29 41\n8 -27 -43\n8 34 9\n-11 -3 33",
"output": "NO"
},
{
"input": "10\n-6 21 18\n20 -11 -8\n37 -11 41\n-5 8 33\n29 23 32\n30 -33 -11\n39 -49 -36\n28 34 -49\n22 29 -34\n-18 -6 7",
"output": "NO"
},
{
"input": "10\n47 -2 -27\n0 26 -14\n5 -12 33\n2 18 3\n45 -30 -49\n4 -18 8\n-46 -44 -41\n-22 -10 -40\n-35 -21 26\n33 20 38",
"output": "NO"
},
{
"input": "13\n-3 -36 -46\n-11 -50 37\n42 -11 -15\n9 42 44\n-29 -12 24\n3 9 -40\n-35 13 50\n14 43 18\n-13 8 24\n-48 -15 10\n50 9 -50\n21 0 -50\n0 0 -6",
"output": "YES"
},
{
"input": "14\n43 23 17\n4 17 44\n5 -5 -16\n-43 -7 -6\n47 -48 12\n50 47 -45\n2 14 43\n37 -30 15\n4 -17 -11\n17 9 -45\n-50 -3 -8\n-50 0 0\n-50 0 0\n-16 0 0",
"output": "YES"
},
{
"input": "13\n29 49 -11\n38 -11 -20\n25 1 -40\n-11 28 11\n23 -19 1\n45 -41 -17\n-3 0 -19\n-13 -33 49\n-30 0 28\n34 17 45\n-50 9 -27\n-50 0 0\n-37 0 0",
"output": "YES"
},
{
"input": "12\n3 28 -35\n-32 -44 -17\n9 -25 -6\n-42 -22 20\n-19 15 38\n-21 38 48\n-1 -37 -28\n-10 -13 -50\n-5 21 29\n34 28 50\n50 11 -49\n34 0 0",
"output": "YES"
},
{
"input": "37\n-64 -79 26\n-22 59 93\n-5 39 -12\n77 -9 76\n55 -86 57\n83 100 -97\n-70 94 84\n-14 46 -94\n26 72 35\n14 78 -62\n17 82 92\n-57 11 91\n23 15 92\n-80 -1 1\n12 39 18\n-23 -99 -75\n-34 50 19\n-39 84 -7\n45 -30 -39\n-60 49 37\n45 -16 -72\n33 -51 -56\n-48 28 5\n97 91 88\n45 -82 -11\n-21 -15 -90\n-53 73 -26\n-74 85 -90\n-40 23 38\n100 -13 49\n32 -100 -100\n0 -100 -70\n0 -100 0\n0 -100 0\n0 -100 0\n0 -100 0\n0 -37 0",
"output": "YES"
},
{
"input": "4\n68 3 100\n68 21 -100\n-100 -24 0\n-36 0 0",
"output": "YES"
},
{
"input": "33\n-1 -46 -12\n45 -16 -21\n-11 45 -21\n-60 -42 -93\n-22 -45 93\n37 96 85\n-76 26 83\n-4 9 55\n7 -52 -9\n66 8 -85\n-100 -54 11\n-29 59 74\n-24 12 2\n-56 81 85\n-92 69 -52\n-26 -97 91\n54 59 -51\n58 21 -57\n7 68 56\n-47 -20 -51\n-59 77 -13\n-85 27 91\n79 60 -56\n66 -80 5\n21 -99 42\n-31 -29 98\n66 93 76\n-49 45 61\n100 -100 -100\n100 -100 -100\n66 -75 -100\n0 0 -100\n0 0 -87",
"output": "YES"
},
{
"input": "3\n1 2 3\n3 2 1\n0 0 0",
"output": "NO"
},
{
"input": "2\n5 -23 12\n0 0 0",
"output": "NO"
},
{
"input": "1\n0 0 0",
"output": "YES"
},
{
"input": "1\n1 -2 0",
"output": "NO"
},
{
"input": "2\n-23 77 -86\n23 -77 86",
"output": "YES"
},
{
"input": "26\n86 7 20\n-57 -64 39\n-45 6 -93\n-44 -21 100\n-11 -49 21\n73 -71 -80\n-2 -89 56\n-65 -2 7\n5 14 84\n57 41 13\n-12 69 54\n40 -25 27\n-17 -59 0\n64 -91 -30\n-53 9 42\n-54 -8 14\n-35 82 27\n-48 -59 -80\n88 70 79\n94 57 97\n44 63 25\n84 -90 -40\n-100 100 -100\n-92 100 -100\n0 10 -100\n0 0 -82",
"output": "YES"
},
{
"input": "42\n11 27 92\n-18 -56 -57\n1 71 81\n33 -92 30\n82 83 49\n-87 -61 -1\n-49 45 49\n73 26 15\n-22 22 -77\n29 -93 87\n-68 44 -90\n-4 -84 20\n85 67 -6\n-39 26 77\n-28 -64 20\n65 -97 24\n-72 -39 51\n35 -75 -91\n39 -44 -8\n-25 -27 -57\n91 8 -46\n-98 -94 56\n94 -60 59\n-9 -95 18\n-53 -37 98\n-8 -94 -84\n-52 55 60\n15 -14 37\n65 -43 -25\n94 12 66\n-8 -19 -83\n29 81 -78\n-58 57 33\n24 86 -84\n-53 32 -88\n-14 7 3\n89 97 -53\n-5 -28 -91\n-100 100 -6\n-84 100 0\n0 100 0\n0 70 0",
"output": "YES"
},
{
"input": "3\n96 49 -12\n2 -66 28\n-98 17 -16",
"output": "YES"
},
{
"input": "5\n70 -46 86\n-100 94 24\n-27 63 -63\n57 -100 -47\n0 -11 0",
"output": "YES"
},
{
"input": "18\n-86 -28 70\n-31 -89 42\n31 -48 -55\n95 -17 -43\n24 -95 -85\n-21 -14 31\n68 -18 81\n13 31 60\n-15 28 99\n-42 15 9\n28 -61 -62\n-16 71 29\n-28 75 -48\n-77 -67 36\n-100 83 89\n100 100 -100\n57 34 -100\n0 0 -53",
"output": "YES"
},
{
"input": "44\n52 -54 -29\n-82 -5 -94\n-54 43 43\n91 16 71\n7 80 -91\n3 15 29\n-99 -6 -77\n-3 -77 -64\n73 67 34\n25 -10 -18\n-29 91 63\n-72 86 -16\n-68 85 -81\n-3 36 44\n-74 -14 -80\n34 -96 -97\n-76 -78 -33\n-24 44 -58\n98 12 77\n95 -63 -6\n-51 3 -90\n-92 -10 72\n7 3 -68\n57 -53 71\n29 57 -48\n35 -60 10\n79 -70 -61\n-20 77 55\n-86 -15 -35\n84 -88 -18\n100 -42 77\n-20 46 8\n-41 -43 -65\n38 -98 -23\n-100 65 45\n-7 -91 -63\n46 88 -85\n48 59 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 1",
"output": "YES"
},
{
"input": "18\n-14 -64 -91\n-8 -66 -86\n-23 92 -40\n6 -3 -53\n57 41 78\n-79 42 -22\n-88 -17 45\n4 -45 44\n83 -18 -25\n34 86 -92\n75 -30 12\n44 99 11\n-67 -13 72\n22 83 -56\n-37 71 72\n-9 -100 100\n0 -100 31\n0 -58 0",
"output": "YES"
},
{
"input": "23\n-70 37 78\n42 84 6\n28 -94 -24\n-49 76 95\n-67 18 84\n-53 78 -5\n65 -63 -64\n-66 -64 -69\n81 -93 95\n10 52 -79\n-89 -61 -64\n-64 -47 43\n-81 -35 55\n80 82 73\n-60 -81 -18\n49 -10 -19\n-58 70 12\n-24 -15 -93\n98 -93 -54\n-28 -75 11\n100 100 -63\n100 100 0\n56 34 0",
"output": "YES"
},
{
"input": "3\n0 2 -2\n1 -1 3\n-3 0 0",
"output": "NO"
}
] | 1,693,122,838
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 81
| 92
| 0
|
n = int(input()) # Read the number of force vectors
total_force = [0, 0, 0] # Initialize the total force as (0, 0, 0)
# Read and accumulate the force vectors
for _ in range(n):
x, y, z = map(int, input().split())
total_force[0] += x
total_force[1] += y
total_force[2] += z
# Check if the total force is (0, 0, 0) and print the result
if total_force == [0, 0, 0]:
print("YES")
else:
print("NO")
|
Title: Young Physicist
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100).
Output Specification:
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
Demo Input:
['3\n4 1 7\n-2 4 -1\n1 -5 -3\n', '3\n3 -1 7\n-5 2 -4\n2 -1 -3\n']
Demo Output:
['NO', 'YES']
Note:
none
|
```python
n = int(input()) # Read the number of force vectors
total_force = [0, 0, 0] # Initialize the total force as (0, 0, 0)
# Read and accumulate the force vectors
for _ in range(n):
x, y, z = map(int, input().split())
total_force[0] += x
total_force[1] += y
total_force[2] += z
# Check if the total force is (0, 0, 0) and print the result
if total_force == [0, 0, 0]:
print("YES")
else:
print("NO")
```
| 3.977
|
518
|
A
|
Vitaly and Strings
|
PROGRAMMING
| 1,600
|
[
"constructive algorithms",
"strings"
] | null | null |
Vitaly is a diligent student who never missed a lesson in his five years of studying in the university. He always does his homework on time and passes his exams in time.
During the last lesson the teacher has provided two strings *s* and *t* to Vitaly. The strings have the same length, they consist of lowercase English letters, string *s* is lexicographically smaller than string *t*. Vitaly wondered if there is such string that is lexicographically larger than string *s* and at the same is lexicographically smaller than string *t*. This string should also consist of lowercase English letters and have the length equal to the lengths of strings *s* and *t*.
Let's help Vitaly solve this easy problem!
|
The first line contains string *s* (1<=≤<=|*s*|<=≤<=100), consisting of lowercase English letters. Here, |*s*| denotes the length of the string.
The second line contains string *t* (|*t*|<==<=|*s*|), consisting of lowercase English letters.
It is guaranteed that the lengths of strings *s* and *t* are the same and string *s* is lexicographically less than string *t*.
|
If the string that meets the given requirements doesn't exist, print a single string "No such string" (without the quotes).
If such string exists, print it. If there are multiple valid strings, you may print any of them.
|
[
"a\nc\n",
"aaa\nzzz\n",
"abcdefg\nabcdefh\n"
] |
[
"b\n",
"kkk\n",
"No such string\n"
] |
String *s* = *s*<sub class="lower-index">1</sub>*s*<sub class="lower-index">2</sub>... *s*<sub class="lower-index">*n*</sub> is said to be lexicographically smaller than *t* = *t*<sub class="lower-index">1</sub>*t*<sub class="lower-index">2</sub>... *t*<sub class="lower-index">*n*</sub>, if there exists such *i*, that *s*<sub class="lower-index">1</sub> = *t*<sub class="lower-index">1</sub>, *s*<sub class="lower-index">2</sub> = *t*<sub class="lower-index">2</sub>, ... *s*<sub class="lower-index">*i* - 1</sub> = *t*<sub class="lower-index">*i* - 1</sub>, *s*<sub class="lower-index">*i*</sub> < *t*<sub class="lower-index">*i*</sub>.
| 500
|
[
{
"input": "a\nc",
"output": "b"
},
{
"input": "aaa\nzzz",
"output": "kkk"
},
{
"input": "abcdefg\nabcdefh",
"output": "No such string"
},
{
"input": "abcdefg\nabcfefg",
"output": "abcdefh"
},
{
"input": "frt\nfru",
"output": "No such string"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa\nzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz",
"output": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaab"
},
{
"input": "zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzx\nzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz",
"output": "zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzy"
},
{
"input": "q\nz",
"output": "r"
},
{
"input": "pnzcl\npnzdf",
"output": "pnzcm"
},
{
"input": "vklldrxnfgyorgfpfezvhbouyzzzzz\nvklldrxnfgyorgfpfezvhbouzaaadv",
"output": "vklldrxnfgyorgfpfezvhbouzaaaaa"
},
{
"input": "pkjlxzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz\npkjlyaaaaaaaaaaaaaaaaaaaaaaaaaaaahr",
"output": "pkjlyaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "exoudpymnspkocwszzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz\nexoudpymnspkocwtaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaabml",
"output": "exoudpymnspkocwtaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "anarzvsklmwvovozwnmhklkpcseeogdgauoppmzrukynbjjoxytuvsiecuzfquxnowewebhtuoxepocyeamqfrblpwqiokbcubil\nanarzvsklmwvovozwnmhklkpcseeogdgauoppmzrukynbjjoxytuvsiecuzfquxnowewebhtuoxepocyeamqfrblpwqiokbcubim",
"output": "No such string"
},
{
"input": "uqyugulumzwlxsjnxxkutzqayskrbjoaaekbhckjryhjjllzzz\nuqyugulumzwlxsjnxxkutzqayskrbjoaaekbhckjryhjjlmaaa",
"output": "No such string"
},
{
"input": "esfaeyxpblcrriizhnhfrxnbopqvhwtetgjqavlqdlxexaifgvkqfwzneibhxxdacbzzzzzzzzzzzzzz\nesfaeyxpblcrriizhnhfrxnbopqvhwtetgjqavlqdlxexaifgvkqfwzneibhxxdaccaaaaaaaaaaaatf",
"output": "esfaeyxpblcrriizhnhfrxnbopqvhwtetgjqavlqdlxexaifgvkqfwzneibhxxdaccaaaaaaaaaaaaaa"
},
{
"input": "oisjtilteipnzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz\noisjtilteipoaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaao",
"output": "oisjtilteipoaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "svpoxbsudndfnnpugbouawegyxgtmvqzbewxpcwhopdbwscimgzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz\nsvpoxbsudndfnnpugbouawegyxgtmvqzbewxpcwhopdbwscimhaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "No such string"
},
{
"input": "ddzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz\ndeaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaao",
"output": "deaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "xqzbhslocdbifnyzyjenlpctocieaccsycmwlcebkqqkeibatfvylbqlutvjijgjhdetqsjqnoipqbmjhhzxggdobyvpczdavdzz\nxqzbhslocdbifnyzyjenlpctocieaccsycmwlcebkqqkeibatfvylbqlutvjijgjhdetqsjqnoipqbmjhhzxggdobyvpczdavilj",
"output": "xqzbhslocdbifnyzyjenlpctocieaccsycmwlcebkqqkeibatfvylbqlutvjijgjhdetqsjqnoipqbmjhhzxggdobyvpczdaveaa"
},
{
"input": "poflpxucohdobeisxfsnkbdzwizjjhgngufssqhmfgmydmmrnuminrvxxamoebhczlwsfefdtnchaisfxkfcovxmvppxnrfawfoq\npoflpxucohdobeisxfsnkbdzwizjjhgngufssqhmfgmydmmrnuminrvxxamoebhczlwsfefdtnchaisfxkfcovxmvppxnrfawujg",
"output": "poflpxucohdobeisxfsnkbdzwizjjhgngufssqhmfgmydmmrnuminrvxxamoebhczlwsfefdtnchaisfxkfcovxmvppxnrfawfor"
},
{
"input": "vonggnmokmvmguwtobkxoqgxkuxtyjmxrygyliohlhwxuxjmlkqcfuxboxjnzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz\nvonggnmokmvmguwtobkxoqgxkuxtyjmxrygyliohlhwxuxjmlkqcfuxboxjoaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaac",
"output": "vonggnmokmvmguwtobkxoqgxkuxtyjmxrygyliohlhwxuxjmlkqcfuxboxjoaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "bqycw\nquhod",
"output": "bqycx"
},
{
"input": "hceslswecf\nnmxshuymaa",
"output": "hceslswecg"
},
{
"input": "awqtzslxowuaefe\nvujscakjpvxviki",
"output": "awqtzslxowuaeff"
},
{
"input": "lerlcnaogdravnogfogcyoxgi\nojrbithvjdqtempegvqxmgmmw",
"output": "lerlcnaogdravnogfogcyoxgj"
},
{
"input": "jbrhvicytqaivheqeourrlosvnsujsxdinryyawgalidsaufxv\noevvkhujmhagaholrmsatdjjyfmyblvgetpnxgjcilugjsncjs",
"output": "jbrhvicytqaivheqeourrlosvnsujsxdinryyawgalidsaufxw"
},
{
"input": "jrpogrcuhqdpmyzpuabuhaptlxaeiqjxhqkmuzsjbhqxvdtoocrkusaeasqdwlunomwzww\nspvgaswympzlscnumemgiznngnxqgccbubmxgqmaakbnyngkxlxjjsafricchhpecdjgxw",
"output": "jrpogrcuhqdpmyzpuabuhaptlxaeiqjxhqkmuzsjbhqxvdtoocrkusaeasqdwlunomwzwx"
},
{
"input": "mzmhjmfxaxaplzjmjkbyadeweltagyyuzpvrmnyvirjpdmebxyzjvdoezhnayfrvtnccryhkvhcvakcf\nohhhhkujfpjbgouebtmmbzizuhuumvrsqfniwpmxdtzhyiaivdyxhywnqzagicydixjtvbqbevhbqttu",
"output": "mzmhjmfxaxaplzjmjkbyadeweltagyyuzpvrmnyvirjpdmebxyzjvdoezhnayfrvtnccryhkvhcvakcg"
},
{
"input": "cdmwmzutsicpzhcokbbhwktqbomozxvvjlhwdgtiledgurxsfreisgczdwgupzxmjnfyjxcpdwzkggludkcmgppndl\nuvuqvyrnhtyubpevizhjxdvmpueittksrnosmfuuzbimnqussasdjufrthrgjbyzomauaxbvwferfvtmydmwmjaoxg",
"output": "cdmwmzutsicpzhcokbbhwktqbomozxvvjlhwdgtiledgurxsfreisgczdwgupzxmjnfyjxcpdwzkggludkcmgppndm"
},
{
"input": "dpnmrwpbgzvcmrcodwgvvfwpyagdwlngmhrazyvalszhruprxzmwltftxmujfyrrnwzvphgqlcphreumqkytswxziugburwrlyay\nqibcfxdfovoejutaeetbbwrgexdrvqywwmhipxgfrvhzovxkfawpfnpjvlhkyahessodqcclangxefcaixysqijnitevwmpalkzd",
"output": "dpnmrwpbgzvcmrcodwgvvfwpyagdwlngmhrazyvalszhruprxzmwltftxmujfyrrnwzvphgqlcphreumqkytswxziugburwrlyaz"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaab",
"output": "No such string"
},
{
"input": "phdvmuwqmvzyurtnshitcypuzbhpceovkibzbhhjwxkdtvqmbpoumeoiztxtvkvsjrlnhowsdmgftuiulzebdigmun\nphdvmuwqmvzyurtnshitcypuzbhpceovkibzbhhjwxkdtvqmbpoumeoiztxtvkvsjrlnhowsdmgftuiulzebdigmuo",
"output": "No such string"
},
{
"input": "hrsantdquixzjyjtqytcmnflnyehzbibkbgkqffgqpkgeuqmbmxzhbjwsnfkizvbcyoghyvnxxjavoahlqjxomtsouzoog\nhrsantdquixzjyjtqytcmnflnyehzbibkbgkqffgqpkgeuqmbmxzhbjwsnfkizvbcyoghyvnxxjavoahlqjxomtsouzooh",
"output": "No such string"
},
{
"input": "kexdbtpkjbwwyibjndbtmwqzolopqitgkomqggojevoankiepxirrcidxldlzsppehmoazdywltmjbxgsxgihwnwpmczjrcwpywl\nkexdbtpkjbwwyibjndbtmwqzolopqitgkomqggojevoankiepxirrcidxldlzsppehmoazdywltmjbxgsxgihwnwpmczjrcwpywm",
"output": "No such string"
},
{
"input": "yijdysvzfcnaedvnecswgoylhzgguxecmucepgstjbdkbjyfdlxxxejkrrxfiuwjpdmdhhqhlqeqzjwudtdryrfkpwfxdjlkowmk\nyijdysvzfcnaedvnecswgoylhzgguxecmucepgstjbdkbjyfdlxxxejkrrxfiuwjpdmdhhqhlqeqzjwudtdryrfkpwfxdjlkowml",
"output": "No such string"
},
{
"input": "a\nb",
"output": "No such string"
},
{
"input": "x\nz",
"output": "y"
},
{
"input": "a\nz",
"output": "b"
},
{
"input": "y\nz",
"output": "No such string"
},
{
"input": "zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzy\nzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz",
"output": "No such string"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaac",
"output": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaab"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa\nzaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaab"
},
{
"input": "yzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz\nzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz",
"output": "zaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "yyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyy\nzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz",
"output": "yyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyz"
},
{
"input": "aaa\naac",
"output": "aab"
},
{
"input": "aaa\naab",
"output": "No such string"
},
{
"input": "zzx\nzzz",
"output": "zzy"
},
{
"input": "zzy\nzzz",
"output": "No such string"
},
{
"input": "aba\naca",
"output": "abb"
},
{
"input": "aaa\nbbb",
"output": "aab"
},
{
"input": "azzz\ncaaa",
"output": "baaa"
}
] | 1,642,311,066
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 51
| 93
| 0
|
s=input()
t=input()
a=[]
b=[]
for i in s:
a.append(i)
for i in t:
b.append(i)
if len(s)==s.count('z'):
ans='a'*(len(s)+1)
if ans!=t:
print(ans)
else:
print("No such string")
else:
pos=-1
for i in range(len(s)-1,-1,-1):
if a[i]!='z':
pos=i
break
val=ord(a[pos])+1
a[pos]=chr(val)
for i in range(pos+1,len(s)):
a[i]='a'
if a!=b:
print(*a,sep="")
elif a==b:
print("No such string")
|
Title: Vitaly and Strings
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vitaly is a diligent student who never missed a lesson in his five years of studying in the university. He always does his homework on time and passes his exams in time.
During the last lesson the teacher has provided two strings *s* and *t* to Vitaly. The strings have the same length, they consist of lowercase English letters, string *s* is lexicographically smaller than string *t*. Vitaly wondered if there is such string that is lexicographically larger than string *s* and at the same is lexicographically smaller than string *t*. This string should also consist of lowercase English letters and have the length equal to the lengths of strings *s* and *t*.
Let's help Vitaly solve this easy problem!
Input Specification:
The first line contains string *s* (1<=≤<=|*s*|<=≤<=100), consisting of lowercase English letters. Here, |*s*| denotes the length of the string.
The second line contains string *t* (|*t*|<==<=|*s*|), consisting of lowercase English letters.
It is guaranteed that the lengths of strings *s* and *t* are the same and string *s* is lexicographically less than string *t*.
Output Specification:
If the string that meets the given requirements doesn't exist, print a single string "No such string" (without the quotes).
If such string exists, print it. If there are multiple valid strings, you may print any of them.
Demo Input:
['a\nc\n', 'aaa\nzzz\n', 'abcdefg\nabcdefh\n']
Demo Output:
['b\n', 'kkk\n', 'No such string\n']
Note:
String *s* = *s*<sub class="lower-index">1</sub>*s*<sub class="lower-index">2</sub>... *s*<sub class="lower-index">*n*</sub> is said to be lexicographically smaller than *t* = *t*<sub class="lower-index">1</sub>*t*<sub class="lower-index">2</sub>... *t*<sub class="lower-index">*n*</sub>, if there exists such *i*, that *s*<sub class="lower-index">1</sub> = *t*<sub class="lower-index">1</sub>, *s*<sub class="lower-index">2</sub> = *t*<sub class="lower-index">2</sub>, ... *s*<sub class="lower-index">*i* - 1</sub> = *t*<sub class="lower-index">*i* - 1</sub>, *s*<sub class="lower-index">*i*</sub> < *t*<sub class="lower-index">*i*</sub>.
|
```python
s=input()
t=input()
a=[]
b=[]
for i in s:
a.append(i)
for i in t:
b.append(i)
if len(s)==s.count('z'):
ans='a'*(len(s)+1)
if ans!=t:
print(ans)
else:
print("No such string")
else:
pos=-1
for i in range(len(s)-1,-1,-1):
if a[i]!='z':
pos=i
break
val=ord(a[pos])+1
a[pos]=chr(val)
for i in range(pos+1,len(s)):
a[i]='a'
if a!=b:
print(*a,sep="")
elif a==b:
print("No such string")
```
| 3
|
|
879
|
B
|
Table Tennis
|
PROGRAMMING
| 1,200
|
[
"data structures",
"implementation"
] | null | null |
*n* people are standing in a line to play table tennis. At first, the first two players in the line play a game. Then the loser goes to the end of the line, and the winner plays with the next person from the line, and so on. They play until someone wins *k* games in a row. This player becomes the winner.
For each of the participants, you know the power to play table tennis, and for all players these values are different. In a game the player with greater power always wins. Determine who will be the winner.
|
The first line contains two integers: *n* and *k* (2<=≤<=*n*<=≤<=500, 2<=≤<=*k*<=≤<=1012) — the number of people and the number of wins.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*n*) — powers of the player. It's guaranteed that this line contains a valid permutation, i.e. all *a**i* are distinct.
|
Output a single integer — power of the winner.
|
[
"2 2\n1 2\n",
"4 2\n3 1 2 4\n",
"6 2\n6 5 3 1 2 4\n",
"2 10000000000\n2 1\n"
] |
[
"2 ",
"3 ",
"6 ",
"2\n"
] |
Games in the second sample:
3 plays with 1. 3 wins. 1 goes to the end of the line.
3 plays with 2. 3 wins. He wins twice in a row. He becomes the winner.
| 1,000
|
[
{
"input": "2 2\n1 2",
"output": "2 "
},
{
"input": "4 2\n3 1 2 4",
"output": "3 "
},
{
"input": "6 2\n6 5 3 1 2 4",
"output": "6 "
},
{
"input": "2 10000000000\n2 1",
"output": "2"
},
{
"input": "4 4\n1 3 4 2",
"output": "4 "
},
{
"input": "2 2147483648\n2 1",
"output": "2"
},
{
"input": "3 2\n1 3 2",
"output": "3 "
},
{
"input": "3 3\n1 2 3",
"output": "3 "
},
{
"input": "5 2\n2 1 3 4 5",
"output": "5 "
},
{
"input": "10 2\n7 10 5 8 9 3 4 6 1 2",
"output": "10 "
},
{
"input": "100 2\n62 70 29 14 12 87 94 78 39 92 84 91 61 49 60 33 69 37 19 82 42 8 45 97 81 43 54 67 1 22 77 58 65 17 18 28 25 57 16 90 40 13 4 21 68 35 15 76 73 93 56 95 79 47 74 75 30 71 66 99 41 24 88 83 5 6 31 96 38 80 27 46 51 53 2 86 32 9 20 100 26 36 63 7 52 55 23 3 50 59 48 89 85 44 34 64 10 72 11 98",
"output": "70 "
},
{
"input": "4 10\n2 1 3 4",
"output": "4"
},
{
"input": "10 2\n1 2 3 4 5 6 7 8 9 10",
"output": "10 "
},
{
"input": "10 2\n10 9 8 7 6 5 4 3 2 1",
"output": "10 "
},
{
"input": "4 1000000000000\n3 4 1 2",
"output": "4"
},
{
"input": "100 10\n19 55 91 50 31 23 60 84 38 1 22 51 27 76 28 98 11 44 61 63 15 93 52 3 66 16 53 36 18 62 35 85 78 37 73 64 87 74 46 26 82 69 49 33 83 89 56 67 71 25 39 94 96 17 21 6 47 68 34 42 57 81 13 10 54 2 48 80 20 77 4 5 59 30 90 95 45 75 8 88 24 41 40 14 97 32 7 9 65 70 100 99 72 58 92 29 79 12 86 43",
"output": "91 "
},
{
"input": "100 50\n2 4 82 12 47 63 52 91 87 45 53 1 17 25 64 50 9 13 22 54 21 30 43 24 38 33 68 11 41 78 99 23 28 18 58 67 79 10 71 56 49 61 26 29 59 20 90 74 5 75 89 8 39 95 72 42 66 98 44 32 88 35 92 3 97 55 65 51 77 27 81 76 84 69 73 85 19 46 62 100 60 37 7 36 57 6 14 83 40 48 16 70 96 15 31 93 80 86 94 34",
"output": "100 "
},
{
"input": "2 1000000000000\n1 2",
"output": "2"
},
{
"input": "5 2\n1 4 3 5 2",
"output": "4 "
},
{
"input": "5 2\n1 3 2 4 5",
"output": "3 "
},
{
"input": "4 1000000000000\n3 1 2 4",
"output": "4"
},
{
"input": "4 2\n1 3 2 4",
"output": "3 "
},
{
"input": "10 3\n8 1 9 2 3 10 4 5 6 7",
"output": "9 "
},
{
"input": "5 2\n2 1 4 3 5",
"output": "4 "
},
{
"input": "3 4294967297\n2 1 3",
"output": "3"
},
{
"input": "4 4294967297\n3 2 1 4",
"output": "4"
},
{
"input": "5 4294967298\n3 2 1 4 5",
"output": "5"
},
{
"input": "10 4\n5 4 7 1 2 9 3 6 8 10",
"output": "9 "
},
{
"input": "11 21474836489\n10 1 2 3 4 5 6 7 8 9 11",
"output": "11"
}
] | 1,521,486,067
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 43
| 78
| 7,065,600
|
IL = lambda: list(map(int, input().split()))
I = lambda: int(input())
n, k = IL()
a = IL()
ans = 0
score = 0
for i in range(n-1):
if a[0] > a[1]:
score += 1
else:
score = 1
if score == k:
ans = a[0]
break
p1, p2 = a[:2]
a.pop(0)
a[0] = max(p1, p2)
a.append(min(p1, p2))
if ans==0:
ans = max(a)
print(ans)
|
Title: Table Tennis
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
*n* people are standing in a line to play table tennis. At first, the first two players in the line play a game. Then the loser goes to the end of the line, and the winner plays with the next person from the line, and so on. They play until someone wins *k* games in a row. This player becomes the winner.
For each of the participants, you know the power to play table tennis, and for all players these values are different. In a game the player with greater power always wins. Determine who will be the winner.
Input Specification:
The first line contains two integers: *n* and *k* (2<=≤<=*n*<=≤<=500, 2<=≤<=*k*<=≤<=1012) — the number of people and the number of wins.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*n*) — powers of the player. It's guaranteed that this line contains a valid permutation, i.e. all *a**i* are distinct.
Output Specification:
Output a single integer — power of the winner.
Demo Input:
['2 2\n1 2\n', '4 2\n3 1 2 4\n', '6 2\n6 5 3 1 2 4\n', '2 10000000000\n2 1\n']
Demo Output:
['2 ', '3 ', '6 ', '2\n']
Note:
Games in the second sample:
3 plays with 1. 3 wins. 1 goes to the end of the line.
3 plays with 2. 3 wins. He wins twice in a row. He becomes the winner.
|
```python
IL = lambda: list(map(int, input().split()))
I = lambda: int(input())
n, k = IL()
a = IL()
ans = 0
score = 0
for i in range(n-1):
if a[0] > a[1]:
score += 1
else:
score = 1
if score == k:
ans = a[0]
break
p1, p2 = a[:2]
a.pop(0)
a[0] = max(p1, p2)
a.append(min(p1, p2))
if ans==0:
ans = max(a)
print(ans)
```
| 3
|
|
426
|
A
|
Sereja and Mugs
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Sereja showed an interesting game to his friends. The game goes like that. Initially, there is a table with an empty cup and *n* water mugs on it. Then all players take turns to move. During a move, a player takes a non-empty mug of water and pours all water from it into the cup. If the cup overfills, then we assume that this player lost.
As soon as Sereja's friends heard of the game, they wanted to play it. Sereja, on the other hand, wanted to find out whether his friends can play the game in such a way that there are no losers. You are given the volumes of all mugs and the cup. Also, you know that Sereja has (*n*<=-<=1) friends. Determine if Sereja's friends can play the game so that nobody loses.
|
The first line contains integers *n* and *s* (2<=≤<=*n*<=≤<=100; 1<=≤<=*s*<=≤<=1000) — the number of mugs and the volume of the cup. The next line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=10). Number *a**i* means the volume of the *i*-th mug.
|
In a single line, print "YES" (without the quotes) if his friends can play in the described manner, and "NO" (without the quotes) otherwise.
|
[
"3 4\n1 1 1\n",
"3 4\n3 1 3\n",
"3 4\n4 4 4\n"
] |
[
"YES\n",
"YES\n",
"NO\n"
] |
none
| 500
|
[
{
"input": "3 4\n1 1 1",
"output": "YES"
},
{
"input": "3 4\n3 1 3",
"output": "YES"
},
{
"input": "3 4\n4 4 4",
"output": "NO"
},
{
"input": "2 1\n1 10",
"output": "YES"
},
{
"input": "3 12\n5 6 6",
"output": "YES"
},
{
"input": "4 10\n6 3 8 7",
"output": "NO"
},
{
"input": "5 16\n3 3 2 7 9",
"output": "YES"
},
{
"input": "6 38\n9 10 3 8 10 6",
"output": "YES"
},
{
"input": "7 12\n4 4 5 2 2 4 9",
"output": "NO"
},
{
"input": "8 15\n8 10 4 2 10 9 7 6",
"output": "NO"
},
{
"input": "9 22\n1 3 5 9 7 6 1 10 1",
"output": "NO"
},
{
"input": "10 30\n9 10 4 5 5 7 1 7 7 2",
"output": "NO"
},
{
"input": "38 83\n9 9 3 10 2 4 6 10 9 5 1 8 7 4 7 2 6 5 3 1 10 8 4 8 3 7 1 2 7 6 8 6 5 2 3 1 1 2",
"output": "NO"
},
{
"input": "84 212\n6 2 3 1 2 7 5 1 7 2 9 10 9 5 2 5 4 10 9 9 1 9 8 8 9 4 9 4 8 2 1 8 4 5 10 7 6 2 1 10 10 7 9 4 5 9 5 10 10 3 6 6 4 4 4 8 5 4 9 1 9 9 1 7 9 2 10 9 10 8 3 3 9 3 9 10 1 8 9 2 6 9 7 2",
"output": "NO"
},
{
"input": "8 50\n8 8 8 4 4 6 10 10",
"output": "YES"
},
{
"input": "7 24\n1 4 9 1 2 3 6",
"output": "YES"
},
{
"input": "47 262\n3 7 6 4 10 3 5 7 2 9 3 2 2 10 8 7 3 10 6 3 1 1 4 10 2 9 2 10 6 4 3 6 3 6 9 7 8 8 3 3 10 5 2 10 7 10 9",
"output": "YES"
},
{
"input": "42 227\n3 6 1 9 4 10 4 10 7 8 10 10 8 7 10 4 6 8 7 7 6 9 3 6 5 5 2 7 2 7 4 4 6 6 4 3 9 3 6 4 7 2",
"output": "NO"
},
{
"input": "97 65\n3 10 2 6 1 4 7 5 10 3 10 4 5 5 1 6 10 7 4 5 3 9 9 8 6 9 2 3 6 8 5 5 5 5 5 3 10 4 1 8 8 9 8 4 1 4 9 3 6 3 1 4 8 3 10 8 6 4 5 4 3 2 2 4 3 6 4 6 2 3 3 3 7 5 1 8 1 4 5 1 1 6 4 2 1 7 8 6 1 1 5 6 5 10 6 7 5",
"output": "NO"
},
{
"input": "94 279\n2 5 9 5 10 3 1 8 1 7 1 8 1 6 7 8 4 9 5 10 3 7 6 8 8 5 6 8 10 9 4 1 3 3 4 7 8 2 6 6 5 1 3 7 1 7 2 2 2 8 4 1 1 5 9 4 1 2 3 10 1 4 9 9 6 8 8 1 9 10 4 1 8 5 8 9 4 8 2 1 1 9 4 5 6 1 2 5 6 7 3 1 4 6",
"output": "NO"
},
{
"input": "58 70\n8 2 10 2 7 3 8 3 8 7 6 2 4 10 10 6 10 3 7 6 4 3 5 5 5 3 8 10 3 4 8 4 2 6 8 9 6 9 4 3 5 2 2 6 10 6 2 1 7 5 6 4 1 9 10 2 4 5",
"output": "NO"
},
{
"input": "6 14\n3 9 2 1 4 2",
"output": "YES"
},
{
"input": "78 400\n5 9 3 4 7 4 1 4 6 3 9 1 8 3 3 6 10 2 1 9 6 1 8 10 1 6 4 5 2 1 5 9 6 10 3 6 5 2 4 10 6 9 3 8 10 7 2 8 8 2 10 1 4 5 2 8 6 4 4 3 5 2 3 10 1 9 8 5 6 7 9 1 8 8 5 4 2 4",
"output": "YES"
},
{
"input": "41 181\n5 3 10 4 2 5 9 3 1 6 6 10 4 3 9 8 5 9 2 5 4 6 6 3 7 9 10 3 10 6 10 5 6 1 6 9 9 1 2 4 3",
"output": "NO"
},
{
"input": "2 4\n4 4",
"output": "YES"
},
{
"input": "29 71\n4 8 9 4 8 10 4 10 2 9 3 9 1 2 9 5 9 7 1 10 4 1 1 9 8 7 4 6 7",
"output": "NO"
},
{
"input": "49 272\n4 10 8 7 5 6 9 7 2 6 6 2 10 7 5 6 5 3 6 4 3 7 9 3 7 7 4 10 5 6 7 3 6 4 6 7 7 2 5 5 7 3 7 9 3 6 6 2 1",
"output": "YES"
},
{
"input": "91 486\n1 3 5 4 4 7 3 9 3 4 5 4 5 4 7 9 5 8 4 10 9 1 1 9 9 1 6 2 5 4 7 4 10 3 2 10 9 3 4 5 1 3 4 2 10 9 10 9 10 2 4 6 2 5 3 6 4 9 10 3 9 8 1 2 5 9 2 10 4 6 10 8 10 9 1 2 5 8 6 6 6 1 10 3 9 3 5 6 1 5 5",
"output": "YES"
},
{
"input": "80 78\n1 9 4 9 8 3 7 10 4 9 2 1 4 4 9 5 9 1 2 6 5 2 4 8 4 6 9 6 7 10 1 9 10 4 7 1 7 10 8 9 10 5 2 6 7 7 7 7 7 8 2 5 1 7 2 3 2 5 10 6 3 4 5 2 6 3 4 2 7 9 9 3 8 8 2 3 7 1 5 10",
"output": "NO"
},
{
"input": "53 245\n5 6 9 9 2 3 2 5 10 9 3 5 6 3 10 10 9 4 9 7 10 9 7 7 3 4 9 3 7 3 8 6 8 9 3 8 9 1 3 1 9 10 3 9 3 1 6 6 3 8 7 8 9",
"output": "NO"
},
{
"input": "50 271\n6 9 10 1 1 1 8 3 6 6 3 2 5 9 7 5 7 9 10 9 4 6 6 2 6 6 9 5 1 6 5 8 3 2 5 10 10 1 4 1 4 6 1 8 7 8 9 4 7 5",
"output": "YES"
},
{
"input": "38 214\n5 8 4 5 1 9 9 2 6 3 4 3 5 7 7 7 3 10 1 5 10 4 2 2 10 10 6 6 6 7 1 6 10 5 7 4 5 10",
"output": "YES"
},
{
"input": "100 1\n10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10",
"output": "NO"
},
{
"input": "100 100\n10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10",
"output": "NO"
},
{
"input": "100 989\n10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10",
"output": "NO"
},
{
"input": "100 990\n10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10",
"output": "YES"
},
{
"input": "100 1000\n10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10",
"output": "YES"
},
{
"input": "2 1\n1 1",
"output": "YES"
},
{
"input": "2 1\n2 2",
"output": "NO"
},
{
"input": "2 1000\n1 1",
"output": "YES"
},
{
"input": "10 89\n10 10 10 10 10 10 10 10 10 10",
"output": "NO"
},
{
"input": "10 44\n1 10 2 3 4 5 6 7 8 9",
"output": "NO"
}
] | 1,587,486,136
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 42
| 140
| 0
|
def canWin(c, arr):
return sum(arr) - max(arr) <= c
n, c = map(int, input().split())
nums = list(map(int, input().split()))
print("YES" if canWin(c, nums) else "NO")
|
Title: Sereja and Mugs
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Sereja showed an interesting game to his friends. The game goes like that. Initially, there is a table with an empty cup and *n* water mugs on it. Then all players take turns to move. During a move, a player takes a non-empty mug of water and pours all water from it into the cup. If the cup overfills, then we assume that this player lost.
As soon as Sereja's friends heard of the game, they wanted to play it. Sereja, on the other hand, wanted to find out whether his friends can play the game in such a way that there are no losers. You are given the volumes of all mugs and the cup. Also, you know that Sereja has (*n*<=-<=1) friends. Determine if Sereja's friends can play the game so that nobody loses.
Input Specification:
The first line contains integers *n* and *s* (2<=≤<=*n*<=≤<=100; 1<=≤<=*s*<=≤<=1000) — the number of mugs and the volume of the cup. The next line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=10). Number *a**i* means the volume of the *i*-th mug.
Output Specification:
In a single line, print "YES" (without the quotes) if his friends can play in the described manner, and "NO" (without the quotes) otherwise.
Demo Input:
['3 4\n1 1 1\n', '3 4\n3 1 3\n', '3 4\n4 4 4\n']
Demo Output:
['YES\n', 'YES\n', 'NO\n']
Note:
none
|
```python
def canWin(c, arr):
return sum(arr) - max(arr) <= c
n, c = map(int, input().split())
nums = list(map(int, input().split()))
print("YES" if canWin(c, nums) else "NO")
```
| 3
|
|
518
|
C
|
Anya and Smartphone
|
PROGRAMMING
| 1,600
|
[
"constructive algorithms",
"data structures",
"implementation"
] | null | null |
Anya has bought a new smartphone that uses Berdroid operating system. The smartphone menu has exactly *n* applications, each application has its own icon. The icons are located on different screens, one screen contains *k* icons. The icons from the first to the *k*-th one are located on the first screen, from the (*k*<=+<=1)-th to the 2*k*-th ones are on the second screen and so on (the last screen may be partially empty).
Initially the smartphone menu is showing the screen number 1. To launch the application with the icon located on the screen *t*, Anya needs to make the following gestures: first she scrolls to the required screen number *t*, by making *t*<=-<=1 gestures (if the icon is on the screen *t*), and then make another gesture — press the icon of the required application exactly once to launch it.
After the application is launched, the menu returns to the first screen. That is, to launch the next application you need to scroll through the menu again starting from the screen number 1.
All applications are numbered from 1 to *n*. We know a certain order in which the icons of the applications are located in the menu at the beginning, but it changes as long as you use the operating system. Berdroid is intelligent system, so it changes the order of the icons by moving the more frequently used icons to the beginning of the list. Formally, right after an application is launched, Berdroid swaps the application icon and the icon of a preceding application (that is, the icon of an application on the position that is smaller by one in the order of menu). The preceding icon may possibly be located on the adjacent screen. The only exception is when the icon of the launched application already occupies the first place, in this case the icon arrangement doesn't change.
Anya has planned the order in which she will launch applications. How many gestures should Anya make to launch the applications in the planned order?
Note that one application may be launched multiple times.
|
The first line of the input contains three numbers *n*,<=*m*,<=*k* (1<=≤<=*n*,<=*m*,<=*k*<=≤<=105) — the number of applications that Anya has on her smartphone, the number of applications that will be launched and the number of icons that are located on the same screen.
The next line contains *n* integers, permutation *a*1,<=*a*2,<=...,<=*a**n* — the initial order of icons from left to right in the menu (from the first to the last one), *a**i* — is the id of the application, whose icon goes *i*-th in the menu. Each integer from 1 to *n* occurs exactly once among *a**i*.
The third line contains *m* integers *b*1,<=*b*2,<=...,<=*b**m*(1<=≤<=*b**i*<=≤<=*n*) — the ids of the launched applications in the planned order. One application may be launched multiple times.
|
Print a single number — the number of gestures that Anya needs to make to launch all the applications in the desired order.
|
[
"8 3 3\n1 2 3 4 5 6 7 8\n7 8 1\n",
"5 4 2\n3 1 5 2 4\n4 4 4 4\n"
] |
[
"7\n",
"8\n"
] |
In the first test the initial configuration looks like (123)(456)(78), that is, the first screen contains icons of applications 1, 2, 3, the second screen contains icons 4, 5, 6, the third screen contains icons 7, 8.
After application 7 is launched, we get the new arrangement of the icons — (123)(457)(68). To launch it Anya makes 3 gestures.
After application 8 is launched, we get configuration (123)(457)(86). To launch it Anya makes 3 gestures.
After application 1 is launched, the arrangement of icons in the menu doesn't change. To launch it Anya makes 1 gesture.
In total, Anya makes 7 gestures.
| 1,500
|
[
{
"input": "8 3 3\n1 2 3 4 5 6 7 8\n7 8 1",
"output": "7"
},
{
"input": "5 4 2\n3 1 5 2 4\n4 4 4 4",
"output": "8"
},
{
"input": "10 10 3\n1 2 3 4 5 6 7 8 9 10\n2 3 4 5 6 7 8 9 10 1",
"output": "25"
},
{
"input": "10 12 3\n6 1 2 9 3 10 8 5 7 4\n3 9 9 4 8 2 3 8 10 8 3 4",
"output": "29"
},
{
"input": "16 12 8\n6 11 12 5 1 10 16 3 13 15 9 4 8 2 14 7\n6 13 8 12 8 15 12 16 8 11 3 16",
"output": "18"
},
{
"input": "4 3 5\n3 1 4 2\n4 2 4",
"output": "3"
},
{
"input": "4 2 15\n2 4 3 1\n1 2",
"output": "2"
},
{
"input": "71 96 75634\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71\n27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26 27 26",
"output": "96"
},
{
"input": "71 81 2149\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 29 28 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71\n29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29 28 29",
"output": "81"
}
] | 1,583,345,788
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 43
| 421
| 19,865,600
|
import sys
input = sys.stdin.readline
from collections import defaultdict
n, m, k = map(int, input().split())
a = list(map(int, input().split()))
a = list(map(lambda x: x-1, a))
b = list(map(int, input().split()))
b = list(map(lambda x: x-1, b))
page = [-1]*n
pos = [-1]*n
d = defaultdict(int)
for i in range(n):
page[a[i]] = i//k
pos[a[i]] = i%k
d[i//k*10**6+i%k] = a[i]
ans = 0
for bi in b:
ans += page[bi]+1
if page[bi]==0 and pos[bi]==0:
continue
page1 = page[bi]
pos1 = pos[bi]
if pos1==0:
page2 = page1-1
pos2 = k-1
else:
page2 = page1
pos2 = pos1-1
ci = d[page2*10**6+pos2]
page[bi] = page2
pos[bi] = pos2
page[ci] = page1
pos[ci] = pos1
d[page1*10**6+pos1] = ci
d[page2*10**6+pos2] = bi
print(ans)
|
Title: Anya and Smartphone
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Anya has bought a new smartphone that uses Berdroid operating system. The smartphone menu has exactly *n* applications, each application has its own icon. The icons are located on different screens, one screen contains *k* icons. The icons from the first to the *k*-th one are located on the first screen, from the (*k*<=+<=1)-th to the 2*k*-th ones are on the second screen and so on (the last screen may be partially empty).
Initially the smartphone menu is showing the screen number 1. To launch the application with the icon located on the screen *t*, Anya needs to make the following gestures: first she scrolls to the required screen number *t*, by making *t*<=-<=1 gestures (if the icon is on the screen *t*), and then make another gesture — press the icon of the required application exactly once to launch it.
After the application is launched, the menu returns to the first screen. That is, to launch the next application you need to scroll through the menu again starting from the screen number 1.
All applications are numbered from 1 to *n*. We know a certain order in which the icons of the applications are located in the menu at the beginning, but it changes as long as you use the operating system. Berdroid is intelligent system, so it changes the order of the icons by moving the more frequently used icons to the beginning of the list. Formally, right after an application is launched, Berdroid swaps the application icon and the icon of a preceding application (that is, the icon of an application on the position that is smaller by one in the order of menu). The preceding icon may possibly be located on the adjacent screen. The only exception is when the icon of the launched application already occupies the first place, in this case the icon arrangement doesn't change.
Anya has planned the order in which she will launch applications. How many gestures should Anya make to launch the applications in the planned order?
Note that one application may be launched multiple times.
Input Specification:
The first line of the input contains three numbers *n*,<=*m*,<=*k* (1<=≤<=*n*,<=*m*,<=*k*<=≤<=105) — the number of applications that Anya has on her smartphone, the number of applications that will be launched and the number of icons that are located on the same screen.
The next line contains *n* integers, permutation *a*1,<=*a*2,<=...,<=*a**n* — the initial order of icons from left to right in the menu (from the first to the last one), *a**i* — is the id of the application, whose icon goes *i*-th in the menu. Each integer from 1 to *n* occurs exactly once among *a**i*.
The third line contains *m* integers *b*1,<=*b*2,<=...,<=*b**m*(1<=≤<=*b**i*<=≤<=*n*) — the ids of the launched applications in the planned order. One application may be launched multiple times.
Output Specification:
Print a single number — the number of gestures that Anya needs to make to launch all the applications in the desired order.
Demo Input:
['8 3 3\n1 2 3 4 5 6 7 8\n7 8 1\n', '5 4 2\n3 1 5 2 4\n4 4 4 4\n']
Demo Output:
['7\n', '8\n']
Note:
In the first test the initial configuration looks like (123)(456)(78), that is, the first screen contains icons of applications 1, 2, 3, the second screen contains icons 4, 5, 6, the third screen contains icons 7, 8.
After application 7 is launched, we get the new arrangement of the icons — (123)(457)(68). To launch it Anya makes 3 gestures.
After application 8 is launched, we get configuration (123)(457)(86). To launch it Anya makes 3 gestures.
After application 1 is launched, the arrangement of icons in the menu doesn't change. To launch it Anya makes 1 gesture.
In total, Anya makes 7 gestures.
|
```python
import sys
input = sys.stdin.readline
from collections import defaultdict
n, m, k = map(int, input().split())
a = list(map(int, input().split()))
a = list(map(lambda x: x-1, a))
b = list(map(int, input().split()))
b = list(map(lambda x: x-1, b))
page = [-1]*n
pos = [-1]*n
d = defaultdict(int)
for i in range(n):
page[a[i]] = i//k
pos[a[i]] = i%k
d[i//k*10**6+i%k] = a[i]
ans = 0
for bi in b:
ans += page[bi]+1
if page[bi]==0 and pos[bi]==0:
continue
page1 = page[bi]
pos1 = pos[bi]
if pos1==0:
page2 = page1-1
pos2 = k-1
else:
page2 = page1
pos2 = pos1-1
ci = d[page2*10**6+pos2]
page[bi] = page2
pos[bi] = pos2
page[ci] = page1
pos[ci] = pos1
d[page1*10**6+pos1] = ci
d[page2*10**6+pos2] = bi
print(ans)
```
| 3
|
|
727
|
A
|
Transformation: from A to B
|
PROGRAMMING
| 1,000
|
[
"brute force",
"dfs and similar",
"math"
] | null | null |
Vasily has a number *a*, which he wants to turn into a number *b*. For this purpose, he can do two types of operations:
- multiply the current number by 2 (that is, replace the number *x* by 2·*x*); - append the digit 1 to the right of current number (that is, replace the number *x* by 10·*x*<=+<=1).
You need to help Vasily to transform the number *a* into the number *b* using only the operations described above, or find that it is impossible.
Note that in this task you are not required to minimize the number of operations. It suffices to find any way to transform *a* into *b*.
|
The first line contains two positive integers *a* and *b* (1<=≤<=*a*<=<<=*b*<=≤<=109) — the number which Vasily has and the number he wants to have.
|
If there is no way to get *b* from *a*, print "NO" (without quotes).
Otherwise print three lines. On the first line print "YES" (without quotes). The second line should contain single integer *k* — the length of the transformation sequence. On the third line print the sequence of transformations *x*1,<=*x*2,<=...,<=*x**k*, where:
- *x*1 should be equal to *a*, - *x**k* should be equal to *b*, - *x**i* should be obtained from *x**i*<=-<=1 using any of two described operations (1<=<<=*i*<=≤<=*k*).
If there are multiple answers, print any of them.
|
[
"2 162\n",
"4 42\n",
"100 40021\n"
] |
[
"YES\n5\n2 4 8 81 162 \n",
"NO\n",
"YES\n5\n100 200 2001 4002 40021 \n"
] |
none
| 1,000
|
[
{
"input": "2 162",
"output": "YES\n5\n2 4 8 81 162 "
},
{
"input": "4 42",
"output": "NO"
},
{
"input": "100 40021",
"output": "YES\n5\n100 200 2001 4002 40021 "
},
{
"input": "1 111111111",
"output": "YES\n9\n1 11 111 1111 11111 111111 1111111 11111111 111111111 "
},
{
"input": "1 1000000000",
"output": "NO"
},
{
"input": "999999999 1000000000",
"output": "NO"
},
{
"input": "1 2",
"output": "YES\n2\n1 2 "
},
{
"input": "1 536870912",
"output": "YES\n30\n1 2 4 8 16 32 64 128 256 512 1024 2048 4096 8192 16384 32768 65536 131072 262144 524288 1048576 2097152 4194304 8388608 16777216 33554432 67108864 134217728 268435456 536870912 "
},
{
"input": "11111 11111111",
"output": "YES\n4\n11111 111111 1111111 11111111 "
},
{
"input": "59139 946224",
"output": "YES\n5\n59139 118278 236556 473112 946224 "
},
{
"input": "9859 19718",
"output": "YES\n2\n9859 19718 "
},
{
"input": "25987 51974222",
"output": "YES\n5\n25987 259871 2598711 25987111 51974222 "
},
{
"input": "9411 188222222",
"output": "YES\n6\n9411 94111 941111 9411111 94111111 188222222 "
},
{
"input": "25539 510782222",
"output": "YES\n6\n25539 255391 2553911 25539111 255391111 510782222 "
},
{
"input": "76259 610072",
"output": "YES\n4\n76259 152518 305036 610072 "
},
{
"input": "92387 184774",
"output": "YES\n2\n92387 184774 "
},
{
"input": "8515 85151111",
"output": "YES\n5\n8515 85151 851511 8515111 85151111 "
},
{
"input": "91939 9193911",
"output": "YES\n3\n91939 919391 9193911 "
},
{
"input": "30518 610361",
"output": "YES\n3\n30518 61036 610361 "
},
{
"input": "46646 373168844",
"output": "YES\n7\n46646 466461 932922 9329221 93292211 186584422 373168844 "
},
{
"input": "30070 300701",
"output": "YES\n2\n30070 300701 "
},
{
"input": "13494 1079528",
"output": "YES\n5\n13494 134941 269882 539764 1079528 "
},
{
"input": "96918 775344422",
"output": "YES\n7\n96918 193836 1938361 3876722 38767221 387672211 775344422 "
},
{
"input": "13046 260921",
"output": "YES\n3\n13046 26092 260921 "
},
{
"input": "29174 5834811",
"output": "YES\n4\n29174 58348 583481 5834811 "
},
{
"input": "79894 319576421",
"output": "YES\n6\n79894 798941 1597882 15978821 31957642 319576421 "
},
{
"input": "96022 1920442",
"output": "YES\n3\n96022 960221 1920442 "
},
{
"input": "79446 6355681",
"output": "YES\n5\n79446 158892 317784 635568 6355681 "
},
{
"input": "5440 27853056",
"output": "YES\n11\n5440 10880 108801 217602 435204 870408 1740816 3481632 6963264 13926528 27853056 "
},
{
"input": "250000000 705032705",
"output": "NO"
},
{
"input": "17 35",
"output": "NO"
},
{
"input": "1 3",
"output": "NO"
},
{
"input": "2 11",
"output": "NO"
}
] | 1,618,144,702
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 33
| 140
| 4,915,200
|
def get_user_input():
user_input = input().split()
x = int(user_input[0])
n = int(user_input[1])
return x, n
def is_valid(x, n):
return x <= n
results = []
def transformation(x, n, seq, i):
if x == n:
results.append((i, seq))
else:
value_append = (10 * x) + 1
value_double = 2 * x
if is_valid(value_append, n):
transformation(value_append, n, [*seq, str(value_append)], i + 1)
if is_valid(value_double, n):
transformation(value_double, n, [*seq, str(value_double)], i + 1)
x, n = get_user_input()
transformation(x, n, [str(x)], 1)
if len(results) > 0:
print("YES")
for result in results:
print(result[0])
print(" ".join(result[1]))
else:
print("NO")
|
Title: Transformation: from A to B
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasily has a number *a*, which he wants to turn into a number *b*. For this purpose, he can do two types of operations:
- multiply the current number by 2 (that is, replace the number *x* by 2·*x*); - append the digit 1 to the right of current number (that is, replace the number *x* by 10·*x*<=+<=1).
You need to help Vasily to transform the number *a* into the number *b* using only the operations described above, or find that it is impossible.
Note that in this task you are not required to minimize the number of operations. It suffices to find any way to transform *a* into *b*.
Input Specification:
The first line contains two positive integers *a* and *b* (1<=≤<=*a*<=<<=*b*<=≤<=109) — the number which Vasily has and the number he wants to have.
Output Specification:
If there is no way to get *b* from *a*, print "NO" (without quotes).
Otherwise print three lines. On the first line print "YES" (without quotes). The second line should contain single integer *k* — the length of the transformation sequence. On the third line print the sequence of transformations *x*1,<=*x*2,<=...,<=*x**k*, where:
- *x*1 should be equal to *a*, - *x**k* should be equal to *b*, - *x**i* should be obtained from *x**i*<=-<=1 using any of two described operations (1<=<<=*i*<=≤<=*k*).
If there are multiple answers, print any of them.
Demo Input:
['2 162\n', '4 42\n', '100 40021\n']
Demo Output:
['YES\n5\n2 4 8 81 162 \n', 'NO\n', 'YES\n5\n100 200 2001 4002 40021 \n']
Note:
none
|
```python
def get_user_input():
user_input = input().split()
x = int(user_input[0])
n = int(user_input[1])
return x, n
def is_valid(x, n):
return x <= n
results = []
def transformation(x, n, seq, i):
if x == n:
results.append((i, seq))
else:
value_append = (10 * x) + 1
value_double = 2 * x
if is_valid(value_append, n):
transformation(value_append, n, [*seq, str(value_append)], i + 1)
if is_valid(value_double, n):
transformation(value_double, n, [*seq, str(value_double)], i + 1)
x, n = get_user_input()
transformation(x, n, [str(x)], 1)
if len(results) > 0:
print("YES")
for result in results:
print(result[0])
print(" ".join(result[1]))
else:
print("NO")
```
| 3
|
|
984
|
A
|
Game
|
PROGRAMMING
| 800
|
[
"sortings"
] | null | null |
Two players play a game.
Initially there are $n$ integers $a_1, a_2, \ldots, a_n$ written on the board. Each turn a player selects one number and erases it from the board. This continues until there is only one number left on the board, i. e. $n - 1$ turns are made. The first player makes the first move, then players alternate turns.
The first player wants to minimize the last number that would be left on the board, while the second player wants to maximize it.
You want to know what number will be left on the board after $n - 1$ turns if both players make optimal moves.
|
The first line contains one integer $n$ ($1 \le n \le 1000$) — the number of numbers on the board.
The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^6$).
|
Print one number that will be left on the board.
|
[
"3\n2 1 3\n",
"3\n2 2 2\n"
] |
[
"2",
"2"
] |
In the first sample, the first player erases $3$ and the second erases $1$. $2$ is left on the board.
In the second sample, $2$ is left on the board regardless of the actions of the players.
| 500
|
[
{
"input": "3\n2 1 3",
"output": "2"
},
{
"input": "3\n2 2 2",
"output": "2"
},
{
"input": "9\n44 53 51 80 5 27 74 79 94",
"output": "53"
},
{
"input": "10\n38 82 23 37 96 4 81 60 67 86",
"output": "60"
},
{
"input": "10\n58 26 77 15 53 81 68 48 22 65",
"output": "53"
},
{
"input": "1\n124",
"output": "124"
},
{
"input": "2\n2 1",
"output": "1"
},
{
"input": "3\n1 1 1000",
"output": "1"
},
{
"input": "2\n322 322",
"output": "322"
},
{
"input": "3\n9 92 12",
"output": "12"
},
{
"input": "3\n1 2 2",
"output": "2"
}
] | 1,596,261,005
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 35
| 109
| 6,963,200
|
n=int(input())
list=[int(i) for i in input().split()]
list.sort()
if(n%2):
print(list[int(n/2)])
else:
print(list[int(n/2)-1])
|
Title: Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Two players play a game.
Initially there are $n$ integers $a_1, a_2, \ldots, a_n$ written on the board. Each turn a player selects one number and erases it from the board. This continues until there is only one number left on the board, i. e. $n - 1$ turns are made. The first player makes the first move, then players alternate turns.
The first player wants to minimize the last number that would be left on the board, while the second player wants to maximize it.
You want to know what number will be left on the board after $n - 1$ turns if both players make optimal moves.
Input Specification:
The first line contains one integer $n$ ($1 \le n \le 1000$) — the number of numbers on the board.
The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^6$).
Output Specification:
Print one number that will be left on the board.
Demo Input:
['3\n2 1 3\n', '3\n2 2 2\n']
Demo Output:
['2', '2']
Note:
In the first sample, the first player erases $3$ and the second erases $1$. $2$ is left on the board.
In the second sample, $2$ is left on the board regardless of the actions of the players.
|
```python
n=int(input())
list=[int(i) for i in input().split()]
list.sort()
if(n%2):
print(list[int(n/2)])
else:
print(list[int(n/2)-1])
```
| 3
|
|
964
|
A
|
Splits
|
PROGRAMMING
| 800
|
[
"math"
] | null | null |
Let's define a split of $n$ as a nonincreasing sequence of positive integers, the sum of which is $n$.
For example, the following sequences are splits of $8$: $[4, 4]$, $[3, 3, 2]$, $[2, 2, 1, 1, 1, 1]$, $[5, 2, 1]$.
The following sequences aren't splits of $8$: $[1, 7]$, $[5, 4]$, $[11, -3]$, $[1, 1, 4, 1, 1]$.
The weight of a split is the number of elements in the split that are equal to the first element. For example, the weight of the split $[1, 1, 1, 1, 1]$ is $5$, the weight of the split $[5, 5, 3, 3, 3]$ is $2$ and the weight of the split $[9]$ equals $1$.
For a given $n$, find out the number of different weights of its splits.
|
The first line contains one integer $n$ ($1 \leq n \leq 10^9$).
|
Output one integer — the answer to the problem.
|
[
"7\n",
"8\n",
"9\n"
] |
[
"4\n",
"5\n",
"5\n"
] |
In the first sample, there are following possible weights of splits of $7$:
Weight 1: [$\textbf 7$]
Weight 2: [$\textbf 3$, $\textbf 3$, 1]
Weight 3: [$\textbf 2$, $\textbf 2$, $\textbf 2$, 1]
Weight 7: [$\textbf 1$, $\textbf 1$, $\textbf 1$, $\textbf 1$, $\textbf 1$, $\textbf 1$, $\textbf 1$]
| 500
|
[
{
"input": "7",
"output": "4"
},
{
"input": "8",
"output": "5"
},
{
"input": "9",
"output": "5"
},
{
"input": "1",
"output": "1"
},
{
"input": "286",
"output": "144"
},
{
"input": "48",
"output": "25"
},
{
"input": "941",
"output": "471"
},
{
"input": "45154",
"output": "22578"
},
{
"input": "60324",
"output": "30163"
},
{
"input": "91840",
"output": "45921"
},
{
"input": "41909",
"output": "20955"
},
{
"input": "58288",
"output": "29145"
},
{
"input": "91641",
"output": "45821"
},
{
"input": "62258",
"output": "31130"
},
{
"input": "79811",
"output": "39906"
},
{
"input": "88740",
"output": "44371"
},
{
"input": "12351",
"output": "6176"
},
{
"input": "1960",
"output": "981"
},
{
"input": "29239",
"output": "14620"
},
{
"input": "85801",
"output": "42901"
},
{
"input": "43255",
"output": "21628"
},
{
"input": "13439",
"output": "6720"
},
{
"input": "35668",
"output": "17835"
},
{
"input": "19122",
"output": "9562"
},
{
"input": "60169",
"output": "30085"
},
{
"input": "50588",
"output": "25295"
},
{
"input": "2467",
"output": "1234"
},
{
"input": "39315",
"output": "19658"
},
{
"input": "29950",
"output": "14976"
},
{
"input": "17286",
"output": "8644"
},
{
"input": "7359066",
"output": "3679534"
},
{
"input": "1016391",
"output": "508196"
},
{
"input": "7928871",
"output": "3964436"
},
{
"input": "3968891",
"output": "1984446"
},
{
"input": "2636452",
"output": "1318227"
},
{
"input": "5076901",
"output": "2538451"
},
{
"input": "9870265",
"output": "4935133"
},
{
"input": "2453786",
"output": "1226894"
},
{
"input": "7263670",
"output": "3631836"
},
{
"input": "1890845",
"output": "945423"
},
{
"input": "574128507",
"output": "287064254"
},
{
"input": "648476655",
"output": "324238328"
},
{
"input": "97349542",
"output": "48674772"
},
{
"input": "716489761",
"output": "358244881"
},
{
"input": "858771038",
"output": "429385520"
},
{
"input": "520778784",
"output": "260389393"
},
{
"input": "439004204",
"output": "219502103"
},
{
"input": "589992198",
"output": "294996100"
},
{
"input": "371106544",
"output": "185553273"
},
{
"input": "894241590",
"output": "447120796"
},
{
"input": "123957268",
"output": "61978635"
},
{
"input": "234149297",
"output": "117074649"
},
{
"input": "789954052",
"output": "394977027"
},
{
"input": "667978920",
"output": "333989461"
},
{
"input": "154647261",
"output": "77323631"
},
{
"input": "751453521",
"output": "375726761"
},
{
"input": "848862308",
"output": "424431155"
},
{
"input": "323926781",
"output": "161963391"
},
{
"input": "576768825",
"output": "288384413"
},
{
"input": "31293802",
"output": "15646902"
},
{
"input": "2",
"output": "2"
},
{
"input": "1000000000",
"output": "500000001"
},
{
"input": "3",
"output": "2"
}
] | 1,596,390,006
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 63
| 156
| 20,172,800
|
# import sys;sys.stdin = open("in.txt", "r");sys.stdout = open("out.txt", "w")
print(int(input())//2 + 1)
|
Title: Splits
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Let's define a split of $n$ as a nonincreasing sequence of positive integers, the sum of which is $n$.
For example, the following sequences are splits of $8$: $[4, 4]$, $[3, 3, 2]$, $[2, 2, 1, 1, 1, 1]$, $[5, 2, 1]$.
The following sequences aren't splits of $8$: $[1, 7]$, $[5, 4]$, $[11, -3]$, $[1, 1, 4, 1, 1]$.
The weight of a split is the number of elements in the split that are equal to the first element. For example, the weight of the split $[1, 1, 1, 1, 1]$ is $5$, the weight of the split $[5, 5, 3, 3, 3]$ is $2$ and the weight of the split $[9]$ equals $1$.
For a given $n$, find out the number of different weights of its splits.
Input Specification:
The first line contains one integer $n$ ($1 \leq n \leq 10^9$).
Output Specification:
Output one integer — the answer to the problem.
Demo Input:
['7\n', '8\n', '9\n']
Demo Output:
['4\n', '5\n', '5\n']
Note:
In the first sample, there are following possible weights of splits of $7$:
Weight 1: [$\textbf 7$]
Weight 2: [$\textbf 3$, $\textbf 3$, 1]
Weight 3: [$\textbf 2$, $\textbf 2$, $\textbf 2$, 1]
Weight 7: [$\textbf 1$, $\textbf 1$, $\textbf 1$, $\textbf 1$, $\textbf 1$, $\textbf 1$, $\textbf 1$]
|
```python
# import sys;sys.stdin = open("in.txt", "r");sys.stdout = open("out.txt", "w")
print(int(input())//2 + 1)
```
| 3
|
|
120
|
F
|
Spiders
|
PROGRAMMING
| 1,400
|
[
"dp",
"greedy",
"trees"
] | null | null |
One day mum asked Petya to sort his toys and get rid of some of them. Petya found a whole box of toy spiders. They were quite dear to him and the boy didn't want to throw them away. Petya conjured a cunning plan: he will glue all the spiders together and attach them to the ceiling. Besides, Petya knows that the lower the spiders will hang, the more mum is going to like it and then she won't throw his favourite toys away. Help Petya carry out the plan.
A spider consists of *k* beads tied together by *k*<=-<=1 threads. Each thread connects two different beads, at that any pair of beads that make up a spider is either directly connected by a thread, or is connected via some chain of threads and beads.
Petya may glue spiders together directly gluing their beads. The length of each thread equals 1. The sizes of the beads can be neglected. That's why we can consider that gluing spiders happens by identifying some of the beads (see the picture). Besides, the construction resulting from the gluing process should also represent a spider, that is, it should have the given features.
After Petya glues all spiders together, he measures the length of the resulting toy. The distance between a pair of beads is identified as the total length of the threads that connect these two beads. The length of the resulting construction is the largest distance between all pairs of beads. Petya wants to make the spider whose length is as much as possible.
The picture two shows two spiders from the second sample. We can glue to the bead number 2 of the first spider the bead number 1 of the second spider. The threads in the spiders that form the sequence of threads of maximum lengths are highlighted on the picture.
|
The first input file line contains one integer *n* (1<=≤<=*n*<=≤<=100) — the number of spiders. Next *n* lines contain the descriptions of each spider: integer *n**i* (2<=≤<=*n**i*<=≤<=100) — the number of beads, then *n**i*<=-<=1 pairs of numbers denoting the numbers of the beads connected by threads. The beads that make up each spider are numbered from 1 to *n**i*.
|
Print a single number — the length of the required construction.
|
[
"1\n3 1 2 2 3\n",
"2\n3 1 2 1 3\n4 1 2 2 3 2 4\n",
"2\n5 1 2 2 3 3 4 3 5\n7 3 4 1 2 2 4 4 6 2 7 6 5\n"
] |
[
"2\n",
"4\n",
"7\n"
] |
none
| 0
|
[
{
"input": "1\n3 1 2 2 3",
"output": "2"
},
{
"input": "2\n3 1 2 1 3\n4 1 2 2 3 2 4",
"output": "4"
},
{
"input": "2\n5 1 2 2 3 3 4 3 5\n7 3 4 1 2 2 4 4 6 2 7 6 5",
"output": "7"
},
{
"input": "3\n3 1 2 2 3\n5 2 5 5 3 3 4 5 1\n9 6 5 5 9 4 8 4 7 2 1 2 6 2 4 6 3",
"output": "10"
},
{
"input": "7\n2 2 1\n4 1 4 2 3 1 2\n3 3 1 3 2\n5 1 4 3 5 1 2 1 3\n6 4 5 1 3 4 2 3 6 5 1\n7 1 3 3 6 7 4 7 1 5 2 3 5\n10 6 8 2 6 6 3 2 7 2 4 6 10 3 1 6 5 6 9",
"output": "23"
},
{
"input": "10\n3 1 2 1 3\n3 1 2 1 3\n7 1 2 1 3 3 4 7 5 1 6 5 1\n2 1 2\n4 4 3 3 1 4 2\n3 3 1 3 2\n5 4 2 5 1 3 5 3 4\n6 1 6 2 4 6 2 4 3 5 1\n7 2 4 4 6 7 3 3 1 3 5 2 7\n10 3 5 5 6 1 9 5 2 7 8 8 1 6 10 4 3 4 7",
"output": "36"
},
{
"input": "7\n4 2 3 4 1 2 4\n4 4 3 2 1 3 2\n3 2 1 2 3\n5 5 4 1 5 1 2 2 3\n6 1 3 4 5 2 6 3 2 1 4\n7 6 4 4 7 6 2 6 3 3 1 6 5\n10 8 10 4 8 5 9 5 6 3 4 3 1 5 3 4 7 1 2",
"output": "26"
},
{
"input": "7\n2 1 2\n4 4 1 1 2 4 3\n3 3 2 2 1\n5 4 1 1 5 4 3 1 2\n6 4 2 3 1 3 4 3 5 3 6\n8 7 4 6 2 6 7 4 5 4 1 1 3 6 8\n10 4 1 8 9 7 8 2 4 8 6 6 5 2 7 8 3 7 10",
"output": "23"
},
{
"input": "3\n4 3 2 3 1 1 4\n4 3 1 2 4 3 2\n4 1 4 2 1 4 3",
"output": "9"
},
{
"input": "3\n10 7 3 10 9 7 10 4 7 8 6 8 2 4 8 8 5 5 1\n12 10 3 11 4 11 9 12 1 10 12 8 7 8 11 6 5 10 6 10 2 6 8\n13 3 7 10 4 3 8 3 1 8 5 4 12 9 2 8 6 10 9 1 10 10 11 4 13",
"output": "18"
},
{
"input": "4\n5 3 2 3 5 4 1 4 3\n6 6 4 1 2 2 3 2 6 6 5\n7 6 1 6 4 4 5 1 7 4 3 2 6\n8 2 1 6 8 4 5 8 3 4 7 4 2 2 6",
"output": "15"
},
{
"input": "2\n10 8 1 6 7 10 8 7 5 3 4 5 3 9 2 6 9 2 10\n10 9 7 6 1 8 10 6 4 8 3 9 2 3 5 8 9 3 6",
"output": "14"
},
{
"input": "6\n3 1 3 1 2\n3 2 1 3 2\n5 2 5 5 1 2 3 2 4\n4 2 3 2 4 1 2\n5 2 3 2 1 3 5 2 4\n5 3 5 2 3 5 1 4 2",
"output": "16"
},
{
"input": "1\n50 4 37 49 41 40 10 23 14 35 3 39 46 40 8 11 35 32 17 3 7 25 47 17 27 49 1 28 9 6 38 41 29 22 13 38 16 4 48 4 21 13 23 49 30 35 44 25 34 25 18 31 50 21 6 32 19 19 28 22 42 39 25 22 26 21 40 21 33 21 39 14 43 3 15 17 2 19 12 26 24 28 5 22 11 4 20 10 45 48 36 4 22 4 32 4 31 39 49",
"output": "10"
},
{
"input": "4\n10 9 5 9 7 7 6 7 2 1 4 9 1 7 3 2 10 3 8\n10 9 10 2 5 3 6 5 7 9 3 9 1 9 2 2 8 2 4\n10 5 3 2 5 1 6 6 8 3 9 3 10 5 1 8 4 3 7\n10 7 6 3 1 8 10 2 3 2 7 5 9 10 5 1 4 2 8",
"output": "23"
}
] | 1,650,452,793
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 30
| 404
| 10,137,600
|
import abc
import itertools
import math
from math import gcd as gcd
import sys
import queue
import itertools
from heapq import heappop, heappush
import random
def solve():
def bfs(g, s):
used = [-1 for i in range(len(g))]
used[s] = 0
q = [s]
it = 0
while it < len(q):
f = q[it]
it += 1
for u in g[f]:
if used[u] == -1:
used[u] = used[f] + 1
q.append(u)
return max(used)
f = open("input.txt", "r")
n = int(f.readline())
res = 0
for i in range(n):
k, *a = map(int, f.readline().split())
g = [[] for i in range(k)]
for j in range(k - 1):
v, u = a[j * 2] - 1, a[j * 2 + 1] - 1
g[v].append(u)
g[u].append(v)
mx = 0
for i in range(k):
mx = max(mx, bfs(g, i))
res += mx
f.close()
f = open("output.txt", "w")
f.write(str(res))
f.close()
if __name__ == '__main__':
multi_test = 0
if multi_test == 1:
t = int(sys.stdin.readline())
for _ in range(t):
solve()
else:
solve()
|
Title: Spiders
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day mum asked Petya to sort his toys and get rid of some of them. Petya found a whole box of toy spiders. They were quite dear to him and the boy didn't want to throw them away. Petya conjured a cunning plan: he will glue all the spiders together and attach them to the ceiling. Besides, Petya knows that the lower the spiders will hang, the more mum is going to like it and then she won't throw his favourite toys away. Help Petya carry out the plan.
A spider consists of *k* beads tied together by *k*<=-<=1 threads. Each thread connects two different beads, at that any pair of beads that make up a spider is either directly connected by a thread, or is connected via some chain of threads and beads.
Petya may glue spiders together directly gluing their beads. The length of each thread equals 1. The sizes of the beads can be neglected. That's why we can consider that gluing spiders happens by identifying some of the beads (see the picture). Besides, the construction resulting from the gluing process should also represent a spider, that is, it should have the given features.
After Petya glues all spiders together, he measures the length of the resulting toy. The distance between a pair of beads is identified as the total length of the threads that connect these two beads. The length of the resulting construction is the largest distance between all pairs of beads. Petya wants to make the spider whose length is as much as possible.
The picture two shows two spiders from the second sample. We can glue to the bead number 2 of the first spider the bead number 1 of the second spider. The threads in the spiders that form the sequence of threads of maximum lengths are highlighted on the picture.
Input Specification:
The first input file line contains one integer *n* (1<=≤<=*n*<=≤<=100) — the number of spiders. Next *n* lines contain the descriptions of each spider: integer *n**i* (2<=≤<=*n**i*<=≤<=100) — the number of beads, then *n**i*<=-<=1 pairs of numbers denoting the numbers of the beads connected by threads. The beads that make up each spider are numbered from 1 to *n**i*.
Output Specification:
Print a single number — the length of the required construction.
Demo Input:
['1\n3 1 2 2 3\n', '2\n3 1 2 1 3\n4 1 2 2 3 2 4\n', '2\n5 1 2 2 3 3 4 3 5\n7 3 4 1 2 2 4 4 6 2 7 6 5\n']
Demo Output:
['2\n', '4\n', '7\n']
Note:
none
|
```python
import abc
import itertools
import math
from math import gcd as gcd
import sys
import queue
import itertools
from heapq import heappop, heappush
import random
def solve():
def bfs(g, s):
used = [-1 for i in range(len(g))]
used[s] = 0
q = [s]
it = 0
while it < len(q):
f = q[it]
it += 1
for u in g[f]:
if used[u] == -1:
used[u] = used[f] + 1
q.append(u)
return max(used)
f = open("input.txt", "r")
n = int(f.readline())
res = 0
for i in range(n):
k, *a = map(int, f.readline().split())
g = [[] for i in range(k)]
for j in range(k - 1):
v, u = a[j * 2] - 1, a[j * 2 + 1] - 1
g[v].append(u)
g[u].append(v)
mx = 0
for i in range(k):
mx = max(mx, bfs(g, i))
res += mx
f.close()
f = open("output.txt", "w")
f.write(str(res))
f.close()
if __name__ == '__main__':
multi_test = 0
if multi_test == 1:
t = int(sys.stdin.readline())
for _ in range(t):
solve()
else:
solve()
```
| 3
|
|
102
|
B
|
Sum of Digits
|
PROGRAMMING
| 1,000
|
[
"implementation"
] |
B. Sum of Digits
|
2
|
265
|
Having watched the last Harry Potter film, little Gerald also decided to practice magic. He found in his father's magical book a spell that turns any number in the sum of its digits. At the moment Gerald learned that, he came across a number *n*. How many times can Gerald put a spell on it until the number becomes one-digit?
|
The first line contains the only integer *n* (0<=≤<=*n*<=≤<=10100000). It is guaranteed that *n* doesn't contain any leading zeroes.
|
Print the number of times a number can be replaced by the sum of its digits until it only contains one digit.
|
[
"0\n",
"10\n",
"991\n"
] |
[
"0\n",
"1\n",
"3\n"
] |
In the first sample the number already is one-digit — Herald can't cast a spell.
The second test contains number 10. After one casting of a spell it becomes 1, and here the process is completed. Thus, Gerald can only cast the spell once.
The third test contains number 991. As one casts a spell the following transformations take place: 991 → 19 → 10 → 1. After three transformations the number becomes one-digit.
| 1,000
|
[
{
"input": "0",
"output": "0"
},
{
"input": "10",
"output": "1"
},
{
"input": "991",
"output": "3"
},
{
"input": "99",
"output": "2"
},
{
"input": "100",
"output": "1"
},
{
"input": "123456789",
"output": "2"
},
{
"input": "32",
"output": "1"
},
{
"input": "86",
"output": "2"
},
{
"input": "2",
"output": "0"
},
{
"input": "8",
"output": "0"
},
{
"input": "34",
"output": "1"
},
{
"input": "13",
"output": "1"
},
{
"input": "28",
"output": "2"
},
{
"input": "23",
"output": "1"
},
{
"input": "57",
"output": "2"
},
{
"input": "29",
"output": "2"
},
{
"input": "353869285",
"output": "3"
},
{
"input": "549548646",
"output": "2"
},
{
"input": "858893424",
"output": "2"
},
{
"input": "644818852",
"output": "3"
},
{
"input": "360322525",
"output": "3"
},
{
"input": "122937520",
"output": "2"
},
{
"input": "288403032",
"output": "2"
},
{
"input": "677257481",
"output": "3"
},
{
"input": "58059561",
"output": "3"
},
{
"input": "211288847",
"output": "2"
},
{
"input": "339900034079539584",
"output": "2"
},
{
"input": "784084029310269952",
"output": "3"
},
{
"input": "774730557264864000",
"output": "3"
},
{
"input": "478233807148352256",
"output": "3"
},
{
"input": "165646874056812544",
"output": "3"
},
{
"input": "477533739511673792",
"output": "3"
},
{
"input": "660119944377002240",
"output": "3"
},
{
"input": "448375218506078080",
"output": "3"
},
{
"input": "77213334785795616",
"output": "3"
},
{
"input": "165767221702271872",
"output": "3"
},
{
"input": "1",
"output": "0"
},
{
"input": "5",
"output": "0"
},
{
"input": "8",
"output": "0"
},
{
"input": "156161456146164104103460452045416165146141414651641564105461465156445621465146514562146145465165145145614561465145614561",
"output": "3"
},
{
"input": "9",
"output": "0"
}
] | 1,665,670,623
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 51
| 184
| 1,843,200
|
import math
def sum_digit(n):
sum = 0
for i in n:
sum += int(i)
return str(sum)
n = input()
result = 0
while (len(n) > 1):
n = sum_digit(n)
# print(n)
result += 1
print(result)
|
Title: Sum of Digits
Time Limit: 2 seconds
Memory Limit: 265 megabytes
Problem Description:
Having watched the last Harry Potter film, little Gerald also decided to practice magic. He found in his father's magical book a spell that turns any number in the sum of its digits. At the moment Gerald learned that, he came across a number *n*. How many times can Gerald put a spell on it until the number becomes one-digit?
Input Specification:
The first line contains the only integer *n* (0<=≤<=*n*<=≤<=10100000). It is guaranteed that *n* doesn't contain any leading zeroes.
Output Specification:
Print the number of times a number can be replaced by the sum of its digits until it only contains one digit.
Demo Input:
['0\n', '10\n', '991\n']
Demo Output:
['0\n', '1\n', '3\n']
Note:
In the first sample the number already is one-digit — Herald can't cast a spell.
The second test contains number 10. After one casting of a spell it becomes 1, and here the process is completed. Thus, Gerald can only cast the spell once.
The third test contains number 991. As one casts a spell the following transformations take place: 991 → 19 → 10 → 1. After three transformations the number becomes one-digit.
|
```python
import math
def sum_digit(n):
sum = 0
for i in n:
sum += int(i)
return str(sum)
n = input()
result = 0
while (len(n) > 1):
n = sum_digit(n)
# print(n)
result += 1
print(result)
```
| 3.950683
|
34
|
B
|
Sale
|
PROGRAMMING
| 900
|
[
"greedy",
"sortings"
] |
B. Sale
|
2
|
256
|
Once Bob got to a sale of old TV sets. There were *n* TV sets at that sale. TV set with index *i* costs *a**i* bellars. Some TV sets have a negative price — their owners are ready to pay Bob if he buys their useless apparatus. Bob can «buy» any TV sets he wants. Though he's very strong, Bob can carry at most *m* TV sets, and he has no desire to go to the sale for the second time. Please, help Bob find out the maximum sum of money that he can earn.
|
The first line contains two space-separated integers *n* and *m* (1<=≤<=*m*<=≤<=*n*<=≤<=100) — amount of TV sets at the sale, and amount of TV sets that Bob can carry. The following line contains *n* space-separated integers *a**i* (<=-<=1000<=≤<=*a**i*<=≤<=1000) — prices of the TV sets.
|
Output the only number — the maximum sum of money that Bob can earn, given that he can carry at most *m* TV sets.
|
[
"5 3\n-6 0 35 -2 4\n",
"4 2\n7 0 0 -7\n"
] |
[
"8\n",
"7\n"
] |
none
| 1,000
|
[
{
"input": "5 3\n-6 0 35 -2 4",
"output": "8"
},
{
"input": "4 2\n7 0 0 -7",
"output": "7"
},
{
"input": "6 6\n756 -611 251 -66 572 -818",
"output": "1495"
},
{
"input": "5 5\n976 437 937 788 518",
"output": "0"
},
{
"input": "5 3\n-2 -2 -2 -2 -2",
"output": "6"
},
{
"input": "5 1\n998 997 985 937 998",
"output": "0"
},
{
"input": "2 2\n-742 -187",
"output": "929"
},
{
"input": "3 3\n522 597 384",
"output": "0"
},
{
"input": "4 2\n-215 -620 192 647",
"output": "835"
},
{
"input": "10 6\n557 605 685 231 910 633 130 838 -564 -85",
"output": "649"
},
{
"input": "20 14\n932 442 960 943 624 624 955 998 631 910 850 517 715 123 1000 155 -10 961 966 59",
"output": "10"
},
{
"input": "30 5\n991 997 996 967 977 999 991 986 1000 965 984 997 998 1000 958 983 974 1000 991 999 1000 978 961 992 990 998 998 978 998 1000",
"output": "0"
},
{
"input": "50 20\n-815 -947 -946 -993 -992 -846 -884 -954 -963 -733 -940 -746 -766 -930 -821 -937 -937 -999 -914 -938 -936 -975 -939 -981 -977 -952 -925 -901 -952 -978 -994 -957 -946 -896 -905 -836 -994 -951 -887 -939 -859 -953 -985 -988 -946 -829 -956 -842 -799 -886",
"output": "19441"
},
{
"input": "88 64\n999 999 1000 1000 999 996 995 1000 1000 999 1000 997 998 1000 999 1000 997 1000 993 998 994 999 998 996 1000 997 1000 1000 1000 997 1000 998 997 1000 1000 998 1000 998 999 1000 996 999 999 999 996 995 999 1000 998 999 1000 999 999 1000 1000 1000 996 1000 1000 1000 997 1000 1000 997 999 1000 1000 1000 1000 1000 999 999 1000 1000 996 999 1000 1000 995 999 1000 996 1000 998 999 999 1000 999",
"output": "0"
},
{
"input": "99 17\n-993 -994 -959 -989 -991 -995 -976 -997 -990 -1000 -996 -994 -999 -995 -1000 -983 -979 -1000 -989 -968 -994 -992 -962 -993 -999 -983 -991 -979 -995 -993 -973 -999 -995 -995 -999 -993 -995 -992 -947 -1000 -999 -998 -982 -988 -979 -993 -963 -988 -980 -990 -979 -976 -995 -999 -981 -988 -998 -999 -970 -1000 -983 -994 -943 -975 -998 -977 -973 -997 -959 -999 -983 -985 -950 -977 -977 -991 -998 -973 -987 -985 -985 -986 -984 -994 -978 -998 -989 -989 -988 -970 -985 -974 -997 -981 -962 -972 -995 -988 -993",
"output": "16984"
},
{
"input": "100 37\n205 19 -501 404 912 -435 -322 -469 -655 880 -804 -470 793 312 -108 586 -642 -928 906 605 -353 -800 745 -440 -207 752 -50 -28 498 -800 -62 -195 602 -833 489 352 536 404 -775 23 145 -512 524 759 651 -461 -427 -557 684 -366 62 592 -563 -811 64 418 -881 -308 591 -318 -145 -261 -321 -216 -18 595 -202 960 -4 219 226 -238 -882 -963 425 970 -434 -160 243 -672 -4 873 8 -633 904 -298 -151 -377 -61 -72 -677 -66 197 -716 3 -870 -30 152 -469 981",
"output": "21743"
},
{
"input": "100 99\n-931 -806 -830 -828 -916 -962 -660 -867 -952 -966 -820 -906 -724 -982 -680 -717 -488 -741 -897 -613 -986 -797 -964 -939 -808 -932 -810 -860 -641 -916 -858 -628 -821 -929 -917 -976 -664 -985 -778 -665 -624 -928 -940 -958 -884 -757 -878 -896 -634 -526 -514 -873 -990 -919 -988 -878 -650 -973 -774 -783 -733 -648 -756 -895 -833 -974 -832 -725 -841 -748 -806 -613 -924 -867 -881 -943 -864 -991 -809 -926 -777 -817 -998 -682 -910 -996 -241 -722 -964 -904 -821 -920 -835 -699 -805 -632 -779 -317 -915 -654",
"output": "81283"
},
{
"input": "100 14\n995 994 745 684 510 737 984 690 979 977 542 933 871 603 758 653 962 997 747 974 773 766 975 770 527 960 841 989 963 865 974 967 950 984 757 685 986 809 982 959 931 880 978 867 805 562 970 900 834 782 616 885 910 608 974 918 576 700 871 980 656 941 978 759 767 840 573 859 841 928 693 853 716 927 976 851 962 962 627 797 707 873 869 988 993 533 665 887 962 880 929 980 877 887 572 790 721 883 848 782",
"output": "0"
},
{
"input": "100 84\n768 946 998 752 931 912 826 1000 991 910 875 962 901 952 958 733 959 908 872 840 923 826 952 980 974 980 947 955 959 822 997 963 966 933 829 923 971 999 926 932 865 984 974 858 994 855 949 941 992 861 951 949 991 711 763 728 935 485 716 907 869 952 960 859 909 963 978 942 968 933 923 909 997 962 687 764 924 774 875 1000 961 951 987 974 848 921 966 859 995 997 974 931 886 941 974 986 906 978 998 823",
"output": "0"
},
{
"input": "100 80\n-795 -994 -833 -930 -974 -980 -950 -940 -788 -927 -583 -956 -945 -949 -809 -974 -957 -736 -967 -908 -975 -961 -986 -983 -963 -771 -952 -847 -751 -741 -982 -959 -925 -931 -839 -937 -880 -914 -858 -998 -812 -911 -862 -965 -943 -984 -738 -920 -950 -998 -909 -998 -781 -901 -677 -940 -985 -951 -675 -952 -967 -949 -882 -641 -969 -937 -975 -993 -913 -941 -807 -851 -832 -960 -939 -943 -895 -929 -528 -880 -823 -930 -888 -862 -948 -966 -962 -857 -799 -969 -833 -998 -952 -878 -946 -971 -976 -974 -723 -992",
"output": "75068"
},
{
"input": "1 1\n0",
"output": "0"
},
{
"input": "1 1\n1",
"output": "0"
},
{
"input": "1 1\n555",
"output": "0"
},
{
"input": "1 1\n-1",
"output": "1"
},
{
"input": "1 1\n-24",
"output": "24"
}
] | 1,585,072,607
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 25
| 280
| 0
|
n,m=map(int,input().split());p=0
a=sorted(list(map(int,input().split())))
for i in range(m):
if a[i]>=0:break
else:p=p+abs(a[i])
print(p)
|
Title: Sale
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Once Bob got to a sale of old TV sets. There were *n* TV sets at that sale. TV set with index *i* costs *a**i* bellars. Some TV sets have a negative price — their owners are ready to pay Bob if he buys their useless apparatus. Bob can «buy» any TV sets he wants. Though he's very strong, Bob can carry at most *m* TV sets, and he has no desire to go to the sale for the second time. Please, help Bob find out the maximum sum of money that he can earn.
Input Specification:
The first line contains two space-separated integers *n* and *m* (1<=≤<=*m*<=≤<=*n*<=≤<=100) — amount of TV sets at the sale, and amount of TV sets that Bob can carry. The following line contains *n* space-separated integers *a**i* (<=-<=1000<=≤<=*a**i*<=≤<=1000) — prices of the TV sets.
Output Specification:
Output the only number — the maximum sum of money that Bob can earn, given that he can carry at most *m* TV sets.
Demo Input:
['5 3\n-6 0 35 -2 4\n', '4 2\n7 0 0 -7\n']
Demo Output:
['8\n', '7\n']
Note:
none
|
```python
n,m=map(int,input().split());p=0
a=sorted(list(map(int,input().split())))
for i in range(m):
if a[i]>=0:break
else:p=p+abs(a[i])
print(p)
```
| 3.93
|
631
|
A
|
Interview
|
PROGRAMMING
| 900
|
[
"brute force",
"implementation"
] | null | null |
Blake is a CEO of a large company called "Blake Technologies". He loves his company very much and he thinks that his company should be the best. That is why every candidate needs to pass through the interview that consists of the following problem.
We define function *f*(*x*,<=*l*,<=*r*) as a bitwise OR of integers *x**l*,<=*x**l*<=+<=1,<=...,<=*x**r*, where *x**i* is the *i*-th element of the array *x*. You are given two arrays *a* and *b* of length *n*. You need to determine the maximum value of sum *f*(*a*,<=*l*,<=*r*)<=+<=*f*(*b*,<=*l*,<=*r*) among all possible 1<=≤<=*l*<=≤<=*r*<=≤<=*n*.
|
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the length of the arrays.
The second line contains *n* integers *a**i* (0<=≤<=*a**i*<=≤<=109).
The third line contains *n* integers *b**i* (0<=≤<=*b**i*<=≤<=109).
|
Print a single integer — the maximum value of sum *f*(*a*,<=*l*,<=*r*)<=+<=*f*(*b*,<=*l*,<=*r*) among all possible 1<=≤<=*l*<=≤<=*r*<=≤<=*n*.
|
[
"5\n1 2 4 3 2\n2 3 3 12 1\n",
"10\n13 2 7 11 8 4 9 8 5 1\n5 7 18 9 2 3 0 11 8 6\n"
] |
[
"22",
"46"
] |
Bitwise OR of two non-negative integers *a* and *b* is the number *c* = *a* *OR* *b*, such that each of its digits in binary notation is 1 if and only if at least one of *a* or *b* have 1 in the corresponding position in binary notation.
In the first sample, one of the optimal answers is *l* = 2 and *r* = 4, because *f*(*a*, 2, 4) + *f*(*b*, 2, 4) = (2 *OR* 4 *OR* 3) + (3 *OR* 3 *OR* 12) = 7 + 15 = 22. Other ways to get maximum value is to choose *l* = 1 and *r* = 4, *l* = 1 and *r* = 5, *l* = 2 and *r* = 4, *l* = 2 and *r* = 5, *l* = 3 and *r* = 4, or *l* = 3 and *r* = 5.
In the second sample, the maximum value is obtained for *l* = 1 and *r* = 9.
| 500
|
[
{
"input": "5\n1 2 4 3 2\n2 3 3 12 1",
"output": "22"
},
{
"input": "10\n13 2 7 11 8 4 9 8 5 1\n5 7 18 9 2 3 0 11 8 6",
"output": "46"
},
{
"input": "25\n12 30 38 109 81 124 80 33 38 48 29 78 96 48 96 27 80 77 102 65 80 113 31 118 35\n25 64 95 13 12 6 111 80 85 16 61 119 23 65 73 65 20 95 124 18 28 79 125 106 116",
"output": "254"
},
{
"input": "20\n64 64 64 64 64 64 64 64 64 64 64 64 64 64 64 64 64 64 64 64\n64 64 64 64 64 64 64 64 64 64 64 64 64 64 64 64 64 64 64 64",
"output": "128"
},
{
"input": "1\n1000000000\n1000000000",
"output": "2000000000"
},
{
"input": "1\n0\n0",
"output": "0"
},
{
"input": "2\n7 16\n16 7",
"output": "46"
},
{
"input": "4\n6 0 0 0\n0 0 0 1",
"output": "7"
},
{
"input": "8\n1 2 4 8 16 32 64 128\n1 2 4 8 16 32 64 128",
"output": "510"
},
{
"input": "1\n2\n3",
"output": "5"
},
{
"input": "1\n4\n3",
"output": "7"
},
{
"input": "1\n1\n1",
"output": "2"
}
] | 1,660,206,010
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 27
| 187
| 0
|
n = int(input())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
maxx = 0
for i in range(n):
res1 = 0
res2 = 0
for j in range(i, n):
res1 |= a[j]
res2 |= b[j]
maxx = max(maxx, res1 + res2)
print(maxx)
|
Title: Interview
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Blake is a CEO of a large company called "Blake Technologies". He loves his company very much and he thinks that his company should be the best. That is why every candidate needs to pass through the interview that consists of the following problem.
We define function *f*(*x*,<=*l*,<=*r*) as a bitwise OR of integers *x**l*,<=*x**l*<=+<=1,<=...,<=*x**r*, where *x**i* is the *i*-th element of the array *x*. You are given two arrays *a* and *b* of length *n*. You need to determine the maximum value of sum *f*(*a*,<=*l*,<=*r*)<=+<=*f*(*b*,<=*l*,<=*r*) among all possible 1<=≤<=*l*<=≤<=*r*<=≤<=*n*.
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the length of the arrays.
The second line contains *n* integers *a**i* (0<=≤<=*a**i*<=≤<=109).
The third line contains *n* integers *b**i* (0<=≤<=*b**i*<=≤<=109).
Output Specification:
Print a single integer — the maximum value of sum *f*(*a*,<=*l*,<=*r*)<=+<=*f*(*b*,<=*l*,<=*r*) among all possible 1<=≤<=*l*<=≤<=*r*<=≤<=*n*.
Demo Input:
['5\n1 2 4 3 2\n2 3 3 12 1\n', '10\n13 2 7 11 8 4 9 8 5 1\n5 7 18 9 2 3 0 11 8 6\n']
Demo Output:
['22', '46']
Note:
Bitwise OR of two non-negative integers *a* and *b* is the number *c* = *a* *OR* *b*, such that each of its digits in binary notation is 1 if and only if at least one of *a* or *b* have 1 in the corresponding position in binary notation.
In the first sample, one of the optimal answers is *l* = 2 and *r* = 4, because *f*(*a*, 2, 4) + *f*(*b*, 2, 4) = (2 *OR* 4 *OR* 3) + (3 *OR* 3 *OR* 12) = 7 + 15 = 22. Other ways to get maximum value is to choose *l* = 1 and *r* = 4, *l* = 1 and *r* = 5, *l* = 2 and *r* = 4, *l* = 2 and *r* = 5, *l* = 3 and *r* = 4, or *l* = 3 and *r* = 5.
In the second sample, the maximum value is obtained for *l* = 1 and *r* = 9.
|
```python
n = int(input())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
maxx = 0
for i in range(n):
res1 = 0
res2 = 0
for j in range(i, n):
res1 |= a[j]
res2 |= b[j]
maxx = max(maxx, res1 + res2)
print(maxx)
```
| 3
|
|
903
|
C
|
Boxes Packing
|
PROGRAMMING
| 1,200
|
[
"greedy"
] | null | null |
Mishka has got *n* empty boxes. For every *i* (1<=≤<=*i*<=≤<=*n*), *i*-th box is a cube with side length *a**i*.
Mishka can put a box *i* into another box *j* if the following conditions are met:
- *i*-th box is not put into another box; - *j*-th box doesn't contain any other boxes; - box *i* is smaller than box *j* (*a**i*<=<<=*a**j*).
Mishka can put boxes into each other an arbitrary number of times. He wants to minimize the number of visible boxes. A box is called visible iff it is not put into some another box.
Help Mishka to determine the minimum possible number of visible boxes!
|
The first line contains one integer *n* (1<=≤<=*n*<=≤<=5000) — the number of boxes Mishka has got.
The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=109), where *a**i* is the side length of *i*-th box.
|
Print the minimum possible number of visible boxes.
|
[
"3\n1 2 3\n",
"4\n4 2 4 3\n"
] |
[
"1\n",
"2\n"
] |
In the first example it is possible to put box 1 into box 2, and 2 into 3.
In the second example Mishka can put box 2 into box 3, and box 4 into box 1.
| 0
|
[
{
"input": "3\n1 2 3",
"output": "1"
},
{
"input": "4\n4 2 4 3",
"output": "2"
},
{
"input": "10\n58 58 58 58 58 58 58 58 58 58",
"output": "10"
},
{
"input": "10\n86 89 89 86 86 89 86 86 89 89",
"output": "5"
},
{
"input": "100\n981 288 186 186 292 876 341 288 981 360 783 907 292 186 341 292 360 876 360 360 981 398 783 288 292 398 876 981 398 907 783 360 288 981 907 186 360 288 186 981 186 288 907 876 288 907 876 360 341 292 907 783 907 783 292 981 907 292 876 398 783 876 398 341 876 186 288 186 981 341 398 360 907 981 341 186 292 981 292 398 876 783 292 186 360 292 288 292 876 398 288 292 341 288 398 360 360 292 981 360",
"output": "14"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "1\n9",
"output": "1"
},
{
"input": "1\n5",
"output": "1"
},
{
"input": "1\n2",
"output": "1"
},
{
"input": "1\n131",
"output": "1"
},
{
"input": "9\n1 1 1 1 1 1 1 1 1",
"output": "9"
},
{
"input": "11\n1 1 1 1 1 1 1 1 1 1 1",
"output": "11"
},
{
"input": "8\n1 2 3 4 5 6 7 8",
"output": "1"
},
{
"input": "8\n1 1 1 1 1 1 1 1",
"output": "8"
},
{
"input": "5\n1 1 1 1 1",
"output": "5"
}
] | 1,665,035,030
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 24
| 62
| 1,843,200
|
'''
probably I am not getting the box packing rules
'''
if __name__ == "__main__":
n=int(input())
nums=[int(i) for i in input().split()]
freq_hashmap={}
for num in nums:
freq_hashmap[num]=freq_hashmap.get(num,0)+1
print(max(freq_hashmap.values()))
|
Title: Boxes Packing
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mishka has got *n* empty boxes. For every *i* (1<=≤<=*i*<=≤<=*n*), *i*-th box is a cube with side length *a**i*.
Mishka can put a box *i* into another box *j* if the following conditions are met:
- *i*-th box is not put into another box; - *j*-th box doesn't contain any other boxes; - box *i* is smaller than box *j* (*a**i*<=<<=*a**j*).
Mishka can put boxes into each other an arbitrary number of times. He wants to minimize the number of visible boxes. A box is called visible iff it is not put into some another box.
Help Mishka to determine the minimum possible number of visible boxes!
Input Specification:
The first line contains one integer *n* (1<=≤<=*n*<=≤<=5000) — the number of boxes Mishka has got.
The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=109), where *a**i* is the side length of *i*-th box.
Output Specification:
Print the minimum possible number of visible boxes.
Demo Input:
['3\n1 2 3\n', '4\n4 2 4 3\n']
Demo Output:
['1\n', '2\n']
Note:
In the first example it is possible to put box 1 into box 2, and 2 into 3.
In the second example Mishka can put box 2 into box 3, and box 4 into box 1.
|
```python
'''
probably I am not getting the box packing rules
'''
if __name__ == "__main__":
n=int(input())
nums=[int(i) for i in input().split()]
freq_hashmap={}
for num in nums:
freq_hashmap[num]=freq_hashmap.get(num,0)+1
print(max(freq_hashmap.values()))
```
| 3
|
|
386
|
A
|
Second-Price Auction
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
In this problem we consider a special type of an auction, which is called the second-price auction. As in regular auction *n* bidders place a bid which is price a bidder ready to pay. The auction is closed, that is, each bidder secretly informs the organizer of the auction price he is willing to pay. After that, the auction winner is the participant who offered the highest price. However, he pay not the price he offers, but the highest price among the offers of other participants (hence the name: the second-price auction).
Write a program that reads prices offered by bidders and finds the winner and the price he will pay. Consider that all of the offered prices are different.
|
The first line of the input contains *n* (2<=≤<=*n*<=≤<=1000) — number of bidders. The second line contains *n* distinct integer numbers *p*1,<=*p*2,<=... *p**n*, separated by single spaces (1<=≤<=*p**i*<=≤<=10000), where *p**i* stands for the price offered by the *i*-th bidder.
|
The single output line should contain two integers: index of the winner and the price he will pay. Indices are 1-based.
|
[
"2\n5 7\n",
"3\n10 2 8\n",
"6\n3 8 2 9 4 14\n"
] |
[
"2 5\n",
"1 8\n",
"6 9\n"
] |
none
| 500
|
[
{
"input": "2\n5 7",
"output": "2 5"
},
{
"input": "3\n10 2 8",
"output": "1 8"
},
{
"input": "6\n3 8 2 9 4 14",
"output": "6 9"
},
{
"input": "4\n4707 7586 4221 5842",
"output": "2 5842"
},
{
"input": "5\n3304 4227 4869 6937 6002",
"output": "4 6002"
},
{
"input": "6\n5083 3289 7708 5362 9031 7458",
"output": "5 7708"
},
{
"input": "7\n9038 6222 3392 1706 3778 1807 2657",
"output": "1 6222"
},
{
"input": "8\n7062 2194 4481 3864 7470 1814 8091 733",
"output": "7 7470"
},
{
"input": "9\n2678 5659 9199 2628 7906 7496 4524 2663 3408",
"output": "3 7906"
},
{
"input": "2\n3458 1504",
"output": "1 1504"
},
{
"input": "50\n9237 3904 407 9052 6657 9229 9752 3888 7732 2512 4614 1055 2355 7108 6506 6849 2529 8862 159 8630 7906 7941 960 8470 333 8659 54 9475 3163 5625 6393 6814 2656 3388 169 7918 4881 8468 9983 6281 6340 280 5108 2996 101 7617 3313 8172 326 1991",
"output": "39 9752"
},
{
"input": "100\n2515 3324 7975 6171 4240 1217 4829 5203 8603 6900 3031 4699 4732 6070 4221 3228 6497 7359 9130 4346 4619 1109 3945 5442 3271 16 9711 2045 6410 2301 3406 8125 6003 1892 1260 9661 3940 6692 4708 7027 4930 6925 1979 5361 4263 3144 867 8639 6230 5562 9714 3676 4231 3347 4733 4920 4881 3431 1059 7313 8912 3038 9308 72 9583 7009 3034 7425 2398 6141 3245 2495 2933 6710 8289 9806 1226 8393 7349 6462 1618 9613 3546 6012 2964 9995 1578 210 2123 4874 1252 8625 348 8020 803 7244 9080 5088 706 2602",
"output": "86 9806"
},
{
"input": "2\n2 1",
"output": "1 1"
},
{
"input": "2\n1 2",
"output": "2 1"
},
{
"input": "3\n10 20 30",
"output": "3 20"
},
{
"input": "3\n10 30 20",
"output": "2 20"
},
{
"input": "3\n20 10 30",
"output": "3 20"
},
{
"input": "3\n20 30 10",
"output": "2 20"
},
{
"input": "3\n30 10 20",
"output": "1 20"
},
{
"input": "3\n30 20 10",
"output": "1 20"
},
{
"input": "2\n1 10000",
"output": "2 1"
},
{
"input": "2\n10000 999",
"output": "1 999"
},
{
"input": "3\n3 4 1",
"output": "2 3"
},
{
"input": "6\n1 2 24 6 5 7",
"output": "3 7"
}
] | 1,635,495,531
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 42
| 61
| 4,505,600
|
p = int(input())
str_price = str(input())
list_price = list(map(lambda x: int(x), str_price.split()))
winner = list_price.index(max(list_price)) + 1
val = sorted(list_price)[-2]
print(winner, val)
|
Title: Second-Price Auction
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In this problem we consider a special type of an auction, which is called the second-price auction. As in regular auction *n* bidders place a bid which is price a bidder ready to pay. The auction is closed, that is, each bidder secretly informs the organizer of the auction price he is willing to pay. After that, the auction winner is the participant who offered the highest price. However, he pay not the price he offers, but the highest price among the offers of other participants (hence the name: the second-price auction).
Write a program that reads prices offered by bidders and finds the winner and the price he will pay. Consider that all of the offered prices are different.
Input Specification:
The first line of the input contains *n* (2<=≤<=*n*<=≤<=1000) — number of bidders. The second line contains *n* distinct integer numbers *p*1,<=*p*2,<=... *p**n*, separated by single spaces (1<=≤<=*p**i*<=≤<=10000), where *p**i* stands for the price offered by the *i*-th bidder.
Output Specification:
The single output line should contain two integers: index of the winner and the price he will pay. Indices are 1-based.
Demo Input:
['2\n5 7\n', '3\n10 2 8\n', '6\n3 8 2 9 4 14\n']
Demo Output:
['2 5\n', '1 8\n', '6 9\n']
Note:
none
|
```python
p = int(input())
str_price = str(input())
list_price = list(map(lambda x: int(x), str_price.split()))
winner = list_price.index(max(list_price)) + 1
val = sorted(list_price)[-2]
print(winner, val)
```
| 3
|
|
801
|
B
|
Valued Keys
|
PROGRAMMING
| 900
|
[
"constructive algorithms",
"greedy",
"strings"
] | null | null |
You found a mysterious function *f*. The function takes two strings *s*1 and *s*2. These strings must consist only of lowercase English letters, and must be the same length.
The output of the function *f* is another string of the same length. The *i*-th character of the output is equal to the minimum of the *i*-th character of *s*1 and the *i*-th character of *s*2.
For example, *f*("ab", "ba") = "aa", and *f*("nzwzl", "zizez") = "niwel".
You found two strings *x* and *y* of the same length and consisting of only lowercase English letters. Find any string *z* such that *f*(*x*,<=*z*)<==<=*y*, or print -1 if no such string *z* exists.
|
The first line of input contains the string *x*.
The second line of input contains the string *y*.
Both *x* and *y* consist only of lowercase English letters, *x* and *y* have same length and this length is between 1 and 100.
|
If there is no string *z* such that *f*(*x*,<=*z*)<==<=*y*, print -1.
Otherwise, print a string *z* such that *f*(*x*,<=*z*)<==<=*y*. If there are multiple possible answers, print any of them. The string *z* should be the same length as *x* and *y* and consist only of lowercase English letters.
|
[
"ab\naa\n",
"nzwzl\nniwel\n",
"ab\nba\n"
] |
[
"ba\n",
"xiyez\n",
"-1\n"
] |
The first case is from the statement.
Another solution for the second case is "zizez"
There is no solution for the third case. That is, there is no *z* such that *f*("ab", *z*) = "ba".
| 1,000
|
[
{
"input": "ab\naa",
"output": "ba"
},
{
"input": "nzwzl\nniwel",
"output": "xiyez"
},
{
"input": "ab\nba",
"output": "-1"
},
{
"input": "r\nl",
"output": "l"
},
{
"input": "d\ny",
"output": "-1"
},
{
"input": "yvowz\ncajav",
"output": "cajav"
},
{
"input": "lwzjp\ninjit",
"output": "-1"
},
{
"input": "epqnlxmiicdidyscjaxqznwur\neodnlemiicdedmkcgavqbnqmm",
"output": "eodnlemiicdedmkcgavqbnqmm"
},
{
"input": "qqdabbsxiibnnjgsgxllfvdqj\nuxmypqtwfdezewdxfgplannrs",
"output": "-1"
},
{
"input": "aanerbaqslfmqmuciqbxyznkevukvznpkmxlcorpmrenwxhzfgbmlfpxtkqpxdrmcqcmbf\naanebbaqkgfiimcciqbaoznkeqqkrgapdillccrfeienwbcvfgbmlfbimkqchcrmclcmbf",
"output": "aanebbaqkgfiimcciqbaoznkeqqkrgapdillccrfeienwbcvfgbmlfbimkqchcrmclcmbf"
},
{
"input": "mbyrkhjctrcrayisflptgfudwgrtegidhqicsjqafvdloritbjhciyxuwavxknezwwudnk\nvvixsutlbdewqoabqhpuerfkzrddcqptfwmxdlxwbvsaqfjoxztlddvwgflcteqbwaiaen",
"output": "-1"
},
{
"input": "eufycwztywhbjrpqobvknwfqmnboqcfdiahkagykeibbsqpljcghhmsgfmswwsanzyiwtvuirwmppfivtekaywkzskyydfvkjgxb\necfwavookadbcilfobojnweqinbcpcfdiahkabwkeibbacpljcghhksgfajgmianfnivmhfifogpffiheegayfkxkkcmdfvihgdb",
"output": "ecfwavookadbcilfobojnweqinbcpcfdiahkabwkeibbacpljcghhksgfajgmianfnivmhfifogpffiheegayfkxkkcmdfvihgdb"
},
{
"input": "qvpltcffyeghtbdhjyhfteojezyzziardduzrbwuxmzzkkoehfnxecafizxglboauhynfbawlfxenmykquyhrxswhjuovvogntok\nchvkcvzxptbcepdjfezcpuvtehewbnvqeoezlcnzhpfwujbmhafoeqmjhtwisnobauinkzyigrvahpuetkgpdjfgbzficsmuqnym",
"output": "-1"
},
{
"input": "nmuwjdihouqrnsuahimssnrbxdpwvxiyqtenahtrlshjkmnfuttnpqhgcagoptinnaptxaccptparldzrhpgbyrzedghudtsswxi\nnilhbdghosqnbebafimconrbvdodjsipqmekahhrllhjkemeketapfhgcagopfidnahtlaccpfpafedqicpcbvfgedghudhddwib",
"output": "nilhbdghosqnbebafimconrbvdodjsipqmekahhrllhjkemeketapfhgcagopfidnahtlaccpfpafedqicpcbvfgedghudhddwib"
},
{
"input": "dyxgwupoauwqtcfoyfjdotzirwztdfrueqiypxoqvkmhiehdppwtdoxrbfvtairdbuvlqohjflznggjpifhwjrshcrfbjtklpykx\ngzqlnoizhxolnditjdhlhptjsbczehicudoybzilwnshmywozwnwuipcgirgzldtvtowdsokfeafggwserzdazkxyddjttiopeew",
"output": "-1"
},
{
"input": "hbgwuqzougqzlxemvyjpeizjfwhgugrfnhbrlxkmkdalikfyunppwgdzmalbwewybnjzqsohwhjkdcyhhzmysflambvhpsjilsyv\nfbdjdqjojdafarakvcjpeipjfehgfgrfehbolxkmkdagikflunnpvadocalbkedibhbflmohnhjkdcthhaigsfjaibqhbcjelirv",
"output": "fbdjdqjojdafarakvcjpeipjfehgfgrfehbolxkmkdagikflunnpvadocalbkedibhbflmohnhjkdcthhaigsfjaibqhbcjelirv"
},
{
"input": "xnjjhjfuhgyxqhpzmvgbaohqarugdoaczcfecofltwemieyxolswkcwhlfagfrgmoiqrgftokbqwtxgxzweozzlikrvafiabivlk\npjfosalbsitcnqiazhmepfifjxvmazvdgffcnozmnqubhonwjldmpdsjagmamniylzjdbklcyrzivjyzgnogahobpkwpwpvraqns",
"output": "-1"
},
{
"input": "zrvzedssbsrfldqvjpgmsefrmsatspzoitwvymahiptphiystjlsauzquzqqbmljobdhijcpdvatorwmyojqgnezvzlgjibxepcf\npesoedmqbmffldqsjggmhefkadaesijointrkmahapaahiysfjdiaupqujngbjhjobdhiecadeatgjvelojjgnepvajgeibfepaf",
"output": "pesoedmqbmffldqsjggmhefkadaesijointrkmahapaahiysfjdiaupqujngbjhjobdhiecadeatgjvelojjgnepvajgeibfepaf"
},
{
"input": "pdvkuwyzntzfqpblzmbynknyhlnqbxijuqaincviugxohcsrofozrrsategwkbwxcvkyzxhurokefpbdnmcfogfhsojayysqbrow\nbvxruombdrywlcjkrltyayaazwpauuhbtgwfzdrmfwwucgffucwelzvpsdgtapogchblzahsrfymjlaghkbmbssghrpxalkslcvp",
"output": "-1"
},
{
"input": "tgharsjyihroiiahwgbjezlxvlterxivdhtzjcqegzmtigqmrehvhiyjeywegxaseoyoacouijudbiruoghgxvxadwzgdxtnxlds\ntghaksjsdhkoiiahegbjexlfrctercipdhmvjbgegxdtggqdpbhvhiseehhegnaseoooacnsijubbirjnghgsvpadhaadrtimfdp",
"output": "tghaksjsdhkoiiahegbjexlfrctercipdhmvjbgegxdtggqdpbhvhiseehhegnaseoooacnsijubbirjnghgsvpadhaadrtimfdp"
},
{
"input": "jsinejpfwhzloulxndzvzftgogfdagrsscxmatldssqsgaknnbkcvhptebjjpkjhrjegrotzwcdosezkedzxeoyibmyzunkguoqj\nkfmvybobocdpipiripysioruqvloopvbggpjksgmwzyqwyxnesmvhsawnbbmntulspvsysfkjqwpvoelliopbaukyagedextzoej",
"output": "-1"
},
{
"input": "nttdcfceptruiomtmwzestrfchnqpgqeztpcvthzelfyggjgqadylzubpvbrlgndrcsursczpxlnoyoadxezncqalupfzmjeqihe\nkttdcfceohrjiaahmoldanpfchnfpgheqpdahqhxecfpbgigqadrkjubjfbrlgndbcgcgmcjpeleinaadretncqaiqpfkmjeqihe",
"output": "kttdcfceohrjiaahmoldanpfchnfpgheqpdahqhxecfpbgigqadrkjubjfbrlgndbcgcgmcjpeleinaadretncqaiqpfkmjeqihe"
},
{
"input": "diuopwglduasnaxgduwslbzoyayoypzznqspljcyqehweydhlwifcvnjmaowuvyqfwynjghecqvxdvuquuwpvwrjljozocaxnktv\ntrdydprdzmjhgbhzytelrfjpgsebijicsigmwhynmcyjtqrvojcndodchzxfcvyqjxqzwibccdvsjqhsnectdjyrrhzkeamukang",
"output": "-1"
},
{
"input": "ftfr\nftfr",
"output": "ftfr"
},
{
"input": "ftr\nftr",
"output": "ftr"
},
{
"input": "shftr\nshftr",
"output": "shftr"
},
{
"input": "vkvkkv\nvkvkkv",
"output": "vkvkkv"
},
{
"input": "ftrd\nftrd",
"output": "ftrd"
},
{
"input": "fztr\nfztr",
"output": "fztr"
},
{
"input": "frtr\nfrtr",
"output": "frtr"
},
{
"input": "shdftr\nshdftr",
"output": "shdftr"
},
{
"input": "zzz\nzzz",
"output": "zzz"
},
{
"input": "shtr\nshtr",
"output": "shtr"
},
{
"input": "aaaaa\nzzzzz",
"output": "-1"
},
{
"input": "efr\nefr",
"output": "efr"
}
] | 1,648,380,611
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 35
| 46
| 0
|
a,b=list(input()),list(input());t=''
for x,y in zip(a,b):
if x>=y:t+=y
else:t=-1;break
print(t)
|
Title: Valued Keys
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You found a mysterious function *f*. The function takes two strings *s*1 and *s*2. These strings must consist only of lowercase English letters, and must be the same length.
The output of the function *f* is another string of the same length. The *i*-th character of the output is equal to the minimum of the *i*-th character of *s*1 and the *i*-th character of *s*2.
For example, *f*("ab", "ba") = "aa", and *f*("nzwzl", "zizez") = "niwel".
You found two strings *x* and *y* of the same length and consisting of only lowercase English letters. Find any string *z* such that *f*(*x*,<=*z*)<==<=*y*, or print -1 if no such string *z* exists.
Input Specification:
The first line of input contains the string *x*.
The second line of input contains the string *y*.
Both *x* and *y* consist only of lowercase English letters, *x* and *y* have same length and this length is between 1 and 100.
Output Specification:
If there is no string *z* such that *f*(*x*,<=*z*)<==<=*y*, print -1.
Otherwise, print a string *z* such that *f*(*x*,<=*z*)<==<=*y*. If there are multiple possible answers, print any of them. The string *z* should be the same length as *x* and *y* and consist only of lowercase English letters.
Demo Input:
['ab\naa\n', 'nzwzl\nniwel\n', 'ab\nba\n']
Demo Output:
['ba\n', 'xiyez\n', '-1\n']
Note:
The first case is from the statement.
Another solution for the second case is "zizez"
There is no solution for the third case. That is, there is no *z* such that *f*("ab", *z*) = "ba".
|
```python
a,b=list(input()),list(input());t=''
for x,y in zip(a,b):
if x>=y:t+=y
else:t=-1;break
print(t)
```
| 3
|
|
965
|
C
|
Greedy Arkady
|
PROGRAMMING
| 2,000
|
[
"math"
] | null | null |
$k$ people want to split $n$ candies between them. Each candy should be given to exactly one of them or be thrown away.
The people are numbered from $1$ to $k$, and Arkady is the first of them. To split the candies, Arkady will choose an integer $x$ and then give the first $x$ candies to himself, the next $x$ candies to the second person, the next $x$ candies to the third person and so on in a cycle. The leftover (the remainder that is not divisible by $x$) will be thrown away.
Arkady can't choose $x$ greater than $M$ as it is considered greedy. Also, he can't choose such a small $x$ that some person will receive candies more than $D$ times, as it is considered a slow splitting.
Please find what is the maximum number of candies Arkady can receive by choosing some valid $x$.
|
The only line contains four integers $n$, $k$, $M$ and $D$ ($2 \le n \le 10^{18}$, $2 \le k \le n$, $1 \le M \le n$, $1 \le D \le \min{(n, 1000)}$, $M \cdot D \cdot k \ge n$) — the number of candies, the number of people, the maximum number of candies given to a person at once, the maximum number of times a person can receive candies.
|
Print a single integer — the maximum possible number of candies Arkady can give to himself.
Note that it is always possible to choose some valid $x$.
|
[
"20 4 5 2\n",
"30 9 4 1\n"
] |
[
"8\n",
"4\n"
] |
In the first example Arkady should choose $x = 4$. He will give $4$ candies to himself, $4$ candies to the second person, $4$ candies to the third person, then $4$ candies to the fourth person and then again $4$ candies to himself. No person is given candies more than $2$ times, and Arkady receives $8$ candies in total.
Note that if Arkady chooses $x = 5$, he will receive only $5$ candies, and if he chooses $x = 3$, he will receive only $3 + 3 = 6$ candies as well as the second person, the third and the fourth persons will receive $3$ candies, and $2$ candies will be thrown away. He can't choose $x = 1$ nor $x = 2$ because in these cases he will receive candies more than $2$ times.
In the second example Arkady has to choose $x = 4$, because any smaller value leads to him receiving candies more than $1$ time.
| 1,500
|
[
{
"input": "20 4 5 2",
"output": "8"
},
{
"input": "30 9 4 1",
"output": "4"
},
{
"input": "2 2 1 1",
"output": "1"
},
{
"input": "42 20 5 29",
"output": "5"
},
{
"input": "1000000000000000000 135 1000000000000000 1000",
"output": "8325624421831635"
},
{
"input": "100 33 100 100",
"output": "100"
},
{
"input": "1000000000 1000000000 1000000000 1000",
"output": "1000000000"
},
{
"input": "1000000000 32428 1000000000 1000",
"output": "1000000000"
},
{
"input": "1000000000 324934 1000 1000",
"output": "4000"
},
{
"input": "1000000000000000000 32400093004 10000000 1000",
"output": "40000000"
},
{
"input": "885 2 160 842",
"output": "504"
},
{
"input": "216 137 202 208",
"output": "202"
},
{
"input": "72 66 28 9",
"output": "28"
},
{
"input": "294 4 13 8",
"output": "80"
},
{
"input": "9 2 2 3",
"output": "4"
},
{
"input": "31 3 2 8",
"output": "10"
},
{
"input": "104 2 5 11",
"output": "50"
},
{
"input": "1000000000000000000 1000000000000000000 1000 1000",
"output": "1000"
},
{
"input": "1000000000000000000 100000000000000000 1 1000",
"output": "10"
},
{
"input": "23925738098196565 23925738098196565 23925738098196565 1000",
"output": "23925738098196565"
},
{
"input": "576460752303423488 576460752303423488 351082447248993993 1000",
"output": "351082447248993993"
},
{
"input": "962768465676381898 72057594037927936 586039918340257175 256",
"output": "586039918340257175"
},
{
"input": "1000000000000000000 1000000000000000000 10 1000",
"output": "10"
},
{
"input": "23925738098196565 23925738098196565 1 1000",
"output": "1"
},
{
"input": "1000000000000000000 1000000000000000000 1 1000",
"output": "1"
}
] | 1,524,680,013
| 2,313
|
Python 3
|
OK
|
TESTS
| 25
| 93
| 7,065,600
|
n, k, m, d = map(int, input().split())
res = 0
for a in range(1, d + 1):
x = n // (a * k - k + 1)
x = min(x, m)
res = max(res, x * a)
print(res)
|
Title: Greedy Arkady
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
$k$ people want to split $n$ candies between them. Each candy should be given to exactly one of them or be thrown away.
The people are numbered from $1$ to $k$, and Arkady is the first of them. To split the candies, Arkady will choose an integer $x$ and then give the first $x$ candies to himself, the next $x$ candies to the second person, the next $x$ candies to the third person and so on in a cycle. The leftover (the remainder that is not divisible by $x$) will be thrown away.
Arkady can't choose $x$ greater than $M$ as it is considered greedy. Also, he can't choose such a small $x$ that some person will receive candies more than $D$ times, as it is considered a slow splitting.
Please find what is the maximum number of candies Arkady can receive by choosing some valid $x$.
Input Specification:
The only line contains four integers $n$, $k$, $M$ and $D$ ($2 \le n \le 10^{18}$, $2 \le k \le n$, $1 \le M \le n$, $1 \le D \le \min{(n, 1000)}$, $M \cdot D \cdot k \ge n$) — the number of candies, the number of people, the maximum number of candies given to a person at once, the maximum number of times a person can receive candies.
Output Specification:
Print a single integer — the maximum possible number of candies Arkady can give to himself.
Note that it is always possible to choose some valid $x$.
Demo Input:
['20 4 5 2\n', '30 9 4 1\n']
Demo Output:
['8\n', '4\n']
Note:
In the first example Arkady should choose $x = 4$. He will give $4$ candies to himself, $4$ candies to the second person, $4$ candies to the third person, then $4$ candies to the fourth person and then again $4$ candies to himself. No person is given candies more than $2$ times, and Arkady receives $8$ candies in total.
Note that if Arkady chooses $x = 5$, he will receive only $5$ candies, and if he chooses $x = 3$, he will receive only $3 + 3 = 6$ candies as well as the second person, the third and the fourth persons will receive $3$ candies, and $2$ candies will be thrown away. He can't choose $x = 1$ nor $x = 2$ because in these cases he will receive candies more than $2$ times.
In the second example Arkady has to choose $x = 4$, because any smaller value leads to him receiving candies more than $1$ time.
|
```python
n, k, m, d = map(int, input().split())
res = 0
for a in range(1, d + 1):
x = n // (a * k - k + 1)
x = min(x, m)
res = max(res, x * a)
print(res)
```
| 3
|
|
281
|
A
|
Word Capitalization
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] | null | null |
Capitalization is writing a word with its first letter as a capital letter. Your task is to capitalize the given word.
Note, that during capitalization all the letters except the first one remains unchanged.
|
A single line contains a non-empty word. This word consists of lowercase and uppercase English letters. The length of the word will not exceed 103.
|
Output the given word after capitalization.
|
[
"ApPLe\n",
"konjac\n"
] |
[
"ApPLe\n",
"Konjac\n"
] |
none
| 500
|
[
{
"input": "ApPLe",
"output": "ApPLe"
},
{
"input": "konjac",
"output": "Konjac"
},
{
"input": "a",
"output": "A"
},
{
"input": "A",
"output": "A"
},
{
"input": "z",
"output": "Z"
},
{
"input": "ABACABA",
"output": "ABACABA"
},
{
"input": "xYaPxPxHxGePfGtQySlNrLxSjDtNnTaRaEpAhPaQpWnDzMqGgRgEwJxGiBdZnMtHxFbObCaGiCeZkUqIgBhHtNvAqAlHpMnQhNeQbMyZrCdElVwHtKrPpJjIaHuIlYwHaRkAkUpPlOhNlBtXwDsKzPyHrPiUwNlXtTaPuMwTqYtJySgFoXvLiHbQwMjSvXsQfKhVlOxGdQkWjBhEyQvBjPoFkThNeRhTuIzFjInJtEfPjOlOsJpJuLgLzFnZmKvFgFrNsOnVqFcNiMfCqTpKnVyLwNqFiTySpWeTdFnWuTwDkRjVxNyQvTrOoEiExYiFaIrLoFmJfZcDkHuWjYfCeEqCvEsZiWnJaEmFbMjDvYwEeJeGcKbVbChGsIzNlExHzHiTlHcSaKxLuZxX",
"output": "XYaPxPxHxGePfGtQySlNrLxSjDtNnTaRaEpAhPaQpWnDzMqGgRgEwJxGiBdZnMtHxFbObCaGiCeZkUqIgBhHtNvAqAlHpMnQhNeQbMyZrCdElVwHtKrPpJjIaHuIlYwHaRkAkUpPlOhNlBtXwDsKzPyHrPiUwNlXtTaPuMwTqYtJySgFoXvLiHbQwMjSvXsQfKhVlOxGdQkWjBhEyQvBjPoFkThNeRhTuIzFjInJtEfPjOlOsJpJuLgLzFnZmKvFgFrNsOnVqFcNiMfCqTpKnVyLwNqFiTySpWeTdFnWuTwDkRjVxNyQvTrOoEiExYiFaIrLoFmJfZcDkHuWjYfCeEqCvEsZiWnJaEmFbMjDvYwEeJeGcKbVbChGsIzNlExHzHiTlHcSaKxLuZxX"
},
{
"input": "rZhIcQlXpNcPgXrOjTiOlMoTgXgIhCfMwZfWoFzGhEkQlOoMjIuShPlZfWkNnMyQfYdUhVgQuSmYoElEtZpDyHtOxXgCpWbZqSbYnPqBcNqRtPgCnJnAyIvNsAhRbNeVlMwZyRyJnFgIsCnSbOdLvUyIeOzQvRpMoMoHfNhHwKvTcHuYnYySfPmAiNwAiWdZnWlLvGfBbRbRrCrBqIgIdWkWiBsNyYkKdNxZdGaToSsDnXpRaGrKxBpQsCzBdQgZzBkGeHgGxNrIyQlSzWsTmSnZwOcHqQpNcQvJlPvKaPiQaMaYsQjUeCqQdCjPgUbDmWiJmNiXgExLqOcCtSwSePnUxIuZfIfBeWbEiVbXnUsPwWyAiXyRbZgKwOqFfCtQuKxEmVeRlAkOeXkO",
"output": "RZhIcQlXpNcPgXrOjTiOlMoTgXgIhCfMwZfWoFzGhEkQlOoMjIuShPlZfWkNnMyQfYdUhVgQuSmYoElEtZpDyHtOxXgCpWbZqSbYnPqBcNqRtPgCnJnAyIvNsAhRbNeVlMwZyRyJnFgIsCnSbOdLvUyIeOzQvRpMoMoHfNhHwKvTcHuYnYySfPmAiNwAiWdZnWlLvGfBbRbRrCrBqIgIdWkWiBsNyYkKdNxZdGaToSsDnXpRaGrKxBpQsCzBdQgZzBkGeHgGxNrIyQlSzWsTmSnZwOcHqQpNcQvJlPvKaPiQaMaYsQjUeCqQdCjPgUbDmWiJmNiXgExLqOcCtSwSePnUxIuZfIfBeWbEiVbXnUsPwWyAiXyRbZgKwOqFfCtQuKxEmVeRlAkOeXkO"
},
{
"input": "hDgZlUmLhYbLkLcNcKeOwJwTePbOvLaRvNzQbSbLsPeHqLhUqWtUbNdQfQqFfXeJqJwWuOrFnDdZiPxIkDyVmHbHvXfIlFqSgAcSyWbOlSlRuPhWdEpEzEeLnXwCtWuVcHaUeRgCiYsIvOaIgDnFuDbRnMoCmPrZfLeFpSjQaTfHgZwZvAzDuSeNwSoWuJvLqKqAuUxFaCxFfRcEjEsJpOfCtDiVrBqNsNwPuGoRgPzRpLpYnNyQxKaNnDnYiJrCrVcHlOxPiPcDbEgKfLwBjLhKcNeMgJhJmOiJvPfOaPaEuGqWvRbErKrIpDkEoQnKwJnTlStLyNsHyOjZfKoIjXwUvRrWpSyYhRpQdLqGmErAiNcGqAqIrTeTiMuPmCrEkHdBrLyCxPtYpRqD",
"output": "HDgZlUmLhYbLkLcNcKeOwJwTePbOvLaRvNzQbSbLsPeHqLhUqWtUbNdQfQqFfXeJqJwWuOrFnDdZiPxIkDyVmHbHvXfIlFqSgAcSyWbOlSlRuPhWdEpEzEeLnXwCtWuVcHaUeRgCiYsIvOaIgDnFuDbRnMoCmPrZfLeFpSjQaTfHgZwZvAzDuSeNwSoWuJvLqKqAuUxFaCxFfRcEjEsJpOfCtDiVrBqNsNwPuGoRgPzRpLpYnNyQxKaNnDnYiJrCrVcHlOxPiPcDbEgKfLwBjLhKcNeMgJhJmOiJvPfOaPaEuGqWvRbErKrIpDkEoQnKwJnTlStLyNsHyOjZfKoIjXwUvRrWpSyYhRpQdLqGmErAiNcGqAqIrTeTiMuPmCrEkHdBrLyCxPtYpRqD"
},
{
"input": "qUdLgGrJeGmIzIeZrCjUtBpYfRvNdXdRpGsThIsEmJjTiMqEwRxBeBaSxEuWrNvExKePjPnXhPzBpWnHiDhTvZhBuIjDnZpTcEkCvRkAcTmMuXhGgErWgFyGyToOyVwYlCuQpTfJkVdWmFyBqQhJjYtXrBbFdHzDlGsFbHmHbFgXgFhIyDhZyEqEiEwNxSeByBwLiVeSnCxIdHbGjOjJrZeVkOzGeMmQrJkVyGhDtCzOlPeAzGrBlWwEnAdUfVaIjNrRyJjCnHkUvFuKuKeKbLzSbEmUcXtVkZzXzKlOrPgQiDmCcCvIyAdBwOeUuLbRmScNcWxIkOkJuIsBxTrIqXhDzLcYdVtPgZdZfAxTmUtByGiTsJkSySjXdJvEwNmSmNoWsChPdAzJrBoW",
"output": "QUdLgGrJeGmIzIeZrCjUtBpYfRvNdXdRpGsThIsEmJjTiMqEwRxBeBaSxEuWrNvExKePjPnXhPzBpWnHiDhTvZhBuIjDnZpTcEkCvRkAcTmMuXhGgErWgFyGyToOyVwYlCuQpTfJkVdWmFyBqQhJjYtXrBbFdHzDlGsFbHmHbFgXgFhIyDhZyEqEiEwNxSeByBwLiVeSnCxIdHbGjOjJrZeVkOzGeMmQrJkVyGhDtCzOlPeAzGrBlWwEnAdUfVaIjNrRyJjCnHkUvFuKuKeKbLzSbEmUcXtVkZzXzKlOrPgQiDmCcCvIyAdBwOeUuLbRmScNcWxIkOkJuIsBxTrIqXhDzLcYdVtPgZdZfAxTmUtByGiTsJkSySjXdJvEwNmSmNoWsChPdAzJrBoW"
},
{
"input": "kHbApGoBcLmIwUlXkVgUmWzYeLoDbGaOkWbIuXoRwMfKuOoMzAoXrBoTvYxGrMbRjDuRxAbGsTnErIiHnHoLeRnTbFiRfDdOkNlWiAcOsChLdLqFqXlDpDoDtPxXqAmSvYgPvOcCpOlWtOjYwFkGkHuCaHwZcFdOfHjBmIxTeSiHkWjXyFcCtOlSuJsZkDxUgPeZkJwMmNpErUlBcGuMlJwKkWnOzFeFiSiPsEvMmQiCsYeHlLuHoMgBjFoZkXlObDkSoQcVyReTmRsFzRhTuIvCeBqVsQdQyTyZjStGrTyDcEcAgTgMiIcVkLbZbGvWeHtXwEqWkXfTcPyHhHjYwIeVxLyVmHmMkUsGiHmNnQuMsXaFyPpVqNrBhOiWmNkBbQuHvQdOjPjKiZcL",
"output": "KHbApGoBcLmIwUlXkVgUmWzYeLoDbGaOkWbIuXoRwMfKuOoMzAoXrBoTvYxGrMbRjDuRxAbGsTnErIiHnHoLeRnTbFiRfDdOkNlWiAcOsChLdLqFqXlDpDoDtPxXqAmSvYgPvOcCpOlWtOjYwFkGkHuCaHwZcFdOfHjBmIxTeSiHkWjXyFcCtOlSuJsZkDxUgPeZkJwMmNpErUlBcGuMlJwKkWnOzFeFiSiPsEvMmQiCsYeHlLuHoMgBjFoZkXlObDkSoQcVyReTmRsFzRhTuIvCeBqVsQdQyTyZjStGrTyDcEcAgTgMiIcVkLbZbGvWeHtXwEqWkXfTcPyHhHjYwIeVxLyVmHmMkUsGiHmNnQuMsXaFyPpVqNrBhOiWmNkBbQuHvQdOjPjKiZcL"
},
{
"input": "aHmRbLgNuWkLxLnWvUbYwTeZeYiOlLhTuOvKfLnVmCiPcMkSgVrYjZiLuRjCiXhAnVzVcTlVeJdBvPdDfFvHkTuIhCdBjEsXbVmGcLrPfNvRdFsZkSdNpYsJeIhIcNqSoLkOjUlYlDmXsOxPbQtIoUxFjGnRtBhFaJvBeEzHsAtVoQbAfYjJqReBiKeUwRqYrUjPjBoHkOkPzDwEwUgTxQxAvKzUpMhKyOhPmEhYhItQwPeKsKaKlUhGuMcTtSwFtXfJsDsFlTtOjVvVfGtBtFlQyIcBaMsPaJlPqUcUvLmReZiFbXxVtRhTzJkLkAjVqTyVuFeKlTyQgUzMsXjOxQnVfTaWmThEnEoIhZeZdStBkKeLpAhJnFoJvQyGwDiStLjEwGfZwBuWsEfC",
"output": "AHmRbLgNuWkLxLnWvUbYwTeZeYiOlLhTuOvKfLnVmCiPcMkSgVrYjZiLuRjCiXhAnVzVcTlVeJdBvPdDfFvHkTuIhCdBjEsXbVmGcLrPfNvRdFsZkSdNpYsJeIhIcNqSoLkOjUlYlDmXsOxPbQtIoUxFjGnRtBhFaJvBeEzHsAtVoQbAfYjJqReBiKeUwRqYrUjPjBoHkOkPzDwEwUgTxQxAvKzUpMhKyOhPmEhYhItQwPeKsKaKlUhGuMcTtSwFtXfJsDsFlTtOjVvVfGtBtFlQyIcBaMsPaJlPqUcUvLmReZiFbXxVtRhTzJkLkAjVqTyVuFeKlTyQgUzMsXjOxQnVfTaWmThEnEoIhZeZdStBkKeLpAhJnFoJvQyGwDiStLjEwGfZwBuWsEfC"
},
{
"input": "sLlZkDiDmEdNaXuUuJwHqYvRtOdGfTiTpEpAoSqAbJaChOiCvHgSwZwEuPkMmXiLcKdXqSsEyViEbZpZsHeZpTuXoGcRmOiQfBfApPjDqSqElWeSeOhUyWjLyNoRuYeGfGwNqUsQoTyVvWeNgNdZfDxGwGfLsDjIdInSqDlMuNvFaHbScZkTlVwNcJpEjMaPaOtFgJjBjOcLlLmDnQrShIrJhOcUmPnZhTxNeClQsZaEaVaReLyQpLwEqJpUwYhLiRzCzKfOoFeTiXzPiNbOsZaZaLgCiNnMkBcFwGgAwPeNyTxJcCtBgXcToKlWaWcBaIvBpNxPeClQlWeQqRyEtAkJdBtSrFdDvAbUlKyLdCuTtXxFvRcKnYnWzVdYqDeCmOqPxUaFjQdTdCtN",
"output": "SLlZkDiDmEdNaXuUuJwHqYvRtOdGfTiTpEpAoSqAbJaChOiCvHgSwZwEuPkMmXiLcKdXqSsEyViEbZpZsHeZpTuXoGcRmOiQfBfApPjDqSqElWeSeOhUyWjLyNoRuYeGfGwNqUsQoTyVvWeNgNdZfDxGwGfLsDjIdInSqDlMuNvFaHbScZkTlVwNcJpEjMaPaOtFgJjBjOcLlLmDnQrShIrJhOcUmPnZhTxNeClQsZaEaVaReLyQpLwEqJpUwYhLiRzCzKfOoFeTiXzPiNbOsZaZaLgCiNnMkBcFwGgAwPeNyTxJcCtBgXcToKlWaWcBaIvBpNxPeClQlWeQqRyEtAkJdBtSrFdDvAbUlKyLdCuTtXxFvRcKnYnWzVdYqDeCmOqPxUaFjQdTdCtN"
},
{
"input": "iRuStKvVhJdJbQwRoIuLiVdTpKaOqKfYlYwAzIpPtUwUtMeKyCaOlXmVrKwWeImYmVuXdLkRlHwFxKqZbZtTzNgOzDbGqTfZnKmUzAcIjDcEmQgYyFbEfWzRpKvCkDmAqDiIiRcLvMxWaJqCgYqXgIcLdNaZlBnXtJyKaMnEaWfXfXwTbDnAiYnWqKbAtDpYdUbZrCzWgRnHzYxFgCdDbOkAgTqBuLqMeStHcDxGnVhSgMzVeTaZoTfLjMxQfRuPcFqVlRyYdHyOdJsDoCeWrUuJyIiAqHwHyVpEeEoMaJwAoUfPtBeJqGhMaHiBjKwAlXoZpUsDhHgMxBkVbLcEvNtJbGnPsUwAvXrAkTlXwYvEnOpNeWyIkRnEnTrIyAcLkRgMyYcKrGiDaAyE",
"output": "IRuStKvVhJdJbQwRoIuLiVdTpKaOqKfYlYwAzIpPtUwUtMeKyCaOlXmVrKwWeImYmVuXdLkRlHwFxKqZbZtTzNgOzDbGqTfZnKmUzAcIjDcEmQgYyFbEfWzRpKvCkDmAqDiIiRcLvMxWaJqCgYqXgIcLdNaZlBnXtJyKaMnEaWfXfXwTbDnAiYnWqKbAtDpYdUbZrCzWgRnHzYxFgCdDbOkAgTqBuLqMeStHcDxGnVhSgMzVeTaZoTfLjMxQfRuPcFqVlRyYdHyOdJsDoCeWrUuJyIiAqHwHyVpEeEoMaJwAoUfPtBeJqGhMaHiBjKwAlXoZpUsDhHgMxBkVbLcEvNtJbGnPsUwAvXrAkTlXwYvEnOpNeWyIkRnEnTrIyAcLkRgMyYcKrGiDaAyE"
},
{
"input": "cRtJkOxHzUbJcDdHzJtLbVmSoWuHoTkVrPqQaVmXeBrHxJbQfNrQbAaMrEhVdQnPxNyCjErKxPoEdWkVrBbDeNmEgBxYiBtWdAfHiLuSwIxJuHpSkAxPoYdNkGoLySsNhUmGoZhDzAfWhJdPlJzQkZbOnMtTkClIoCqOlIcJcMlGjUyOiEmHdYfIcPtTgQhLlLcPqQjAnQnUzHpCaQsCnYgQsBcJrQwBnWsIwFfSfGuYgTzQmShFpKqEeRlRkVfMuZbUsDoFoPrNuNwTtJqFkRiXxPvKyElDzLoUnIwAaBaOiNxMpEvPzSpGpFhMtGhGdJrFnZmNiMcUfMtBnDuUnXqDcMsNyGoLwLeNnLfRsIwRfBtXkHrFcPsLdXaAoYaDzYnZuQeVcZrElWmP",
"output": "CRtJkOxHzUbJcDdHzJtLbVmSoWuHoTkVrPqQaVmXeBrHxJbQfNrQbAaMrEhVdQnPxNyCjErKxPoEdWkVrBbDeNmEgBxYiBtWdAfHiLuSwIxJuHpSkAxPoYdNkGoLySsNhUmGoZhDzAfWhJdPlJzQkZbOnMtTkClIoCqOlIcJcMlGjUyOiEmHdYfIcPtTgQhLlLcPqQjAnQnUzHpCaQsCnYgQsBcJrQwBnWsIwFfSfGuYgTzQmShFpKqEeRlRkVfMuZbUsDoFoPrNuNwTtJqFkRiXxPvKyElDzLoUnIwAaBaOiNxMpEvPzSpGpFhMtGhGdJrFnZmNiMcUfMtBnDuUnXqDcMsNyGoLwLeNnLfRsIwRfBtXkHrFcPsLdXaAoYaDzYnZuQeVcZrElWmP"
},
{
"input": "wVaCsGxZrBbFnTbKsCoYlAvUkIpBaYpYmJkMlPwCaFvUkDxAiJgIqWsFqZlFvTtAnGzEwXbYiBdFfFxRiDoUkLmRfAwOlKeOlKgXdUnVqLkTuXtNdQpBpXtLvZxWoBeNePyHcWmZyRiUkPlRqYiQdGeXwOhHbCqVjDcEvJmBkRwWnMqPjXpUsIyXqGjHsEsDwZiFpIbTkQaUlUeFxMwJzSaHdHnDhLaLdTuYgFuJsEcMmDvXyPjKsSeBaRwNtPuOuBtNeOhQdVgKzPzOdYtPjPfDzQzHoWcYjFbSvRgGdGsCmGnQsErToBkCwGeQaCbBpYkLhHxTbUvRnJpZtXjKrHdRiUmUbSlJyGaLnWsCrJbBnSjFaZrIzIrThCmGhQcMsTtOxCuUcRaEyPaG",
"output": "WVaCsGxZrBbFnTbKsCoYlAvUkIpBaYpYmJkMlPwCaFvUkDxAiJgIqWsFqZlFvTtAnGzEwXbYiBdFfFxRiDoUkLmRfAwOlKeOlKgXdUnVqLkTuXtNdQpBpXtLvZxWoBeNePyHcWmZyRiUkPlRqYiQdGeXwOhHbCqVjDcEvJmBkRwWnMqPjXpUsIyXqGjHsEsDwZiFpIbTkQaUlUeFxMwJzSaHdHnDhLaLdTuYgFuJsEcMmDvXyPjKsSeBaRwNtPuOuBtNeOhQdVgKzPzOdYtPjPfDzQzHoWcYjFbSvRgGdGsCmGnQsErToBkCwGeQaCbBpYkLhHxTbUvRnJpZtXjKrHdRiUmUbSlJyGaLnWsCrJbBnSjFaZrIzIrThCmGhQcMsTtOxCuUcRaEyPaG"
},
{
"input": "kEiLxLmPjGzNoGkJdBlAfXhThYhMsHmZoZbGyCvNiUoLoZdAxUbGyQiEfXvPzZzJrPbEcMpHsMjIkRrVvDvQtHuKmXvGpQtXbPzJpFjJdUgWcPdFxLjLtXgVpEiFhImHnKkGiWnZbJqRjCyEwHsNbYfYfTyBaEuKlCtWnOqHmIgGrFmQiYrBnLiFcGuZxXlMfEuVoCxPkVrQvZoIpEhKsYtXrPxLcSfQqXsWaDgVlOnAzUvAhOhMrJfGtWcOwQfRjPmGhDyAeXrNqBvEiDfCiIvWxPjTwPlXpVsMjVjUnCkXgBuWnZaDyJpWkCfBrWnHxMhJgItHdRqNrQaEeRjAuUwRkUdRhEeGlSqVqGmOjNcUhFfXjCmWzBrGvIuZpRyWkWiLyUwFpYjNmNfV",
"output": "KEiLxLmPjGzNoGkJdBlAfXhThYhMsHmZoZbGyCvNiUoLoZdAxUbGyQiEfXvPzZzJrPbEcMpHsMjIkRrVvDvQtHuKmXvGpQtXbPzJpFjJdUgWcPdFxLjLtXgVpEiFhImHnKkGiWnZbJqRjCyEwHsNbYfYfTyBaEuKlCtWnOqHmIgGrFmQiYrBnLiFcGuZxXlMfEuVoCxPkVrQvZoIpEhKsYtXrPxLcSfQqXsWaDgVlOnAzUvAhOhMrJfGtWcOwQfRjPmGhDyAeXrNqBvEiDfCiIvWxPjTwPlXpVsMjVjUnCkXgBuWnZaDyJpWkCfBrWnHxMhJgItHdRqNrQaEeRjAuUwRkUdRhEeGlSqVqGmOjNcUhFfXjCmWzBrGvIuZpRyWkWiLyUwFpYjNmNfV"
},
{
"input": "eIhDoLmDeReKqXsHcVgFxUqNfScAiQnFrTlCgSuTtXiYvBxKaPaGvUeYfSgHqEaWcHxKpFaSlCxGqAmNeFcIzFcZsBiVoZhUjXaDaIcKoBzYdIlEnKfScRqSkYpPtVsVhXsBwUsUfAqRoCkBxWbHgDiCkRtPvUwVgDjOzObYwNiQwXlGnAqEkHdSqLgUkOdZiWaHqQnOhUnDhIzCiQtVcJlGoRfLuVlFjWqSuMsLgLwOdZvKtWdRuRqDoBoInKqPbJdXpIqLtFlMlDaWgSiKbFpCxOnQeNeQzXeKsBzIjCyPxCmBnYuHzQoYxZgGzSgGtZiTeQmUeWlNzZeKiJbQmEjIiDhPeSyZlNdHpZnIkPdJzSeJpPiXxToKyBjJfPwNzZpWzIzGySqPxLtI",
"output": "EIhDoLmDeReKqXsHcVgFxUqNfScAiQnFrTlCgSuTtXiYvBxKaPaGvUeYfSgHqEaWcHxKpFaSlCxGqAmNeFcIzFcZsBiVoZhUjXaDaIcKoBzYdIlEnKfScRqSkYpPtVsVhXsBwUsUfAqRoCkBxWbHgDiCkRtPvUwVgDjOzObYwNiQwXlGnAqEkHdSqLgUkOdZiWaHqQnOhUnDhIzCiQtVcJlGoRfLuVlFjWqSuMsLgLwOdZvKtWdRuRqDoBoInKqPbJdXpIqLtFlMlDaWgSiKbFpCxOnQeNeQzXeKsBzIjCyPxCmBnYuHzQoYxZgGzSgGtZiTeQmUeWlNzZeKiJbQmEjIiDhPeSyZlNdHpZnIkPdJzSeJpPiXxToKyBjJfPwNzZpWzIzGySqPxLtI"
},
{
"input": "uOoQzIeTwYeKpJtGoUdNiXbPgEwVsZkAnJcArHxIpEnEhZwQhZvAiOuLeMkVqLeDsAyKeYgFxGmRoLaRsZjAeXgNfYhBkHeDrHdPuTuYhKmDlAvYzYxCdYgYfVaYlGeVqTeSfBxQePbQrKsTaIkGzMjFrQlJuYaMxWpQkLdEcDsIiMnHnDtThRvAcKyGwBsHqKdXpJfIeTeZtYjFbMeUoXoXzGrShTwSwBpQlKeDrZdCjRqNtXoTsIzBkWbMsObTtDvYaPhUeLeHqHeMpZmTaCcIqXzAmGnPfNdDaFhOqWqDrWuFiBpRjZrQmAdViOuMbFfRyXyWfHgRkGpPnDrEqQcEmHcKpEvWlBrOtJbUaXbThJaSxCbVoGvTmHvZrHvXpCvLaYbRiHzYuQyX",
"output": "UOoQzIeTwYeKpJtGoUdNiXbPgEwVsZkAnJcArHxIpEnEhZwQhZvAiOuLeMkVqLeDsAyKeYgFxGmRoLaRsZjAeXgNfYhBkHeDrHdPuTuYhKmDlAvYzYxCdYgYfVaYlGeVqTeSfBxQePbQrKsTaIkGzMjFrQlJuYaMxWpQkLdEcDsIiMnHnDtThRvAcKyGwBsHqKdXpJfIeTeZtYjFbMeUoXoXzGrShTwSwBpQlKeDrZdCjRqNtXoTsIzBkWbMsObTtDvYaPhUeLeHqHeMpZmTaCcIqXzAmGnPfNdDaFhOqWqDrWuFiBpRjZrQmAdViOuMbFfRyXyWfHgRkGpPnDrEqQcEmHcKpEvWlBrOtJbUaXbThJaSxCbVoGvTmHvZrHvXpCvLaYbRiHzYuQyX"
},
{
"input": "lZqBqKeGvNdSeYuWxRiVnFtYbKuJwQtUcKnVtQhAlOeUzMaAuTaEnDdPfDcNyHgEoBmYjZyFePeJrRiKyAzFnBfAuGiUyLrIeLrNhBeBdVcEeKgCcBrQzDsPwGcNnZvTsEaYmFfMeOmMdNuZbUtDoQoNcGwDqEkEjIdQaPwAxJbXeNxOgKgXoEbZiIsVkRrNpNyAkLeHkNfEpLuQvEcMbIoGaDzXbEtNsLgGfOkZaFiUsOvEjVeCaMcZqMzKeAdXxJsVeCrZaFpJtZxInQxFaSmGgSsVyGeLlFgFqTpIbAvPkIfJrVcJeBxSdEvPyVwIjHpYrLrKqLnAmCuGmPoZrSbOtGaLaTmBmSuUyAmAsRiMqOtRjJhPhAfXaJnTpLbFqPmJgFcBxImTqIiJ",
"output": "LZqBqKeGvNdSeYuWxRiVnFtYbKuJwQtUcKnVtQhAlOeUzMaAuTaEnDdPfDcNyHgEoBmYjZyFePeJrRiKyAzFnBfAuGiUyLrIeLrNhBeBdVcEeKgCcBrQzDsPwGcNnZvTsEaYmFfMeOmMdNuZbUtDoQoNcGwDqEkEjIdQaPwAxJbXeNxOgKgXoEbZiIsVkRrNpNyAkLeHkNfEpLuQvEcMbIoGaDzXbEtNsLgGfOkZaFiUsOvEjVeCaMcZqMzKeAdXxJsVeCrZaFpJtZxInQxFaSmGgSsVyGeLlFgFqTpIbAvPkIfJrVcJeBxSdEvPyVwIjHpYrLrKqLnAmCuGmPoZrSbOtGaLaTmBmSuUyAmAsRiMqOtRjJhPhAfXaJnTpLbFqPmJgFcBxImTqIiJ"
},
{
"input": "P",
"output": "P"
},
{
"input": "Xyzzy",
"output": "Xyzzy"
},
{
"input": "Zzz",
"output": "Zzz"
},
{
"input": "Zp",
"output": "Zp"
}
] | 1,697,303,046
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 25
| 154
| 0
|
l=input()
print(l[0].capitalize()+l[1:])
|
Title: Word Capitalization
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Capitalization is writing a word with its first letter as a capital letter. Your task is to capitalize the given word.
Note, that during capitalization all the letters except the first one remains unchanged.
Input Specification:
A single line contains a non-empty word. This word consists of lowercase and uppercase English letters. The length of the word will not exceed 103.
Output Specification:
Output the given word after capitalization.
Demo Input:
['ApPLe\n', 'konjac\n']
Demo Output:
['ApPLe\n', 'Konjac\n']
Note:
none
|
```python
l=input()
print(l[0].capitalize()+l[1:])
```
| 3
|
|
989
|
B
|
A Tide of Riverscape
|
PROGRAMMING
| 1,200
|
[
"constructive algorithms",
"strings"
] | null | null |
"Time," Mino thinks aloud.
"What?"
"Time and tide wait for no man," explains Mino. "My name, taken from the river, always reminds me of this."
"And what are you recording?"
"You see it, tide. Everything has its own period, and I think I've figured out this one," says Mino with confidence.
Doubtfully, Kanno peeks at Mino's records.
The records are expressed as a string $s$ of characters '0', '1' and '.', where '0' denotes a low tide, '1' denotes a high tide, and '.' denotes an unknown one (either high or low).
You are to help Mino determine whether it's possible that after replacing each '.' independently with '0' or '1', a given integer $p$ is not a period of the resulting string. In case the answer is yes, please also show such a replacement to Mino.
In this problem, a positive integer $p$ is considered a period of string $s$, if for all $1 \leq i \leq \lvert s \rvert - p$, the $i$-th and $(i + p)$-th characters of $s$ are the same. Here $\lvert s \rvert$ is the length of $s$.
|
The first line contains two space-separated integers $n$ and $p$ ($1 \leq p \leq n \leq 2000$) — the length of the given string and the supposed period, respectively.
The second line contains a string $s$ of $n$ characters — Mino's records. $s$ only contains characters '0', '1' and '.', and contains at least one '.' character.
|
Output one line — if it's possible that $p$ is not a period of the resulting string, output any one of such strings; otherwise output "No" (without quotes, you can print letters in any case (upper or lower)).
|
[
"10 7\n1.0.1.0.1.\n",
"10 6\n1.0.1.1000\n",
"10 9\n1........1\n"
] |
[
"1000100010\n",
"1001101000\n",
"No\n"
] |
In the first example, $7$ is not a period of the resulting string because the $1$-st and $8$-th characters of it are different.
In the second example, $6$ is not a period of the resulting string because the $4$-th and $10$-th characters of it are different.
In the third example, $9$ is always a period because the only constraint that the first and last characters are the same is already satisfied.
Note that there are multiple acceptable answers for the first two examples, you can print any of them.
| 1,000
|
[
{
"input": "10 7\n1.0.1.0.1.",
"output": "1000100010"
},
{
"input": "10 6\n1.0.1.1000",
"output": "1001101000"
},
{
"input": "10 9\n1........1",
"output": "No"
},
{
"input": "1 1\n.",
"output": "No"
},
{
"input": "5 1\n0...1",
"output": "00001"
},
{
"input": "17 10\n..1.100..1..0.100",
"output": "00101000010000100"
},
{
"input": "2 1\n0.",
"output": "01"
},
{
"input": "2 1\n..",
"output": "01"
},
{
"input": "3 1\n.0.",
"output": "001"
},
{
"input": "3 1\n00.",
"output": "001"
},
{
"input": "3 2\n0..",
"output": "001"
},
{
"input": "3 2\n0.0",
"output": "No"
},
{
"input": "3 2\n1..",
"output": "100"
},
{
"input": "3 2\n.1.",
"output": "011"
},
{
"input": "3 2\n1.0",
"output": "100"
},
{
"input": "3 3\n1..",
"output": "No"
},
{
"input": "3 3\n.00",
"output": "No"
},
{
"input": "5 3\n0.000",
"output": "01000"
},
{
"input": "10 6\n10010.1001",
"output": "No"
},
{
"input": "75 38\n00.0.1.0.0110.1.00010..100.1110..110..00.0.1.0.0110.1.00010..100.1110..110.",
"output": "000001000011001000010001000111000110000000010000110010000100010001110001101"
},
{
"input": "128 108\n01100.110...000.0001.1.11.11.010010.01100.0.1.01.0.0011.11001.000101...1.0.0..100.0110.0110.0.0101.0.0.0001.01100.110...100.0001",
"output": "01100011000000000001010110110010010001100000100100000110110010000101000100000010000110001100000101000000001001100011000010000001"
},
{
"input": "5 4\n.101.",
"output": "01011"
},
{
"input": "4 2\n101.",
"output": "1011"
},
{
"input": "5 4\n.1011",
"output": "01011"
},
{
"input": "2 1\n..",
"output": "01"
},
{
"input": "5 3\n00.11",
"output": "00011"
},
{
"input": "10 8\n1111.00000",
"output": "1111000000"
},
{
"input": "10 3\n11111111.1",
"output": "1111111101"
},
{
"input": "3 2\n1.0",
"output": "100"
},
{
"input": "6 4\n11..10",
"output": "110010"
},
{
"input": "4 2\n.111",
"output": "0111"
},
{
"input": "3 2\n01.",
"output": "011"
},
{
"input": "5 4\n10.00",
"output": "10000"
},
{
"input": "10 9\n1........0",
"output": "1000000000"
},
{
"input": "2 1\n0.",
"output": "01"
},
{
"input": "8 4\n111111..",
"output": "11111100"
},
{
"input": "3 2\n0.1",
"output": "001"
},
{
"input": "4 1\n111.",
"output": "1110"
},
{
"input": "3 1\n01.",
"output": "010"
},
{
"input": "10 7\n000....111",
"output": "0000000111"
}
] | 1,528,974,549
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 59
| 93
| 0
|
n,p=map(int,input().split())
s=input()
ans=s.replace('.','0')
if all(ans[i]==ans[i+p] for i in range(n-p)):
i=0
while i<n and s[i]!='.': i+=1
if i+p<n:
ans=ans[:i]+'1'+ans[i+1:]
print(ans)
exit()
i=n-1
while i>=0 and s[i]!='.':i-=1
if i-p>=0:
ans=ans[:i]+'1'+ans[i+1:]
print(ans)
exit()
print("No")
else:
print(ans)
|
Title: A Tide of Riverscape
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
"Time," Mino thinks aloud.
"What?"
"Time and tide wait for no man," explains Mino. "My name, taken from the river, always reminds me of this."
"And what are you recording?"
"You see it, tide. Everything has its own period, and I think I've figured out this one," says Mino with confidence.
Doubtfully, Kanno peeks at Mino's records.
The records are expressed as a string $s$ of characters '0', '1' and '.', where '0' denotes a low tide, '1' denotes a high tide, and '.' denotes an unknown one (either high or low).
You are to help Mino determine whether it's possible that after replacing each '.' independently with '0' or '1', a given integer $p$ is not a period of the resulting string. In case the answer is yes, please also show such a replacement to Mino.
In this problem, a positive integer $p$ is considered a period of string $s$, if for all $1 \leq i \leq \lvert s \rvert - p$, the $i$-th and $(i + p)$-th characters of $s$ are the same. Here $\lvert s \rvert$ is the length of $s$.
Input Specification:
The first line contains two space-separated integers $n$ and $p$ ($1 \leq p \leq n \leq 2000$) — the length of the given string and the supposed period, respectively.
The second line contains a string $s$ of $n$ characters — Mino's records. $s$ only contains characters '0', '1' and '.', and contains at least one '.' character.
Output Specification:
Output one line — if it's possible that $p$ is not a period of the resulting string, output any one of such strings; otherwise output "No" (without quotes, you can print letters in any case (upper or lower)).
Demo Input:
['10 7\n1.0.1.0.1.\n', '10 6\n1.0.1.1000\n', '10 9\n1........1\n']
Demo Output:
['1000100010\n', '1001101000\n', 'No\n']
Note:
In the first example, $7$ is not a period of the resulting string because the $1$-st and $8$-th characters of it are different.
In the second example, $6$ is not a period of the resulting string because the $4$-th and $10$-th characters of it are different.
In the third example, $9$ is always a period because the only constraint that the first and last characters are the same is already satisfied.
Note that there are multiple acceptable answers for the first two examples, you can print any of them.
|
```python
n,p=map(int,input().split())
s=input()
ans=s.replace('.','0')
if all(ans[i]==ans[i+p] for i in range(n-p)):
i=0
while i<n and s[i]!='.': i+=1
if i+p<n:
ans=ans[:i]+'1'+ans[i+1:]
print(ans)
exit()
i=n-1
while i>=0 and s[i]!='.':i-=1
if i-p>=0:
ans=ans[:i]+'1'+ans[i+1:]
print(ans)
exit()
print("No")
else:
print(ans)
```
| 3
|
|
472
|
A
|
Design Tutorial: Learn from Math
|
PROGRAMMING
| 800
|
[
"math",
"number theory"
] | null | null |
One way to create a task is to learn from math. You can generate some random math statement or modify some theorems to get something new and build a new task from that.
For example, there is a statement called the "Goldbach's conjecture". It says: "each even number no less than four can be expressed as the sum of two primes". Let's modify it. How about a statement like that: "each integer no less than 12 can be expressed as the sum of two composite numbers." Not like the Goldbach's conjecture, I can prove this theorem.
You are given an integer *n* no less than 12, express it as a sum of two composite numbers.
|
The only line contains an integer *n* (12<=≤<=*n*<=≤<=106).
|
Output two composite integers *x* and *y* (1<=<<=*x*,<=*y*<=<<=*n*) such that *x*<=+<=*y*<==<=*n*. If there are multiple solutions, you can output any of them.
|
[
"12\n",
"15\n",
"23\n",
"1000000\n"
] |
[
"4 8\n",
"6 9\n",
"8 15\n",
"500000 500000\n"
] |
In the first example, 12 = 4 + 8 and both 4, 8 are composite numbers. You can output "6 6" or "8 4" as well.
In the second example, 15 = 6 + 9. Note that you can't output "1 14" because 1 is not a composite number.
| 500
|
[
{
"input": "12",
"output": "4 8"
},
{
"input": "15",
"output": "6 9"
},
{
"input": "23",
"output": "8 15"
},
{
"input": "1000000",
"output": "500000 500000"
},
{
"input": "63874",
"output": "4 63870"
},
{
"input": "14568",
"output": "4 14564"
},
{
"input": "192",
"output": "4 188"
},
{
"input": "86",
"output": "4 82"
},
{
"input": "46220",
"output": "4 46216"
},
{
"input": "57114",
"output": "4 57110"
},
{
"input": "869",
"output": "4 865"
},
{
"input": "738457",
"output": "4 738453"
},
{
"input": "58113",
"output": "6 58107"
},
{
"input": "4864",
"output": "4 4860"
},
{
"input": "15",
"output": "6 9"
},
{
"input": "74752",
"output": "4 74748"
},
{
"input": "6073",
"output": "4 6069"
},
{
"input": "1289",
"output": "4 1285"
},
{
"input": "20",
"output": "4 16"
},
{
"input": "58134",
"output": "4 58130"
},
{
"input": "57756",
"output": "4 57752"
},
{
"input": "765",
"output": "6 759"
},
{
"input": "59",
"output": "4 55"
},
{
"input": "991666",
"output": "4 991662"
},
{
"input": "70761",
"output": "4 70757"
},
{
"input": "13",
"output": "4 9"
},
{
"input": "999999",
"output": "4 999995"
},
{
"input": "17",
"output": "8 9"
},
{
"input": "21",
"output": "6 15"
},
{
"input": "19",
"output": "4 15"
},
{
"input": "100007",
"output": "6 100001"
},
{
"input": "999987",
"output": "6 999981"
},
{
"input": "22",
"output": "4 18"
}
] | 1,677,934,622
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 33
| 62
| 0
|
n = int(input())
if (n%2 == 0):
print("4", n-4)
else:
print("9", n-9)
|
Title: Design Tutorial: Learn from Math
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One way to create a task is to learn from math. You can generate some random math statement or modify some theorems to get something new and build a new task from that.
For example, there is a statement called the "Goldbach's conjecture". It says: "each even number no less than four can be expressed as the sum of two primes". Let's modify it. How about a statement like that: "each integer no less than 12 can be expressed as the sum of two composite numbers." Not like the Goldbach's conjecture, I can prove this theorem.
You are given an integer *n* no less than 12, express it as a sum of two composite numbers.
Input Specification:
The only line contains an integer *n* (12<=≤<=*n*<=≤<=106).
Output Specification:
Output two composite integers *x* and *y* (1<=<<=*x*,<=*y*<=<<=*n*) such that *x*<=+<=*y*<==<=*n*. If there are multiple solutions, you can output any of them.
Demo Input:
['12\n', '15\n', '23\n', '1000000\n']
Demo Output:
['4 8\n', '6 9\n', '8 15\n', '500000 500000\n']
Note:
In the first example, 12 = 4 + 8 and both 4, 8 are composite numbers. You can output "6 6" or "8 4" as well.
In the second example, 15 = 6 + 9. Note that you can't output "1 14" because 1 is not a composite number.
|
```python
n = int(input())
if (n%2 == 0):
print("4", n-4)
else:
print("9", n-9)
```
| 3
|
|
208
|
A
|
Dubstep
|
PROGRAMMING
| 900
|
[
"strings"
] | null | null |
Vasya works as a DJ in the best Berland nightclub, and he often uses dubstep music in his performance. Recently, he has decided to take a couple of old songs and make dubstep remixes from them.
Let's assume that a song consists of some number of words. To make the dubstep remix of this song, Vasya inserts a certain number of words "WUB" before the first word of the song (the number may be zero), after the last word (the number may be zero), and between words (at least one between any pair of neighbouring words), and then the boy glues together all the words, including "WUB", in one string and plays the song at the club.
For example, a song with words "I AM X" can transform into a dubstep remix as "WUBWUBIWUBAMWUBWUBX" and cannot transform into "WUBWUBIAMWUBX".
Recently, Petya has heard Vasya's new dubstep track, but since he isn't into modern music, he decided to find out what was the initial song that Vasya remixed. Help Petya restore the original song.
|
The input consists of a single non-empty string, consisting only of uppercase English letters, the string's length doesn't exceed 200 characters. It is guaranteed that before Vasya remixed the song, no word contained substring "WUB" in it; Vasya didn't change the word order. It is also guaranteed that initially the song had at least one word.
|
Print the words of the initial song that Vasya used to make a dubsteb remix. Separate the words with a space.
|
[
"WUBWUBABCWUB\n",
"WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB\n"
] |
[
"ABC ",
"WE ARE THE CHAMPIONS MY FRIEND "
] |
In the first sample: "WUBWUBABCWUB" = "WUB" + "WUB" + "ABC" + "WUB". That means that the song originally consisted of a single word "ABC", and all words "WUB" were added by Vasya.
In the second sample Vasya added a single word "WUB" between all neighbouring words, in the beginning and in the end, except for words "ARE" and "THE" — between them Vasya added two "WUB".
| 500
|
[
{
"input": "WUBWUBABCWUB",
"output": "ABC "
},
{
"input": "WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB",
"output": "WE ARE THE CHAMPIONS MY FRIEND "
},
{
"input": "WUBWUBWUBSR",
"output": "SR "
},
{
"input": "RWUBWUBWUBLWUB",
"output": "R L "
},
{
"input": "ZJWUBWUBWUBJWUBWUBWUBL",
"output": "ZJ J L "
},
{
"input": "CWUBBWUBWUBWUBEWUBWUBWUBQWUBWUBWUB",
"output": "C B E Q "
},
{
"input": "WUBJKDWUBWUBWBIRAQKFWUBWUBYEWUBWUBWUBWVWUBWUB",
"output": "JKD WBIRAQKF YE WV "
},
{
"input": "WUBKSDHEMIXUJWUBWUBRWUBWUBWUBSWUBWUBWUBHWUBWUBWUB",
"output": "KSDHEMIXUJ R S H "
},
{
"input": "OGWUBWUBWUBXWUBWUBWUBIWUBWUBWUBKOWUBWUB",
"output": "OG X I KO "
},
{
"input": "QWUBQQWUBWUBWUBIWUBWUBWWWUBWUBWUBJOPJPBRH",
"output": "Q QQ I WW JOPJPBRH "
},
{
"input": "VSRNVEATZTLGQRFEGBFPWUBWUBWUBAJWUBWUBWUBPQCHNWUBCWUB",
"output": "VSRNVEATZTLGQRFEGBFP AJ PQCHN C "
},
{
"input": "WUBWUBEWUBWUBWUBIQMJNIQWUBWUBWUBGZZBQZAUHYPWUBWUBWUBPMRWUBWUBWUBDCV",
"output": "E IQMJNIQ GZZBQZAUHYP PMR DCV "
},
{
"input": "WUBWUBWUBFVWUBWUBWUBBPSWUBWUBWUBRXNETCJWUBWUBWUBJDMBHWUBWUBWUBBWUBWUBVWUBWUBB",
"output": "FV BPS RXNETCJ JDMBH B V B "
},
{
"input": "WUBWUBWUBFBQWUBWUBWUBIDFSYWUBWUBWUBCTWDMWUBWUBWUBSXOWUBWUBWUBQIWUBWUBWUBL",
"output": "FBQ IDFSY CTWDM SXO QI L "
},
{
"input": "IWUBWUBQLHDWUBYIIKZDFQWUBWUBWUBCXWUBWUBUWUBWUBWUBKWUBWUBWUBNL",
"output": "I QLHD YIIKZDFQ CX U K NL "
},
{
"input": "KWUBUPDYXGOKUWUBWUBWUBAGOAHWUBIZDWUBWUBWUBIYWUBWUBWUBVWUBWUBWUBPWUBWUBWUBE",
"output": "K UPDYXGOKU AGOAH IZD IY V P E "
},
{
"input": "WUBWUBOWUBWUBWUBIPVCQAFWYWUBWUBWUBQWUBWUBWUBXHDKCPYKCTWWYWUBWUBWUBVWUBWUBWUBFZWUBWUB",
"output": "O IPVCQAFWY Q XHDKCPYKCTWWY V FZ "
},
{
"input": "PAMJGYWUBWUBWUBXGPQMWUBWUBWUBTKGSXUYWUBWUBWUBEWUBWUBWUBNWUBWUBWUBHWUBWUBWUBEWUBWUB",
"output": "PAMJGY XGPQM TKGSXUY E N H E "
},
{
"input": "WUBYYRTSMNWUWUBWUBWUBCWUBWUBWUBCWUBWUBWUBFSYUINDWOBVWUBWUBWUBFWUBWUBWUBAUWUBWUBWUBVWUBWUBWUBJB",
"output": "YYRTSMNWU C C FSYUINDWOBV F AU V JB "
},
{
"input": "WUBWUBYGPYEYBNRTFKOQCWUBWUBWUBUYGRTQEGWLFYWUBWUBWUBFVWUBHPWUBWUBWUBXZQWUBWUBWUBZDWUBWUBWUBM",
"output": "YGPYEYBNRTFKOQC UYGRTQEGWLFY FV HP XZQ ZD M "
},
{
"input": "WUBZVMJWUBWUBWUBFOIMJQWKNZUBOFOFYCCWUBWUBWUBAUWWUBRDRADWUBWUBWUBCHQVWUBWUBWUBKFTWUBWUBWUBW",
"output": "ZVMJ FOIMJQWKNZUBOFOFYCC AUW RDRAD CHQV KFT W "
},
{
"input": "WUBWUBZBKOKHQLGKRVIMZQMQNRWUBWUBWUBDACWUBWUBNZHFJMPEYKRVSWUBWUBWUBPPHGAVVPRZWUBWUBWUBQWUBWUBAWUBG",
"output": "ZBKOKHQLGKRVIMZQMQNR DAC NZHFJMPEYKRVS PPHGAVVPRZ Q A G "
},
{
"input": "WUBWUBJWUBWUBWUBNFLWUBWUBWUBGECAWUBYFKBYJWTGBYHVSSNTINKWSINWSMAWUBWUBWUBFWUBWUBWUBOVWUBWUBLPWUBWUBWUBN",
"output": "J NFL GECA YFKBYJWTGBYHVSSNTINKWSINWSMA F OV LP N "
},
{
"input": "WUBWUBLCWUBWUBWUBZGEQUEATJVIXETVTWUBWUBWUBEXMGWUBWUBWUBRSWUBWUBWUBVWUBWUBWUBTAWUBWUBWUBCWUBWUBWUBQG",
"output": "LC ZGEQUEATJVIXETVT EXMG RS V TA C QG "
},
{
"input": "WUBMPWUBWUBWUBORWUBWUBDLGKWUBWUBWUBVVZQCAAKVJTIKWUBWUBWUBTJLUBZJCILQDIFVZWUBWUBYXWUBWUBWUBQWUBWUBWUBLWUB",
"output": "MP OR DLGK VVZQCAAKVJTIK TJLUBZJCILQDIFVZ YX Q L "
},
{
"input": "WUBNXOLIBKEGXNWUBWUBWUBUWUBGITCNMDQFUAOVLWUBWUBWUBAIJDJZJHFMPVTPOXHPWUBWUBWUBISCIOWUBWUBWUBGWUBWUBWUBUWUB",
"output": "NXOLIBKEGXN U GITCNMDQFUAOVL AIJDJZJHFMPVTPOXHP ISCIO G U "
},
{
"input": "WUBWUBNMMWCZOLYPNBELIYVDNHJUNINWUBWUBWUBDXLHYOWUBWUBWUBOJXUWUBWUBWUBRFHTGJCEFHCGWARGWUBWUBWUBJKWUBWUBSJWUBWUB",
"output": "NMMWCZOLYPNBELIYVDNHJUNIN DXLHYO OJXU RFHTGJCEFHCGWARG JK SJ "
},
{
"input": "SGWLYSAUJOJBNOXNWUBWUBWUBBOSSFWKXPDPDCQEWUBWUBWUBDIRZINODWUBWUBWUBWWUBWUBWUBPPHWUBWUBWUBRWUBWUBWUBQWUBWUBWUBJWUB",
"output": "SGWLYSAUJOJBNOXN BOSSFWKXPDPDCQE DIRZINOD W PPH R Q J "
},
{
"input": "TOWUBWUBWUBGBTBNWUBWUBWUBJVIOJBIZFUUYHUAIEBQLQXPQKZJMPTCWBKPOSAWUBWUBWUBSWUBWUBWUBTOLVXWUBWUBWUBNHWUBWUBWUBO",
"output": "TO GBTBN JVIOJBIZFUUYHUAIEBQLQXPQKZJMPTCWBKPOSA S TOLVX NH O "
},
{
"input": "WUBWUBWSPLAYSZSAUDSWUBWUBWUBUWUBWUBWUBKRWUBWUBWUBRSOKQMZFIYZQUWUBWUBWUBELSHUWUBWUBWUBUKHWUBWUBWUBQXEUHQWUBWUBWUBBWUBWUBWUBR",
"output": "WSPLAYSZSAUDS U KR RSOKQMZFIYZQU ELSHU UKH QXEUHQ B R "
},
{
"input": "WUBXEMWWVUHLSUUGRWUBWUBWUBAWUBXEGILZUNKWUBWUBWUBJDHHKSWUBWUBWUBDTSUYSJHWUBWUBWUBPXFWUBMOHNJWUBWUBWUBZFXVMDWUBWUBWUBZMWUBWUB",
"output": "XEMWWVUHLSUUGR A XEGILZUNK JDHHKS DTSUYSJH PXF MOHNJ ZFXVMD ZM "
},
{
"input": "BMBWUBWUBWUBOQKWUBWUBWUBPITCIHXHCKLRQRUGXJWUBWUBWUBVWUBWUBWUBJCWUBWUBWUBQJPWUBWUBWUBBWUBWUBWUBBMYGIZOOXWUBWUBWUBTAGWUBWUBHWUB",
"output": "BMB OQK PITCIHXHCKLRQRUGXJ V JC QJP B BMYGIZOOX TAG H "
},
{
"input": "CBZNWUBWUBWUBNHWUBWUBWUBYQSYWUBWUBWUBMWUBWUBWUBXRHBTMWUBWUBWUBPCRCWUBWUBWUBTZUYLYOWUBWUBWUBCYGCWUBWUBWUBCLJWUBWUBWUBSWUBWUBWUB",
"output": "CBZN NH YQSY M XRHBTM PCRC TZUYLYO CYGC CLJ S "
},
{
"input": "DPDWUBWUBWUBEUQKWPUHLTLNXHAEKGWUBRRFYCAYZFJDCJLXBAWUBWUBWUBHJWUBOJWUBWUBWUBNHBJEYFWUBWUBWUBRWUBWUBWUBSWUBWWUBWUBWUBXDWUBWUBWUBJWUB",
"output": "DPD EUQKWPUHLTLNXHAEKG RRFYCAYZFJDCJLXBA HJ OJ NHBJEYF R S W XD J "
},
{
"input": "WUBWUBWUBISERPQITVIYERSCNWUBWUBWUBQWUBWUBWUBDGSDIPWUBWUBWUBCAHKDZWEXBIBJVVSKKVQJWUBWUBWUBKIWUBWUBWUBCWUBWUBWUBAWUBWUBWUBPWUBWUBWUBHWUBWUBWUBF",
"output": "ISERPQITVIYERSCN Q DGSDIP CAHKDZWEXBIBJVVSKKVQJ KI C A P H F "
},
{
"input": "WUBWUBWUBIWUBWUBLIKNQVWUBWUBWUBPWUBWUBWUBHWUBWUBWUBMWUBWUBWUBDPRSWUBWUBWUBBSAGYLQEENWXXVWUBWUBWUBXMHOWUBWUBWUBUWUBWUBWUBYRYWUBWUBWUBCWUBWUBWUBY",
"output": "I LIKNQV P H M DPRS BSAGYLQEENWXXV XMHO U YRY C Y "
},
{
"input": "WUBWUBWUBMWUBWUBWUBQWUBWUBWUBITCFEYEWUBWUBWUBHEUWGNDFNZGWKLJWUBWUBWUBMZPWUBWUBWUBUWUBWUBWUBBWUBWUBWUBDTJWUBHZVIWUBWUBWUBPWUBFNHHWUBWUBWUBVTOWUB",
"output": "M Q ITCFEYE HEUWGNDFNZGWKLJ MZP U B DTJ HZVI P FNHH VTO "
},
{
"input": "WUBWUBNDNRFHYJAAUULLHRRDEDHYFSRXJWUBWUBWUBMUJVDTIRSGYZAVWKRGIFWUBWUBWUBHMZWUBWUBWUBVAIWUBWUBWUBDDKJXPZRGWUBWUBWUBSGXWUBWUBWUBIFKWUBWUBWUBUWUBWUBWUBW",
"output": "NDNRFHYJAAUULLHRRDEDHYFSRXJ MUJVDTIRSGYZAVWKRGIF HMZ VAI DDKJXPZRG SGX IFK U W "
},
{
"input": "WUBOJMWRSLAXXHQRTPMJNCMPGWUBWUBWUBNYGMZIXNLAKSQYWDWUBWUBWUBXNIWUBWUBWUBFWUBWUBWUBXMBWUBWUBWUBIWUBWUBWUBINWUBWUBWUBWDWUBWUBWUBDDWUBWUBWUBD",
"output": "OJMWRSLAXXHQRTPMJNCMPG NYGMZIXNLAKSQYWD XNI F XMB I IN WD DD D "
},
{
"input": "WUBWUBWUBREHMWUBWUBWUBXWUBWUBWUBQASNWUBWUBWUBNLSMHLCMTICWUBWUBWUBVAWUBWUBWUBHNWUBWUBWUBNWUBWUBWUBUEXLSFOEULBWUBWUBWUBXWUBWUBWUBJWUBWUBWUBQWUBWUBWUBAWUBWUB",
"output": "REHM X QASN NLSMHLCMTIC VA HN N UEXLSFOEULB X J Q A "
},
{
"input": "WUBWUBWUBSTEZTZEFFIWUBWUBWUBSWUBWUBWUBCWUBFWUBHRJPVWUBWUBWUBDYJUWUBWUBWUBPWYDKCWUBWUBWUBCWUBWUBWUBUUEOGCVHHBWUBWUBWUBEXLWUBWUBWUBVCYWUBWUBWUBMWUBWUBWUBYWUB",
"output": "STEZTZEFFI S C F HRJPV DYJU PWYDKC C UUEOGCVHHB EXL VCY M Y "
},
{
"input": "WPPNMSQOQIWUBWUBWUBPNQXWUBWUBWUBHWUBWUBWUBNFLWUBWUBWUBGWSGAHVJFNUWUBWUBWUBFWUBWUBWUBWCMLRICFSCQQQTNBWUBWUBWUBSWUBWUBWUBKGWUBWUBWUBCWUBWUBWUBBMWUBWUBWUBRWUBWUB",
"output": "WPPNMSQOQI PNQX H NFL GWSGAHVJFNU F WCMLRICFSCQQQTNB S KG C BM R "
},
{
"input": "YZJOOYITZRARKVFYWUBWUBRZQGWUBWUBWUBUOQWUBWUBWUBIWUBWUBWUBNKVDTBOLETKZISTWUBWUBWUBWLWUBQQFMMGSONZMAWUBZWUBWUBWUBQZUXGCWUBWUBWUBIRZWUBWUBWUBLTTVTLCWUBWUBWUBY",
"output": "YZJOOYITZRARKVFY RZQG UOQ I NKVDTBOLETKZIST WL QQFMMGSONZMA Z QZUXGC IRZ LTTVTLC Y "
},
{
"input": "WUBCAXNCKFBVZLGCBWCOAWVWOFKZVQYLVTWUBWUBWUBNLGWUBWUBWUBAMGDZBDHZMRMQMDLIRMIWUBWUBWUBGAJSHTBSWUBWUBWUBCXWUBWUBWUBYWUBZLXAWWUBWUBWUBOHWUBWUBWUBZWUBWUBWUBGBWUBWUBWUBE",
"output": "CAXNCKFBVZLGCBWCOAWVWOFKZVQYLVT NLG AMGDZBDHZMRMQMDLIRMI GAJSHTBS CX Y ZLXAW OH Z GB E "
},
{
"input": "WUBWUBCHXSOWTSQWUBWUBWUBCYUZBPBWUBWUBWUBSGWUBWUBWKWORLRRLQYUUFDNWUBWUBWUBYYGOJNEVEMWUBWUBWUBRWUBWUBWUBQWUBWUBWUBIHCKWUBWUBWUBKTWUBWUBWUBRGSNTGGWUBWUBWUBXCXWUBWUBWUBS",
"output": "CHXSOWTSQ CYUZBPB SG WKWORLRRLQYUUFDN YYGOJNEVEM R Q IHCK KT RGSNTGG XCX S "
},
{
"input": "WUBWUBWUBHJHMSBURXTHXWSCHNAIJOWBHLZGJZDHEDSPWBWACCGQWUBWUBWUBXTZKGIITWUBWUBWUBAWUBWUBWUBVNCXPUBCQWUBWUBWUBIDPNAWUBWUBWUBOWUBWUBWUBYGFWUBWUBWUBMQOWUBWUBWUBKWUBWUBWUBAZVWUBWUBWUBEP",
"output": "HJHMSBURXTHXWSCHNAIJOWBHLZGJZDHEDSPWBWACCGQ XTZKGIIT A VNCXPUBCQ IDPNA O YGF MQO K AZV EP "
},
{
"input": "WUBKYDZOYWZSNGMKJSWAXFDFLTHDHEOGTDBNZMSMKZTVWUBWUBWUBLRMIIWUBWUBWUBGWUBWUBWUBADPSWUBWUBWUBANBWUBWUBPCWUBWUBWUBPWUBWUBWUBGPVNLSWIRFORYGAABUXMWUBWUBWUBOWUBWUBWUBNWUBWUBWUBYWUBWUB",
"output": "KYDZOYWZSNGMKJSWAXFDFLTHDHEOGTDBNZMSMKZTV LRMII G ADPS ANB PC P GPVNLSWIRFORYGAABUXM O N Y "
},
{
"input": "REWUBWUBWUBJDWUBWUBWUBNWUBWUBWUBTWWUBWUBWUBWZDOCKKWUBWUBWUBLDPOVBFRCFWUBWUBAKZIBQKEUAZEEWUBWUBWUBLQYPNPFWUBYEWUBWUBWUBFWUBWUBWUBBPWUBWUBWUBAWWUBWUBWUBQWUBWUBWUBBRWUBWUBWUBXJL",
"output": "RE JD N TW WZDOCKK LDPOVBFRCF AKZIBQKEUAZEE LQYPNPF YE F BP AW Q BR XJL "
},
{
"input": "CUFGJDXGMWUBWUBWUBOMWUBWUBWUBSIEWUBWUBWUBJJWKNOWUBWUBWUBYBHVNRNORGYWUBWUBWUBOAGCAWUBWUBWUBSBLBKTPFKPBIWUBWUBWUBJBWUBWUBWUBRMFCJPGWUBWUBWUBDWUBWUBWUBOJOWUBWUBWUBZPWUBWUBWUBMWUBRWUBWUBWUBFXWWUBWUBWUBO",
"output": "CUFGJDXGM OM SIE JJWKNO YBHVNRNORGY OAGCA SBLBKTPFKPBI JB RMFCJPG D OJO ZP M R FXW O "
},
{
"input": "WUBJZGAEXFMFEWMAKGQLUWUBWUBWUBICYTPQWGENELVYWANKUOJYWUBWUBWUBGWUBWUBWUBHYCJVLPHTUPNEGKCDGQWUBWUBWUBOFWUBWUBWUBCPGSOGZBRPRPVJJEWUBWUBWUBDQBCWUBWUBWUBHWUBWUBWUBMHOHYBMATWUBWUBWUBVWUBWUBWUBSWUBWUBWUBKOWU",
"output": "JZGAEXFMFEWMAKGQLU ICYTPQWGENELVYWANKUOJY G HYCJVLPHTUPNEGKCDGQ OF CPGSOGZBRPRPVJJE DQBC H MHOHYBMAT V S KOWU "
},
{
"input": "A",
"output": "A "
},
{
"input": "WUBA",
"output": "A "
},
{
"input": "WUBWUBA",
"output": "A "
},
{
"input": "AWUBWUBWUB",
"output": "A "
},
{
"input": "AWUBBWUBCWUBD",
"output": "A B C D "
},
{
"input": "WUBWWUBWUBWUBUWUBWUBBWUB",
"output": "W U B "
},
{
"input": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA",
"output": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA "
},
{
"input": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAWUBAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA",
"output": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA "
},
{
"input": "WUWUBBWWUBUB",
"output": "WU BW UB "
},
{
"input": "WUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUABWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUB",
"output": "WUAB "
},
{
"input": "ZWUB",
"output": "Z "
},
{
"input": "WU",
"output": "WU "
},
{
"input": "UB",
"output": "UB "
},
{
"input": "U",
"output": "U "
},
{
"input": "WUBW",
"output": "W "
},
{
"input": "WUBWU",
"output": "WU "
},
{
"input": "WUWUB",
"output": "WU "
},
{
"input": "UBWUB",
"output": "UB "
},
{
"input": "WUWUBUBWUBUWUB",
"output": "WU UB U "
},
{
"input": "WUBWWUBAWUB",
"output": "W A "
},
{
"input": "WUUUUU",
"output": "WUUUUU "
}
] | 1,697,566,125
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 71
| 92
| 0
|
# Read the input string
remix = input()
# Split the remix string by "WUB" to get a list of words
words = remix.split("WUB")
# Filter out any empty strings from the list
words = [word for word in words if word]
# Join the words to restore the initial song
initial_song = " ".join(words)
# Print the initial song
print(initial_song)
|
Title: Dubstep
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya works as a DJ in the best Berland nightclub, and he often uses dubstep music in his performance. Recently, he has decided to take a couple of old songs and make dubstep remixes from them.
Let's assume that a song consists of some number of words. To make the dubstep remix of this song, Vasya inserts a certain number of words "WUB" before the first word of the song (the number may be zero), after the last word (the number may be zero), and between words (at least one between any pair of neighbouring words), and then the boy glues together all the words, including "WUB", in one string and plays the song at the club.
For example, a song with words "I AM X" can transform into a dubstep remix as "WUBWUBIWUBAMWUBWUBX" and cannot transform into "WUBWUBIAMWUBX".
Recently, Petya has heard Vasya's new dubstep track, but since he isn't into modern music, he decided to find out what was the initial song that Vasya remixed. Help Petya restore the original song.
Input Specification:
The input consists of a single non-empty string, consisting only of uppercase English letters, the string's length doesn't exceed 200 characters. It is guaranteed that before Vasya remixed the song, no word contained substring "WUB" in it; Vasya didn't change the word order. It is also guaranteed that initially the song had at least one word.
Output Specification:
Print the words of the initial song that Vasya used to make a dubsteb remix. Separate the words with a space.
Demo Input:
['WUBWUBABCWUB\n', 'WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB\n']
Demo Output:
['ABC ', 'WE ARE THE CHAMPIONS MY FRIEND ']
Note:
In the first sample: "WUBWUBABCWUB" = "WUB" + "WUB" + "ABC" + "WUB". That means that the song originally consisted of a single word "ABC", and all words "WUB" were added by Vasya.
In the second sample Vasya added a single word "WUB" between all neighbouring words, in the beginning and in the end, except for words "ARE" and "THE" — between them Vasya added two "WUB".
|
```python
# Read the input string
remix = input()
# Split the remix string by "WUB" to get a list of words
words = remix.split("WUB")
# Filter out any empty strings from the list
words = [word for word in words if word]
# Join the words to restore the initial song
initial_song = " ".join(words)
# Print the initial song
print(initial_song)
```
| 3
|
|
343
|
B
|
Alternating Current
|
PROGRAMMING
| 1,600
|
[
"data structures",
"greedy",
"implementation"
] | null | null |
Mad scientist Mike has just finished constructing a new device to search for extraterrestrial intelligence! He was in such a hurry to launch it for the first time that he plugged in the power wires without giving it a proper glance and started experimenting right away. After a while Mike observed that the wires ended up entangled and now have to be untangled again.
The device is powered by two wires "plus" and "minus". The wires run along the floor from the wall (on the left) to the device (on the right). Both the wall and the device have two contacts in them on the same level, into which the wires are plugged in some order. The wires are considered entangled if there are one or more places where one wire runs above the other one. For example, the picture below has four such places (top view):
Mike knows the sequence in which the wires run above each other. Mike also noticed that on the left side, the "plus" wire is always plugged into the top contact (as seen on the picture). He would like to untangle the wires without unplugging them and without moving the device. Determine if it is possible to do that. A wire can be freely moved and stretched on the floor, but cannot be cut.
To understand the problem better please read the notes to the test samples.
|
The single line of the input contains a sequence of characters "+" and "-" of length *n* (1<=≤<=*n*<=≤<=100000). The *i*-th (1<=≤<=*i*<=≤<=*n*) position of the sequence contains the character "+", if on the *i*-th step from the wall the "plus" wire runs above the "minus" wire, and the character "-" otherwise.
|
Print either "Yes" (without the quotes) if the wires can be untangled or "No" (without the quotes) if the wires cannot be untangled.
|
[
"-++-\n",
"+-\n",
"++\n",
"-\n"
] |
[
"Yes\n",
"No\n",
"Yes\n",
"No\n"
] |
The first testcase corresponds to the picture in the statement. To untangle the wires, one can first move the "plus" wire lower, thus eliminating the two crosses in the middle, and then draw it under the "minus" wire, eliminating also the remaining two crosses.
In the second testcase the "plus" wire makes one full revolution around the "minus" wire. Thus the wires cannot be untangled:
In the third testcase the "plus" wire simply runs above the "minus" wire twice in sequence. The wires can be untangled by lifting "plus" and moving it higher:
In the fourth testcase the "minus" wire runs above the "plus" wire once. The wires cannot be untangled without moving the device itself:
| 1,000
|
[
{
"input": "-++-",
"output": "Yes"
},
{
"input": "+-",
"output": "No"
},
{
"input": "++",
"output": "Yes"
},
{
"input": "-",
"output": "No"
},
{
"input": "+-+-",
"output": "No"
},
{
"input": "-+-",
"output": "No"
},
{
"input": "-++-+--+",
"output": "Yes"
},
{
"input": "+",
"output": "No"
},
{
"input": "-+",
"output": "No"
},
{
"input": "--",
"output": "Yes"
},
{
"input": "+++",
"output": "No"
},
{
"input": "--+",
"output": "No"
},
{
"input": "++--++",
"output": "Yes"
},
{
"input": "+-++-+",
"output": "Yes"
},
{
"input": "+-+--+",
"output": "No"
},
{
"input": "--++-+",
"output": "No"
},
{
"input": "-+-+--",
"output": "No"
},
{
"input": "+-+++-",
"output": "No"
},
{
"input": "-+-+-+",
"output": "No"
},
{
"input": "-++-+--++--+-++-",
"output": "Yes"
},
{
"input": "+-----+-++---+------+++-++++",
"output": "No"
},
{
"input": "-+-++--+++-++++---+--+----+--+-+-+++-+++-+---++-++++-+--+--+--+-+-++-+-+-++++++---++--+++++-+--++--+-+--++-----+--+-++---+++---++----+++-++++--++-++-",
"output": "No"
},
{
"input": "-+-----++++--++-+-++",
"output": "Yes"
},
{
"input": "+--+--+------+++++++-+-+++--++---+--+-+---+--+++-+++-------+++++-+-++++--+-+-+++++++----+----+++----+-+++-+++-----+++-+-++-+-+++++-+--++----+--+-++-----+-+-++++---+++---+-+-+-++++--+--+++---+++++-+---+-----+++-++--+++---++-++-+-+++-+-+-+---+++--+--++++-+-+--++-------+--+---++-----+++--+-+++--++-+-+++-++--+++-++++++++++-++-++++++-+++--+--++-+++--+++-++++----+++---+-+----++++-+-+",
"output": "Yes"
},
{
"input": "-+-+-++-+-+-",
"output": "Yes"
},
{
"input": "-+-++-+-",
"output": "Yes"
},
{
"input": "-+-++-+-+-",
"output": "No"
},
{
"input": "++-+-+-+-+--+",
"output": "No"
},
{
"input": "+++---",
"output": "No"
},
{
"input": "+-+-+-+-+--+-+-+-+-++--++--+",
"output": "Yes"
},
{
"input": "+-+-++",
"output": "No"
},
{
"input": "-++--+--+++-+-+-+-+-",
"output": "No"
},
{
"input": "+---+-+-",
"output": "No"
},
{
"input": "+-+--+-+",
"output": "Yes"
},
{
"input": "+++---+++---",
"output": "No"
},
{
"input": "-+++++",
"output": "No"
},
{
"input": "-+-+-+-+-+-+-++-+-+-+-+-+-+-",
"output": "Yes"
},
{
"input": "-+++--",
"output": "No"
},
{
"input": "+---+",
"output": "No"
},
{
"input": "-++",
"output": "No"
},
{
"input": "-+--+-",
"output": "Yes"
},
{
"input": "+---++--++",
"output": "No"
},
{
"input": "+++-",
"output": "No"
},
{
"input": "--+++",
"output": "No"
},
{
"input": "++-+",
"output": "No"
}
] | 1,579,946,898
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 62
| 280
| 5,017,600
|
a = input()
stack = []
for i in a:
if not stack:
stack.append(i)
elif stack and stack[-1]==i:
stack.pop()
else:
stack.append(i)
if stack:
print('No')
else:
print('Yes')
|
Title: Alternating Current
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mad scientist Mike has just finished constructing a new device to search for extraterrestrial intelligence! He was in such a hurry to launch it for the first time that he plugged in the power wires without giving it a proper glance and started experimenting right away. After a while Mike observed that the wires ended up entangled and now have to be untangled again.
The device is powered by two wires "plus" and "minus". The wires run along the floor from the wall (on the left) to the device (on the right). Both the wall and the device have two contacts in them on the same level, into which the wires are plugged in some order. The wires are considered entangled if there are one or more places where one wire runs above the other one. For example, the picture below has four such places (top view):
Mike knows the sequence in which the wires run above each other. Mike also noticed that on the left side, the "plus" wire is always plugged into the top contact (as seen on the picture). He would like to untangle the wires without unplugging them and without moving the device. Determine if it is possible to do that. A wire can be freely moved and stretched on the floor, but cannot be cut.
To understand the problem better please read the notes to the test samples.
Input Specification:
The single line of the input contains a sequence of characters "+" and "-" of length *n* (1<=≤<=*n*<=≤<=100000). The *i*-th (1<=≤<=*i*<=≤<=*n*) position of the sequence contains the character "+", if on the *i*-th step from the wall the "plus" wire runs above the "minus" wire, and the character "-" otherwise.
Output Specification:
Print either "Yes" (without the quotes) if the wires can be untangled or "No" (without the quotes) if the wires cannot be untangled.
Demo Input:
['-++-\n', '+-\n', '++\n', '-\n']
Demo Output:
['Yes\n', 'No\n', 'Yes\n', 'No\n']
Note:
The first testcase corresponds to the picture in the statement. To untangle the wires, one can first move the "plus" wire lower, thus eliminating the two crosses in the middle, and then draw it under the "minus" wire, eliminating also the remaining two crosses.
In the second testcase the "plus" wire makes one full revolution around the "minus" wire. Thus the wires cannot be untangled:
In the third testcase the "plus" wire simply runs above the "minus" wire twice in sequence. The wires can be untangled by lifting "plus" and moving it higher:
In the fourth testcase the "minus" wire runs above the "plus" wire once. The wires cannot be untangled without moving the device itself:
|
```python
a = input()
stack = []
for i in a:
if not stack:
stack.append(i)
elif stack and stack[-1]==i:
stack.pop()
else:
stack.append(i)
if stack:
print('No')
else:
print('Yes')
```
| 3
|
|
146
|
A
|
Lucky Ticket
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya loves tickets very much. As we know, each ticket has a number that is a positive integer. Its length equals *n* (*n* is always even). Petya calls a ticket lucky if the ticket's number is a lucky number and the sum of digits in the first half (the sum of the first *n*<=/<=2 digits) equals the sum of digits in the second half (the sum of the last *n*<=/<=2 digits). Check if the given ticket is lucky.
|
The first line contains an even integer *n* (2<=≤<=*n*<=≤<=50) — the length of the ticket number that needs to be checked. The second line contains an integer whose length equals exactly *n* — the ticket number. The number may contain leading zeros.
|
On the first line print "YES" if the given ticket number is lucky. Otherwise, print "NO" (without the quotes).
|
[
"2\n47\n",
"4\n4738\n",
"4\n4774\n"
] |
[
"NO\n",
"NO\n",
"YES\n"
] |
In the first sample the sum of digits in the first half does not equal the sum of digits in the second half (4 ≠ 7).
In the second sample the ticket number is not the lucky number.
| 500
|
[
{
"input": "2\n47",
"output": "NO"
},
{
"input": "4\n4738",
"output": "NO"
},
{
"input": "4\n4774",
"output": "YES"
},
{
"input": "4\n4570",
"output": "NO"
},
{
"input": "6\n477477",
"output": "YES"
},
{
"input": "6\n777777",
"output": "YES"
},
{
"input": "20\n44444444444444444444",
"output": "YES"
},
{
"input": "2\n44",
"output": "YES"
},
{
"input": "10\n4745474547",
"output": "NO"
},
{
"input": "14\n77770004444444",
"output": "NO"
},
{
"input": "10\n4747777744",
"output": "YES"
},
{
"input": "10\n1234567890",
"output": "NO"
},
{
"input": "50\n44444444444444444444444444444444444444444444444444",
"output": "YES"
},
{
"input": "50\n44444444444444444444444444444444444444444444444447",
"output": "NO"
},
{
"input": "50\n74444444444444444444444444444444444444444444444444",
"output": "NO"
},
{
"input": "50\n07777777777777777777777777777777777777777777777770",
"output": "NO"
},
{
"input": "50\n77777777777777777777777777777777777777777777777777",
"output": "YES"
},
{
"input": "50\n44747747774474747747747447777447774747447477444474",
"output": "YES"
},
{
"input": "48\n447474444777444474747747744774447444747474774474",
"output": "YES"
},
{
"input": "32\n74474474777444474444747774474774",
"output": "YES"
},
{
"input": "40\n4747777444447747777447447747447474774777",
"output": "YES"
},
{
"input": "10\n4477477444",
"output": "YES"
},
{
"input": "18\n447747474447744747",
"output": "YES"
},
{
"input": "26\n44747744444774744774474447",
"output": "YES"
},
{
"input": "50\n44707747774474747747747447777447774747447477444474",
"output": "NO"
},
{
"input": "40\n4747777444447737777447447747447474774777",
"output": "NO"
},
{
"input": "36\n764477744747444444447747747474744444",
"output": "NO"
},
{
"input": "22\n4477407474777477744447",
"output": "NO"
},
{
"input": "32\n74274474777444474444747774474774",
"output": "NO"
},
{
"input": "16\n4744447974444747",
"output": "NO"
},
{
"input": "2\n11",
"output": "NO"
},
{
"input": "2\n22",
"output": "NO"
},
{
"input": "2\n33",
"output": "NO"
},
{
"input": "2\n74",
"output": "NO"
},
{
"input": "2\n55",
"output": "NO"
},
{
"input": "2\n66",
"output": "NO"
},
{
"input": "2\n77",
"output": "YES"
},
{
"input": "2\n88",
"output": "NO"
},
{
"input": "2\n99",
"output": "NO"
},
{
"input": "4\n4004",
"output": "NO"
},
{
"input": "14\n00077774444444",
"output": "NO"
},
{
"input": "6\n004444",
"output": "NO"
},
{
"input": "4\n0044",
"output": "NO"
},
{
"input": "14\n77771114444444",
"output": "NO"
},
{
"input": "14\n44444447777000",
"output": "NO"
},
{
"input": "6\n004774",
"output": "NO"
}
] | 1,641,684,640
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 46
| 216
| 0
|
n=int(input())
number=input()
while True:
if '1' in number or '2' in number or '3' in number or '5' in number or '6' in number or '8' in number or '9' in number or '0' in number:
print('NO')
break
number = [int(i) for i in number]
if sum(number[0:n//2])==sum(number[n//2:]):
print('YES')
break
else:
print('NO')
break
|
Title: Lucky Ticket
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya loves tickets very much. As we know, each ticket has a number that is a positive integer. Its length equals *n* (*n* is always even). Petya calls a ticket lucky if the ticket's number is a lucky number and the sum of digits in the first half (the sum of the first *n*<=/<=2 digits) equals the sum of digits in the second half (the sum of the last *n*<=/<=2 digits). Check if the given ticket is lucky.
Input Specification:
The first line contains an even integer *n* (2<=≤<=*n*<=≤<=50) — the length of the ticket number that needs to be checked. The second line contains an integer whose length equals exactly *n* — the ticket number. The number may contain leading zeros.
Output Specification:
On the first line print "YES" if the given ticket number is lucky. Otherwise, print "NO" (without the quotes).
Demo Input:
['2\n47\n', '4\n4738\n', '4\n4774\n']
Demo Output:
['NO\n', 'NO\n', 'YES\n']
Note:
In the first sample the sum of digits in the first half does not equal the sum of digits in the second half (4 ≠ 7).
In the second sample the ticket number is not the lucky number.
|
```python
n=int(input())
number=input()
while True:
if '1' in number or '2' in number or '3' in number or '5' in number or '6' in number or '8' in number or '9' in number or '0' in number:
print('NO')
break
number = [int(i) for i in number]
if sum(number[0:n//2])==sum(number[n//2:]):
print('YES')
break
else:
print('NO')
break
```
| 3
|
|
271
|
A
|
Beautiful Year
|
PROGRAMMING
| 800
|
[
"brute force"
] | null | null |
It seems like the year of 2013 came only yesterday. Do you know a curious fact? The year of 2013 is the first year after the old 1987 with only distinct digits.
Now you are suggested to solve the following problem: given a year number, find the minimum year number which is strictly larger than the given one and has only distinct digits.
|
The single line contains integer *y* (1000<=≤<=*y*<=≤<=9000) — the year number.
|
Print a single integer — the minimum year number that is strictly larger than *y* and all it's digits are distinct. It is guaranteed that the answer exists.
|
[
"1987\n",
"2013\n"
] |
[
"2013\n",
"2014\n"
] |
none
| 500
|
[
{
"input": "1987",
"output": "2013"
},
{
"input": "2013",
"output": "2014"
},
{
"input": "1000",
"output": "1023"
},
{
"input": "1001",
"output": "1023"
},
{
"input": "1234",
"output": "1235"
},
{
"input": "5555",
"output": "5601"
},
{
"input": "9000",
"output": "9012"
},
{
"input": "1111",
"output": "1203"
},
{
"input": "8999",
"output": "9012"
},
{
"input": "4572",
"output": "4573"
},
{
"input": "6666",
"output": "6701"
},
{
"input": "2001",
"output": "2013"
},
{
"input": "3000",
"output": "3012"
},
{
"input": "7712",
"output": "7801"
},
{
"input": "8088",
"output": "8091"
},
{
"input": "1594",
"output": "1596"
},
{
"input": "6016",
"output": "6017"
},
{
"input": "8800",
"output": "8901"
},
{
"input": "3331",
"output": "3401"
},
{
"input": "5090",
"output": "5091"
},
{
"input": "2342",
"output": "2345"
},
{
"input": "2334",
"output": "2340"
},
{
"input": "1123",
"output": "1203"
},
{
"input": "8989",
"output": "9012"
},
{
"input": "8977",
"output": "9012"
},
{
"input": "6869",
"output": "6870"
},
{
"input": "8999",
"output": "9012"
}
] | 1,695,552,712
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 27
| 92
| 0
|
# Read the year
y = int(input())
# Function to check if a year has distinct digits
def has_distinct_digits(year):
return len(set(str(year))) == 4
# Start iterating from y+1
next_year = y + 1
while not has_distinct_digits(next_year):
next_year += 1
# Output the minimum year with distinct digits
print(next_year)
|
Title: Beautiful Year
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
It seems like the year of 2013 came only yesterday. Do you know a curious fact? The year of 2013 is the first year after the old 1987 with only distinct digits.
Now you are suggested to solve the following problem: given a year number, find the minimum year number which is strictly larger than the given one and has only distinct digits.
Input Specification:
The single line contains integer *y* (1000<=≤<=*y*<=≤<=9000) — the year number.
Output Specification:
Print a single integer — the minimum year number that is strictly larger than *y* and all it's digits are distinct. It is guaranteed that the answer exists.
Demo Input:
['1987\n', '2013\n']
Demo Output:
['2013\n', '2014\n']
Note:
none
|
```python
# Read the year
y = int(input())
# Function to check if a year has distinct digits
def has_distinct_digits(year):
return len(set(str(year))) == 4
# Start iterating from y+1
next_year = y + 1
while not has_distinct_digits(next_year):
next_year += 1
# Output the minimum year with distinct digits
print(next_year)
```
| 3
|
|
915
|
A
|
Garden
|
PROGRAMMING
| 900
|
[
"implementation"
] | null | null |
Luba thinks about watering her garden. The garden can be represented as a segment of length *k*. Luba has got *n* buckets, the *i*-th bucket allows her to water some continuous subsegment of garden of length exactly *a**i* each hour. Luba can't water any parts of the garden that were already watered, also she can't water the ground outside the garden.
Luba has to choose one of the buckets in order to water the garden as fast as possible (as mentioned above, each hour she will water some continuous subsegment of length *a**i* if she chooses the *i*-th bucket). Help her to determine the minimum number of hours she has to spend watering the garden. It is guaranteed that Luba can always choose a bucket so it is possible water the garden.
See the examples for better understanding.
|
The first line of input contains two integer numbers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=100) — the number of buckets and the length of the garden, respectively.
The second line of input contains *n* integer numbers *a**i* (1<=≤<=*a**i*<=≤<=100) — the length of the segment that can be watered by the *i*-th bucket in one hour.
It is guaranteed that there is at least one bucket such that it is possible to water the garden in integer number of hours using only this bucket.
|
Print one integer number — the minimum number of hours required to water the garden.
|
[
"3 6\n2 3 5\n",
"6 7\n1 2 3 4 5 6\n"
] |
[
"2\n",
"7\n"
] |
In the first test the best option is to choose the bucket that allows to water the segment of length 3. We can't choose the bucket that allows to water the segment of length 5 because then we can't water the whole garden.
In the second test we can choose only the bucket that allows us to water the segment of length 1.
| 0
|
[
{
"input": "3 6\n2 3 5",
"output": "2"
},
{
"input": "6 7\n1 2 3 4 5 6",
"output": "7"
},
{
"input": "5 97\n1 10 50 97 2",
"output": "1"
},
{
"input": "5 97\n1 10 50 100 2",
"output": "97"
},
{
"input": "100 100\n2 46 24 18 86 90 31 38 84 49 58 28 15 80 14 24 87 56 62 87 41 87 55 71 87 32 41 56 91 32 24 75 43 42 35 30 72 53 31 26 54 61 87 85 36 75 44 31 7 38 77 57 61 54 70 77 45 96 39 57 11 8 91 42 52 15 42 30 92 41 27 26 34 27 3 80 32 86 26 97 63 91 30 75 14 7 19 23 45 11 8 43 44 73 11 56 3 55 63 16",
"output": "50"
},
{
"input": "100 91\n13 13 62 96 74 47 81 46 78 21 20 42 4 73 25 30 76 74 58 28 25 52 42 48 74 40 82 9 25 29 17 22 46 64 57 95 81 39 47 86 40 95 97 35 31 98 45 98 47 78 52 63 58 14 89 97 17 95 28 22 20 36 68 38 95 16 2 26 54 47 42 31 31 81 21 21 65 40 82 53 60 71 75 33 96 98 6 22 95 12 5 48 18 27 58 62 5 96 36 75",
"output": "7"
},
{
"input": "8 8\n8 7 6 5 4 3 2 1",
"output": "1"
},
{
"input": "3 8\n4 3 2",
"output": "2"
},
{
"input": "3 8\n2 4 2",
"output": "2"
},
{
"input": "3 6\n1 3 2",
"output": "2"
},
{
"input": "3 6\n3 2 5",
"output": "2"
},
{
"input": "3 8\n4 2 1",
"output": "2"
},
{
"input": "5 6\n2 3 5 1 2",
"output": "2"
},
{
"input": "2 6\n5 3",
"output": "2"
},
{
"input": "4 12\n6 4 3 1",
"output": "2"
},
{
"input": "3 18\n1 9 6",
"output": "2"
},
{
"input": "3 9\n3 2 1",
"output": "3"
},
{
"input": "3 6\n5 3 2",
"output": "2"
},
{
"input": "2 10\n5 2",
"output": "2"
},
{
"input": "2 18\n6 3",
"output": "3"
},
{
"input": "4 12\n1 2 12 3",
"output": "1"
},
{
"input": "3 7\n3 2 1",
"output": "7"
},
{
"input": "3 6\n3 2 1",
"output": "2"
},
{
"input": "5 10\n5 4 3 2 1",
"output": "2"
},
{
"input": "5 16\n8 4 2 1 7",
"output": "2"
},
{
"input": "6 7\n6 5 4 3 7 1",
"output": "1"
},
{
"input": "2 6\n3 2",
"output": "2"
},
{
"input": "2 4\n4 1",
"output": "1"
},
{
"input": "6 8\n2 4 1 3 5 7",
"output": "2"
},
{
"input": "6 8\n6 5 4 3 2 1",
"output": "2"
},
{
"input": "6 15\n5 2 3 6 4 3",
"output": "3"
},
{
"input": "4 8\n2 4 8 1",
"output": "1"
},
{
"input": "2 5\n5 1",
"output": "1"
},
{
"input": "4 18\n3 1 1 2",
"output": "6"
},
{
"input": "2 1\n2 1",
"output": "1"
},
{
"input": "3 10\n2 10 5",
"output": "1"
},
{
"input": "5 12\n12 4 4 4 3",
"output": "1"
},
{
"input": "3 6\n6 3 2",
"output": "1"
},
{
"input": "2 2\n2 1",
"output": "1"
},
{
"input": "3 18\n1 9 3",
"output": "2"
},
{
"input": "3 8\n7 2 4",
"output": "2"
},
{
"input": "2 100\n99 1",
"output": "100"
},
{
"input": "4 12\n1 3 4 2",
"output": "3"
},
{
"input": "3 6\n2 3 1",
"output": "2"
},
{
"input": "4 6\n3 2 5 12",
"output": "2"
},
{
"input": "4 97\n97 1 50 10",
"output": "1"
},
{
"input": "3 12\n1 12 2",
"output": "1"
},
{
"input": "4 12\n1 4 3 2",
"output": "3"
},
{
"input": "1 1\n1",
"output": "1"
},
{
"input": "3 19\n7 1 1",
"output": "19"
},
{
"input": "5 12\n12 4 3 4 4",
"output": "1"
},
{
"input": "3 8\n8 4 2",
"output": "1"
},
{
"input": "3 3\n3 2 1",
"output": "1"
},
{
"input": "5 6\n3 2 4 2 2",
"output": "2"
},
{
"input": "2 16\n8 4",
"output": "2"
},
{
"input": "3 6\n10 2 3",
"output": "2"
},
{
"input": "5 3\n2 4 5 3 6",
"output": "1"
},
{
"input": "11 99\n1 2 3 6 5 4 7 8 99 33 66",
"output": "1"
},
{
"input": "3 12\n3 12 2",
"output": "1"
},
{
"input": "5 25\n24 5 15 25 23",
"output": "1"
},
{
"input": "2 4\n8 1",
"output": "4"
},
{
"input": "4 100\n2 50 4 1",
"output": "2"
},
{
"input": "3 28\n7 14 1",
"output": "2"
},
{
"input": "4 8\n2 8 4 1",
"output": "1"
},
{
"input": "4 6\n6 1 2 3",
"output": "1"
},
{
"input": "2 12\n4 3",
"output": "3"
},
{
"input": "4 12\n1 2 4 3",
"output": "3"
},
{
"input": "5 12\n2 3 12 6 4",
"output": "1"
},
{
"input": "4 4\n1 2 2 4",
"output": "1"
},
{
"input": "3 6\n2 3 2",
"output": "2"
},
{
"input": "4 21\n21 20 21 2",
"output": "1"
},
{
"input": "3 8\n3 4 2",
"output": "2"
},
{
"input": "1 25\n25",
"output": "1"
},
{
"input": "99 12\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99",
"output": "1"
},
{
"input": "98 12\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98",
"output": "1"
},
{
"input": "79 12\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79",
"output": "1"
},
{
"input": "4 32\n1 1 1 1",
"output": "32"
},
{
"input": "1 100\n1",
"output": "100"
},
{
"input": "2 100\n7 1",
"output": "100"
},
{
"input": "7 24\n1 3 6 4 5 2 7",
"output": "4"
},
{
"input": "6 87\n1 2 8 4 5 7",
"output": "87"
},
{
"input": "1 88\n1",
"output": "88"
},
{
"input": "1 89\n1",
"output": "89"
}
] | 1,654,667,414
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 83
| 46
| 0
|
n,k=map(int, input().split())
a=list(map(int, input().split()))
x=0
for i in range(n):
if k%a[i]==0 and a[i]>x:
x=a[i]
print(k//x)
|
Title: Garden
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Luba thinks about watering her garden. The garden can be represented as a segment of length *k*. Luba has got *n* buckets, the *i*-th bucket allows her to water some continuous subsegment of garden of length exactly *a**i* each hour. Luba can't water any parts of the garden that were already watered, also she can't water the ground outside the garden.
Luba has to choose one of the buckets in order to water the garden as fast as possible (as mentioned above, each hour she will water some continuous subsegment of length *a**i* if she chooses the *i*-th bucket). Help her to determine the minimum number of hours she has to spend watering the garden. It is guaranteed that Luba can always choose a bucket so it is possible water the garden.
See the examples for better understanding.
Input Specification:
The first line of input contains two integer numbers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=100) — the number of buckets and the length of the garden, respectively.
The second line of input contains *n* integer numbers *a**i* (1<=≤<=*a**i*<=≤<=100) — the length of the segment that can be watered by the *i*-th bucket in one hour.
It is guaranteed that there is at least one bucket such that it is possible to water the garden in integer number of hours using only this bucket.
Output Specification:
Print one integer number — the minimum number of hours required to water the garden.
Demo Input:
['3 6\n2 3 5\n', '6 7\n1 2 3 4 5 6\n']
Demo Output:
['2\n', '7\n']
Note:
In the first test the best option is to choose the bucket that allows to water the segment of length 3. We can't choose the bucket that allows to water the segment of length 5 because then we can't water the whole garden.
In the second test we can choose only the bucket that allows us to water the segment of length 1.
|
```python
n,k=map(int, input().split())
a=list(map(int, input().split()))
x=0
for i in range(n):
if k%a[i]==0 and a[i]>x:
x=a[i]
print(k//x)
```
| 3
|
|
659
|
A
|
Round House
|
PROGRAMMING
| 1,000
|
[
"implementation",
"math"
] | null | null |
Vasya lives in a round building, whose entrances are numbered sequentially by integers from 1 to *n*. Entrance *n* and entrance 1 are adjacent.
Today Vasya got bored and decided to take a walk in the yard. Vasya lives in entrance *a* and he decided that during his walk he will move around the house *b* entrances in the direction of increasing numbers (in this order entrance *n* should be followed by entrance 1). The negative value of *b* corresponds to moving |*b*| entrances in the order of decreasing numbers (in this order entrance 1 is followed by entrance *n*). If *b*<==<=0, then Vasya prefers to walk beside his entrance.
Help Vasya to determine the number of the entrance, near which he will be at the end of his walk.
|
The single line of the input contains three space-separated integers *n*, *a* and *b* (1<=≤<=*n*<=≤<=100,<=1<=≤<=*a*<=≤<=*n*,<=<=-<=100<=≤<=*b*<=≤<=100) — the number of entrances at Vasya's place, the number of his entrance and the length of his walk, respectively.
|
Print a single integer *k* (1<=≤<=*k*<=≤<=*n*) — the number of the entrance where Vasya will be at the end of his walk.
|
[
"6 2 -5\n",
"5 1 3\n",
"3 2 7\n"
] |
[
"3\n",
"4\n",
"3\n"
] |
The first example is illustrated by the picture in the statements.
| 500
|
[
{
"input": "6 2 -5",
"output": "3"
},
{
"input": "5 1 3",
"output": "4"
},
{
"input": "3 2 7",
"output": "3"
},
{
"input": "1 1 0",
"output": "1"
},
{
"input": "1 1 -1",
"output": "1"
},
{
"input": "1 1 1",
"output": "1"
},
{
"input": "100 1 -1",
"output": "100"
},
{
"input": "100 54 100",
"output": "54"
},
{
"input": "100 37 -100",
"output": "37"
},
{
"input": "99 41 0",
"output": "41"
},
{
"input": "97 37 -92",
"output": "42"
},
{
"input": "99 38 59",
"output": "97"
},
{
"input": "35 34 1",
"output": "35"
},
{
"input": "48 1 -1",
"output": "48"
},
{
"input": "87 65 -76",
"output": "76"
},
{
"input": "76 26 29",
"output": "55"
},
{
"input": "100 65 0",
"output": "65"
},
{
"input": "2 1 100",
"output": "1"
},
{
"input": "3 2 -100",
"output": "1"
},
{
"input": "1 1 100",
"output": "1"
},
{
"input": "1 1 -100",
"output": "1"
},
{
"input": "3 1 -100",
"output": "3"
},
{
"input": "4 3 -100",
"output": "3"
},
{
"input": "3 2 -12",
"output": "2"
},
{
"input": "2 2 -100",
"output": "2"
},
{
"input": "3 2 -90",
"output": "2"
},
{
"input": "6 2 -10",
"output": "4"
},
{
"input": "3 3 -100",
"output": "2"
},
{
"input": "5 2 4",
"output": "1"
},
{
"input": "6 4 5",
"output": "3"
},
{
"input": "3 2 -6",
"output": "2"
},
{
"input": "5 1 -99",
"output": "2"
},
{
"input": "6 2 5",
"output": "1"
},
{
"input": "10 1 -100",
"output": "1"
},
{
"input": "2 2 1",
"output": "1"
},
{
"input": "3 3 1",
"output": "1"
},
{
"input": "6 4 4",
"output": "2"
},
{
"input": "17 17 2",
"output": "2"
},
{
"input": "6 6 1",
"output": "1"
},
{
"input": "5 3 -2",
"output": "1"
},
{
"input": "6 2 -100",
"output": "4"
},
{
"input": "5 3 -100",
"output": "3"
},
{
"input": "5 4 3",
"output": "2"
},
{
"input": "3 2 2",
"output": "1"
},
{
"input": "5 5 2",
"output": "2"
},
{
"input": "3 2 5",
"output": "1"
},
{
"input": "5 5 -1",
"output": "4"
},
{
"input": "5 3 3",
"output": "1"
},
{
"input": "4 2 3",
"output": "1"
},
{
"input": "88 76 74",
"output": "62"
}
] | 1,564,452,450
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 50
| 124
| 0
|
n,a,b=map(int,input().split())
print(((a+b)%n+n-1)%n+1)
|
Title: Round House
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya lives in a round building, whose entrances are numbered sequentially by integers from 1 to *n*. Entrance *n* and entrance 1 are adjacent.
Today Vasya got bored and decided to take a walk in the yard. Vasya lives in entrance *a* and he decided that during his walk he will move around the house *b* entrances in the direction of increasing numbers (in this order entrance *n* should be followed by entrance 1). The negative value of *b* corresponds to moving |*b*| entrances in the order of decreasing numbers (in this order entrance 1 is followed by entrance *n*). If *b*<==<=0, then Vasya prefers to walk beside his entrance.
Help Vasya to determine the number of the entrance, near which he will be at the end of his walk.
Input Specification:
The single line of the input contains three space-separated integers *n*, *a* and *b* (1<=≤<=*n*<=≤<=100,<=1<=≤<=*a*<=≤<=*n*,<=<=-<=100<=≤<=*b*<=≤<=100) — the number of entrances at Vasya's place, the number of his entrance and the length of his walk, respectively.
Output Specification:
Print a single integer *k* (1<=≤<=*k*<=≤<=*n*) — the number of the entrance where Vasya will be at the end of his walk.
Demo Input:
['6 2 -5\n', '5 1 3\n', '3 2 7\n']
Demo Output:
['3\n', '4\n', '3\n']
Note:
The first example is illustrated by the picture in the statements.
|
```python
n,a,b=map(int,input().split())
print(((a+b)%n+n-1)%n+1)
```
| 3
|
|
1,008
|
A
|
Romaji
|
PROGRAMMING
| 900
|
[
"implementation",
"strings"
] | null | null |
Vitya has just started learning Berlanese language. It is known that Berlanese uses the Latin alphabet. Vowel letters are "a", "o", "u", "i", and "e". Other letters are consonant.
In Berlanese, there has to be a vowel after every consonant, but there can be any letter after any vowel. The only exception is a consonant "n"; after this letter, there can be any letter (not only a vowel) or there can be no letter at all. For example, the words "harakiri", "yupie", "man", and "nbo" are Berlanese while the words "horse", "king", "my", and "nz" are not.
Help Vitya find out if a word $s$ is Berlanese.
|
The first line of the input contains the string $s$ consisting of $|s|$ ($1\leq |s|\leq 100$) lowercase Latin letters.
|
Print "YES" (without quotes) if there is a vowel after every consonant except "n", otherwise print "NO".
You can print each letter in any case (upper or lower).
|
[
"sumimasen\n",
"ninja\n",
"codeforces\n"
] |
[
"YES\n",
"YES\n",
"NO\n"
] |
In the first and second samples, a vowel goes after each consonant except "n", so the word is Berlanese.
In the third sample, the consonant "c" goes after the consonant "r", and the consonant "s" stands on the end, so the word is not Berlanese.
| 500
|
[
{
"input": "sumimasen",
"output": "YES"
},
{
"input": "ninja",
"output": "YES"
},
{
"input": "codeforces",
"output": "NO"
},
{
"input": "auuaoonntanonnuewannnnpuuinniwoonennyolonnnvienonpoujinndinunnenannmuveoiuuhikucuziuhunnnmunzancenen",
"output": "YES"
},
{
"input": "n",
"output": "YES"
},
{
"input": "necnei",
"output": "NO"
},
{
"input": "nternn",
"output": "NO"
},
{
"input": "aucunuohja",
"output": "NO"
},
{
"input": "a",
"output": "YES"
},
{
"input": "b",
"output": "NO"
},
{
"input": "nn",
"output": "YES"
},
{
"input": "nnnzaaa",
"output": "YES"
},
{
"input": "zn",
"output": "NO"
},
{
"input": "ab",
"output": "NO"
},
{
"input": "aaaaaaaaaa",
"output": "YES"
},
{
"input": "aaaaaaaaab",
"output": "NO"
},
{
"input": "aaaaaaaaan",
"output": "YES"
},
{
"input": "baaaaaaaaa",
"output": "YES"
},
{
"input": "naaaaaaaaa",
"output": "YES"
},
{
"input": "nbaaaaaaaa",
"output": "YES"
},
{
"input": "bbaaaaaaaa",
"output": "NO"
},
{
"input": "bnaaaaaaaa",
"output": "NO"
},
{
"input": "eonwonojannonnufimiiniewuqaienokacevecinfuqihatenhunliquuyebayiaenifuexuanenuaounnboancaeowonu",
"output": "YES"
},
{
"input": "uixinnepnlinqaingieianndeakuniooudidonnnqeaituioeneiroionxuowudiooonayenfeonuino",
"output": "NO"
},
{
"input": "nnnnnyigaveteononnnnxaalenxuiiwannntoxonyoqonlejuoxuoconnnentoinnul",
"output": "NO"
},
{
"input": "ndonneasoiunhomuunnhuitonnntunntoanerekonoupunanuauenu",
"output": "YES"
},
{
"input": "anujemogawautiedoneobninnibonuunaoennnyoorufonxionntinimiboonununnnnnleenqunminzayoutanlalo",
"output": "NO"
},
{
"input": "y",
"output": "NO"
},
{
"input": "by",
"output": "NO"
},
{
"input": "yy",
"output": "NO"
},
{
"input": "nbn",
"output": "NO"
},
{
"input": "nz",
"output": "NO"
},
{
"input": "king",
"output": "NO"
},
{
"input": "g",
"output": "NO"
},
{
"input": "az",
"output": "NO"
},
{
"input": "x",
"output": "NO"
},
{
"input": "z",
"output": "NO"
},
{
"input": "yn",
"output": "NO"
},
{
"input": "aeo",
"output": "YES"
},
{
"input": "nb",
"output": "NO"
},
{
"input": "npn",
"output": "NO"
},
{
"input": "kini",
"output": "YES"
},
{
"input": "pya",
"output": "NO"
},
{
"input": "m",
"output": "NO"
},
{
"input": "p",
"output": "NO"
},
{
"input": "aaaaaaaak",
"output": "NO"
},
{
"input": "aab",
"output": "NO"
},
{
"input": "d",
"output": "NO"
},
{
"input": "at",
"output": "NO"
},
{
"input": "aaaaaak",
"output": "NO"
},
{
"input": "aaz",
"output": "NO"
},
{
"input": "aaab",
"output": "NO"
},
{
"input": "s",
"output": "NO"
},
{
"input": "nzzen",
"output": "NO"
},
{
"input": "aeionnhhhn",
"output": "NO"
},
{
"input": "h",
"output": "NO"
},
{
"input": "ny",
"output": "NO"
}
] | 1,636,799,175
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 57
| 46
| 4,505,600
|
a=input()
c=0
if a[-1] in 'aeioun':
for i in range(0,len(a)-1):
if a[i] not in 'aeioun':
if a[i+1] not in 'aeiou':
print('NO')
c=c+1
break
else:
print('NO')
c=c+1
if c==0:
print('YES')
|
Title: Romaji
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vitya has just started learning Berlanese language. It is known that Berlanese uses the Latin alphabet. Vowel letters are "a", "o", "u", "i", and "e". Other letters are consonant.
In Berlanese, there has to be a vowel after every consonant, but there can be any letter after any vowel. The only exception is a consonant "n"; after this letter, there can be any letter (not only a vowel) or there can be no letter at all. For example, the words "harakiri", "yupie", "man", and "nbo" are Berlanese while the words "horse", "king", "my", and "nz" are not.
Help Vitya find out if a word $s$ is Berlanese.
Input Specification:
The first line of the input contains the string $s$ consisting of $|s|$ ($1\leq |s|\leq 100$) lowercase Latin letters.
Output Specification:
Print "YES" (without quotes) if there is a vowel after every consonant except "n", otherwise print "NO".
You can print each letter in any case (upper or lower).
Demo Input:
['sumimasen\n', 'ninja\n', 'codeforces\n']
Demo Output:
['YES\n', 'YES\n', 'NO\n']
Note:
In the first and second samples, a vowel goes after each consonant except "n", so the word is Berlanese.
In the third sample, the consonant "c" goes after the consonant "r", and the consonant "s" stands on the end, so the word is not Berlanese.
|
```python
a=input()
c=0
if a[-1] in 'aeioun':
for i in range(0,len(a)-1):
if a[i] not in 'aeioun':
if a[i+1] not in 'aeiou':
print('NO')
c=c+1
break
else:
print('NO')
c=c+1
if c==0:
print('YES')
```
| 3
|
|
919
|
F
|
A Game With Numbers
|
PROGRAMMING
| 2,600
|
[
"games",
"graphs",
"shortest paths"
] | null | null |
Imagine that Alice is playing a card game with her friend Bob. They both have exactly $8$ cards and there is an integer on each card, ranging from $0$ to $4$. In each round, Alice or Bob in turns choose two cards from different players, let them be $a$ and $b$, where $a$ is the number on the player's card, and $b$ is the number on the opponent's card. It is necessary that $a \cdot b \ne 0$. Then they calculate $c = (a + b) \bmod 5$ and replace the number $a$ with $c$. The player who ends up with numbers on all $8$ cards being $0$, wins.
Now Alice wants to know who wins in some situations. She will give you her cards' numbers, Bob's cards' numbers and the person playing the first round. Your task is to determine who wins if both of them choose the best operation in their rounds.
|
The first line contains one positive integer $T$ ($1 \leq T \leq 100\,000$), denoting the number of situations you need to consider.
The following lines describe those $T$ situations. For each situation:
- The first line contains a non-negative integer $f$ ($0 \leq f \leq 1$), where $f = 0$ means that Alice plays first and $f = 1$ means Bob plays first. - The second line contains $8$ non-negative integers $a_1, a_2, \ldots, a_8$ ($0 \leq a_i \leq 4$), describing Alice's cards. - The third line contains $8$ non-negative integers $b_1, b_2, \ldots, b_8$ ($0 \leq b_i \leq 4$), describing Bob's cards.
We guarantee that if $f=0$, we have $\sum_{i=1}^{8}a_i \ne 0$. Also when $f=1$, $\sum_{i=1}^{8}b_i \ne 0$ holds.
|
Output $T$ lines. For each situation, determine who wins. Output
- "Alice" (without quotes) if Alice wins. - "Bob" (without quotes) if Bob wins. - "Deal" (without quotes) if it gets into a deal, i.e. no one wins.
|
[
"4\n1\n0 0 0 0 0 0 0 0\n1 2 3 4 1 2 3 4\n1\n0 0 0 1 0 0 0 0\n0 0 0 0 4 0 0 0\n0\n1 0 0 0 0 0 0 0\n0 0 0 4 0 0 2 0\n1\n1 1 1 1 1 1 1 1\n1 1 1 1 1 1 1 1\n"
] |
[
"Alice\nBob\nAlice\nDeal\n"
] |
In the first situation, Alice has all her numbers $0$. So she wins immediately.
In the second situation, Bob picks the numbers $4$ and $1$. Because we have $(4 + 1) \bmod 5 = 0$, Bob wins after this operation.
In the third situation, Alice picks the numbers $1$ and $4$. She wins after this operation.
In the fourth situation, we can prove that it falls into a loop.
| 2,500
|
[
{
"input": "4\n1\n0 0 0 0 0 0 0 0\n1 2 3 4 1 2 3 4\n1\n0 0 0 1 0 0 0 0\n0 0 0 0 4 0 0 0\n0\n1 0 0 0 0 0 0 0\n0 0 0 4 0 0 2 0\n1\n1 1 1 1 1 1 1 1\n1 1 1 1 1 1 1 1",
"output": "Alice\nBob\nAlice\nDeal"
},
{
"input": "1\n0\n0 2 2 0 1 2 1 2\n1 2 4 3 2 1 1 0",
"output": "Alice"
}
] | 1,631,749,582
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 20
| 3,649
| 34,918,400
|
# My Solution for CF 919/F.
import sys
import itertools
from collections import deque
t = int(input())
hand_size = 8
num_values = 5
res = 1
for x in range(hand_size + 1, hand_size + num_values):
res *= x
for x in range(1, num_values):
res //= x
# res = (handsize + numvalues - 1) choose (handsize)
counts = [0] * res * 5
# Precompute the possible winning/losing locations.
# First, generate the possible hand values, and turn them into integers.
m = {}
for i, v in enumerate(itertools.combinations(range(hand_size + num_values - 1), num_values - 1)):
prev = -1
ind = 0
for x in v:
counts[i * 5 + ind] = x - prev - 1
prev = x
ind += 1
counts[i * 5 + ind] = hand_size + num_values - prev - 2
m[tuple(counts[i*5:i*5+5])] = i
# 0 = DEAL
# 1 = WIN
# -1 = LOSE
CURRENT_STATUS = [0] * (res * res)
DEG = [0] * (res * res)
# Now, any vertex is just two numbers, ranging from 0 to res-1.
# We assume for a particular node that it is my turn.
# Then in the recursion we simply invert at each opportunity.
decided_queue = deque()
win_index = m[(hand_size, 0, 0, 0, 0)]
for i in range(res):
if i != win_index:
decided_queue.append(win_index * res + i)
CURRENT_STATUS[win_index * res + i] = 1
decided_queue.append(i * res + win_index)
CURRENT_STATUS[i * res + win_index] = -1
for i in range(res * res):
id1 = i // res
id2 = i % res
# What choices can I make as id1?
n_options_l = 0
for x in range(1, 5):
if counts[id1 * 5 + x] > 0:
n_options_l += 1
n_options_r = 0
for x in range(1, 5):
if counts[id2 * 5 + x] > 0:
n_options_r += 1
DEG[i] = n_options_l * n_options_r
p = [False] * (res * res)
parent = [-1] * (res * res)
while len(decided_queue) > 0:
index = decided_queue.pop()
if p[index]:
raise ValueError("Cycle bad.")
p[index] = True
# Check all possible inroads to this state.
# Which of the cards could've p2 just made?
id1 = index // res
id2 = index % res
for x in range(5):
if counts[id2 * 5 + x] > 0:
# We could've made this card.
# But what would the card have been / what of my cards should they have picked?
for y in range(1, 5):
if counts[id1 * 5 + y] > 0:
# They could've picked this card.
# x = y + z mod 5. So z = x - y mod 5.
z = (x - y) % 5
if z == 0:
continue
n_count = [counts[id2 * 5 + i] for i in range(5)]
n_count[x] -= 1
n_count[z] += 1
old_id2 = m[tuple(n_count)]
# Remember to flip
old_index = old_id2 * res + id1
if CURRENT_STATUS[old_index] == 0:
DEG[old_index] -= 1
if CURRENT_STATUS[index] == -1:
# If the other player is winning here, they can move here to win.
CURRENT_STATUS[old_index] = 1
decided_queue.append(old_index)
parent[old_index] = index
else:
if DEG[old_index] == 0:
# We've tried every edge and all of them are winning for me, so losing for the person above.
CURRENT_STATUS[old_index] = -1
decided_queue.append(old_index)
parent[old_index] = index
for case in range(t):
f = int(input())
alice_hand = tuple(map(int, input().split()))
c = [0] * num_values
for h in alice_hand:
c[h] += 1
alice = m[tuple(c)]
bob_hand = tuple(map(int, input().split()))
c = [0] * num_values
for h in bob_hand:
c[h] += 1
bob = m[tuple(c)]
if f == 0:
index = alice * res + bob
if CURRENT_STATUS[index] == 0:
print("Deal")
elif CURRENT_STATUS[index] == 1:
print("Alice")
else:
print("Bob")
else:
index = bob * res + alice
if CURRENT_STATUS[index] == 0:
print("Deal")
elif CURRENT_STATUS[index] == 1:
print("Bob")
else:
print("Alice")
|
Title: A Game With Numbers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Imagine that Alice is playing a card game with her friend Bob. They both have exactly $8$ cards and there is an integer on each card, ranging from $0$ to $4$. In each round, Alice or Bob in turns choose two cards from different players, let them be $a$ and $b$, where $a$ is the number on the player's card, and $b$ is the number on the opponent's card. It is necessary that $a \cdot b \ne 0$. Then they calculate $c = (a + b) \bmod 5$ and replace the number $a$ with $c$. The player who ends up with numbers on all $8$ cards being $0$, wins.
Now Alice wants to know who wins in some situations. She will give you her cards' numbers, Bob's cards' numbers and the person playing the first round. Your task is to determine who wins if both of them choose the best operation in their rounds.
Input Specification:
The first line contains one positive integer $T$ ($1 \leq T \leq 100\,000$), denoting the number of situations you need to consider.
The following lines describe those $T$ situations. For each situation:
- The first line contains a non-negative integer $f$ ($0 \leq f \leq 1$), where $f = 0$ means that Alice plays first and $f = 1$ means Bob plays first. - The second line contains $8$ non-negative integers $a_1, a_2, \ldots, a_8$ ($0 \leq a_i \leq 4$), describing Alice's cards. - The third line contains $8$ non-negative integers $b_1, b_2, \ldots, b_8$ ($0 \leq b_i \leq 4$), describing Bob's cards.
We guarantee that if $f=0$, we have $\sum_{i=1}^{8}a_i \ne 0$. Also when $f=1$, $\sum_{i=1}^{8}b_i \ne 0$ holds.
Output Specification:
Output $T$ lines. For each situation, determine who wins. Output
- "Alice" (without quotes) if Alice wins. - "Bob" (without quotes) if Bob wins. - "Deal" (without quotes) if it gets into a deal, i.e. no one wins.
Demo Input:
['4\n1\n0 0 0 0 0 0 0 0\n1 2 3 4 1 2 3 4\n1\n0 0 0 1 0 0 0 0\n0 0 0 0 4 0 0 0\n0\n1 0 0 0 0 0 0 0\n0 0 0 4 0 0 2 0\n1\n1 1 1 1 1 1 1 1\n1 1 1 1 1 1 1 1\n']
Demo Output:
['Alice\nBob\nAlice\nDeal\n']
Note:
In the first situation, Alice has all her numbers $0$. So she wins immediately.
In the second situation, Bob picks the numbers $4$ and $1$. Because we have $(4 + 1) \bmod 5 = 0$, Bob wins after this operation.
In the third situation, Alice picks the numbers $1$ and $4$. She wins after this operation.
In the fourth situation, we can prove that it falls into a loop.
|
```python
# My Solution for CF 919/F.
import sys
import itertools
from collections import deque
t = int(input())
hand_size = 8
num_values = 5
res = 1
for x in range(hand_size + 1, hand_size + num_values):
res *= x
for x in range(1, num_values):
res //= x
# res = (handsize + numvalues - 1) choose (handsize)
counts = [0] * res * 5
# Precompute the possible winning/losing locations.
# First, generate the possible hand values, and turn them into integers.
m = {}
for i, v in enumerate(itertools.combinations(range(hand_size + num_values - 1), num_values - 1)):
prev = -1
ind = 0
for x in v:
counts[i * 5 + ind] = x - prev - 1
prev = x
ind += 1
counts[i * 5 + ind] = hand_size + num_values - prev - 2
m[tuple(counts[i*5:i*5+5])] = i
# 0 = DEAL
# 1 = WIN
# -1 = LOSE
CURRENT_STATUS = [0] * (res * res)
DEG = [0] * (res * res)
# Now, any vertex is just two numbers, ranging from 0 to res-1.
# We assume for a particular node that it is my turn.
# Then in the recursion we simply invert at each opportunity.
decided_queue = deque()
win_index = m[(hand_size, 0, 0, 0, 0)]
for i in range(res):
if i != win_index:
decided_queue.append(win_index * res + i)
CURRENT_STATUS[win_index * res + i] = 1
decided_queue.append(i * res + win_index)
CURRENT_STATUS[i * res + win_index] = -1
for i in range(res * res):
id1 = i // res
id2 = i % res
# What choices can I make as id1?
n_options_l = 0
for x in range(1, 5):
if counts[id1 * 5 + x] > 0:
n_options_l += 1
n_options_r = 0
for x in range(1, 5):
if counts[id2 * 5 + x] > 0:
n_options_r += 1
DEG[i] = n_options_l * n_options_r
p = [False] * (res * res)
parent = [-1] * (res * res)
while len(decided_queue) > 0:
index = decided_queue.pop()
if p[index]:
raise ValueError("Cycle bad.")
p[index] = True
# Check all possible inroads to this state.
# Which of the cards could've p2 just made?
id1 = index // res
id2 = index % res
for x in range(5):
if counts[id2 * 5 + x] > 0:
# We could've made this card.
# But what would the card have been / what of my cards should they have picked?
for y in range(1, 5):
if counts[id1 * 5 + y] > 0:
# They could've picked this card.
# x = y + z mod 5. So z = x - y mod 5.
z = (x - y) % 5
if z == 0:
continue
n_count = [counts[id2 * 5 + i] for i in range(5)]
n_count[x] -= 1
n_count[z] += 1
old_id2 = m[tuple(n_count)]
# Remember to flip
old_index = old_id2 * res + id1
if CURRENT_STATUS[old_index] == 0:
DEG[old_index] -= 1
if CURRENT_STATUS[index] == -1:
# If the other player is winning here, they can move here to win.
CURRENT_STATUS[old_index] = 1
decided_queue.append(old_index)
parent[old_index] = index
else:
if DEG[old_index] == 0:
# We've tried every edge and all of them are winning for me, so losing for the person above.
CURRENT_STATUS[old_index] = -1
decided_queue.append(old_index)
parent[old_index] = index
for case in range(t):
f = int(input())
alice_hand = tuple(map(int, input().split()))
c = [0] * num_values
for h in alice_hand:
c[h] += 1
alice = m[tuple(c)]
bob_hand = tuple(map(int, input().split()))
c = [0] * num_values
for h in bob_hand:
c[h] += 1
bob = m[tuple(c)]
if f == 0:
index = alice * res + bob
if CURRENT_STATUS[index] == 0:
print("Deal")
elif CURRENT_STATUS[index] == 1:
print("Alice")
else:
print("Bob")
else:
index = bob * res + alice
if CURRENT_STATUS[index] == 0:
print("Deal")
elif CURRENT_STATUS[index] == 1:
print("Bob")
else:
print("Alice")
```
| 3
|
|
534
|
A
|
Exam
|
PROGRAMMING
| 1,100
|
[
"constructive algorithms",
"implementation",
"math"
] | null | null |
An exam for *n* students will take place in a long and narrow room, so the students will sit in a line in some order. The teacher suspects that students with adjacent numbers (*i* and *i*<=+<=1) always studied side by side and became friends and if they take an exam sitting next to each other, they will help each other for sure.
Your task is to choose the maximum number of students and make such an arrangement of students in the room that no two students with adjacent numbers sit side by side.
|
A single line contains integer *n* (1<=≤<=*n*<=≤<=5000) — the number of students at an exam.
|
In the first line print integer *k* — the maximum number of students who can be seated so that no two students with adjacent numbers sit next to each other.
In the second line print *k* distinct integers *a*1,<=*a*2,<=...,<=*a**k* (1<=≤<=*a**i*<=≤<=*n*), where *a**i* is the number of the student on the *i*-th position. The students on adjacent positions mustn't have adjacent numbers. Formally, the following should be true: |*a**i*<=-<=*a**i*<=+<=1|<=≠<=1 for all *i* from 1 to *k*<=-<=1.
If there are several possible answers, output any of them.
|
[
"6",
"3\n"
] |
[
"6\n1 5 3 6 2 4",
"2\n1 3"
] |
none
| 500
|
[
{
"input": "6",
"output": "6\n5 3 1 6 4 2 "
},
{
"input": "3",
"output": "2\n1 3"
},
{
"input": "1",
"output": "1\n1 "
},
{
"input": "2",
"output": "1\n1"
},
{
"input": "4",
"output": "4\n3 1 4 2 "
},
{
"input": "5",
"output": "5\n5 3 1 4 2 "
},
{
"input": "7",
"output": "7\n7 5 3 1 6 4 2 "
},
{
"input": "8",
"output": "8\n7 5 3 1 8 6 4 2 "
},
{
"input": "9",
"output": "9\n9 7 5 3 1 8 6 4 2 "
},
{
"input": "10",
"output": "10\n9 7 5 3 1 10 8 6 4 2 "
},
{
"input": "13",
"output": "13\n13 11 9 7 5 3 1 12 10 8 6 4 2 "
},
{
"input": "16",
"output": "16\n15 13 11 9 7 5 3 1 16 14 12 10 8 6 4 2 "
},
{
"input": "25",
"output": "25\n25 23 21 19 17 15 13 11 9 7 5 3 1 24 22 20 18 16 14 12 10 8 6 4 2 "
},
{
"input": "29",
"output": "29\n29 27 25 23 21 19 17 15 13 11 9 7 5 3 1 28 26 24 22 20 18 16 14 12 10 8 6 4 2 "
},
{
"input": "120",
"output": "120\n119 117 115 113 111 109 107 105 103 101 99 97 95 93 91 89 87 85 83 81 79 77 75 73 71 69 67 65 63 61 59 57 55 53 51 49 47 45 43 41 39 37 35 33 31 29 27 25 23 21 19 17 15 13 11 9 7 5 3 1 120 118 116 114 112 110 108 106 104 102 100 98 96 94 92 90 88 86 84 82 80 78 76 74 72 70 68 66 64 62 60 58 56 54 52 50 48 46 44 42 40 38 36 34 32 30 28 26 24 22 20 18 16 14 12 10 8 6 4 2 "
},
{
"input": "128",
"output": "128\n127 125 123 121 119 117 115 113 111 109 107 105 103 101 99 97 95 93 91 89 87 85 83 81 79 77 75 73 71 69 67 65 63 61 59 57 55 53 51 49 47 45 43 41 39 37 35 33 31 29 27 25 23 21 19 17 15 13 11 9 7 5 3 1 128 126 124 122 120 118 116 114 112 110 108 106 104 102 100 98 96 94 92 90 88 86 84 82 80 78 76 74 72 70 68 66 64 62 60 58 56 54 52 50 48 46 44 42 40 38 36 34 32 30 28 26 24 22 20 18 16 14 12 10 8 6 4 2 "
},
{
"input": "216",
"output": "216\n215 213 211 209 207 205 203 201 199 197 195 193 191 189 187 185 183 181 179 177 175 173 171 169 167 165 163 161 159 157 155 153 151 149 147 145 143 141 139 137 135 133 131 129 127 125 123 121 119 117 115 113 111 109 107 105 103 101 99 97 95 93 91 89 87 85 83 81 79 77 75 73 71 69 67 65 63 61 59 57 55 53 51 49 47 45 43 41 39 37 35 33 31 29 27 25 23 21 19 17 15 13 11 9 7 5 3 1 216 214 212 210 208 206 204 202 200 198 196 194 192 190 188 186 184 182 180 178 176 174 172 170 168 166 164 162 160 158 156 154 1..."
},
{
"input": "729",
"output": "729\n729 727 725 723 721 719 717 715 713 711 709 707 705 703 701 699 697 695 693 691 689 687 685 683 681 679 677 675 673 671 669 667 665 663 661 659 657 655 653 651 649 647 645 643 641 639 637 635 633 631 629 627 625 623 621 619 617 615 613 611 609 607 605 603 601 599 597 595 593 591 589 587 585 583 581 579 577 575 573 571 569 567 565 563 561 559 557 555 553 551 549 547 545 543 541 539 537 535 533 531 529 527 525 523 521 519 517 515 513 511 509 507 505 503 501 499 497 495 493 491 489 487 485 483 481 479 47..."
},
{
"input": "1111",
"output": "1111\n1111 1109 1107 1105 1103 1101 1099 1097 1095 1093 1091 1089 1087 1085 1083 1081 1079 1077 1075 1073 1071 1069 1067 1065 1063 1061 1059 1057 1055 1053 1051 1049 1047 1045 1043 1041 1039 1037 1035 1033 1031 1029 1027 1025 1023 1021 1019 1017 1015 1013 1011 1009 1007 1005 1003 1001 999 997 995 993 991 989 987 985 983 981 979 977 975 973 971 969 967 965 963 961 959 957 955 953 951 949 947 945 943 941 939 937 935 933 931 929 927 925 923 921 919 917 915 913 911 909 907 905 903 901 899 897 895 893 891 889 8..."
},
{
"input": "1597",
"output": "1597\n1597 1595 1593 1591 1589 1587 1585 1583 1581 1579 1577 1575 1573 1571 1569 1567 1565 1563 1561 1559 1557 1555 1553 1551 1549 1547 1545 1543 1541 1539 1537 1535 1533 1531 1529 1527 1525 1523 1521 1519 1517 1515 1513 1511 1509 1507 1505 1503 1501 1499 1497 1495 1493 1491 1489 1487 1485 1483 1481 1479 1477 1475 1473 1471 1469 1467 1465 1463 1461 1459 1457 1455 1453 1451 1449 1447 1445 1443 1441 1439 1437 1435 1433 1431 1429 1427 1425 1423 1421 1419 1417 1415 1413 1411 1409 1407 1405 1403 1401 1399 1397 ..."
},
{
"input": "1777",
"output": "1777\n1777 1775 1773 1771 1769 1767 1765 1763 1761 1759 1757 1755 1753 1751 1749 1747 1745 1743 1741 1739 1737 1735 1733 1731 1729 1727 1725 1723 1721 1719 1717 1715 1713 1711 1709 1707 1705 1703 1701 1699 1697 1695 1693 1691 1689 1687 1685 1683 1681 1679 1677 1675 1673 1671 1669 1667 1665 1663 1661 1659 1657 1655 1653 1651 1649 1647 1645 1643 1641 1639 1637 1635 1633 1631 1629 1627 1625 1623 1621 1619 1617 1615 1613 1611 1609 1607 1605 1603 1601 1599 1597 1595 1593 1591 1589 1587 1585 1583 1581 1579 1577 ..."
},
{
"input": "2048",
"output": "2048\n2047 2045 2043 2041 2039 2037 2035 2033 2031 2029 2027 2025 2023 2021 2019 2017 2015 2013 2011 2009 2007 2005 2003 2001 1999 1997 1995 1993 1991 1989 1987 1985 1983 1981 1979 1977 1975 1973 1971 1969 1967 1965 1963 1961 1959 1957 1955 1953 1951 1949 1947 1945 1943 1941 1939 1937 1935 1933 1931 1929 1927 1925 1923 1921 1919 1917 1915 1913 1911 1909 1907 1905 1903 1901 1899 1897 1895 1893 1891 1889 1887 1885 1883 1881 1879 1877 1875 1873 1871 1869 1867 1865 1863 1861 1859 1857 1855 1853 1851 1849 1847 ..."
},
{
"input": "2999",
"output": "2999\n2999 2997 2995 2993 2991 2989 2987 2985 2983 2981 2979 2977 2975 2973 2971 2969 2967 2965 2963 2961 2959 2957 2955 2953 2951 2949 2947 2945 2943 2941 2939 2937 2935 2933 2931 2929 2927 2925 2923 2921 2919 2917 2915 2913 2911 2909 2907 2905 2903 2901 2899 2897 2895 2893 2891 2889 2887 2885 2883 2881 2879 2877 2875 2873 2871 2869 2867 2865 2863 2861 2859 2857 2855 2853 2851 2849 2847 2845 2843 2841 2839 2837 2835 2833 2831 2829 2827 2825 2823 2821 2819 2817 2815 2813 2811 2809 2807 2805 2803 2801 2799 ..."
},
{
"input": "3001",
"output": "3001\n3001 2999 2997 2995 2993 2991 2989 2987 2985 2983 2981 2979 2977 2975 2973 2971 2969 2967 2965 2963 2961 2959 2957 2955 2953 2951 2949 2947 2945 2943 2941 2939 2937 2935 2933 2931 2929 2927 2925 2923 2921 2919 2917 2915 2913 2911 2909 2907 2905 2903 2901 2899 2897 2895 2893 2891 2889 2887 2885 2883 2881 2879 2877 2875 2873 2871 2869 2867 2865 2863 2861 2859 2857 2855 2853 2851 2849 2847 2845 2843 2841 2839 2837 2835 2833 2831 2829 2827 2825 2823 2821 2819 2817 2815 2813 2811 2809 2807 2805 2803 2801 ..."
},
{
"input": "4181",
"output": "4181\n4181 4179 4177 4175 4173 4171 4169 4167 4165 4163 4161 4159 4157 4155 4153 4151 4149 4147 4145 4143 4141 4139 4137 4135 4133 4131 4129 4127 4125 4123 4121 4119 4117 4115 4113 4111 4109 4107 4105 4103 4101 4099 4097 4095 4093 4091 4089 4087 4085 4083 4081 4079 4077 4075 4073 4071 4069 4067 4065 4063 4061 4059 4057 4055 4053 4051 4049 4047 4045 4043 4041 4039 4037 4035 4033 4031 4029 4027 4025 4023 4021 4019 4017 4015 4013 4011 4009 4007 4005 4003 4001 3999 3997 3995 3993 3991 3989 3987 3985 3983 3981 ..."
},
{
"input": "4990",
"output": "4990\n4989 4987 4985 4983 4981 4979 4977 4975 4973 4971 4969 4967 4965 4963 4961 4959 4957 4955 4953 4951 4949 4947 4945 4943 4941 4939 4937 4935 4933 4931 4929 4927 4925 4923 4921 4919 4917 4915 4913 4911 4909 4907 4905 4903 4901 4899 4897 4895 4893 4891 4889 4887 4885 4883 4881 4879 4877 4875 4873 4871 4869 4867 4865 4863 4861 4859 4857 4855 4853 4851 4849 4847 4845 4843 4841 4839 4837 4835 4833 4831 4829 4827 4825 4823 4821 4819 4817 4815 4813 4811 4809 4807 4805 4803 4801 4799 4797 4795 4793 4791 4789 ..."
},
{
"input": "4991",
"output": "4991\n4991 4989 4987 4985 4983 4981 4979 4977 4975 4973 4971 4969 4967 4965 4963 4961 4959 4957 4955 4953 4951 4949 4947 4945 4943 4941 4939 4937 4935 4933 4931 4929 4927 4925 4923 4921 4919 4917 4915 4913 4911 4909 4907 4905 4903 4901 4899 4897 4895 4893 4891 4889 4887 4885 4883 4881 4879 4877 4875 4873 4871 4869 4867 4865 4863 4861 4859 4857 4855 4853 4851 4849 4847 4845 4843 4841 4839 4837 4835 4833 4831 4829 4827 4825 4823 4821 4819 4817 4815 4813 4811 4809 4807 4805 4803 4801 4799 4797 4795 4793 4791 ..."
},
{
"input": "4992",
"output": "4992\n4991 4989 4987 4985 4983 4981 4979 4977 4975 4973 4971 4969 4967 4965 4963 4961 4959 4957 4955 4953 4951 4949 4947 4945 4943 4941 4939 4937 4935 4933 4931 4929 4927 4925 4923 4921 4919 4917 4915 4913 4911 4909 4907 4905 4903 4901 4899 4897 4895 4893 4891 4889 4887 4885 4883 4881 4879 4877 4875 4873 4871 4869 4867 4865 4863 4861 4859 4857 4855 4853 4851 4849 4847 4845 4843 4841 4839 4837 4835 4833 4831 4829 4827 4825 4823 4821 4819 4817 4815 4813 4811 4809 4807 4805 4803 4801 4799 4797 4795 4793 4791 ..."
},
{
"input": "4993",
"output": "4993\n4993 4991 4989 4987 4985 4983 4981 4979 4977 4975 4973 4971 4969 4967 4965 4963 4961 4959 4957 4955 4953 4951 4949 4947 4945 4943 4941 4939 4937 4935 4933 4931 4929 4927 4925 4923 4921 4919 4917 4915 4913 4911 4909 4907 4905 4903 4901 4899 4897 4895 4893 4891 4889 4887 4885 4883 4881 4879 4877 4875 4873 4871 4869 4867 4865 4863 4861 4859 4857 4855 4853 4851 4849 4847 4845 4843 4841 4839 4837 4835 4833 4831 4829 4827 4825 4823 4821 4819 4817 4815 4813 4811 4809 4807 4805 4803 4801 4799 4797 4795 4793 ..."
},
{
"input": "4994",
"output": "4994\n4993 4991 4989 4987 4985 4983 4981 4979 4977 4975 4973 4971 4969 4967 4965 4963 4961 4959 4957 4955 4953 4951 4949 4947 4945 4943 4941 4939 4937 4935 4933 4931 4929 4927 4925 4923 4921 4919 4917 4915 4913 4911 4909 4907 4905 4903 4901 4899 4897 4895 4893 4891 4889 4887 4885 4883 4881 4879 4877 4875 4873 4871 4869 4867 4865 4863 4861 4859 4857 4855 4853 4851 4849 4847 4845 4843 4841 4839 4837 4835 4833 4831 4829 4827 4825 4823 4821 4819 4817 4815 4813 4811 4809 4807 4805 4803 4801 4799 4797 4795 4793 ..."
},
{
"input": "4995",
"output": "4995\n4995 4993 4991 4989 4987 4985 4983 4981 4979 4977 4975 4973 4971 4969 4967 4965 4963 4961 4959 4957 4955 4953 4951 4949 4947 4945 4943 4941 4939 4937 4935 4933 4931 4929 4927 4925 4923 4921 4919 4917 4915 4913 4911 4909 4907 4905 4903 4901 4899 4897 4895 4893 4891 4889 4887 4885 4883 4881 4879 4877 4875 4873 4871 4869 4867 4865 4863 4861 4859 4857 4855 4853 4851 4849 4847 4845 4843 4841 4839 4837 4835 4833 4831 4829 4827 4825 4823 4821 4819 4817 4815 4813 4811 4809 4807 4805 4803 4801 4799 4797 4795 ..."
},
{
"input": "4996",
"output": "4996\n4995 4993 4991 4989 4987 4985 4983 4981 4979 4977 4975 4973 4971 4969 4967 4965 4963 4961 4959 4957 4955 4953 4951 4949 4947 4945 4943 4941 4939 4937 4935 4933 4931 4929 4927 4925 4923 4921 4919 4917 4915 4913 4911 4909 4907 4905 4903 4901 4899 4897 4895 4893 4891 4889 4887 4885 4883 4881 4879 4877 4875 4873 4871 4869 4867 4865 4863 4861 4859 4857 4855 4853 4851 4849 4847 4845 4843 4841 4839 4837 4835 4833 4831 4829 4827 4825 4823 4821 4819 4817 4815 4813 4811 4809 4807 4805 4803 4801 4799 4797 4795 ..."
},
{
"input": "4997",
"output": "4997\n4997 4995 4993 4991 4989 4987 4985 4983 4981 4979 4977 4975 4973 4971 4969 4967 4965 4963 4961 4959 4957 4955 4953 4951 4949 4947 4945 4943 4941 4939 4937 4935 4933 4931 4929 4927 4925 4923 4921 4919 4917 4915 4913 4911 4909 4907 4905 4903 4901 4899 4897 4895 4893 4891 4889 4887 4885 4883 4881 4879 4877 4875 4873 4871 4869 4867 4865 4863 4861 4859 4857 4855 4853 4851 4849 4847 4845 4843 4841 4839 4837 4835 4833 4831 4829 4827 4825 4823 4821 4819 4817 4815 4813 4811 4809 4807 4805 4803 4801 4799 4797 ..."
},
{
"input": "4998",
"output": "4998\n4997 4995 4993 4991 4989 4987 4985 4983 4981 4979 4977 4975 4973 4971 4969 4967 4965 4963 4961 4959 4957 4955 4953 4951 4949 4947 4945 4943 4941 4939 4937 4935 4933 4931 4929 4927 4925 4923 4921 4919 4917 4915 4913 4911 4909 4907 4905 4903 4901 4899 4897 4895 4893 4891 4889 4887 4885 4883 4881 4879 4877 4875 4873 4871 4869 4867 4865 4863 4861 4859 4857 4855 4853 4851 4849 4847 4845 4843 4841 4839 4837 4835 4833 4831 4829 4827 4825 4823 4821 4819 4817 4815 4813 4811 4809 4807 4805 4803 4801 4799 4797 ..."
},
{
"input": "4999",
"output": "4999\n4999 4997 4995 4993 4991 4989 4987 4985 4983 4981 4979 4977 4975 4973 4971 4969 4967 4965 4963 4961 4959 4957 4955 4953 4951 4949 4947 4945 4943 4941 4939 4937 4935 4933 4931 4929 4927 4925 4923 4921 4919 4917 4915 4913 4911 4909 4907 4905 4903 4901 4899 4897 4895 4893 4891 4889 4887 4885 4883 4881 4879 4877 4875 4873 4871 4869 4867 4865 4863 4861 4859 4857 4855 4853 4851 4849 4847 4845 4843 4841 4839 4837 4835 4833 4831 4829 4827 4825 4823 4821 4819 4817 4815 4813 4811 4809 4807 4805 4803 4801 4799 ..."
},
{
"input": "5000",
"output": "5000\n4999 4997 4995 4993 4991 4989 4987 4985 4983 4981 4979 4977 4975 4973 4971 4969 4967 4965 4963 4961 4959 4957 4955 4953 4951 4949 4947 4945 4943 4941 4939 4937 4935 4933 4931 4929 4927 4925 4923 4921 4919 4917 4915 4913 4911 4909 4907 4905 4903 4901 4899 4897 4895 4893 4891 4889 4887 4885 4883 4881 4879 4877 4875 4873 4871 4869 4867 4865 4863 4861 4859 4857 4855 4853 4851 4849 4847 4845 4843 4841 4839 4837 4835 4833 4831 4829 4827 4825 4823 4821 4819 4817 4815 4813 4811 4809 4807 4805 4803 4801 4799 ..."
}
] | 1,477,562,599
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 36
| 77
| 921,600
|
import functools
n=int(input())
if n==1 or n==2:
print(1)
print(1)
exit(0)
elif n==3:
print(2)
print(1,3)
exit(0)
odds=list(filter(lambda x:True if x%2==1 else False, range(1,n+1)))
evens=list(filter(lambda x:True if x%2==0 else False, range(1,n+1)))
odds.reverse()
evens.reverse()
od=' '.join(map(str,odds))
ev=' '.join(map(str,evens))
print(n)
print(od+" "+ev)
|
Title: Exam
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
An exam for *n* students will take place in a long and narrow room, so the students will sit in a line in some order. The teacher suspects that students with adjacent numbers (*i* and *i*<=+<=1) always studied side by side and became friends and if they take an exam sitting next to each other, they will help each other for sure.
Your task is to choose the maximum number of students and make such an arrangement of students in the room that no two students with adjacent numbers sit side by side.
Input Specification:
A single line contains integer *n* (1<=≤<=*n*<=≤<=5000) — the number of students at an exam.
Output Specification:
In the first line print integer *k* — the maximum number of students who can be seated so that no two students with adjacent numbers sit next to each other.
In the second line print *k* distinct integers *a*1,<=*a*2,<=...,<=*a**k* (1<=≤<=*a**i*<=≤<=*n*), where *a**i* is the number of the student on the *i*-th position. The students on adjacent positions mustn't have adjacent numbers. Formally, the following should be true: |*a**i*<=-<=*a**i*<=+<=1|<=≠<=1 for all *i* from 1 to *k*<=-<=1.
If there are several possible answers, output any of them.
Demo Input:
['6', '3\n']
Demo Output:
['6\n1 5 3 6 2 4', '2\n1 3']
Note:
none
|
```python
import functools
n=int(input())
if n==1 or n==2:
print(1)
print(1)
exit(0)
elif n==3:
print(2)
print(1,3)
exit(0)
odds=list(filter(lambda x:True if x%2==1 else False, range(1,n+1)))
evens=list(filter(lambda x:True if x%2==0 else False, range(1,n+1)))
odds.reverse()
evens.reverse()
od=' '.join(map(str,odds))
ev=' '.join(map(str,evens))
print(n)
print(od+" "+ev)
```
| 3
|
|
39
|
D
|
Cubical Planet
|
PROGRAMMING
| 1,100
|
[
"math"
] |
D. Cubical Planet
|
2
|
64
|
You can find anything whatsoever in our Galaxy! A cubical planet goes round an icosahedral star. Let us introduce a system of axes so that the edges of the cubical planet are parallel to the coordinate axes and two opposite vertices lay in the points (0,<=0,<=0) and (1,<=1,<=1). Two flies live on the planet. At the moment they are sitting on two different vertices of the cubical planet. Your task is to determine whether they see each other or not. The flies see each other when the vertices they occupy lie on the same face of the cube.
|
The first line contains three space-separated integers (0 or 1) — the coordinates of the first fly, the second line analogously contains the coordinates of the second fly.
|
Output "YES" (without quotes) if the flies see each other. Otherwise, output "NO".
|
[
"0 0 0\n0 1 0\n",
"1 1 0\n0 1 0\n",
"0 0 0\n1 1 1\n"
] |
[
"YES\n",
"YES\n",
"NO\n"
] |
none
| 0
|
[
{
"input": "0 0 0\n0 1 0",
"output": "YES"
},
{
"input": "1 1 0\n0 1 0",
"output": "YES"
},
{
"input": "0 0 0\n1 1 1",
"output": "NO"
},
{
"input": "0 0 0\n1 0 0",
"output": "YES"
},
{
"input": "0 0 0\n0 1 0",
"output": "YES"
},
{
"input": "0 0 0\n1 1 0",
"output": "YES"
},
{
"input": "0 0 0\n0 0 1",
"output": "YES"
},
{
"input": "0 0 0\n1 0 1",
"output": "YES"
},
{
"input": "0 0 0\n0 1 1",
"output": "YES"
},
{
"input": "0 0 0\n1 1 1",
"output": "NO"
},
{
"input": "1 0 0\n0 0 0",
"output": "YES"
},
{
"input": "1 0 0\n0 1 0",
"output": "YES"
},
{
"input": "1 0 0\n1 1 0",
"output": "YES"
},
{
"input": "1 0 0\n0 0 1",
"output": "YES"
},
{
"input": "1 0 0\n1 0 1",
"output": "YES"
},
{
"input": "1 0 0\n0 1 1",
"output": "NO"
},
{
"input": "1 0 0\n1 1 1",
"output": "YES"
},
{
"input": "0 1 0\n0 0 0",
"output": "YES"
},
{
"input": "0 1 0\n1 0 0",
"output": "YES"
},
{
"input": "0 1 0\n1 1 0",
"output": "YES"
},
{
"input": "0 1 0\n0 0 1",
"output": "YES"
},
{
"input": "0 1 0\n1 0 1",
"output": "NO"
},
{
"input": "0 1 0\n0 1 1",
"output": "YES"
},
{
"input": "0 1 0\n1 1 1",
"output": "YES"
},
{
"input": "1 1 0\n0 0 0",
"output": "YES"
},
{
"input": "1 1 0\n1 0 0",
"output": "YES"
},
{
"input": "1 1 0\n0 1 0",
"output": "YES"
},
{
"input": "1 1 0\n0 0 1",
"output": "NO"
},
{
"input": "1 1 0\n1 0 1",
"output": "YES"
},
{
"input": "1 1 0\n0 1 1",
"output": "YES"
},
{
"input": "1 1 0\n1 1 1",
"output": "YES"
},
{
"input": "0 0 1\n0 0 0",
"output": "YES"
},
{
"input": "0 0 1\n1 0 0",
"output": "YES"
},
{
"input": "0 0 1\n0 1 0",
"output": "YES"
},
{
"input": "0 0 1\n1 1 0",
"output": "NO"
},
{
"input": "0 0 1\n1 0 1",
"output": "YES"
},
{
"input": "0 0 1\n0 1 1",
"output": "YES"
},
{
"input": "0 0 1\n1 1 1",
"output": "YES"
},
{
"input": "1 0 1\n0 0 0",
"output": "YES"
},
{
"input": "1 0 1\n1 0 0",
"output": "YES"
},
{
"input": "1 0 1\n0 1 0",
"output": "NO"
},
{
"input": "1 0 1\n1 1 0",
"output": "YES"
},
{
"input": "1 0 1\n0 0 1",
"output": "YES"
},
{
"input": "1 0 1\n0 1 1",
"output": "YES"
},
{
"input": "1 0 1\n1 1 1",
"output": "YES"
},
{
"input": "0 1 1\n0 0 0",
"output": "YES"
},
{
"input": "0 1 1\n1 0 0",
"output": "NO"
},
{
"input": "0 1 1\n0 1 0",
"output": "YES"
},
{
"input": "0 1 1\n1 1 0",
"output": "YES"
},
{
"input": "0 1 1\n0 0 1",
"output": "YES"
},
{
"input": "0 1 1\n1 0 1",
"output": "YES"
},
{
"input": "0 1 1\n1 1 1",
"output": "YES"
},
{
"input": "1 1 1\n0 0 0",
"output": "NO"
},
{
"input": "1 1 1\n1 0 0",
"output": "YES"
},
{
"input": "1 1 1\n0 1 0",
"output": "YES"
},
{
"input": "1 1 1\n1 1 0",
"output": "YES"
},
{
"input": "1 1 1\n0 0 1",
"output": "YES"
},
{
"input": "1 1 1\n1 0 1",
"output": "YES"
},
{
"input": "1 1 1\n0 1 1",
"output": "YES"
}
] | 1,523,603,745
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 59
| 186
| 7,372,800
|
# from dust i have come dust i will be
'''
if the coordinates lie in the same plane then yes
'''
a,b,c=map(int,input().split())
x,y,z=map(int,input().split())
x=a-x
y=b-y
z=c-z
dist=x**2+y**2+z**2
if dist==3:
print('NO')
else:
print('YES')
|
Title: Cubical Planet
Time Limit: 2 seconds
Memory Limit: 64 megabytes
Problem Description:
You can find anything whatsoever in our Galaxy! A cubical planet goes round an icosahedral star. Let us introduce a system of axes so that the edges of the cubical planet are parallel to the coordinate axes and two opposite vertices lay in the points (0,<=0,<=0) and (1,<=1,<=1). Two flies live on the planet. At the moment they are sitting on two different vertices of the cubical planet. Your task is to determine whether they see each other or not. The flies see each other when the vertices they occupy lie on the same face of the cube.
Input Specification:
The first line contains three space-separated integers (0 or 1) — the coordinates of the first fly, the second line analogously contains the coordinates of the second fly.
Output Specification:
Output "YES" (without quotes) if the flies see each other. Otherwise, output "NO".
Demo Input:
['0 0 0\n0 1 0\n', '1 1 0\n0 1 0\n', '0 0 0\n1 1 1\n']
Demo Output:
['YES\n', 'YES\n', 'NO\n']
Note:
none
|
```python
# from dust i have come dust i will be
'''
if the coordinates lie in the same plane then yes
'''
a,b,c=map(int,input().split())
x,y,z=map(int,input().split())
x=a-x
y=b-y
z=c-z
dist=x**2+y**2+z**2
if dist==3:
print('NO')
else:
print('YES')
```
| 3.898568
|
53
|
C
|
Little Frog
|
PROGRAMMING
| 1,200
|
[
"constructive algorithms"
] |
C. Little Frog
|
2
|
256
|
Once upon a time a little frog whose name was Vasya decided to travel around his home swamp. Overall there are *n* mounds on the swamp, located on one line. The distance between the neighboring mounds is one meter. Vasya wants to visit all the mounds in one day; besides, he wants to visit each one exactly once. For that he makes a route plan, to decide the order in which to jump on the mounds. Vasya can pick any mound as the first one. He thinks it boring to jump two times at the same distance. That's why he wants any two jumps on his route to have different lengths. Help Vasya the Frog and make the plan for him.
|
The single line contains a number *n* (1<=≤<=*n*<=≤<=104) which is the number of mounds.
|
Print *n* integers *p**i* (1<=≤<=*p**i*<=≤<=*n*) which are the frog's route plan.
- All the *p**i*'s should be mutually different. - All the |*p**i*–*p**i*<=+<=1|'s should be mutually different (1<=≤<=*i*<=≤<=*n*<=-<=1).
If there are several solutions, output any.
|
[
"2\n",
"3\n"
] |
[
"1 2 ",
"1 3 2 "
] |
none
| 1,500
|
[
{
"input": "2",
"output": "1 2 "
},
{
"input": "3",
"output": "1 3 2 "
},
{
"input": "4",
"output": "1 4 2 3 "
},
{
"input": "5",
"output": "1 5 2 4 3 "
},
{
"input": "6",
"output": "1 6 2 5 3 4 "
},
{
"input": "1",
"output": "1 "
},
{
"input": "9149",
"output": "1 9149 2 9148 3 9147 4 9146 5 9145 6 9144 7 9143 8 9142 9 9141 10 9140 11 9139 12 9138 13 9137 14 9136 15 9135 16 9134 17 9133 18 9132 19 9131 20 9130 21 9129 22 9128 23 9127 24 9126 25 9125 26 9124 27 9123 28 9122 29 9121 30 9120 31 9119 32 9118 33 9117 34 9116 35 9115 36 9114 37 9113 38 9112 39 9111 40 9110 41 9109 42 9108 43 9107 44 9106 45 9105 46 9104 47 9103 48 9102 49 9101 50 9100 51 9099 52 9098 53 9097 54 9096 55 9095 56 9094 57 9093 58 9092 59 9091 60 9090 61 9089 62 9088 63 9087 64 9086 65 9085 ..."
},
{
"input": "2877",
"output": "1 2877 2 2876 3 2875 4 2874 5 2873 6 2872 7 2871 8 2870 9 2869 10 2868 11 2867 12 2866 13 2865 14 2864 15 2863 16 2862 17 2861 18 2860 19 2859 20 2858 21 2857 22 2856 23 2855 24 2854 25 2853 26 2852 27 2851 28 2850 29 2849 30 2848 31 2847 32 2846 33 2845 34 2844 35 2843 36 2842 37 2841 38 2840 39 2839 40 2838 41 2837 42 2836 43 2835 44 2834 45 2833 46 2832 47 2831 48 2830 49 2829 50 2828 51 2827 52 2826 53 2825 54 2824 55 2823 56 2822 57 2821 58 2820 59 2819 60 2818 61 2817 62 2816 63 2815 64 2814 65 2813 ..."
},
{
"input": "2956",
"output": "1 2956 2 2955 3 2954 4 2953 5 2952 6 2951 7 2950 8 2949 9 2948 10 2947 11 2946 12 2945 13 2944 14 2943 15 2942 16 2941 17 2940 18 2939 19 2938 20 2937 21 2936 22 2935 23 2934 24 2933 25 2932 26 2931 27 2930 28 2929 29 2928 30 2927 31 2926 32 2925 33 2924 34 2923 35 2922 36 2921 37 2920 38 2919 39 2918 40 2917 41 2916 42 2915 43 2914 44 2913 45 2912 46 2911 47 2910 48 2909 49 2908 50 2907 51 2906 52 2905 53 2904 54 2903 55 2902 56 2901 57 2900 58 2899 59 2898 60 2897 61 2896 62 2895 63 2894 64 2893 65 2892 ..."
},
{
"input": "3035",
"output": "1 3035 2 3034 3 3033 4 3032 5 3031 6 3030 7 3029 8 3028 9 3027 10 3026 11 3025 12 3024 13 3023 14 3022 15 3021 16 3020 17 3019 18 3018 19 3017 20 3016 21 3015 22 3014 23 3013 24 3012 25 3011 26 3010 27 3009 28 3008 29 3007 30 3006 31 3005 32 3004 33 3003 34 3002 35 3001 36 3000 37 2999 38 2998 39 2997 40 2996 41 2995 42 2994 43 2993 44 2992 45 2991 46 2990 47 2989 48 2988 49 2987 50 2986 51 2985 52 2984 53 2983 54 2982 55 2981 56 2980 57 2979 58 2978 59 2977 60 2976 61 2975 62 2974 63 2973 64 2972 65 2971 ..."
},
{
"input": "3114",
"output": "1 3114 2 3113 3 3112 4 3111 5 3110 6 3109 7 3108 8 3107 9 3106 10 3105 11 3104 12 3103 13 3102 14 3101 15 3100 16 3099 17 3098 18 3097 19 3096 20 3095 21 3094 22 3093 23 3092 24 3091 25 3090 26 3089 27 3088 28 3087 29 3086 30 3085 31 3084 32 3083 33 3082 34 3081 35 3080 36 3079 37 3078 38 3077 39 3076 40 3075 41 3074 42 3073 43 3072 44 3071 45 3070 46 3069 47 3068 48 3067 49 3066 50 3065 51 3064 52 3063 53 3062 54 3061 55 3060 56 3059 57 3058 58 3057 59 3056 60 3055 61 3054 62 3053 63 3052 64 3051 65 3050 ..."
},
{
"input": "3193",
"output": "1 3193 2 3192 3 3191 4 3190 5 3189 6 3188 7 3187 8 3186 9 3185 10 3184 11 3183 12 3182 13 3181 14 3180 15 3179 16 3178 17 3177 18 3176 19 3175 20 3174 21 3173 22 3172 23 3171 24 3170 25 3169 26 3168 27 3167 28 3166 29 3165 30 3164 31 3163 32 3162 33 3161 34 3160 35 3159 36 3158 37 3157 38 3156 39 3155 40 3154 41 3153 42 3152 43 3151 44 3150 45 3149 46 3148 47 3147 48 3146 49 3145 50 3144 51 3143 52 3142 53 3141 54 3140 55 3139 56 3138 57 3137 58 3136 59 3135 60 3134 61 3133 62 3132 63 3131 64 3130 65 3129 ..."
},
{
"input": "3273",
"output": "1 3273 2 3272 3 3271 4 3270 5 3269 6 3268 7 3267 8 3266 9 3265 10 3264 11 3263 12 3262 13 3261 14 3260 15 3259 16 3258 17 3257 18 3256 19 3255 20 3254 21 3253 22 3252 23 3251 24 3250 25 3249 26 3248 27 3247 28 3246 29 3245 30 3244 31 3243 32 3242 33 3241 34 3240 35 3239 36 3238 37 3237 38 3236 39 3235 40 3234 41 3233 42 3232 43 3231 44 3230 45 3229 46 3228 47 3227 48 3226 49 3225 50 3224 51 3223 52 3222 53 3221 54 3220 55 3219 56 3218 57 3217 58 3216 59 3215 60 3214 61 3213 62 3212 63 3211 64 3210 65 3209 ..."
},
{
"input": "7000",
"output": "1 7000 2 6999 3 6998 4 6997 5 6996 6 6995 7 6994 8 6993 9 6992 10 6991 11 6990 12 6989 13 6988 14 6987 15 6986 16 6985 17 6984 18 6983 19 6982 20 6981 21 6980 22 6979 23 6978 24 6977 25 6976 26 6975 27 6974 28 6973 29 6972 30 6971 31 6970 32 6969 33 6968 34 6967 35 6966 36 6965 37 6964 38 6963 39 6962 40 6961 41 6960 42 6959 43 6958 44 6957 45 6956 46 6955 47 6954 48 6953 49 6952 50 6951 51 6950 52 6949 53 6948 54 6947 55 6946 56 6945 57 6944 58 6943 59 6942 60 6941 61 6940 62 6939 63 6938 64 6937 65 6936 ..."
},
{
"input": "7079",
"output": "1 7079 2 7078 3 7077 4 7076 5 7075 6 7074 7 7073 8 7072 9 7071 10 7070 11 7069 12 7068 13 7067 14 7066 15 7065 16 7064 17 7063 18 7062 19 7061 20 7060 21 7059 22 7058 23 7057 24 7056 25 7055 26 7054 27 7053 28 7052 29 7051 30 7050 31 7049 32 7048 33 7047 34 7046 35 7045 36 7044 37 7043 38 7042 39 7041 40 7040 41 7039 42 7038 43 7037 44 7036 45 7035 46 7034 47 7033 48 7032 49 7031 50 7030 51 7029 52 7028 53 7027 54 7026 55 7025 56 7024 57 7023 58 7022 59 7021 60 7020 61 7019 62 7018 63 7017 64 7016 65 7015 ..."
},
{
"input": "4653",
"output": "1 4653 2 4652 3 4651 4 4650 5 4649 6 4648 7 4647 8 4646 9 4645 10 4644 11 4643 12 4642 13 4641 14 4640 15 4639 16 4638 17 4637 18 4636 19 4635 20 4634 21 4633 22 4632 23 4631 24 4630 25 4629 26 4628 27 4627 28 4626 29 4625 30 4624 31 4623 32 4622 33 4621 34 4620 35 4619 36 4618 37 4617 38 4616 39 4615 40 4614 41 4613 42 4612 43 4611 44 4610 45 4609 46 4608 47 4607 48 4606 49 4605 50 4604 51 4603 52 4602 53 4601 54 4600 55 4599 56 4598 57 4597 58 4596 59 4595 60 4594 61 4593 62 4592 63 4591 64 4590 65 4589 ..."
},
{
"input": "9995",
"output": "1 9995 2 9994 3 9993 4 9992 5 9991 6 9990 7 9989 8 9988 9 9987 10 9986 11 9985 12 9984 13 9983 14 9982 15 9981 16 9980 17 9979 18 9978 19 9977 20 9976 21 9975 22 9974 23 9973 24 9972 25 9971 26 9970 27 9969 28 9968 29 9967 30 9966 31 9965 32 9964 33 9963 34 9962 35 9961 36 9960 37 9959 38 9958 39 9957 40 9956 41 9955 42 9954 43 9953 44 9952 45 9951 46 9950 47 9949 48 9948 49 9947 50 9946 51 9945 52 9944 53 9943 54 9942 55 9941 56 9940 57 9939 58 9938 59 9937 60 9936 61 9935 62 9934 63 9933 64 9932 65 9931 ..."
},
{
"input": "9996",
"output": "1 9996 2 9995 3 9994 4 9993 5 9992 6 9991 7 9990 8 9989 9 9988 10 9987 11 9986 12 9985 13 9984 14 9983 15 9982 16 9981 17 9980 18 9979 19 9978 20 9977 21 9976 22 9975 23 9974 24 9973 25 9972 26 9971 27 9970 28 9969 29 9968 30 9967 31 9966 32 9965 33 9964 34 9963 35 9962 36 9961 37 9960 38 9959 39 9958 40 9957 41 9956 42 9955 43 9954 44 9953 45 9952 46 9951 47 9950 48 9949 49 9948 50 9947 51 9946 52 9945 53 9944 54 9943 55 9942 56 9941 57 9940 58 9939 59 9938 60 9937 61 9936 62 9935 63 9934 64 9933 65 9932 ..."
},
{
"input": "9997",
"output": "1 9997 2 9996 3 9995 4 9994 5 9993 6 9992 7 9991 8 9990 9 9989 10 9988 11 9987 12 9986 13 9985 14 9984 15 9983 16 9982 17 9981 18 9980 19 9979 20 9978 21 9977 22 9976 23 9975 24 9974 25 9973 26 9972 27 9971 28 9970 29 9969 30 9968 31 9967 32 9966 33 9965 34 9964 35 9963 36 9962 37 9961 38 9960 39 9959 40 9958 41 9957 42 9956 43 9955 44 9954 45 9953 46 9952 47 9951 48 9950 49 9949 50 9948 51 9947 52 9946 53 9945 54 9944 55 9943 56 9942 57 9941 58 9940 59 9939 60 9938 61 9937 62 9936 63 9935 64 9934 65 9933 ..."
},
{
"input": "9998",
"output": "1 9998 2 9997 3 9996 4 9995 5 9994 6 9993 7 9992 8 9991 9 9990 10 9989 11 9988 12 9987 13 9986 14 9985 15 9984 16 9983 17 9982 18 9981 19 9980 20 9979 21 9978 22 9977 23 9976 24 9975 25 9974 26 9973 27 9972 28 9971 29 9970 30 9969 31 9968 32 9967 33 9966 34 9965 35 9964 36 9963 37 9962 38 9961 39 9960 40 9959 41 9958 42 9957 43 9956 44 9955 45 9954 46 9953 47 9952 48 9951 49 9950 50 9949 51 9948 52 9947 53 9946 54 9945 55 9944 56 9943 57 9942 58 9941 59 9940 60 9939 61 9938 62 9937 63 9936 64 9935 65 9934 ..."
},
{
"input": "9999",
"output": "1 9999 2 9998 3 9997 4 9996 5 9995 6 9994 7 9993 8 9992 9 9991 10 9990 11 9989 12 9988 13 9987 14 9986 15 9985 16 9984 17 9983 18 9982 19 9981 20 9980 21 9979 22 9978 23 9977 24 9976 25 9975 26 9974 27 9973 28 9972 29 9971 30 9970 31 9969 32 9968 33 9967 34 9966 35 9965 36 9964 37 9963 38 9962 39 9961 40 9960 41 9959 42 9958 43 9957 44 9956 45 9955 46 9954 47 9953 48 9952 49 9951 50 9950 51 9949 52 9948 53 9947 54 9946 55 9945 56 9944 57 9943 58 9942 59 9941 60 9940 61 9939 62 9938 63 9937 64 9936 65 9935 ..."
},
{
"input": "10000",
"output": "1 10000 2 9999 3 9998 4 9997 5 9996 6 9995 7 9994 8 9993 9 9992 10 9991 11 9990 12 9989 13 9988 14 9987 15 9986 16 9985 17 9984 18 9983 19 9982 20 9981 21 9980 22 9979 23 9978 24 9977 25 9976 26 9975 27 9974 28 9973 29 9972 30 9971 31 9970 32 9969 33 9968 34 9967 35 9966 36 9965 37 9964 38 9963 39 9962 40 9961 41 9960 42 9959 43 9958 44 9957 45 9956 46 9955 47 9954 48 9953 49 9952 50 9951 51 9950 52 9949 53 9948 54 9947 55 9946 56 9945 57 9944 58 9943 59 9942 60 9941 61 9940 62 9939 63 9938 64 9937 65 9936..."
}
] | 1,607,918,081
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 22
| 372
| 4,403,200
|
import sys
input = sys.stdin.readline
for _ in range(1):
n=int(input())
l,r=1,n
while l<r:
print(l,end=' ')
print(r,end=' ')
l+=1
r-=1
if l==r:
print(l)
|
Title: Little Frog
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Once upon a time a little frog whose name was Vasya decided to travel around his home swamp. Overall there are *n* mounds on the swamp, located on one line. The distance between the neighboring mounds is one meter. Vasya wants to visit all the mounds in one day; besides, he wants to visit each one exactly once. For that he makes a route plan, to decide the order in which to jump on the mounds. Vasya can pick any mound as the first one. He thinks it boring to jump two times at the same distance. That's why he wants any two jumps on his route to have different lengths. Help Vasya the Frog and make the plan for him.
Input Specification:
The single line contains a number *n* (1<=≤<=*n*<=≤<=104) which is the number of mounds.
Output Specification:
Print *n* integers *p**i* (1<=≤<=*p**i*<=≤<=*n*) which are the frog's route plan.
- All the *p**i*'s should be mutually different. - All the |*p**i*–*p**i*<=+<=1|'s should be mutually different (1<=≤<=*i*<=≤<=*n*<=-<=1).
If there are several solutions, output any.
Demo Input:
['2\n', '3\n']
Demo Output:
['1 2 ', '1 3 2 ']
Note:
none
|
```python
import sys
input = sys.stdin.readline
for _ in range(1):
n=int(input())
l,r=1,n
while l<r:
print(l,end=' ')
print(r,end=' ')
l+=1
r-=1
if l==r:
print(l)
```
| 3.898798
|
989
|
A
|
A Blend of Springtime
|
PROGRAMMING
| 900
|
[
"implementation",
"strings"
] | null | null |
"What a pity it's already late spring," sighs Mino with regret, "one more drizzling night and they'd be gone."
"But these blends are at their best, aren't they?" Absorbed in the landscape, Kanno remains optimistic.
The landscape can be expressed as a row of consecutive cells, each of which either contains a flower of colour amber or buff or canary yellow, or is empty.
When a flower withers, it disappears from the cell that it originally belonged to, and it spreads petals of its colour in its two neighbouring cells (or outside the field if the cell is on the side of the landscape). In case petals fall outside the given cells, they simply become invisible.
You are to help Kanno determine whether it's possible that after some (possibly none or all) flowers shed their petals, at least one of the cells contains all three colours, considering both petals and flowers. Note that flowers can wither in arbitrary order.
|
The first and only line of input contains a non-empty string $s$ consisting of uppercase English letters 'A', 'B', 'C' and characters '.' (dots) only ($\lvert s \rvert \leq 100$) — denoting cells containing an amber flower, a buff one, a canary yellow one, and no flowers, respectively.
|
Output "Yes" if it's possible that all three colours appear in some cell, and "No" otherwise.
You can print each letter in any case (upper or lower).
|
[
".BAC.\n",
"AA..CB\n"
] |
[
"Yes\n",
"No\n"
] |
In the first example, the buff and canary yellow flowers can leave their petals in the central cell, blending all three colours in it.
In the second example, it's impossible to satisfy the requirement because there is no way that amber and buff meet in any cell.
| 500
|
[
{
"input": ".BAC.",
"output": "Yes"
},
{
"input": "AA..CB",
"output": "No"
},
{
"input": ".",
"output": "No"
},
{
"input": "ACB.AAAAAA",
"output": "Yes"
},
{
"input": "B.BC.BBBCA",
"output": "Yes"
},
{
"input": "BA..CAB..B",
"output": "Yes"
},
{
"input": "CACCBAA.BC",
"output": "Yes"
},
{
"input": ".CAACCBBA.CBB.AC..BABCCBCCB..B.BC..CBC.CA.CC.C.CC.B.A.CC.BBCCBB..ACAACAC.CBCCB.AABAAC.CBCC.BA..CCBC.",
"output": "Yes"
},
{
"input": "A",
"output": "No"
},
{
"input": "..",
"output": "No"
},
{
"input": "BC",
"output": "No"
},
{
"input": "CAB",
"output": "Yes"
},
{
"input": "A.CB",
"output": "No"
},
{
"input": "B.ACAA.CA..CBCBBAA.B.CCBCB.CAC.ABC...BC.BCCC.BC.CB",
"output": "Yes"
},
{
"input": "B.B...CC.B..CCCB.CB..CBCB..CBCC.CCBC.B.CB..CA.C.C.",
"output": "No"
},
{
"input": "AA.CBAABABCCC..B..B.ABBABAB.B.B.CCA..CB.B...A..CBC",
"output": "Yes"
},
{
"input": "CA.ABB.CC.B.C.BBBABAAB.BBBAACACAAA.C.AACA.AAC.C.BCCB.CCBC.C..CCACA.CBCCB.CCAABAAB.AACAA..A.AAA.",
"output": "No"
},
{
"input": "CBC...AC.BBBB.BBABABA.CAAACC.AAABB..A.BA..BC.CBBBC.BBBBCCCAA.ACCBB.AB.C.BA..CC..AAAC...AB.A.AAABBA.A",
"output": "No"
},
{
"input": "CC.AAAC.BA.BBB.AABABBCCAA.A.CBCCB.B.BC.ABCBCBBAA.CACA.CCCA.CB.CCB.A.BCCCB...C.A.BCCBC..B.ABABB.C.BCB",
"output": "Yes"
},
{
"input": "CCC..A..CACACCA.CA.ABAAB.BBA..C.AAA...ACB.ACA.CA.B.AB.A..C.BC.BC.A.C....ABBCCACCCBCC.BBBAA.ACCACB.BB",
"output": "Yes"
},
{
"input": "BC.ABACAACC..AC.A..CCCAABBCCACAC.AA.CC.BAABABABBCBB.BA..C.C.C.A.BBA.C..BC.ACACCC.AAAACCCCC.AAC.AC.AB",
"output": "Yes"
},
{
"input": "ACAC.BAA.C..CAAC..ABBAACC..BAA...CC...ACCBBCA.BAABABAACCAC.A.BBCACCC..BCB.BABAAAACCBCB.BCAABBC.C.BBB",
"output": "Yes"
},
{
"input": "CCAC.BCBC.A.ABBAB.C.C.BC.CCABBCBCCBC..B.AA.C.BC...B..BAA.ACCCCBBB.AAAACA.CAACCB.CCB.CC.BCCAB.BBBBABB",
"output": "Yes"
},
{
"input": ".AACAA.AAAAC.BBBB.BC...CCACCACAAA.A..CCA..BCC.AB.ABAAB..AABA...B.C.CBAB.BAAB.A.C.AAC.BBBA.ACAAA.BB.C",
"output": "Yes"
},
{
"input": "CC.ACCC.BCCCCAA.BBAACB.ABABAAAA.A.CBAB.CBACBBC..C.CA.AAA..AA..ABBB.A.C..CBBCAAACC.B..CC.AC..CAABACB.",
"output": "Yes"
},
{
"input": ".BAB.",
"output": "No"
},
{
"input": "BBBBBBBBB",
"output": "No"
},
{
"input": "..AAC..",
"output": "No"
},
{
"input": ".AAABBBCCC.",
"output": "No"
},
{
"input": "AAABC",
"output": "Yes"
},
{
"input": "BBB",
"output": "No"
},
{
"input": "AAAAABABAAAAA",
"output": "No"
},
{
"input": "AABBCC",
"output": "No"
},
{
"input": ".BA",
"output": "No"
},
{
"input": "CAAAAB",
"output": "No"
}
] | 1,528,724,514
| 414
|
Python 3
|
OK
|
TESTS
| 37
| 92
| 0
|
r=input()
a=['A','B','C']
le=len(r)
for i in range(le-2):
l=list(r[i:i+3])
l.sort()
if l==a:
print ("Yes")
exit(0)
print ("No")
|
Title: A Blend of Springtime
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
"What a pity it's already late spring," sighs Mino with regret, "one more drizzling night and they'd be gone."
"But these blends are at their best, aren't they?" Absorbed in the landscape, Kanno remains optimistic.
The landscape can be expressed as a row of consecutive cells, each of which either contains a flower of colour amber or buff or canary yellow, or is empty.
When a flower withers, it disappears from the cell that it originally belonged to, and it spreads petals of its colour in its two neighbouring cells (or outside the field if the cell is on the side of the landscape). In case petals fall outside the given cells, they simply become invisible.
You are to help Kanno determine whether it's possible that after some (possibly none or all) flowers shed their petals, at least one of the cells contains all three colours, considering both petals and flowers. Note that flowers can wither in arbitrary order.
Input Specification:
The first and only line of input contains a non-empty string $s$ consisting of uppercase English letters 'A', 'B', 'C' and characters '.' (dots) only ($\lvert s \rvert \leq 100$) — denoting cells containing an amber flower, a buff one, a canary yellow one, and no flowers, respectively.
Output Specification:
Output "Yes" if it's possible that all three colours appear in some cell, and "No" otherwise.
You can print each letter in any case (upper or lower).
Demo Input:
['.BAC.\n', 'AA..CB\n']
Demo Output:
['Yes\n', 'No\n']
Note:
In the first example, the buff and canary yellow flowers can leave their petals in the central cell, blending all three colours in it.
In the second example, it's impossible to satisfy the requirement because there is no way that amber and buff meet in any cell.
|
```python
r=input()
a=['A','B','C']
le=len(r)
for i in range(le-2):
l=list(r[i:i+3])
l.sort()
if l==a:
print ("Yes")
exit(0)
print ("No")
```
| 3
|
|
205
|
A
|
Little Elephant and Rozdil
|
PROGRAMMING
| 900
|
[
"brute force",
"implementation"
] | null | null |
The Little Elephant loves Ukraine very much. Most of all he loves town Rozdol (ukr. "Rozdil").
However, Rozdil is dangerous to settle, so the Little Elephant wants to go to some other town. The Little Elephant doesn't like to spend much time on travelling, so for his journey he will choose a town that needs minimum time to travel to. If there are multiple such cities, then the Little Elephant won't go anywhere.
For each town except for Rozdil you know the time needed to travel to this town. Find the town the Little Elephant will go to or print "Still Rozdil", if he stays in Rozdil.
|
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105) — the number of cities. The next line contains *n* integers, separated by single spaces: the *i*-th integer represents the time needed to go from town Rozdil to the *i*-th town. The time values are positive integers, not exceeding 109.
You can consider the cities numbered from 1 to *n*, inclusive. Rozdil is not among the numbered cities.
|
Print the answer on a single line — the number of the town the Little Elephant will go to. If there are multiple cities with minimum travel time, print "Still Rozdil" (without the quotes).
|
[
"2\n7 4\n",
"7\n7 4 47 100 4 9 12\n"
] |
[
"2\n",
"Still Rozdil\n"
] |
In the first sample there are only two cities where the Little Elephant can go. The travel time for the first town equals 7, to the second one — 4. The town which is closest to Rodzil (the only one) is the second one, so the answer is 2.
In the second sample the closest cities are cities two and five, the travelling time to both of them equals 4, so the answer is "Still Rozdil".
| 500
|
[
{
"input": "2\n7 4",
"output": "2"
},
{
"input": "7\n7 4 47 100 4 9 12",
"output": "Still Rozdil"
},
{
"input": "1\n47",
"output": "1"
},
{
"input": "2\n1000000000 1000000000",
"output": "Still Rozdil"
},
{
"input": "7\n7 6 5 4 3 2 1",
"output": "7"
},
{
"input": "10\n1 1 1 1 1 1 1 1 1 1",
"output": "Still Rozdil"
},
{
"input": "4\n1000000000 100000000 1000000 1000000",
"output": "Still Rozdil"
},
{
"input": "20\n7 1 1 2 1 1 8 7 7 8 4 3 7 10 5 3 10 5 10 6",
"output": "Still Rozdil"
},
{
"input": "20\n3 3 6 9 8 2 4 1 7 3 2 9 7 7 9 7 2 6 2 7",
"output": "8"
},
{
"input": "47\n35 79 84 56 67 95 80 34 77 68 14 55 95 32 40 89 58 79 96 66 50 79 35 86 31 74 91 35 22 72 84 38 11 59 73 51 65 11 11 62 30 12 32 71 69 15 11",
"output": "Still Rozdil"
},
{
"input": "47\n3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3",
"output": "32"
},
{
"input": "25\n73 56 65 70 57 57 49 50 65 68 54 48 66 55 55 66 60 70 47 62 63 70 62 70 63",
"output": "19"
},
{
"input": "100\n922 938 103 689 526 118 314 496 373 47 224 544 495 656 804 754 269 356 354 536 817 597 787 178 516 659 790 993 786 169 866 779 997 350 674 437 219 546 564 893 402 449 751 628 238 686 392 951 716 439 332 563 215 146 746 311 866 773 548 675 731 183 890 802 488 944 473 25 662 559 732 534 777 144 502 593 839 692 84 337 679 718 1005 663 709 940 86 469 310 960 902 233 606 761 116 231 662 903 68 622",
"output": "68"
},
{
"input": "5\n5 5 2 3 1",
"output": "5"
},
{
"input": "1\n1000000000",
"output": "1"
},
{
"input": "3\n7 7 4",
"output": "3"
},
{
"input": "3\n2 2 1",
"output": "3"
},
{
"input": "3\n4 4 3",
"output": "3"
},
{
"input": "2\n10000001 10000002",
"output": "1"
},
{
"input": "4\n3 3 2 1",
"output": "4"
},
{
"input": "3\n1 2 3",
"output": "1"
},
{
"input": "2\n1000000000 100000000",
"output": "2"
},
{
"input": "1\n99999999",
"output": "1"
},
{
"input": "3\n5 5 3",
"output": "3"
},
{
"input": "3\n7 7 5",
"output": "3"
},
{
"input": "3\n3 3 1",
"output": "3"
},
{
"input": "4\n10 10 10 1",
"output": "4"
}
] | 1,629,523,163
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 45
| 404
| 30,924,800
|
n,ct=int(input()),list(map(int,input().split()))
print('Still Rozdil' if ct.count(min(ct))>1 else ct.index(min(ct))+1)
|
Title: Little Elephant and Rozdil
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Little Elephant loves Ukraine very much. Most of all he loves town Rozdol (ukr. "Rozdil").
However, Rozdil is dangerous to settle, so the Little Elephant wants to go to some other town. The Little Elephant doesn't like to spend much time on travelling, so for his journey he will choose a town that needs minimum time to travel to. If there are multiple such cities, then the Little Elephant won't go anywhere.
For each town except for Rozdil you know the time needed to travel to this town. Find the town the Little Elephant will go to or print "Still Rozdil", if he stays in Rozdil.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105) — the number of cities. The next line contains *n* integers, separated by single spaces: the *i*-th integer represents the time needed to go from town Rozdil to the *i*-th town. The time values are positive integers, not exceeding 109.
You can consider the cities numbered from 1 to *n*, inclusive. Rozdil is not among the numbered cities.
Output Specification:
Print the answer on a single line — the number of the town the Little Elephant will go to. If there are multiple cities with minimum travel time, print "Still Rozdil" (without the quotes).
Demo Input:
['2\n7 4\n', '7\n7 4 47 100 4 9 12\n']
Demo Output:
['2\n', 'Still Rozdil\n']
Note:
In the first sample there are only two cities where the Little Elephant can go. The travel time for the first town equals 7, to the second one — 4. The town which is closest to Rodzil (the only one) is the second one, so the answer is 2.
In the second sample the closest cities are cities two and five, the travelling time to both of them equals 4, so the answer is "Still Rozdil".
|
```python
n,ct=int(input()),list(map(int,input().split()))
print('Still Rozdil' if ct.count(min(ct))>1 else ct.index(min(ct))+1)
```
| 3
|
|
851
|
A
|
Arpa and a research in Mexican wave
|
PROGRAMMING
| 800
|
[
"implementation",
"math"
] | null | null |
Arpa is researching the Mexican wave.
There are *n* spectators in the stadium, labeled from 1 to *n*. They start the Mexican wave at time 0.
- At time 1, the first spectator stands. - At time 2, the second spectator stands. - ... - At time *k*, the *k*-th spectator stands. - At time *k*<=+<=1, the (*k*<=+<=1)-th spectator stands and the first spectator sits. - At time *k*<=+<=2, the (*k*<=+<=2)-th spectator stands and the second spectator sits. - ... - At time *n*, the *n*-th spectator stands and the (*n*<=-<=*k*)-th spectator sits. - At time *n*<=+<=1, the (*n*<=+<=1<=-<=*k*)-th spectator sits. - ... - At time *n*<=+<=*k*, the *n*-th spectator sits.
Arpa wants to know how many spectators are standing at time *t*.
|
The first line contains three integers *n*, *k*, *t* (1<=≤<=*n*<=≤<=109, 1<=≤<=*k*<=≤<=*n*, 1<=≤<=*t*<=<<=*n*<=+<=*k*).
|
Print single integer: how many spectators are standing at time *t*.
|
[
"10 5 3\n",
"10 5 7\n",
"10 5 12\n"
] |
[
"3\n",
"5\n",
"3\n"
] |
In the following a sitting spectator is represented as -, a standing spectator is represented as ^.
- At *t* = 0 ---------- <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 0. - At *t* = 1 ^--------- <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 1. - At *t* = 2 ^^-------- <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 2. - At *t* = 3 ^^^------- <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 3. - At *t* = 4 ^^^^------ <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 4. - At *t* = 5 ^^^^^----- <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 5. - At *t* = 6 -^^^^^---- <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 5. - At *t* = 7 --^^^^^--- <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 5. - At *t* = 8 ---^^^^^-- <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 5. - At *t* = 9 ----^^^^^- <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 5. - At *t* = 10 -----^^^^^ <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 5. - At *t* = 11 ------^^^^ <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 4. - At *t* = 12 -------^^^ <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 3. - At *t* = 13 --------^^ <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 2. - At *t* = 14 ---------^ <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 1. - At *t* = 15 ---------- <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 0.
| 500
|
[
{
"input": "10 5 3",
"output": "3"
},
{
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{
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"output": "4"
}
] | 1,659,535,691
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 166
| 62
| 0
|
import sys
n, k, t = (int(el) for el in input().split())
if t <= k:
print(t)
sys.exit()
if t >= k and t <= n:
print(k)
sys.exit()
print(k - (t - n))
|
Title: Arpa and a research in Mexican wave
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Arpa is researching the Mexican wave.
There are *n* spectators in the stadium, labeled from 1 to *n*. They start the Mexican wave at time 0.
- At time 1, the first spectator stands. - At time 2, the second spectator stands. - ... - At time *k*, the *k*-th spectator stands. - At time *k*<=+<=1, the (*k*<=+<=1)-th spectator stands and the first spectator sits. - At time *k*<=+<=2, the (*k*<=+<=2)-th spectator stands and the second spectator sits. - ... - At time *n*, the *n*-th spectator stands and the (*n*<=-<=*k*)-th spectator sits. - At time *n*<=+<=1, the (*n*<=+<=1<=-<=*k*)-th spectator sits. - ... - At time *n*<=+<=*k*, the *n*-th spectator sits.
Arpa wants to know how many spectators are standing at time *t*.
Input Specification:
The first line contains three integers *n*, *k*, *t* (1<=≤<=*n*<=≤<=109, 1<=≤<=*k*<=≤<=*n*, 1<=≤<=*t*<=<<=*n*<=+<=*k*).
Output Specification:
Print single integer: how many spectators are standing at time *t*.
Demo Input:
['10 5 3\n', '10 5 7\n', '10 5 12\n']
Demo Output:
['3\n', '5\n', '3\n']
Note:
In the following a sitting spectator is represented as -, a standing spectator is represented as ^.
- At *t* = 0 ---------- <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 0. - At *t* = 1 ^--------- <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 1. - At *t* = 2 ^^-------- <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 2. - At *t* = 3 ^^^------- <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 3. - At *t* = 4 ^^^^------ <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 4. - At *t* = 5 ^^^^^----- <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 5. - At *t* = 6 -^^^^^---- <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 5. - At *t* = 7 --^^^^^--- <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 5. - At *t* = 8 ---^^^^^-- <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 5. - At *t* = 9 ----^^^^^- <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 5. - At *t* = 10 -----^^^^^ <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 5. - At *t* = 11 ------^^^^ <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 4. - At *t* = 12 -------^^^ <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 3. - At *t* = 13 --------^^ <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 2. - At *t* = 14 ---------^ <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 1. - At *t* = 15 ---------- <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4d97e684117250a9afe9be022ab8a63653dd15aa.png" style="max-width: 100.0%;max-height: 100.0%;"/> number of standing spectators = 0.
|
```python
import sys
n, k, t = (int(el) for el in input().split())
if t <= k:
print(t)
sys.exit()
if t >= k and t <= n:
print(k)
sys.exit()
print(k - (t - n))
```
| 3
|
|
294
|
A
|
Shaass and Oskols
|
PROGRAMMING
| 800
|
[
"implementation",
"math"
] | null | null |
Shaass has decided to hunt some birds. There are *n* horizontal electricity wires aligned parallel to each other. Wires are numbered 1 to *n* from top to bottom. On each wire there are some oskols sitting next to each other. Oskol is the name of a delicious kind of birds in Shaass's territory. Supposed there are *a**i* oskols sitting on the *i*-th wire.
Sometimes Shaass shots one of the birds and the bird dies (suppose that this bird sat at the *i*-th wire). Consequently all the birds on the *i*-th wire to the left of the dead bird get scared and jump up on the wire number *i*<=-<=1, if there exists no upper wire they fly away. Also all the birds to the right of the dead bird jump down on wire number *i*<=+<=1, if there exists no such wire they fly away.
Shaass has shot *m* birds. You're given the initial number of birds on each wire, tell him how many birds are sitting on each wire after the shots.
|
The first line of the input contains an integer *n*, (1<=≤<=*n*<=≤<=100). The next line contains a list of space-separated integers *a*1,<=*a*2,<=...,<=*a**n*, (0<=≤<=*a**i*<=≤<=100).
The third line contains an integer *m*, (0<=≤<=*m*<=≤<=100). Each of the next *m* lines contains two integers *x**i* and *y**i*. The integers mean that for the *i*-th time Shaass shoot the *y**i*-th (from left) bird on the *x**i*-th wire, (1<=≤<=*x**i*<=≤<=*n*,<=1<=≤<=*y**i*). It's guaranteed there will be at least *y**i* birds on the *x**i*-th wire at that moment.
|
On the *i*-th line of the output print the number of birds on the *i*-th wire.
|
[
"5\n10 10 10 10 10\n5\n2 5\n3 13\n2 12\n1 13\n4 6\n",
"3\n2 4 1\n1\n2 2\n"
] |
[
"0\n12\n5\n0\n16\n",
"3\n0\n3\n"
] |
none
| 500
|
[
{
"input": "5\n10 10 10 10 10\n5\n2 5\n3 13\n2 12\n1 13\n4 6",
"output": "0\n12\n5\n0\n16"
},
{
"input": "3\n2 4 1\n1\n2 2",
"output": "3\n0\n3"
},
{
"input": "5\n58 51 45 27 48\n5\n4 9\n5 15\n4 5\n5 8\n1 43",
"output": "0\n66\n57\n7\n0"
},
{
"input": "10\n48 53 10 28 91 56 81 2 67 52\n2\n2 40\n6 51",
"output": "87\n0\n23\n28\n141\n0\n86\n2\n67\n52"
},
{
"input": "2\n72 45\n6\n1 69\n2 41\n1 19\n2 7\n1 5\n2 1",
"output": "0\n0"
},
{
"input": "10\n95 54 36 39 98 30 19 24 14 12\n3\n9 5\n8 15\n7 5",
"output": "95\n54\n36\n39\n98\n34\n0\n28\n13\n21"
},
{
"input": "100\n95 15 25 18 64 62 23 59 70 84 50 26 87 35 75 86 0 22 77 60 66 41 21 9 75 50 25 3 69 14 39 68 64 46 59 99 2 0 21 76 90 12 61 42 6 91 36 39 47 41 93 81 66 57 70 36 68 89 52 1 19 93 67 22 76 20 8 81 98 18 100 73 61 93 75 80 53 72 40 20 2 86 33 59 27 16 11 26 55 44 47 55 94 58 2 55 82 22 9 95\n0",
"output": "95\n15\n25\n18\n64\n62\n23\n59\n70\n84\n50\n26\n87\n35\n75\n86\n0\n22\n77\n60\n66\n41\n21\n9\n75\n50\n25\n3\n69\n14\n39\n68\n64\n46\n59\n99\n2\n0\n21\n76\n90\n12\n61\n42\n6\n91\n36\n39\n47\n41\n93\n81\n66\n57\n70\n36\n68\n89\n52\n1\n19\n93\n67\n22\n76\n20\n8\n81\n98\n18\n100\n73\n61\n93\n75\n80\n53\n72\n40\n20\n2\n86\n33\n59\n27\n16\n11\n26\n55\n44\n47\n55\n94\n58\n2\n55\n82\n22\n9\n95"
},
{
"input": "1\n100\n1\n1 100",
"output": "0"
},
{
"input": "1\n100\n1\n1 1",
"output": "0"
},
{
"input": "1\n50\n1\n1 25",
"output": "0"
},
{
"input": "2\n50 0\n1\n1 1",
"output": "0\n49"
},
{
"input": "1\n10\n0",
"output": "10"
}
] | 1,676,639,194
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 31
| 124
| 0
|
n = int(input())
list_a = [int(x) for x in input().split(' ')]
list_a.append(0)
m = int(input())
for i in range(m):
list_m = [int(x) for x in input().split(' ')]
pos = list_m[0] - 1
val = list_m[1]
list_a[pos + 1] += list_a[pos] - val
list_a[pos - 1] += val - 1
list_a[pos] = 0
for j in range(n):
print(list_a[j])
|
Title: Shaass and Oskols
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Shaass has decided to hunt some birds. There are *n* horizontal electricity wires aligned parallel to each other. Wires are numbered 1 to *n* from top to bottom. On each wire there are some oskols sitting next to each other. Oskol is the name of a delicious kind of birds in Shaass's territory. Supposed there are *a**i* oskols sitting on the *i*-th wire.
Sometimes Shaass shots one of the birds and the bird dies (suppose that this bird sat at the *i*-th wire). Consequently all the birds on the *i*-th wire to the left of the dead bird get scared and jump up on the wire number *i*<=-<=1, if there exists no upper wire they fly away. Also all the birds to the right of the dead bird jump down on wire number *i*<=+<=1, if there exists no such wire they fly away.
Shaass has shot *m* birds. You're given the initial number of birds on each wire, tell him how many birds are sitting on each wire after the shots.
Input Specification:
The first line of the input contains an integer *n*, (1<=≤<=*n*<=≤<=100). The next line contains a list of space-separated integers *a*1,<=*a*2,<=...,<=*a**n*, (0<=≤<=*a**i*<=≤<=100).
The third line contains an integer *m*, (0<=≤<=*m*<=≤<=100). Each of the next *m* lines contains two integers *x**i* and *y**i*. The integers mean that for the *i*-th time Shaass shoot the *y**i*-th (from left) bird on the *x**i*-th wire, (1<=≤<=*x**i*<=≤<=*n*,<=1<=≤<=*y**i*). It's guaranteed there will be at least *y**i* birds on the *x**i*-th wire at that moment.
Output Specification:
On the *i*-th line of the output print the number of birds on the *i*-th wire.
Demo Input:
['5\n10 10 10 10 10\n5\n2 5\n3 13\n2 12\n1 13\n4 6\n', '3\n2 4 1\n1\n2 2\n']
Demo Output:
['0\n12\n5\n0\n16\n', '3\n0\n3\n']
Note:
none
|
```python
n = int(input())
list_a = [int(x) for x in input().split(' ')]
list_a.append(0)
m = int(input())
for i in range(m):
list_m = [int(x) for x in input().split(' ')]
pos = list_m[0] - 1
val = list_m[1]
list_a[pos + 1] += list_a[pos] - val
list_a[pos - 1] += val - 1
list_a[pos] = 0
for j in range(n):
print(list_a[j])
```
| 3
|
|
928
|
A
|
Login Verification
|
PROGRAMMING
| 1,200
|
[
"*special",
"strings"
] | null | null |
When registering in a social network, users are allowed to create their own convenient login to make it easier to share contacts, print it on business cards, etc.
Login is an arbitrary sequence of lower and uppercase latin letters, digits and underline symbols («_»). However, in order to decrease the number of frauds and user-inattention related issues, it is prohibited to register a login if it is similar with an already existing login. More precisely, two logins *s* and *t* are considered similar if we can transform *s* to *t* via a sequence of operations of the following types:
- transform lowercase letters to uppercase and vice versa; - change letter «O» (uppercase latin letter) to digit «0» and vice versa; - change digit «1» (one) to any letter among «l» (lowercase latin «L»), «I» (uppercase latin «i») and vice versa, or change one of these letters to other.
For example, logins «Codeforces» and «codef0rces» as well as «OO0OOO00O0OOO0O00OOO0OO_lol» and «OO0OOO0O00OOO0O00OO0OOO_1oI» are considered similar whereas «Codeforces» and «Code_forces» are not.
You're given a list of existing logins with no two similar amonst and a newly created user login. Check whether this new login is similar with any of the existing ones.
|
The first line contains a non-empty string *s* consisting of lower and uppercase latin letters, digits and underline symbols («_») with length not exceeding 50 — the login itself.
The second line contains a single integer *n* (1<=≤<=*n*<=≤<=1<=000) — the number of existing logins.
The next *n* lines describe the existing logins, following the same constraints as the user login (refer to the first line of the input). It's guaranteed that no two existing logins are similar.
|
Print «Yes» (without quotes), if user can register via this login, i.e. none of the existing logins is similar with it.
Otherwise print «No» (without quotes).
|
[
"1_wat\n2\n2_wat\nwat_1\n",
"000\n3\n00\nooA\noOo\n",
"_i_\n3\n__i_\n_1_\nI\n",
"La0\n3\n2a0\nLa1\n1a0\n",
"abc\n1\naBc\n",
"0Lil\n2\nLIL0\n0Ril\n"
] |
[
"Yes\n",
"No\n",
"No\n",
"No\n",
"No\n",
"Yes\n"
] |
In the second sample case the user wants to create a login consisting of three zeros. It's impossible due to collision with the third among the existing.
In the third sample case the new login is similar with the second one.
| 500
|
[
{
"input": "1_wat\n2\n2_wat\nwat_1",
"output": "Yes"
},
{
"input": "000\n3\n00\nooA\noOo",
"output": "No"
},
{
"input": "_i_\n3\n__i_\n_1_\nI",
"output": "No"
},
{
"input": "La0\n3\n2a0\nLa1\n1a0",
"output": "No"
},
{
"input": "abc\n1\naBc",
"output": "No"
},
{
"input": "0Lil\n2\nLIL0\n0Ril",
"output": "Yes"
},
{
"input": "iloO\n3\niIl0\noIl0\nIooO",
"output": "Yes"
},
{
"input": "L1il0o1L1\n5\niLLoLL\noOI1Io10il\nIoLLoO\nO01ilOoI\nI10l0o",
"output": "Yes"
},
{
"input": "ELioO1lOoOIOiLoooi1iolul1O\n7\nOoEIuOIl1ui1010uiooOoi0Oio001L0EoEolO0\nOLIoOEuoE11u1u1iLOI0oO\nuEOuO0uIOOlO01OlEI0E1Oo0IO1LI0uE0LILO0\nEOo0Il11iIOOOIiuOiIiiLOLEOOII001EE\niOoO0LOulioE0OLIIIulli01OoiuOOOoOlEiI0EiiElIIu0\nlE1LOE1Oil\n1u0EOliIiIOl1u110il0l1O0u",
"output": "Yes"
},
{
"input": "0blo7X\n20\n1oobb6\nXIXIO2X\n2iYI2\n607XXol\n2I6io22\nOl10I\nbXX0Lo\nolOOb7X\n07LlXL\nlXY17\n12iIX2\n7lL70\nbOo11\n17Y6b62\n0O6L7\n1lX2L\n2iYl6lI\n7bXIi1o\niLIY2\n0OIo1X",
"output": "Yes"
},
{
"input": "lkUL\n25\nIIfL\nokl\nfoo\ni0U\noko\niIoU\nUUv\nvli\nv0Uk\n0Of\niill\n1vkl\nUIf\nUfOO\nlvLO\nUUo0\nIOf1\nlovL\nIkk\noIv\nLvfU\n0UI\nkol\n1OO0\n1OOi",
"output": "Yes"
},
{
"input": "L1lo\n3\nOOo1\nL1lo\n0lOl",
"output": "No"
},
{
"input": "LIoooiLO\n5\nLIoooiLO\nl0o01I00\n0OOl0lLO01\nil10i0\noiloi",
"output": "No"
},
{
"input": "1i1lQI\n7\nuLg1uLLigIiOLoggu\nLLLgIuQIQIIloiQuIIoIO0l0o000\n0u1LQu11oIuooIl0OooLg0i0IQu1O1lloI1\nQuQgIQi0LOIliLOuuuioLQou1l\nlLIO00QLi01LogOliOIggII1\no0Ll1uIOQl10IL0IILQ\n1i1lQI",
"output": "No"
},
{
"input": "oIzz1\n20\n1TTl0O\nloF0LT\n1lLzo\noi0Ov\nFlIF1zT\nzoITzx\n0TIFlT\nl1vllil\nOviix1F\nLFvI1lL\nLIl0loz\nixz1v\n1i1vFi\nTIFTol\noIzz1\nIvTl0o\nxv1U0O\niiiioF\n1oiLUlO\nxToxv1",
"output": "No"
},
{
"input": "00L0\n25\n0il\nIlkZ\nL0I\n00L0\nBd0\nZLd\n0d1k\nddk\nIdl\nkBd\nkBOL\nZ1lI\nkBL\nLOko\noZ0i\nZ1lO\nLiOk\niBld\nLO0d\ndIo\nZ10\n1k1i\n0o0L\nIoBd\ni0B0",
"output": "No"
},
{
"input": "Z\n1\nz",
"output": "No"
},
{
"input": "0\n1\no",
"output": "No"
},
{
"input": "0\n1\nO",
"output": "No"
},
{
"input": "o\n1\n0",
"output": "No"
},
{
"input": "o\n1\nO",
"output": "No"
},
{
"input": "o\n1\no",
"output": "No"
},
{
"input": "O\n1\no",
"output": "No"
},
{
"input": "O\n1\n0",
"output": "No"
},
{
"input": "1\n1\nl",
"output": "No"
},
{
"input": "1\n1\nL",
"output": "No"
},
{
"input": "1\n1\ni",
"output": "No"
},
{
"input": "1\n1\nI",
"output": "No"
},
{
"input": "1\n1\no",
"output": "Yes"
},
{
"input": "i\n1\n1",
"output": "No"
},
{
"input": "i\n1\nL",
"output": "No"
},
{
"input": "i\n1\nl",
"output": "No"
},
{
"input": "I\n1\nL",
"output": "No"
},
{
"input": "I\n1\nl",
"output": "No"
},
{
"input": "I\n1\n1",
"output": "No"
},
{
"input": "l\n1\n1",
"output": "No"
},
{
"input": "l\n1\nL",
"output": "No"
},
{
"input": "l\n1\ni",
"output": "No"
},
{
"input": "l\n1\nI",
"output": "No"
},
{
"input": "L\n1\nl",
"output": "No"
},
{
"input": "L\n1\n1",
"output": "No"
},
{
"input": "L\n1\ni",
"output": "No"
},
{
"input": "L\n1\nI",
"output": "No"
}
] | 1,519,489,514
| 3,014
|
Python 3
|
OK
|
TESTS
| 73
| 62
| 5,632,000
|
new_login = input()
def is_similar(login1, login2):
return login1.replace('O', '0').replace('o', '0').replace('l', '1').replace('L', '1').replace('I', '1').replace('i', '1').lower()==login2.replace('O', '0').replace('o', '0').replace('l', '1').replace('L', '1').replace('I', '1').replace('i', '1').lower()
for i in range(int(input())):
if is_similar(input(), new_login):
print("No")
quit()
print("Yes")
|
Title: Login Verification
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
When registering in a social network, users are allowed to create their own convenient login to make it easier to share contacts, print it on business cards, etc.
Login is an arbitrary sequence of lower and uppercase latin letters, digits and underline symbols («_»). However, in order to decrease the number of frauds and user-inattention related issues, it is prohibited to register a login if it is similar with an already existing login. More precisely, two logins *s* and *t* are considered similar if we can transform *s* to *t* via a sequence of operations of the following types:
- transform lowercase letters to uppercase and vice versa; - change letter «O» (uppercase latin letter) to digit «0» and vice versa; - change digit «1» (one) to any letter among «l» (lowercase latin «L»), «I» (uppercase latin «i») and vice versa, or change one of these letters to other.
For example, logins «Codeforces» and «codef0rces» as well as «OO0OOO00O0OOO0O00OOO0OO_lol» and «OO0OOO0O00OOO0O00OO0OOO_1oI» are considered similar whereas «Codeforces» and «Code_forces» are not.
You're given a list of existing logins with no two similar amonst and a newly created user login. Check whether this new login is similar with any of the existing ones.
Input Specification:
The first line contains a non-empty string *s* consisting of lower and uppercase latin letters, digits and underline symbols («_») with length not exceeding 50 — the login itself.
The second line contains a single integer *n* (1<=≤<=*n*<=≤<=1<=000) — the number of existing logins.
The next *n* lines describe the existing logins, following the same constraints as the user login (refer to the first line of the input). It's guaranteed that no two existing logins are similar.
Output Specification:
Print «Yes» (without quotes), if user can register via this login, i.e. none of the existing logins is similar with it.
Otherwise print «No» (without quotes).
Demo Input:
['1_wat\n2\n2_wat\nwat_1\n', '000\n3\n00\nooA\noOo\n', '_i_\n3\n__i_\n_1_\nI\n', 'La0\n3\n2a0\nLa1\n1a0\n', 'abc\n1\naBc\n', '0Lil\n2\nLIL0\n0Ril\n']
Demo Output:
['Yes\n', 'No\n', 'No\n', 'No\n', 'No\n', 'Yes\n']
Note:
In the second sample case the user wants to create a login consisting of three zeros. It's impossible due to collision with the third among the existing.
In the third sample case the new login is similar with the second one.
|
```python
new_login = input()
def is_similar(login1, login2):
return login1.replace('O', '0').replace('o', '0').replace('l', '1').replace('L', '1').replace('I', '1').replace('i', '1').lower()==login2.replace('O', '0').replace('o', '0').replace('l', '1').replace('L', '1').replace('I', '1').replace('i', '1').lower()
for i in range(int(input())):
if is_similar(input(), new_login):
print("No")
quit()
print("Yes")
```
| 3
|
|
920
|
A
|
Water The Garden
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
It is winter now, and Max decided it's about time he watered the garden.
The garden can be represented as *n* consecutive garden beds, numbered from 1 to *n*. *k* beds contain water taps (*i*-th tap is located in the bed *x**i*), which, if turned on, start delivering water to neighbouring beds. If the tap on the bed *x**i* is turned on, then after one second has passed, the bed *x**i* will be watered; after two seconds have passed, the beds from the segment [*x**i*<=-<=1,<=*x**i*<=+<=1] will be watered (if they exist); after *j* seconds have passed (*j* is an integer number), the beds from the segment [*x**i*<=-<=(*j*<=-<=1),<=*x**i*<=+<=(*j*<=-<=1)] will be watered (if they exist). Nothing changes during the seconds, so, for example, we can't say that the segment [*x**i*<=-<=2.5,<=*x**i*<=+<=2.5] will be watered after 2.5 seconds have passed; only the segment [*x**i*<=-<=2,<=*x**i*<=+<=2] will be watered at that moment.
Max wants to turn on all the water taps at the same moment, and now he wonders, what is the minimum number of seconds that have to pass after he turns on some taps until the whole garden is watered. Help him to find the answer!
|
The first line contains one integer *t* — the number of test cases to solve (1<=≤<=*t*<=≤<=200).
Then *t* test cases follow. The first line of each test case contains two integers *n* and *k* (1<=≤<=*n*<=≤<=200, 1<=≤<=*k*<=≤<=*n*) — the number of garden beds and water taps, respectively.
Next line contains *k* integers *x**i* (1<=≤<=*x**i*<=≤<=*n*) — the location of *i*-th water tap. It is guaranteed that for each condition *x**i*<=-<=1<=<<=*x**i* holds.
It is guaranteed that the sum of *n* over all test cases doesn't exceed 200.
Note that in hacks you have to set *t*<==<=1.
|
For each test case print one integer — the minimum number of seconds that have to pass after Max turns on some of the water taps, until the whole garden is watered.
|
[
"3\n5 1\n3\n3 3\n1 2 3\n4 1\n1\n"
] |
[
"3\n1\n4\n"
] |
The first example consists of 3 tests:
1. There are 5 garden beds, and a water tap in the bed 3. If we turn it on, then after 1 second passes, only bed 3 will be watered; after 2 seconds pass, beds [1, 3] will be watered, and after 3 seconds pass, everything will be watered. 1. There are 3 garden beds, and there is a water tap in each one. If we turn all of them on, then everything will be watered after 1 second passes. 1. There are 4 garden beds, and only one tap in the bed 1. It will take 4 seconds to water, for example, bed 4.
| 0
|
[
{
"input": "3\n5 1\n3\n3 3\n1 2 3\n4 1\n1",
"output": "3\n1\n4"
},
{
"input": "26\n1 1\n1\n2 1\n2\n2 1\n1\n2 2\n1 2\n3 1\n3\n3 1\n2\n3 2\n2 3\n3 1\n1\n3 2\n1 3\n3 2\n1 2\n3 3\n1 2 3\n4 1\n4\n4 1\n3\n4 2\n3 4\n4 1\n2\n4 2\n2 4\n4 2\n2 3\n4 3\n2 3 4\n4 1\n1\n4 2\n1 4\n4 2\n1 3\n4 3\n1 3 4\n4 2\n1 2\n4 3\n1 2 4\n4 3\n1 2 3\n4 4\n1 2 3 4",
"output": "1\n2\n2\n1\n3\n2\n2\n3\n2\n2\n1\n4\n3\n3\n3\n2\n2\n2\n4\n2\n2\n2\n3\n2\n2\n1"
},
{
"input": "31\n5 1\n5\n5 1\n4\n5 2\n4 5\n5 1\n3\n5 2\n3 5\n5 2\n3 4\n5 3\n3 4 5\n5 1\n2\n5 2\n2 5\n5 2\n2 4\n5 3\n2 4 5\n5 2\n2 3\n5 3\n2 3 5\n5 3\n2 3 4\n5 4\n2 3 4 5\n5 1\n1\n5 2\n1 5\n5 2\n1 4\n5 3\n1 4 5\n5 2\n1 3\n5 3\n1 3 5\n5 3\n1 3 4\n5 4\n1 3 4 5\n5 2\n1 2\n5 3\n1 2 5\n5 3\n1 2 4\n5 4\n1 2 4 5\n5 3\n1 2 3\n5 4\n1 2 3 5\n5 4\n1 2 3 4\n5 5\n1 2 3 4 5",
"output": "5\n4\n4\n3\n3\n3\n3\n4\n2\n2\n2\n3\n2\n2\n2\n5\n3\n2\n2\n3\n2\n2\n2\n4\n2\n2\n2\n3\n2\n2\n1"
},
{
"input": "1\n200 1\n200",
"output": "200"
},
{
"input": "1\n5 1\n5",
"output": "5"
},
{
"input": "1\n177 99\n1 4 7 10 11 13 14 15 16 17 19 21 22 24 25 26 27 28 32 34 35 38 39 40 42 45 46 52 54 55 57 58 59 60 62 64 65 67 70 71 74 77 78 79 80 81 83 84 88 92 93 94 95 100 101 102 104 106 107 108 109 110 112 113 114 115 116 118 122 123 124 125 127 128 129 130 134 135 137 138 139 140 142 146 148 149 154 158 160 161 162 165 166 167 169 171 172 173 176",
"output": "4"
},
{
"input": "1\n69 12\n5 7 10 11 12 18 20 27 28 31 47 67",
"output": "11"
},
{
"input": "1\n74 7\n19 39 40 47 55 57 61",
"output": "19"
},
{
"input": "1\n170 11\n14 18 37 39 80 83 103 112 124 127 131",
"output": "40"
},
{
"input": "1\n200 1\n8",
"output": "193"
},
{
"input": "1\n155 53\n2 3 7 9 10 11 12 20 24 26 28 31 38 39 40 51 53 56 58 63 65 66 69 70 72 74 79 81 83 88 90 92 100 103 104 106 111 113 114 115 116 121 124 126 127 128 133 136 142 143 145 148 150",
"output": "6"
},
{
"input": "1\n161 69\n2 5 8 11 12 13 17 18 23 25 28 29 30 33 34 35 36 38 39 44 45 49 52 53 56 57 58 60 62 70 71 74 76 77 82 83 86 90 94 95 97 104 105 108 109 112 113 118 120 123 126 127 132 135 137 139 140 141 142 143 144 146 147 148 151 152 153 154 161",
"output": "5"
},
{
"input": "1\n8 4\n1 2 3 7",
"output": "3"
},
{
"input": "1\n12 2\n5 12",
"output": "5"
},
{
"input": "1\n13 2\n6 12",
"output": "6"
},
{
"input": "1\n13 2\n8 13",
"output": "8"
},
{
"input": "1\n10 4\n1 2 3 5",
"output": "6"
},
{
"input": "1\n200 2\n50 150",
"output": "51"
},
{
"input": "1\n110 2\n1 110",
"output": "55"
}
] | 1,517,583,000
| 900
|
Python 3
|
OK
|
TESTS
| 21
| 93
| 5,632,000
|
# A
import math
t = int(input())
ans = []
for _ in range(t):
n, k = map(int, input().split())
taps = input().split()
for i in range(k):
taps[i] = int(taps[i])
max_time = 0
for i in range(n):
min_time = 1000
for j in range(k):
min_time = min(min_time, int(math.fabs(i+1 - taps[j])))
max_time = max(max_time, min_time)
ans.append(max_time)
for i in range(t):
print(ans[i]+1)
|
Title: Water The Garden
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
It is winter now, and Max decided it's about time he watered the garden.
The garden can be represented as *n* consecutive garden beds, numbered from 1 to *n*. *k* beds contain water taps (*i*-th tap is located in the bed *x**i*), which, if turned on, start delivering water to neighbouring beds. If the tap on the bed *x**i* is turned on, then after one second has passed, the bed *x**i* will be watered; after two seconds have passed, the beds from the segment [*x**i*<=-<=1,<=*x**i*<=+<=1] will be watered (if they exist); after *j* seconds have passed (*j* is an integer number), the beds from the segment [*x**i*<=-<=(*j*<=-<=1),<=*x**i*<=+<=(*j*<=-<=1)] will be watered (if they exist). Nothing changes during the seconds, so, for example, we can't say that the segment [*x**i*<=-<=2.5,<=*x**i*<=+<=2.5] will be watered after 2.5 seconds have passed; only the segment [*x**i*<=-<=2,<=*x**i*<=+<=2] will be watered at that moment.
Max wants to turn on all the water taps at the same moment, and now he wonders, what is the minimum number of seconds that have to pass after he turns on some taps until the whole garden is watered. Help him to find the answer!
Input Specification:
The first line contains one integer *t* — the number of test cases to solve (1<=≤<=*t*<=≤<=200).
Then *t* test cases follow. The first line of each test case contains two integers *n* and *k* (1<=≤<=*n*<=≤<=200, 1<=≤<=*k*<=≤<=*n*) — the number of garden beds and water taps, respectively.
Next line contains *k* integers *x**i* (1<=≤<=*x**i*<=≤<=*n*) — the location of *i*-th water tap. It is guaranteed that for each condition *x**i*<=-<=1<=<<=*x**i* holds.
It is guaranteed that the sum of *n* over all test cases doesn't exceed 200.
Note that in hacks you have to set *t*<==<=1.
Output Specification:
For each test case print one integer — the minimum number of seconds that have to pass after Max turns on some of the water taps, until the whole garden is watered.
Demo Input:
['3\n5 1\n3\n3 3\n1 2 3\n4 1\n1\n']
Demo Output:
['3\n1\n4\n']
Note:
The first example consists of 3 tests:
1. There are 5 garden beds, and a water tap in the bed 3. If we turn it on, then after 1 second passes, only bed 3 will be watered; after 2 seconds pass, beds [1, 3] will be watered, and after 3 seconds pass, everything will be watered. 1. There are 3 garden beds, and there is a water tap in each one. If we turn all of them on, then everything will be watered after 1 second passes. 1. There are 4 garden beds, and only one tap in the bed 1. It will take 4 seconds to water, for example, bed 4.
|
```python
# A
import math
t = int(input())
ans = []
for _ in range(t):
n, k = map(int, input().split())
taps = input().split()
for i in range(k):
taps[i] = int(taps[i])
max_time = 0
for i in range(n):
min_time = 1000
for j in range(k):
min_time = min(min_time, int(math.fabs(i+1 - taps[j])))
max_time = max(max_time, min_time)
ans.append(max_time)
for i in range(t):
print(ans[i]+1)
```
| 3
|
|
49
|
A
|
Sleuth
|
PROGRAMMING
| 800
|
[
"implementation"
] |
A. Sleuth
|
2
|
256
|
Vasya plays the sleuth with his friends. The rules of the game are as follows: those who play for the first time, that is Vasya is the sleuth, he should investigate a "crime" and find out what is happening. He can ask any questions whatsoever that can be answered with "Yes" or "No". All the rest agree beforehand to answer the questions like that: if the question’s last letter is a vowel, they answer "Yes" and if the last letter is a consonant, they answer "No". Of course, the sleuth knows nothing about it and his task is to understand that.
Unfortunately, Vasya is not very smart. After 5 hours of endless stupid questions everybody except Vasya got bored. That’s why Vasya’s friends ask you to write a program that would give answers instead of them.
The English alphabet vowels are: A, E, I, O, U, Y
The English alphabet consonants are: B, C, D, F, G, H, J, K, L, M, N, P, Q, R, S, T, V, W, X, Z
|
The single line contains a question represented by a non-empty line consisting of large and small Latin letters, spaces and a question mark. The line length does not exceed 100. It is guaranteed that the question mark occurs exactly once in the line — as the last symbol and that the line contains at least one letter.
|
Print answer for the question in a single line: YES if the answer is "Yes", NO if the answer is "No".
Remember that in the reply to the question the last letter, not the last character counts. I. e. the spaces and the question mark do not count as letters.
|
[
"Is it a melon?\n",
"Is it an apple?\n",
"Is it a banana ?\n",
"Is it an apple and a banana simultaneouSLY?\n"
] |
[
"NO\n",
"YES\n",
"YES\n",
"YES\n"
] |
none
| 500
|
[
{
"input": "Is it a melon?",
"output": "NO"
},
{
"input": "Is it an apple?",
"output": "YES"
},
{
"input": " Is it a banana ?",
"output": "YES"
},
{
"input": "Is it an apple and a banana simultaneouSLY?",
"output": "YES"
},
{
"input": "oHtSbDwzHb?",
"output": "NO"
},
{
"input": "sZecYdUvZHrXx?",
"output": "NO"
},
{
"input": "uMtXK?",
"output": "NO"
},
{
"input": "U?",
"output": "YES"
},
{
"input": "aqFDkCUKeHMyvZFcAyWlMUSQTFomtaWjoKLVyxLCw vcufPBFbaljOuHWiDCROYTcmbgzbaqHXKPOYEbuEtRqqoxBbOETCsQzhw?",
"output": "NO"
},
{
"input": "dJcNqQiFXzcbsj fItCpBLyXOnrSBPebwyFHlxUJHqCUzzCmcAvMiKL NunwOXnKeIxUZmBVwiCUfPkjRAkTPbkYCmwRRnDSLaz?",
"output": "NO"
},
{
"input": "gxzXbdcAQMuFKuuiPohtMgeypr wpDIoDSyOYTdvylcg SoEBZjnMHHYZGEqKgCgBeTbyTwyGuPZxkxsnSczotBdYyfcQsOVDVC?",
"output": "NO"
},
{
"input": "FQXBisXaJFMiHFQlXjixBDMaQuIbyqSBKGsBfTmBKCjszlGVZxEOqYYqRTUkGpSDDAoOXyXcQbHcPaegeOUBNeSD JiKOdECPOF?",
"output": "NO"
},
{
"input": "YhCuZnrWUBEed?",
"output": "NO"
},
{
"input": "hh?",
"output": "NO"
},
{
"input": "whU?",
"output": "YES"
},
{
"input": "fgwg?",
"output": "NO"
},
{
"input": "GlEmEPKrYcOnBNJUIFjszWUyVdvWw DGDjoCMtRJUburkPToCyDrOtMr?",
"output": "NO"
},
{
"input": "n?",
"output": "NO"
},
{
"input": "BueDOlxgzeNlxrzRrMbKiQdmGujEKmGxclvaPpTuHmTqBp?",
"output": "NO"
},
{
"input": "iehvZNQXDGCuVmJPOEysLyUryTdfaIxIuTzTadDbqRQGoCLXkxnyfWSGoLXebNnQQNTqAQJebbyYvHOfpUnXeWdjx?",
"output": "NO"
},
{
"input": " J ?",
"output": "NO"
},
{
"input": " j ?",
"output": "NO"
},
{
"input": " o ?",
"output": "YES"
},
{
"input": " T ?",
"output": "NO"
},
{
"input": " q ?",
"output": "NO"
},
{
"input": " j ?",
"output": "NO"
},
{
"input": " c ?",
"output": "NO"
},
{
"input": " B ?",
"output": "NO"
},
{
"input": "LuhxDHVwMPTtUIUMIQTuQETgXCOQPsfdFlyHvpfOVedjUTpGLAZGOHloIjJJtOLAlHPivzA?",
"output": "YES"
},
{
"input": "wmztmzFfwbGyOmNHENUFMTsFEMWYA?",
"output": "YES"
},
{
"input": "wGsfZCSwN PEUhNUrLfABrxA?",
"output": "YES"
},
{
"input": "mCDHENXjYbgMdBimAdPnewaHfpGWowjWrVAdvWczjw iDcUbyzMsmsnwbviiKiAyGVA?",
"output": "YES"
},
{
"input": "ARIWnwqFqxsQXsXXzHqvFjxOCttAGPUzDtWzsenPYdNXuFOIUGYZsLLK IaoxiyjBBRThoelwdPTkuCQfcBLUEJpCPIrVZlvUWA?",
"output": "YES"
},
{
"input": " PslvVpgpN BXkMFBEVXsyZFIQbBEFxGkYTeXKrOdcmhbiTUatYRUoYAayrchqbksswIlfIjerZPqptvCGnMUhyrQSvwltRhFzA?",
"output": "YES"
},
{
"input": "HpBkttwSjBXDmyleGiRWNUMPaAIE uzTrp KJDzaUiCdsMYOoWKHoUhWUoecCPmACymMUUbGav UMRpCytPETwNFAObZJA?",
"output": "YES"
}
] | 1,660,795,828
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 35
| 92
| 0
|
s = input()
for i in s[::-1]:
if i.isalpha():
if i.lower() in ["a", "e", "o","u","y", "i"]:
print("YES")
else:
print("NO")
break
|
Title: Sleuth
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya plays the sleuth with his friends. The rules of the game are as follows: those who play for the first time, that is Vasya is the sleuth, he should investigate a "crime" and find out what is happening. He can ask any questions whatsoever that can be answered with "Yes" or "No". All the rest agree beforehand to answer the questions like that: if the question’s last letter is a vowel, they answer "Yes" and if the last letter is a consonant, they answer "No". Of course, the sleuth knows nothing about it and his task is to understand that.
Unfortunately, Vasya is not very smart. After 5 hours of endless stupid questions everybody except Vasya got bored. That’s why Vasya’s friends ask you to write a program that would give answers instead of them.
The English alphabet vowels are: A, E, I, O, U, Y
The English alphabet consonants are: B, C, D, F, G, H, J, K, L, M, N, P, Q, R, S, T, V, W, X, Z
Input Specification:
The single line contains a question represented by a non-empty line consisting of large and small Latin letters, spaces and a question mark. The line length does not exceed 100. It is guaranteed that the question mark occurs exactly once in the line — as the last symbol and that the line contains at least one letter.
Output Specification:
Print answer for the question in a single line: YES if the answer is "Yes", NO if the answer is "No".
Remember that in the reply to the question the last letter, not the last character counts. I. e. the spaces and the question mark do not count as letters.
Demo Input:
['Is it a melon?\n', 'Is it an apple?\n', 'Is it a banana ?\n', 'Is it an apple and a banana simultaneouSLY?\n']
Demo Output:
['NO\n', 'YES\n', 'YES\n', 'YES\n']
Note:
none
|
```python
s = input()
for i in s[::-1]:
if i.isalpha():
if i.lower() in ["a", "e", "o","u","y", "i"]:
print("YES")
else:
print("NO")
break
```
| 3.977
|
630
|
A
|
Again Twenty Five!
|
PROGRAMMING
| 800
|
[
"number theory"
] | null | null |
The HR manager was disappointed again. The last applicant failed the interview the same way as 24 previous ones. "Do I give such a hard task?" — the HR manager thought. "Just raise number 5 to the power of *n* and get last two digits of the number. Yes, of course, *n* can be rather big, and one cannot find the power using a calculator, but we need people who are able to think, not just follow the instructions."
Could you pass the interview in the machine vision company in IT City?
|
The only line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=2·1018) — the power in which you need to raise number 5.
|
Output the last two digits of 5*n* without spaces between them.
|
[
"2\n"
] |
[
"25"
] |
none
| 0
|
[
{
"input": "2",
"output": "25"
},
{
"input": "7",
"output": "25"
},
{
"input": "1000000000000000000",
"output": "25"
},
{
"input": "2000000000000000000",
"output": "25"
},
{
"input": "987654321012345678",
"output": "25"
}
] | 1,698,053,923
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 5
| 31
| 0
|
n = input();print('25')
|
Title: Again Twenty Five!
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The HR manager was disappointed again. The last applicant failed the interview the same way as 24 previous ones. "Do I give such a hard task?" — the HR manager thought. "Just raise number 5 to the power of *n* and get last two digits of the number. Yes, of course, *n* can be rather big, and one cannot find the power using a calculator, but we need people who are able to think, not just follow the instructions."
Could you pass the interview in the machine vision company in IT City?
Input Specification:
The only line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=2·1018) — the power in which you need to raise number 5.
Output Specification:
Output the last two digits of 5*n* without spaces between them.
Demo Input:
['2\n']
Demo Output:
['25']
Note:
none
|
```python
n = input();print('25')
```
| 3
|
|
258
|
A
|
Little Elephant and Bits
|
PROGRAMMING
| 1,100
|
[
"greedy",
"math"
] | null | null |
The Little Elephant has an integer *a*, written in the binary notation. He wants to write this number on a piece of paper.
To make sure that the number *a* fits on the piece of paper, the Little Elephant ought to delete exactly one any digit from number *a* in the binary record. At that a new number appears. It consists of the remaining binary digits, written in the corresponding order (possible, with leading zeroes).
The Little Elephant wants the number he is going to write on the paper to be as large as possible. Help him find the maximum number that he can obtain after deleting exactly one binary digit and print it in the binary notation.
|
The single line contains integer *a*, written in the binary notation without leading zeroes. This number contains more than 1 and at most 105 digits.
|
In the single line print the number that is written without leading zeroes in the binary notation — the answer to the problem.
|
[
"101\n",
"110010\n"
] |
[
"11\n",
"11010\n"
] |
In the first sample the best strategy is to delete the second digit. That results in number 11<sub class="lower-index">2</sub> = 3<sub class="lower-index">10</sub>.
In the second sample the best strategy is to delete the third or fourth digits — that results in number 11010<sub class="lower-index">2</sub> = 26<sub class="lower-index">10</sub>.
| 500
|
[
{
"input": "101",
"output": "11"
},
{
"input": "110010",
"output": "11010"
},
{
"input": "10000",
"output": "1000"
},
{
"input": "1111111110",
"output": "111111111"
},
{
"input": "10100101011110101",
"output": "1100101011110101"
},
{
"input": "111010010111",
"output": "11110010111"
},
{
"input": "11110111011100000000",
"output": "1111111011100000000"
},
{
"input": "11110010010100001110110101110011110110100111101",
"output": "1111010010100001110110101110011110110100111101"
},
{
"input": "1001011111010010100111111",
"output": "101011111010010100111111"
},
{
"input": "1111111111",
"output": "111111111"
},
{
"input": "1111111111111111111100111101001110110111111000001111110101001101001110011000001011001111111000110101",
"output": "111111111111111111110111101001110110111111000001111110101001101001110011000001011001111111000110101"
},
{
"input": "11010110000100100101111110111001001010011000011011000010010100111010101000111010011101101111110001111000101000001100011101110100",
"output": "1110110000100100101111110111001001010011000011011000010010100111010101000111010011101101111110001111000101000001100011101110100"
},
{
"input": "11111111111111111111111110110111001101100111010010101101101001011100011011000111010011110010101100010001011101011010010100001000011100001101101001100010100001001010010100100001111110100110011000101100001111111011010111001011111110111101000100101001001011",
"output": "1111111111111111111111111110111001101100111010010101101101001011100011011000111010011110010101100010001011101011010010100001000011100001101101001100010100001001010010100100001111110100110011000101100001111111011010111001011111110111101000100101001001011"
},
{
"input": "11100010010010000110101101101100111111001010001101101001001111010110010111001011010000001100110101000101111000001111101111110010000010101110011110101101010110001100011101111011100010011101100111110010111111100110101000000111101000000000110100100101111101000110101010101101001110001110000101011010101100011100100111100010001011010010001100011111110010011010011000111000100111100010110100011010010101011011011111110100001110000011011",
"output": "1110010010010000110101101101100111111001010001101101001001111010110010111001011010000001100110101000101111000001111101111110010000010101110011110101101010110001100011101111011100010011101100111110010111111100110101000000111101000000000110100100101111101000110101010101101001110001110000101011010101100011100100111100010001011010010001100011111110010011010011000111000100111100010110100011010010101011011011111110100001110000011011"
},
{
"input": "11",
"output": "1"
},
{
"input": "111",
"output": "11"
},
{
"input": "111111",
"output": "11111"
},
{
"input": "11111",
"output": "1111"
},
{
"input": "1111",
"output": "111"
}
] | 1,567,277,240
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 37
| 310
| 3,379,200
|
string=list(input())
if '0' in string:
string.remove('0')
print(''.join(string))
else:
string.remove('1')
print(''.join(string))
|
Title: Little Elephant and Bits
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Little Elephant has an integer *a*, written in the binary notation. He wants to write this number on a piece of paper.
To make sure that the number *a* fits on the piece of paper, the Little Elephant ought to delete exactly one any digit from number *a* in the binary record. At that a new number appears. It consists of the remaining binary digits, written in the corresponding order (possible, with leading zeroes).
The Little Elephant wants the number he is going to write on the paper to be as large as possible. Help him find the maximum number that he can obtain after deleting exactly one binary digit and print it in the binary notation.
Input Specification:
The single line contains integer *a*, written in the binary notation without leading zeroes. This number contains more than 1 and at most 105 digits.
Output Specification:
In the single line print the number that is written without leading zeroes in the binary notation — the answer to the problem.
Demo Input:
['101\n', '110010\n']
Demo Output:
['11\n', '11010\n']
Note:
In the first sample the best strategy is to delete the second digit. That results in number 11<sub class="lower-index">2</sub> = 3<sub class="lower-index">10</sub>.
In the second sample the best strategy is to delete the third or fourth digits — that results in number 11010<sub class="lower-index">2</sub> = 26<sub class="lower-index">10</sub>.
|
```python
string=list(input())
if '0' in string:
string.remove('0')
print(''.join(string))
else:
string.remove('1')
print(''.join(string))
```
| 3
|
|
200
|
B
|
Drinks
|
PROGRAMMING
| 800
|
[
"implementation",
"math"
] | null | null |
Little Vasya loves orange juice very much. That's why any food and drink in his kitchen necessarily contains orange juice. There are *n* drinks in his fridge, the volume fraction of orange juice in the *i*-th drink equals *p**i* percent.
One day Vasya decided to make himself an orange cocktail. He took equal proportions of each of the *n* drinks and mixed them. Then he wondered, how much orange juice the cocktail has.
Find the volume fraction of orange juice in the final drink.
|
The first input line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of orange-containing drinks in Vasya's fridge. The second line contains *n* integers *p**i* (0<=≤<=*p**i*<=≤<=100) — the volume fraction of orange juice in the *i*-th drink, in percent. The numbers are separated by a space.
|
Print the volume fraction in percent of orange juice in Vasya's cocktail. The answer will be considered correct if the absolute or relative error does not exceed 10<=<=-<=4.
|
[
"3\n50 50 100\n",
"4\n0 25 50 75\n"
] |
[
"66.666666666667\n",
"37.500000000000\n"
] |
Note to the first sample: let's assume that Vasya takes *x* milliliters of each drink from the fridge. Then the volume of pure juice in the cocktail will equal <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/c1fac6e64d3a8ee6a5ac138cbe51e60039b22473.png" style="max-width: 100.0%;max-height: 100.0%;"/> milliliters. The total cocktail's volume equals 3·*x* milliliters, so the volume fraction of the juice in the cocktail equals <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/ceb0664e55a1f9f5fa1243ec74680a4665a4d58d.png" style="max-width: 100.0%;max-height: 100.0%;"/>, that is, 66.(6) percent.
| 500
|
[
{
"input": "3\n50 50 100",
"output": "66.666666666667"
},
{
"input": "4\n0 25 50 75",
"output": "37.500000000000"
},
{
"input": "3\n0 1 8",
"output": "3.000000000000"
},
{
"input": "5\n96 89 93 95 70",
"output": "88.600000000000"
},
{
"input": "7\n62 41 78 4 38 39 75",
"output": "48.142857142857"
},
{
"input": "13\n2 22 7 0 1 17 3 17 11 2 21 26 22",
"output": "11.615384615385"
},
{
"input": "21\n5 4 11 7 0 5 45 21 0 14 51 6 0 16 10 19 8 9 7 12 18",
"output": "12.761904761905"
},
{
"input": "26\n95 70 93 74 94 70 91 70 39 79 80 57 87 75 37 93 48 67 51 90 85 26 23 64 66 84",
"output": "69.538461538462"
},
{
"input": "29\n84 99 72 96 83 92 95 98 97 93 76 84 99 93 81 76 93 99 99 100 95 100 96 95 97 100 71 98 94",
"output": "91.551724137931"
},
{
"input": "33\n100 99 100 100 99 99 99 100 100 100 99 99 99 100 100 100 100 99 100 99 100 100 97 100 100 100 100 100 100 100 98 98 100",
"output": "99.515151515152"
},
{
"input": "34\n14 9 10 5 4 26 18 23 0 1 0 20 18 15 2 2 3 5 14 1 9 4 2 15 7 1 7 19 10 0 0 11 0 2",
"output": "8.147058823529"
},
{
"input": "38\n99 98 100 100 99 92 99 99 98 84 88 94 86 99 93 100 98 99 65 98 85 84 64 97 96 89 79 96 91 84 99 93 72 96 94 97 96 93",
"output": "91.921052631579"
},
{
"input": "52\n100 94 99 98 99 99 99 95 97 97 98 100 100 98 97 100 98 90 100 99 97 94 90 98 100 100 90 99 100 95 98 95 94 85 97 94 96 94 99 99 99 98 100 100 94 99 99 100 98 87 100 100",
"output": "97.019230769231"
},
{
"input": "58\n10 70 12 89 1 82 100 53 40 100 21 69 92 91 67 66 99 77 25 48 8 63 93 39 46 79 82 14 44 42 1 79 0 69 56 73 67 17 59 4 65 80 20 60 77 52 3 61 16 76 33 18 46 100 28 59 9 6",
"output": "50.965517241379"
},
{
"input": "85\n7 8 1 16 0 15 1 7 0 11 15 6 2 12 2 8 9 8 2 0 3 7 15 7 1 8 5 7 2 26 0 3 11 1 8 10 31 0 7 6 1 8 1 0 9 14 4 8 7 16 9 1 0 16 10 9 6 1 1 4 2 7 4 5 4 1 20 6 16 16 1 1 10 17 8 12 14 19 3 8 1 7 10 23 10",
"output": "7.505882352941"
},
{
"input": "74\n5 3 0 7 13 10 12 10 18 5 0 18 2 13 7 17 2 7 5 2 40 19 0 2 2 3 0 45 4 20 0 4 2 8 1 19 3 9 17 1 15 0 16 1 9 4 0 9 32 2 6 18 11 18 1 15 16 12 7 19 5 3 9 28 26 8 3 10 33 29 4 13 28 6",
"output": "10.418918918919"
},
{
"input": "98\n42 9 21 11 9 11 22 12 52 20 10 6 56 9 26 27 1 29 29 14 38 17 41 21 7 45 15 5 29 4 51 20 6 8 34 17 13 53 30 45 0 10 16 41 4 5 6 4 14 2 31 6 0 11 13 3 3 43 13 36 51 0 7 16 28 23 8 36 30 22 8 54 21 45 39 4 50 15 1 30 17 8 18 10 2 20 16 50 6 68 15 6 38 7 28 8 29 41",
"output": "20.928571428571"
},
{
"input": "99\n60 65 40 63 57 44 30 84 3 10 39 53 40 45 72 20 76 11 61 32 4 26 97 55 14 57 86 96 34 69 52 22 26 79 31 4 21 35 82 47 81 28 72 70 93 84 40 4 69 39 83 58 30 7 32 73 74 12 92 23 61 88 9 58 70 32 75 40 63 71 46 55 39 36 14 97 32 16 95 41 28 20 85 40 5 50 50 50 75 6 10 64 38 19 77 91 50 72 96",
"output": "49.191919191919"
},
{
"input": "99\n100 88 40 30 81 80 91 98 69 73 88 96 79 58 14 100 87 84 52 91 83 88 72 83 99 35 54 80 46 79 52 72 85 32 99 39 79 79 45 83 88 50 75 75 50 59 65 75 97 63 92 58 89 46 93 80 89 33 69 86 99 99 66 85 72 74 79 98 85 95 46 63 77 97 49 81 89 39 70 76 68 91 90 56 31 93 51 87 73 95 74 69 87 95 57 68 49 95 92",
"output": "73.484848484848"
},
{
"input": "100\n18 15 17 0 3 3 0 4 1 8 2 22 7 21 5 0 0 8 3 16 1 0 2 9 9 3 10 8 17 20 5 4 8 12 2 3 1 1 3 2 23 0 1 0 5 7 4 0 1 3 3 4 25 2 2 14 8 4 9 3 0 11 0 3 12 3 14 16 7 7 14 1 17 9 0 35 42 12 3 1 25 9 3 8 5 3 2 8 22 14 11 6 3 9 6 8 7 7 4 6",
"output": "7.640000000000"
},
{
"input": "100\n88 77 65 87 100 63 91 96 92 89 77 95 76 80 84 83 100 71 85 98 26 54 74 78 69 59 96 86 88 91 95 26 52 88 64 70 84 81 76 84 94 82 100 66 97 98 43 94 59 94 100 80 98 73 69 83 94 70 74 79 91 31 62 88 69 55 62 97 40 64 62 83 87 85 50 90 69 72 67 49 100 51 69 96 81 90 83 91 86 34 79 69 100 66 97 98 47 97 74 100",
"output": "77.660000000000"
},
{
"input": "100\n91 92 90 91 98 84 85 96 83 98 99 87 94 70 87 75 86 90 89 88 82 83 91 94 88 86 90 99 100 98 97 75 95 99 95 100 91 92 76 93 95 97 88 93 95 81 96 89 88 100 98 87 90 96 100 99 58 90 96 77 92 82 100 100 93 93 98 99 79 88 97 95 98 66 96 83 96 100 99 92 98 98 92 93 100 97 98 100 98 97 100 100 94 90 99 100 98 79 80 81",
"output": "91.480000000000"
},
{
"input": "1\n0",
"output": "0.000000000000"
},
{
"input": "1\n100",
"output": "100.000000000000"
},
{
"input": "1\n78",
"output": "78.000000000000"
},
{
"input": "2\n0 100",
"output": "50.000000000000"
},
{
"input": "2\n100 100",
"output": "100.000000000000"
},
{
"input": "5\n0 0 0 0 1",
"output": "0.200000000000"
},
{
"input": "100\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 99",
"output": "99.990000000000"
},
{
"input": "100\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1",
"output": "0.010000000000"
},
{
"input": "5\n100 100 100 100 100",
"output": "100.000000000000"
}
] | 1,685,090,845
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 31
| 92
| 0
|
num = int(input())
juise = input().split(' ')
total = 0.000000000000
for i in range(num):
piece = float(juise[i])/100
total += piece
present = (total/num)*100
print(round(present, 12))
|
Title: Drinks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Vasya loves orange juice very much. That's why any food and drink in his kitchen necessarily contains orange juice. There are *n* drinks in his fridge, the volume fraction of orange juice in the *i*-th drink equals *p**i* percent.
One day Vasya decided to make himself an orange cocktail. He took equal proportions of each of the *n* drinks and mixed them. Then he wondered, how much orange juice the cocktail has.
Find the volume fraction of orange juice in the final drink.
Input Specification:
The first input line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of orange-containing drinks in Vasya's fridge. The second line contains *n* integers *p**i* (0<=≤<=*p**i*<=≤<=100) — the volume fraction of orange juice in the *i*-th drink, in percent. The numbers are separated by a space.
Output Specification:
Print the volume fraction in percent of orange juice in Vasya's cocktail. The answer will be considered correct if the absolute or relative error does not exceed 10<=<=-<=4.
Demo Input:
['3\n50 50 100\n', '4\n0 25 50 75\n']
Demo Output:
['66.666666666667\n', '37.500000000000\n']
Note:
Note to the first sample: let's assume that Vasya takes *x* milliliters of each drink from the fridge. Then the volume of pure juice in the cocktail will equal <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/c1fac6e64d3a8ee6a5ac138cbe51e60039b22473.png" style="max-width: 100.0%;max-height: 100.0%;"/> milliliters. The total cocktail's volume equals 3·*x* milliliters, so the volume fraction of the juice in the cocktail equals <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/ceb0664e55a1f9f5fa1243ec74680a4665a4d58d.png" style="max-width: 100.0%;max-height: 100.0%;"/>, that is, 66.(6) percent.
|
```python
num = int(input())
juise = input().split(' ')
total = 0.000000000000
for i in range(num):
piece = float(juise[i])/100
total += piece
present = (total/num)*100
print(round(present, 12))
```
| 3
|
|
306
|
A
|
Candies
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Polycarpus has got *n* candies and *m* friends (*n*<=≥<=*m*). He wants to make a New Year present with candies to each friend. Polycarpus is planning to present all candies and he wants to do this in the fairest (that is, most equal) manner. He wants to choose such *a**i*, where *a**i* is the number of candies in the *i*-th friend's present, that the maximum *a**i* differs from the least *a**i* as little as possible.
For example, if *n* is divisible by *m*, then he is going to present the same number of candies to all his friends, that is, the maximum *a**i* won't differ from the minimum one.
|
The single line of the input contains a pair of space-separated positive integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=100;*n*<=≥<=*m*) — the number of candies and the number of Polycarpus's friends.
|
Print the required sequence *a*1,<=*a*2,<=...,<=*a**m*, where *a**i* is the number of candies in the *i*-th friend's present. All numbers *a**i* must be positive integers, total up to *n*, the maximum one should differ from the minimum one by the smallest possible value.
|
[
"12 3\n",
"15 4\n",
"18 7\n"
] |
[
"4 4 4 ",
"3 4 4 4 ",
"2 2 2 3 3 3 3 "
] |
Print *a*<sub class="lower-index">*i*</sub> in any order, separate the numbers by spaces.
| 500
|
[
{
"input": "12 4",
"output": "3 3 3 3 "
},
{
"input": "15 6",
"output": "2 2 2 3 3 3 "
},
{
"input": "18 8",
"output": "2 2 2 2 2 2 3 3 "
},
{
"input": "1 1",
"output": "1 "
},
{
"input": "2 1",
"output": "2 "
},
{
"input": "100 1",
"output": "100 "
},
{
"input": "100 100",
"output": "1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 "
},
{
"input": "100 10",
"output": "10 10 10 10 10 10 10 10 10 10 "
},
{
"input": "64 8",
"output": "8 8 8 8 8 8 8 8 "
},
{
"input": "81 27",
"output": "3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 "
},
{
"input": "100 99",
"output": "1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 "
},
{
"input": "100 51",
"output": "1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 "
},
{
"input": "100 49",
"output": "2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 3 "
},
{
"input": "100 13",
"output": "7 7 7 7 8 8 8 8 8 8 8 8 8 "
},
{
"input": "97 11",
"output": "8 8 9 9 9 9 9 9 9 9 9 "
},
{
"input": "19 3",
"output": "6 6 7 "
},
{
"input": "99 10",
"output": "9 10 10 10 10 10 10 10 10 10 "
},
{
"input": "88 5",
"output": "17 17 18 18 18 "
},
{
"input": "87 34",
"output": "2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 "
},
{
"input": "97 55",
"output": "1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 "
}
] | 1,631,647,926
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 20
| 154
| 6,758,400
|
n,m = map(int, input().split())
l = [n//m]*(m-n%m) + [n//m+1]*(n%m)
print(*l)
|
Title: Candies
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarpus has got *n* candies and *m* friends (*n*<=≥<=*m*). He wants to make a New Year present with candies to each friend. Polycarpus is planning to present all candies and he wants to do this in the fairest (that is, most equal) manner. He wants to choose such *a**i*, where *a**i* is the number of candies in the *i*-th friend's present, that the maximum *a**i* differs from the least *a**i* as little as possible.
For example, if *n* is divisible by *m*, then he is going to present the same number of candies to all his friends, that is, the maximum *a**i* won't differ from the minimum one.
Input Specification:
The single line of the input contains a pair of space-separated positive integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=100;*n*<=≥<=*m*) — the number of candies and the number of Polycarpus's friends.
Output Specification:
Print the required sequence *a*1,<=*a*2,<=...,<=*a**m*, where *a**i* is the number of candies in the *i*-th friend's present. All numbers *a**i* must be positive integers, total up to *n*, the maximum one should differ from the minimum one by the smallest possible value.
Demo Input:
['12 3\n', '15 4\n', '18 7\n']
Demo Output:
['4 4 4 ', '3 4 4 4 ', '2 2 2 3 3 3 3 ']
Note:
Print *a*<sub class="lower-index">*i*</sub> in any order, separate the numbers by spaces.
|
```python
n,m = map(int, input().split())
l = [n//m]*(m-n%m) + [n//m+1]*(n%m)
print(*l)
```
| 3
|
|
1
|
A
|
Theatre Square
|
PROGRAMMING
| 1,000
|
[
"math"
] |
A. Theatre Square
|
1
|
256
|
Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=×<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=×<=*a*.
What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square.
|
The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109).
|
Write the needed number of flagstones.
|
[
"6 6 4\n"
] |
[
"4\n"
] |
none
| 0
|
[
{
"input": "6 6 4",
"output": "4"
},
{
"input": "1 1 1",
"output": "1"
},
{
"input": "2 1 1",
"output": "2"
},
{
"input": "1 2 1",
"output": "2"
},
{
"input": "2 2 1",
"output": "4"
},
{
"input": "2 1 2",
"output": "1"
},
{
"input": "1 1 3",
"output": "1"
},
{
"input": "2 3 4",
"output": "1"
},
{
"input": "1000000000 1000000000 1",
"output": "1000000000000000000"
},
{
"input": "12 13 4",
"output": "12"
},
{
"input": "222 332 5",
"output": "3015"
},
{
"input": "1000 1000 10",
"output": "10000"
},
{
"input": "1001 1000 10",
"output": "10100"
},
{
"input": "100 10001 1000000000",
"output": "1"
},
{
"input": "1000000000 1000000000 1000000000",
"output": "1"
},
{
"input": "1000000000 1000000000 999999999",
"output": "4"
},
{
"input": "1000000000 1000000000 192",
"output": "27126743055556"
},
{
"input": "1000000000 987654321 1",
"output": "987654321000000000"
},
{
"input": "456784567 1000000000 51",
"output": "175618850864484"
},
{
"input": "39916800 134217728 40320",
"output": "3295710"
}
] | 1,692,886,717
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 20
| 46
| 0
|
import math
n,m,a=map(int,input().split())
ans=math.ceil(n/a)*math.ceil(m/a)
print(ans) #23
|
Title: Theatre Square
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=×<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=×<=*a*.
What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square.
Input Specification:
The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109).
Output Specification:
Write the needed number of flagstones.
Demo Input:
['6 6 4\n']
Demo Output:
['4\n']
Note:
none
|
```python
import math
n,m,a=map(int,input().split())
ans=math.ceil(n/a)*math.ceil(m/a)
print(ans) #23
```
| 3.977
|
999
|
A
|
Mishka and Contest
|
PROGRAMMING
| 800
|
[
"brute force",
"implementation"
] | null | null |
Mishka started participating in a programming contest. There are $n$ problems in the contest. Mishka's problem-solving skill is equal to $k$.
Mishka arranges all problems from the contest into a list. Because of his weird principles, Mishka only solves problems from one of the ends of the list. Every time, he chooses which end (left or right) he will solve the next problem from. Thus, each problem Mishka solves is either the leftmost or the rightmost problem in the list.
Mishka cannot solve a problem with difficulty greater than $k$. When Mishka solves the problem, it disappears from the list, so the length of the list decreases by $1$. Mishka stops when he is unable to solve any problem from any end of the list.
How many problems can Mishka solve?
|
The first line of input contains two integers $n$ and $k$ ($1 \le n, k \le 100$) — the number of problems in the contest and Mishka's problem-solving skill.
The second line of input contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 100$), where $a_i$ is the difficulty of the $i$-th problem. The problems are given in order from the leftmost to the rightmost in the list.
|
Print one integer — the maximum number of problems Mishka can solve.
|
[
"8 4\n4 2 3 1 5 1 6 4\n",
"5 2\n3 1 2 1 3\n",
"5 100\n12 34 55 43 21\n"
] |
[
"5\n",
"0\n",
"5\n"
] |
In the first example, Mishka can solve problems in the following order: $[4, 2, 3, 1, 5, 1, 6, 4] \rightarrow [2, 3, 1, 5, 1, 6, 4] \rightarrow [2, 3, 1, 5, 1, 6] \rightarrow [3, 1, 5, 1, 6] \rightarrow [1, 5, 1, 6] \rightarrow [5, 1, 6]$, so the number of solved problems will be equal to $5$.
In the second example, Mishka can't solve any problem because the difficulties of problems from both ends are greater than $k$.
In the third example, Mishka's solving skill is so amazing that he can solve all the problems.
| 0
|
[
{
"input": "8 4\n4 2 3 1 5 1 6 4",
"output": "5"
},
{
"input": "5 2\n3 1 2 1 3",
"output": "0"
},
{
"input": "5 100\n12 34 55 43 21",
"output": "5"
},
{
"input": "100 100\n44 47 36 83 76 94 86 69 31 2 22 77 37 51 10 19 25 78 53 25 1 29 48 95 35 53 22 72 49 86 60 38 13 91 89 18 54 19 71 2 25 33 65 49 53 5 95 90 100 68 25 5 87 48 45 72 34 14 100 44 94 75 80 26 25 7 57 82 49 73 55 43 42 60 34 8 51 11 71 41 81 23 20 89 12 72 68 26 96 92 32 63 13 47 19 9 35 56 79 62",
"output": "100"
},
{
"input": "100 99\n84 82 43 4 71 3 30 92 15 47 76 43 2 17 76 4 1 33 24 96 44 98 75 99 59 11 73 27 67 17 8 88 69 41 44 22 91 48 4 46 42 21 21 67 85 51 57 84 11 100 100 59 39 72 89 82 74 19 98 14 37 97 20 78 38 52 44 83 19 83 69 32 56 6 93 13 98 80 80 2 33 71 11 15 55 51 98 58 16 91 39 32 83 58 77 79 88 81 17 98",
"output": "98"
},
{
"input": "100 69\n80 31 12 89 16 35 8 28 39 12 32 51 42 67 64 53 17 88 63 97 29 41 57 28 51 33 82 75 93 79 57 86 32 100 83 82 99 33 1 27 86 22 65 15 60 100 42 37 38 85 26 43 90 62 91 13 1 92 16 20 100 19 28 30 23 6 5 69 24 22 9 1 10 14 28 14 25 9 32 8 67 4 39 7 10 57 15 7 8 35 62 6 53 59 62 13 24 7 53 2",
"output": "39"
},
{
"input": "100 2\n2 2 2 2 1 1 1 2 1 2 2 2 1 2 2 2 2 1 2 1 2 1 1 1 2 1 2 1 2 1 1 2 2 2 2 2 1 2 1 2 1 1 2 1 2 1 1 2 1 2 1 2 2 1 2 1 2 1 1 2 1 2 2 1 1 2 2 2 1 1 2 1 1 2 2 2 1 1 1 2 2 2 1 2 1 2 1 1 1 1 1 1 1 1 1 1 1 2 2 16",
"output": "99"
},
{
"input": "100 3\n86 53 82 40 2 20 59 2 46 63 75 49 24 81 70 22 9 9 93 72 47 23 29 77 78 51 17 59 19 71 35 3 20 60 70 9 11 96 71 94 91 19 88 93 50 49 72 19 53 30 38 67 62 71 81 86 5 26 5 32 63 98 1 97 22 32 87 65 96 55 43 85 56 37 56 67 12 100 98 58 77 54 18 20 33 53 21 66 24 64 42 71 59 32 51 69 49 79 10 1",
"output": "1"
},
{
"input": "13 7\n1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "13"
},
{
"input": "1 5\n4",
"output": "1"
},
{
"input": "3 2\n1 4 1",
"output": "2"
},
{
"input": "1 2\n100",
"output": "0"
},
{
"input": "7 4\n4 2 3 4 4 2 3",
"output": "7"
},
{
"input": "1 2\n1",
"output": "1"
},
{
"input": "1 2\n15",
"output": "0"
},
{
"input": "2 1\n1 1",
"output": "2"
},
{
"input": "5 3\n3 4 3 2 1",
"output": "4"
},
{
"input": "1 1\n2",
"output": "0"
},
{
"input": "1 5\n1",
"output": "1"
},
{
"input": "6 6\n7 1 1 1 1 1",
"output": "5"
},
{
"input": "5 5\n6 5 5 5 5",
"output": "4"
},
{
"input": "1 4\n2",
"output": "1"
},
{
"input": "9 4\n1 2 1 2 4 2 1 2 1",
"output": "9"
},
{
"input": "1 1\n1",
"output": "1"
},
{
"input": "1 10\n5",
"output": "1"
},
{
"input": "5 5\n1 1 1 1 1",
"output": "5"
},
{
"input": "100 10\n2 5 1 10 10 2 7 7 9 4 1 8 1 1 8 4 7 9 10 5 7 9 5 6 7 2 7 5 3 2 1 82 4 80 9 8 6 1 10 7 5 7 1 5 6 7 19 4 2 4 6 2 1 8 31 6 2 2 57 42 3 2 7 1 9 5 10 8 5 4 10 8 3 5 8 7 2 7 6 5 3 3 4 10 6 7 10 8 7 10 7 2 4 6 8 10 10 2 6 4",
"output": "71"
},
{
"input": "100 90\n17 16 5 51 17 62 24 45 49 41 90 30 19 78 67 66 59 34 28 47 42 8 33 77 90 41 61 16 86 33 43 71 90 95 23 9 56 41 24 90 31 12 77 36 90 67 47 15 92 50 79 88 42 19 21 79 86 60 41 26 47 4 70 62 44 90 82 89 84 91 54 16 90 53 29 69 21 44 18 28 88 74 56 43 12 76 10 22 34 24 27 52 28 76 90 75 5 29 50 90",
"output": "63"
},
{
"input": "100 10\n6 4 8 4 1 9 4 8 5 2 2 5 2 6 10 2 2 5 3 5 2 3 10 5 2 9 1 1 6 1 5 9 16 42 33 49 26 31 81 27 53 63 81 90 55 97 70 51 87 21 79 62 60 91 54 95 26 26 30 61 87 79 47 11 59 34 40 82 37 40 81 2 7 1 8 4 10 7 1 10 8 7 3 5 2 8 3 3 9 2 1 1 5 7 8 7 1 10 9 8",
"output": "61"
},
{
"input": "100 90\n45 57 52 69 17 81 85 60 59 39 55 14 87 90 90 31 41 57 35 89 74 20 53 4 33 49 71 11 46 90 71 41 71 90 63 74 51 13 99 92 99 91 100 97 93 40 93 96 100 99 100 92 98 96 78 91 91 91 91 100 94 97 95 97 96 95 17 13 45 35 54 26 2 74 6 51 20 3 73 90 90 42 66 43 86 28 84 70 37 27 90 30 55 80 6 58 57 51 10 22",
"output": "72"
},
{
"input": "100 10\n10 2 10 10 10 10 10 10 10 7 10 10 10 10 10 10 9 10 10 10 10 10 10 10 10 7 9 10 10 10 37 10 4 10 10 10 59 5 95 10 10 10 10 39 10 10 10 10 10 10 10 5 10 10 10 10 10 10 10 10 10 10 10 10 66 10 10 10 10 10 5 10 10 10 10 10 10 44 10 10 10 10 10 10 10 10 10 10 10 7 10 10 10 10 10 10 10 10 10 2",
"output": "52"
},
{
"input": "100 90\n57 90 90 90 90 90 90 90 81 90 3 90 39 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 92 90 90 90 90 90 90 90 90 98 90 90 90 90 90 90 90 90 90 90 90 90 90 54 90 90 90 90 90 62 90 90 91 90 90 90 90 90 90 91 90 90 90 90 90 90 90 3 90 90 90 90 90 90 90 2 90 90 90 90 90 90 90 90 90 2 90 90 90 90 90",
"output": "60"
},
{
"input": "100 10\n10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 6 10 10 10 10 10 10 78 90 61 40 87 39 91 50 64 30 10 24 10 55 28 11 28 35 26 26 10 57 45 67 14 99 96 51 67 79 59 11 21 55 70 33 10 16 92 70 38 50 66 52 5 10 10 10 2 4 10 10 10 10 10 10 10 10 10 6 10 10 10 10 10 10 10 10 10 10 8 10 10 10 10 10",
"output": "56"
},
{
"input": "100 90\n90 90 90 90 90 90 55 21 90 90 90 90 90 90 90 90 90 90 69 83 90 90 90 90 90 90 90 90 93 95 92 98 92 97 91 92 92 91 91 95 94 95 100 100 96 97 94 93 90 90 95 95 97 99 90 95 98 91 94 96 99 99 94 95 95 97 99 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 12 90 3 90 90 90 90 90 90 90",
"output": "61"
},
{
"input": "100 49\n71 25 14 36 36 48 36 49 28 40 49 49 49 38 40 49 33 22 49 49 14 46 8 44 49 11 37 49 40 49 2 49 3 49 37 49 49 11 25 49 49 32 49 11 49 30 16 21 49 49 23 24 30 49 49 49 49 49 49 27 49 42 49 49 20 32 30 29 35 49 30 49 9 49 27 25 5 49 49 42 49 20 49 35 49 22 15 49 49 49 19 49 29 28 13 49 22 7 6 24",
"output": "99"
},
{
"input": "100 50\n38 68 9 6 50 18 19 50 50 20 33 34 43 50 24 50 50 2 50 50 50 50 50 21 30 50 41 40 50 50 50 50 50 7 50 21 19 23 1 50 24 50 50 50 25 50 50 50 50 50 50 50 7 24 28 18 50 5 43 50 20 50 13 50 50 16 50 3 2 24 50 50 18 5 50 4 50 50 38 50 33 49 12 33 11 14 50 50 50 33 50 50 50 50 50 50 7 4 50 50",
"output": "99"
},
{
"input": "100 48\n8 6 23 47 29 48 48 48 48 48 48 26 24 48 48 48 3 48 27 28 41 45 9 29 48 48 48 48 48 48 48 48 48 48 47 23 48 48 48 5 48 22 40 48 48 48 20 48 48 57 48 32 19 48 33 2 4 19 48 48 39 48 16 48 48 44 48 48 48 48 29 14 25 43 46 7 48 19 30 48 18 8 39 48 30 47 35 18 48 45 48 48 30 13 48 48 48 17 9 48",
"output": "99"
},
{
"input": "100 57\n57 9 57 4 43 57 57 57 57 26 57 18 57 57 57 57 57 57 57 47 33 57 57 43 57 57 55 57 14 57 57 4 1 57 57 57 57 57 46 26 57 57 57 57 57 57 57 39 57 57 57 5 57 12 11 57 57 57 25 37 34 57 54 18 29 57 39 57 5 57 56 34 57 24 7 57 57 57 2 57 57 57 57 1 55 39 19 57 57 57 57 21 3 40 13 3 57 57 62 57",
"output": "99"
},
{
"input": "100 51\n51 51 38 51 51 45 51 51 51 18 51 36 51 19 51 26 37 51 11 51 45 34 51 21 51 51 33 51 6 51 51 51 21 47 51 13 51 51 30 29 50 51 51 51 51 51 51 45 14 51 2 51 51 23 9 51 50 23 51 29 34 51 40 32 1 36 31 51 11 51 51 47 51 51 51 51 51 51 51 50 39 51 14 4 4 12 3 11 51 51 51 51 41 51 51 51 49 37 5 93",
"output": "99"
},
{
"input": "100 50\n87 91 95 73 50 50 16 97 39 24 58 50 33 89 42 37 50 50 12 71 3 55 50 50 80 10 76 50 52 36 88 44 66 69 86 71 77 50 72 50 21 55 50 50 78 61 75 89 65 2 50 69 62 47 11 92 97 77 41 31 55 29 35 51 36 48 50 91 92 86 50 36 50 94 51 74 4 27 55 63 50 36 87 50 67 7 65 75 20 96 88 50 41 73 35 51 66 21 29 33",
"output": "3"
},
{
"input": "100 50\n50 37 28 92 7 76 50 50 50 76 100 57 50 50 50 32 76 50 8 72 14 8 50 91 67 50 55 82 50 50 24 97 88 50 59 61 68 86 44 15 61 67 88 50 40 50 36 99 1 23 63 50 88 59 76 82 99 76 68 50 50 30 31 68 57 98 71 12 15 60 35 79 90 6 67 50 50 50 50 68 13 6 50 50 16 87 84 50 67 67 50 64 50 58 50 50 77 51 50 51",
"output": "3"
},
{
"input": "100 50\n43 50 50 91 97 67 6 50 86 50 76 60 50 59 4 56 11 38 49 50 37 50 50 20 60 47 33 54 95 58 22 50 77 77 72 9 57 40 81 57 95 50 81 63 62 76 13 87 50 39 74 69 50 99 63 1 11 62 84 31 97 99 56 73 70 36 45 100 28 91 93 9 19 52 73 50 83 58 84 52 86 12 50 44 64 52 97 50 12 71 97 52 87 66 83 66 86 50 9 49",
"output": "6"
},
{
"input": "88 10\n10 8 1 10 10 1 3 7 10 5 8 8 10 2 7 10 10 10 10 10 1 10 10 10 10 1 2 9 10 9 10 10 10 64 100 25 10 12 9 52 13 8 10 56 10 4 10 7 10 3 10 79 74 8 73 10 10 10 9 10 3 5 10 10 10 5 1 10 10 4 3 10 10 10 4 10 6 4 10 10 10 10 3 3 8 5 6 8",
"output": "66"
},
{
"input": "100 50\n80 39 33 69 75 50 23 88 50 50 67 90 87 50 29 15 55 32 60 50 50 50 38 95 62 50 50 88 8 97 45 50 42 12 22 93 49 50 24 50 50 71 60 4 50 72 57 57 50 50 50 83 69 17 1 31 72 55 50 11 50 80 93 41 91 94 20 60 50 50 51 48 53 56 76 73 50 72 19 98 50 50 50 50 50 28 48 45 62 11 16 67 93 88 63 50 50 66 48 95",
"output": "0"
},
{
"input": "100 50\n70 50 38 50 38 50 32 30 50 31 26 42 50 33 34 50 50 50 28 21 50 44 50 47 50 50 9 40 50 50 50 50 50 42 50 50 16 50 50 3 24 50 50 50 4 26 50 2 50 50 33 1 27 50 50 50 8 29 50 23 33 50 6 29 50 50 15 50 50 50 32 50 43 50 50 50 31 50 4 50 50 31 50 50 31 16 50 17 50 17 31 13 25 16 50 10 50 47 50 66",
"output": "0"
},
{
"input": "2 8\n8 8",
"output": "2"
},
{
"input": "1 6\n3",
"output": "1"
},
{
"input": "1 5\n5",
"output": "1"
}
] | 1,617,276,532
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 48
| 77
| 0
|
n,k = map(int,input().split())
li = list(map(int,input().split()))
i = l = 0
j =n-1
count = 0
while i<=j:
if li[i]<=k:
count+=1
i+=1
elif li[j]<=k:
j-=1
count+=1
else:
break
print(count)
|
Title: Mishka and Contest
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mishka started participating in a programming contest. There are $n$ problems in the contest. Mishka's problem-solving skill is equal to $k$.
Mishka arranges all problems from the contest into a list. Because of his weird principles, Mishka only solves problems from one of the ends of the list. Every time, he chooses which end (left or right) he will solve the next problem from. Thus, each problem Mishka solves is either the leftmost or the rightmost problem in the list.
Mishka cannot solve a problem with difficulty greater than $k$. When Mishka solves the problem, it disappears from the list, so the length of the list decreases by $1$. Mishka stops when he is unable to solve any problem from any end of the list.
How many problems can Mishka solve?
Input Specification:
The first line of input contains two integers $n$ and $k$ ($1 \le n, k \le 100$) — the number of problems in the contest and Mishka's problem-solving skill.
The second line of input contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 100$), where $a_i$ is the difficulty of the $i$-th problem. The problems are given in order from the leftmost to the rightmost in the list.
Output Specification:
Print one integer — the maximum number of problems Mishka can solve.
Demo Input:
['8 4\n4 2 3 1 5 1 6 4\n', '5 2\n3 1 2 1 3\n', '5 100\n12 34 55 43 21\n']
Demo Output:
['5\n', '0\n', '5\n']
Note:
In the first example, Mishka can solve problems in the following order: $[4, 2, 3, 1, 5, 1, 6, 4] \rightarrow [2, 3, 1, 5, 1, 6, 4] \rightarrow [2, 3, 1, 5, 1, 6] \rightarrow [3, 1, 5, 1, 6] \rightarrow [1, 5, 1, 6] \rightarrow [5, 1, 6]$, so the number of solved problems will be equal to $5$.
In the second example, Mishka can't solve any problem because the difficulties of problems from both ends are greater than $k$.
In the third example, Mishka's solving skill is so amazing that he can solve all the problems.
|
```python
n,k = map(int,input().split())
li = list(map(int,input().split()))
i = l = 0
j =n-1
count = 0
while i<=j:
if li[i]<=k:
count+=1
i+=1
elif li[j]<=k:
j-=1
count+=1
else:
break
print(count)
```
| 3
|
|
981
|
A
|
Antipalindrome
|
PROGRAMMING
| 900
|
[
"brute force",
"implementation",
"strings"
] | null | null |
A string is a palindrome if it reads the same from the left to the right and from the right to the left. For example, the strings "kek", "abacaba", "r" and "papicipap" are palindromes, while the strings "abb" and "iq" are not.
A substring $s[l \ldots r]$ ($1<=\leq<=l<=\leq<=r<=\leq<=|s|$) of a string $s<==<=s_{1}s_{2} \ldots s_{|s|}$ is the string $s_{l}s_{l<=+<=1} \ldots s_{r}$.
Anna does not like palindromes, so she makes her friends call her Ann. She also changes all the words she reads in a similar way. Namely, each word $s$ is changed into its longest substring that is not a palindrome. If all the substrings of $s$ are palindromes, she skips the word at all.
Some time ago Ann read the word $s$. What is the word she changed it into?
|
The first line contains a non-empty string $s$ with length at most $50$ characters, containing lowercase English letters only.
|
If there is such a substring in $s$ that is not a palindrome, print the maximum length of such a substring. Otherwise print $0$.
Note that there can be multiple longest substrings that are not palindromes, but their length is unique.
|
[
"mew\n",
"wuffuw\n",
"qqqqqqqq\n"
] |
[
"3\n",
"5\n",
"0\n"
] |
"mew" is not a palindrome, so the longest substring of it that is not a palindrome, is the string "mew" itself. Thus, the answer for the first example is $3$.
The string "uffuw" is one of the longest non-palindrome substrings (of length $5$) of the string "wuffuw", so the answer for the second example is $5$.
All substrings of the string "qqqqqqqq" consist of equal characters so they are palindromes. This way, there are no non-palindrome substrings. Thus, the answer for the third example is $0$.
| 500
|
[
{
"input": "mew",
"output": "3"
},
{
"input": "wuffuw",
"output": "5"
},
{
"input": "qqqqqqqq",
"output": "0"
},
{
"input": "ijvji",
"output": "4"
},
{
"input": "iiiiiii",
"output": "0"
},
{
"input": "wobervhvvkihcuyjtmqhaaigvvgiaahqmtjyuchikvvhvrebow",
"output": "49"
},
{
"input": "wwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww",
"output": "0"
},
{
"input": "wobervhvvkihcuyjtmqhaaigvahheoqleromusrartldojsjvy",
"output": "50"
},
{
"input": "ijvxljt",
"output": "7"
},
{
"input": "fyhcncnchyf",
"output": "10"
},
{
"input": "ffffffffffff",
"output": "0"
},
{
"input": "fyhcncfsepqj",
"output": "12"
},
{
"input": "ybejrrlbcinttnicblrrjeby",
"output": "23"
},
{
"input": "yyyyyyyyyyyyyyyyyyyyyyyyy",
"output": "0"
},
{
"input": "ybejrrlbcintahovgjddrqatv",
"output": "25"
},
{
"input": "oftmhcmclgyqaojljoaqyglcmchmtfo",
"output": "30"
},
{
"input": "oooooooooooooooooooooooooooooooo",
"output": "0"
},
{
"input": "oftmhcmclgyqaojllbotztajglsmcilv",
"output": "32"
},
{
"input": "gxandbtgpbknxvnkjaajknvxnkbpgtbdnaxg",
"output": "35"
},
{
"input": "gggggggggggggggggggggggggggggggggggg",
"output": "0"
},
{
"input": "gxandbtgpbknxvnkjaygommzqitqzjfalfkk",
"output": "36"
},
{
"input": "fcliblymyqckxvieotjooojtoeivxkcqymylbilcf",
"output": "40"
},
{
"input": "fffffffffffffffffffffffffffffffffffffffffff",
"output": "0"
},
{
"input": "fcliblymyqckxvieotjootiqwtyznhhvuhbaixwqnsy",
"output": "43"
},
{
"input": "rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr",
"output": "0"
},
{
"input": "rajccqwqnqmshmerpvjyfepxwpxyldzpzhctqjnstxyfmlhiy",
"output": "49"
},
{
"input": "a",
"output": "0"
},
{
"input": "abca",
"output": "4"
},
{
"input": "aaaaabaaaaa",
"output": "10"
},
{
"input": "aba",
"output": "2"
},
{
"input": "asaa",
"output": "4"
},
{
"input": "aabaa",
"output": "4"
},
{
"input": "aabbaa",
"output": "5"
},
{
"input": "abcdaaa",
"output": "7"
},
{
"input": "aaholaa",
"output": "7"
},
{
"input": "abcdefghijka",
"output": "12"
},
{
"input": "aaadcba",
"output": "7"
},
{
"input": "aaaabaaaa",
"output": "8"
},
{
"input": "abaa",
"output": "4"
},
{
"input": "abcbaa",
"output": "6"
},
{
"input": "ab",
"output": "2"
},
{
"input": "l",
"output": "0"
},
{
"input": "aaaabcaaaa",
"output": "10"
},
{
"input": "abbaaaaaabba",
"output": "11"
},
{
"input": "abaaa",
"output": "5"
},
{
"input": "baa",
"output": "3"
},
{
"input": "aaaaaaabbba",
"output": "11"
},
{
"input": "ccbcc",
"output": "4"
},
{
"input": "bbbaaab",
"output": "7"
},
{
"input": "abaaaaaaaa",
"output": "10"
},
{
"input": "abaaba",
"output": "5"
},
{
"input": "aabsdfaaaa",
"output": "10"
},
{
"input": "aaaba",
"output": "5"
},
{
"input": "aaabaaa",
"output": "6"
},
{
"input": "baaabbb",
"output": "7"
},
{
"input": "ccbbabbcc",
"output": "8"
},
{
"input": "cabc",
"output": "4"
},
{
"input": "aabcd",
"output": "5"
},
{
"input": "abcdea",
"output": "6"
},
{
"input": "bbabb",
"output": "4"
},
{
"input": "aaaaabababaaaaa",
"output": "14"
},
{
"input": "bbabbb",
"output": "6"
},
{
"input": "aababd",
"output": "6"
},
{
"input": "abaaaa",
"output": "6"
},
{
"input": "aaaaaaaabbba",
"output": "12"
},
{
"input": "aabca",
"output": "5"
},
{
"input": "aaabccbaaa",
"output": "9"
},
{
"input": "aaaaaaaaaaaaaaaaaaaab",
"output": "21"
},
{
"input": "babb",
"output": "4"
},
{
"input": "abcaa",
"output": "5"
},
{
"input": "qwqq",
"output": "4"
},
{
"input": "aaaaaaaaaaabbbbbbbbbbbbbbbaaaaaaaaaaaaaaaaaaaaaa",
"output": "48"
},
{
"input": "aaab",
"output": "4"
},
{
"input": "aaaaaabaaaaa",
"output": "12"
},
{
"input": "wwuww",
"output": "4"
},
{
"input": "aaaaabcbaaaaa",
"output": "12"
},
{
"input": "aaabbbaaa",
"output": "8"
},
{
"input": "aabcbaa",
"output": "6"
},
{
"input": "abccdefccba",
"output": "11"
},
{
"input": "aabbcbbaa",
"output": "8"
},
{
"input": "aaaabbaaaa",
"output": "9"
},
{
"input": "aabcda",
"output": "6"
},
{
"input": "abbca",
"output": "5"
},
{
"input": "aaaaaabbaaa",
"output": "11"
},
{
"input": "sssssspssssss",
"output": "12"
},
{
"input": "sdnmsdcs",
"output": "8"
},
{
"input": "aaabbbccbbbaaa",
"output": "13"
},
{
"input": "cbdbdc",
"output": "6"
},
{
"input": "abb",
"output": "3"
},
{
"input": "abcdefaaaa",
"output": "10"
},
{
"input": "abbbaaa",
"output": "7"
},
{
"input": "v",
"output": "0"
},
{
"input": "abccbba",
"output": "7"
},
{
"input": "axyza",
"output": "5"
},
{
"input": "abcdefgaaaa",
"output": "11"
},
{
"input": "aaabcdaaa",
"output": "9"
},
{
"input": "aaaacaaaa",
"output": "8"
},
{
"input": "aaaaaaaaaaaaaaaaaaaabaaaaaaaaaaaaaaaaaaaaa",
"output": "42"
},
{
"input": "abbbaa",
"output": "6"
},
{
"input": "abcdee",
"output": "6"
},
{
"input": "oom",
"output": "3"
},
{
"input": "aabcaa",
"output": "6"
},
{
"input": "abba",
"output": "3"
},
{
"input": "aaca",
"output": "4"
},
{
"input": "aacbca",
"output": "6"
},
{
"input": "ababa",
"output": "4"
},
{
"input": "abcda",
"output": "5"
},
{
"input": "cccaaccc",
"output": "7"
},
{
"input": "aaabcda",
"output": "7"
},
{
"input": "aa",
"output": "0"
},
{
"input": "aabaaaa",
"output": "7"
},
{
"input": "abbaaaa",
"output": "7"
},
{
"input": "aaabcbaaa",
"output": "8"
},
{
"input": "aabba",
"output": "5"
},
{
"input": "xyxx",
"output": "4"
},
{
"input": "aaaaaaaaaaaabc",
"output": "14"
},
{
"input": "bbaaaabb",
"output": "7"
},
{
"input": "aaabaa",
"output": "6"
},
{
"input": "sssssabsssss",
"output": "12"
},
{
"input": "bbbaaaabbb",
"output": "9"
},
{
"input": "abbbbaaaa",
"output": "9"
},
{
"input": "wwufuww",
"output": "6"
},
{
"input": "oowoo",
"output": "4"
},
{
"input": "cccaccc",
"output": "6"
},
{
"input": "aaa",
"output": "0"
},
{
"input": "bbbcc",
"output": "5"
},
{
"input": "abcdef",
"output": "6"
},
{
"input": "abbba",
"output": "4"
},
{
"input": "aab",
"output": "3"
},
{
"input": "aaba",
"output": "4"
},
{
"input": "azbyaaa",
"output": "7"
},
{
"input": "oooooiooooo",
"output": "10"
},
{
"input": "aabbbbbaaaaaa",
"output": "13"
}
] | 1,674,916,285
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 133
| 46
| 0
|
str=input()
if(str==str[::-1]):
y=str[0]
flag=0
for i in range(1,len(str)):
if(str[i]!=str[0]):
flag=1
print(len(str)-1)
break
if(flag==0):
print("0")
else:
print(len(str))
|
Title: Antipalindrome
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A string is a palindrome if it reads the same from the left to the right and from the right to the left. For example, the strings "kek", "abacaba", "r" and "papicipap" are palindromes, while the strings "abb" and "iq" are not.
A substring $s[l \ldots r]$ ($1<=\leq<=l<=\leq<=r<=\leq<=|s|$) of a string $s<==<=s_{1}s_{2} \ldots s_{|s|}$ is the string $s_{l}s_{l<=+<=1} \ldots s_{r}$.
Anna does not like palindromes, so she makes her friends call her Ann. She also changes all the words she reads in a similar way. Namely, each word $s$ is changed into its longest substring that is not a palindrome. If all the substrings of $s$ are palindromes, she skips the word at all.
Some time ago Ann read the word $s$. What is the word she changed it into?
Input Specification:
The first line contains a non-empty string $s$ with length at most $50$ characters, containing lowercase English letters only.
Output Specification:
If there is such a substring in $s$ that is not a palindrome, print the maximum length of such a substring. Otherwise print $0$.
Note that there can be multiple longest substrings that are not palindromes, but their length is unique.
Demo Input:
['mew\n', 'wuffuw\n', 'qqqqqqqq\n']
Demo Output:
['3\n', '5\n', '0\n']
Note:
"mew" is not a palindrome, so the longest substring of it that is not a palindrome, is the string "mew" itself. Thus, the answer for the first example is $3$.
The string "uffuw" is one of the longest non-palindrome substrings (of length $5$) of the string "wuffuw", so the answer for the second example is $5$.
All substrings of the string "qqqqqqqq" consist of equal characters so they are palindromes. This way, there are no non-palindrome substrings. Thus, the answer for the third example is $0$.
|
```python
str=input()
if(str==str[::-1]):
y=str[0]
flag=0
for i in range(1,len(str)):
if(str[i]!=str[0]):
flag=1
print(len(str)-1)
break
if(flag==0):
print("0")
else:
print(len(str))
```
| 3
|
|
929
|
B
|
Места в самолёте
|
PROGRAMMING
| 1,300
|
[
"*special",
"implementation"
] | null | null |
В самолёте есть *n* рядов мест. Если смотреть на ряды сверху, то в каждом ряду есть 3 места слева, затем проход между рядами, затем 4 центральных места, затем ещё один проход между рядами, а затем ещё 3 места справа.
Известно, что некоторые места уже заняты пассажирами. Всего есть два вида пассажиров — статусные (те, которые часто летают) и обычные.
Перед вами стоит задача рассадить ещё *k* обычных пассажиров так, чтобы суммарное число соседей у статусных пассажиров было минимально возможным. Два пассажира считаются соседями, если они сидят в одном ряду и между ними нет других мест и прохода между рядами. Если пассажир является соседним пассажиром для двух статусных пассажиров, то его следует учитывать в сумме соседей дважды.
|
В первой строке следуют два целых числа *n* и *k* (1<=≤<=*n*<=≤<=100, 1<=≤<=*k*<=≤<=10·*n*) — количество рядов мест в самолёте и количество пассажиров, которых нужно рассадить.
Далее следует описание рядов мест самолёта по одному ряду в строке. Если очередной символ равен '-', то это проход между рядами. Если очередной символ равен '.', то это свободное место. Если очередной символ равен 'S', то на текущем месте будет сидеть статусный пассажир. Если очередной символ равен 'P', то на текущем месте будет сидеть обычный пассажир.
Гарантируется, что количество свободных мест не меньше *k*. Гарантируется, что все ряды удовлетворяют описанному в условии формату.
|
В первую строку выведите минимальное суммарное число соседей у статусных пассажиров.
Далее выведите план рассадки пассажиров, который минимизирует суммарное количество соседей у статусных пассажиров, в том же формате, что и во входных данных. Если в свободное место нужно посадить одного из *k* пассажиров, выведите строчную букву 'x' вместо символа '.'.
|
[
"1 2\nSP.-SS.S-S.S\n",
"4 9\nPP.-PPPS-S.S\nPSP-PPSP-.S.\n.S.-S..P-SS.\nP.S-P.PP-PSP\n"
] |
[
"5\nSPx-SSxS-S.S\n",
"15\nPPx-PPPS-S.S\nPSP-PPSP-xSx\nxSx-SxxP-SSx\nP.S-PxPP-PSP\n"
] |
В первом примере нужно посадить ещё двух обычных пассажиров. Для минимизации соседей у статусных пассажиров, нужно посадить первого из них на третье слева место, а второго на любое из оставшихся двух мест, так как независимо от выбора места он станет соседом двух статусных пассажиров.
Изначально, у статусного пассажира, который сидит на самом левом месте уже есть сосед. Также на четвёртом и пятом местах слева сидят статусные пассажиры, являющиеся соседями друг для друга (что добавляет к сумме 2).
Таким образом, после посадки ещё двух обычных пассажиров, итоговое суммарное количество соседей у статусных пассажиров станет равно пяти.
| 1,000
|
[
{
"input": "1 2\nSP.-SS.S-S.S",
"output": "5\nSPx-SSxS-S.S"
},
{
"input": "4 9\nPP.-PPPS-S.S\nPSP-PPSP-.S.\n.S.-S..P-SS.\nP.S-P.PP-PSP",
"output": "15\nPPx-PPPS-S.S\nPSP-PPSP-xSx\nxSx-SxxP-SSx\nP.S-PxPP-PSP"
},
{
"input": "3 7\n.S.-SSSP-..S\nS..-.SPP-S.P\n.S.-PPPP-PSP",
"output": "13\nxSx-SSSP-xxS\nSxx-xSPP-S.P\n.S.-PPPP-PSP"
},
{
"input": "5 6\nPP.-PS.P-P..\nPPS-SP..-P.P\nP.P-....-S..\nSPP-.P.S-.S.\nSP.-S.PS-PPP",
"output": "6\nPPx-PS.P-Pxx\nPPS-SPxx-PxP\nP.P-....-S..\nSPP-.P.S-.S.\nSP.-S.PS-PPP"
},
{
"input": "1 1\n..S-PS..-.PP",
"output": "1\nx.S-PS..-.PP"
},
{
"input": "2 2\nPP.-S.SS-.S.\nSSP-SSSS-S.S",
"output": "12\nPPx-S.SS-xS.\nSSP-SSSS-S.S"
},
{
"input": "30 1\nPPP-PPP.-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP",
"output": "0\nPPP-PPPx-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP"
},
{
"input": "1 1\nSPS-....-P.P",
"output": "2\nSPS-x...-P.P"
},
{
"input": "2 1\nSSS-S.S.-SSS\nSSP-.PP.-S.S",
"output": "11\nSSS-S.S.-SSS\nSSP-xPP.-S.S"
},
{
"input": "30 1\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPP.-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP",
"output": "0\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPx-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP\nPPP-PPPP-PPP"
}
] | 1,520,079,229
| 74,329
|
Python 3
|
OK
|
TESTS
| 47
| 77
| 6,451,200
|
s = []
indxs = []
n, k = map(int, input().split())
t = 0
for i in range(n):
row = list(input().upper())
ind = []
for j in range(len(row)):
if row[j] == '.':
ind.append(j)
elif row[j] == 'S':
if j > 0 and (row[j - 1] == 'P' or row[j - 1] == 'S'): t += 1
if j < 11 and (row[j + 1] == 'P' or row[j + 1] == 'S'): t += 1
row[3] = 'P'
row[8] = 'P'
indxs.append(ind)
s.append(row)
for i in range(n):
j = 0
while True:
if k == 0 or j == len(indxs[i]): break
if indxs[i][j] > 0:
if s[i][indxs[i][j] - 1] == 'S':
j += 1
continue
if indxs[i][j] < 11:
if s[i][indxs[i][j] + 1] == 'S':
j += 1
continue
s[i][indxs[i][j]] = 'x'
indxs[i].pop(j)
k -= 1
for i in range(n):
j = 0
while True:
if k == 0 or j == len(indxs[i]): break
if indxs[i][j] > 0 and indxs[i][j] < 11:
if (s[i][indxs[i][j] - 1] == 'S' and s[i][indxs[i][j] + 1] == 'S'):
j += 1
continue
t += 1
k -= 1
s[i][indxs[i][j]] = 'x'
indxs[i].pop(j)
for i in range(n):
for j in indxs[i]:
if k == 0: break
s[i][j] = 'x'
t += 2
k -= 1
print(t)
for row in s:
row[3] = '-'
row[8] = '-'
for i in row:
print(i, end='')
print()
|
Title: Места в самолёте
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
В самолёте есть *n* рядов мест. Если смотреть на ряды сверху, то в каждом ряду есть 3 места слева, затем проход между рядами, затем 4 центральных места, затем ещё один проход между рядами, а затем ещё 3 места справа.
Известно, что некоторые места уже заняты пассажирами. Всего есть два вида пассажиров — статусные (те, которые часто летают) и обычные.
Перед вами стоит задача рассадить ещё *k* обычных пассажиров так, чтобы суммарное число соседей у статусных пассажиров было минимально возможным. Два пассажира считаются соседями, если они сидят в одном ряду и между ними нет других мест и прохода между рядами. Если пассажир является соседним пассажиром для двух статусных пассажиров, то его следует учитывать в сумме соседей дважды.
Input Specification:
В первой строке следуют два целых числа *n* и *k* (1<=≤<=*n*<=≤<=100, 1<=≤<=*k*<=≤<=10·*n*) — количество рядов мест в самолёте и количество пассажиров, которых нужно рассадить.
Далее следует описание рядов мест самолёта по одному ряду в строке. Если очередной символ равен '-', то это проход между рядами. Если очередной символ равен '.', то это свободное место. Если очередной символ равен 'S', то на текущем месте будет сидеть статусный пассажир. Если очередной символ равен 'P', то на текущем месте будет сидеть обычный пассажир.
Гарантируется, что количество свободных мест не меньше *k*. Гарантируется, что все ряды удовлетворяют описанному в условии формату.
Output Specification:
В первую строку выведите минимальное суммарное число соседей у статусных пассажиров.
Далее выведите план рассадки пассажиров, который минимизирует суммарное количество соседей у статусных пассажиров, в том же формате, что и во входных данных. Если в свободное место нужно посадить одного из *k* пассажиров, выведите строчную букву 'x' вместо символа '.'.
Demo Input:
['1 2\nSP.-SS.S-S.S\n', '4 9\nPP.-PPPS-S.S\nPSP-PPSP-.S.\n.S.-S..P-SS.\nP.S-P.PP-PSP\n']
Demo Output:
['5\nSPx-SSxS-S.S\n', '15\nPPx-PPPS-S.S\nPSP-PPSP-xSx\nxSx-SxxP-SSx\nP.S-PxPP-PSP\n']
Note:
В первом примере нужно посадить ещё двух обычных пассажиров. Для минимизации соседей у статусных пассажиров, нужно посадить первого из них на третье слева место, а второго на любое из оставшихся двух мест, так как независимо от выбора места он станет соседом двух статусных пассажиров.
Изначально, у статусного пассажира, который сидит на самом левом месте уже есть сосед. Также на четвёртом и пятом местах слева сидят статусные пассажиры, являющиеся соседями друг для друга (что добавляет к сумме 2).
Таким образом, после посадки ещё двух обычных пассажиров, итоговое суммарное количество соседей у статусных пассажиров станет равно пяти.
|
```python
s = []
indxs = []
n, k = map(int, input().split())
t = 0
for i in range(n):
row = list(input().upper())
ind = []
for j in range(len(row)):
if row[j] == '.':
ind.append(j)
elif row[j] == 'S':
if j > 0 and (row[j - 1] == 'P' or row[j - 1] == 'S'): t += 1
if j < 11 and (row[j + 1] == 'P' or row[j + 1] == 'S'): t += 1
row[3] = 'P'
row[8] = 'P'
indxs.append(ind)
s.append(row)
for i in range(n):
j = 0
while True:
if k == 0 or j == len(indxs[i]): break
if indxs[i][j] > 0:
if s[i][indxs[i][j] - 1] == 'S':
j += 1
continue
if indxs[i][j] < 11:
if s[i][indxs[i][j] + 1] == 'S':
j += 1
continue
s[i][indxs[i][j]] = 'x'
indxs[i].pop(j)
k -= 1
for i in range(n):
j = 0
while True:
if k == 0 or j == len(indxs[i]): break
if indxs[i][j] > 0 and indxs[i][j] < 11:
if (s[i][indxs[i][j] - 1] == 'S' and s[i][indxs[i][j] + 1] == 'S'):
j += 1
continue
t += 1
k -= 1
s[i][indxs[i][j]] = 'x'
indxs[i].pop(j)
for i in range(n):
for j in indxs[i]:
if k == 0: break
s[i][j] = 'x'
t += 2
k -= 1
print(t)
for row in s:
row[3] = '-'
row[8] = '-'
for i in row:
print(i, end='')
print()
```
| 3
|
|
52
|
A
|
123-sequence
|
PROGRAMMING
| 900
|
[
"implementation"
] |
A. 123-sequence
|
2
|
256
|
There is a given sequence of integers *a*1,<=*a*2,<=...,<=*a**n*, where every number is from 1 to 3 inclusively. You have to replace the minimum number of numbers in it so that all the numbers in the sequence are equal to each other.
|
The first line contains an integer *n* (1<=≤<=*n*<=≤<=106). The second line contains a sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=3).
|
Print the minimum number of replacements needed to be performed to make all the numbers in the sequence equal.
|
[
"9\n1 3 2 2 2 1 1 2 3\n"
] |
[
"5\n"
] |
In the example all the numbers equal to 1 and 3 should be replaced by 2.
| 500
|
[
{
"input": "9\n1 3 2 2 2 1 1 2 3",
"output": "5"
},
{
"input": "6\n3 3 2 2 1 3",
"output": "3"
},
{
"input": "12\n3 1 3 1 2 1 3 2 2 1 2 1",
"output": "7"
},
{
"input": "15\n3 2 1 1 1 1 3 2 2 3 3 1 2 3 2",
"output": "10"
},
{
"input": "2\n2 1",
"output": "1"
},
{
"input": "2\n3 2",
"output": "1"
},
{
"input": "2\n3 1",
"output": "1"
},
{
"input": "18\n2 3 2 1 2 3 2 1 2 3 3 3 1 2 3 3 3 2",
"output": "10"
},
{
"input": "30\n2 1 3 2 3 2 2 2 2 3 2 2 3 2 1 1 3 1 3 2 1 2 3 1 1 3 3 1 3 1",
"output": "19"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "1\n2",
"output": "0"
},
{
"input": "1\n3",
"output": "0"
}
] | 1,678,571,081
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 23
| 498
| 14,233,600
|
a = int(input());*q, = map(int, input().split());print(a - max(q.count(1), q.count(2), q.count(3)))
|
Title: 123-sequence
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
There is a given sequence of integers *a*1,<=*a*2,<=...,<=*a**n*, where every number is from 1 to 3 inclusively. You have to replace the minimum number of numbers in it so that all the numbers in the sequence are equal to each other.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=106). The second line contains a sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=3).
Output Specification:
Print the minimum number of replacements needed to be performed to make all the numbers in the sequence equal.
Demo Input:
['9\n1 3 2 2 2 1 1 2 3\n']
Demo Output:
['5\n']
Note:
In the example all the numbers equal to 1 and 3 should be replaced by 2.
|
```python
a = int(input());*q, = map(int, input().split());print(a - max(q.count(1), q.count(2), q.count(3)))
```
| 3.848988
|
381
|
A
|
Sereja and Dima
|
PROGRAMMING
| 800
|
[
"greedy",
"implementation",
"two pointers"
] | null | null |
Sereja and Dima play a game. The rules of the game are very simple. The players have *n* cards in a row. Each card contains a number, all numbers on the cards are distinct. The players take turns, Sereja moves first. During his turn a player can take one card: either the leftmost card in a row, or the rightmost one. The game ends when there is no more cards. The player who has the maximum sum of numbers on his cards by the end of the game, wins.
Sereja and Dima are being greedy. Each of them chooses the card with the larger number during his move.
Inna is a friend of Sereja and Dima. She knows which strategy the guys are using, so she wants to determine the final score, given the initial state of the game. Help her.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=1000) — the number of cards on the table. The second line contains space-separated numbers on the cards from left to right. The numbers on the cards are distinct integers from 1 to 1000.
|
On a single line, print two integers. The first number is the number of Sereja's points at the end of the game, the second number is the number of Dima's points at the end of the game.
|
[
"4\n4 1 2 10\n",
"7\n1 2 3 4 5 6 7\n"
] |
[
"12 5\n",
"16 12\n"
] |
In the first sample Sereja will take cards with numbers 10 and 2, so Sereja's sum is 12. Dima will take cards with numbers 4 and 1, so Dima's sum is 5.
| 500
|
[
{
"input": "4\n4 1 2 10",
"output": "12 5"
},
{
"input": "7\n1 2 3 4 5 6 7",
"output": "16 12"
},
{
"input": "42\n15 29 37 22 16 5 26 31 6 32 19 3 45 36 33 14 25 20 48 7 42 11 24 28 9 18 8 21 47 17 38 40 44 4 35 1 43 39 41 27 12 13",
"output": "613 418"
},
{
"input": "43\n32 1 15 48 38 26 25 14 20 44 11 30 3 42 49 19 18 46 5 45 10 23 34 9 29 41 2 52 6 17 35 4 50 22 33 51 7 28 47 13 39 37 24",
"output": "644 500"
},
{
"input": "1\n3",
"output": "3 0"
},
{
"input": "45\n553 40 94 225 415 471 126 190 647 394 515 303 189 159 308 6 139 132 326 78 455 75 85 295 135 613 360 614 351 228 578 259 258 591 444 29 33 463 561 174 368 183 140 168 646",
"output": "6848 6568"
},
{
"input": "44\n849 373 112 307 479 608 856 769 526 82 168 143 573 762 115 501 688 36 214 450 396 496 236 309 287 786 397 43 811 141 745 846 350 270 276 677 420 459 403 722 267 54 394 727",
"output": "9562 9561"
},
{
"input": "35\n10 15 18 1 28 16 2 33 6 22 23 4 9 25 35 8 7 26 3 20 30 14 31 19 27 32 11 5 29 24 21 34 13 17 12",
"output": "315 315"
},
{
"input": "17\n580 376 191 496 73 44 520 357 483 149 81 178 514 300 216 598 304",
"output": "3238 2222"
},
{
"input": "30\n334 443 223 424 168 549 189 303 429 559 516 220 459 134 344 346 316 446 209 148 487 526 69 286 102 366 518 280 392 325",
"output": "5246 4864"
},
{
"input": "95\n122 29 188 265 292 287 183 225 222 187 155 256 64 148 173 278 218 136 290 17 31 130 2 87 57 283 255 280 68 166 174 142 102 39 116 206 288 154 26 78 296 172 184 232 77 91 277 8 249 186 94 93 207 251 257 195 101 299 193 124 293 65 58 35 24 302 220 189 252 125 27 284 247 182 141 103 198 97 234 83 281 216 85 180 267 236 109 143 149 239 79 300 191 244 71",
"output": "8147 7807"
},
{
"input": "1\n1",
"output": "1 0"
}
] | 1,679,288,037
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 34
| 77
| 0
|
n = int(input())
strokes = [int(i) for i in input().split()]
l, r = 0, n - 1;
ans = [0, 0]
for i in range(0, n):
if strokes[l] > strokes[r]:
ans[i % 2] += strokes[l]
l += 1
else:
ans[i % 2] += strokes[r]
r -= 1
print(*ans)
|
Title: Sereja and Dima
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Sereja and Dima play a game. The rules of the game are very simple. The players have *n* cards in a row. Each card contains a number, all numbers on the cards are distinct. The players take turns, Sereja moves first. During his turn a player can take one card: either the leftmost card in a row, or the rightmost one. The game ends when there is no more cards. The player who has the maximum sum of numbers on his cards by the end of the game, wins.
Sereja and Dima are being greedy. Each of them chooses the card with the larger number during his move.
Inna is a friend of Sereja and Dima. She knows which strategy the guys are using, so she wants to determine the final score, given the initial state of the game. Help her.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=1000) — the number of cards on the table. The second line contains space-separated numbers on the cards from left to right. The numbers on the cards are distinct integers from 1 to 1000.
Output Specification:
On a single line, print two integers. The first number is the number of Sereja's points at the end of the game, the second number is the number of Dima's points at the end of the game.
Demo Input:
['4\n4 1 2 10\n', '7\n1 2 3 4 5 6 7\n']
Demo Output:
['12 5\n', '16 12\n']
Note:
In the first sample Sereja will take cards with numbers 10 and 2, so Sereja's sum is 12. Dima will take cards with numbers 4 and 1, so Dima's sum is 5.
|
```python
n = int(input())
strokes = [int(i) for i in input().split()]
l, r = 0, n - 1;
ans = [0, 0]
for i in range(0, n):
if strokes[l] > strokes[r]:
ans[i % 2] += strokes[l]
l += 1
else:
ans[i % 2] += strokes[r]
r -= 1
print(*ans)
```
| 3
|
|
47
|
A
|
Triangular numbers
|
PROGRAMMING
| 800
|
[
"brute force",
"math"
] |
A. Triangular numbers
|
2
|
256
|
A triangular number is the number of dots in an equilateral triangle uniformly filled with dots. For example, three dots can be arranged in a triangle; thus three is a triangular number. The *n*-th triangular number is the number of dots in a triangle with *n* dots on a side. . You can learn more about these numbers from Wikipedia (http://en.wikipedia.org/wiki/Triangular_number).
Your task is to find out if a given integer is a triangular number.
|
The first line contains the single number *n* (1<=≤<=*n*<=≤<=500) — the given integer.
|
If the given integer is a triangular number output YES, otherwise output NO.
|
[
"1\n",
"2\n",
"3\n"
] |
[
"YES\n",
"NO\n",
"YES\n"
] |
none
| 500
|
[
{
"input": "1",
"output": "YES"
},
{
"input": "2",
"output": "NO"
},
{
"input": "3",
"output": "YES"
},
{
"input": "4",
"output": "NO"
},
{
"input": "5",
"output": "NO"
},
{
"input": "6",
"output": "YES"
},
{
"input": "7",
"output": "NO"
},
{
"input": "8",
"output": "NO"
},
{
"input": "12",
"output": "NO"
},
{
"input": "10",
"output": "YES"
},
{
"input": "11",
"output": "NO"
},
{
"input": "9",
"output": "NO"
},
{
"input": "14",
"output": "NO"
},
{
"input": "15",
"output": "YES"
},
{
"input": "16",
"output": "NO"
},
{
"input": "20",
"output": "NO"
},
{
"input": "21",
"output": "YES"
},
{
"input": "22",
"output": "NO"
},
{
"input": "121",
"output": "NO"
},
{
"input": "135",
"output": "NO"
},
{
"input": "136",
"output": "YES"
},
{
"input": "137",
"output": "NO"
},
{
"input": "152",
"output": "NO"
},
{
"input": "153",
"output": "YES"
},
{
"input": "154",
"output": "NO"
},
{
"input": "171",
"output": "YES"
},
{
"input": "189",
"output": "NO"
},
{
"input": "190",
"output": "YES"
},
{
"input": "191",
"output": "NO"
},
{
"input": "210",
"output": "YES"
},
{
"input": "211",
"output": "NO"
},
{
"input": "231",
"output": "YES"
},
{
"input": "232",
"output": "NO"
},
{
"input": "252",
"output": "NO"
},
{
"input": "253",
"output": "YES"
},
{
"input": "254",
"output": "NO"
},
{
"input": "275",
"output": "NO"
},
{
"input": "276",
"output": "YES"
},
{
"input": "277",
"output": "NO"
},
{
"input": "299",
"output": "NO"
},
{
"input": "300",
"output": "YES"
},
{
"input": "301",
"output": "NO"
},
{
"input": "324",
"output": "NO"
},
{
"input": "325",
"output": "YES"
},
{
"input": "326",
"output": "NO"
},
{
"input": "350",
"output": "NO"
},
{
"input": "351",
"output": "YES"
},
{
"input": "352",
"output": "NO"
},
{
"input": "377",
"output": "NO"
},
{
"input": "378",
"output": "YES"
},
{
"input": "406",
"output": "YES"
},
{
"input": "434",
"output": "NO"
},
{
"input": "435",
"output": "YES"
},
{
"input": "464",
"output": "NO"
},
{
"input": "465",
"output": "YES"
},
{
"input": "495",
"output": "NO"
},
{
"input": "496",
"output": "YES"
},
{
"input": "497",
"output": "NO"
},
{
"input": "331",
"output": "NO"
},
{
"input": "413",
"output": "NO"
},
{
"input": "177",
"output": "NO"
},
{
"input": "122",
"output": "NO"
},
{
"input": "478",
"output": "NO"
},
{
"input": "188",
"output": "NO"
},
{
"input": "21",
"output": "YES"
},
{
"input": "474",
"output": "NO"
},
{
"input": "21",
"output": "YES"
},
{
"input": "48",
"output": "NO"
},
{
"input": "498",
"output": "NO"
},
{
"input": "499",
"output": "NO"
},
{
"input": "500",
"output": "NO"
}
] | 1,620,993,022
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 71
| 154
| 6,758,400
|
import math
def solve(x):
x = (-1 + math.sqrt(1+8*x)) / 2
if int(x) == x:
return "YES"
return "NO"
# m,n = map(int,input().split())
# for _ in range(int(input())):
a = int(input())
print(solve(a))
|
Title: Triangular numbers
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A triangular number is the number of dots in an equilateral triangle uniformly filled with dots. For example, three dots can be arranged in a triangle; thus three is a triangular number. The *n*-th triangular number is the number of dots in a triangle with *n* dots on a side. . You can learn more about these numbers from Wikipedia (http://en.wikipedia.org/wiki/Triangular_number).
Your task is to find out if a given integer is a triangular number.
Input Specification:
The first line contains the single number *n* (1<=≤<=*n*<=≤<=500) — the given integer.
Output Specification:
If the given integer is a triangular number output YES, otherwise output NO.
Demo Input:
['1\n', '2\n', '3\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n']
Note:
none
|
```python
import math
def solve(x):
x = (-1 + math.sqrt(1+8*x)) / 2
if int(x) == x:
return "YES"
return "NO"
# m,n = map(int,input().split())
# for _ in range(int(input())):
a = int(input())
print(solve(a))
```
| 3.948911
|
25
|
A
|
IQ test
|
PROGRAMMING
| 1,300
|
[
"brute force"
] |
A. IQ test
|
2
|
256
|
Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given *n* numbers differs from the others. Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given *n* numbers finds one that is different in evenness.
|
The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of numbers in the task. The second line contains *n* space-separated natural numbers, not exceeding 100. It is guaranteed, that exactly one of these numbers differs from the others in evenness.
|
Output index of number that differs from the others in evenness. Numbers are numbered from 1 in the input order.
|
[
"5\n2 4 7 8 10\n",
"4\n1 2 1 1\n"
] |
[
"3\n",
"2\n"
] |
none
| 0
|
[
{
"input": "5\n2 4 7 8 10",
"output": "3"
},
{
"input": "4\n1 2 1 1",
"output": "2"
},
{
"input": "3\n1 2 2",
"output": "1"
},
{
"input": "3\n100 99 100",
"output": "2"
},
{
"input": "3\n5 3 2",
"output": "3"
},
{
"input": "4\n43 28 1 91",
"output": "2"
},
{
"input": "4\n75 13 94 77",
"output": "3"
},
{
"input": "4\n97 8 27 3",
"output": "2"
},
{
"input": "10\n95 51 12 91 85 3 1 31 25 7",
"output": "3"
},
{
"input": "20\n88 96 66 51 14 88 2 92 18 72 18 88 20 30 4 82 90 100 24 46",
"output": "4"
},
{
"input": "30\n20 94 56 50 10 98 52 32 14 22 24 60 4 8 98 46 34 68 82 82 98 90 50 20 78 49 52 94 64 36",
"output": "26"
},
{
"input": "50\n79 27 77 57 37 45 27 49 65 33 57 21 71 19 75 85 65 61 23 97 85 9 23 1 9 3 99 77 77 21 79 69 15 37 15 7 93 81 13 89 91 31 45 93 15 97 55 80 85 83",
"output": "48"
},
{
"input": "60\n46 11 73 65 3 69 3 53 43 53 97 47 55 93 31 75 35 3 9 73 23 31 3 81 91 79 61 21 15 11 11 11 81 7 83 75 39 87 83 59 89 55 93 27 49 67 67 29 1 93 11 17 9 19 35 21 63 31 31 25",
"output": "1"
},
{
"input": "70\n28 42 42 92 64 54 22 38 38 78 62 38 4 38 14 66 4 92 66 58 94 26 4 44 41 88 48 82 44 26 74 44 48 4 16 92 34 38 26 64 94 4 30 78 50 54 12 90 8 16 80 98 28 100 74 50 36 42 92 18 76 98 8 22 2 50 58 50 64 46",
"output": "25"
},
{
"input": "100\n43 35 79 53 13 91 91 45 65 83 57 9 42 39 85 45 71 51 61 59 31 13 63 39 25 21 79 39 91 67 21 61 97 75 93 83 29 79 59 97 11 37 63 51 39 55 91 23 21 17 47 23 35 75 49 5 69 99 5 7 41 17 25 89 15 79 21 63 53 81 43 91 59 91 69 99 85 15 91 51 49 37 65 7 89 81 21 93 61 63 97 93 45 17 13 69 57 25 75 73",
"output": "13"
},
{
"input": "100\n50 24 68 60 70 30 52 22 18 74 68 98 20 82 4 46 26 68 100 78 84 58 74 98 38 88 68 86 64 80 82 100 20 22 98 98 52 6 94 10 48 68 2 18 38 22 22 82 44 20 66 72 36 58 64 6 36 60 4 96 76 64 12 90 10 58 64 60 74 28 90 26 24 60 40 58 2 16 76 48 58 36 82 60 24 44 4 78 28 38 8 12 40 16 38 6 66 24 31 76",
"output": "99"
},
{
"input": "100\n47 48 94 48 14 18 94 36 96 22 12 30 94 20 48 98 40 58 2 94 8 36 98 18 98 68 2 60 76 38 18 100 8 72 100 68 2 86 92 72 58 16 48 14 6 58 72 76 6 88 80 66 20 28 74 62 86 68 90 86 2 56 34 38 56 90 4 8 76 44 32 86 12 98 38 34 54 92 70 94 10 24 82 66 90 58 62 2 32 58 100 22 58 72 2 22 68 72 42 14",
"output": "1"
},
{
"input": "99\n38 20 68 60 84 16 28 88 60 48 80 28 4 92 70 60 46 46 20 34 12 100 76 2 40 10 8 86 6 80 50 66 12 34 14 28 26 70 46 64 34 96 10 90 98 96 56 88 50 74 70 94 2 94 24 66 68 46 22 30 6 10 64 32 88 14 98 100 64 58 50 18 50 50 8 38 8 16 54 2 60 54 62 84 92 98 4 72 66 26 14 88 99 16 10 6 88 56 22",
"output": "93"
},
{
"input": "99\n50 83 43 89 53 47 69 1 5 37 63 87 95 15 55 95 75 89 33 53 89 75 93 75 11 85 49 29 11 97 49 67 87 11 25 37 97 73 67 49 87 43 53 97 43 29 53 33 45 91 37 73 39 49 59 5 21 43 87 35 5 63 89 57 63 47 29 99 19 85 13 13 3 13 43 19 5 9 61 51 51 57 15 89 13 97 41 13 99 79 13 27 97 95 73 33 99 27 23",
"output": "1"
},
{
"input": "98\n61 56 44 30 58 14 20 24 88 28 46 56 96 52 58 42 94 50 46 30 46 80 72 88 68 16 6 60 26 90 10 98 76 20 56 40 30 16 96 20 88 32 62 30 74 58 36 76 60 4 24 36 42 54 24 92 28 14 2 74 86 90 14 52 34 82 40 76 8 64 2 56 10 8 78 16 70 86 70 42 70 74 22 18 76 98 88 28 62 70 36 72 20 68 34 48 80 98",
"output": "1"
},
{
"input": "98\n66 26 46 42 78 32 76 42 26 82 8 12 4 10 24 26 64 44 100 46 94 64 30 18 88 28 8 66 30 82 82 28 74 52 62 80 80 60 94 86 64 32 44 88 92 20 12 74 94 28 34 58 4 22 16 10 94 76 82 58 40 66 22 6 30 32 92 54 16 76 74 98 18 48 48 30 92 2 16 42 84 74 30 60 64 52 50 26 16 86 58 96 79 60 20 62 82 94",
"output": "93"
},
{
"input": "95\n9 31 27 93 17 77 75 9 9 53 89 39 51 99 5 1 11 39 27 49 91 17 27 79 81 71 37 75 35 13 93 4 99 55 85 11 23 57 5 43 5 61 15 35 23 91 3 81 99 85 43 37 39 27 5 67 7 33 75 59 13 71 51 27 15 93 51 63 91 53 43 99 25 47 17 71 81 15 53 31 59 83 41 23 73 25 91 91 13 17 25 13 55 57 29",
"output": "32"
},
{
"input": "100\n91 89 81 45 53 1 41 3 77 93 55 97 55 97 87 27 69 95 73 41 93 21 75 35 53 56 5 51 87 59 91 67 33 3 99 45 83 17 97 47 75 97 7 89 17 99 23 23 81 25 55 97 27 35 69 5 77 35 93 19 55 59 37 21 31 37 49 41 91 53 73 69 7 37 37 39 17 71 7 97 55 17 47 23 15 73 31 39 57 37 9 5 61 41 65 57 77 79 35 47",
"output": "26"
},
{
"input": "99\n38 56 58 98 80 54 26 90 14 16 78 92 52 74 40 30 84 14 44 80 16 90 98 68 26 24 78 72 42 16 84 40 14 44 2 52 50 2 12 96 58 66 8 80 44 52 34 34 72 98 74 4 66 74 56 21 8 38 76 40 10 22 48 32 98 34 12 62 80 68 64 82 22 78 58 74 20 22 48 56 12 38 32 72 6 16 74 24 94 84 26 38 18 24 76 78 98 94 72",
"output": "56"
},
{
"input": "100\n44 40 6 40 56 90 98 8 36 64 76 86 98 76 36 92 6 30 98 70 24 98 96 60 24 82 88 68 86 96 34 42 58 10 40 26 56 10 88 58 70 32 24 28 14 82 52 12 62 36 70 60 52 34 74 30 78 76 10 16 42 94 66 90 70 38 52 12 58 22 98 96 14 68 24 70 4 30 84 98 8 50 14 52 66 34 100 10 28 100 56 48 38 12 38 14 91 80 70 86",
"output": "97"
},
{
"input": "100\n96 62 64 20 90 46 56 90 68 36 30 56 70 28 16 64 94 34 6 32 34 50 94 22 90 32 40 2 72 10 88 38 28 92 20 26 56 80 4 100 100 90 16 74 74 84 8 2 30 20 80 32 16 46 92 56 42 12 96 64 64 42 64 58 50 42 74 28 2 4 36 32 70 50 54 92 70 16 45 76 28 16 18 50 48 2 62 94 4 12 52 52 4 100 70 60 82 62 98 42",
"output": "79"
},
{
"input": "99\n14 26 34 68 90 58 50 36 8 16 18 6 2 74 54 20 36 84 32 50 52 2 26 24 3 64 20 10 54 26 66 44 28 72 4 96 78 90 96 86 68 28 94 4 12 46 100 32 22 36 84 32 44 94 76 94 4 52 12 30 74 4 34 64 58 72 44 16 70 56 54 8 14 74 8 6 58 62 98 54 14 40 80 20 36 72 28 98 20 58 40 52 90 64 22 48 54 70 52",
"output": "25"
},
{
"input": "95\n82 86 30 78 6 46 80 66 74 72 16 24 18 52 52 38 60 36 86 26 62 28 22 46 96 26 94 84 20 46 66 88 76 32 12 86 74 18 34 88 4 48 94 6 58 6 100 82 4 24 88 32 54 98 34 48 6 76 42 88 42 28 100 4 22 2 10 66 82 54 98 20 60 66 38 98 32 47 86 58 6 100 12 46 2 42 8 84 78 28 24 70 34 28 86",
"output": "78"
},
{
"input": "90\n40 50 8 42 76 24 58 42 26 68 20 48 54 12 34 84 14 36 32 88 6 50 96 56 20 92 48 16 40 34 96 46 20 84 30 50 20 98 8 44 96 42 8 76 70 38 84 30 40 88 84 72 2 22 52 58 16 62 100 66 80 40 50 32 14 62 88 72 22 99 76 50 84 82 8 82 98 46 26 40 2 98 18 78 30 72 70 18 34 68",
"output": "70"
},
{
"input": "80\n81 43 87 1 55 43 53 61 27 19 43 13 89 9 33 83 75 55 97 71 91 37 95 5 21 69 81 93 95 69 31 83 55 7 97 7 79 57 8 61 27 85 49 1 15 97 63 79 29 73 41 85 5 41 31 93 67 11 63 59 15 99 91 77 43 69 23 23 81 73 19 1 67 51 1 75 99 67 3 81",
"output": "39"
},
{
"input": "98\n13 83 61 27 35 1 85 95 97 73 95 65 73 45 5 43 27 83 91 19 11 3 85 59 9 39 69 23 45 7 51 85 5 71 5 95 1 51 75 3 43 57 3 11 33 71 21 99 47 41 87 39 71 87 31 85 91 49 83 5 49 85 47 91 55 99 33 23 31 23 23 73 29 77 55 31 25 5 81 49 91 15 15 39 87 5 9 40 69 47 29 33 11 21 49 79 51 83",
"output": "88"
},
{
"input": "3\n100 100 1",
"output": "3"
}
] | 1,485,833,288
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 32
| 124
| 4,608,000
|
n = int(input())
a = list(map(int,input().split()))
res = 0
for i in a:
if i%2==0:
n-=1
if n==1:
for i in a:
if i%2==1:
res = a.index(i)
else:
for i in a:
if i%2==0:
res = a.index(i)
print(res+1)
|
Title: IQ test
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given *n* numbers differs from the others. Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given *n* numbers finds one that is different in evenness.
Input Specification:
The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of numbers in the task. The second line contains *n* space-separated natural numbers, not exceeding 100. It is guaranteed, that exactly one of these numbers differs from the others in evenness.
Output Specification:
Output index of number that differs from the others in evenness. Numbers are numbered from 1 in the input order.
Demo Input:
['5\n2 4 7 8 10\n', '4\n1 2 1 1\n']
Demo Output:
['3\n', '2\n']
Note:
none
|
```python
n = int(input())
a = list(map(int,input().split()))
res = 0
for i in a:
if i%2==0:
n-=1
if n==1:
for i in a:
if i%2==1:
res = a.index(i)
else:
for i in a:
if i%2==0:
res = a.index(i)
print(res+1)
```
| 3.960417
|
199
|
A
|
Hexadecimal's theorem
|
PROGRAMMING
| 900
|
[
"brute force",
"constructive algorithms",
"implementation",
"number theory"
] | null | null |
Recently, a chaotic virus Hexadecimal advanced a new theorem which will shake the Universe. She thinks that each Fibonacci number can be represented as sum of three not necessary different Fibonacci numbers.
Let's remember how Fibonacci numbers can be calculated. *F*0<==<=0, *F*1<==<=1, and all the next numbers are *F**i*<==<=*F**i*<=-<=2<=+<=*F**i*<=-<=1.
So, Fibonacci numbers make a sequence of numbers: 0, 1, 1, 2, 3, 5, 8, 13, ...
If you haven't run away from the PC in fear, you have to help the virus. Your task is to divide given Fibonacci number *n* by three not necessary different Fibonacci numbers or say that it is impossible.
|
The input contains of a single integer *n* (0<=≤<=*n*<=<<=109) — the number that should be represented by the rules described above. It is guaranteed that *n* is a Fibonacci number.
|
Output three required numbers: *a*, *b* and *c*. If there is no answer for the test you have to print "I'm too stupid to solve this problem" without the quotes.
If there are multiple answers, print any of them.
|
[
"3\n",
"13\n"
] |
[
"1 1 1\n",
"2 3 8\n"
] |
none
| 500
|
[
{
"input": "3",
"output": "1 1 1"
},
{
"input": "13",
"output": "2 3 8"
},
{
"input": "0",
"output": "0 0 0"
},
{
"input": "1",
"output": "1 0 0"
},
{
"input": "2",
"output": "1 1 0"
},
{
"input": "1597",
"output": "233 377 987"
},
{
"input": "0",
"output": "0 0 0"
},
{
"input": "1",
"output": "1 0 0"
},
{
"input": "1",
"output": "1 0 0"
},
{
"input": "2",
"output": "1 1 0"
},
{
"input": "3",
"output": "1 1 1"
},
{
"input": "5",
"output": "1 1 3"
},
{
"input": "8",
"output": "1 2 5"
},
{
"input": "13",
"output": "2 3 8"
},
{
"input": "21",
"output": "3 5 13"
},
{
"input": "34",
"output": "5 8 21"
},
{
"input": "55",
"output": "8 13 34"
},
{
"input": "89",
"output": "13 21 55"
},
{
"input": "144",
"output": "21 34 89"
},
{
"input": "233",
"output": "34 55 144"
},
{
"input": "377",
"output": "55 89 233"
},
{
"input": "610",
"output": "89 144 377"
},
{
"input": "987",
"output": "144 233 610"
},
{
"input": "1597",
"output": "233 377 987"
},
{
"input": "2584",
"output": "377 610 1597"
},
{
"input": "4181",
"output": "610 987 2584"
},
{
"input": "6765",
"output": "987 1597 4181"
},
{
"input": "10946",
"output": "1597 2584 6765"
},
{
"input": "17711",
"output": "2584 4181 10946"
},
{
"input": "28657",
"output": "4181 6765 17711"
},
{
"input": "46368",
"output": "6765 10946 28657"
},
{
"input": "75025",
"output": "10946 17711 46368"
},
{
"input": "121393",
"output": "17711 28657 75025"
},
{
"input": "196418",
"output": "28657 46368 121393"
},
{
"input": "317811",
"output": "46368 75025 196418"
},
{
"input": "514229",
"output": "75025 121393 317811"
},
{
"input": "832040",
"output": "121393 196418 514229"
},
{
"input": "1346269",
"output": "196418 317811 832040"
},
{
"input": "2178309",
"output": "317811 514229 1346269"
},
{
"input": "3524578",
"output": "514229 832040 2178309"
},
{
"input": "5702887",
"output": "832040 1346269 3524578"
},
{
"input": "9227465",
"output": "1346269 2178309 5702887"
},
{
"input": "14930352",
"output": "2178309 3524578 9227465"
},
{
"input": "24157817",
"output": "3524578 5702887 14930352"
},
{
"input": "39088169",
"output": "5702887 9227465 24157817"
},
{
"input": "63245986",
"output": "9227465 14930352 39088169"
},
{
"input": "102334155",
"output": "14930352 24157817 63245986"
},
{
"input": "165580141",
"output": "24157817 39088169 102334155"
},
{
"input": "267914296",
"output": "39088169 63245986 165580141"
},
{
"input": "433494437",
"output": "63245986 102334155 267914296"
},
{
"input": "701408733",
"output": "102334155 165580141 433494437"
},
{
"input": "701408733",
"output": "102334155 165580141 433494437"
},
{
"input": "102334155",
"output": "14930352 24157817 63245986"
},
{
"input": "63245986",
"output": "9227465 14930352 39088169"
}
] | 1,671,313,043
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 54
| 124
| 0
|
n = int(input())
if n <= 3:
print(0, 0, n)
else:
a, b, c = 2, 3, 5
while c != n:
a, b, c = b, c, b + c
print(0, a, b)
|
Title: Hexadecimal's theorem
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Recently, a chaotic virus Hexadecimal advanced a new theorem which will shake the Universe. She thinks that each Fibonacci number can be represented as sum of three not necessary different Fibonacci numbers.
Let's remember how Fibonacci numbers can be calculated. *F*0<==<=0, *F*1<==<=1, and all the next numbers are *F**i*<==<=*F**i*<=-<=2<=+<=*F**i*<=-<=1.
So, Fibonacci numbers make a sequence of numbers: 0, 1, 1, 2, 3, 5, 8, 13, ...
If you haven't run away from the PC in fear, you have to help the virus. Your task is to divide given Fibonacci number *n* by three not necessary different Fibonacci numbers or say that it is impossible.
Input Specification:
The input contains of a single integer *n* (0<=≤<=*n*<=<<=109) — the number that should be represented by the rules described above. It is guaranteed that *n* is a Fibonacci number.
Output Specification:
Output three required numbers: *a*, *b* and *c*. If there is no answer for the test you have to print "I'm too stupid to solve this problem" without the quotes.
If there are multiple answers, print any of them.
Demo Input:
['3\n', '13\n']
Demo Output:
['1 1 1\n', '2 3 8\n']
Note:
none
|
```python
n = int(input())
if n <= 3:
print(0, 0, n)
else:
a, b, c = 2, 3, 5
while c != n:
a, b, c = b, c, b + c
print(0, a, b)
```
| 3
|
|
61
|
A
|
Ultra-Fast Mathematician
|
PROGRAMMING
| 800
|
[
"implementation"
] |
A. Ultra-Fast Mathematician
|
2
|
256
|
Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate.
|
There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.
|
Write one line — the corresponding answer. Do not omit the leading 0s.
|
[
"1010100\n0100101\n",
"000\n111\n",
"1110\n1010\n",
"01110\n01100\n"
] |
[
"1110001\n",
"111\n",
"0100\n",
"00010\n"
] |
none
| 500
|
[
{
"input": "1010100\n0100101",
"output": "1110001"
},
{
"input": "000\n111",
"output": "111"
},
{
"input": "1110\n1010",
"output": "0100"
},
{
"input": "01110\n01100",
"output": "00010"
},
{
"input": "011101\n000001",
"output": "011100"
},
{
"input": "10\n01",
"output": "11"
},
{
"input": "00111111\n11011101",
"output": "11100010"
},
{
"input": "011001100\n101001010",
"output": "110000110"
},
{
"input": "1100100001\n0110101100",
"output": "1010001101"
},
{
"input": "00011101010\n10010100101",
"output": "10001001111"
},
{
"input": "100000101101\n111010100011",
"output": "011010001110"
},
{
"input": "1000001111010\n1101100110001",
"output": "0101101001011"
},
{
"input": "01011111010111\n10001110111010",
"output": "11010001101101"
},
{
"input": "110010000111100\n001100101011010",
"output": "111110101100110"
},
{
"input": "0010010111110000\n0000000011010110",
"output": "0010010100100110"
},
{
"input": "00111110111110000\n01111100001100000",
"output": "01000010110010000"
},
{
"input": "101010101111010001\n001001111101111101",
"output": "100011010010101100"
},
{
"input": "0110010101111100000\n0011000101000000110",
"output": "0101010000111100110"
},
{
"input": "11110100011101010111\n00001000011011000000",
"output": "11111100000110010111"
},
{
"input": "101010101111101101001\n111010010010000011111",
"output": "010000111101101110110"
},
{
"input": "0000111111100011000010\n1110110110110000001010",
"output": "1110001001010011001000"
},
{
"input": "10010010101000110111000\n00101110100110111000111",
"output": "10111100001110001111111"
},
{
"input": "010010010010111100000111\n100100111111100011001110",
"output": "110110101101011111001001"
},
{
"input": "0101110100100111011010010\n0101100011010111001010001",
"output": "0000010111110000010000011"
},
{
"input": "10010010100011110111111011\n10000110101100000001000100",
"output": "00010100001111110110111111"
},
{
"input": "000001111000000100001000000\n011100111101111001110110001",
"output": "011101000101111101111110001"
},
{
"input": "0011110010001001011001011100\n0000101101000011101011001010",
"output": "0011011111001010110010010110"
},
{
"input": "11111000000000010011001101111\n11101110011001010100010000000",
"output": "00010110011001000111011101111"
},
{
"input": "011001110000110100001100101100\n001010000011110000001000101001",
"output": "010011110011000100000100000101"
},
{
"input": "1011111010001100011010110101111\n1011001110010000000101100010101",
"output": "0000110100011100011111010111010"
},
{
"input": "10111000100001000001010110000001\n10111000001100101011011001011000",
"output": "00000000101101101010001111011001"
},
{
"input": "000001010000100001000000011011100\n111111111001010100100001100000111",
"output": "111110101001110101100001111011011"
},
{
"input": "1101000000000010011011101100000110\n1110000001100010011010000011011110",
"output": "0011000001100000000001101111011000"
},
{
"input": "01011011000010100001100100011110001\n01011010111000001010010100001110000",
"output": "00000001111010101011110000010000001"
},
{
"input": "000011111000011001000110111100000100\n011011000110000111101011100111000111",
"output": "011000111110011110101101011011000011"
},
{
"input": "1001000010101110001000000011111110010\n0010001011010111000011101001010110000",
"output": "1011001001111001001011101010101000010"
},
{
"input": "00011101011001100101111111000000010101\n10010011011011001011111000000011101011",
"output": "10001110000010101110000111000011111110"
},
{
"input": "111011100110001001101111110010111001010\n111111101101111001110010000101101000100",
"output": "000100001011110000011101110111010001110"
},
{
"input": "1111001001101000001000000010010101001010\n0010111100111110001011000010111110111001",
"output": "1101110101010110000011000000101011110011"
},
{
"input": "00100101111000000101011111110010100011010\n11101110001010010101001000111110101010100",
"output": "11001011110010010000010111001100001001110"
},
{
"input": "101011001110110100101001000111010101101111\n100111100110101011010100111100111111010110",
"output": "001100101000011111111101111011101010111001"
},
{
"input": "1111100001100101000111101001001010011100001\n1000110011000011110010001011001110001000001",
"output": "0111010010100110110101100010000100010100000"
},
{
"input": "01100111011111010101000001101110000001110101\n10011001011111110000000101011001001101101100",
"output": "11111110000000100101000100110111001100011001"
},
{
"input": "110010100111000100100101100000011100000011001\n011001111011100110000110111001110110100111011",
"output": "101011011100100010100011011001101010100100010"
},
{
"input": "0001100111111011010110100100111000000111000110\n1100101011000000000001010010010111001100110001",
"output": "1101001100111011010111110110101111001011110111"
},
{
"input": "00000101110110110001110010100001110100000100000\n10010000110011110001101000111111101010011010001",
"output": "10010101000101000000011010011110011110011110001"
},
{
"input": "110000100101011100100011001111110011111110010001\n101011111001011100110110111101110011010110101100",
"output": "011011011100000000010101110010000000101000111101"
},
{
"input": "0101111101011111010101011101000011101100000000111\n0000101010110110001110101011011110111001010100100",
"output": "0101010111101001011011110110011101010101010100011"
},
{
"input": "11000100010101110011101000011111001010110111111100\n00001111000111001011111110000010101110111001000011",
"output": "11001011010010111000010110011101100100001110111111"
},
{
"input": "101000001101111101101111111000001110110010101101010\n010011100111100001100000010001100101000000111011011",
"output": "111011101010011100001111101001101011110010010110001"
},
{
"input": "0011111110010001010100010110111000110011001101010100\n0111000000100010101010000100101000000100101000111001",
"output": "0100111110110011111110010010010000110111100101101101"
},
{
"input": "11101010000110000011011010000001111101000111011111100\n10110011110001010100010110010010101001010111100100100",
"output": "01011001110111010111001100010011010100010000111011000"
},
{
"input": "011000100001000001101000010110100110011110100111111011\n111011001000001001110011001111011110111110110011011111",
"output": "100011101001001000011011011001111000100000010100100100"
},
{
"input": "0111010110010100000110111011010110100000000111110110000\n1011100100010001101100000100111111101001110010000100110",
"output": "1100110010000101101010111111101001001001110101110010110"
},
{
"input": "10101000100111000111010001011011011011110100110101100011\n11101111000000001100100011111000100100000110011001101110",
"output": "01000111100111001011110010100011111111110010101100001101"
},
{
"input": "000000111001010001000000110001001011100010011101010011011\n110001101000010010000101000100001111101001100100001010010",
"output": "110001010001000011000101110101000100001011111001011001001"
},
{
"input": "0101011100111010000111110010101101111111000000111100011100\n1011111110000010101110111001000011100000100111111111000111",
"output": "1110100010111000101001001011101110011111100111000011011011"
},
{
"input": "11001000001100100111100111100100101011000101001111001001101\n10111110100010000011010100110100100011101001100000001110110",
"output": "01110110101110100100110011010000001000101100101111000111011"
},
{
"input": "010111011011101000000110000110100110001110100001110110111011\n101011110011101011101101011111010100100001100111100100111011",
"output": "111100101000000011101011011001110010101111000110010010000000"
},
{
"input": "1001011110110110000100011001010110000100011010010111010101110\n1101111100001000010111110011010101111010010100000001000010111",
"output": "0100100010111110010011101010000011111110001110010110010111001"
},
{
"input": "10000010101111100111110101111000010100110111101101111111111010\n10110110101100101010011001011010100110111011101100011001100111",
"output": "00110100000011001101101100100010110010001100000001100110011101"
},
{
"input": "011111010011111000001010101001101001000010100010111110010100001\n011111001011000011111001000001111001010110001010111101000010011",
"output": "000000011000111011110011101000010000010100101000000011010110010"
},
{
"input": "1111000000110001011101000100100100001111011100001111001100011111\n1101100110000101100001100000001001011011111011010101000101001010",
"output": "0010100110110100111100100100101101010100100111011010001001010101"
},
{
"input": "01100000101010010011001110100110110010000110010011011001100100011\n10110110010110111100100111000111000110010000000101101110000010111",
"output": "11010110111100101111101001100001110100010110010110110111100110100"
},
{
"input": "001111111010000100001100001010011001111110011110010111110001100111\n110000101001011000100010101100100110000111100000001101001110010111",
"output": "111111010011011100101110100110111111111001111110011010111111110000"
},
{
"input": "1011101011101101011110101101011101011000010011100101010101000100110\n0001000001001111010111100100111101100000000001110001000110000000110",
"output": "1010101010100010001001001001100000111000010010010100010011000100000"
},
{
"input": "01000001011001010011011100010000100100110101111011011011110000001110\n01011110000110011011000000000011000111100001010000000011111001110000",
"output": "00011111011111001000011100010011100011010100101011011000001001111110"
},
{
"input": "110101010100110101000001111110110100010010000100111110010100110011100\n111010010111111011100110101011001011001110110111110100000110110100111",
"output": "001111000011001110100111010101111111011100110011001010010010000111011"
},
{
"input": "1001101011000001011111100110010010000011010001001111011100010100110001\n1111100111110101001111010001010000011001001001010110001111000000100101",
"output": "0110001100110100010000110111000010011010011000011001010011010100010100"
},
{
"input": "00000111110010110001110110001010010101000111011001111111100110011110010\n00010111110100000100110101000010010001100001100011100000001100010100010",
"output": "00010000000110110101000011001000000100100110111010011111101010001010000"
},
{
"input": "100101011100101101000011010001011001101110101110001100010001010111001110\n100001111100101011011111110000001111000111001011111110000010101110111001",
"output": "000100100000000110011100100001010110101001100101110010010011111001110111"
},
{
"input": "1101100001000111001101001011101000111000011110000001001101101001111011010\n0101011101010100011011010110101000010010110010011110101100000110110001000",
"output": "1000111100010011010110011101000000101010101100011111100001101111001010010"
},
{
"input": "01101101010011110101100001110101111011100010000010001101111000011110111111\n00101111001101001100111010000101110000100101101111100111101110010100011011",
"output": "01000010011110111001011011110000001011000111101101101010010110001010100100"
},
{
"input": "101100101100011001101111110110110010100110110010100001110010110011001101011\n000001011010101011110011111101001110000111000010001101000010010000010001101",
"output": "101101110110110010011100001011111100100001110000101100110000100011011100110"
},
{
"input": "0010001011001010001100000010010011110110011000100000000100110000101111001110\n1100110100111000110100001110111001011101001100001010100001010011100110110001",
"output": "1110111111110010111000001100101010101011010100101010100101100011001001111111"
},
{
"input": "00101101010000000101011001101011001100010001100000101011101110000001111001000\n10010110010111000000101101000011101011001010000011011101101011010000000011111",
"output": "10111011000111000101110100101000100111011011100011110110000101010001111010111"
},
{
"input": "111100000100100000101001100001001111001010001000001000000111010000010101101011\n001000100010100101111011111011010110101100001111011000010011011011100010010110",
"output": "110100100110000101010010011010011001100110000111010000010100001011110111111101"
},
{
"input": "0110001101100100001111110101101000100101010010101010011001101001001101110000000\n0111011000000010010111011110010000000001000110001000011001101000000001110100111",
"output": "0001010101100110011000101011111000100100010100100010000000000001001100000100111"
},
{
"input": "10001111111001000101001011110101111010100001011010101100111001010001010010001000\n10000111010010011110111000111010101100000011110001101111001000111010100000000001",
"output": "00001000101011011011110011001111010110100010101011000011110001101011110010001001"
},
{
"input": "100110001110110000100101001110000011110110000110000000100011110100110110011001101\n110001110101110000000100101001101011111100100100001001000110000001111100011110110",
"output": "010111111011000000100001100111101000001010100010001001100101110101001010000111011"
},
{
"input": "0000010100100000010110111100011111111010011101000000100000011001001101101100111010\n0100111110011101010110101011110110010111001111000110101100101110111100101000111111",
"output": "0100101010111101000000010111101001101101010010000110001100110111110001000100000101"
},
{
"input": "11000111001010100001110000001001011010010010110000001110100101000001010101100110111\n11001100100100100001101010110100000111100011101110011010110100001001000011011011010",
"output": "00001011101110000000011010111101011101110001011110010100010001001000010110111101101"
},
{
"input": "010110100010001000100010101001101010011010111110100001000100101000111011100010100001\n110000011111101101010011111000101010111010100001001100001001100101000000111000000000",
"output": "100110111101100101110001010001000000100000011111101101001101001101111011011010100001"
},
{
"input": "0000011110101110010101110110110101100001011001101010101001000010000010000000101001101\n1100111111011100000110000111101110011111100111110001011001000010011111100001001100011",
"output": "1100100001110010010011110001011011111110111110011011110000000000011101100001100101110"
},
{
"input": "10100000101101110001100010010010100101100011010010101000110011100000101010110010000000\n10001110011011010010111011011101101111000111110000111000011010010101001100000001010011",
"output": "00101110110110100011011001001111001010100100100010010000101001110101100110110011010011"
},
{
"input": "001110000011111101101010011111000101010111010100001001100001001100101000000111000000000\n111010000000000000101001110011001000111011001100101010011001000011101001001011110000011",
"output": "110100000011111101000011101100001101101100011000100011111000001111000001001100110000011"
},
{
"input": "1110111100111011010101011011001110001010010010110011110010011111000010011111010101100001\n1001010101011001001010100010101100000110111101011000100010101111111010111100001110010010",
"output": "0111101001100010011111111001100010001100101111101011010000110000111000100011011011110011"
},
{
"input": "11100010001100010011001100001100010011010001101110011110100101110010101101011101000111111\n01110000000110111010110100001010000101011110100101010011000110101110101101110111011110001",
"output": "10010010001010101001111000000110010110001111001011001101100011011100000000101010011001110"
},
{
"input": "001101011001100101101100110000111000101011001001100100000100101000100000110100010111111101\n101001111110000010111101111110001001111001111101111010000110111000100100110010010001011111",
"output": "100100100111100111010001001110110001010010110100011110000010010000000100000110000110100010"
},
{
"input": "1010110110010101000110010010110101011101010100011001101011000110000000100011100100011000000\n0011011111100010001111101101000111001011101110100000110111100100101111010110101111011100011",
"output": "1001101001110111001001111111110010010110111010111001011100100010101111110101001011000100011"
},
{
"input": "10010010000111010111011111110010100101100000001100011100111011100010000010010001011100001100\n00111010100010110010000100010111010001111110100100100011101000101111111111001101101100100100",
"output": "10101000100101100101011011100101110100011110101000111111010011001101111101011100110000101000"
},
{
"input": "010101110001010101100000010111010000000111110011001101100011001000000011001111110000000010100\n010010111011100101010101111110110000000111000100001101101001001000001100101110001010000100001",
"output": "000111001010110000110101101001100000000000110111000000001010000000001111100001111010000110101"
},
{
"input": "1100111110011001000111101001001011000110011010111111100010111111001100111111011101100111101011\n1100000011001000110100110111000001011001010111101000010010100011000001100100111101101000010110",
"output": "0000111101010001110011011110001010011111001101010111110000011100001101011011100000001111111101"
},
{
"input": "00011000100100110111100101100100000000010011110111110010101110110011100001010111010011110100101\n00011011111011111011100101100111100101001110010111000010000111000100100100000001110101111011011",
"output": "00000011011111001100000000000011100101011101100000110000101001110111000101010110100110001111110"
},
{
"input": "000101011001001100000111100010110101111011110101111101000110001101011010111110110011100100000001\n011000101010011111011000111000100000000011011000000001111110001000001111101010110000011100001111",
"output": "011101110011010011011111011010010101111000101101111100111000000101010101010100000011111000001110"
},
{
"input": "1000101001011010000100100100010010011101011001110101111011101111111110010101001101010001010101001\n0110110010011100011111011111110111000000010001110100001010111110101011010011111011111110000110000",
"output": "1110011011000110011011111011100101011101001000000001110001010001010101000110110110101111010011001"
},
{
"input": "01111010010101100111110111111011011010100001011101010000111100101101101110111011001100101011100111\n00001100110110100001111011000010001001001100000010111101000001111011100000010111010010000011000010",
"output": "01110110100011000110001100111001010011101101011111101101111101010110001110101100011110101000100101"
},
{
"input": "111101011101111000001011001101111010110111001101110100100011111011011101001101010101011010111000110\n101111100101101101001100110011000001111010011101110111110110101110011011110011111100001001110101101",
"output": "010010111000010101000111111110111011001101010000000011010101010101000110111110101001010011001101011"
},
{
"input": "1010000011010110011101001101110001110010000111011101110010110110111111001001110100101100010101010001\n0111101000111100101100000101111010100100001000011101010100110011100011010011010101000100101011100011",
"output": "1101101011101010110001001000001011010110001111000000100110000101011100011010100001101000111110110010"
},
{
"input": "0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001\n1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111010",
"output": "1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111011"
},
{
"input": "0\n0",
"output": "0"
},
{
"input": "0\n1",
"output": "1"
}
] | 1,591,599,981
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 102
| 124
| 0
|
num1 = input()
num2 = input()
answer = []
for i in range(len(num1)):
if num1[i] != num2[i]:
answer.append(1)
else:
answer.append(0)
print(''.join([str(v) for v in answer]))
|
Title: Ultra-Fast Mathematician
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate.
Input Specification:
There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.
Output Specification:
Write one line — the corresponding answer. Do not omit the leading 0s.
Demo Input:
['1010100\n0100101\n', '000\n111\n', '1110\n1010\n', '01110\n01100\n']
Demo Output:
['1110001\n', '111\n', '0100\n', '00010\n']
Note:
none
|
```python
num1 = input()
num2 = input()
answer = []
for i in range(len(num1)):
if num1[i] != num2[i]:
answer.append(1)
else:
answer.append(0)
print(''.join([str(v) for v in answer]))
```
| 3.969
|
439
|
A
|
Devu, the Singer and Churu, the Joker
|
PROGRAMMING
| 900
|
[
"greedy",
"implementation"
] | null | null |
Devu is a renowned classical singer. He is invited to many big functions/festivals. Recently he was invited to "All World Classical Singing Festival". Other than Devu, comedian Churu was also invited.
Devu has provided organizers a list of the songs and required time for singing them. He will sing *n* songs, *i**th* song will take *t**i* minutes exactly.
The Comedian, Churu will crack jokes. All his jokes are of 5 minutes exactly.
People have mainly come to listen Devu. But you know that he needs rest of 10 minutes after each song. On the other hand, Churu being a very active person, doesn't need any rest.
You as one of the organizers should make an optimal sсhedule for the event. For some reasons you must follow the conditions:
- The duration of the event must be no more than *d* minutes; - Devu must complete all his songs; - With satisfying the two previous conditions the number of jokes cracked by Churu should be as many as possible.
If it is not possible to find a way to conduct all the songs of the Devu, output -1. Otherwise find out maximum number of jokes that Churu can crack in the grand event.
|
The first line contains two space separated integers *n*, *d* (1<=≤<=*n*<=≤<=100; 1<=≤<=*d*<=≤<=10000). The second line contains *n* space-separated integers: *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=100).
|
If there is no way to conduct all the songs of Devu, output -1. Otherwise output the maximum number of jokes that Churu can crack in the grand event.
|
[
"3 30\n2 2 1\n",
"3 20\n2 1 1\n"
] |
[
"5\n",
"-1\n"
] |
Consider the first example. The duration of the event is 30 minutes. There could be maximum 5 jokes in the following way:
- First Churu cracks a joke in 5 minutes. - Then Devu performs the first song for 2 minutes. - Then Churu cracks 2 jokes in 10 minutes. - Now Devu performs second song for 2 minutes. - Then Churu cracks 2 jokes in 10 minutes. - Now finally Devu will perform his last song in 1 minutes.
Total time spent is 5 + 2 + 10 + 2 + 10 + 1 = 30 minutes.
Consider the second example. There is no way of organizing Devu's all songs. Hence the answer is -1.
| 500
|
[
{
"input": "3 30\n2 2 1",
"output": "5"
},
{
"input": "3 20\n2 1 1",
"output": "-1"
},
{
"input": "50 10000\n5 4 10 9 9 6 7 7 7 3 3 7 7 4 7 4 10 10 1 7 10 3 1 4 5 7 2 10 10 10 2 3 4 7 6 1 8 4 7 3 8 8 4 10 1 1 9 2 6 1",
"output": "1943"
},
{
"input": "50 10000\n4 7 15 9 11 12 20 9 14 14 10 13 6 13 14 17 6 8 20 12 10 15 13 17 5 12 13 11 7 5 5 2 3 15 13 7 14 14 19 2 13 14 5 15 3 19 15 16 4 1",
"output": "1891"
},
{
"input": "100 9000\n5 2 3 1 1 3 4 9 9 6 7 10 10 10 2 10 6 8 8 6 7 9 9 5 6 2 1 10 10 9 4 5 9 2 4 3 8 5 6 1 1 5 3 6 2 6 6 6 5 8 3 6 7 3 1 10 9 1 8 3 10 9 5 6 3 4 1 1 10 10 2 3 4 8 10 10 5 1 5 3 6 8 10 6 10 2 1 8 10 1 7 6 9 10 5 2 3 5 3 2",
"output": "1688"
},
{
"input": "100 8007\n5 19 14 18 9 6 15 8 1 14 11 20 3 17 7 12 2 6 3 17 7 20 1 14 20 17 2 10 13 7 18 18 9 10 16 8 1 11 11 9 13 18 9 20 12 12 7 15 12 17 11 5 11 15 9 2 15 1 18 3 18 16 15 4 10 5 18 13 13 12 3 8 17 2 12 2 13 3 1 13 2 4 9 10 18 10 14 4 4 17 12 19 2 9 6 5 5 20 18 12",
"output": "1391"
},
{
"input": "39 2412\n1 1 1 1 1 1 26 1 1 1 99 1 1 1 1 1 1 1 1 1 1 88 7 1 1 1 1 76 1 1 1 93 40 1 13 1 68 1 32",
"output": "368"
},
{
"input": "39 2617\n47 1 1 1 63 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 70 1 99 63 1 1 1 1 1 1 1 1 64 1 1",
"output": "435"
},
{
"input": "39 3681\n83 77 1 94 85 47 1 98 29 16 1 1 1 71 96 85 31 97 96 93 40 50 98 1 60 51 1 96 100 72 1 1 1 89 1 93 1 92 100",
"output": "326"
},
{
"input": "45 894\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 28 28 1 1 1 1 1 1 1 1 1 1 1 1 1 1 99 3 1 1",
"output": "139"
},
{
"input": "45 4534\n1 99 65 99 4 46 54 80 51 30 96 1 28 30 44 70 78 1 1 100 1 62 1 1 1 85 1 1 1 61 1 46 75 1 61 77 97 26 67 1 1 63 81 85 86",
"output": "514"
},
{
"input": "72 3538\n52 1 8 1 1 1 7 1 1 1 1 48 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 40 1 1 38 1 1 1 1 1 1 1 1 1 1 1 35 1 93 79 1 1 1 1 1 1 1 1 1 51 1 1 1 1 1 1 1 1 1 1 1 1 96 1",
"output": "586"
},
{
"input": "81 2200\n1 59 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 93 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 50 1 1 1 1 1 1 1 1 1 1 1",
"output": "384"
},
{
"input": "81 2577\n85 91 1 1 2 1 1 100 1 80 1 1 17 86 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 37 1 66 24 1 1 96 49 1 66 1 44 1 1 1 1 98 1 1 1 1 35 1 37 3 35 1 1 87 64 1 24 1 58 1 1 42 83 5 1 1 1 1 1 95 1 94 1 50 1 1",
"output": "174"
},
{
"input": "81 4131\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 16 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "807"
},
{
"input": "81 6315\n1 1 67 100 1 99 36 1 92 5 1 96 42 12 1 57 91 1 1 66 41 30 74 95 1 37 1 39 91 69 1 52 77 47 65 1 1 93 96 74 90 35 85 76 71 92 92 1 1 67 92 74 1 1 86 76 35 1 56 16 27 57 37 95 1 40 20 100 51 1 80 60 45 79 95 1 46 1 25 100 96",
"output": "490"
},
{
"input": "96 1688\n1 1 1 1 1 1 1 1 1 1 1 1 1 2 1 1 45 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 25 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 71 1 1 1 30 1 1 1",
"output": "284"
},
{
"input": "96 8889\n1 1 18 1 1 1 1 1 1 1 1 1 99 1 1 1 1 88 1 45 1 1 1 1 1 1 1 1 1 1 1 1 1 1 96 1 1 1 1 21 1 1 1 1 1 1 1 73 1 1 1 1 1 10 1 1 1 1 1 1 1 46 43 1 1 1 1 1 98 1 1 1 1 1 1 6 1 1 1 1 1 74 1 25 1 55 1 1 1 13 1 1 54 1 1 1",
"output": "1589"
},
{
"input": "10 100\n1 1 1 1 1 1 1 1 1 1",
"output": "18"
},
{
"input": "100 10000\n54 46 72 94 79 83 91 54 73 3 24 55 54 31 28 20 19 6 25 19 47 23 1 70 15 87 51 39 54 77 55 5 60 3 15 99 56 88 22 78 79 21 38 27 28 86 7 88 12 59 55 70 25 1 70 49 1 45 69 72 50 17 4 56 8 100 90 34 35 20 61 76 88 79 4 74 65 68 75 26 40 72 59 94 10 67 96 85 29 90 47 24 44 1 66 93 55 36 1 99",
"output": "1017"
},
{
"input": "100 6000\n41 31 23 17 24 78 26 96 93 48 46 2 49 33 35 9 73 100 34 48 83 36 33 69 43 24 3 74 8 81 27 33 94 38 77 9 76 90 62 90 21 67 22 22 12 2 17 27 61 18 72 85 59 65 71 38 90 75 74 66 60 47 58 50 90 95 75 10 5 100 97 29 83 88 65 26 93 90 22 98 36 55 70 38 50 92 88 72 99 96 25 14 74 16 25 92 67 94 77 96",
"output": "-1"
},
{
"input": "1 1\n1",
"output": "0"
},
{
"input": "1 6\n1",
"output": "1"
},
{
"input": "1 5\n1",
"output": "0"
},
{
"input": "1 3\n4",
"output": "-1"
},
{
"input": "3 24\n2 1 2",
"output": "-1"
}
] | 1,627,495,844
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 26
| 77
| 6,963,200
|
from sys import exit
info = input().split()
n = int(info[0])
d = int(info[1])
songs = input().split()
sum = 0
for i in range(n):
songs[i] = int(songs[i])
sum += songs[i]
if sum + 10*(n-1) > d:
print(-1)
else:
jokenum = (d - sum)/5
print(int(jokenum))
|
Title: Devu, the Singer and Churu, the Joker
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Devu is a renowned classical singer. He is invited to many big functions/festivals. Recently he was invited to "All World Classical Singing Festival". Other than Devu, comedian Churu was also invited.
Devu has provided organizers a list of the songs and required time for singing them. He will sing *n* songs, *i**th* song will take *t**i* minutes exactly.
The Comedian, Churu will crack jokes. All his jokes are of 5 minutes exactly.
People have mainly come to listen Devu. But you know that he needs rest of 10 minutes after each song. On the other hand, Churu being a very active person, doesn't need any rest.
You as one of the organizers should make an optimal sсhedule for the event. For some reasons you must follow the conditions:
- The duration of the event must be no more than *d* minutes; - Devu must complete all his songs; - With satisfying the two previous conditions the number of jokes cracked by Churu should be as many as possible.
If it is not possible to find a way to conduct all the songs of the Devu, output -1. Otherwise find out maximum number of jokes that Churu can crack in the grand event.
Input Specification:
The first line contains two space separated integers *n*, *d* (1<=≤<=*n*<=≤<=100; 1<=≤<=*d*<=≤<=10000). The second line contains *n* space-separated integers: *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=100).
Output Specification:
If there is no way to conduct all the songs of Devu, output -1. Otherwise output the maximum number of jokes that Churu can crack in the grand event.
Demo Input:
['3 30\n2 2 1\n', '3 20\n2 1 1\n']
Demo Output:
['5\n', '-1\n']
Note:
Consider the first example. The duration of the event is 30 minutes. There could be maximum 5 jokes in the following way:
- First Churu cracks a joke in 5 minutes. - Then Devu performs the first song for 2 minutes. - Then Churu cracks 2 jokes in 10 minutes. - Now Devu performs second song for 2 minutes. - Then Churu cracks 2 jokes in 10 minutes. - Now finally Devu will perform his last song in 1 minutes.
Total time spent is 5 + 2 + 10 + 2 + 10 + 1 = 30 minutes.
Consider the second example. There is no way of organizing Devu's all songs. Hence the answer is -1.
|
```python
from sys import exit
info = input().split()
n = int(info[0])
d = int(info[1])
songs = input().split()
sum = 0
for i in range(n):
songs[i] = int(songs[i])
sum += songs[i]
if sum + 10*(n-1) > d:
print(-1)
else:
jokenum = (d - sum)/5
print(int(jokenum))
```
| 3
|
|
279
|
B
|
Books
|
PROGRAMMING
| 1,400
|
[
"binary search",
"brute force",
"implementation",
"two pointers"
] | null | null |
When Valera has got some free time, he goes to the library to read some books. Today he's got *t* free minutes to read. That's why Valera took *n* books in the library and for each book he estimated the time he is going to need to read it. Let's number the books by integers from 1 to *n*. Valera needs *a**i* minutes to read the *i*-th book.
Valera decided to choose an arbitrary book with number *i* and read the books one by one, starting from this book. In other words, he will first read book number *i*, then book number *i*<=+<=1, then book number *i*<=+<=2 and so on. He continues the process until he either runs out of the free time or finishes reading the *n*-th book. Valera reads each book up to the end, that is, he doesn't start reading the book if he doesn't have enough free time to finish reading it.
Print the maximum number of books Valera can read.
|
The first line contains two integers *n* and *t* (1<=≤<=*n*<=≤<=105; 1<=≤<=*t*<=≤<=109) — the number of books and the number of free minutes Valera's got. The second line contains a sequence of *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=104), where number *a**i* shows the number of minutes that the boy needs to read the *i*-th book.
|
Print a single integer — the maximum number of books Valera can read.
|
[
"4 5\n3 1 2 1\n",
"3 3\n2 2 3\n"
] |
[
"3\n",
"1\n"
] |
none
| 1,000
|
[
{
"input": "4 5\n3 1 2 1",
"output": "3"
},
{
"input": "3 3\n2 2 3",
"output": "1"
},
{
"input": "1 3\n5",
"output": "0"
},
{
"input": "1 10\n4",
"output": "1"
},
{
"input": "2 10\n6 4",
"output": "2"
},
{
"input": "6 10\n2 3 4 2 1 1",
"output": "4"
},
{
"input": "7 13\n6 8 14 9 4 11 10",
"output": "2"
},
{
"input": "10 15\n10 9 1 1 5 10 5 3 7 2",
"output": "3"
},
{
"input": "20 30\n8 1 2 6 9 4 1 9 9 10 4 7 8 9 5 7 1 8 7 4",
"output": "6"
},
{
"input": "30 60\n16 13 22 38 13 35 17 17 20 38 12 19 9 22 20 3 35 34 34 21 35 40 22 3 27 19 12 4 8 19",
"output": "4"
},
{
"input": "100 100\n75 92 18 6 81 67 7 92 100 65 82 32 50 67 85 31 80 91 84 63 39 52 92 81 1 98 24 12 43 48 17 86 51 72 48 95 45 50 12 66 19 79 49 89 34 1 97 75 20 33 96 27 42 23 73 71 93 1 85 19 66 14 17 61 20 39 36 33 42 61 56 64 23 91 80 99 40 74 13 18 98 85 74 39 62 84 46 74 50 23 38 11 79 14 9 25 66 100 25 52",
"output": "3"
},
{
"input": "10 1\n4418 7528 8170 1736 1317 3205 8183 4995 8039 4708",
"output": "0"
},
{
"input": "50 2\n124 214 63 73 996 760 38 571 451 300 970 1 706 937 837 494 619 88 851 411 957 990 842 613 821 649 627 34 693 678 734 116 816 985 705 940 499 493 922 967 854 439 112 644 961 438 189 572 655 550",
"output": "1"
}
] | 1,691,913,799
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 38
| 216
| 13,824,000
|
def max_books_to_read(n, t, a):
left = 0
right = 0
current_sum = 0
max_books = 0
while right < n:
current_sum += a[right]
while current_sum > t:
current_sum -= a[left]
left += 1
max_books = max(max_books, right - left + 1)
right += 1
return max_books
# Read input
n, t = map(int, input().split())
a = list(map(int, input().split()))
# Calculate and print the result
result = max_books_to_read(n, t, a)
print(result)
|
Title: Books
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
When Valera has got some free time, he goes to the library to read some books. Today he's got *t* free minutes to read. That's why Valera took *n* books in the library and for each book he estimated the time he is going to need to read it. Let's number the books by integers from 1 to *n*. Valera needs *a**i* minutes to read the *i*-th book.
Valera decided to choose an arbitrary book with number *i* and read the books one by one, starting from this book. In other words, he will first read book number *i*, then book number *i*<=+<=1, then book number *i*<=+<=2 and so on. He continues the process until he either runs out of the free time or finishes reading the *n*-th book. Valera reads each book up to the end, that is, he doesn't start reading the book if he doesn't have enough free time to finish reading it.
Print the maximum number of books Valera can read.
Input Specification:
The first line contains two integers *n* and *t* (1<=≤<=*n*<=≤<=105; 1<=≤<=*t*<=≤<=109) — the number of books and the number of free minutes Valera's got. The second line contains a sequence of *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=104), where number *a**i* shows the number of minutes that the boy needs to read the *i*-th book.
Output Specification:
Print a single integer — the maximum number of books Valera can read.
Demo Input:
['4 5\n3 1 2 1\n', '3 3\n2 2 3\n']
Demo Output:
['3\n', '1\n']
Note:
none
|
```python
def max_books_to_read(n, t, a):
left = 0
right = 0
current_sum = 0
max_books = 0
while right < n:
current_sum += a[right]
while current_sum > t:
current_sum -= a[left]
left += 1
max_books = max(max_books, right - left + 1)
right += 1
return max_books
# Read input
n, t = map(int, input().split())
a = list(map(int, input().split()))
# Calculate and print the result
result = max_books_to_read(n, t, a)
print(result)
```
| 3
|
|
725
|
A
|
Jumping Ball
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
In a new version of the famous Pinball game, one of the most important parts of the game field is a sequence of *n* bumpers. The bumpers are numbered with integers from 1 to *n* from left to right. There are two types of bumpers. They are denoted by the characters '<' and '>'. When the ball hits the bumper at position *i* it goes one position to the right (to the position *i*<=+<=1) if the type of this bumper is '>', or one position to the left (to *i*<=-<=1) if the type of the bumper at position *i* is '<'. If there is no such position, in other words if *i*<=-<=1<=<<=1 or *i*<=+<=1<=><=*n*, the ball falls from the game field.
Depending on the ball's starting position, the ball may eventually fall from the game field or it may stay there forever. You are given a string representing the bumpers' types. Calculate the number of positions such that the ball will eventually fall from the game field if it starts at that position.
|
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=200<=000) — the length of the sequence of bumpers. The second line contains the string, which consists of the characters '<' and '>'. The character at the *i*-th position of this string corresponds to the type of the *i*-th bumper.
|
Print one integer — the number of positions in the sequence such that the ball will eventually fall from the game field if it starts at that position.
|
[
"4\n<<><\n",
"5\n>>>>>\n",
"4\n>><<\n"
] |
[
"2",
"5",
"0"
] |
In the first sample, the ball will fall from the field if starts at position 1 or position 2.
In the second sample, any starting position will result in the ball falling from the field.
| 500
|
[
{
"input": "4\n<<><",
"output": "2"
},
{
"input": "5\n>>>>>",
"output": "5"
},
{
"input": "4\n>><<",
"output": "0"
},
{
"input": "3\n<<>",
"output": "3"
},
{
"input": "3\n<<<",
"output": "3"
},
{
"input": "3\n><<",
"output": "0"
},
{
"input": "1\n<",
"output": "1"
},
{
"input": "2\n<>",
"output": "2"
},
{
"input": "3\n<>>",
"output": "3"
},
{
"input": "3\n><>",
"output": "1"
},
{
"input": "2\n><",
"output": "0"
},
{
"input": "2\n>>",
"output": "2"
},
{
"input": "2\n<<",
"output": "2"
},
{
"input": "1\n>",
"output": "1"
},
{
"input": "3\n>><",
"output": "0"
},
{
"input": "3\n>>>",
"output": "3"
},
{
"input": "3\n<><",
"output": "1"
},
{
"input": "10\n<<<><<<>>>",
"output": "6"
},
{
"input": "20\n><><<><<<>>>>>>>>>>>",
"output": "11"
},
{
"input": "20\n<<<<<<<<<<><<<<>>>>>",
"output": "15"
},
{
"input": "50\n<<<<<<<<<<<<<<<<<<<<<<<<<>>>>>>>>>>>>>>>>>>>>>>>>>",
"output": "50"
},
{
"input": "100\n<<<<<<<<<<<<<<<<<<<<<<<<>><<>><<<<<>><>><<<>><><<>>><<>>><<<<><><><<><<<<><>>>>>>>>>>>>>>>>>>>>>>>>>",
"output": "49"
},
{
"input": "100\n<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<>>>><<>><>><>><<><><><><>>>><><<<>>>><<<>>>>>>><><",
"output": "50"
},
{
"input": "100\n<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<",
"output": "100"
},
{
"input": "100\n>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>",
"output": "100"
},
{
"input": "12\n<<>><<>><<>>",
"output": "4"
},
{
"input": "6\n<<><>>",
"output": "4"
},
{
"input": "6\n><>>>>",
"output": "4"
},
{
"input": "8\n>>>><<<>",
"output": "1"
},
{
"input": "4\n<><>",
"output": "2"
},
{
"input": "4\n><><",
"output": "0"
},
{
"input": "7\n<<>>><>",
"output": "3"
},
{
"input": "10\n><><>>>>>>",
"output": "6"
},
{
"input": "5\n<><>>",
"output": "3"
},
{
"input": "12\n<><<<<>>>>>>",
"output": "7"
},
{
"input": "6\n<>><<>",
"output": "2"
},
{
"input": "6\n>>><>>",
"output": "2"
},
{
"input": "10\n><><>>>><>",
"output": "1"
},
{
"input": "5\n><>>>",
"output": "3"
},
{
"input": "5\n<<><>",
"output": "3"
},
{
"input": "5\n<><<<",
"output": "1"
},
{
"input": "4\n<><<",
"output": "1"
},
{
"input": "8\n<<>><<>>",
"output": "4"
},
{
"input": "7\n<<><>>>",
"output": "5"
},
{
"input": "5\n><<>>",
"output": "2"
},
{
"input": "10\n<<<<<>>>>>",
"output": "10"
},
{
"input": "6\n><<<<<",
"output": "0"
},
{
"input": "8\n<<><><>>",
"output": "4"
},
{
"input": "10\n<<<<><<<><",
"output": "4"
},
{
"input": "12\n<<<>>>><<>>>",
"output": "6"
},
{
"input": "4\n><>>",
"output": "2"
},
{
"input": "11\n<<><<>><<>>",
"output": "4"
}
] | 1,619,091,583
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 78
| 124
| 4,710,400
|
n = int(input())
s = input()
left, right = [0] * n, [0] * n
cur = 0
for i in range(n):
if s[i] == '<':
cur -= 1
else:
cur += 1
left[i] = cur
cur = 0
for i in reversed(range(n)):
if s[i] == '<':
cur -= 1
else:
cur += 1
right[i] = cur
ans = 0
for i in range(n):
if s[i] == '<':
if -left[i] == i + 1:
ans += 1
else:
if right[i] == n - i:
ans += 1
print(ans)
|
Title: Jumping Ball
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In a new version of the famous Pinball game, one of the most important parts of the game field is a sequence of *n* bumpers. The bumpers are numbered with integers from 1 to *n* from left to right. There are two types of bumpers. They are denoted by the characters '<' and '>'. When the ball hits the bumper at position *i* it goes one position to the right (to the position *i*<=+<=1) if the type of this bumper is '>', or one position to the left (to *i*<=-<=1) if the type of the bumper at position *i* is '<'. If there is no such position, in other words if *i*<=-<=1<=<<=1 or *i*<=+<=1<=><=*n*, the ball falls from the game field.
Depending on the ball's starting position, the ball may eventually fall from the game field or it may stay there forever. You are given a string representing the bumpers' types. Calculate the number of positions such that the ball will eventually fall from the game field if it starts at that position.
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=200<=000) — the length of the sequence of bumpers. The second line contains the string, which consists of the characters '<' and '>'. The character at the *i*-th position of this string corresponds to the type of the *i*-th bumper.
Output Specification:
Print one integer — the number of positions in the sequence such that the ball will eventually fall from the game field if it starts at that position.
Demo Input:
['4\n<<><\n', '5\n>>>>>\n', '4\n>><<\n']
Demo Output:
['2', '5', '0']
Note:
In the first sample, the ball will fall from the field if starts at position 1 or position 2.
In the second sample, any starting position will result in the ball falling from the field.
|
```python
n = int(input())
s = input()
left, right = [0] * n, [0] * n
cur = 0
for i in range(n):
if s[i] == '<':
cur -= 1
else:
cur += 1
left[i] = cur
cur = 0
for i in reversed(range(n)):
if s[i] == '<':
cur -= 1
else:
cur += 1
right[i] = cur
ans = 0
for i in range(n):
if s[i] == '<':
if -left[i] == i + 1:
ans += 1
else:
if right[i] == n - i:
ans += 1
print(ans)
```
| 3
|
|
493
|
D
|
Vasya and Chess
|
PROGRAMMING
| 1,700
|
[
"constructive algorithms",
"games",
"math"
] | null | null |
Vasya decided to learn to play chess. Classic chess doesn't seem interesting to him, so he plays his own sort of chess.
The queen is the piece that captures all squares on its vertical, horizontal and diagonal lines. If the cell is located on the same vertical, horizontal or diagonal line with queen, and the cell contains a piece of the enemy color, the queen is able to move to this square. After that the enemy's piece is removed from the board. The queen cannot move to a cell containing an enemy piece if there is some other piece between it and the queen.
There is an *n*<=×<=*n* chessboard. We'll denote a cell on the intersection of the *r*-th row and *c*-th column as (*r*,<=*c*). The square (1,<=1) contains the white queen and the square (1,<=*n*) contains the black queen. All other squares contain green pawns that don't belong to anyone.
The players move in turns. The player that moves first plays for the white queen, his opponent plays for the black queen.
On each move the player has to capture some piece with his queen (that is, move to a square that contains either a green pawn or the enemy queen). The player loses if either he cannot capture any piece during his move or the opponent took his queen during the previous move.
Help Vasya determine who wins if both players play with an optimal strategy on the board *n*<=×<=*n*.
|
The input contains a single number *n* (2<=≤<=*n*<=≤<=109) — the size of the board.
|
On the first line print the answer to problem — string "white" or string "black", depending on who wins if the both players play optimally.
If the answer is "white", then you should also print two integers *r* and *c* representing the cell (*r*,<=*c*), where the first player should make his first move to win. If there are multiple such cells, print the one with the minimum *r*. If there are still multiple squares, print the one with the minimum *c*.
|
[
"2\n",
"3\n"
] |
[
"white\n1 2\n",
"black\n"
] |
In the first sample test the white queen can capture the black queen at the first move, so the white player wins.
In the second test from the statement if the white queen captures the green pawn located on the central vertical line, then it will be captured by the black queen during the next move. So the only move for the white player is to capture the green pawn located at (2, 1).
Similarly, the black queen doesn't have any other options but to capture the green pawn located at (2, 3), otherwise if it goes to the middle vertical line, it will be captured by the white queen.
During the next move the same thing happens — neither the white, nor the black queen has other options rather than to capture green pawns situated above them. Thus, the white queen ends up on square (3, 1), and the black queen ends up on square (3, 3).
In this situation the white queen has to capture any of the green pawns located on the middle vertical line, after that it will be captured by the black queen. Thus, the player who plays for the black queen wins.
| 1,500
|
[
{
"input": "2",
"output": "white\n1 2"
},
{
"input": "3",
"output": "black"
},
{
"input": "4",
"output": "white\n1 2"
},
{
"input": "6",
"output": "white\n1 2"
},
{
"input": "10",
"output": "white\n1 2"
},
{
"input": "16",
"output": "white\n1 2"
},
{
"input": "100",
"output": "white\n1 2"
},
{
"input": "10006",
"output": "white\n1 2"
},
{
"input": "99966246",
"output": "white\n1 2"
},
{
"input": "1000000000",
"output": "white\n1 2"
},
{
"input": "999999999",
"output": "black"
},
{
"input": "999999997",
"output": "black"
},
{
"input": "900001",
"output": "black"
},
{
"input": "775681",
"output": "black"
},
{
"input": "666666",
"output": "white\n1 2"
},
{
"input": "12345",
"output": "black"
},
{
"input": "111111",
"output": "black"
},
{
"input": "346367",
"output": "black"
},
{
"input": "13",
"output": "black"
},
{
"input": "11",
"output": "black"
},
{
"input": "9",
"output": "black"
},
{
"input": "7",
"output": "black"
},
{
"input": "5",
"output": "black"
},
{
"input": "19",
"output": "black"
},
{
"input": "939698497",
"output": "black"
},
{
"input": "999999996",
"output": "white\n1 2"
}
] | 1,417,876,557
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 26
| 46
| 0
|
n=int(input())
print("white\n1 2" if (n % 2==0) else "black");
|
Title: Vasya and Chess
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya decided to learn to play chess. Classic chess doesn't seem interesting to him, so he plays his own sort of chess.
The queen is the piece that captures all squares on its vertical, horizontal and diagonal lines. If the cell is located on the same vertical, horizontal or diagonal line with queen, and the cell contains a piece of the enemy color, the queen is able to move to this square. After that the enemy's piece is removed from the board. The queen cannot move to a cell containing an enemy piece if there is some other piece between it and the queen.
There is an *n*<=×<=*n* chessboard. We'll denote a cell on the intersection of the *r*-th row and *c*-th column as (*r*,<=*c*). The square (1,<=1) contains the white queen and the square (1,<=*n*) contains the black queen. All other squares contain green pawns that don't belong to anyone.
The players move in turns. The player that moves first plays for the white queen, his opponent plays for the black queen.
On each move the player has to capture some piece with his queen (that is, move to a square that contains either a green pawn or the enemy queen). The player loses if either he cannot capture any piece during his move or the opponent took his queen during the previous move.
Help Vasya determine who wins if both players play with an optimal strategy on the board *n*<=×<=*n*.
Input Specification:
The input contains a single number *n* (2<=≤<=*n*<=≤<=109) — the size of the board.
Output Specification:
On the first line print the answer to problem — string "white" or string "black", depending on who wins if the both players play optimally.
If the answer is "white", then you should also print two integers *r* and *c* representing the cell (*r*,<=*c*), where the first player should make his first move to win. If there are multiple such cells, print the one with the minimum *r*. If there are still multiple squares, print the one with the minimum *c*.
Demo Input:
['2\n', '3\n']
Demo Output:
['white\n1 2\n', 'black\n']
Note:
In the first sample test the white queen can capture the black queen at the first move, so the white player wins.
In the second test from the statement if the white queen captures the green pawn located on the central vertical line, then it will be captured by the black queen during the next move. So the only move for the white player is to capture the green pawn located at (2, 1).
Similarly, the black queen doesn't have any other options but to capture the green pawn located at (2, 3), otherwise if it goes to the middle vertical line, it will be captured by the white queen.
During the next move the same thing happens — neither the white, nor the black queen has other options rather than to capture green pawns situated above them. Thus, the white queen ends up on square (3, 1), and the black queen ends up on square (3, 3).
In this situation the white queen has to capture any of the green pawns located on the middle vertical line, after that it will be captured by the black queen. Thus, the player who plays for the black queen wins.
|
```python
n=int(input())
print("white\n1 2" if (n % 2==0) else "black");
```
| 3
|
|
0
|
none
|
none
|
none
| 0
|
[
"none"
] | null | null |
You are given an array *a*1,<=*a*2,<=...,<=*a**n* consisting of *n* integers, and an integer *k*. You have to split the array into exactly *k* non-empty subsegments. You'll then compute the minimum integer on each subsegment, and take the maximum integer over the *k* obtained minimums. What is the maximum possible integer you can get?
Definitions of subsegment and array splitting are given in notes.
|
The first line contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=<=105) — the size of the array *a* and the number of subsegments you have to split the array to.
The second line contains *n* integers *a*1,<=<=*a*2,<=<=...,<=<=*a**n* (<=-<=109<=<=≤<=<=*a**i*<=≤<=<=109).
|
Print single integer — the maximum possible integer you can get if you split the array into *k* non-empty subsegments and take maximum of minimums on the subsegments.
|
[
"5 2\n1 2 3 4 5\n",
"5 1\n-4 -5 -3 -2 -1\n"
] |
[
"5\n",
"-5\n"
] |
A subsegment [*l*, *r*] (*l* ≤ *r*) of array *a* is the sequence *a*<sub class="lower-index">*l*</sub>, *a*<sub class="lower-index">*l* + 1</sub>, ..., *a*<sub class="lower-index">*r*</sub>.
Splitting of array *a* of *n* elements into *k* subsegments [*l*<sub class="lower-index">1</sub>, *r*<sub class="lower-index">1</sub>], [*l*<sub class="lower-index">2</sub>, *r*<sub class="lower-index">2</sub>], ..., [*l*<sub class="lower-index">*k*</sub>, *r*<sub class="lower-index">*k*</sub>] (*l*<sub class="lower-index">1</sub> = 1, *r*<sub class="lower-index">*k*</sub> = *n*, *l*<sub class="lower-index">*i*</sub> = *r*<sub class="lower-index">*i* - 1</sub> + 1 for all *i* > 1) is *k* sequences (*a*<sub class="lower-index">*l*<sub class="lower-index">1</sub></sub>, ..., *a*<sub class="lower-index">*r*<sub class="lower-index">1</sub></sub>), ..., (*a*<sub class="lower-index">*l*<sub class="lower-index">*k*</sub></sub>, ..., *a*<sub class="lower-index">*r*<sub class="lower-index">*k*</sub></sub>).
In the first example you should split the array into subsegments [1, 4] and [5, 5] that results in sequences (1, 2, 3, 4) and (5). The minimums are *min*(1, 2, 3, 4) = 1 and *min*(5) = 5. The resulting maximum is *max*(1, 5) = 5. It is obvious that you can't reach greater result.
In the second example the only option you have is to split the array into one subsegment [1, 5], that results in one sequence ( - 4, - 5, - 3, - 2, - 1). The only minimum is *min*( - 4, - 5, - 3, - 2, - 1) = - 5. The resulting maximum is - 5.
| 0
|
[
{
"input": "5 2\n1 2 3 4 5",
"output": "5"
},
{
"input": "5 1\n-4 -5 -3 -2 -1",
"output": "-5"
},
{
"input": "10 2\n10 9 1 -9 -7 -9 3 8 -10 5",
"output": "10"
},
{
"input": "10 4\n-8 -1 2 -3 9 -8 4 -3 5 9",
"output": "9"
},
{
"input": "1 1\n504262064",
"output": "504262064"
},
{
"input": "3 3\n-54481850 -878017339 -486296116",
"output": "-54481850"
},
{
"input": "2 2\n-333653905 224013643",
"output": "224013643"
},
{
"input": "14 2\n-14 84 44 46 -75 -75 77 -49 44 -82 -74 -51 -9 -50",
"output": "-14"
},
{
"input": "88 71\n-497 -488 182 104 40 183 201 282 -384 44 -29 494 224 -80 -491 -197 157 130 -52 233 -426 252 -61 -51 203 -50 195 -442 -38 385 232 -243 -49 163 340 -200 406 -254 -29 227 -194 193 487 -325 230 146 421 158 20 447 -97 479 493 -130 164 -471 -198 -330 -152 359 -554 319 544 -444 235 281 -467 337 -385 227 -366 -210 266 69 -261 525 526 -234 -355 177 109 275 -301 7 -41 553 -284 540",
"output": "553"
},
{
"input": "39 1\n676941771 -923780377 -163050076 -230110947 -208029500 329620771 13954060 158950156 -252501602 926390671 -678745080 -921892226 -100127643 610420285 602175224 -839193819 471391946 910035173 777969600 -736144413 -489685522 60986249 830784148 278642552 -375298304 197973611 -354482364 187294011 636628282 25350767 636184407 -550869740 53830680 -42049274 -451383278 900048257 93225803 877923341 -279506435",
"output": "-923780377"
},
{
"input": "3 2\n1 5 3",
"output": "3"
},
{
"input": "5 2\n1 2 5 4 3",
"output": "3"
},
{
"input": "3 2\n1 3 2",
"output": "2"
},
{
"input": "3 2\n1 3 1",
"output": "1"
},
{
"input": "5 3\n-2 -2 -2 -2 -2",
"output": "-2"
},
{
"input": "5 2\n1 2 3 5 4",
"output": "4"
},
{
"input": "5 2\n1 1 11 1 1",
"output": "1"
},
{
"input": "3 3\n3 8 4",
"output": "8"
},
{
"input": "6 3\n4 3 1 5 6 2",
"output": "6"
},
{
"input": "2 1\n1 2",
"output": "1"
},
{
"input": "5 2\n2 5 4 3 1",
"output": "2"
},
{
"input": "5 2\n-1 1 5 4 3",
"output": "3"
},
{
"input": "5 2\n5 2 1 9 3",
"output": "5"
},
{
"input": "2 1\n1000000000 1000000000",
"output": "1000000000"
},
{
"input": "1 1\n1000000000",
"output": "1000000000"
},
{
"input": "5 2\n1 5 3 4 1",
"output": "1"
},
{
"input": "3 2\n-1000000000 -1000000000 -1000000000",
"output": "-1000000000"
},
{
"input": "2 2\n5 2",
"output": "5"
},
{
"input": "7 3\n1 1 1 10 1 1 1",
"output": "10"
},
{
"input": "9 3\n1 2 1 1 5 1 1 1 2",
"output": "5"
},
{
"input": "9 3\n2 2 2 2 9 2 2 2 2",
"output": "9"
},
{
"input": "3 3\n-1000000000 -1000000000 -1000000000",
"output": "-1000000000"
}
] | 1,508,057,993
| 3,293
|
Python 3
|
OK
|
TESTS
| 55
| 124
| 8,601,600
|
from sys import stdin, stdout
n, k = [int(x) for x in stdin.readline().split()]
a = [int(x) for x in stdin.readline().split()]
if k == 1:
stdout.write('{}\n'.format(min(a)))
elif k == 2:
m = 10**9 + 1
mm = -m
for i in range(len(a)):
if a[i] < m:
m = a[i]
if m > mm:
mm = m
m = 10**9 + 1
for i in range(len(a) -1, -1, -1):
if a[i] < m:
m = a[i]
if m > mm:
mm = m
stdout.write('{}\n'.format(mm))
else:
stdout.write('{}\n'.format(max(a)))
|
Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given an array *a*1,<=*a*2,<=...,<=*a**n* consisting of *n* integers, and an integer *k*. You have to split the array into exactly *k* non-empty subsegments. You'll then compute the minimum integer on each subsegment, and take the maximum integer over the *k* obtained minimums. What is the maximum possible integer you can get?
Definitions of subsegment and array splitting are given in notes.
Input Specification:
The first line contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=<=105) — the size of the array *a* and the number of subsegments you have to split the array to.
The second line contains *n* integers *a*1,<=<=*a*2,<=<=...,<=<=*a**n* (<=-<=109<=<=≤<=<=*a**i*<=≤<=<=109).
Output Specification:
Print single integer — the maximum possible integer you can get if you split the array into *k* non-empty subsegments and take maximum of minimums on the subsegments.
Demo Input:
['5 2\n1 2 3 4 5\n', '5 1\n-4 -5 -3 -2 -1\n']
Demo Output:
['5\n', '-5\n']
Note:
A subsegment [*l*, *r*] (*l* ≤ *r*) of array *a* is the sequence *a*<sub class="lower-index">*l*</sub>, *a*<sub class="lower-index">*l* + 1</sub>, ..., *a*<sub class="lower-index">*r*</sub>.
Splitting of array *a* of *n* elements into *k* subsegments [*l*<sub class="lower-index">1</sub>, *r*<sub class="lower-index">1</sub>], [*l*<sub class="lower-index">2</sub>, *r*<sub class="lower-index">2</sub>], ..., [*l*<sub class="lower-index">*k*</sub>, *r*<sub class="lower-index">*k*</sub>] (*l*<sub class="lower-index">1</sub> = 1, *r*<sub class="lower-index">*k*</sub> = *n*, *l*<sub class="lower-index">*i*</sub> = *r*<sub class="lower-index">*i* - 1</sub> + 1 for all *i* > 1) is *k* sequences (*a*<sub class="lower-index">*l*<sub class="lower-index">1</sub></sub>, ..., *a*<sub class="lower-index">*r*<sub class="lower-index">1</sub></sub>), ..., (*a*<sub class="lower-index">*l*<sub class="lower-index">*k*</sub></sub>, ..., *a*<sub class="lower-index">*r*<sub class="lower-index">*k*</sub></sub>).
In the first example you should split the array into subsegments [1, 4] and [5, 5] that results in sequences (1, 2, 3, 4) and (5). The minimums are *min*(1, 2, 3, 4) = 1 and *min*(5) = 5. The resulting maximum is *max*(1, 5) = 5. It is obvious that you can't reach greater result.
In the second example the only option you have is to split the array into one subsegment [1, 5], that results in one sequence ( - 4, - 5, - 3, - 2, - 1). The only minimum is *min*( - 4, - 5, - 3, - 2, - 1) = - 5. The resulting maximum is - 5.
|
```python
from sys import stdin, stdout
n, k = [int(x) for x in stdin.readline().split()]
a = [int(x) for x in stdin.readline().split()]
if k == 1:
stdout.write('{}\n'.format(min(a)))
elif k == 2:
m = 10**9 + 1
mm = -m
for i in range(len(a)):
if a[i] < m:
m = a[i]
if m > mm:
mm = m
m = 10**9 + 1
for i in range(len(a) -1, -1, -1):
if a[i] < m:
m = a[i]
if m > mm:
mm = m
stdout.write('{}\n'.format(mm))
else:
stdout.write('{}\n'.format(max(a)))
```
| 3
|
|
25
|
A
|
IQ test
|
PROGRAMMING
| 1,300
|
[
"brute force"
] |
A. IQ test
|
2
|
256
|
Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given *n* numbers differs from the others. Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given *n* numbers finds one that is different in evenness.
|
The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of numbers in the task. The second line contains *n* space-separated natural numbers, not exceeding 100. It is guaranteed, that exactly one of these numbers differs from the others in evenness.
|
Output index of number that differs from the others in evenness. Numbers are numbered from 1 in the input order.
|
[
"5\n2 4 7 8 10\n",
"4\n1 2 1 1\n"
] |
[
"3\n",
"2\n"
] |
none
| 0
|
[
{
"input": "5\n2 4 7 8 10",
"output": "3"
},
{
"input": "4\n1 2 1 1",
"output": "2"
},
{
"input": "3\n1 2 2",
"output": "1"
},
{
"input": "3\n100 99 100",
"output": "2"
},
{
"input": "3\n5 3 2",
"output": "3"
},
{
"input": "4\n43 28 1 91",
"output": "2"
},
{
"input": "4\n75 13 94 77",
"output": "3"
},
{
"input": "4\n97 8 27 3",
"output": "2"
},
{
"input": "10\n95 51 12 91 85 3 1 31 25 7",
"output": "3"
},
{
"input": "20\n88 96 66 51 14 88 2 92 18 72 18 88 20 30 4 82 90 100 24 46",
"output": "4"
},
{
"input": "30\n20 94 56 50 10 98 52 32 14 22 24 60 4 8 98 46 34 68 82 82 98 90 50 20 78 49 52 94 64 36",
"output": "26"
},
{
"input": "50\n79 27 77 57 37 45 27 49 65 33 57 21 71 19 75 85 65 61 23 97 85 9 23 1 9 3 99 77 77 21 79 69 15 37 15 7 93 81 13 89 91 31 45 93 15 97 55 80 85 83",
"output": "48"
},
{
"input": "60\n46 11 73 65 3 69 3 53 43 53 97 47 55 93 31 75 35 3 9 73 23 31 3 81 91 79 61 21 15 11 11 11 81 7 83 75 39 87 83 59 89 55 93 27 49 67 67 29 1 93 11 17 9 19 35 21 63 31 31 25",
"output": "1"
},
{
"input": "70\n28 42 42 92 64 54 22 38 38 78 62 38 4 38 14 66 4 92 66 58 94 26 4 44 41 88 48 82 44 26 74 44 48 4 16 92 34 38 26 64 94 4 30 78 50 54 12 90 8 16 80 98 28 100 74 50 36 42 92 18 76 98 8 22 2 50 58 50 64 46",
"output": "25"
},
{
"input": "100\n43 35 79 53 13 91 91 45 65 83 57 9 42 39 85 45 71 51 61 59 31 13 63 39 25 21 79 39 91 67 21 61 97 75 93 83 29 79 59 97 11 37 63 51 39 55 91 23 21 17 47 23 35 75 49 5 69 99 5 7 41 17 25 89 15 79 21 63 53 81 43 91 59 91 69 99 85 15 91 51 49 37 65 7 89 81 21 93 61 63 97 93 45 17 13 69 57 25 75 73",
"output": "13"
},
{
"input": "100\n50 24 68 60 70 30 52 22 18 74 68 98 20 82 4 46 26 68 100 78 84 58 74 98 38 88 68 86 64 80 82 100 20 22 98 98 52 6 94 10 48 68 2 18 38 22 22 82 44 20 66 72 36 58 64 6 36 60 4 96 76 64 12 90 10 58 64 60 74 28 90 26 24 60 40 58 2 16 76 48 58 36 82 60 24 44 4 78 28 38 8 12 40 16 38 6 66 24 31 76",
"output": "99"
},
{
"input": "100\n47 48 94 48 14 18 94 36 96 22 12 30 94 20 48 98 40 58 2 94 8 36 98 18 98 68 2 60 76 38 18 100 8 72 100 68 2 86 92 72 58 16 48 14 6 58 72 76 6 88 80 66 20 28 74 62 86 68 90 86 2 56 34 38 56 90 4 8 76 44 32 86 12 98 38 34 54 92 70 94 10 24 82 66 90 58 62 2 32 58 100 22 58 72 2 22 68 72 42 14",
"output": "1"
},
{
"input": "99\n38 20 68 60 84 16 28 88 60 48 80 28 4 92 70 60 46 46 20 34 12 100 76 2 40 10 8 86 6 80 50 66 12 34 14 28 26 70 46 64 34 96 10 90 98 96 56 88 50 74 70 94 2 94 24 66 68 46 22 30 6 10 64 32 88 14 98 100 64 58 50 18 50 50 8 38 8 16 54 2 60 54 62 84 92 98 4 72 66 26 14 88 99 16 10 6 88 56 22",
"output": "93"
},
{
"input": "99\n50 83 43 89 53 47 69 1 5 37 63 87 95 15 55 95 75 89 33 53 89 75 93 75 11 85 49 29 11 97 49 67 87 11 25 37 97 73 67 49 87 43 53 97 43 29 53 33 45 91 37 73 39 49 59 5 21 43 87 35 5 63 89 57 63 47 29 99 19 85 13 13 3 13 43 19 5 9 61 51 51 57 15 89 13 97 41 13 99 79 13 27 97 95 73 33 99 27 23",
"output": "1"
},
{
"input": "98\n61 56 44 30 58 14 20 24 88 28 46 56 96 52 58 42 94 50 46 30 46 80 72 88 68 16 6 60 26 90 10 98 76 20 56 40 30 16 96 20 88 32 62 30 74 58 36 76 60 4 24 36 42 54 24 92 28 14 2 74 86 90 14 52 34 82 40 76 8 64 2 56 10 8 78 16 70 86 70 42 70 74 22 18 76 98 88 28 62 70 36 72 20 68 34 48 80 98",
"output": "1"
},
{
"input": "98\n66 26 46 42 78 32 76 42 26 82 8 12 4 10 24 26 64 44 100 46 94 64 30 18 88 28 8 66 30 82 82 28 74 52 62 80 80 60 94 86 64 32 44 88 92 20 12 74 94 28 34 58 4 22 16 10 94 76 82 58 40 66 22 6 30 32 92 54 16 76 74 98 18 48 48 30 92 2 16 42 84 74 30 60 64 52 50 26 16 86 58 96 79 60 20 62 82 94",
"output": "93"
},
{
"input": "95\n9 31 27 93 17 77 75 9 9 53 89 39 51 99 5 1 11 39 27 49 91 17 27 79 81 71 37 75 35 13 93 4 99 55 85 11 23 57 5 43 5 61 15 35 23 91 3 81 99 85 43 37 39 27 5 67 7 33 75 59 13 71 51 27 15 93 51 63 91 53 43 99 25 47 17 71 81 15 53 31 59 83 41 23 73 25 91 91 13 17 25 13 55 57 29",
"output": "32"
},
{
"input": "100\n91 89 81 45 53 1 41 3 77 93 55 97 55 97 87 27 69 95 73 41 93 21 75 35 53 56 5 51 87 59 91 67 33 3 99 45 83 17 97 47 75 97 7 89 17 99 23 23 81 25 55 97 27 35 69 5 77 35 93 19 55 59 37 21 31 37 49 41 91 53 73 69 7 37 37 39 17 71 7 97 55 17 47 23 15 73 31 39 57 37 9 5 61 41 65 57 77 79 35 47",
"output": "26"
},
{
"input": "99\n38 56 58 98 80 54 26 90 14 16 78 92 52 74 40 30 84 14 44 80 16 90 98 68 26 24 78 72 42 16 84 40 14 44 2 52 50 2 12 96 58 66 8 80 44 52 34 34 72 98 74 4 66 74 56 21 8 38 76 40 10 22 48 32 98 34 12 62 80 68 64 82 22 78 58 74 20 22 48 56 12 38 32 72 6 16 74 24 94 84 26 38 18 24 76 78 98 94 72",
"output": "56"
},
{
"input": "100\n44 40 6 40 56 90 98 8 36 64 76 86 98 76 36 92 6 30 98 70 24 98 96 60 24 82 88 68 86 96 34 42 58 10 40 26 56 10 88 58 70 32 24 28 14 82 52 12 62 36 70 60 52 34 74 30 78 76 10 16 42 94 66 90 70 38 52 12 58 22 98 96 14 68 24 70 4 30 84 98 8 50 14 52 66 34 100 10 28 100 56 48 38 12 38 14 91 80 70 86",
"output": "97"
},
{
"input": "100\n96 62 64 20 90 46 56 90 68 36 30 56 70 28 16 64 94 34 6 32 34 50 94 22 90 32 40 2 72 10 88 38 28 92 20 26 56 80 4 100 100 90 16 74 74 84 8 2 30 20 80 32 16 46 92 56 42 12 96 64 64 42 64 58 50 42 74 28 2 4 36 32 70 50 54 92 70 16 45 76 28 16 18 50 48 2 62 94 4 12 52 52 4 100 70 60 82 62 98 42",
"output": "79"
},
{
"input": "99\n14 26 34 68 90 58 50 36 8 16 18 6 2 74 54 20 36 84 32 50 52 2 26 24 3 64 20 10 54 26 66 44 28 72 4 96 78 90 96 86 68 28 94 4 12 46 100 32 22 36 84 32 44 94 76 94 4 52 12 30 74 4 34 64 58 72 44 16 70 56 54 8 14 74 8 6 58 62 98 54 14 40 80 20 36 72 28 98 20 58 40 52 90 64 22 48 54 70 52",
"output": "25"
},
{
"input": "95\n82 86 30 78 6 46 80 66 74 72 16 24 18 52 52 38 60 36 86 26 62 28 22 46 96 26 94 84 20 46 66 88 76 32 12 86 74 18 34 88 4 48 94 6 58 6 100 82 4 24 88 32 54 98 34 48 6 76 42 88 42 28 100 4 22 2 10 66 82 54 98 20 60 66 38 98 32 47 86 58 6 100 12 46 2 42 8 84 78 28 24 70 34 28 86",
"output": "78"
},
{
"input": "90\n40 50 8 42 76 24 58 42 26 68 20 48 54 12 34 84 14 36 32 88 6 50 96 56 20 92 48 16 40 34 96 46 20 84 30 50 20 98 8 44 96 42 8 76 70 38 84 30 40 88 84 72 2 22 52 58 16 62 100 66 80 40 50 32 14 62 88 72 22 99 76 50 84 82 8 82 98 46 26 40 2 98 18 78 30 72 70 18 34 68",
"output": "70"
},
{
"input": "80\n81 43 87 1 55 43 53 61 27 19 43 13 89 9 33 83 75 55 97 71 91 37 95 5 21 69 81 93 95 69 31 83 55 7 97 7 79 57 8 61 27 85 49 1 15 97 63 79 29 73 41 85 5 41 31 93 67 11 63 59 15 99 91 77 43 69 23 23 81 73 19 1 67 51 1 75 99 67 3 81",
"output": "39"
},
{
"input": "98\n13 83 61 27 35 1 85 95 97 73 95 65 73 45 5 43 27 83 91 19 11 3 85 59 9 39 69 23 45 7 51 85 5 71 5 95 1 51 75 3 43 57 3 11 33 71 21 99 47 41 87 39 71 87 31 85 91 49 83 5 49 85 47 91 55 99 33 23 31 23 23 73 29 77 55 31 25 5 81 49 91 15 15 39 87 5 9 40 69 47 29 33 11 21 49 79 51 83",
"output": "88"
},
{
"input": "3\n100 100 1",
"output": "3"
}
] | 1,520,927,295
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 32
| 186
| 5,632,000
|
n=int(input())
input_string=input().split(" ")
even_number_count=0
odd_number_count=0
for i in range(n):
number=int(input_string[i])
if number%2==0:
even_number_count+=1
else:
odd_number_count+=1
if even_number_count>odd_number_count:
for i in range(n):
number=int(input_string[i])
if number%2!=0:
print("%d" %(i+1))
raise exit()
else:
for i in range(n):
number=int(input_string[i])
if number%2==0:
print("%d" %(i+1))
raise exit()
|
Title: IQ test
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given *n* numbers differs from the others. Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given *n* numbers finds one that is different in evenness.
Input Specification:
The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of numbers in the task. The second line contains *n* space-separated natural numbers, not exceeding 100. It is guaranteed, that exactly one of these numbers differs from the others in evenness.
Output Specification:
Output index of number that differs from the others in evenness. Numbers are numbered from 1 in the input order.
Demo Input:
['5\n2 4 7 8 10\n', '4\n1 2 1 1\n']
Demo Output:
['3\n', '2\n']
Note:
none
|
```python
n=int(input())
input_string=input().split(" ")
even_number_count=0
odd_number_count=0
for i in range(n):
number=int(input_string[i])
if number%2==0:
even_number_count+=1
else:
odd_number_count+=1
if even_number_count>odd_number_count:
for i in range(n):
number=int(input_string[i])
if number%2!=0:
print("%d" %(i+1))
raise exit()
else:
for i in range(n):
number=int(input_string[i])
if number%2==0:
print("%d" %(i+1))
raise exit()
```
| 3.94301
|
978
|
B
|
File Name
|
PROGRAMMING
| 800
|
[
"greedy",
"strings"
] | null | null |
You can not just take the file and send it. When Polycarp trying to send a file in the social network "Codehorses", he encountered an unexpected problem. If the name of the file contains three or more "x" (lowercase Latin letters "x") in a row, the system considers that the file content does not correspond to the social network topic. In this case, the file is not sent and an error message is displayed.
Determine the minimum number of characters to remove from the file name so after that the name does not contain "xxx" as a substring. Print 0 if the file name does not initially contain a forbidden substring "xxx".
You can delete characters in arbitrary positions (not necessarily consecutive). If you delete a character, then the length of a string is reduced by $1$. For example, if you delete the character in the position $2$ from the string "exxxii", then the resulting string is "exxii".
|
The first line contains integer $n$ $(3 \le n \le 100)$ — the length of the file name.
The second line contains a string of length $n$ consisting of lowercase Latin letters only — the file name.
|
Print the minimum number of characters to remove from the file name so after that the name does not contain "xxx" as a substring. If initially the file name dost not contain a forbidden substring "xxx", print 0.
|
[
"6\nxxxiii\n",
"5\nxxoxx\n",
"10\nxxxxxxxxxx\n"
] |
[
"1\n",
"0\n",
"8\n"
] |
In the first example Polycarp tried to send a file with name contains number $33$, written in Roman numerals. But he can not just send the file, because it name contains three letters "x" in a row. To send the file he needs to remove any one of this letters.
| 0
|
[
{
"input": "6\nxxxiii",
"output": "1"
},
{
"input": "5\nxxoxx",
"output": "0"
},
{
"input": "10\nxxxxxxxxxx",
"output": "8"
},
{
"input": "100\nxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx",
"output": "98"
},
{
"input": "99\nxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxaxxa",
"output": "0"
},
{
"input": "3\nxxx",
"output": "1"
},
{
"input": "77\naaabbbcccdddeeefffggghhhiiijjjkkklllmmmnnnooopppqqqrrrssstttuuuvvvwwwxxyyyzzz",
"output": "0"
},
{
"input": "100\nuxxxxxlmexxxxxxxwnxxexxxxxcxxfydxxxxxxvmdxxxxxxisxxxxxxxxidkxxxpxxxxxxxxmnuxxxxjxxxqcxxwmxxxxxxxxmrx",
"output": "41"
},
{
"input": "100\nxxxxxxxxxxxjtxxxxxxxxcxxxxxxcfxxxxzxxxxxxgxxxxxbxxxxbxxxxxxxxdycxxxxokixxxkizxxgcxxxxxxxxexxxxxfxxxc",
"output": "49"
},
{
"input": "100\nuxxxxxlmexxxxxxxwnxxexxxxxcxxfydxxxxxxvmdxxxxxxisxxxxxxxxidkxxxpxxxxxxxxmnuxxxxjxxxqcxxwmxxxxxwxxxxx",
"output": "41"
},
{
"input": "34\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "0"
},
{
"input": "5\nfcyju",
"output": "0"
},
{
"input": "100\nihygyvdvyeifomhxhkhdkimquvgallbqharcyriyqkidnwykozuhvkwdldlztpabgyuflikychqpdenwzgtlzotyumjgdsrbxxxx",
"output": "2"
}
] | 1,687,619,547
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 13
| 31
| 0
|
n = int(input())
txt = str(input())
copy = txt
while txt.find("xxx") != -1:
txt = txt.replace("xxx", "xx")
print(len(copy) - len(txt))
|
Title: File Name
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You can not just take the file and send it. When Polycarp trying to send a file in the social network "Codehorses", he encountered an unexpected problem. If the name of the file contains three or more "x" (lowercase Latin letters "x") in a row, the system considers that the file content does not correspond to the social network topic. In this case, the file is not sent and an error message is displayed.
Determine the minimum number of characters to remove from the file name so after that the name does not contain "xxx" as a substring. Print 0 if the file name does not initially contain a forbidden substring "xxx".
You can delete characters in arbitrary positions (not necessarily consecutive). If you delete a character, then the length of a string is reduced by $1$. For example, if you delete the character in the position $2$ from the string "exxxii", then the resulting string is "exxii".
Input Specification:
The first line contains integer $n$ $(3 \le n \le 100)$ — the length of the file name.
The second line contains a string of length $n$ consisting of lowercase Latin letters only — the file name.
Output Specification:
Print the minimum number of characters to remove from the file name so after that the name does not contain "xxx" as a substring. If initially the file name dost not contain a forbidden substring "xxx", print 0.
Demo Input:
['6\nxxxiii\n', '5\nxxoxx\n', '10\nxxxxxxxxxx\n']
Demo Output:
['1\n', '0\n', '8\n']
Note:
In the first example Polycarp tried to send a file with name contains number $33$, written in Roman numerals. But he can not just send the file, because it name contains three letters "x" in a row. To send the file he needs to remove any one of this letters.
|
```python
n = int(input())
txt = str(input())
copy = txt
while txt.find("xxx") != -1:
txt = txt.replace("xxx", "xx")
print(len(copy) - len(txt))
```
| 3
|
|
101
|
A
|
Homework
|
PROGRAMMING
| 1,200
|
[
"greedy"
] |
A. Homework
|
2
|
256
|
Once when Gerald studied in the first year at school, his teacher gave the class the following homework. She offered the students a string consisting of *n* small Latin letters; the task was to learn the way the letters that the string contains are written. However, as Gerald is too lazy, he has no desire whatsoever to learn those letters. That's why he decided to lose some part of the string (not necessarily a connected part). The lost part can consist of any number of segments of any length, at any distance from each other. However, Gerald knows that if he loses more than *k* characters, it will be very suspicious.
Find the least number of distinct characters that can remain in the string after no more than *k* characters are deleted. You also have to find any possible way to delete the characters.
|
The first input data line contains a string whose length is equal to *n* (1<=≤<=*n*<=≤<=105). The string consists of lowercase Latin letters. The second line contains the number *k* (0<=≤<=*k*<=≤<=105).
|
Print on the first line the only number *m* — the least possible number of different characters that could remain in the given string after it loses no more than *k* characters.
Print on the second line the string that Gerald can get after some characters are lost. The string should have exactly *m* distinct characters. The final string should be the subsequence of the initial string. If Gerald can get several different strings with exactly *m* distinct characters, print any of them.
|
[
"aaaaa\n4\n",
"abacaba\n4\n",
"abcdefgh\n10\n"
] |
[
"1\naaaaa\n",
"1\naaaa\n",
"0\n\n"
] |
In the first sample the string consists of five identical letters but you are only allowed to delete 4 of them so that there was at least one letter left. Thus, the right answer is 1 and any string consisting of characters "a" from 1 to 5 in length.
In the second sample you are allowed to delete 4 characters. You cannot delete all the characters, because the string has length equal to 7. However, you can delete all characters apart from "a" (as they are no more than four), which will result in the "aaaa" string.
In the third sample you are given a line whose length is equal to 8, and *k* = 10, so that the whole line can be deleted. The correct answer is 0 and an empty string.
| 500
|
[
{
"input": "aaaaa\n4",
"output": "1\naaaaa"
},
{
"input": "abacaba\n4",
"output": "1\naaaa"
},
{
"input": "abcdefgh\n10",
"output": "0"
},
{
"input": "aaaaaaaaaaaaaaaaaaaa\n19",
"output": "1\naaaaaaaaaaaaaaaaaaaa"
},
{
"input": "abcdefghijjihgedcba\n0",
"output": "10\nabcdefghijjihgedcba"
},
{
"input": "aababcabcdabcde\n9",
"output": "2\naabababab"
},
{
"input": "xyzuvwxyz\n4",
"output": "3\nxyzxyz"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa\n99",
"output": "1\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx\n0",
"output": "1\nxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx"
},
{
"input": "abcdefghijklmnopqrstuvwxyz\n17",
"output": "9\nrstuvwxyz"
},
{
"input": "abcdefghijklmnopqrstuvwxyz\n0",
"output": "26\nabcdefghijklmnopqrstuvwxyz"
},
{
"input": "abcdefghijklmnopqrsttsrqponmlkjihgfedcba\n0",
"output": "20\nabcdefghijklmnopqrsttsrqponmlkjihgfedcba"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaeeeeeeeeeeeeeeeeeeee\n20",
"output": "1\naaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "xyxjvqrbehasypiekxwjhurlrnegtkiplbogkgxwubzhlyvjwj\n24",
"output": "8\nxyxjrhykxwjhrlrklkxwhlyjwj"
},
{
"input": "clpdaxnimfkubdxtpjwtjkqh\n21",
"output": "2\nxxtt"
},
{
"input": "jeliuewohkqtghdneuuhcputwiddnmkbhhnlxxbfjunhcd\n50",
"output": "0"
},
{
"input": "zgwmpjfeiwtfagp\n62",
"output": "0"
},
{
"input": "halasouqgfxfcrwhqgllaqiphaxekljz\n87",
"output": "0"
},
{
"input": "zimxucbrzojfqvizcopkplrpnvihveqpgvzszkubftoozrydxijokjxfhdfjracjonqupmnhadtsotxrxmwgno\n51",
"output": "7\nzxrzojvzopprpvvpvzzoozrxjojxjrjopoxrxo"
},
{
"input": "geovcaxzjyhxbpnbkbsxfpkyofopxquzzxeigdflfumisevzsjdywehxconimkkbvjyxbqlnmaphvnngcjqoefqkfzmiruubbcmv\n24",
"output": "16\neovxzjyxbpnbkbxfpkyofopxquzzxeiffumievzjyexonimkkbvjyxbqnmpvnnjqoefqkfzmiuubbmv"
},
{
"input": "jsreqtehsewsiwzqbpniwuhbgcrrkxlgbhuobphjigfuinffvvatrcxnzbcxjazrrxyiwxncuiifzndpvqahwpdfo\n67",
"output": "4\nrwiwiwrrxiirxxrrxiwxiiw"
},
{
"input": "uwvkcydkhbmbqyfjuryqnxcxhoanwnjubuvpgfipdeserodhh\n65",
"output": "0"
},
{
"input": "xclfxmeqhfjwurwmazpysafoxepb\n26",
"output": "1\nxxx"
},
{
"input": "hlhugwawagrnpojcmzfiqtffrzuqfovcmxnfqukgzxilglfbtsqgtsweblymqrdskcxjtuytodzujgtivkmiktvskvoqpegoiw\n27",
"output": "15\nlugwwgomzfiqtffzuqfovmxfqukgzxilglftsqgtswlmqskxtutozugtivkmiktvskvoqgoiw"
},
{
"input": "cky\n79",
"output": "0"
},
{
"input": "oodcvb\n16",
"output": "0"
},
{
"input": "lfbfwdoeggorzdsxqnpophbcjcatphjsewamrgzjszf\n20",
"output": "8\nffwoggozspopjpjswgzjszf"
},
{
"input": "ksnizygvqy\n42",
"output": "0"
},
{
"input": "myenljgyxkwcfyzjcpffsucstschcevbzh\n44",
"output": "0"
},
{
"input": "yumufcicodkpuhvifnvi\n36",
"output": "0"
},
{
"input": "fntrmjfquczybyjllywsqwllsxdmqynmyfcqhakftitvvfbxtqktbfsvvvanjbkqubyxu\n63",
"output": "1\nyyyyyy"
},
{
"input": "smiclwubkoobnapkkletsnbbsvihqbvikochzteaewjonkzvsqrbjkywsfcvczwretmhscowapcraof\n45",
"output": "6\nscwbkoobkksbbsbkocwoksbkwsccwscowco"
},
{
"input": "lwkjydpagifuvbhifryskegmzuexfksazfurlsnzfrgvuxcazitfchimmvomdnbdirzccstmuvlpghwskinayvucodiwn\n16",
"output": "17\nlwkydagifuvifryskgmzufksazfurlsnzfrgvucazifcimmvmdndirzccsmuvlgwskinayvucdiwn"
},
{
"input": "a\n0",
"output": "1\na"
},
{
"input": "bbb\n100000",
"output": "0"
},
{
"input": "aa\n2",
"output": "0"
},
{
"input": "a\n1",
"output": "0"
},
{
"input": "aaaa\n4",
"output": "0"
}
] | 1,655,917,296
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 48
| 186
| 3,686,400
|
import sys
input = sys.stdin.readline
from collections import Counter
s = input()[:-1]
n = len(s)
k = int(input())
if k >= n:
print(0)
print('')
else:
c = Counter()
for i in s:
c[i] += 1
x = c.most_common()
i = len(x) - 1
while k > 0:
k -= x[i][1]
if k < 0:
break
s = s.replace(x[i][0],'')
i -= 1
print(i+1)
print(s)
|
Title: Homework
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Once when Gerald studied in the first year at school, his teacher gave the class the following homework. She offered the students a string consisting of *n* small Latin letters; the task was to learn the way the letters that the string contains are written. However, as Gerald is too lazy, he has no desire whatsoever to learn those letters. That's why he decided to lose some part of the string (not necessarily a connected part). The lost part can consist of any number of segments of any length, at any distance from each other. However, Gerald knows that if he loses more than *k* characters, it will be very suspicious.
Find the least number of distinct characters that can remain in the string after no more than *k* characters are deleted. You also have to find any possible way to delete the characters.
Input Specification:
The first input data line contains a string whose length is equal to *n* (1<=≤<=*n*<=≤<=105). The string consists of lowercase Latin letters. The second line contains the number *k* (0<=≤<=*k*<=≤<=105).
Output Specification:
Print on the first line the only number *m* — the least possible number of different characters that could remain in the given string after it loses no more than *k* characters.
Print on the second line the string that Gerald can get after some characters are lost. The string should have exactly *m* distinct characters. The final string should be the subsequence of the initial string. If Gerald can get several different strings with exactly *m* distinct characters, print any of them.
Demo Input:
['aaaaa\n4\n', 'abacaba\n4\n', 'abcdefgh\n10\n']
Demo Output:
['1\naaaaa\n', '1\naaaa\n', '0\n\n']
Note:
In the first sample the string consists of five identical letters but you are only allowed to delete 4 of them so that there was at least one letter left. Thus, the right answer is 1 and any string consisting of characters "a" from 1 to 5 in length.
In the second sample you are allowed to delete 4 characters. You cannot delete all the characters, because the string has length equal to 7. However, you can delete all characters apart from "a" (as they are no more than four), which will result in the "aaaa" string.
In the third sample you are given a line whose length is equal to 8, and *k* = 10, so that the whole line can be deleted. The correct answer is 0 and an empty string.
|
```python
import sys
input = sys.stdin.readline
from collections import Counter
s = input()[:-1]
n = len(s)
k = int(input())
if k >= n:
print(0)
print('')
else:
c = Counter()
for i in s:
c[i] += 1
x = c.most_common()
i = len(x) - 1
while k > 0:
k -= x[i][1]
if k < 0:
break
s = s.replace(x[i][0],'')
i -= 1
print(i+1)
print(s)
```
| 3.946634
|
524
|
A
|
Возможно, вы знаете этих людей?
|
PROGRAMMING
| 1,600
|
[
"implementation"
] | null | null |
Основой любой социальной сети является отношение дружбы между двумя пользователями в том или ином смысле. В одной известной социальной сети дружба симметрична, то есть если *a* является другом *b*, то *b* также является другом *a*.
В этой же сети есть функция, которая демонстрирует множество людей, имеющих высокую вероятность быть знакомыми для пользователя. Эта функция работает следующим образом. Зафиксируем пользователя *x*. Пусть некоторый другой человек *y*, не являющийся другом *x* на текущий момент, является другом не менее, чем для *k*% друзей *x*. Тогда он является предполагаемым другом для *x*.
У каждого человека в социальной сети есть свой уникальный идентификатор — это целое число от 1 до 109. Вам дан список пар пользователей, являющихся друзьями. Определите для каждого упомянутого пользователя множество его предполагаемых друзей.
|
В первой строке следуют два целых числа *m* и *k* (1<=≤<=*m*<=≤<=100, 0<=≤<=*k*<=≤<=100) — количество пар друзей и необходимый процент общих друзей для того, чтобы считаться предполагаемым другом.
В последующих *m* строках записано по два числа *a**i*,<=*b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=109, *a**i*<=≠<=*b**i*), обозначающих идентификаторы пользователей, являющихся друзьями.
Гарантируется, что каждая пара людей фигурирует в списке не более одного раза.
|
Для всех упомянутых людей в порядке возрастания id выведите информацию о предполагаемых друзьях. Информация должна иметь вид "*id*:<= *k* *id*1 *id*2 ... *id**k*", где *id* — это id самого человека, *k* — количество его предполагаемых друзей, а *id*1, *id*2, ..., *id**k* — идентификаторы его предполагаемых друзей в возрастающем порядке.
|
[
"5 51\n10 23\n23 42\n39 42\n10 39\n39 58\n",
"5 100\n1 2\n1 3\n1 4\n2 3\n2 4\n"
] |
[
"10: 1 42\n23: 1 39\n39: 1 23\n42: 1 10\n58: 2 10 42\n",
"1: 0\n2: 0\n3: 1 4\n4: 1 3\n"
] |
none
| 500
|
[
{
"input": "5 51\n10 23\n23 42\n39 42\n10 39\n39 58",
"output": "10: 1 42\n23: 1 39\n39: 1 23\n42: 1 10\n58: 2 10 42"
},
{
"input": "5 100\n1 2\n1 3\n1 4\n2 3\n2 4",
"output": "1: 0\n2: 0\n3: 1 4\n4: 1 3"
},
{
"input": "4 1\n1 2\n1 3\n2 3\n4 5",
"output": "1: 0\n2: 0\n3: 0\n4: 0\n5: 0"
},
{
"input": "10 0\n648169314 459970755\n973677547 255163231\n982998000 498743911\n959912791 891928\n404623428 891928\n474720235 271683165\n709045873 539751127\n973677547 179588015\n629049356 622519100\n624998275 958914560",
"output": "891928: 15 179588015 255163231 271683165 459970755 474720235 498743911 539751127 622519100 624998275 629049356 648169314 709045873 958914560 973677547 982998000\n179588015: 16 891928 255163231 271683165 404623428 459970755 474720235 498743911 539751127 622519100 624998275 629049356 648169314 709045873 958914560 959912791 982998000\n255163231: 16 891928 179588015 271683165 404623428 459970755 474720235 498743911 539751127 622519100 624998275 629049356 648169314 709045873 958914560 959912791 982998000\n27168..."
},
{
"input": "10 100\n60976680 603454792\n575754027 696647370\n7534463 570826751\n117972518 472387015\n35713567 439985965\n439985965 928160845\n443596853 828504858\n689509731 117972518\n909843480 592765058\n251752353 490387136",
"output": "7534463: 0\n35713567: 1 928160845\n60976680: 0\n117972518: 0\n251752353: 0\n439985965: 0\n443596853: 0\n472387015: 1 689509731\n490387136: 0\n570826751: 0\n575754027: 0\n592765058: 0\n603454792: 0\n689509731: 1 472387015\n696647370: 0\n828504858: 0\n909843480: 0\n928160845: 1 35713567"
},
{
"input": "10 50\n389900784 512305545\n839319681 243581524\n653226215 616982889\n448655722 826601897\n681021965 23289895\n719595063 481480420\n919744525 839319681\n231872856 784056465\n971842495 248017394\n653226215 297224467",
"output": "23289895: 0\n231872856: 0\n243581524: 1 919744525\n248017394: 0\n297224467: 1 616982889\n389900784: 0\n448655722: 0\n481480420: 0\n512305545: 0\n616982889: 1 297224467\n653226215: 0\n681021965: 0\n719595063: 0\n784056465: 0\n826601897: 0\n839319681: 0\n919744525: 1 243581524\n971842495: 0"
},
{
"input": "10 0\n180745113 666631448\n362104151 349631376\n214251560 538865550\n562805929 576329835\n64121410 646478528\n283223383 861810719\n773038401 214251560\n64208401 693054606\n493180926 960545197\n159614568 831490031",
"output": "64121410: 17 64208401 159614568 180745113 214251560 283223383 349631376 362104151 493180926 538865550 562805929 576329835 666631448 693054606 773038401 831490031 861810719 960545197\n64208401: 17 64121410 159614568 180745113 214251560 283223383 349631376 362104151 493180926 538865550 562805929 576329835 646478528 666631448 773038401 831490031 861810719 960545197\n159614568: 17 64121410 64208401 180745113 214251560 283223383 349631376 362104151 493180926 538865550 562805929 576329835 646478528 666631448 693..."
},
{
"input": "10 50\n946010975 207263044\n923545573 749203275\n862015642 426425906\n749203275 839134958\n910721783 289091881\n827003531 333726912\n49704846 538788252\n382891592 207263044\n333726912 438209022\n974360048 49704846",
"output": "49704846: 0\n207263044: 0\n289091881: 0\n333726912: 0\n382891592: 1 946010975\n426425906: 0\n438209022: 1 827003531\n538788252: 1 974360048\n749203275: 0\n827003531: 1 438209022\n839134958: 1 923545573\n862015642: 0\n910721783: 0\n923545573: 1 839134958\n946010975: 1 382891592\n974360048: 1 538788252"
},
{
"input": "10 100\n570936669 651631651\n508122950 793810569\n374422919 757639639\n395046911 359158844\n544971368 55608511\n554227847 109843524\n199021332 421407912\n82125712 395046911\n923097829 637659245\n754413496 971876441",
"output": "55608511: 0\n82125712: 1 359158844\n109843524: 0\n199021332: 0\n359158844: 1 82125712\n374422919: 0\n395046911: 0\n421407912: 0\n508122950: 0\n544971368: 0\n554227847: 0\n570936669: 0\n637659245: 0\n651631651: 0\n754413496: 0\n757639639: 0\n793810569: 0\n923097829: 0\n971876441: 0"
},
{
"input": "1 0\n42 23",
"output": "23: 0\n42: 0"
},
{
"input": "1 1\n42 23",
"output": "23: 0\n42: 0"
},
{
"input": "1 50\n42 23",
"output": "23: 0\n42: 0"
},
{
"input": "1 99\n42 23",
"output": "23: 0\n42: 0"
},
{
"input": "1 100\n42 23",
"output": "23: 0\n42: 0"
},
{
"input": "2 49\n42 23\n23 14",
"output": "14: 1 42\n23: 0\n42: 1 14"
},
{
"input": "2 50\n42 23\n23 19",
"output": "19: 1 42\n23: 0\n42: 1 19"
},
{
"input": "2 51\n42 23\n23 19",
"output": "19: 1 42\n23: 0\n42: 1 19"
},
{
"input": "3 49\n42 23\n23 19\n32 23",
"output": "19: 2 32 42\n23: 0\n32: 2 19 42\n42: 2 19 32"
},
{
"input": "3 50\n42 23\n23 19\n32 23",
"output": "19: 2 32 42\n23: 0\n32: 2 19 42\n42: 2 19 32"
},
{
"input": "3 51\n42 23\n23 19\n32 23",
"output": "19: 2 32 42\n23: 0\n32: 2 19 42\n42: 2 19 32"
},
{
"input": "10 50\n642733947 618790811\n508838679 118350938\n175990043 144671010\n246628250 434416712\n77433126 913934904\n414906480 399777199\n252618318 930317425\n316103842 356219969\n530311152 441130575\n15047025 839165125",
"output": "15047025: 0\n77433126: 0\n118350938: 0\n144671010: 0\n175990043: 0\n246628250: 0\n252618318: 0\n316103842: 0\n356219969: 0\n399777199: 0\n414906480: 0\n434416712: 0\n441130575: 0\n508838679: 0\n530311152: 0\n618790811: 0\n642733947: 0\n839165125: 0\n913934904: 0\n930317425: 0"
},
{
"input": "10 0\n106531296 450097353\n947110486 953723914\n774225709 111023810\n774225709 642354614\n559826213 258125349\n768234906 870027419\n4234645 388060649\n870027419 545107061\n484961505 497401821\n76024092 367527096",
"output": "4234645: 16 76024092 106531296 111023810 258125349 367527096 450097353 484961505 497401821 545107061 559826213 642354614 768234906 774225709 870027419 947110486 953723914\n76024092: 16 4234645 106531296 111023810 258125349 388060649 450097353 484961505 497401821 545107061 559826213 642354614 768234906 774225709 870027419 947110486 953723914\n106531296: 16 4234645 76024092 111023810 258125349 367527096 388060649 484961505 497401821 545107061 559826213 642354614 768234906 774225709 870027419 947110486 953723..."
},
{
"input": "10 50\n384319275 425419607\n201879842 153061155\n308316219 268723666\n20837191 401468340\n969142307 78803322\n55247385 365896022\n479817129 222255243\n980789245 697120853\n550086907 652472194\n203728124 229637404",
"output": "20837191: 0\n55247385: 0\n78803322: 0\n153061155: 0\n201879842: 0\n203728124: 0\n222255243: 0\n229637404: 0\n268723666: 0\n308316219: 0\n365896022: 0\n384319275: 0\n401468340: 0\n425419607: 0\n479817129: 0\n550086907: 0\n652472194: 0\n697120853: 0\n969142307: 0\n980789245: 0"
},
{
"input": "10 100\n620468113 665248777\n541840309 963681159\n144393749 136223789\n800116851 648535048\n730845154 277782209\n142473309 2838660\n14940106 355463174\n745034887 545886019\n570717131 701899093\n250611530 857683655",
"output": "2838660: 0\n14940106: 0\n136223789: 0\n142473309: 0\n144393749: 0\n250611530: 0\n277782209: 0\n355463174: 0\n541840309: 0\n545886019: 0\n570717131: 0\n620468113: 0\n648535048: 0\n665248777: 0\n701899093: 0\n730845154: 0\n745034887: 0\n800116851: 0\n857683655: 0\n963681159: 0"
},
{
"input": "5 66\n4242 1\n4242 2\n4242 3\n2323 1\n2323 2",
"output": "1: 1 2\n2: 1 1\n3: 2 1 2\n2323: 1 4242\n4242: 1 2323"
},
{
"input": "5 67\n4242 1\n4242 2\n4242 3\n2323 1\n2323 2",
"output": "1: 1 2\n2: 1 1\n3: 2 1 2\n2323: 1 4242\n4242: 0"
},
{
"input": "6 49\n4242 1\n4242 2\n4242 3\n4242 4\n2323 1\n2323 2",
"output": "1: 3 2 3 4\n2: 3 1 3 4\n3: 3 1 2 4\n4: 3 1 2 3\n2323: 1 4242\n4242: 1 2323"
},
{
"input": "6 50\n4242 1\n4242 2\n4242 3\n4242 4\n2323 1\n2323 2",
"output": "1: 3 2 3 4\n2: 3 1 3 4\n3: 3 1 2 4\n4: 3 1 2 3\n2323: 1 4242\n4242: 1 2323"
},
{
"input": "6 51\n4242 1\n4242 2\n4242 3\n4242 4\n2323 1\n2323 2",
"output": "1: 1 2\n2: 1 1\n3: 3 1 2 4\n4: 3 1 2 3\n2323: 1 4242\n4242: 0"
},
{
"input": "11 12\n4242 1\n4242 2\n4242 3\n4242 4\n4242 5\n4242 6\n4242 7\n4242 8\n2323 1\n2323 2\n2323 3",
"output": "1: 7 2 3 4 5 6 7 8\n2: 7 1 3 4 5 6 7 8\n3: 7 1 2 4 5 6 7 8\n4: 7 1 2 3 5 6 7 8\n5: 7 1 2 3 4 6 7 8\n6: 7 1 2 3 4 5 7 8\n7: 7 1 2 3 4 5 6 8\n8: 7 1 2 3 4 5 6 7\n2323: 1 4242\n4242: 1 2323"
},
{
"input": "11 13\n4242 1\n4242 2\n4242 3\n4242 4\n4242 5\n4242 6\n4242 7\n4242 8\n2323 1\n2323 2\n2323 3",
"output": "1: 7 2 3 4 5 6 7 8\n2: 7 1 3 4 5 6 7 8\n3: 7 1 2 4 5 6 7 8\n4: 7 1 2 3 5 6 7 8\n5: 7 1 2 3 4 6 7 8\n6: 7 1 2 3 4 5 7 8\n7: 7 1 2 3 4 5 6 8\n8: 7 1 2 3 4 5 6 7\n2323: 1 4242\n4242: 1 2323"
},
{
"input": "79 58\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34\n1 35\n1 36\n1 37\n1 38\n1 39\n1 40\n1 41\n1 42\n1 43\n1 44\n1 45\n1 46\n1 47\n1 48\n1 49\n1 50\n1 51\n1 52\n2 3\n2 4\n2 5\n2 6\n2 7\n2 8\n2 9\n2 10\n2 11\n2 12\n2 13\n2 14\n2 15\n2 16\n2 17\n2 18\n2 19\n2 20\n2 21\n2 22\n2 23\n2 24\n2 25\n2 26\n2 27\n2 28\n2 29\n2 30\n2 31",
"output": "1: 1 2\n2: 1 1\n3: 28 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31\n4: 28 3 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31\n5: 28 3 4 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31\n6: 28 3 4 5 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31\n7: 28 3 4 5 6 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31\n8: 28 3 4 5 6 7 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 2..."
}
] | 1,426,946,960
| 560
|
Python 3
|
OK
|
TESTS
| 73
| 171
| 512,000
|
m, k = [int(x) for x in input().split()]
seen = 0
users = set()
friendsof = {}
for i in range(m):
a, b = [int(x) for x in input().split()]
users.add(a)
users.add(b)
if a in friendsof:
friendsof[a].add(b)
else:
friendsof[a] = set((b,))
if b in friendsof:
friendsof[b].add(a)
else:
friendsof[b] = set((a,))
users_sorted = list(users)
users_sorted.sort()
for u in users_sorted:
possible = []
this_friends = friendsof[u]
for v in users_sorted:
if v in this_friends: continue
if v == u: continue
common = friendsof[v].intersection(this_friends)
if len(common) * 100 >= k * len(this_friends):
possible.append(v)
print('{}:'.format(u), len(possible), *possible)
|
Title: Возможно, вы знаете этих людей?
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Основой любой социальной сети является отношение дружбы между двумя пользователями в том или ином смысле. В одной известной социальной сети дружба симметрична, то есть если *a* является другом *b*, то *b* также является другом *a*.
В этой же сети есть функция, которая демонстрирует множество людей, имеющих высокую вероятность быть знакомыми для пользователя. Эта функция работает следующим образом. Зафиксируем пользователя *x*. Пусть некоторый другой человек *y*, не являющийся другом *x* на текущий момент, является другом не менее, чем для *k*% друзей *x*. Тогда он является предполагаемым другом для *x*.
У каждого человека в социальной сети есть свой уникальный идентификатор — это целое число от 1 до 109. Вам дан список пар пользователей, являющихся друзьями. Определите для каждого упомянутого пользователя множество его предполагаемых друзей.
Input Specification:
В первой строке следуют два целых числа *m* и *k* (1<=≤<=*m*<=≤<=100, 0<=≤<=*k*<=≤<=100) — количество пар друзей и необходимый процент общих друзей для того, чтобы считаться предполагаемым другом.
В последующих *m* строках записано по два числа *a**i*,<=*b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=109, *a**i*<=≠<=*b**i*), обозначающих идентификаторы пользователей, являющихся друзьями.
Гарантируется, что каждая пара людей фигурирует в списке не более одного раза.
Output Specification:
Для всех упомянутых людей в порядке возрастания id выведите информацию о предполагаемых друзьях. Информация должна иметь вид "*id*:<= *k* *id*1 *id*2 ... *id**k*", где *id* — это id самого человека, *k* — количество его предполагаемых друзей, а *id*1, *id*2, ..., *id**k* — идентификаторы его предполагаемых друзей в возрастающем порядке.
Demo Input:
['5 51\n10 23\n23 42\n39 42\n10 39\n39 58\n', '5 100\n1 2\n1 3\n1 4\n2 3\n2 4\n']
Demo Output:
['10: 1 42\n23: 1 39\n39: 1 23\n42: 1 10\n58: 2 10 42\n', '1: 0\n2: 0\n3: 1 4\n4: 1 3\n']
Note:
none
|
```python
m, k = [int(x) for x in input().split()]
seen = 0
users = set()
friendsof = {}
for i in range(m):
a, b = [int(x) for x in input().split()]
users.add(a)
users.add(b)
if a in friendsof:
friendsof[a].add(b)
else:
friendsof[a] = set((b,))
if b in friendsof:
friendsof[b].add(a)
else:
friendsof[b] = set((a,))
users_sorted = list(users)
users_sorted.sort()
for u in users_sorted:
possible = []
this_friends = friendsof[u]
for v in users_sorted:
if v in this_friends: continue
if v == u: continue
common = friendsof[v].intersection(this_friends)
if len(common) * 100 >= k * len(this_friends):
possible.append(v)
print('{}:'.format(u), len(possible), *possible)
```
| 3
|
|
0
|
none
|
none
|
none
| 0
|
[
"none"
] | null | null |
Andryusha is an orderly boy and likes to keep things in their place.
Today he faced a problem to put his socks in the wardrobe. He has *n* distinct pairs of socks which are initially in a bag. The pairs are numbered from 1 to *n*. Andryusha wants to put paired socks together and put them in the wardrobe. He takes the socks one by one from the bag, and for each sock he looks whether the pair of this sock has been already took out of the bag, or not. If not (that means the pair of this sock is still in the bag), he puts the current socks on the table in front of him. Otherwise, he puts both socks from the pair to the wardrobe.
Andryusha remembers the order in which he took the socks from the bag. Can you tell him what is the maximum number of socks that were on the table at the same time?
|
The first line contains the single integer *n* (1<=≤<=*n*<=≤<=105) — the number of sock pairs.
The second line contains 2*n* integers *x*1,<=*x*2,<=...,<=*x*2*n* (1<=≤<=*x**i*<=≤<=*n*), which describe the order in which Andryusha took the socks from the bag. More precisely, *x**i* means that the *i*-th sock Andryusha took out was from pair *x**i*.
It is guaranteed that Andryusha took exactly two socks of each pair.
|
Print single integer — the maximum number of socks that were on the table at the same time.
|
[
"1\n1 1\n",
"3\n2 1 1 3 2 3\n"
] |
[
"1\n",
"2\n"
] |
In the first example Andryusha took a sock from the first pair and put it on the table. Then he took the next sock which is from the first pair as well, so he immediately puts both socks to the wardrobe. Thus, at most one sock was on the table at the same time.
In the second example Andryusha behaved as follows:
- Initially the table was empty, he took out a sock from pair 2 and put it on the table. - Sock (2) was on the table. Andryusha took out a sock from pair 1 and put it on the table. - Socks (1, 2) were on the table. Andryusha took out a sock from pair 1, and put this pair into the wardrobe. - Sock (2) was on the table. Andryusha took out a sock from pair 3 and put it on the table. - Socks (2, 3) were on the table. Andryusha took out a sock from pair 2, and put this pair into the wardrobe. - Sock (3) was on the table. Andryusha took out a sock from pair 3 and put this pair into the wardrobe.
| 0
|
[
{
"input": "1\n1 1",
"output": "1"
},
{
"input": "3\n2 1 1 3 2 3",
"output": "2"
},
{
"input": "5\n5 1 3 2 4 3 1 2 4 5",
"output": "5"
},
{
"input": "10\n4 2 6 3 4 8 7 1 1 5 2 10 6 8 3 5 10 9 9 7",
"output": "6"
},
{
"input": "50\n30 47 31 38 37 50 36 43 9 23 2 2 15 31 14 49 9 16 6 44 27 14 5 6 3 47 25 26 1 35 3 15 24 19 8 46 49 41 4 26 40 28 42 11 34 35 46 18 7 28 18 40 19 42 4 41 38 48 50 12 29 39 33 17 25 22 22 21 36 45 27 30 20 7 13 29 39 44 21 8 37 45 34 1 20 10 11 17 33 12 43 13 10 16 48 24 32 5 23 32",
"output": "25"
},
{
"input": "50\n1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 11 11 12 12 13 13 14 14 15 15 16 16 17 17 18 18 19 19 20 20 21 21 22 22 23 23 24 24 25 25 26 26 27 27 28 28 29 29 30 30 31 31 32 32 33 33 34 34 35 35 36 36 37 37 38 38 39 39 40 40 41 41 42 42 43 43 44 44 45 45 46 46 47 47 48 48 49 49 50 50",
"output": "1"
},
{
"input": "50\n50 50 49 49 48 48 47 47 46 46 45 45 44 44 43 43 42 42 41 41 40 40 39 39 38 38 37 37 36 36 35 35 34 34 33 33 32 32 31 31 30 30 29 29 28 28 27 27 26 26 25 25 24 24 23 23 22 22 21 21 20 20 19 19 18 18 17 17 16 16 15 15 14 14 13 13 12 12 11 11 10 10 9 9 8 8 7 7 6 6 5 5 4 4 3 3 2 2 1 1",
"output": "1"
},
{
"input": "50\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50",
"output": "50"
},
{
"input": "50\n50 49 48 47 46 45 44 43 42 41 40 39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 50 49 48 47 46 45 44 43 42 41 40 39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1",
"output": "50"
},
{
"input": "10\n2 9 4 1 6 7 10 3 1 5 8 6 2 3 10 7 4 8 5 9",
"output": "9"
}
] | 1,529,628,848
| 68
|
PyPy 3
|
OK
|
TESTS
| 56
| 248
| 19,353,600
|
input = __import__('sys').stdin.readline
MIS = lambda: map(int,input().split())
n = int(input())
table = set()
ans = 0
for x in MIS():
if x in table: table.remove(x)
else: table.add(x)
ans = max(ans, len(table))
print(ans)
|
Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Andryusha is an orderly boy and likes to keep things in their place.
Today he faced a problem to put his socks in the wardrobe. He has *n* distinct pairs of socks which are initially in a bag. The pairs are numbered from 1 to *n*. Andryusha wants to put paired socks together and put them in the wardrobe. He takes the socks one by one from the bag, and for each sock he looks whether the pair of this sock has been already took out of the bag, or not. If not (that means the pair of this sock is still in the bag), he puts the current socks on the table in front of him. Otherwise, he puts both socks from the pair to the wardrobe.
Andryusha remembers the order in which he took the socks from the bag. Can you tell him what is the maximum number of socks that were on the table at the same time?
Input Specification:
The first line contains the single integer *n* (1<=≤<=*n*<=≤<=105) — the number of sock pairs.
The second line contains 2*n* integers *x*1,<=*x*2,<=...,<=*x*2*n* (1<=≤<=*x**i*<=≤<=*n*), which describe the order in which Andryusha took the socks from the bag. More precisely, *x**i* means that the *i*-th sock Andryusha took out was from pair *x**i*.
It is guaranteed that Andryusha took exactly two socks of each pair.
Output Specification:
Print single integer — the maximum number of socks that were on the table at the same time.
Demo Input:
['1\n1 1\n', '3\n2 1 1 3 2 3\n']
Demo Output:
['1\n', '2\n']
Note:
In the first example Andryusha took a sock from the first pair and put it on the table. Then he took the next sock which is from the first pair as well, so he immediately puts both socks to the wardrobe. Thus, at most one sock was on the table at the same time.
In the second example Andryusha behaved as follows:
- Initially the table was empty, he took out a sock from pair 2 and put it on the table. - Sock (2) was on the table. Andryusha took out a sock from pair 1 and put it on the table. - Socks (1, 2) were on the table. Andryusha took out a sock from pair 1, and put this pair into the wardrobe. - Sock (2) was on the table. Andryusha took out a sock from pair 3 and put it on the table. - Socks (2, 3) were on the table. Andryusha took out a sock from pair 2, and put this pair into the wardrobe. - Sock (3) was on the table. Andryusha took out a sock from pair 3 and put this pair into the wardrobe.
|
```python
input = __import__('sys').stdin.readline
MIS = lambda: map(int,input().split())
n = int(input())
table = set()
ans = 0
for x in MIS():
if x in table: table.remove(x)
else: table.add(x)
ans = max(ans, len(table))
print(ans)
```
| 3
|
|
588
|
A
|
Duff and Meat
|
PROGRAMMING
| 900
|
[
"greedy"
] | null | null |
Duff is addicted to meat! Malek wants to keep her happy for *n* days. In order to be happy in *i*-th day, she needs to eat exactly *a**i* kilograms of meat.
There is a big shop uptown and Malek wants to buy meat for her from there. In *i*-th day, they sell meat for *p**i* dollars per kilogram. Malek knows all numbers *a*1,<=...,<=*a**n* and *p*1,<=...,<=*p**n*. In each day, he can buy arbitrary amount of meat, also he can keep some meat he has for the future.
Malek is a little tired from cooking meat, so he asked for your help. Help him to minimize the total money he spends to keep Duff happy for *n* days.
|
The first line of input contains integer *n* (1<=≤<=*n*<=≤<=105), the number of days.
In the next *n* lines, *i*-th line contains two integers *a**i* and *p**i* (1<=≤<=*a**i*,<=*p**i*<=≤<=100), the amount of meat Duff needs and the cost of meat in that day.
|
Print the minimum money needed to keep Duff happy for *n* days, in one line.
|
[
"3\n1 3\n2 2\n3 1\n",
"3\n1 3\n2 1\n3 2\n"
] |
[
"10\n",
"8\n"
] |
In the first sample case: An optimal way would be to buy 1 kg on the first day, 2 kg on the second day and 3 kg on the third day.
In the second sample case: An optimal way would be to buy 1 kg on the first day and 5 kg (needed meat for the second and third day) on the second day.
| 750
|
[
{
"input": "3\n1 3\n2 2\n3 1",
"output": "10"
},
{
"input": "3\n1 3\n2 1\n3 2",
"output": "8"
},
{
"input": "1\n39 52",
"output": "2028"
},
{
"input": "2\n25 56\n94 17",
"output": "2998"
},
{
"input": "5\n39 21\n95 89\n73 90\n9 55\n85 32",
"output": "6321"
},
{
"input": "12\n70 11\n74 27\n32 11\n26 83\n57 18\n97 28\n75 43\n75 21\n84 29\n16 2\n89 63\n21 88",
"output": "6742"
},
{
"input": "2\n100 68\n38 80",
"output": "9384"
}
] | 1,513,487,040
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 30
| 358
| 5,632,000
|
n = int(input())
low = 101
ans = 0
for i in range(n):
a, b = input().split()
a, b = int(a), int(b)
if b < low:
low = b
ans += a * low
print(ans)
|
Title: Duff and Meat
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Duff is addicted to meat! Malek wants to keep her happy for *n* days. In order to be happy in *i*-th day, she needs to eat exactly *a**i* kilograms of meat.
There is a big shop uptown and Malek wants to buy meat for her from there. In *i*-th day, they sell meat for *p**i* dollars per kilogram. Malek knows all numbers *a*1,<=...,<=*a**n* and *p*1,<=...,<=*p**n*. In each day, he can buy arbitrary amount of meat, also he can keep some meat he has for the future.
Malek is a little tired from cooking meat, so he asked for your help. Help him to minimize the total money he spends to keep Duff happy for *n* days.
Input Specification:
The first line of input contains integer *n* (1<=≤<=*n*<=≤<=105), the number of days.
In the next *n* lines, *i*-th line contains two integers *a**i* and *p**i* (1<=≤<=*a**i*,<=*p**i*<=≤<=100), the amount of meat Duff needs and the cost of meat in that day.
Output Specification:
Print the minimum money needed to keep Duff happy for *n* days, in one line.
Demo Input:
['3\n1 3\n2 2\n3 1\n', '3\n1 3\n2 1\n3 2\n']
Demo Output:
['10\n', '8\n']
Note:
In the first sample case: An optimal way would be to buy 1 kg on the first day, 2 kg on the second day and 3 kg on the third day.
In the second sample case: An optimal way would be to buy 1 kg on the first day and 5 kg (needed meat for the second and third day) on the second day.
|
```python
n = int(input())
low = 101
ans = 0
for i in range(n):
a, b = input().split()
a, b = int(a), int(b)
if b < low:
low = b
ans += a * low
print(ans)
```
| 3
|
|
798
|
B
|
Mike and strings
|
PROGRAMMING
| 1,300
|
[
"brute force",
"dp",
"strings"
] | null | null |
Mike has *n* strings *s*1,<=*s*2,<=...,<=*s**n* each consisting of lowercase English letters. In one move he can choose a string *s**i*, erase the first character and append it to the end of the string. For example, if he has the string "coolmike", in one move he can transform it into the string "oolmikec".
Now Mike asks himself: what is minimal number of moves that he needs to do in order to make all the strings equal?
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=50) — the number of strings.
This is followed by *n* lines which contain a string each. The *i*-th line corresponding to string *s**i*. Lengths of strings are equal. Lengths of each string is positive and don't exceed 50.
|
Print the minimal number of moves Mike needs in order to make all the strings equal or print <=-<=1 if there is no solution.
|
[
"4\nxzzwo\nzwoxz\nzzwox\nxzzwo\n",
"2\nmolzv\nlzvmo\n",
"3\nkc\nkc\nkc\n",
"3\naa\naa\nab\n"
] |
[
"5\n",
"2\n",
"0\n",
"-1\n"
] |
In the first sample testcase the optimal scenario is to perform operations in such a way as to transform all strings into "zwoxz".
| 1,000
|
[
{
"input": "4\nxzzwo\nzwoxz\nzzwox\nxzzwo",
"output": "5"
},
{
"input": "2\nmolzv\nlzvmo",
"output": "2"
},
{
"input": "3\nkc\nkc\nkc",
"output": "0"
},
{
"input": "3\naa\naa\nab",
"output": "-1"
},
{
"input": "3\nkwkb\nkbkw\nbkwk",
"output": "3"
},
{
"input": "1\na",
"output": "0"
},
{
"input": "2\nnjtazaab\nabnjtaza",
"output": "2"
},
{
"input": "38\nkmlzdcnm\nmlzdcnmk\nlzdcnmkm\nkmlzdcnm\nlzdcnmkm\nzdcnmkml\nzdcnmkml\nmlzdcnmk\nzdcnmkml\nmlzdcnmk\nlzdcnmkm\nzdcnmkml\nkmlzdcnm\nlzdcnmkm\nzdcnmkml\nmlzdcnmk\nkmlzdcnm\nmkmlzdcn\nlzdcnmkm\nnmkmlzdc\nzdcnmkml\nnmkmlzdc\nkmlzdcnm\nmlzdcnmk\nmkmlzdcn\ndcnmkmlz\ncnmkmlzd\ncnmkmlzd\nmkmlzdcn\ncnmkmlzd\ndcnmkmlz\nkmlzdcnm\nnmkmlzdc\nnmkmlzdc\nkmlzdcnm\nkmlzdcnm\nlzdcnmkm\nzdcnmkml",
"output": "104"
},
{
"input": "4\nxwppaubrphxjwmwfwypvwwjzotyobpiynyka\nubrphxjwmwfwypvwwjzotyobpiynykaxwppa\nwjzotyobpiynykaxwppaubrphxjwmwfwypvw\ntyobpiynykaxwppaubrphxjwmwfwypvwwjzo",
"output": "41"
},
{
"input": "15\ngnizfqwqmimtgmtf\nmtgmtfgnizfqwqmi\ngmtfgnizfqwqmimt\nzfqwqmimtgmtfgni\nzfqwqmimtgmtfgni\nfqwqmimtgmtfgniz\nimtgmtfgnizfqwqm\nfgnizfqwqmimtgmt\ngmtfgnizfqwqmimt\nmtgmtfgnizfqwqmi\nqwqmimtgmtfgnizf\nizfqwqmimtgmtfgn\nmtfgnizfqwqmimtg\ntgmtfgnizfqwqmim\nmtfgnizfqwqmimtg",
"output": "89"
},
{
"input": "33\nnkgcmrfvxe\nvxenkgcmrf\nrfvxenkgcm\nvxenkgcmrf\nxenkgcmrfv\nenkgcmrfvx\nenkgcmrfvx\nnkgcmrfvxe\nkgcmrfvxen\ncmrfvxenkg\ncmrfvxenkg\nxenkgcmrfv\nrfvxenkgcm\nrfvxenkgcm\nnkgcmrfvxe\nxenkgcmrfv\nrfvxenkgcm\nxenkgcmrfv\nxenkgcmrfv\ngcmrfvxenk\nmrfvxenkgc\nfvxenkgcmr\nvxenkgcmrf\nenkgcmrfvx\ncmrfvxenkg\ncmrfvxenkg\nmrfvxenkgc\nkgcmrfvxen\nvxenkgcmrf\nenkgcmrfvx\ncmrfvxenkg\ncmrfvxenkg\ngcmrfvxenk",
"output": "135"
},
{
"input": "11\nxdngtxuqjalamqvotuhx\notuhxxdngtxuqjalamqv\ngtxuqjalamqvotuhxxdn\ndngtxuqjalamqvotuhxx\nvotuhxxdngtxuqjalamq\nxxdngtxuqjalamqvotuh\nalamqvotuhxxdngtxuqj\nuqjalamqvotuhxxdngtx\nqjalamqvotuhxxdngtxu\nhxxdngtxuqjalamqvotu\njalamqvotuhxxdngtxuq",
"output": "79"
},
{
"input": "2\noiadfnwpdcxxhbwwqbrcdujcusgtkqdjmintwjlb\nbrcdujcusgtkqdjmintwjlboiadfnwpdcxxhbwwq",
"output": "17"
},
{
"input": "20\ncynedh\nnedhcy\nhcyned\ncynedh\nynedhc\nynedhc\nnedhcy\nnedhcy\nnedhcy\nhcyned\nnedhcy\nhcyned\nnedhcy\ndhcyne\nynedhc\nedhcyn\ndhcyne\nynedhc\ncynedh\ncynedh",
"output": "34"
},
{
"input": "9\nrgycrkgcjktfdjkffcnlnhiawq\nawqrgycrkgcjktfdjkffcnlnhi\nrkgcjktfdjkffcnlnhiawqrgyc\njktfdjkffcnlnhiawqrgycrkgc\ncjktfdjkffcnlnhiawqrgycrkg\nfdjkffcnlnhiawqrgycrkgcjkt\nffcnlnhiawqrgycrkgcjktfdjk\nktfdjkffcnlnhiawqrgycrkgcj\nwqrgycrkgcjktfdjkffcnlnhia",
"output": "76"
},
{
"input": "2\ndzlisvouhbqogzusikmkuvkql\nqogzusikmkuvkqldzlisvouhb",
"output": "10"
},
{
"input": "2\nsfotivvfgbdfcnvaybxhstavaoktatflelpyi\nsfotivvfgbdfcnvaybxhstavaoktatflelpyi",
"output": "0"
},
{
"input": "1\numwnrjtcytnquvdmqfiqt",
"output": "0"
},
{
"input": "4\nzumixjfqhbkeg\nkegzumixjfqhb\nhbkegzumixjfq\ngzumixjfqhbke",
"output": "9"
},
{
"input": "12\nktwwduoopsnkhfklrskdxakbmqhl\nlktwwduoopsnkhfklrskdxakbmqh\nduoopsnkhfklrskdxakbmqhlktww\nklrskdxakbmqhlktwwduoopsnkhf\noopsnkhfklrskdxakbmqhlktwwdu\nopsnkhfklrskdxakbmqhlktwwduo\nkbmqhlktwwduoopsnkhfklrskdxa\nlrskdxakbmqhlktwwduoopsnkhfk\nwduoopsnkhfklrskdxakbmqhlktw\nklrskdxakbmqhlktwwduoopsnkhf\nhfklrskdxakbmqhlktwwduoopsnk\ndxakbmqhlktwwduoopsnkhfklrsk",
"output": "121"
},
{
"input": "12\naaaaaaaaaaaaaaaaaaaaaaaaaaa\naaaaaaaaaaaaaaaaaaaaaaaaaaa\naaaaaaaaaaaaaaaaaaaaaaaaaaa\naaaaaaaaaaaaaaaaaaaaaaaaaaa\naaaaaaaaaaaaaaaaaaaaaaaaaaa\naaaaaaaaaaaaaaaaaaaaaaaaaaa\naaaaaaaaaaaaaaaaaaaaaaaaaaa\naaaaaaaaaaaaaaaaaaaaaaaaaaa\naaaaaaaaaaaaaaaaaaaaaaaaaaa\naaaaaaaaaaaaaaaaaaaaaaaaaaa\naaaaaaaaaaaaaaaaaaaaaaaaaaa\naaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "0"
},
{
"input": "15\nkknrrejishjz\nhilbaftsfcaq\nlncsgtjqgwjz\nathvctulbhmb\nnfvsjyiulmmr\nhxjnvumwnwtr\nrncsxqvkvqeg\nqoabapuhodxk\nylinhbhyqjsn\ncnzxgdgytgav\nxufmacyangpv\nhwvzionkdmjl\nspklymjxiolk\nqjkfrccaayak\nonwrbgfvxrjx",
"output": "-1"
},
{
"input": "2\nadam\nmdaa",
"output": "-1"
},
{
"input": "2\naabc\nacab",
"output": "-1"
},
{
"input": "2\nabc\ncba",
"output": "-1"
},
{
"input": "5\naaaa\naaaa\naaaa\naaaa\naaaa",
"output": "0"
},
{
"input": "2\na\nb",
"output": "-1"
},
{
"input": "2\nabab\naabb",
"output": "-1"
},
{
"input": "2\nbac\nabc",
"output": "-1"
},
{
"input": "2\naabb\nabab",
"output": "-1"
},
{
"input": "3\naa\naa\naa",
"output": "0"
},
{
"input": "2\nabc\nacb",
"output": "-1"
},
{
"input": "3\naaaa\naaaa\naaaa",
"output": "0"
},
{
"input": "2\naa\naa",
"output": "0"
},
{
"input": "2\nab\naa",
"output": "-1"
},
{
"input": "2\nxyxy\nxxyy",
"output": "-1"
},
{
"input": "2\nabc\nbac",
"output": "-1"
},
{
"input": "2\naaabb\nbaaba",
"output": "-1"
},
{
"input": "2\nabcde\ndcabe",
"output": "-1"
},
{
"input": "2\nabcd\nbdca",
"output": "-1"
},
{
"input": "5\naaaaa\naabaa\naaaaa\naaaaa\naaaaa",
"output": "-1"
},
{
"input": "3\naab\nabb\nbab",
"output": "-1"
},
{
"input": "2\nnzxv\nzvnx",
"output": "-1"
},
{
"input": "10\nab\nbc\ncd\nde\nef\ngh\nhi\nij\nik\nmn",
"output": "-1"
},
{
"input": "3\naaa\naaa\naaa",
"output": "0"
},
{
"input": "2\nac\nbb",
"output": "-1"
},
{
"input": "2\nabcd\nbdac",
"output": "-1"
},
{
"input": "2\nabcabc\ncabcab",
"output": "1"
},
{
"input": "7\naaa\naab\naba\nabb\nbaa\nbab\nbba",
"output": "-1"
},
{
"input": "4\naa\naa\nbb\nbb",
"output": "-1"
},
{
"input": "2\nabcd\ncabd",
"output": "-1"
},
{
"input": "4\nabcabcabc\nbcabcabca\ncabcabcab\ncabcabcab",
"output": "3"
},
{
"input": "3\nabcabc\nbcabca\nbcabca",
"output": "1"
},
{
"input": "2\nabbc\nabcc",
"output": "-1"
},
{
"input": "2\naaabb\nababa",
"output": "-1"
},
{
"input": "3\naabbbaba\nabaabbab\nbbbaaaba",
"output": "-1"
},
{
"input": "2\naabaab\nbaabaa",
"output": "1"
}
] | 1,640,592,589
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 99
| 108
| 1,945,600
|
n=int(input())
s=list()
for _ in range(n):
x=input()
s.append(x)
l=len(s[0])
flag=True
c=1e9+10
for i in range(0,n):
if(len(s[i])!=l):
flag=False
break
t=s[i]*2
if(s[0] not in t):
flag=False
break
f=0
for j in range(n):
x=l+1
for k in range(l):
if(s[i]==s[j][k:l]+s[j][0:k]):
x=min(x,k)
f+=x
c=min(c,f)
if(flag):
print(c)
else:
print(-1)
|
Title: Mike and strings
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mike has *n* strings *s*1,<=*s*2,<=...,<=*s**n* each consisting of lowercase English letters. In one move he can choose a string *s**i*, erase the first character and append it to the end of the string. For example, if he has the string "coolmike", in one move he can transform it into the string "oolmikec".
Now Mike asks himself: what is minimal number of moves that he needs to do in order to make all the strings equal?
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=50) — the number of strings.
This is followed by *n* lines which contain a string each. The *i*-th line corresponding to string *s**i*. Lengths of strings are equal. Lengths of each string is positive and don't exceed 50.
Output Specification:
Print the minimal number of moves Mike needs in order to make all the strings equal or print <=-<=1 if there is no solution.
Demo Input:
['4\nxzzwo\nzwoxz\nzzwox\nxzzwo\n', '2\nmolzv\nlzvmo\n', '3\nkc\nkc\nkc\n', '3\naa\naa\nab\n']
Demo Output:
['5\n', '2\n', '0\n', '-1\n']
Note:
In the first sample testcase the optimal scenario is to perform operations in such a way as to transform all strings into "zwoxz".
|
```python
n=int(input())
s=list()
for _ in range(n):
x=input()
s.append(x)
l=len(s[0])
flag=True
c=1e9+10
for i in range(0,n):
if(len(s[i])!=l):
flag=False
break
t=s[i]*2
if(s[0] not in t):
flag=False
break
f=0
for j in range(n):
x=l+1
for k in range(l):
if(s[i]==s[j][k:l]+s[j][0:k]):
x=min(x,k)
f+=x
c=min(c,f)
if(flag):
print(c)
else:
print(-1)
```
| 3
|
|
32
|
B
|
Borze
|
PROGRAMMING
| 800
|
[
"expression parsing",
"implementation"
] |
B. Borze
|
2
|
256
|
Ternary numeric notation is quite popular in Berland. To telegraph the ternary number the Borze alphabet is used. Digit 0 is transmitted as «.», 1 as «-.» and 2 as «--». You are to decode the Borze code, i.e. to find out the ternary number given its representation in Borze alphabet.
|
The first line contains a number in Borze code. The length of the string is between 1 and 200 characters. It's guaranteed that the given string is a valid Borze code of some ternary number (this number can have leading zeroes).
|
Output the decoded ternary number. It can have leading zeroes.
|
[
".-.--\n",
"--.\n",
"-..-.--\n"
] |
[
"012",
"20",
"1012"
] |
none
| 1,000
|
[
{
"input": ".-.--",
"output": "012"
},
{
"input": "--.",
"output": "20"
},
{
"input": "-..-.--",
"output": "1012"
},
{
"input": "---..",
"output": "210"
},
{
"input": "..--.---..",
"output": "0020210"
},
{
"input": "-.....----.",
"output": "10000220"
},
{
"input": ".",
"output": "0"
},
{
"input": "-.",
"output": "1"
},
{
"input": "--",
"output": "2"
},
{
"input": "..",
"output": "00"
},
{
"input": "--.",
"output": "20"
},
{
"input": ".--.",
"output": "020"
},
{
"input": ".-.-..",
"output": "0110"
},
{
"input": "----.-.",
"output": "2201"
},
{
"input": "-..--.-.",
"output": "10201"
},
{
"input": "..--..--.",
"output": "0020020"
},
{
"input": "-.-.---.--..-..-.-.-..-..-.--.",
"output": "112120010111010120"
},
{
"input": "---.-.-.------..-..-..-..-.-..-.--.-.-..-.-.-----..-.-.",
"output": "21112220010101011012011011221011"
},
{
"input": "-.-..--.-.-.-.-.-..-.-.-.---------.--.---..--...--.-----.-.-.-...--.-.-.---.------.--..-.--.-----.-...-..------",
"output": "11020111110111222212021020002022111100201121222020012022110010222"
},
{
"input": "-.-..-.--.---..---.-..---.-...-.-.----..-.---.-.---..-.--.---.-.-------.---.--....----.-.---.---.---.----.-----..---.-.-.-.-----.--.-------.-..",
"output": "110120210211021100112200121121012021122212120000220121212122022102111122120222110"
},
{
"input": ".-..-.-.---.-----.--.---...-.--.-.-....-..",
"output": "01011212212021001201100010"
},
{
"input": ".------.-.---..--...-..-..-.-.-.--.--.-..-.--...-.-.---.-.-.------..--..-.---..----.-..-.--.---.-.----.-.---...-.-.-.-----.-.-.---.---.-.....-.-...-----.-...-.---.-..-.-----.--...---.-.-..-.--.-.---..",
"output": "022201210200010101112020101200011211122200200121022010120211220121001112211121211000011002211001211012212000211101201210"
},
{
"input": ".-.--.---.-----.-.-----.-.-..-----..-..----..--.-.--.----..---.---..-.-.-----..-------.----..----.-..---...-----..-..-----...-..-.-.-----....---..---..-.-----...-.--...--.-.---.-.-.-.-.-...---..----.",
"output": "01202122112211102210102200201202200212101122102221220022010210022101022100101122100021021012210012000201211111100210220"
},
{
"input": "..-.-.-.---.-.-.-..-.-..-.-.---.-------.---..-----.---....-.---.--.--.-.---.---------.-..---.-.-.--..---.---.-.---.-.-..-.-..-.-.-.----.--.-....--------.-.---..----.------.-.-.--.--.-----.-----.----",
"output": "0011121111011011212221210221210001212020121222211021112002121121110110111220201000222201210220222011202022122122"
},
{
"input": "-..-------.------.-..--.-.-..--.-.-..-----..-.-.-..-..-..--.---..-----..---..-..--.-..-.-.---...-.....-------.---.-----.-...-.-...-.-.---.---.-----.--.--...-.--..-.-..-...-.-.-.-.---..---.-..-.-.-.-..",
"output": "102221222010201102011022101110101020210221021010201011210010000222121221100110011212122120200012001101001111210211011110"
},
{
"input": ".-.----.-.--..-.-.-.-..----..-.-...--.-.---.---.-------..-.--..-......--.------.--.----.--...-.--.--..-----..-.....--.--.-.-.------..--------.----------..-.---.----.---.-..--..-.....-..------.--.",
"output": "012201200111102200110020121212221012001000002022202022020001202002210100002020112220022220222220012122021102001000010222020"
},
{
"input": "------.-----.-....--.-.----.-.---.-.-..---.-.---.-----..-...-.-.---..-.-.-..-.-.-...-.-.-.----..--.------.----.-..-.--...-.-------...-.-..-.-.--.--.---..--..--------.--.-.-.---.-.-.-...----.--..-.--..",
"output": "222022110002012201211102112122101001121011101110011122002022202201012000122210011011202021020022220201121111002202001200"
},
{
"input": "-.---...----...--.--...-.--.----",
"output": "121002200020200012022"
},
{
"input": "--.--.--.---.--.-.---.-.-..-..--.-..---.-.....-..---.-----.--...-.-.-------.-.--.-.----.-..-.------.",
"output": "202020212012111010201021100001021221200011222112012201012220"
}
] | 1,608,952,200
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 30
| 218
| 0
|
n=str(input())
n=n.replace('--', '2')
n=n.replace('-.', '1')
n=n.replace('.','0')
print(n)
|
Title: Borze
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Ternary numeric notation is quite popular in Berland. To telegraph the ternary number the Borze alphabet is used. Digit 0 is transmitted as «.», 1 as «-.» and 2 as «--». You are to decode the Borze code, i.e. to find out the ternary number given its representation in Borze alphabet.
Input Specification:
The first line contains a number in Borze code. The length of the string is between 1 and 200 characters. It's guaranteed that the given string is a valid Borze code of some ternary number (this number can have leading zeroes).
Output Specification:
Output the decoded ternary number. It can have leading zeroes.
Demo Input:
['.-.--\n', '--.\n', '-..-.--\n']
Demo Output:
['012', '20', '1012']
Note:
none
|
```python
n=str(input())
n=n.replace('--', '2')
n=n.replace('-.', '1')
n=n.replace('.','0')
print(n)
```
| 3.9455
|
722
|
B
|
Verse Pattern
|
PROGRAMMING
| 1,200
|
[
"implementation",
"strings"
] | null | null |
You are given a text consisting of *n* lines. Each line contains some space-separated words, consisting of lowercase English letters.
We define a syllable as a string that contains exactly one vowel and any arbitrary number (possibly none) of consonants. In English alphabet following letters are considered to be vowels: 'a', 'e', 'i', 'o', 'u' and 'y'.
Each word of the text that contains at least one vowel can be divided into syllables. Each character should be a part of exactly one syllable. For example, the word "mamma" can be divided into syllables as "ma" and "mma", "mam" and "ma", and "mamm" and "a". Words that consist of only consonants should be ignored.
The verse patterns for the given text is a sequence of *n* integers *p*1,<=*p*2,<=...,<=*p**n*. Text matches the given verse pattern if for each *i* from 1 to *n* one can divide words of the *i*-th line in syllables in such a way that the total number of syllables is equal to *p**i*.
You are given the text and the verse pattern. Check, if the given text matches the given verse pattern.
|
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of lines in the text.
The second line contains integers *p*1,<=...,<=*p**n* (0<=≤<=*p**i*<=≤<=100) — the verse pattern.
Next *n* lines contain the text itself. Text consists of lowercase English letters and spaces. It's guaranteed that all lines are non-empty, each line starts and ends with a letter and words are separated by exactly one space. The length of each line doesn't exceed 100 characters.
|
If the given text matches the given verse pattern, then print "YES" (without quotes) in the only line of the output. Otherwise, print "NO" (without quotes).
|
[
"3\n2 2 3\nintel\ncode\nch allenge\n",
"4\n1 2 3 1\na\nbcdefghi\njklmnopqrstu\nvwxyz\n",
"4\n13 11 15 15\nto be or not to be that is the question\nwhether tis nobler in the mind to suffer\nthe slings and arrows of outrageous fortune\nor to take arms against a sea of troubles\n"
] |
[
"YES\n",
"NO\n",
"YES\n"
] |
In the first sample, one can split words into syllables in the following way:
Since the word "ch" in the third line doesn't contain vowels, we can ignore it. As the result we get 2 syllabels in first two lines and 3 syllables in the third one.
| 500
|
[
{
"input": "3\n2 2 3\nintel\ncode\nch allenge",
"output": "YES"
},
{
"input": "4\n1 2 3 1\na\nbcdefghi\njklmnopqrstu\nvwxyz",
"output": "NO"
},
{
"input": "4\n13 11 15 15\nto be or not to be that is the question\nwhether tis nobler in the mind to suffer\nthe slings and arrows of outrageous fortune\nor to take arms against a sea of troubles",
"output": "YES"
},
{
"input": "5\n2 2 1 1 1\nfdbie\naaj\ni\ni n\nshi",
"output": "YES"
},
{
"input": "5\n2 11 10 7 9\nhy of\nyur pjyacbatdoylojayu\nemd ibweioiimyxya\nyocpyivudobua\nuiraueect impxqhzpty e",
"output": "NO"
},
{
"input": "5\n6 9 7 3 10\nabtbdaa\nom auhz ub iaravozegs\ncieulibsdhj ufki\nadu pnpurt\nh naony i jaysjsjxpwuuc",
"output": "NO"
},
{
"input": "2\n26 35\ngouojxaoobw iu bkaadyo degnjkubeabt kbap thwki dyebailrhnoh ooa\npiaeaebaocptyswuc wezesazipu osebhaonouygasjrciyiqaejtqsioubiuakg umynbsvw xpfqdwxo",
"output": "NO"
},
{
"input": "5\n1 0 0 1 1\ngqex\nw\nh\nzsvu\nqcqd",
"output": "NO"
},
{
"input": "5\n0 0 0 0 0\njtv\nl\nqg\ntp\nfgd",
"output": "YES"
},
{
"input": "10\n0 0 0 0 0 0 0 0 0 0\nj t fr\nn\nnhcgx\np\nmb hmhtz\ndbjc\ncwdxj\nn j whkbt\nzk m cwh\nqr n",
"output": "YES"
},
{
"input": "5\n4 5 1 0 0\noa\nqfohq\ni l\naik\nx",
"output": "NO"
},
{
"input": "10\n2 9 0 3 2 4 1 2 4 2\nxtwl oy\nafgeju fi\nr hy\nddsowagw\nxoredo f\nwufnxy k uh\nod\nlejrinw\nsueecohfjl\nedufg",
"output": "NO"
},
{
"input": "10\n1 1 0 0 0 4 0 4 0 0\na bn\nhnwss f\nd s bn\nbdzxzgsxq\nghh v\neimblv i er\nca kn k\nzm ffc zcb\nn\nz hkhvfkwhg",
"output": "NO"
},
{
"input": "5\n0 10 6 6 0\nfgthrxst\nsohnweymewnnmbobj\nj\nfwwt acdtfvkpv khbxokn\nhndovkkgfhnhqod",
"output": "NO"
},
{
"input": "5\n3 2 2 4 2\ni yu\niu\noa\naiio\nuo",
"output": "YES"
},
{
"input": "5\n11 12 11 4 6\nuuuayoiaoiy\nuaiee iai eieu\nooayaayeuee\noii o\noea uuo",
"output": "YES"
},
{
"input": "3\n2 3 2\nintel\ncode\nch allenge",
"output": "NO"
},
{
"input": "2\n1 2\ncode\na",
"output": "NO"
},
{
"input": "2\n1 1\nbababa\nbababa",
"output": "NO"
},
{
"input": "1\n1\naa",
"output": "NO"
},
{
"input": "1\n1\naaa",
"output": "NO"
},
{
"input": "2\n2 3\naee\nae",
"output": "NO"
}
] | 1,475,921,857
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 54
| 77
| 4,608,000
|
n = int(input())
a = [int(i) for i in input().split()]
k = 0
q = 0
for i in range(n):
s = input()
for j in s:
if (j=="a" or j=="e" or j=="i" or j=="o" or j=="u" or j=="y"):
q+=1
if q!=a[i]:
print("NO")
exit(0)
q=0
print("YES")
|
Title: Verse Pattern
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a text consisting of *n* lines. Each line contains some space-separated words, consisting of lowercase English letters.
We define a syllable as a string that contains exactly one vowel and any arbitrary number (possibly none) of consonants. In English alphabet following letters are considered to be vowels: 'a', 'e', 'i', 'o', 'u' and 'y'.
Each word of the text that contains at least one vowel can be divided into syllables. Each character should be a part of exactly one syllable. For example, the word "mamma" can be divided into syllables as "ma" and "mma", "mam" and "ma", and "mamm" and "a". Words that consist of only consonants should be ignored.
The verse patterns for the given text is a sequence of *n* integers *p*1,<=*p*2,<=...,<=*p**n*. Text matches the given verse pattern if for each *i* from 1 to *n* one can divide words of the *i*-th line in syllables in such a way that the total number of syllables is equal to *p**i*.
You are given the text and the verse pattern. Check, if the given text matches the given verse pattern.
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of lines in the text.
The second line contains integers *p*1,<=...,<=*p**n* (0<=≤<=*p**i*<=≤<=100) — the verse pattern.
Next *n* lines contain the text itself. Text consists of lowercase English letters and spaces. It's guaranteed that all lines are non-empty, each line starts and ends with a letter and words are separated by exactly one space. The length of each line doesn't exceed 100 characters.
Output Specification:
If the given text matches the given verse pattern, then print "YES" (without quotes) in the only line of the output. Otherwise, print "NO" (without quotes).
Demo Input:
['3\n2 2 3\nintel\ncode\nch allenge\n', '4\n1 2 3 1\na\nbcdefghi\njklmnopqrstu\nvwxyz\n', '4\n13 11 15 15\nto be or not to be that is the question\nwhether tis nobler in the mind to suffer\nthe slings and arrows of outrageous fortune\nor to take arms against a sea of troubles\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n']
Note:
In the first sample, one can split words into syllables in the following way:
Since the word "ch" in the third line doesn't contain vowels, we can ignore it. As the result we get 2 syllabels in first two lines and 3 syllables in the third one.
|
```python
n = int(input())
a = [int(i) for i in input().split()]
k = 0
q = 0
for i in range(n):
s = input()
for j in s:
if (j=="a" or j=="e" or j=="i" or j=="o" or j=="u" or j=="y"):
q+=1
if q!=a[i]:
print("NO")
exit(0)
q=0
print("YES")
```
| 3
|
|
41
|
A
|
Translation
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] |
A. Translation
|
2
|
256
|
The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly.
|
The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols.
|
If the word *t* is a word *s*, written reversely, print YES, otherwise print NO.
|
[
"code\nedoc\n",
"abb\naba\n",
"code\ncode\n"
] |
[
"YES\n",
"NO\n",
"NO\n"
] |
none
| 500
|
[
{
"input": "code\nedoc",
"output": "YES"
},
{
"input": "abb\naba",
"output": "NO"
},
{
"input": "code\ncode",
"output": "NO"
},
{
"input": "abacaba\nabacaba",
"output": "YES"
},
{
"input": "q\nq",
"output": "YES"
},
{
"input": "asrgdfngfnmfgnhweratgjkk\nasrgdfngfnmfgnhweratgjkk",
"output": "NO"
},
{
"input": "z\na",
"output": "NO"
},
{
"input": "asd\ndsa",
"output": "YES"
},
{
"input": "abcdef\nfecdba",
"output": "NO"
},
{
"input": "ywjjbirapvskozubvxoemscfwl\ngnduubaogtfaiowjizlvjcu",
"output": "NO"
},
{
"input": "mfrmqxtzvgaeuleubcmcxcfqyruwzenguhgrmkuhdgnhgtgkdszwqyd\nmfxufheiperjnhyczclkmzyhcxntdfskzkzdwzzujdinf",
"output": "NO"
},
{
"input": "bnbnemvybqizywlnghlykniaxxxlkhftppbdeqpesrtgkcpoeqowjwhrylpsziiwcldodcoonpimudvrxejjo\ntiynnekmlalogyvrgptbinkoqdwzuiyjlrldxhzjmmp",
"output": "NO"
},
{
"input": "pwlpubwyhzqvcitemnhvvwkmwcaawjvdiwtoxyhbhbxerlypelevasmelpfqwjk\nstruuzebbcenziscuoecywugxncdwzyfozhljjyizpqcgkyonyetarcpwkqhuugsqjuixsxptmbnlfupdcfigacdhhrzb",
"output": "NO"
},
{
"input": "gdvqjoyxnkypfvdxssgrihnwxkeojmnpdeobpecytkbdwujqfjtxsqspxvxpqioyfagzjxupqqzpgnpnpxcuipweunqch\nkkqkiwwasbhezqcfeceyngcyuogrkhqecwsyerdniqiocjehrpkljiljophqhyaiefjpavoom",
"output": "NO"
},
{
"input": "umeszdawsvgkjhlqwzents\nhxqhdungbylhnikwviuh",
"output": "NO"
},
{
"input": "juotpscvyfmgntshcealgbsrwwksgrwnrrbyaqqsxdlzhkbugdyx\nibqvffmfktyipgiopznsqtrtxiijntdbgyy",
"output": "NO"
},
{
"input": "zbwueheveouatecaglziqmudxemhrsozmaujrwlqmppzoumxhamwugedikvkblvmxwuofmpafdprbcftew\nulczwrqhctbtbxrhhodwbcxwimncnexosksujlisgclllxokrsbnozthajnnlilyffmsyko",
"output": "NO"
},
{
"input": "nkgwuugukzcv\nqktnpxedwxpxkrxdvgmfgoxkdfpbzvwsduyiybynbkouonhvmzakeiruhfmvrktghadbfkmwxduoqv",
"output": "NO"
},
{
"input": "incenvizhqpcenhjhehvjvgbsnfixbatrrjstxjzhlmdmxijztphxbrldlqwdfimweepkggzcxsrwelodpnryntepioqpvk\ndhjbjjftlvnxibkklxquwmzhjfvnmwpapdrslioxisbyhhfymyiaqhlgecpxamqnocizwxniubrmpyubvpenoukhcobkdojlybxd",
"output": "NO"
},
{
"input": "w\nw",
"output": "YES"
},
{
"input": "vz\nzv",
"output": "YES"
},
{
"input": "ry\nyr",
"output": "YES"
},
{
"input": "xou\nuox",
"output": "YES"
},
{
"input": "axg\ngax",
"output": "NO"
},
{
"input": "zdsl\nlsdz",
"output": "YES"
},
{
"input": "kudl\nldku",
"output": "NO"
},
{
"input": "zzlzwnqlcl\nlclqnwzlzz",
"output": "YES"
},
{
"input": "vzzgicnzqooejpjzads\nsdazjpjeooqzncigzzv",
"output": "YES"
},
{
"input": "raqhmvmzuwaykjpyxsykr\nxkysrypjkyawuzmvmhqar",
"output": "NO"
},
{
"input": "ngedczubzdcqbxksnxuavdjaqtmdwncjnoaicvmodcqvhfezew\nwezefhvqcdomvciaonjcnwdmtqajdvauxnskxbqcdzbuzcdegn",
"output": "YES"
},
{
"input": "muooqttvrrljcxbroizkymuidvfmhhsjtumksdkcbwwpfqdyvxtrlymofendqvznzlmim\nmimlznzvqdnefomylrtxvydqfpwwbckdskmutjshhmfvdiumykziorbxcjlrrvttqooum",
"output": "YES"
},
{
"input": "vxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaivg\ngviayyikkitmuomcpiakhbxszgbnhvwyzkftwoagzixaearxpjacrnvpvbuzenvovehkmmxvblqyxvctroddksdsgebcmlluqpxv",
"output": "YES"
},
{
"input": "mnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfdc\ncdfmkdgrdptkpewbsqvszipgxvgvuiuzbkkwuowbafkikgvnqdkxnayzdjygvezmtsgywnupocdntipiyiorblqkrzjpzatxahnm",
"output": "NO"
},
{
"input": "dgxmzbqofstzcdgthbaewbwocowvhqpinehpjatnnbrijcolvsatbblsrxabzrpszoiecpwhfjmwuhqrapvtcgvikuxtzbftydkw\nwkdytfbztxukivgctvparqhuwmjfhwpceiozsprzbaxrslbbqasvlocjirbnntajphenipthvwocowbweabhtgdcztsfoqbzmxgd",
"output": "NO"
},
{
"input": "gxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwgeh\nhegwxvocotmzstqfbmpjvijgkcyodlxyjawrpkczpmdspsuhoiruavnnnuwvtwohglkdxjetshkboalvzqbgjgthoteceixioxg",
"output": "YES"
},
{
"input": "sihxuwvmaambplxvjfoskinghzicyfqebjtkysotattkahssumfcgrkheotdxwjckpvapbkaepqrxseyfrwtyaycmrzsrsngkh\nhkgnsrszrmcyaytwrfyesxrqpeakbpavpkcjwxdtoehkrgcfmusshakttatosyktjbeqfycizhgniksofjvxlpbmaamvwuxhis",
"output": "YES"
},
{
"input": "ycnahksbughnonldzrhkysujmylcgcfuludjvjiahtkyzqvkopzqcnwhltbzfugzojqkjjlggmvnultascmygelkiktmfieok\nkoeifmtkiklegkmcsatlunvmggkjjlqjozgufzbtlhwncqzpokvqzykthaijvjdulufcgclymjusyyhrzdlnonhgubskhancy",
"output": "NO"
},
{
"input": "wbqasaehtkfojruzyhrlgwmtyiovmzyfifslvlemhqheyaelzwnthrenjsbmntwaoryzwfbxmscmypvxlfmzpnkkjlvwvmtz\nztmvwvljkknpzmflxvpymcsmxbfwzyroawtnmbsjnerhtnwzleayehqhmelvlsfifyzmvoiytmwglrhyzurjofktheasaqbw",
"output": "YES"
},
{
"input": "imippqurprbhfugngtgifelytadegwrgaefnfhbjjnmzikvjaccotqzemufqieqldgnbmviisgkynzeldlhqxuqphjfmyij\njiymfjhpquxqhldleznykgsiivmbngdlqeiqfumezqtoccajvkizmnjjbhfnfeagrwgedatylefigtgngufhbrpruqppimi",
"output": "YES"
},
{
"input": "bikydffiuisckpvzqlteqfhegsagimodb\nbdomigasgehfqetlqzvpkcsiuiffdykib",
"output": "YES"
}
] | 1,529,524,080
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 40
| 184
| 0
|
s = input()
tt = input()
ss=""
for c in s:
ss = c + ss
if(ss == tt):
print('YES')
else:
print('NO')
|
Title: Translation
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly.
Input Specification:
The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols.
Output Specification:
If the word *t* is a word *s*, written reversely, print YES, otherwise print NO.
Demo Input:
['code\nedoc\n', 'abb\naba\n', 'code\ncode\n']
Demo Output:
['YES\n', 'NO\n', 'NO\n']
Note:
none
|
```python
s = input()
tt = input()
ss=""
for c in s:
ss = c + ss
if(ss == tt):
print('YES')
else:
print('NO')
```
| 3.954
|
788
|
A
|
Functions again
|
PROGRAMMING
| 1,600
|
[
"dp",
"two pointers"
] | null | null |
Something happened in Uzhlyandia again... There are riots on the streets... Famous Uzhlyandian superheroes Shean the Sheep and Stas the Giraffe were called in order to save the situation. Upon the arriving, they found that citizens are worried about maximum values of the Main Uzhlyandian Function *f*, which is defined as follows:
In the above formula, 1<=≤<=*l*<=<<=*r*<=≤<=*n* must hold, where *n* is the size of the Main Uzhlyandian Array *a*, and |*x*| means absolute value of *x*. But the heroes skipped their math lessons in school, so they asked you for help. Help them calculate the maximum value of *f* among all possible values of *l* and *r* for the given array *a*.
|
The first line contains single integer *n* (2<=≤<=*n*<=≤<=105) — the size of the array *a*.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (-109<=≤<=*a**i*<=≤<=109) — the array elements.
|
Print the only integer — the maximum value of *f*.
|
[
"5\n1 4 2 3 1\n",
"4\n1 5 4 7\n"
] |
[
"3",
"6"
] |
In the first sample case, the optimal value of *f* is reached on intervals [1, 2] and [2, 5].
In the second case maximal value of *f* is reachable only on the whole array.
| 500
|
[
{
"input": "5\n1 4 2 3 1",
"output": "3"
},
{
"input": "4\n1 5 4 7",
"output": "6"
},
{
"input": "8\n16 14 12 10 8 100 50 0",
"output": "92"
},
{
"input": "2\n1 1",
"output": "0"
},
{
"input": "50\n-5 -9 0 44 -10 37 34 -49 11 -22 -26 44 8 -13 23 -46 34 12 -24 2 -40 -15 -28 38 -40 -42 -42 7 -43 5 2 -11 10 43 9 49 -13 36 2 24 46 50 -15 -26 -6 -6 8 4 -44 -3",
"output": "208"
},
{
"input": "100\n23 64 60 -45 -36 -64 -59 15 -75 69 -30 -7 -20 17 -77 58 93 -76 -98 -22 -31 16 -50 6 -20 -85 1 64 -88 -8 -15 -6 -57 25 91 10 2 -90 74 -66 -42 73 28 49 -85 59 96 79 -25 49 -59 -89 -75 12 -96 -33 -65 -8 -100 -81 17 99 -91 -5 7 -21 1 85 63 86 -26 85 -31 11 -75 35 -82 15 98 93 -55 66 70 36 -38 8 92 -63 -5 60 -78 -7 -22 -1 4 54 36 16 32 -20",
"output": "632"
},
{
"input": "3\n0 0 0",
"output": "0"
},
{
"input": "3\n0 1 -1",
"output": "2"
},
{
"input": "5\n1000000000 0 0 1000000000 -1000000000",
"output": "2000000000"
},
{
"input": "2\n1000000000 -1000000000",
"output": "2000000000"
},
{
"input": "4\n1000000000 0 0 -1000000000",
"output": "2000000000"
},
{
"input": "4\n1 0 1 0",
"output": "1"
},
{
"input": "5\n0 10 20 30 40",
"output": "10"
},
{
"input": "2\n-1000000000 1000000000",
"output": "2000000000"
},
{
"input": "2\n1 2",
"output": "1"
},
{
"input": "9\n1000000000 0 0 1000000000 1000000000 0 0 1000000000 1000000000",
"output": "4000000000"
},
{
"input": "16\n-1000000000 1000000000 1000000000 -1000000000 -1000000000 1000000000 1000000000 -1000000000 -1000000000 1000000000 1000000000 -1000000000 -1000000000 1000000000 1000000000 -1000000000",
"output": "16000000000"
},
{
"input": "6\n1000000000 0 0 1000000000 1000000000 0",
"output": "3000000000"
},
{
"input": "17\n1000000000 0 0 1000000000 1000000000 0 0 1000000000 1000000000 0 0 1000000000 1000000000 0 0 1000000000 1000000000",
"output": "8000000000"
},
{
"input": "11\n1000000000 0 1 1000000000 999999999 1 0 1000000000 999999999 1 1000000000",
"output": "4999999991"
},
{
"input": "4\n1000000000 -1000000000 -1000000000 1000000000",
"output": "4000000000"
},
{
"input": "10\n1000000000 -1000000000 -1000000000 1000000000 1000000000 -1000000000 -1000000000 1000000000 1000000000 -1000000000",
"output": "10000000000"
}
] | 1,537,927,313
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 52
| 483
| 9,216,000
|
lectura= lambda:map (int, input().split())
n= (list(lectura())[0])
lista= list(lectura())
fDescrita=0
alt1=1
maxV1=0
maxV2=0
C1=0
C2=0
for i in range(0, n -1):
fDescrita= abs(lista[i] - lista[i + 1]) * alt1
maxV1=max(maxV1 + fDescrita, fDescrita)
maxV2 = max(maxV2, maxV1)
alt1 = alt1 * (-1)
#print(fDescrita,maxV1,maxV2)
C1=maxV2
maxV1=0
maxV2=0
alt1=1
for i in range(1, n -1):
fDescrita= abs(lista[i] - lista[i + 1]) * alt1
maxV1=max(maxV1 + fDescrita, fDescrita)
maxV2 = max(maxV2, maxV1)
alt1 = alt1 * (-1)
#print(fDescrita,maxV1,maxV2)
C2=maxV2
print(max(C1,C2))
|
Title: Functions again
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Something happened in Uzhlyandia again... There are riots on the streets... Famous Uzhlyandian superheroes Shean the Sheep and Stas the Giraffe were called in order to save the situation. Upon the arriving, they found that citizens are worried about maximum values of the Main Uzhlyandian Function *f*, which is defined as follows:
In the above formula, 1<=≤<=*l*<=<<=*r*<=≤<=*n* must hold, where *n* is the size of the Main Uzhlyandian Array *a*, and |*x*| means absolute value of *x*. But the heroes skipped their math lessons in school, so they asked you for help. Help them calculate the maximum value of *f* among all possible values of *l* and *r* for the given array *a*.
Input Specification:
The first line contains single integer *n* (2<=≤<=*n*<=≤<=105) — the size of the array *a*.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (-109<=≤<=*a**i*<=≤<=109) — the array elements.
Output Specification:
Print the only integer — the maximum value of *f*.
Demo Input:
['5\n1 4 2 3 1\n', '4\n1 5 4 7\n']
Demo Output:
['3', '6']
Note:
In the first sample case, the optimal value of *f* is reached on intervals [1, 2] and [2, 5].
In the second case maximal value of *f* is reachable only on the whole array.
|
```python
lectura= lambda:map (int, input().split())
n= (list(lectura())[0])
lista= list(lectura())
fDescrita=0
alt1=1
maxV1=0
maxV2=0
C1=0
C2=0
for i in range(0, n -1):
fDescrita= abs(lista[i] - lista[i + 1]) * alt1
maxV1=max(maxV1 + fDescrita, fDescrita)
maxV2 = max(maxV2, maxV1)
alt1 = alt1 * (-1)
#print(fDescrita,maxV1,maxV2)
C1=maxV2
maxV1=0
maxV2=0
alt1=1
for i in range(1, n -1):
fDescrita= abs(lista[i] - lista[i + 1]) * alt1
maxV1=max(maxV1 + fDescrita, fDescrita)
maxV2 = max(maxV2, maxV1)
alt1 = alt1 * (-1)
#print(fDescrita,maxV1,maxV2)
C2=maxV2
print(max(C1,C2))
```
| 3
|
|
520
|
A
|
Pangram
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] | null | null |
A word or a sentence in some language is called a pangram if all the characters of the alphabet of this language appear in it at least once. Pangrams are often used to demonstrate fonts in printing or test the output devices.
You are given a string consisting of lowercase and uppercase Latin letters. Check whether this string is a pangram. We say that the string contains a letter of the Latin alphabet if this letter occurs in the string in uppercase or lowercase.
|
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of characters in the string.
The second line contains the string. The string consists only of uppercase and lowercase Latin letters.
|
Output "YES", if the string is a pangram and "NO" otherwise.
|
[
"12\ntoosmallword\n",
"35\nTheQuickBrownFoxJumpsOverTheLazyDog\n"
] |
[
"NO\n",
"YES\n"
] |
none
| 500
|
[
{
"input": "12\ntoosmallword",
"output": "NO"
},
{
"input": "35\nTheQuickBrownFoxJumpsOverTheLazyDog",
"output": "YES"
},
{
"input": "1\na",
"output": "NO"
},
{
"input": "26\nqwertyuiopasdfghjklzxcvbnm",
"output": "YES"
},
{
"input": "26\nABCDEFGHIJKLMNOPQRSTUVWXYZ",
"output": "YES"
},
{
"input": "48\nthereisasyetinsufficientdataforameaningfulanswer",
"output": "NO"
},
{
"input": "30\nToBeOrNotToBeThatIsTheQuestion",
"output": "NO"
},
{
"input": "30\njackdawslovemybigsphinxofquarz",
"output": "NO"
},
{
"input": "31\nTHEFIVEBOXINGWIZARDSJUMPQUICKLY",
"output": "YES"
},
{
"input": "26\naaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "NO"
},
{
"input": "26\nMGJYIZDKsbhpVeNFlquRTcWoAx",
"output": "YES"
},
{
"input": "26\nfWMOhAPsbIVtyUEZrGNQXDklCJ",
"output": "YES"
},
{
"input": "26\nngPMVFSThiRCwLEuyOAbKxQzDJ",
"output": "YES"
},
{
"input": "25\nnxYTzLFwzNolAumjgcAboyxAj",
"output": "NO"
},
{
"input": "26\npRWdodGdxUESvcScPGbUoooZsC",
"output": "NO"
},
{
"input": "66\nBovdMlDzTaqKllZILFVfxbLGsRnzmtVVTmqiIDTYrossLEPlmsPrkUYtWEsGHVOnFj",
"output": "NO"
},
{
"input": "100\nmKtsiDRJypUieHIkvJaMFkwaKxcCIbBszZQLIyPpCDCjhNpAnYFngLjRpnKWpKWtGnwoSteeZXuFHWQxxxOpFlNeYTwKocsXuCoa",
"output": "YES"
},
{
"input": "26\nEoqxUbsLjPytUHMiFnvcGWZdRK",
"output": "NO"
},
{
"input": "26\nvCUFRKElZOnjmXGylWQaHDiPst",
"output": "NO"
},
{
"input": "26\nWtrPuaHdXLKJMsnvQfgOiJZBEY",
"output": "NO"
},
{
"input": "26\npGiFluRteQwkaVoPszJyNBChxM",
"output": "NO"
},
{
"input": "26\ncTUpqjPmANrdbzSFhlWIoKxgVY",
"output": "NO"
},
{
"input": "26\nLndjgvAEuICHKxPwqYztosrmBN",
"output": "NO"
},
{
"input": "26\nMdaXJrCipnOZLykfqHWEStevbU",
"output": "NO"
},
{
"input": "26\nEjDWsVxfKTqGXRnUMOLYcIzPba",
"output": "NO"
},
{
"input": "26\nxKwzRMpunYaqsdfaBgJcVElTHo",
"output": "NO"
},
{
"input": "26\nnRYUQsTwCPLZkgshfEXvBdoiMa",
"output": "NO"
},
{
"input": "26\nHNCQPfJutyAlDGsvRxZWMEbIdO",
"output": "NO"
},
{
"input": "26\nDaHJIpvKznQcmUyWsTGObXRFDe",
"output": "NO"
},
{
"input": "26\nkqvAnFAiRhzlJbtyuWedXSPcOG",
"output": "NO"
},
{
"input": "26\nhlrvgdwsIOyjcmUZXtAKEqoBpF",
"output": "NO"
},
{
"input": "26\njLfXXiMhBTcAwQVReGnpKzdsYu",
"output": "NO"
},
{
"input": "26\nlNMcVuwItjxRBGAekjhyDsQOzf",
"output": "NO"
},
{
"input": "26\nRkSwbNoYldUGtAZvpFMcxhIJFE",
"output": "NO"
},
{
"input": "26\nDqspXZJTuONYieKgaHLMBwfVSC",
"output": "NO"
},
{
"input": "26\necOyUkqNljFHRVXtIpWabGMLDz",
"output": "NO"
},
{
"input": "26\nEKAvqZhBnPmVCDRlgWJfOusxYI",
"output": "NO"
},
{
"input": "26\naLbgqeYchKdMrsZxIPFvTOWNjA",
"output": "NO"
},
{
"input": "26\nxfpBLsndiqtacOCHGmeWUjRkYz",
"output": "NO"
},
{
"input": "26\nXsbRKtqleZPNIVCdfUhyagAomJ",
"output": "NO"
},
{
"input": "26\nAmVtbrwquEthZcjKPLiyDgSoNF",
"output": "NO"
},
{
"input": "26\nOhvXDcwqAUmSEPRZGnjFLiKtNB",
"output": "NO"
},
{
"input": "26\nEKWJqCFLRmstxVBdYuinpbhaOg",
"output": "NO"
},
{
"input": "26\nmnbvcxxlkjhgfdsapoiuytrewq",
"output": "NO"
},
{
"input": "26\naAbcdefghijklmnopqrstuvwxy",
"output": "NO"
},
{
"input": "30\nABCDEFGHTYRIOPLabcdefghtyriopl",
"output": "NO"
},
{
"input": "25\nabcdefghijklmnopqrstuvwxy",
"output": "NO"
},
{
"input": "26\nabcdefhijklmnopqrstVxyzABC",
"output": "NO"
},
{
"input": "25\nqwertyuiopasdfghjklxcvbnm",
"output": "NO"
},
{
"input": "34\nTheQuickBrownFoxJumpsOverTheLayDog",
"output": "NO"
},
{
"input": "26\nabcdefghigklmnopqrstuvwxyz",
"output": "NO"
},
{
"input": "26\nabcdefghijklmnopqrstuvwxyA",
"output": "NO"
},
{
"input": "50\nqazwsxedcrfvtgbyhnujmikolQWERTYUIOASDFGHJKLZXCVBNM",
"output": "NO"
},
{
"input": "35\nTheQuickBrownFoxJumpsOverTheLasyDog",
"output": "NO"
},
{
"input": "25\nbcdefghijklmnopqrstuvwxyz",
"output": "NO"
},
{
"input": "38\nAbCdEfGhIjKlMnOpQrStVwXyZzzzzzzaaaaaaa",
"output": "NO"
},
{
"input": "26\nabcdefghiklmnopqrstvxyzABC",
"output": "NO"
},
{
"input": "26\nabcdefghijklmnopqrstuvwxzZ",
"output": "NO"
},
{
"input": "50\nabcdefghijklmnopqrstuvwxyABCDEFGHIJKLMNOPQRSTUVWXY",
"output": "NO"
}
] | 1,695,556,402
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 59
| 61
| 0
|
n=int(input())
x=input().lower()
if len(set(x))==26:
print("YES")
else:
print("NO")
|
Title: Pangram
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A word or a sentence in some language is called a pangram if all the characters of the alphabet of this language appear in it at least once. Pangrams are often used to demonstrate fonts in printing or test the output devices.
You are given a string consisting of lowercase and uppercase Latin letters. Check whether this string is a pangram. We say that the string contains a letter of the Latin alphabet if this letter occurs in the string in uppercase or lowercase.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of characters in the string.
The second line contains the string. The string consists only of uppercase and lowercase Latin letters.
Output Specification:
Output "YES", if the string is a pangram and "NO" otherwise.
Demo Input:
['12\ntoosmallword\n', '35\nTheQuickBrownFoxJumpsOverTheLazyDog\n']
Demo Output:
['NO\n', 'YES\n']
Note:
none
|
```python
n=int(input())
x=input().lower()
if len(set(x))==26:
print("YES")
else:
print("NO")
```
| 3
|
|
177
|
B1
|
Rectangular Game
|
PROGRAMMING
| 1,000
|
[
"number theory"
] | null | null |
The Smart Beaver from ABBYY decided to have a day off. But doing nothing the whole day turned out to be too boring, and he decided to play a game with pebbles. Initially, the Beaver has *n* pebbles. He arranges them in *a* equal rows, each row has *b* pebbles (*a*<=><=1). Note that the Beaver must use all the pebbles he has, i. e. *n*<==<=*a*·*b*.
Once the Smart Beaver has arranged the pebbles, he takes back any of the resulting rows (that is, *b* pebbles) and discards all other pebbles. Then he arranges all his pebbles again (possibly choosing other values of *a* and *b*) and takes back one row, and so on. The game continues until at some point the Beaver ends up with exactly one pebble.
The game process can be represented as a finite sequence of integers *c*1,<=...,<=*c**k*, where:
- *c*1<==<=*n* - *c**i*<=+<=1 is the number of pebbles that the Beaver ends up with after the *i*-th move, that is, the number of pebbles in a row after some arrangement of *c**i* pebbles (1<=≤<=*i*<=<<=*k*). Note that *c**i*<=><=*c**i*<=+<=1. - *c**k*<==<=1
The result of the game is the sum of numbers *c**i*. You are given *n*. Find the maximum possible result of the game.
|
The single line of the input contains a single integer *n* — the initial number of pebbles the Smart Beaver has.
The input limitations for getting 30 points are:
- 2<=≤<=*n*<=≤<=50
The input limitations for getting 100 points are:
- 2<=≤<=*n*<=≤<=109
|
Print a single number — the maximum possible result of the game.
|
[
"10\n",
"8\n"
] |
[
"16\n",
"15\n"
] |
Consider the first example (*c*<sub class="lower-index">1</sub> = 10). The possible options for the game development are:
- Arrange the pebbles in 10 rows, one pebble per row. Then *c*<sub class="lower-index">2</sub> = 1, and the game ends after the first move with the result of 11. - Arrange the pebbles in 5 rows, two pebbles per row. Then *c*<sub class="lower-index">2</sub> = 2, and the game continues. During the second move we have two pebbles which can be arranged in a unique way (remember that you are not allowed to put all the pebbles in the same row!) — 2 rows, one pebble per row. *c*<sub class="lower-index">3</sub> = 1, and the game ends with the result of 13. - Finally, arrange the pebbles in two rows, five pebbles per row. The same logic leads us to *c*<sub class="lower-index">2</sub> = 5, *c*<sub class="lower-index">3</sub> = 1, and the game ends with the result of 16 — the maximum possible result.
| 30
|
[
{
"input": "10",
"output": "16"
},
{
"input": "8",
"output": "15"
},
{
"input": "4",
"output": "7"
},
{
"input": "36",
"output": "67"
},
{
"input": "32",
"output": "63"
},
{
"input": "46",
"output": "70"
},
{
"input": "6",
"output": "10"
},
{
"input": "13",
"output": "14"
},
{
"input": "9",
"output": "13"
},
{
"input": "18",
"output": "31"
},
{
"input": "20",
"output": "36"
},
{
"input": "37",
"output": "38"
},
{
"input": "42",
"output": "71"
},
{
"input": "50",
"output": "81"
},
{
"input": "48",
"output": "94"
},
{
"input": "33",
"output": "45"
},
{
"input": "49",
"output": "57"
},
{
"input": "34",
"output": "52"
},
{
"input": "35",
"output": "43"
},
{
"input": "37",
"output": "38"
},
{
"input": "39",
"output": "53"
},
{
"input": "42",
"output": "71"
},
{
"input": "14",
"output": "22"
},
{
"input": "2",
"output": "3"
},
{
"input": "3",
"output": "4"
},
{
"input": "5",
"output": "6"
}
] | 1,694,452,955
| 2,147,483,647
|
Python 3
|
OK
|
TESTS1
| 26
| 92
| 0
|
n = int(input())
total = n
while(n != 1):
for i in range(2, 40000):
if(n % i == 0):
n = n // i
total += n
break
print(total)
|
Title: Rectangular Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Smart Beaver from ABBYY decided to have a day off. But doing nothing the whole day turned out to be too boring, and he decided to play a game with pebbles. Initially, the Beaver has *n* pebbles. He arranges them in *a* equal rows, each row has *b* pebbles (*a*<=><=1). Note that the Beaver must use all the pebbles he has, i. e. *n*<==<=*a*·*b*.
Once the Smart Beaver has arranged the pebbles, he takes back any of the resulting rows (that is, *b* pebbles) and discards all other pebbles. Then he arranges all his pebbles again (possibly choosing other values of *a* and *b*) and takes back one row, and so on. The game continues until at some point the Beaver ends up with exactly one pebble.
The game process can be represented as a finite sequence of integers *c*1,<=...,<=*c**k*, where:
- *c*1<==<=*n* - *c**i*<=+<=1 is the number of pebbles that the Beaver ends up with after the *i*-th move, that is, the number of pebbles in a row after some arrangement of *c**i* pebbles (1<=≤<=*i*<=<<=*k*). Note that *c**i*<=><=*c**i*<=+<=1. - *c**k*<==<=1
The result of the game is the sum of numbers *c**i*. You are given *n*. Find the maximum possible result of the game.
Input Specification:
The single line of the input contains a single integer *n* — the initial number of pebbles the Smart Beaver has.
The input limitations for getting 30 points are:
- 2<=≤<=*n*<=≤<=50
The input limitations for getting 100 points are:
- 2<=≤<=*n*<=≤<=109
Output Specification:
Print a single number — the maximum possible result of the game.
Demo Input:
['10\n', '8\n']
Demo Output:
['16\n', '15\n']
Note:
Consider the first example (*c*<sub class="lower-index">1</sub> = 10). The possible options for the game development are:
- Arrange the pebbles in 10 rows, one pebble per row. Then *c*<sub class="lower-index">2</sub> = 1, and the game ends after the first move with the result of 11. - Arrange the pebbles in 5 rows, two pebbles per row. Then *c*<sub class="lower-index">2</sub> = 2, and the game continues. During the second move we have two pebbles which can be arranged in a unique way (remember that you are not allowed to put all the pebbles in the same row!) — 2 rows, one pebble per row. *c*<sub class="lower-index">3</sub> = 1, and the game ends with the result of 13. - Finally, arrange the pebbles in two rows, five pebbles per row. The same logic leads us to *c*<sub class="lower-index">2</sub> = 5, *c*<sub class="lower-index">3</sub> = 1, and the game ends with the result of 16 — the maximum possible result.
|
```python
n = int(input())
total = n
while(n != 1):
for i in range(2, 40000):
if(n % i == 0):
n = n // i
total += n
break
print(total)
```
| 3
|
|
322
|
A
|
Ciel and Dancing
|
PROGRAMMING
| 1,000
|
[
"greedy"
] | null | null |
Fox Ciel and her friends are in a dancing room. There are *n* boys and *m* girls here, and they never danced before. There will be some songs, during each song, there must be exactly one boy and one girl are dancing. Besides, there is a special rule:
- either the boy in the dancing pair must dance for the first time (so, he didn't dance with anyone before); - or the girl in the dancing pair must dance for the first time.
Help Fox Ciel to make a schedule that they can dance as many songs as possible.
|
The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of boys and girls in the dancing room.
|
In the first line print *k* — the number of songs during which they can dance. Then in the following *k* lines, print the indexes of boys and girls dancing during songs chronologically. You can assume that the boys are indexed from 1 to *n*, and the girls are indexed from 1 to *m*.
|
[
"2 1\n",
"2 2\n"
] |
[
"2\n1 1\n2 1\n",
"3\n1 1\n1 2\n2 2\n"
] |
In test case 1, there are 2 boys and 1 girl. We can have 2 dances: the 1st boy and 1st girl (during the first song), the 2nd boy and 1st girl (during the second song).
And in test case 2, we have 2 boys with 2 girls, the answer is 3.
| 500
|
[
{
"input": "2 1",
"output": "2\n1 1\n2 1"
},
{
"input": "2 2",
"output": "3\n1 1\n1 2\n2 2"
},
{
"input": "1 1",
"output": "1\n1 1"
},
{
"input": "2 3",
"output": "4\n1 1\n1 2\n1 3\n2 3"
},
{
"input": "4 4",
"output": "7\n1 1\n1 2\n1 3\n1 4\n4 4\n3 4\n2 4"
},
{
"input": "1 12",
"output": "12\n1 1\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12"
},
{
"input": "12 1",
"output": "12\n1 1\n12 1\n11 1\n10 1\n9 1\n8 1\n7 1\n6 1\n5 1\n4 1\n3 1\n2 1"
},
{
"input": "100 100",
"output": "199\n1 1\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34\n1 35\n1 36\n1 37\n1 38\n1 39\n1 40\n1 41\n1 42\n1 43\n1 44\n1 45\n1 46\n1 47\n1 48\n1 49\n1 50\n1 51\n1 52\n1 53\n1 54\n1 55\n1 56\n1 57\n1 58\n1 59\n1 60\n1 61\n1 62\n1 63\n1 64\n1 65\n1 66\n1 67\n1 68\n1 69\n1 70\n1 71\n1 72\n1 73\n1 74\n1 75\n1 76\n1 77\n1 78\n1 79\n1 80\n1 81\n1 82\n1 83\n1 84\n1 85\n1 86\n..."
},
{
"input": "24 6",
"output": "29\n1 1\n1 2\n1 3\n1 4\n1 5\n1 6\n24 6\n23 6\n22 6\n21 6\n20 6\n19 6\n18 6\n17 6\n16 6\n15 6\n14 6\n13 6\n12 6\n11 6\n10 6\n9 6\n8 6\n7 6\n6 6\n5 6\n4 6\n3 6\n2 6"
},
{
"input": "7 59",
"output": "65\n1 1\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34\n1 35\n1 36\n1 37\n1 38\n1 39\n1 40\n1 41\n1 42\n1 43\n1 44\n1 45\n1 46\n1 47\n1 48\n1 49\n1 50\n1 51\n1 52\n1 53\n1 54\n1 55\n1 56\n1 57\n1 58\n1 59\n7 59\n6 59\n5 59\n4 59\n3 59\n2 59"
},
{
"input": "26 75",
"output": "100\n1 1\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34\n1 35\n1 36\n1 37\n1 38\n1 39\n1 40\n1 41\n1 42\n1 43\n1 44\n1 45\n1 46\n1 47\n1 48\n1 49\n1 50\n1 51\n1 52\n1 53\n1 54\n1 55\n1 56\n1 57\n1 58\n1 59\n1 60\n1 61\n1 62\n1 63\n1 64\n1 65\n1 66\n1 67\n1 68\n1 69\n1 70\n1 71\n1 72\n1 73\n1 74\n1 75\n26 75\n25 75\n24 75\n23 75\n22 75\n21 75\n20 75\n19 75\n18 75\n17..."
},
{
"input": "32 87",
"output": "118\n1 1\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34\n1 35\n1 36\n1 37\n1 38\n1 39\n1 40\n1 41\n1 42\n1 43\n1 44\n1 45\n1 46\n1 47\n1 48\n1 49\n1 50\n1 51\n1 52\n1 53\n1 54\n1 55\n1 56\n1 57\n1 58\n1 59\n1 60\n1 61\n1 62\n1 63\n1 64\n1 65\n1 66\n1 67\n1 68\n1 69\n1 70\n1 71\n1 72\n1 73\n1 74\n1 75\n1 76\n1 77\n1 78\n1 79\n1 80\n1 81\n1 82\n1 83\n1 84\n1 85\n1 86\n..."
},
{
"input": "42 51",
"output": "92\n1 1\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34\n1 35\n1 36\n1 37\n1 38\n1 39\n1 40\n1 41\n1 42\n1 43\n1 44\n1 45\n1 46\n1 47\n1 48\n1 49\n1 50\n1 51\n42 51\n41 51\n40 51\n39 51\n38 51\n37 51\n36 51\n35 51\n34 51\n33 51\n32 51\n31 51\n30 51\n29 51\n28 51\n27 51\n26 51\n25 51\n24 51\n23 51\n22 51\n21 51\n20 51\n19 51\n18 51\n17 51\n16 51\n15 51\n14 51\n13 51\n..."
},
{
"input": "4 63",
"output": "66\n1 1\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34\n1 35\n1 36\n1 37\n1 38\n1 39\n1 40\n1 41\n1 42\n1 43\n1 44\n1 45\n1 46\n1 47\n1 48\n1 49\n1 50\n1 51\n1 52\n1 53\n1 54\n1 55\n1 56\n1 57\n1 58\n1 59\n1 60\n1 61\n1 62\n1 63\n4 63\n3 63\n2 63"
},
{
"input": "10 79",
"output": "88\n1 1\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34\n1 35\n1 36\n1 37\n1 38\n1 39\n1 40\n1 41\n1 42\n1 43\n1 44\n1 45\n1 46\n1 47\n1 48\n1 49\n1 50\n1 51\n1 52\n1 53\n1 54\n1 55\n1 56\n1 57\n1 58\n1 59\n1 60\n1 61\n1 62\n1 63\n1 64\n1 65\n1 66\n1 67\n1 68\n1 69\n1 70\n1 71\n1 72\n1 73\n1 74\n1 75\n1 76\n1 77\n1 78\n1 79\n10 79\n9 79\n8 79\n7 79\n6 79\n5 79\n4 79\n..."
},
{
"input": "20 95",
"output": "114\n1 1\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34\n1 35\n1 36\n1 37\n1 38\n1 39\n1 40\n1 41\n1 42\n1 43\n1 44\n1 45\n1 46\n1 47\n1 48\n1 49\n1 50\n1 51\n1 52\n1 53\n1 54\n1 55\n1 56\n1 57\n1 58\n1 59\n1 60\n1 61\n1 62\n1 63\n1 64\n1 65\n1 66\n1 67\n1 68\n1 69\n1 70\n1 71\n1 72\n1 73\n1 74\n1 75\n1 76\n1 77\n1 78\n1 79\n1 80\n1 81\n1 82\n1 83\n1 84\n1 85\n1 86\n..."
},
{
"input": "35 55",
"output": "89\n1 1\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34\n1 35\n1 36\n1 37\n1 38\n1 39\n1 40\n1 41\n1 42\n1 43\n1 44\n1 45\n1 46\n1 47\n1 48\n1 49\n1 50\n1 51\n1 52\n1 53\n1 54\n1 55\n35 55\n34 55\n33 55\n32 55\n31 55\n30 55\n29 55\n28 55\n27 55\n26 55\n25 55\n24 55\n23 55\n22 55\n21 55\n20 55\n19 55\n18 55\n17 55\n16 55\n15 55\n14 55\n13 55\n12 55\n11 55\n10 55\n9 55..."
},
{
"input": "45 71",
"output": "115\n1 1\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34\n1 35\n1 36\n1 37\n1 38\n1 39\n1 40\n1 41\n1 42\n1 43\n1 44\n1 45\n1 46\n1 47\n1 48\n1 49\n1 50\n1 51\n1 52\n1 53\n1 54\n1 55\n1 56\n1 57\n1 58\n1 59\n1 60\n1 61\n1 62\n1 63\n1 64\n1 65\n1 66\n1 67\n1 68\n1 69\n1 70\n1 71\n45 71\n44 71\n43 71\n42 71\n41 71\n40 71\n39 71\n38 71\n37 71\n36 71\n35 71\n34 71\n33 71..."
},
{
"input": "7 83",
"output": "89\n1 1\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34\n1 35\n1 36\n1 37\n1 38\n1 39\n1 40\n1 41\n1 42\n1 43\n1 44\n1 45\n1 46\n1 47\n1 48\n1 49\n1 50\n1 51\n1 52\n1 53\n1 54\n1 55\n1 56\n1 57\n1 58\n1 59\n1 60\n1 61\n1 62\n1 63\n1 64\n1 65\n1 66\n1 67\n1 68\n1 69\n1 70\n1 71\n1 72\n1 73\n1 74\n1 75\n1 76\n1 77\n1 78\n1 79\n1 80\n1 81\n1 82\n1 83\n7 83\n6 83\n5 83\n..."
},
{
"input": "32 100",
"output": "131\n1 1\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34\n1 35\n1 36\n1 37\n1 38\n1 39\n1 40\n1 41\n1 42\n1 43\n1 44\n1 45\n1 46\n1 47\n1 48\n1 49\n1 50\n1 51\n1 52\n1 53\n1 54\n1 55\n1 56\n1 57\n1 58\n1 59\n1 60\n1 61\n1 62\n1 63\n1 64\n1 65\n1 66\n1 67\n1 68\n1 69\n1 70\n1 71\n1 72\n1 73\n1 74\n1 75\n1 76\n1 77\n1 78\n1 79\n1 80\n1 81\n1 82\n1 83\n1 84\n1 85\n1 86\n..."
},
{
"input": "42 17",
"output": "58\n1 1\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n42 17\n41 17\n40 17\n39 17\n38 17\n37 17\n36 17\n35 17\n34 17\n33 17\n32 17\n31 17\n30 17\n29 17\n28 17\n27 17\n26 17\n25 17\n24 17\n23 17\n22 17\n21 17\n20 17\n19 17\n18 17\n17 17\n16 17\n15 17\n14 17\n13 17\n12 17\n11 17\n10 17\n9 17\n8 17\n7 17\n6 17\n5 17\n4 17\n3 17\n2 17"
},
{
"input": "1 77",
"output": "77\n1 1\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34\n1 35\n1 36\n1 37\n1 38\n1 39\n1 40\n1 41\n1 42\n1 43\n1 44\n1 45\n1 46\n1 47\n1 48\n1 49\n1 50\n1 51\n1 52\n1 53\n1 54\n1 55\n1 56\n1 57\n1 58\n1 59\n1 60\n1 61\n1 62\n1 63\n1 64\n1 65\n1 66\n1 67\n1 68\n1 69\n1 70\n1 71\n1 72\n1 73\n1 74\n1 75\n1 76\n1 77"
},
{
"input": "19 93",
"output": "111\n1 1\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34\n1 35\n1 36\n1 37\n1 38\n1 39\n1 40\n1 41\n1 42\n1 43\n1 44\n1 45\n1 46\n1 47\n1 48\n1 49\n1 50\n1 51\n1 52\n1 53\n1 54\n1 55\n1 56\n1 57\n1 58\n1 59\n1 60\n1 61\n1 62\n1 63\n1 64\n1 65\n1 66\n1 67\n1 68\n1 69\n1 70\n1 71\n1 72\n1 73\n1 74\n1 75\n1 76\n1 77\n1 78\n1 79\n1 80\n1 81\n1 82\n1 83\n1 84\n1 85\n1 86\n..."
},
{
"input": "25 5",
"output": "29\n1 1\n1 2\n1 3\n1 4\n1 5\n25 5\n24 5\n23 5\n22 5\n21 5\n20 5\n19 5\n18 5\n17 5\n16 5\n15 5\n14 5\n13 5\n12 5\n11 5\n10 5\n9 5\n8 5\n7 5\n6 5\n5 5\n4 5\n3 5\n2 5"
},
{
"input": "35 21",
"output": "55\n1 1\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n35 21\n34 21\n33 21\n32 21\n31 21\n30 21\n29 21\n28 21\n27 21\n26 21\n25 21\n24 21\n23 21\n22 21\n21 21\n20 21\n19 21\n18 21\n17 21\n16 21\n15 21\n14 21\n13 21\n12 21\n11 21\n10 21\n9 21\n8 21\n7 21\n6 21\n5 21\n4 21\n3 21\n2 21"
},
{
"input": "99 99",
"output": "197\n1 1\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34\n1 35\n1 36\n1 37\n1 38\n1 39\n1 40\n1 41\n1 42\n1 43\n1 44\n1 45\n1 46\n1 47\n1 48\n1 49\n1 50\n1 51\n1 52\n1 53\n1 54\n1 55\n1 56\n1 57\n1 58\n1 59\n1 60\n1 61\n1 62\n1 63\n1 64\n1 65\n1 66\n1 67\n1 68\n1 69\n1 70\n1 71\n1 72\n1 73\n1 74\n1 75\n1 76\n1 77\n1 78\n1 79\n1 80\n1 81\n1 82\n1 83\n1 84\n1 85\n1 86\n..."
},
{
"input": "99 100",
"output": "198\n1 1\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34\n1 35\n1 36\n1 37\n1 38\n1 39\n1 40\n1 41\n1 42\n1 43\n1 44\n1 45\n1 46\n1 47\n1 48\n1 49\n1 50\n1 51\n1 52\n1 53\n1 54\n1 55\n1 56\n1 57\n1 58\n1 59\n1 60\n1 61\n1 62\n1 63\n1 64\n1 65\n1 66\n1 67\n1 68\n1 69\n1 70\n1 71\n1 72\n1 73\n1 74\n1 75\n1 76\n1 77\n1 78\n1 79\n1 80\n1 81\n1 82\n1 83\n1 84\n1 85\n1 86\n..."
},
{
"input": "100 99",
"output": "198\n1 1\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19\n1 20\n1 21\n1 22\n1 23\n1 24\n1 25\n1 26\n1 27\n1 28\n1 29\n1 30\n1 31\n1 32\n1 33\n1 34\n1 35\n1 36\n1 37\n1 38\n1 39\n1 40\n1 41\n1 42\n1 43\n1 44\n1 45\n1 46\n1 47\n1 48\n1 49\n1 50\n1 51\n1 52\n1 53\n1 54\n1 55\n1 56\n1 57\n1 58\n1 59\n1 60\n1 61\n1 62\n1 63\n1 64\n1 65\n1 66\n1 67\n1 68\n1 69\n1 70\n1 71\n1 72\n1 73\n1 74\n1 75\n1 76\n1 77\n1 78\n1 79\n1 80\n1 81\n1 82\n1 83\n1 84\n1 85\n1 86\n..."
}
] | 1,372,436,632
| 3,232
|
Python 3
|
OK
|
TESTS
| 28
| 93
| 102,400
|
n,m = map(int,input().split())
ii = set()
jj = set()
ans = []
for i in range(1,n+1):
for j in range(1,m+1):
if i not in ii or j not in jj:
ans.append ((i,j))
ii.add(i)
jj.add(j)
print(len(ans))
for a,b in ans:
print(a,b)
|
Title: Ciel and Dancing
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Fox Ciel and her friends are in a dancing room. There are *n* boys and *m* girls here, and they never danced before. There will be some songs, during each song, there must be exactly one boy and one girl are dancing. Besides, there is a special rule:
- either the boy in the dancing pair must dance for the first time (so, he didn't dance with anyone before); - or the girl in the dancing pair must dance for the first time.
Help Fox Ciel to make a schedule that they can dance as many songs as possible.
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of boys and girls in the dancing room.
Output Specification:
In the first line print *k* — the number of songs during which they can dance. Then in the following *k* lines, print the indexes of boys and girls dancing during songs chronologically. You can assume that the boys are indexed from 1 to *n*, and the girls are indexed from 1 to *m*.
Demo Input:
['2 1\n', '2 2\n']
Demo Output:
['2\n1 1\n2 1\n', '3\n1 1\n1 2\n2 2\n']
Note:
In test case 1, there are 2 boys and 1 girl. We can have 2 dances: the 1st boy and 1st girl (during the first song), the 2nd boy and 1st girl (during the second song).
And in test case 2, we have 2 boys with 2 girls, the answer is 3.
|
```python
n,m = map(int,input().split())
ii = set()
jj = set()
ans = []
for i in range(1,n+1):
for j in range(1,m+1):
if i not in ii or j not in jj:
ans.append ((i,j))
ii.add(i)
jj.add(j)
print(len(ans))
for a,b in ans:
print(a,b)
```
| 3
|
|
994
|
A
|
Fingerprints
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
You are locked in a room with a door that has a keypad with 10 keys corresponding to digits from 0 to 9. To escape from the room, you need to enter a correct code. You also have a sequence of digits.
Some keys on the keypad have fingerprints. You believe the correct code is the longest not necessarily contiguous subsequence of the sequence you have that only contains digits with fingerprints on the corresponding keys. Find such code.
|
The first line contains two integers $n$ and $m$ ($1 \le n, m \le 10$) representing the number of digits in the sequence you have and the number of keys on the keypad that have fingerprints.
The next line contains $n$ distinct space-separated integers $x_1, x_2, \ldots, x_n$ ($0 \le x_i \le 9$) representing the sequence.
The next line contains $m$ distinct space-separated integers $y_1, y_2, \ldots, y_m$ ($0 \le y_i \le 9$) — the keys with fingerprints.
|
In a single line print a space-separated sequence of integers representing the code. If the resulting sequence is empty, both printing nothing and printing a single line break is acceptable.
|
[
"7 3\n3 5 7 1 6 2 8\n1 2 7\n",
"4 4\n3 4 1 0\n0 1 7 9\n"
] |
[
"7 1 2\n",
"1 0\n"
] |
In the first example, the only digits with fingerprints are $1$, $2$ and $7$. All three of them appear in the sequence you know, $7$ first, then $1$ and then $2$. Therefore the output is 7 1 2. Note that the order is important, and shall be the same as the order in the original sequence.
In the second example digits $0$, $1$, $7$ and $9$ have fingerprints, however only $0$ and $1$ appear in the original sequence. $1$ appears earlier, so the output is 1 0. Again, the order is important.
| 500
|
[
{
"input": "7 3\n3 5 7 1 6 2 8\n1 2 7",
"output": "7 1 2"
},
{
"input": "4 4\n3 4 1 0\n0 1 7 9",
"output": "1 0"
},
{
"input": "9 4\n9 8 7 6 5 4 3 2 1\n2 4 6 8",
"output": "8 6 4 2"
},
{
"input": "10 5\n3 7 1 2 4 6 9 0 5 8\n4 3 0 7 9",
"output": "3 7 4 9 0"
},
{
"input": "10 10\n1 2 3 4 5 6 7 8 9 0\n4 5 6 7 1 2 3 0 9 8",
"output": "1 2 3 4 5 6 7 8 9 0"
},
{
"input": "1 1\n4\n4",
"output": "4"
},
{
"input": "3 7\n6 3 4\n4 9 0 1 7 8 6",
"output": "6 4"
},
{
"input": "10 1\n9 0 8 1 7 4 6 5 2 3\n0",
"output": "0"
},
{
"input": "5 10\n6 0 3 8 1\n3 1 0 5 4 7 2 8 9 6",
"output": "6 0 3 8 1"
},
{
"input": "8 2\n7 2 9 6 1 0 3 4\n6 3",
"output": "6 3"
},
{
"input": "5 4\n7 0 1 4 9\n0 9 5 3",
"output": "0 9"
},
{
"input": "10 1\n9 6 2 0 1 8 3 4 7 5\n6",
"output": "6"
},
{
"input": "10 2\n7 1 0 2 4 6 5 9 3 8\n3 2",
"output": "2 3"
},
{
"input": "5 9\n3 7 9 2 4\n3 8 4 5 9 6 1 0 2",
"output": "3 9 2 4"
},
{
"input": "10 6\n7 1 2 3 8 0 6 4 5 9\n1 5 8 2 3 6",
"output": "1 2 3 8 6 5"
},
{
"input": "8 2\n7 4 8 9 2 5 6 1\n6 4",
"output": "4 6"
},
{
"input": "10 2\n1 0 3 5 8 9 4 7 6 2\n0 3",
"output": "0 3"
},
{
"input": "7 6\n9 2 8 6 1 3 7\n4 2 0 3 1 8",
"output": "2 8 1 3"
},
{
"input": "1 6\n3\n6 8 2 4 5 3",
"output": "3"
},
{
"input": "1 8\n0\n9 2 4 8 1 5 0 7",
"output": "0"
},
{
"input": "6 9\n7 3 9 4 1 0\n9 1 5 8 0 6 2 7 4",
"output": "7 9 4 1 0"
},
{
"input": "10 2\n4 9 6 8 3 0 1 5 7 2\n0 1",
"output": "0 1"
},
{
"input": "10 5\n5 2 8 0 9 7 6 1 4 3\n9 6 4 1 2",
"output": "2 9 6 1 4"
},
{
"input": "6 3\n8 3 9 2 7 6\n5 4 3",
"output": "3"
},
{
"input": "4 10\n8 3 9 6\n4 9 6 2 7 0 8 1 3 5",
"output": "8 3 9 6"
},
{
"input": "1 2\n1\n1 0",
"output": "1"
},
{
"input": "3 6\n1 2 3\n4 5 6 1 2 3",
"output": "1 2 3"
},
{
"input": "1 2\n2\n1 2",
"output": "2"
},
{
"input": "1 10\n9\n0 1 2 3 4 5 6 7 8 9",
"output": "9"
}
] | 1,628,097,683
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 31
| 77
| 6,758,400
|
n, k = [int(x) for x in input().split()]
xs = input().split()
ys = set(input().split())
for x in xs:
if x in ys:
print(x, end=' ')
print('')
|
Title: Fingerprints
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are locked in a room with a door that has a keypad with 10 keys corresponding to digits from 0 to 9. To escape from the room, you need to enter a correct code. You also have a sequence of digits.
Some keys on the keypad have fingerprints. You believe the correct code is the longest not necessarily contiguous subsequence of the sequence you have that only contains digits with fingerprints on the corresponding keys. Find such code.
Input Specification:
The first line contains two integers $n$ and $m$ ($1 \le n, m \le 10$) representing the number of digits in the sequence you have and the number of keys on the keypad that have fingerprints.
The next line contains $n$ distinct space-separated integers $x_1, x_2, \ldots, x_n$ ($0 \le x_i \le 9$) representing the sequence.
The next line contains $m$ distinct space-separated integers $y_1, y_2, \ldots, y_m$ ($0 \le y_i \le 9$) — the keys with fingerprints.
Output Specification:
In a single line print a space-separated sequence of integers representing the code. If the resulting sequence is empty, both printing nothing and printing a single line break is acceptable.
Demo Input:
['7 3\n3 5 7 1 6 2 8\n1 2 7\n', '4 4\n3 4 1 0\n0 1 7 9\n']
Demo Output:
['7 1 2\n', '1 0\n']
Note:
In the first example, the only digits with fingerprints are $1$, $2$ and $7$. All three of them appear in the sequence you know, $7$ first, then $1$ and then $2$. Therefore the output is 7 1 2. Note that the order is important, and shall be the same as the order in the original sequence.
In the second example digits $0$, $1$, $7$ and $9$ have fingerprints, however only $0$ and $1$ appear in the original sequence. $1$ appears earlier, so the output is 1 0. Again, the order is important.
|
```python
n, k = [int(x) for x in input().split()]
xs = input().split()
ys = set(input().split())
for x in xs:
if x in ys:
print(x, end=' ')
print('')
```
| 3
|
|
306
|
A
|
Candies
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Polycarpus has got *n* candies and *m* friends (*n*<=≥<=*m*). He wants to make a New Year present with candies to each friend. Polycarpus is planning to present all candies and he wants to do this in the fairest (that is, most equal) manner. He wants to choose such *a**i*, where *a**i* is the number of candies in the *i*-th friend's present, that the maximum *a**i* differs from the least *a**i* as little as possible.
For example, if *n* is divisible by *m*, then he is going to present the same number of candies to all his friends, that is, the maximum *a**i* won't differ from the minimum one.
|
The single line of the input contains a pair of space-separated positive integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=100;*n*<=≥<=*m*) — the number of candies and the number of Polycarpus's friends.
|
Print the required sequence *a*1,<=*a*2,<=...,<=*a**m*, where *a**i* is the number of candies in the *i*-th friend's present. All numbers *a**i* must be positive integers, total up to *n*, the maximum one should differ from the minimum one by the smallest possible value.
|
[
"12 3\n",
"15 4\n",
"18 7\n"
] |
[
"4 4 4 ",
"3 4 4 4 ",
"2 2 2 3 3 3 3 "
] |
Print *a*<sub class="lower-index">*i*</sub> in any order, separate the numbers by spaces.
| 500
|
[
{
"input": "12 4",
"output": "3 3 3 3 "
},
{
"input": "15 6",
"output": "2 2 2 3 3 3 "
},
{
"input": "18 8",
"output": "2 2 2 2 2 2 3 3 "
},
{
"input": "1 1",
"output": "1 "
},
{
"input": "2 1",
"output": "2 "
},
{
"input": "100 1",
"output": "100 "
},
{
"input": "100 100",
"output": "1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 "
},
{
"input": "100 10",
"output": "10 10 10 10 10 10 10 10 10 10 "
},
{
"input": "64 8",
"output": "8 8 8 8 8 8 8 8 "
},
{
"input": "81 27",
"output": "3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 "
},
{
"input": "100 99",
"output": "1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 "
},
{
"input": "100 51",
"output": "1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 "
},
{
"input": "100 49",
"output": "2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 3 "
},
{
"input": "100 13",
"output": "7 7 7 7 8 8 8 8 8 8 8 8 8 "
},
{
"input": "97 11",
"output": "8 8 9 9 9 9 9 9 9 9 9 "
},
{
"input": "19 3",
"output": "6 6 7 "
},
{
"input": "99 10",
"output": "9 10 10 10 10 10 10 10 10 10 "
},
{
"input": "88 5",
"output": "17 17 18 18 18 "
},
{
"input": "87 34",
"output": "2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 "
},
{
"input": "97 55",
"output": "1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 "
}
] | 1,473,442,028
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 20
| 92
| 0
|
n,m=map(int,input().split())
x=n//m
y=n%m
print(' '.join([str(x+1)]*y+[str(x)]*(m-y)))
|
Title: Candies
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarpus has got *n* candies and *m* friends (*n*<=≥<=*m*). He wants to make a New Year present with candies to each friend. Polycarpus is planning to present all candies and he wants to do this in the fairest (that is, most equal) manner. He wants to choose such *a**i*, where *a**i* is the number of candies in the *i*-th friend's present, that the maximum *a**i* differs from the least *a**i* as little as possible.
For example, if *n* is divisible by *m*, then he is going to present the same number of candies to all his friends, that is, the maximum *a**i* won't differ from the minimum one.
Input Specification:
The single line of the input contains a pair of space-separated positive integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=100;*n*<=≥<=*m*) — the number of candies and the number of Polycarpus's friends.
Output Specification:
Print the required sequence *a*1,<=*a*2,<=...,<=*a**m*, where *a**i* is the number of candies in the *i*-th friend's present. All numbers *a**i* must be positive integers, total up to *n*, the maximum one should differ from the minimum one by the smallest possible value.
Demo Input:
['12 3\n', '15 4\n', '18 7\n']
Demo Output:
['4 4 4 ', '3 4 4 4 ', '2 2 2 3 3 3 3 ']
Note:
Print *a*<sub class="lower-index">*i*</sub> in any order, separate the numbers by spaces.
|
```python
n,m=map(int,input().split())
x=n//m
y=n%m
print(' '.join([str(x+1)]*y+[str(x)]*(m-y)))
```
| 3
|
|
208
|
A
|
Dubstep
|
PROGRAMMING
| 900
|
[
"strings"
] | null | null |
Vasya works as a DJ in the best Berland nightclub, and he often uses dubstep music in his performance. Recently, he has decided to take a couple of old songs and make dubstep remixes from them.
Let's assume that a song consists of some number of words. To make the dubstep remix of this song, Vasya inserts a certain number of words "WUB" before the first word of the song (the number may be zero), after the last word (the number may be zero), and between words (at least one between any pair of neighbouring words), and then the boy glues together all the words, including "WUB", in one string and plays the song at the club.
For example, a song with words "I AM X" can transform into a dubstep remix as "WUBWUBIWUBAMWUBWUBX" and cannot transform into "WUBWUBIAMWUBX".
Recently, Petya has heard Vasya's new dubstep track, but since he isn't into modern music, he decided to find out what was the initial song that Vasya remixed. Help Petya restore the original song.
|
The input consists of a single non-empty string, consisting only of uppercase English letters, the string's length doesn't exceed 200 characters. It is guaranteed that before Vasya remixed the song, no word contained substring "WUB" in it; Vasya didn't change the word order. It is also guaranteed that initially the song had at least one word.
|
Print the words of the initial song that Vasya used to make a dubsteb remix. Separate the words with a space.
|
[
"WUBWUBABCWUB\n",
"WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB\n"
] |
[
"ABC ",
"WE ARE THE CHAMPIONS MY FRIEND "
] |
In the first sample: "WUBWUBABCWUB" = "WUB" + "WUB" + "ABC" + "WUB". That means that the song originally consisted of a single word "ABC", and all words "WUB" were added by Vasya.
In the second sample Vasya added a single word "WUB" between all neighbouring words, in the beginning and in the end, except for words "ARE" and "THE" — between them Vasya added two "WUB".
| 500
|
[
{
"input": "WUBWUBABCWUB",
"output": "ABC "
},
{
"input": "WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB",
"output": "WE ARE THE CHAMPIONS MY FRIEND "
},
{
"input": "WUBWUBWUBSR",
"output": "SR "
},
{
"input": "RWUBWUBWUBLWUB",
"output": "R L "
},
{
"input": "ZJWUBWUBWUBJWUBWUBWUBL",
"output": "ZJ J L "
},
{
"input": "CWUBBWUBWUBWUBEWUBWUBWUBQWUBWUBWUB",
"output": "C B E Q "
},
{
"input": "WUBJKDWUBWUBWBIRAQKFWUBWUBYEWUBWUBWUBWVWUBWUB",
"output": "JKD WBIRAQKF YE WV "
},
{
"input": "WUBKSDHEMIXUJWUBWUBRWUBWUBWUBSWUBWUBWUBHWUBWUBWUB",
"output": "KSDHEMIXUJ R S H "
},
{
"input": "OGWUBWUBWUBXWUBWUBWUBIWUBWUBWUBKOWUBWUB",
"output": "OG X I KO "
},
{
"input": "QWUBQQWUBWUBWUBIWUBWUBWWWUBWUBWUBJOPJPBRH",
"output": "Q QQ I WW JOPJPBRH "
},
{
"input": "VSRNVEATZTLGQRFEGBFPWUBWUBWUBAJWUBWUBWUBPQCHNWUBCWUB",
"output": "VSRNVEATZTLGQRFEGBFP AJ PQCHN C "
},
{
"input": "WUBWUBEWUBWUBWUBIQMJNIQWUBWUBWUBGZZBQZAUHYPWUBWUBWUBPMRWUBWUBWUBDCV",
"output": "E IQMJNIQ GZZBQZAUHYP PMR DCV "
},
{
"input": "WUBWUBWUBFVWUBWUBWUBBPSWUBWUBWUBRXNETCJWUBWUBWUBJDMBHWUBWUBWUBBWUBWUBVWUBWUBB",
"output": "FV BPS RXNETCJ JDMBH B V B "
},
{
"input": "WUBWUBWUBFBQWUBWUBWUBIDFSYWUBWUBWUBCTWDMWUBWUBWUBSXOWUBWUBWUBQIWUBWUBWUBL",
"output": "FBQ IDFSY CTWDM SXO QI L "
},
{
"input": "IWUBWUBQLHDWUBYIIKZDFQWUBWUBWUBCXWUBWUBUWUBWUBWUBKWUBWUBWUBNL",
"output": "I QLHD YIIKZDFQ CX U K NL "
},
{
"input": "KWUBUPDYXGOKUWUBWUBWUBAGOAHWUBIZDWUBWUBWUBIYWUBWUBWUBVWUBWUBWUBPWUBWUBWUBE",
"output": "K UPDYXGOKU AGOAH IZD IY V P E "
},
{
"input": "WUBWUBOWUBWUBWUBIPVCQAFWYWUBWUBWUBQWUBWUBWUBXHDKCPYKCTWWYWUBWUBWUBVWUBWUBWUBFZWUBWUB",
"output": "O IPVCQAFWY Q XHDKCPYKCTWWY V FZ "
},
{
"input": "PAMJGYWUBWUBWUBXGPQMWUBWUBWUBTKGSXUYWUBWUBWUBEWUBWUBWUBNWUBWUBWUBHWUBWUBWUBEWUBWUB",
"output": "PAMJGY XGPQM TKGSXUY E N H E "
},
{
"input": "WUBYYRTSMNWUWUBWUBWUBCWUBWUBWUBCWUBWUBWUBFSYUINDWOBVWUBWUBWUBFWUBWUBWUBAUWUBWUBWUBVWUBWUBWUBJB",
"output": "YYRTSMNWU C C FSYUINDWOBV F AU V JB "
},
{
"input": "WUBWUBYGPYEYBNRTFKOQCWUBWUBWUBUYGRTQEGWLFYWUBWUBWUBFVWUBHPWUBWUBWUBXZQWUBWUBWUBZDWUBWUBWUBM",
"output": "YGPYEYBNRTFKOQC UYGRTQEGWLFY FV HP XZQ ZD M "
},
{
"input": "WUBZVMJWUBWUBWUBFOIMJQWKNZUBOFOFYCCWUBWUBWUBAUWWUBRDRADWUBWUBWUBCHQVWUBWUBWUBKFTWUBWUBWUBW",
"output": "ZVMJ FOIMJQWKNZUBOFOFYCC AUW RDRAD CHQV KFT W "
},
{
"input": "WUBWUBZBKOKHQLGKRVIMZQMQNRWUBWUBWUBDACWUBWUBNZHFJMPEYKRVSWUBWUBWUBPPHGAVVPRZWUBWUBWUBQWUBWUBAWUBG",
"output": "ZBKOKHQLGKRVIMZQMQNR DAC NZHFJMPEYKRVS PPHGAVVPRZ Q A G "
},
{
"input": "WUBWUBJWUBWUBWUBNFLWUBWUBWUBGECAWUBYFKBYJWTGBYHVSSNTINKWSINWSMAWUBWUBWUBFWUBWUBWUBOVWUBWUBLPWUBWUBWUBN",
"output": "J NFL GECA YFKBYJWTGBYHVSSNTINKWSINWSMA F OV LP N "
},
{
"input": "WUBWUBLCWUBWUBWUBZGEQUEATJVIXETVTWUBWUBWUBEXMGWUBWUBWUBRSWUBWUBWUBVWUBWUBWUBTAWUBWUBWUBCWUBWUBWUBQG",
"output": "LC ZGEQUEATJVIXETVT EXMG RS V TA C QG "
},
{
"input": "WUBMPWUBWUBWUBORWUBWUBDLGKWUBWUBWUBVVZQCAAKVJTIKWUBWUBWUBTJLUBZJCILQDIFVZWUBWUBYXWUBWUBWUBQWUBWUBWUBLWUB",
"output": "MP OR DLGK VVZQCAAKVJTIK TJLUBZJCILQDIFVZ YX Q L "
},
{
"input": "WUBNXOLIBKEGXNWUBWUBWUBUWUBGITCNMDQFUAOVLWUBWUBWUBAIJDJZJHFMPVTPOXHPWUBWUBWUBISCIOWUBWUBWUBGWUBWUBWUBUWUB",
"output": "NXOLIBKEGXN U GITCNMDQFUAOVL AIJDJZJHFMPVTPOXHP ISCIO G U "
},
{
"input": "WUBWUBNMMWCZOLYPNBELIYVDNHJUNINWUBWUBWUBDXLHYOWUBWUBWUBOJXUWUBWUBWUBRFHTGJCEFHCGWARGWUBWUBWUBJKWUBWUBSJWUBWUB",
"output": "NMMWCZOLYPNBELIYVDNHJUNIN DXLHYO OJXU RFHTGJCEFHCGWARG JK SJ "
},
{
"input": "SGWLYSAUJOJBNOXNWUBWUBWUBBOSSFWKXPDPDCQEWUBWUBWUBDIRZINODWUBWUBWUBWWUBWUBWUBPPHWUBWUBWUBRWUBWUBWUBQWUBWUBWUBJWUB",
"output": "SGWLYSAUJOJBNOXN BOSSFWKXPDPDCQE DIRZINOD W PPH R Q J "
},
{
"input": "TOWUBWUBWUBGBTBNWUBWUBWUBJVIOJBIZFUUYHUAIEBQLQXPQKZJMPTCWBKPOSAWUBWUBWUBSWUBWUBWUBTOLVXWUBWUBWUBNHWUBWUBWUBO",
"output": "TO GBTBN JVIOJBIZFUUYHUAIEBQLQXPQKZJMPTCWBKPOSA S TOLVX NH O "
},
{
"input": "WUBWUBWSPLAYSZSAUDSWUBWUBWUBUWUBWUBWUBKRWUBWUBWUBRSOKQMZFIYZQUWUBWUBWUBELSHUWUBWUBWUBUKHWUBWUBWUBQXEUHQWUBWUBWUBBWUBWUBWUBR",
"output": "WSPLAYSZSAUDS U KR RSOKQMZFIYZQU ELSHU UKH QXEUHQ B R "
},
{
"input": "WUBXEMWWVUHLSUUGRWUBWUBWUBAWUBXEGILZUNKWUBWUBWUBJDHHKSWUBWUBWUBDTSUYSJHWUBWUBWUBPXFWUBMOHNJWUBWUBWUBZFXVMDWUBWUBWUBZMWUBWUB",
"output": "XEMWWVUHLSUUGR A XEGILZUNK JDHHKS DTSUYSJH PXF MOHNJ ZFXVMD ZM "
},
{
"input": "BMBWUBWUBWUBOQKWUBWUBWUBPITCIHXHCKLRQRUGXJWUBWUBWUBVWUBWUBWUBJCWUBWUBWUBQJPWUBWUBWUBBWUBWUBWUBBMYGIZOOXWUBWUBWUBTAGWUBWUBHWUB",
"output": "BMB OQK PITCIHXHCKLRQRUGXJ V JC QJP B BMYGIZOOX TAG H "
},
{
"input": "CBZNWUBWUBWUBNHWUBWUBWUBYQSYWUBWUBWUBMWUBWUBWUBXRHBTMWUBWUBWUBPCRCWUBWUBWUBTZUYLYOWUBWUBWUBCYGCWUBWUBWUBCLJWUBWUBWUBSWUBWUBWUB",
"output": "CBZN NH YQSY M XRHBTM PCRC TZUYLYO CYGC CLJ S "
},
{
"input": "DPDWUBWUBWUBEUQKWPUHLTLNXHAEKGWUBRRFYCAYZFJDCJLXBAWUBWUBWUBHJWUBOJWUBWUBWUBNHBJEYFWUBWUBWUBRWUBWUBWUBSWUBWWUBWUBWUBXDWUBWUBWUBJWUB",
"output": "DPD EUQKWPUHLTLNXHAEKG RRFYCAYZFJDCJLXBA HJ OJ NHBJEYF R S W XD J "
},
{
"input": "WUBWUBWUBISERPQITVIYERSCNWUBWUBWUBQWUBWUBWUBDGSDIPWUBWUBWUBCAHKDZWEXBIBJVVSKKVQJWUBWUBWUBKIWUBWUBWUBCWUBWUBWUBAWUBWUBWUBPWUBWUBWUBHWUBWUBWUBF",
"output": "ISERPQITVIYERSCN Q DGSDIP CAHKDZWEXBIBJVVSKKVQJ KI C A P H F "
},
{
"input": "WUBWUBWUBIWUBWUBLIKNQVWUBWUBWUBPWUBWUBWUBHWUBWUBWUBMWUBWUBWUBDPRSWUBWUBWUBBSAGYLQEENWXXVWUBWUBWUBXMHOWUBWUBWUBUWUBWUBWUBYRYWUBWUBWUBCWUBWUBWUBY",
"output": "I LIKNQV P H M DPRS BSAGYLQEENWXXV XMHO U YRY C Y "
},
{
"input": "WUBWUBWUBMWUBWUBWUBQWUBWUBWUBITCFEYEWUBWUBWUBHEUWGNDFNZGWKLJWUBWUBWUBMZPWUBWUBWUBUWUBWUBWUBBWUBWUBWUBDTJWUBHZVIWUBWUBWUBPWUBFNHHWUBWUBWUBVTOWUB",
"output": "M Q ITCFEYE HEUWGNDFNZGWKLJ MZP U B DTJ HZVI P FNHH VTO "
},
{
"input": "WUBWUBNDNRFHYJAAUULLHRRDEDHYFSRXJWUBWUBWUBMUJVDTIRSGYZAVWKRGIFWUBWUBWUBHMZWUBWUBWUBVAIWUBWUBWUBDDKJXPZRGWUBWUBWUBSGXWUBWUBWUBIFKWUBWUBWUBUWUBWUBWUBW",
"output": "NDNRFHYJAAUULLHRRDEDHYFSRXJ MUJVDTIRSGYZAVWKRGIF HMZ VAI DDKJXPZRG SGX IFK U W "
},
{
"input": "WUBOJMWRSLAXXHQRTPMJNCMPGWUBWUBWUBNYGMZIXNLAKSQYWDWUBWUBWUBXNIWUBWUBWUBFWUBWUBWUBXMBWUBWUBWUBIWUBWUBWUBINWUBWUBWUBWDWUBWUBWUBDDWUBWUBWUBD",
"output": "OJMWRSLAXXHQRTPMJNCMPG NYGMZIXNLAKSQYWD XNI F XMB I IN WD DD D "
},
{
"input": "WUBWUBWUBREHMWUBWUBWUBXWUBWUBWUBQASNWUBWUBWUBNLSMHLCMTICWUBWUBWUBVAWUBWUBWUBHNWUBWUBWUBNWUBWUBWUBUEXLSFOEULBWUBWUBWUBXWUBWUBWUBJWUBWUBWUBQWUBWUBWUBAWUBWUB",
"output": "REHM X QASN NLSMHLCMTIC VA HN N UEXLSFOEULB X J Q A "
},
{
"input": "WUBWUBWUBSTEZTZEFFIWUBWUBWUBSWUBWUBWUBCWUBFWUBHRJPVWUBWUBWUBDYJUWUBWUBWUBPWYDKCWUBWUBWUBCWUBWUBWUBUUEOGCVHHBWUBWUBWUBEXLWUBWUBWUBVCYWUBWUBWUBMWUBWUBWUBYWUB",
"output": "STEZTZEFFI S C F HRJPV DYJU PWYDKC C UUEOGCVHHB EXL VCY M Y "
},
{
"input": "WPPNMSQOQIWUBWUBWUBPNQXWUBWUBWUBHWUBWUBWUBNFLWUBWUBWUBGWSGAHVJFNUWUBWUBWUBFWUBWUBWUBWCMLRICFSCQQQTNBWUBWUBWUBSWUBWUBWUBKGWUBWUBWUBCWUBWUBWUBBMWUBWUBWUBRWUBWUB",
"output": "WPPNMSQOQI PNQX H NFL GWSGAHVJFNU F WCMLRICFSCQQQTNB S KG C BM R "
},
{
"input": "YZJOOYITZRARKVFYWUBWUBRZQGWUBWUBWUBUOQWUBWUBWUBIWUBWUBWUBNKVDTBOLETKZISTWUBWUBWUBWLWUBQQFMMGSONZMAWUBZWUBWUBWUBQZUXGCWUBWUBWUBIRZWUBWUBWUBLTTVTLCWUBWUBWUBY",
"output": "YZJOOYITZRARKVFY RZQG UOQ I NKVDTBOLETKZIST WL QQFMMGSONZMA Z QZUXGC IRZ LTTVTLC Y "
},
{
"input": "WUBCAXNCKFBVZLGCBWCOAWVWOFKZVQYLVTWUBWUBWUBNLGWUBWUBWUBAMGDZBDHZMRMQMDLIRMIWUBWUBWUBGAJSHTBSWUBWUBWUBCXWUBWUBWUBYWUBZLXAWWUBWUBWUBOHWUBWUBWUBZWUBWUBWUBGBWUBWUBWUBE",
"output": "CAXNCKFBVZLGCBWCOAWVWOFKZVQYLVT NLG AMGDZBDHZMRMQMDLIRMI GAJSHTBS CX Y ZLXAW OH Z GB E "
},
{
"input": "WUBWUBCHXSOWTSQWUBWUBWUBCYUZBPBWUBWUBWUBSGWUBWUBWKWORLRRLQYUUFDNWUBWUBWUBYYGOJNEVEMWUBWUBWUBRWUBWUBWUBQWUBWUBWUBIHCKWUBWUBWUBKTWUBWUBWUBRGSNTGGWUBWUBWUBXCXWUBWUBWUBS",
"output": "CHXSOWTSQ CYUZBPB SG WKWORLRRLQYUUFDN YYGOJNEVEM R Q IHCK KT RGSNTGG XCX S "
},
{
"input": "WUBWUBWUBHJHMSBURXTHXWSCHNAIJOWBHLZGJZDHEDSPWBWACCGQWUBWUBWUBXTZKGIITWUBWUBWUBAWUBWUBWUBVNCXPUBCQWUBWUBWUBIDPNAWUBWUBWUBOWUBWUBWUBYGFWUBWUBWUBMQOWUBWUBWUBKWUBWUBWUBAZVWUBWUBWUBEP",
"output": "HJHMSBURXTHXWSCHNAIJOWBHLZGJZDHEDSPWBWACCGQ XTZKGIIT A VNCXPUBCQ IDPNA O YGF MQO K AZV EP "
},
{
"input": "WUBKYDZOYWZSNGMKJSWAXFDFLTHDHEOGTDBNZMSMKZTVWUBWUBWUBLRMIIWUBWUBWUBGWUBWUBWUBADPSWUBWUBWUBANBWUBWUBPCWUBWUBWUBPWUBWUBWUBGPVNLSWIRFORYGAABUXMWUBWUBWUBOWUBWUBWUBNWUBWUBWUBYWUBWUB",
"output": "KYDZOYWZSNGMKJSWAXFDFLTHDHEOGTDBNZMSMKZTV LRMII G ADPS ANB PC P GPVNLSWIRFORYGAABUXM O N Y "
},
{
"input": "REWUBWUBWUBJDWUBWUBWUBNWUBWUBWUBTWWUBWUBWUBWZDOCKKWUBWUBWUBLDPOVBFRCFWUBWUBAKZIBQKEUAZEEWUBWUBWUBLQYPNPFWUBYEWUBWUBWUBFWUBWUBWUBBPWUBWUBWUBAWWUBWUBWUBQWUBWUBWUBBRWUBWUBWUBXJL",
"output": "RE JD N TW WZDOCKK LDPOVBFRCF AKZIBQKEUAZEE LQYPNPF YE F BP AW Q BR XJL "
},
{
"input": "CUFGJDXGMWUBWUBWUBOMWUBWUBWUBSIEWUBWUBWUBJJWKNOWUBWUBWUBYBHVNRNORGYWUBWUBWUBOAGCAWUBWUBWUBSBLBKTPFKPBIWUBWUBWUBJBWUBWUBWUBRMFCJPGWUBWUBWUBDWUBWUBWUBOJOWUBWUBWUBZPWUBWUBWUBMWUBRWUBWUBWUBFXWWUBWUBWUBO",
"output": "CUFGJDXGM OM SIE JJWKNO YBHVNRNORGY OAGCA SBLBKTPFKPBI JB RMFCJPG D OJO ZP M R FXW O "
},
{
"input": "WUBJZGAEXFMFEWMAKGQLUWUBWUBWUBICYTPQWGENELVYWANKUOJYWUBWUBWUBGWUBWUBWUBHYCJVLPHTUPNEGKCDGQWUBWUBWUBOFWUBWUBWUBCPGSOGZBRPRPVJJEWUBWUBWUBDQBCWUBWUBWUBHWUBWUBWUBMHOHYBMATWUBWUBWUBVWUBWUBWUBSWUBWUBWUBKOWU",
"output": "JZGAEXFMFEWMAKGQLU ICYTPQWGENELVYWANKUOJY G HYCJVLPHTUPNEGKCDGQ OF CPGSOGZBRPRPVJJE DQBC H MHOHYBMAT V S KOWU "
},
{
"input": "A",
"output": "A "
},
{
"input": "WUBA",
"output": "A "
},
{
"input": "WUBWUBA",
"output": "A "
},
{
"input": "AWUBWUBWUB",
"output": "A "
},
{
"input": "AWUBBWUBCWUBD",
"output": "A B C D "
},
{
"input": "WUBWWUBWUBWUBUWUBWUBBWUB",
"output": "W U B "
},
{
"input": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA",
"output": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA "
},
{
"input": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAWUBAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA",
"output": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA "
},
{
"input": "WUWUBBWWUBUB",
"output": "WU BW UB "
},
{
"input": "WUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUABWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUBWUB",
"output": "WUAB "
},
{
"input": "ZWUB",
"output": "Z "
},
{
"input": "WU",
"output": "WU "
},
{
"input": "UB",
"output": "UB "
},
{
"input": "U",
"output": "U "
},
{
"input": "WUBW",
"output": "W "
},
{
"input": "WUBWU",
"output": "WU "
},
{
"input": "WUWUB",
"output": "WU "
},
{
"input": "UBWUB",
"output": "UB "
},
{
"input": "WUWUBUBWUBUWUB",
"output": "WU UB U "
},
{
"input": "WUBWWUBAWUB",
"output": "W A "
},
{
"input": "WUUUUU",
"output": "WUUUUU "
}
] | 1,689,712,018
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 71
| 124
| 0
|
text = input().replace("WUB", " ").strip()
print(text)
|
Title: Dubstep
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya works as a DJ in the best Berland nightclub, and he often uses dubstep music in his performance. Recently, he has decided to take a couple of old songs and make dubstep remixes from them.
Let's assume that a song consists of some number of words. To make the dubstep remix of this song, Vasya inserts a certain number of words "WUB" before the first word of the song (the number may be zero), after the last word (the number may be zero), and between words (at least one between any pair of neighbouring words), and then the boy glues together all the words, including "WUB", in one string and plays the song at the club.
For example, a song with words "I AM X" can transform into a dubstep remix as "WUBWUBIWUBAMWUBWUBX" and cannot transform into "WUBWUBIAMWUBX".
Recently, Petya has heard Vasya's new dubstep track, but since he isn't into modern music, he decided to find out what was the initial song that Vasya remixed. Help Petya restore the original song.
Input Specification:
The input consists of a single non-empty string, consisting only of uppercase English letters, the string's length doesn't exceed 200 characters. It is guaranteed that before Vasya remixed the song, no word contained substring "WUB" in it; Vasya didn't change the word order. It is also guaranteed that initially the song had at least one word.
Output Specification:
Print the words of the initial song that Vasya used to make a dubsteb remix. Separate the words with a space.
Demo Input:
['WUBWUBABCWUB\n', 'WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB\n']
Demo Output:
['ABC ', 'WE ARE THE CHAMPIONS MY FRIEND ']
Note:
In the first sample: "WUBWUBABCWUB" = "WUB" + "WUB" + "ABC" + "WUB". That means that the song originally consisted of a single word "ABC", and all words "WUB" were added by Vasya.
In the second sample Vasya added a single word "WUB" between all neighbouring words, in the beginning and in the end, except for words "ARE" and "THE" — between them Vasya added two "WUB".
|
```python
text = input().replace("WUB", " ").strip()
print(text)
```
| 3
|
|
894
|
A
|
QAQ
|
PROGRAMMING
| 800
|
[
"brute force",
"dp"
] | null | null |
"QAQ" is a word to denote an expression of crying. Imagine "Q" as eyes with tears and "A" as a mouth.
Now Diamond has given Bort a string consisting of only uppercase English letters of length *n*. There is a great number of "QAQ" in the string (Diamond is so cute!).
Bort wants to know how many subsequences "QAQ" are in the string Diamond has given. Note that the letters "QAQ" don't have to be consecutive, but the order of letters should be exact.
|
The only line contains a string of length *n* (1<=≤<=*n*<=≤<=100). It's guaranteed that the string only contains uppercase English letters.
|
Print a single integer — the number of subsequences "QAQ" in the string.
|
[
"QAQAQYSYIOIWIN\n",
"QAQQQZZYNOIWIN\n"
] |
[
"4\n",
"3\n"
] |
In the first example there are 4 subsequences "QAQ": "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN".
| 500
|
[
{
"input": "QAQAQYSYIOIWIN",
"output": "4"
},
{
"input": "QAQQQZZYNOIWIN",
"output": "3"
},
{
"input": "QA",
"output": "0"
},
{
"input": "IAQVAQZLQBQVQFTQQQADAQJA",
"output": "24"
},
{
"input": "QQAAQASGAYAAAAKAKAQIQEAQAIAAIAQQQQQ",
"output": "378"
},
{
"input": "AMVFNFJIAVNQJWIVONQOAOOQSNQSONOASONAONQINAONAOIQONANOIQOANOQINAONOQINAONOXJCOIAQOAOQAQAQAQAQWWWAQQAQ",
"output": "1077"
},
{
"input": "AAQQAXBQQBQQXBNQRJAQKQNAQNQVDQASAGGANQQQQTJFFQQQTQQA",
"output": "568"
},
{
"input": "KAZXAVLPJQBQVQQQQQAPAQQGQTQVZQAAAOYA",
"output": "70"
},
{
"input": "W",
"output": "0"
},
{
"input": "DBA",
"output": "0"
},
{
"input": "RQAWNACASAAKAGAAAAQ",
"output": "10"
},
{
"input": "QJAWZAAOAAGIAAAAAOQATASQAEAAAAQFQQHPA",
"output": "111"
},
{
"input": "QQKWQAQAAAAAAAAGAAVAQUEQQUMQMAQQQNQLAMAAAUAEAAEMAAA",
"output": "411"
},
{
"input": "QQUMQAYAUAAGWAAAQSDAVAAQAAAASKQJJQQQQMAWAYYAAAAAAEAJAXWQQ",
"output": "625"
},
{
"input": "QORZOYAQ",
"output": "1"
},
{
"input": "QCQAQAGAWAQQQAQAVQAQQQQAQAQQQAQAAATQAAVAAAQQQQAAAUUQAQQNQQWQQWAQAAQQKQYAQAAQQQAAQRAQQQWBQQQQAPBAQGQA",
"output": "13174"
},
{
"input": "QQAQQAKQFAQLQAAWAMQAZQAJQAAQQOACQQAAAYANAQAQQAQAAQQAOBQQJQAQAQAQQQAAAAABQQQAVNZAQQQQAMQQAFAAEAQAQHQT",
"output": "10420"
},
{
"input": "AQEGQHQQKQAQQPQKAQQQAAAAQQQAQEQAAQAAQAQFSLAAQQAQOQQAVQAAAPQQAWAQAQAFQAXAQQQQTRLOQAQQJQNQXQQQQSQVDQQQ",
"output": "12488"
},
{
"input": "QNQKQQQLASQBAVQQQQAAQQOQRJQQAQQQEQZUOANAADAAQQJAQAQARAAAQQQEQBHTQAAQAAAAQQMKQQQIAOJJQQAQAAADADQUQQQA",
"output": "9114"
},
{
"input": "QQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQ",
"output": "35937"
},
{
"input": "AMQQAAQAAQAAAAAAQQQBOAAANAAKQJCYQAE",
"output": "254"
},
{
"input": "AYQBAEQGAQEOAKGIXLQJAIAKQAAAQPUAJAKAATFWQQAOQQQUFQYAQQMQHOKAAJXGFCARAQSATHAUQQAATQJJQDQRAANQQAE",
"output": "2174"
},
{
"input": "AAQXAAQAYQAAAAGAQHVQYAGIVACADFAAQAAAAQZAAQMAKZAADQAQDAAQDAAAMQQOXYAQQQAKQBAAQQKAXQBJZDDLAAHQQ",
"output": "2962"
},
{
"input": "AYQQYAVAMNIAUAAKBBQVACWKTQSAQZAAQAAASZJAWBCAALAARHACQAKQQAQAARPAQAAQAQAAZQUSHQAMFVFZQQQQSAQQXAA",
"output": "2482"
},
{
"input": "LQMAQQARQAQBJQQQAGAAZQQXALQQAARQAQQQQAAQQAQQQAQQCAQQAQQAYQQQRAAZATQALYQQAAHHAAQHAAAAAAAAQQMAAQNAKQ",
"output": "7768"
},
{
"input": "MAQQWAQOYQMAAAQAQPQZAOAAQAUAQNAAQAAAITQSAQAKAQKAQQWSQAAQQAGUCDQMQWKQUXKWQQAAQQAAQQZQDQQQAABXQUUXQOA",
"output": "5422"
},
{
"input": "QTAAQDAQXAQQJQQQGAAAQQQQSBQZKAQQAQQQQEAQNUQBZCQLYQZQEQQAAQHQVAORKQVAQYQNASZQAARZAAGAAAAOQDCQ",
"output": "3024"
},
{
"input": "QQWAQQGQQUZQQQLZAAQYQXQVAQFQUAQZUQZZQUKBHSHTQYLQAOQXAQQGAQQTQOAQARQADAJRAAQPQAQQUQAUAMAUVQAAAQQAWQ",
"output": "4527"
},
{
"input": "QQAAQQAQVAQZQQQQAOEAQZPQIBQZACQQAFQQLAAQDATZQANHKYQQAQTAAFQRQAIQAJPWQAQTEIRXAEQQAYWAAAUKQQAQAQQQSQQH",
"output": "6416"
},
{
"input": "AQQQQAQAAQQAQAQAAAAAAAAAQAQAAAAAQAQAQQQAQQQAAAQQQAAAAAAAQAAAAQQQQQQQAQQQQAQAAAQAAAAAQAQAAAAAQAQAAAA",
"output": "14270"
},
{
"input": "AQQQQAQAAQQAQAQAAAAAAAAAQAQAAAAAQAQAQQQAQQQAAAQQQAAAAAAAQAAAAQQQQQQQAQQQQAQAAAQAAAAAQAQAAAAAQ",
"output": "13136"
},
{
"input": "AQQQQAQAAQQAQAQAAAAAAAAAQAQAAAAAQAQAQQQAQQQAAAQQQAAAAAAAQAAAAQQQQQQQAQQQQAQAAAQAAAAAQAQAAAAAQAQAA",
"output": "14270"
},
{
"input": "AQQQQAQAAQQAQAQAAAAAAAAAQAQAAAAAQAQAQQQAQQQAAAQQQAAAAAAAQAAAAQQQQQQQAQQQQAQAAAQAAAAAQAQAAAAAQQAA",
"output": "14231"
},
{
"input": "QQQQQAQAAQQAQAQAAAAAAAAAQAQAAAAAQAQAQQQAQQQAAAQQQAAAAAAAQAAAAQQQQQQQAQQQQAQAAAQAAAAAQAQAAAAAQAQAA",
"output": "15296"
},
{
"input": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQ",
"output": "0"
},
{
"input": "QQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA",
"output": "0"
},
{
"input": "QQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQ",
"output": "0"
},
{
"input": "QAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQA",
"output": "20825"
},
{
"input": "AQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQ",
"output": "20825"
},
{
"input": "Q",
"output": "0"
},
{
"input": "A",
"output": "0"
},
{
"input": "FFF",
"output": "0"
},
{
"input": "AAAAAA",
"output": "0"
}
] | 1,625,063,189
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 43
| 109
| 0
|
def findAllSubsequence(string, word):
prefix_arr = [None for _ in range(len(string))]
curr_no = 0
for i in range(len(string)):
if string[i] == word[0]:
curr_no += 1
prefix_arr[i] = curr_no
ans = 0
for i in range(1, len(string)-1):
if string[i] == word[1]:
ans += prefix_arr[i-1] * (prefix_arr[-1] - prefix_arr[i])
return ans
if __name__ == "__main__":
string = input()
#ans = findAllSubsequence(string, "MOM") + findAllSubsequence(string, "DAD")
ans = findAllSubsequence(string, "QAQ")
print(ans)
|
Title: QAQ
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
"QAQ" is a word to denote an expression of crying. Imagine "Q" as eyes with tears and "A" as a mouth.
Now Diamond has given Bort a string consisting of only uppercase English letters of length *n*. There is a great number of "QAQ" in the string (Diamond is so cute!).
Bort wants to know how many subsequences "QAQ" are in the string Diamond has given. Note that the letters "QAQ" don't have to be consecutive, but the order of letters should be exact.
Input Specification:
The only line contains a string of length *n* (1<=≤<=*n*<=≤<=100). It's guaranteed that the string only contains uppercase English letters.
Output Specification:
Print a single integer — the number of subsequences "QAQ" in the string.
Demo Input:
['QAQAQYSYIOIWIN\n', 'QAQQQZZYNOIWIN\n']
Demo Output:
['4\n', '3\n']
Note:
In the first example there are 4 subsequences "QAQ": "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN".
|
```python
def findAllSubsequence(string, word):
prefix_arr = [None for _ in range(len(string))]
curr_no = 0
for i in range(len(string)):
if string[i] == word[0]:
curr_no += 1
prefix_arr[i] = curr_no
ans = 0
for i in range(1, len(string)-1):
if string[i] == word[1]:
ans += prefix_arr[i-1] * (prefix_arr[-1] - prefix_arr[i])
return ans
if __name__ == "__main__":
string = input()
#ans = findAllSubsequence(string, "MOM") + findAllSubsequence(string, "DAD")
ans = findAllSubsequence(string, "QAQ")
print(ans)
```
| 3
|
|
6
|
A
|
Triangle
|
PROGRAMMING
| 900
|
[
"brute force",
"geometry"
] |
A. Triangle
|
2
|
64
|
Johnny has a younger sister Anne, who is very clever and smart. As she came home from the kindergarten, she told his brother about the task that her kindergartener asked her to solve. The task was just to construct a triangle out of four sticks of different colours. Naturally, one of the sticks is extra. It is not allowed to break the sticks or use their partial length. Anne has perfectly solved this task, now she is asking Johnny to do the same.
The boy answered that he would cope with it without any difficulty. However, after a while he found out that different tricky things can occur. It can happen that it is impossible to construct a triangle of a positive area, but it is possible to construct a degenerate triangle. It can be so, that it is impossible to construct a degenerate triangle even. As Johnny is very lazy, he does not want to consider such a big amount of cases, he asks you to help him.
|
The first line of the input contains four space-separated positive integer numbers not exceeding 100 — lengthes of the sticks.
|
Output TRIANGLE if it is possible to construct a non-degenerate triangle. Output SEGMENT if the first case cannot take place and it is possible to construct a degenerate triangle. Output IMPOSSIBLE if it is impossible to construct any triangle. Remember that you are to use three sticks. It is not allowed to break the sticks or use their partial length.
|
[
"4 2 1 3\n",
"7 2 2 4\n",
"3 5 9 1\n"
] |
[
"TRIANGLE\n",
"SEGMENT\n",
"IMPOSSIBLE\n"
] |
none
| 0
|
[
{
"input": "4 2 1 3",
"output": "TRIANGLE"
},
{
"input": "7 2 2 4",
"output": "SEGMENT"
},
{
"input": "3 5 9 1",
"output": "IMPOSSIBLE"
},
{
"input": "3 1 5 1",
"output": "IMPOSSIBLE"
},
{
"input": "10 10 10 10",
"output": "TRIANGLE"
},
{
"input": "11 5 6 11",
"output": "TRIANGLE"
},
{
"input": "1 1 1 1",
"output": "TRIANGLE"
},
{
"input": "10 20 30 40",
"output": "TRIANGLE"
},
{
"input": "45 25 5 15",
"output": "IMPOSSIBLE"
},
{
"input": "20 5 8 13",
"output": "TRIANGLE"
},
{
"input": "10 30 7 20",
"output": "SEGMENT"
},
{
"input": "3 2 3 2",
"output": "TRIANGLE"
},
{
"input": "70 10 100 30",
"output": "SEGMENT"
},
{
"input": "4 8 16 2",
"output": "IMPOSSIBLE"
},
{
"input": "3 3 3 10",
"output": "TRIANGLE"
},
{
"input": "1 5 5 5",
"output": "TRIANGLE"
},
{
"input": "13 25 12 1",
"output": "SEGMENT"
},
{
"input": "10 100 7 3",
"output": "SEGMENT"
},
{
"input": "50 1 50 100",
"output": "TRIANGLE"
},
{
"input": "50 1 100 49",
"output": "SEGMENT"
},
{
"input": "49 51 100 1",
"output": "SEGMENT"
},
{
"input": "5 11 2 25",
"output": "IMPOSSIBLE"
},
{
"input": "91 50 9 40",
"output": "IMPOSSIBLE"
},
{
"input": "27 53 7 97",
"output": "IMPOSSIBLE"
},
{
"input": "51 90 24 8",
"output": "IMPOSSIBLE"
},
{
"input": "3 5 1 1",
"output": "IMPOSSIBLE"
},
{
"input": "13 49 69 15",
"output": "IMPOSSIBLE"
},
{
"input": "16 99 9 35",
"output": "IMPOSSIBLE"
},
{
"input": "27 6 18 53",
"output": "IMPOSSIBLE"
},
{
"input": "57 88 17 8",
"output": "IMPOSSIBLE"
},
{
"input": "95 20 21 43",
"output": "IMPOSSIBLE"
},
{
"input": "6 19 32 61",
"output": "IMPOSSIBLE"
},
{
"input": "100 21 30 65",
"output": "IMPOSSIBLE"
},
{
"input": "85 16 61 9",
"output": "IMPOSSIBLE"
},
{
"input": "5 6 19 82",
"output": "IMPOSSIBLE"
},
{
"input": "1 5 1 3",
"output": "IMPOSSIBLE"
},
{
"input": "65 10 36 17",
"output": "IMPOSSIBLE"
},
{
"input": "81 64 9 7",
"output": "IMPOSSIBLE"
},
{
"input": "11 30 79 43",
"output": "IMPOSSIBLE"
},
{
"input": "1 1 5 3",
"output": "IMPOSSIBLE"
},
{
"input": "21 94 61 31",
"output": "IMPOSSIBLE"
},
{
"input": "49 24 9 74",
"output": "IMPOSSIBLE"
},
{
"input": "11 19 5 77",
"output": "IMPOSSIBLE"
},
{
"input": "52 10 19 71",
"output": "SEGMENT"
},
{
"input": "2 3 7 10",
"output": "SEGMENT"
},
{
"input": "1 2 6 3",
"output": "SEGMENT"
},
{
"input": "2 6 1 8",
"output": "SEGMENT"
},
{
"input": "1 2 4 1",
"output": "SEGMENT"
},
{
"input": "4 10 6 2",
"output": "SEGMENT"
},
{
"input": "2 10 7 3",
"output": "SEGMENT"
},
{
"input": "5 2 3 9",
"output": "SEGMENT"
},
{
"input": "6 1 4 10",
"output": "SEGMENT"
},
{
"input": "10 6 4 1",
"output": "SEGMENT"
},
{
"input": "3 2 9 1",
"output": "SEGMENT"
},
{
"input": "22 80 29 7",
"output": "SEGMENT"
},
{
"input": "2 6 3 9",
"output": "SEGMENT"
},
{
"input": "3 1 2 1",
"output": "SEGMENT"
},
{
"input": "3 4 7 1",
"output": "SEGMENT"
},
{
"input": "8 4 3 1",
"output": "SEGMENT"
},
{
"input": "2 8 3 5",
"output": "SEGMENT"
},
{
"input": "4 1 2 1",
"output": "SEGMENT"
},
{
"input": "8 1 3 2",
"output": "SEGMENT"
},
{
"input": "6 2 1 8",
"output": "SEGMENT"
},
{
"input": "3 3 3 6",
"output": "TRIANGLE"
},
{
"input": "3 6 3 3",
"output": "TRIANGLE"
},
{
"input": "4 10 4 4",
"output": "TRIANGLE"
},
{
"input": "1 1 2 1",
"output": "TRIANGLE"
},
{
"input": "3 3 3 6",
"output": "TRIANGLE"
},
{
"input": "5 4 5 5",
"output": "TRIANGLE"
},
{
"input": "8 7 8 8",
"output": "TRIANGLE"
},
{
"input": "3 3 3 1",
"output": "TRIANGLE"
},
{
"input": "1 1 6 6",
"output": "TRIANGLE"
},
{
"input": "1 9 1 9",
"output": "TRIANGLE"
},
{
"input": "7 2 2 7",
"output": "TRIANGLE"
},
{
"input": "7 2 3 2",
"output": "TRIANGLE"
},
{
"input": "4 4 10 10",
"output": "TRIANGLE"
},
{
"input": "7 7 10 7",
"output": "TRIANGLE"
},
{
"input": "4 4 4 5",
"output": "TRIANGLE"
},
{
"input": "1 10 9 2",
"output": "TRIANGLE"
},
{
"input": "1 8 2 7",
"output": "TRIANGLE"
},
{
"input": "4 3 2 8",
"output": "TRIANGLE"
},
{
"input": "5 9 5 3",
"output": "TRIANGLE"
},
{
"input": "4 10 3 5",
"output": "TRIANGLE"
}
] | 1,520,601,704
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 83
| 124
| 5,632,000
|
arr=[int(x) for x in input().split()]
tri=0
seg=0
for (i,item) in enumerate(arr):
arr2=list(arr)
del arr2[i]
if (arr2[0]<arr2[1]+arr2[2])&(arr2[1]<arr2[0]+arr2[2])&(arr2[2]<arr2[1]+arr2[0]):
tri=1
break
elif (arr2[0]==arr2[1]+arr2[2])|(arr2[1]==arr2[0]+arr2[2])|(arr2[2]==arr2[1]+arr2[0]):
seg=1
if tri==1:
print("TRIANGLE")
elif seg==1:
print("SEGMENT")
else:print("IMPOSSIBLE")
|
Title: Triangle
Time Limit: 2 seconds
Memory Limit: 64 megabytes
Problem Description:
Johnny has a younger sister Anne, who is very clever and smart. As she came home from the kindergarten, she told his brother about the task that her kindergartener asked her to solve. The task was just to construct a triangle out of four sticks of different colours. Naturally, one of the sticks is extra. It is not allowed to break the sticks or use their partial length. Anne has perfectly solved this task, now she is asking Johnny to do the same.
The boy answered that he would cope with it without any difficulty. However, after a while he found out that different tricky things can occur. It can happen that it is impossible to construct a triangle of a positive area, but it is possible to construct a degenerate triangle. It can be so, that it is impossible to construct a degenerate triangle even. As Johnny is very lazy, he does not want to consider such a big amount of cases, he asks you to help him.
Input Specification:
The first line of the input contains four space-separated positive integer numbers not exceeding 100 — lengthes of the sticks.
Output Specification:
Output TRIANGLE if it is possible to construct a non-degenerate triangle. Output SEGMENT if the first case cannot take place and it is possible to construct a degenerate triangle. Output IMPOSSIBLE if it is impossible to construct any triangle. Remember that you are to use three sticks. It is not allowed to break the sticks or use their partial length.
Demo Input:
['4 2 1 3\n', '7 2 2 4\n', '3 5 9 1\n']
Demo Output:
['TRIANGLE\n', 'SEGMENT\n', 'IMPOSSIBLE\n']
Note:
none
|
```python
arr=[int(x) for x in input().split()]
tri=0
seg=0
for (i,item) in enumerate(arr):
arr2=list(arr)
del arr2[i]
if (arr2[0]<arr2[1]+arr2[2])&(arr2[1]<arr2[0]+arr2[2])&(arr2[2]<arr2[1]+arr2[0]):
tri=1
break
elif (arr2[0]==arr2[1]+arr2[2])|(arr2[1]==arr2[0]+arr2[2])|(arr2[2]==arr2[1]+arr2[0]):
seg=1
if tri==1:
print("TRIANGLE")
elif seg==1:
print("SEGMENT")
else:print("IMPOSSIBLE")
```
| 3.927038
|
337
|
A
|
Puzzles
|
PROGRAMMING
| 900
|
[
"greedy"
] | null | null |
The end of the school year is near and Ms. Manana, the teacher, will soon have to say goodbye to a yet another class. She decided to prepare a goodbye present for her *n* students and give each of them a jigsaw puzzle (which, as wikipedia states, is a tiling puzzle that requires the assembly of numerous small, often oddly shaped, interlocking and tessellating pieces).
The shop assistant told the teacher that there are *m* puzzles in the shop, but they might differ in difficulty and size. Specifically, the first jigsaw puzzle consists of *f*1 pieces, the second one consists of *f*2 pieces and so on.
Ms. Manana doesn't want to upset the children, so she decided that the difference between the numbers of pieces in her presents must be as small as possible. Let *A* be the number of pieces in the largest puzzle that the teacher buys and *B* be the number of pieces in the smallest such puzzle. She wants to choose such *n* puzzles that *A*<=-<=*B* is minimum possible. Help the teacher and find the least possible value of *A*<=-<=*B*.
|
The first line contains space-separated integers *n* and *m* (2<=≤<=*n*<=≤<=*m*<=≤<=50). The second line contains *m* space-separated integers *f*1,<=*f*2,<=...,<=*f**m* (4<=≤<=*f**i*<=≤<=1000) — the quantities of pieces in the puzzles sold in the shop.
|
Print a single integer — the least possible difference the teacher can obtain.
|
[
"4 6\n10 12 10 7 5 22\n"
] |
[
"5\n"
] |
Sample 1. The class has 4 students. The shop sells 6 puzzles. If Ms. Manana buys the first four puzzles consisting of 10, 12, 10 and 7 pieces correspondingly, then the difference between the sizes of the largest and the smallest puzzle will be equal to 5. It is impossible to obtain a smaller difference. Note that the teacher can also buy puzzles 1, 3, 4 and 5 to obtain the difference 5.
| 500
|
[
{
"input": "4 6\n10 12 10 7 5 22",
"output": "5"
},
{
"input": "2 2\n4 4",
"output": "0"
},
{
"input": "2 10\n4 5 6 7 8 9 10 11 12 12",
"output": "0"
},
{
"input": "4 5\n818 136 713 59 946",
"output": "759"
},
{
"input": "3 20\n446 852 783 313 549 965 40 88 86 617 479 118 768 34 47 826 366 957 463 903",
"output": "13"
},
{
"input": "2 25\n782 633 152 416 432 825 115 97 386 357 836 310 530 413 354 373 847 882 913 682 729 582 671 674 94",
"output": "3"
},
{
"input": "4 25\n226 790 628 528 114 64 239 279 619 39 894 763 763 847 525 93 882 697 999 643 650 244 159 884 190",
"output": "31"
},
{
"input": "2 50\n971 889 628 39 253 157 925 694 129 516 660 272 738 319 611 816 142 717 514 392 41 105 132 676 958 118 306 768 600 685 103 857 704 346 857 309 23 718 618 161 176 379 846 834 640 468 952 878 164 997",
"output": "0"
},
{
"input": "25 50\n582 146 750 905 313 509 402 21 488 512 32 898 282 64 579 869 37 996 377 929 975 697 666 837 311 205 116 992 533 298 648 268 54 479 792 595 152 69 267 417 184 433 894 603 988 712 24 414 301 176",
"output": "412"
},
{
"input": "49 50\n58 820 826 960 271 294 473 102 925 318 729 672 244 914 796 646 868 6 893 882 726 203 528 498 271 195 355 459 721 680 547 147 631 116 169 804 145 996 133 559 110 257 771 476 576 251 607 314 427 886",
"output": "938"
},
{
"input": "50 50\n374 573 323 744 190 806 485 247 628 336 491 606 702 321 991 678 337 579 86 240 993 208 668 686 855 205 363 177 719 249 896 919 782 434 59 647 787 996 286 216 636 212 546 903 958 559 544 126 608 993",
"output": "937"
},
{
"input": "6 50\n6 8 7 8 5 4 4 5 7 8 6 5 7 4 7 7 7 8 6 4 6 6 8 8 7 7 8 7 5 8 5 4 4 7 8 4 4 6 6 6 8 7 4 7 6 6 5 8 4 7",
"output": "0"
},
{
"input": "37 50\n14 5 11 17 8 20 19 16 20 11 17 20 16 9 14 14 13 18 11 20 8 8 8 5 19 17 6 18 10 20 9 7 12 6 14 17 4 4 10 13 7 4 11 6 20 19 12 12 15 19",
"output": "12"
},
{
"input": "40 50\n4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4",
"output": "0"
},
{
"input": "40 50\n17 20 43 26 41 37 14 8 30 35 30 24 43 8 42 9 41 50 41 35 27 32 35 43 28 36 31 16 5 7 23 16 14 29 8 39 12 16 36 18 49 39 33 37 38 6 6 27 23 17",
"output": "31"
},
{
"input": "2 2\n1000 4",
"output": "996"
},
{
"input": "2 3\n4 502 1000",
"output": "498"
},
{
"input": "3 3\n4 1000 4",
"output": "996"
}
] | 1,676,730,784
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 18
| 92
| 0
|
q,qq = input().split()
q = int(q)
qq = int(qq)
row = input().split()
row_x = []
for _ in row:
row_x.append(int(_))
def foobar(row):
row = sorted(row)
res = []
r = q
g = 0
for _ in range(qq-q+1):
d = row[r-1]-row[g]
r = r +1
g = g +1
res.append(d)
return min(res)
print(foobar(row_x))
|
Title: Puzzles
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The end of the school year is near and Ms. Manana, the teacher, will soon have to say goodbye to a yet another class. She decided to prepare a goodbye present for her *n* students and give each of them a jigsaw puzzle (which, as wikipedia states, is a tiling puzzle that requires the assembly of numerous small, often oddly shaped, interlocking and tessellating pieces).
The shop assistant told the teacher that there are *m* puzzles in the shop, but they might differ in difficulty and size. Specifically, the first jigsaw puzzle consists of *f*1 pieces, the second one consists of *f*2 pieces and so on.
Ms. Manana doesn't want to upset the children, so she decided that the difference between the numbers of pieces in her presents must be as small as possible. Let *A* be the number of pieces in the largest puzzle that the teacher buys and *B* be the number of pieces in the smallest such puzzle. She wants to choose such *n* puzzles that *A*<=-<=*B* is minimum possible. Help the teacher and find the least possible value of *A*<=-<=*B*.
Input Specification:
The first line contains space-separated integers *n* and *m* (2<=≤<=*n*<=≤<=*m*<=≤<=50). The second line contains *m* space-separated integers *f*1,<=*f*2,<=...,<=*f**m* (4<=≤<=*f**i*<=≤<=1000) — the quantities of pieces in the puzzles sold in the shop.
Output Specification:
Print a single integer — the least possible difference the teacher can obtain.
Demo Input:
['4 6\n10 12 10 7 5 22\n']
Demo Output:
['5\n']
Note:
Sample 1. The class has 4 students. The shop sells 6 puzzles. If Ms. Manana buys the first four puzzles consisting of 10, 12, 10 and 7 pieces correspondingly, then the difference between the sizes of the largest and the smallest puzzle will be equal to 5. It is impossible to obtain a smaller difference. Note that the teacher can also buy puzzles 1, 3, 4 and 5 to obtain the difference 5.
|
```python
q,qq = input().split()
q = int(q)
qq = int(qq)
row = input().split()
row_x = []
for _ in row:
row_x.append(int(_))
def foobar(row):
row = sorted(row)
res = []
r = q
g = 0
for _ in range(qq-q+1):
d = row[r-1]-row[g]
r = r +1
g = g +1
res.append(d)
return min(res)
print(foobar(row_x))
```
| 3
|
|
847
|
G
|
University Classes
|
PROGRAMMING
| 900
|
[
"implementation"
] | null | null |
There are *n* student groups at the university. During the study day, each group can take no more than 7 classes. Seven time slots numbered from 1 to 7 are allocated for the classes.
The schedule on Monday is known for each group, i. e. time slots when group will have classes are known.
Your task is to determine the minimum number of rooms needed to hold classes for all groups on Monday. Note that one room can hold at most one group class in a single time slot.
|
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of groups.
Each of the following *n* lines contains a sequence consisting of 7 zeroes and ones — the schedule of classes on Monday for a group. If the symbol in a position equals to 1 then the group has class in the corresponding time slot. In the other case, the group has no class in the corresponding time slot.
|
Print minimum number of rooms needed to hold all groups classes on Monday.
|
[
"2\n0101010\n1010101\n",
"3\n0101011\n0011001\n0110111\n"
] |
[
"1\n",
"3\n"
] |
In the first example one room is enough. It will be occupied in each of the seven time slot by the first group or by the second group.
In the second example three rooms is enough, because in the seventh time slot all three groups have classes.
| 0
|
[
{
"input": "2\n0101010\n1010101",
"output": "1"
},
{
"input": "3\n0101011\n0011001\n0110111",
"output": "3"
},
{
"input": "1\n0111000",
"output": "1"
},
{
"input": "1\n0000000",
"output": "0"
},
{
"input": "1\n1111111",
"output": "1"
},
{
"input": "2\n1000000\n0101000",
"output": "1"
},
{
"input": "3\n0101111\n1101011\n1010011",
"output": "3"
},
{
"input": "5\n0100101\n0000001\n0110000\n0010000\n0011110",
"output": "3"
},
{
"input": "6\n1101110\n1111011\n1101110\n0100011\n1110110\n1110100",
"output": "6"
},
{
"input": "10\n0000000\n0010000\n0000000\n0000010\n0000000\n0100001\n1000000\n0000000\n0000000\n0000000",
"output": "1"
},
{
"input": "20\n1111111\n1101011\n1111011\n0111111\n1111111\n1110111\n1111111\n1111111\n1111111\n1111111\n1110111\n1111111\n0111111\n1011111\n1111111\n1111111\n1101110\n1111111\n1111111\n1111111",
"output": "20"
}
] | 1,542,530,686
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 21
| 109
| 0
|
n = int(input())
l = []
for i in range(n):
li = list(map(int,input()))
l.append(li)
a = len(li)
num = []
for i in range(a):
numm = 0
for j in range(n):
numm += l[j][i]
num.append(numm)
print(max(num))
|
Title: University Classes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* student groups at the university. During the study day, each group can take no more than 7 classes. Seven time slots numbered from 1 to 7 are allocated for the classes.
The schedule on Monday is known for each group, i. e. time slots when group will have classes are known.
Your task is to determine the minimum number of rooms needed to hold classes for all groups on Monday. Note that one room can hold at most one group class in a single time slot.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of groups.
Each of the following *n* lines contains a sequence consisting of 7 zeroes and ones — the schedule of classes on Monday for a group. If the symbol in a position equals to 1 then the group has class in the corresponding time slot. In the other case, the group has no class in the corresponding time slot.
Output Specification:
Print minimum number of rooms needed to hold all groups classes on Monday.
Demo Input:
['2\n0101010\n1010101\n', '3\n0101011\n0011001\n0110111\n']
Demo Output:
['1\n', '3\n']
Note:
In the first example one room is enough. It will be occupied in each of the seven time slot by the first group or by the second group.
In the second example three rooms is enough, because in the seventh time slot all three groups have classes.
|
```python
n = int(input())
l = []
for i in range(n):
li = list(map(int,input()))
l.append(li)
a = len(li)
num = []
for i in range(a):
numm = 0
for j in range(n):
numm += l[j][i]
num.append(numm)
print(max(num))
```
| 3
|
|
50
|
A
|
Domino piling
|
PROGRAMMING
| 800
|
[
"greedy",
"math"
] |
A. Domino piling
|
2
|
256
|
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions.
|
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
|
Output one number — the maximal number of dominoes, which can be placed.
|
[
"2 4\n",
"3 3\n"
] |
[
"4\n",
"4\n"
] |
none
| 500
|
[
{
"input": "2 4",
"output": "4"
},
{
"input": "3 3",
"output": "4"
},
{
"input": "1 5",
"output": "2"
},
{
"input": "1 6",
"output": "3"
},
{
"input": "1 15",
"output": "7"
},
{
"input": "1 16",
"output": "8"
},
{
"input": "2 5",
"output": "5"
},
{
"input": "2 6",
"output": "6"
},
{
"input": "2 7",
"output": "7"
},
{
"input": "2 14",
"output": "14"
},
{
"input": "2 15",
"output": "15"
},
{
"input": "1 4",
"output": "2"
},
{
"input": "2 16",
"output": "16"
},
{
"input": "3 5",
"output": "7"
},
{
"input": "3 6",
"output": "9"
},
{
"input": "3 10",
"output": "15"
},
{
"input": "3 14",
"output": "21"
},
{
"input": "3 15",
"output": "22"
},
{
"input": "3 16",
"output": "24"
},
{
"input": "5 7",
"output": "17"
},
{
"input": "16 16",
"output": "128"
},
{
"input": "15 16",
"output": "120"
},
{
"input": "2 3",
"output": "3"
},
{
"input": "15 15",
"output": "112"
},
{
"input": "14 16",
"output": "112"
},
{
"input": "11 13",
"output": "71"
},
{
"input": "5 16",
"output": "40"
},
{
"input": "8 15",
"output": "60"
},
{
"input": "2 2",
"output": "2"
},
{
"input": "3 4",
"output": "6"
},
{
"input": "4 4",
"output": "8"
},
{
"input": "1 1",
"output": "0"
},
{
"input": "1 2",
"output": "1"
},
{
"input": "1 3",
"output": "1"
},
{
"input": "14 15",
"output": "105"
}
] | 1,635,607,762
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 35
| 92
| 4,300,800
|
a,b = [int(x) for x in input().split()]
print(int((a*b)/2))
|
Title: Domino piling
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions.
Input Specification:
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
Output Specification:
Output one number — the maximal number of dominoes, which can be placed.
Demo Input:
['2 4\n', '3 3\n']
Demo Output:
['4\n', '4\n']
Note:
none
|
```python
a,b = [int(x) for x in input().split()]
print(int((a*b)/2))
```
| 3.968989
|
903
|
D
|
Almost Difference
|
PROGRAMMING
| 2,200
|
[
"data structures",
"math"
] | null | null |
Let's denote a function
You are given an array *a* consisting of *n* integers. You have to calculate the sum of *d*(*a**i*,<=*a**j*) over all pairs (*i*,<=*j*) such that 1<=≤<=*i*<=≤<=*j*<=≤<=*n*.
|
The first line contains one integer *n* (1<=≤<=*n*<=≤<=200000) — the number of elements in *a*.
The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=109) — elements of the array.
|
Print one integer — the sum of *d*(*a**i*,<=*a**j*) over all pairs (*i*,<=*j*) such that 1<=≤<=*i*<=≤<=*j*<=≤<=*n*.
|
[
"5\n1 2 3 1 3\n",
"4\n6 6 5 5\n",
"4\n6 6 4 4\n"
] |
[
"4\n",
"0\n",
"-8\n"
] |
In the first example:
1. *d*(*a*<sub class="lower-index">1</sub>, *a*<sub class="lower-index">2</sub>) = 0; 1. *d*(*a*<sub class="lower-index">1</sub>, *a*<sub class="lower-index">3</sub>) = 2; 1. *d*(*a*<sub class="lower-index">1</sub>, *a*<sub class="lower-index">4</sub>) = 0; 1. *d*(*a*<sub class="lower-index">1</sub>, *a*<sub class="lower-index">5</sub>) = 2; 1. *d*(*a*<sub class="lower-index">2</sub>, *a*<sub class="lower-index">3</sub>) = 0; 1. *d*(*a*<sub class="lower-index">2</sub>, *a*<sub class="lower-index">4</sub>) = 0; 1. *d*(*a*<sub class="lower-index">2</sub>, *a*<sub class="lower-index">5</sub>) = 0; 1. *d*(*a*<sub class="lower-index">3</sub>, *a*<sub class="lower-index">4</sub>) = - 2; 1. *d*(*a*<sub class="lower-index">3</sub>, *a*<sub class="lower-index">5</sub>) = 0; 1. *d*(*a*<sub class="lower-index">4</sub>, *a*<sub class="lower-index">5</sub>) = 2.
| 0
|
[
{
"input": "5\n1 2 3 1 3",
"output": "4"
},
{
"input": "4\n6 6 5 5",
"output": "0"
},
{
"input": "4\n6 6 4 4",
"output": "-8"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "1\n1000000000",
"output": "0"
},
{
"input": "2\n1 1000000000",
"output": "999999999"
},
{
"input": "5\n1 999999996 999999998 999999994 1000000000",
"output": "3999999992"
},
{
"input": "100\n7 4 5 5 10 10 5 8 5 7 4 5 4 6 8 8 2 6 3 3 10 7 10 8 6 2 7 3 9 7 7 2 4 5 2 4 9 5 10 1 10 5 10 4 1 3 4 2 6 9 9 9 10 6 2 5 6 1 8 10 4 10 3 4 10 5 5 4 10 4 5 3 7 10 2 7 3 6 9 6 1 6 5 5 4 6 6 4 4 1 5 1 6 6 6 8 8 6 2 6",
"output": "-1774"
},
{
"input": "100\n591 417 888 251 792 847 685 3 182 461 102 348 555 956 771 901 712 878 580 631 342 333 285 899 525 725 537 718 929 653 84 788 104 355 624 803 253 853 201 995 536 184 65 205 540 652 549 777 248 405 677 950 431 580 600 846 328 429 134 983 526 103 500 963 400 23 276 704 570 757 410 658 507 620 984 244 486 454 802 411 985 303 635 283 96 597 855 775 139 839 839 61 219 986 776 72 729 69 20 917",
"output": "-91018"
},
{
"input": "100\n7 8 5 9 5 6 6 9 7 6 8 7 5 10 7 2 6 1 8 10 7 9 9 8 9 6 8 5 10 6 3 7 5 8 9 7 6 1 9 9 6 9 9 2 10 4 4 6 7 9 7 7 9 10 6 10 8 6 4 7 5 5 8 10 10 7 6 9 8 1 5 1 6 6 2 9 8 4 6 6 9 10 6 1 9 9 9 6 1 8 9 2 8 7 1 10 8 2 4 7",
"output": "-1713"
},
{
"input": "100\n82 81 14 33 78 80 15 60 89 82 79 13 15 17 25 13 21 20 63 26 62 63 79 36 18 21 88 92 27 18 59 64 18 96 28 4 76 43 26 25 89 88 96 33 27 97 52 37 92 80 23 18 78 14 88 5 3 14 85 72 84 75 41 3 51 92 91 79 18 78 19 79 8 35 85 86 78 17 51 36 100 32 49 95 2 100 67 72 55 53 42 3 21 100 12 51 50 79 47 2",
"output": "6076"
},
{
"input": "5\n3 1 1 1 3",
"output": "0"
},
{
"input": "1\n22955",
"output": "0"
},
{
"input": "1\n32955",
"output": "0"
}
] | 1,682,765,554
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 45
| 530
| 29,593,600
|
# LUOGU_RID: 109195401
n=int(input())
a=list(map(int,input().split()))
mp={}
for i in range(n):mp[a[i]]=mp[a[i]+1]=mp[a[i]-1]=0
ans=0
tot=0
for i in range(n):
ans+=i*a[i]-tot
tot+=a[i]
ans+=mp[a[i]+1]
ans-=mp[a[i]-1]
mp[a[i]]+=1
print(ans)
|
Title: Almost Difference
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Let's denote a function
You are given an array *a* consisting of *n* integers. You have to calculate the sum of *d*(*a**i*,<=*a**j*) over all pairs (*i*,<=*j*) such that 1<=≤<=*i*<=≤<=*j*<=≤<=*n*.
Input Specification:
The first line contains one integer *n* (1<=≤<=*n*<=≤<=200000) — the number of elements in *a*.
The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=109) — elements of the array.
Output Specification:
Print one integer — the sum of *d*(*a**i*,<=*a**j*) over all pairs (*i*,<=*j*) such that 1<=≤<=*i*<=≤<=*j*<=≤<=*n*.
Demo Input:
['5\n1 2 3 1 3\n', '4\n6 6 5 5\n', '4\n6 6 4 4\n']
Demo Output:
['4\n', '0\n', '-8\n']
Note:
In the first example:
1. *d*(*a*<sub class="lower-index">1</sub>, *a*<sub class="lower-index">2</sub>) = 0; 1. *d*(*a*<sub class="lower-index">1</sub>, *a*<sub class="lower-index">3</sub>) = 2; 1. *d*(*a*<sub class="lower-index">1</sub>, *a*<sub class="lower-index">4</sub>) = 0; 1. *d*(*a*<sub class="lower-index">1</sub>, *a*<sub class="lower-index">5</sub>) = 2; 1. *d*(*a*<sub class="lower-index">2</sub>, *a*<sub class="lower-index">3</sub>) = 0; 1. *d*(*a*<sub class="lower-index">2</sub>, *a*<sub class="lower-index">4</sub>) = 0; 1. *d*(*a*<sub class="lower-index">2</sub>, *a*<sub class="lower-index">5</sub>) = 0; 1. *d*(*a*<sub class="lower-index">3</sub>, *a*<sub class="lower-index">4</sub>) = - 2; 1. *d*(*a*<sub class="lower-index">3</sub>, *a*<sub class="lower-index">5</sub>) = 0; 1. *d*(*a*<sub class="lower-index">4</sub>, *a*<sub class="lower-index">5</sub>) = 2.
|
```python
# LUOGU_RID: 109195401
n=int(input())
a=list(map(int,input().split()))
mp={}
for i in range(n):mp[a[i]]=mp[a[i]+1]=mp[a[i]-1]=0
ans=0
tot=0
for i in range(n):
ans+=i*a[i]-tot
tot+=a[i]
ans+=mp[a[i]+1]
ans-=mp[a[i]-1]
mp[a[i]]+=1
print(ans)
```
| 3
|
|
129
|
A
|
Cookies
|
PROGRAMMING
| 900
|
[
"implementation"
] | null | null |
Olga came to visit the twins Anna and Maria and saw that they have many cookies. The cookies are distributed into bags. As there are many cookies, Olga decided that it's no big deal if she steals a bag. However, she doesn't want the sisters to quarrel because of nothing when they divide the cookies. That's why Olga wants to steal a bag with cookies so that the number of cookies in the remaining bags was even, that is, so that Anna and Maria could evenly divide it into two (even 0 remaining cookies will do, just as any other even number). How many ways there are to steal exactly one cookie bag so that the total number of cookies in the remaining bags was even?
|
The first line contains the only integer *n* (1<=≤<=*n*<=≤<=100) — the number of cookie bags Anna and Maria have. The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=100) — the number of cookies in the *i*-th bag.
|
Print in the only line the only number — the sought number of ways. If there are no such ways print 0.
|
[
"1\n1\n",
"10\n1 2 2 3 4 4 4 2 2 2\n",
"11\n2 2 2 2 2 2 2 2 2 2 99\n"
] |
[
"1\n",
"8\n",
"1\n"
] |
In the first sample Olga should take the only bag so that the twins ended up with the even number of cookies.
In the second sample Olga can take any of five bags with two cookies or any of three bags with four cookies — 5 + 3 = 8 ways in total.
In the third sample, no matter which bag with two cookies Olga chooses, the twins are left with 2 * 9 + 99 = 117 cookies. Thus, Olga has only one option: to take the bag with 99 cookies.
| 500
|
[
{
"input": "1\n1",
"output": "1"
},
{
"input": "10\n1 2 2 3 4 4 4 2 2 2",
"output": "8"
},
{
"input": "11\n2 2 2 2 2 2 2 2 2 2 99",
"output": "1"
},
{
"input": "2\n1 1",
"output": "0"
},
{
"input": "2\n2 2",
"output": "2"
},
{
"input": "2\n1 2",
"output": "1"
},
{
"input": "7\n7 7 7 7 7 7 7",
"output": "7"
},
{
"input": "8\n1 2 3 4 5 6 7 8",
"output": "4"
},
{
"input": "100\n1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2",
"output": "50"
},
{
"input": "99\n99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99",
"output": "49"
},
{
"input": "82\n43 44 96 33 23 42 33 66 53 87 8 90 43 91 40 88 51 18 48 62 59 10 22 20 54 6 13 63 2 56 31 52 98 42 54 32 26 77 9 24 33 91 16 30 39 34 78 82 73 90 12 15 67 76 30 18 44 86 84 98 65 54 100 79 28 34 40 56 11 43 72 35 86 59 89 40 30 33 7 19 44 15",
"output": "50"
},
{
"input": "17\n50 14 17 77 74 74 38 76 41 27 45 29 66 98 38 73 38",
"output": "7"
},
{
"input": "94\n81 19 90 99 26 11 86 44 78 36 80 59 99 90 78 72 71 20 94 56 42 40 71 84 10 85 10 70 52 27 39 55 90 16 48 25 7 79 99 100 38 10 99 56 3 4 78 9 16 57 14 40 52 54 57 70 30 86 56 84 97 60 59 69 49 66 23 92 90 46 86 73 53 47 1 83 14 20 24 66 13 45 41 14 86 75 55 88 48 95 82 24 47 87",
"output": "39"
},
{
"input": "88\n64 95 12 90 40 65 98 45 52 54 79 7 81 25 98 19 68 82 41 53 35 50 5 22 32 21 8 39 8 6 72 27 81 30 12 79 21 42 60 2 66 87 46 93 62 78 52 71 76 32 78 94 86 85 55 15 34 76 41 20 32 26 94 81 89 45 74 49 11 40 40 39 49 46 80 85 90 23 80 40 86 58 70 26 48 93 23 53",
"output": "37"
},
{
"input": "84\n95 9 43 43 13 84 60 90 1 8 97 99 54 34 59 83 33 15 51 26 40 12 66 65 19 30 29 78 92 60 25 13 19 84 71 73 12 24 54 49 16 41 11 40 57 59 34 40 39 9 71 83 1 77 79 53 94 47 78 55 77 85 29 52 80 90 53 77 97 97 27 79 28 23 83 25 26 22 49 86 63 56 3 32",
"output": "51"
},
{
"input": "47\n61 97 76 94 91 22 2 68 62 73 90 47 16 79 44 71 98 68 43 6 53 52 40 27 68 67 43 96 14 91 60 61 96 24 97 13 32 65 85 96 81 77 34 18 23 14 80",
"output": "21"
},
{
"input": "69\n71 1 78 74 58 89 30 6 100 90 22 61 11 59 14 74 27 25 78 61 45 19 25 33 37 4 52 43 53 38 9 100 56 67 69 38 76 91 63 60 93 52 28 61 9 98 8 14 57 63 89 64 98 51 36 66 36 86 13 82 50 91 52 64 86 78 78 83 81",
"output": "37"
},
{
"input": "52\n38 78 36 75 19 3 56 1 39 97 24 79 84 16 93 55 96 64 12 24 1 86 80 29 12 32 36 36 73 39 76 65 53 98 30 20 28 8 86 43 70 22 75 69 62 65 81 25 53 40 71 59",
"output": "28"
},
{
"input": "74\n81 31 67 97 26 75 69 81 11 13 13 74 77 88 52 20 52 64 66 75 72 28 41 54 26 75 41 91 75 15 18 36 13 83 63 61 14 48 53 63 19 67 35 48 23 65 73 100 44 55 92 88 99 17 73 25 83 7 31 89 12 80 98 39 42 75 14 29 81 35 77 87 33 94",
"output": "47"
},
{
"input": "44\n46 56 31 31 37 71 94 2 14 100 45 72 36 72 80 3 38 54 42 98 50 32 31 42 62 31 45 50 95 100 18 17 64 22 18 25 52 56 70 57 43 40 81 28",
"output": "15"
},
{
"input": "22\n28 57 40 74 51 4 45 84 99 12 95 14 92 60 47 81 84 51 31 91 59 42",
"output": "11"
},
{
"input": "59\n73 45 94 76 41 49 65 13 74 66 36 25 47 75 40 23 92 72 11 32 32 8 81 26 68 56 41 8 76 47 96 55 70 11 84 14 83 18 70 22 30 39 28 100 48 11 92 45 78 69 86 1 54 90 98 91 13 17 35",
"output": "33"
},
{
"input": "63\n20 18 44 94 68 57 16 43 74 55 68 24 21 95 76 84 50 50 47 86 86 12 58 55 28 72 86 18 34 45 81 88 3 72 41 9 60 90 81 93 12 6 9 6 2 41 1 7 9 29 81 14 64 80 20 36 67 54 7 5 35 81 22",
"output": "37"
},
{
"input": "28\n49 84 48 19 44 91 11 82 96 95 88 90 71 82 87 25 31 23 18 13 98 45 26 65 35 12 31 14",
"output": "15"
},
{
"input": "61\n34 18 28 64 28 45 9 77 77 20 63 92 79 16 16 100 86 2 91 91 57 15 31 95 10 88 84 5 82 83 53 98 59 17 97 80 76 80 81 3 91 81 87 93 61 46 10 49 6 22 21 75 63 89 21 81 30 19 67 38 77",
"output": "35"
},
{
"input": "90\n41 90 43 1 28 75 90 50 3 70 76 64 81 63 25 69 83 82 29 91 59 66 21 61 7 55 72 49 38 69 72 20 64 58 30 81 61 29 96 14 39 5 100 20 29 98 75 29 44 78 97 45 26 77 73 59 22 99 41 6 3 96 71 20 9 18 96 18 90 62 34 78 54 5 41 6 73 33 2 54 26 21 18 6 45 57 43 73 95 75",
"output": "42"
},
{
"input": "45\n93 69 4 27 20 14 71 48 79 3 32 26 49 30 57 88 13 56 49 61 37 32 47 41 41 70 45 68 82 18 8 6 25 20 15 13 71 99 28 6 52 34 19 59 26",
"output": "23"
},
{
"input": "33\n29 95 48 49 91 10 83 71 47 25 66 36 51 12 34 10 54 74 41 96 89 26 89 1 42 33 1 62 9 32 49 65 78",
"output": "15"
},
{
"input": "34\n98 24 42 36 41 82 28 58 89 34 77 70 76 44 74 54 66 100 13 79 4 88 21 1 11 45 91 29 87 100 29 54 82 78",
"output": "13"
},
{
"input": "29\n91 84 26 84 9 63 52 9 65 56 90 2 36 7 67 33 91 14 65 38 53 36 81 83 85 14 33 95 51",
"output": "17"
},
{
"input": "100\n2 88 92 82 87 100 78 28 84 43 78 32 43 33 97 19 15 52 29 84 57 72 54 13 99 28 82 79 40 70 34 92 91 53 9 88 27 43 14 92 72 37 26 37 20 95 19 34 49 64 33 37 34 27 80 79 9 54 99 68 25 4 68 73 46 66 24 78 3 87 26 52 50 84 4 95 23 83 39 58 86 36 33 16 98 2 84 19 53 12 69 60 10 11 78 17 79 92 77 59",
"output": "45"
},
{
"input": "100\n2 95 45 73 9 54 20 97 57 82 88 26 18 71 25 27 75 54 31 11 58 85 69 75 72 91 76 5 25 80 45 49 4 73 8 81 81 38 5 12 53 77 7 96 90 35 28 80 73 94 19 69 96 17 94 49 69 9 32 19 5 12 46 29 26 40 59 59 6 95 82 50 72 2 45 69 12 5 72 29 39 72 23 96 81 28 28 56 68 58 37 41 30 1 90 84 15 24 96 43",
"output": "53"
},
{
"input": "100\n27 72 35 91 13 10 35 45 24 55 83 84 63 96 29 79 34 67 63 92 48 83 18 77 28 27 49 66 29 88 55 15 6 58 14 67 94 36 77 7 7 64 61 52 71 18 36 99 76 6 50 67 16 13 41 7 89 73 61 51 78 22 78 32 76 100 3 31 89 71 63 53 15 85 77 54 89 33 68 74 3 23 57 5 43 89 75 35 9 86 90 11 31 46 48 37 74 17 77 8",
"output": "40"
},
{
"input": "100\n69 98 69 88 11 49 55 8 25 91 17 81 47 26 15 73 96 71 18 42 42 61 48 14 92 78 35 72 4 27 62 75 83 79 17 16 46 80 96 90 82 54 37 69 85 21 67 70 96 10 46 63 21 59 56 92 54 88 77 30 75 45 44 29 86 100 51 11 65 69 66 56 82 63 27 1 51 51 13 10 3 55 26 85 34 16 87 72 13 100 81 71 90 95 86 50 83 55 55 54",
"output": "53"
},
{
"input": "100\n34 35 99 64 2 66 78 93 20 48 12 79 19 10 87 7 42 92 60 79 5 2 24 89 57 48 63 92 74 4 16 51 7 12 90 48 87 17 18 73 51 58 97 97 25 38 15 97 96 73 67 91 6 75 14 13 87 79 75 3 15 55 35 95 71 45 10 13 20 37 82 26 2 22 13 83 97 84 39 79 43 100 54 59 98 8 61 34 7 65 75 44 24 77 73 88 34 95 44 77",
"output": "55"
},
{
"input": "100\n15 86 3 1 51 26 74 85 37 87 64 58 10 6 57 26 30 47 85 65 24 72 50 40 12 35 91 47 91 60 47 87 95 34 80 91 26 3 36 39 14 86 28 70 51 44 28 21 72 79 57 61 16 71 100 94 57 67 36 74 24 21 89 85 25 2 97 67 76 53 76 80 97 64 35 13 8 32 21 52 62 61 67 14 74 73 66 44 55 76 24 3 43 42 99 61 36 80 38 66",
"output": "52"
},
{
"input": "100\n45 16 54 54 80 94 74 93 75 85 58 95 79 30 81 2 84 4 57 23 92 64 78 1 50 36 13 27 56 54 10 77 87 1 5 38 85 74 94 82 30 45 72 83 82 30 81 82 82 3 69 82 7 92 39 60 94 42 41 5 3 17 67 21 79 44 79 96 28 3 53 68 79 89 63 83 1 44 4 31 84 15 73 77 19 66 54 6 73 1 67 24 91 11 86 45 96 82 20 89",
"output": "51"
},
{
"input": "100\n84 23 50 32 90 71 92 43 58 70 6 82 7 55 85 19 70 89 12 26 29 56 74 30 2 27 4 39 63 67 91 81 11 33 75 10 82 88 39 43 43 80 68 35 55 67 53 62 73 65 86 74 43 51 14 48 42 92 83 57 22 33 24 99 5 27 78 96 7 28 11 15 8 38 85 67 5 92 24 96 57 59 14 95 91 4 9 18 45 33 74 83 64 85 14 51 51 94 29 2",
"output": "53"
},
{
"input": "100\n77 56 56 45 73 55 32 37 39 50 30 95 79 21 44 34 51 43 86 91 39 30 85 15 35 93 100 14 57 31 80 79 38 40 88 4 91 54 7 95 76 26 62 84 17 33 67 47 6 82 69 51 17 2 59 24 11 12 31 90 12 11 55 38 72 49 30 50 42 46 5 97 9 9 30 45 86 23 19 82 40 42 5 40 35 98 35 32 60 60 5 28 84 35 21 49 68 53 68 23",
"output": "48"
},
{
"input": "100\n78 38 79 61 45 86 83 83 86 90 74 69 2 84 73 39 2 5 20 71 24 80 54 89 58 34 77 40 39 62 2 47 28 53 97 75 88 98 94 96 33 71 44 90 47 36 19 89 87 98 90 87 5 85 34 79 82 3 42 88 89 63 35 7 89 30 40 48 12 41 56 76 83 60 80 80 39 56 77 4 72 96 30 55 57 51 7 19 11 1 66 1 91 87 11 62 95 85 79 25",
"output": "48"
},
{
"input": "100\n5 34 23 20 76 75 19 51 17 82 60 13 83 6 65 16 20 43 66 54 87 10 87 73 50 24 16 98 33 28 80 52 54 82 26 92 14 13 84 92 94 29 61 21 60 20 48 94 24 20 75 70 58 27 68 45 86 89 29 8 67 38 83 48 18 100 11 22 46 84 52 97 70 19 50 75 3 7 52 53 72 41 18 31 1 38 49 53 11 64 99 76 9 87 48 12 100 32 44 71",
"output": "58"
},
{
"input": "100\n76 89 68 78 24 72 73 95 98 72 58 15 2 5 56 32 9 65 50 70 94 31 29 54 89 52 31 93 43 56 26 35 72 95 51 55 78 70 11 92 17 5 54 94 81 31 78 95 73 91 95 37 59 9 53 48 65 55 84 8 45 97 64 37 96 34 36 53 66 17 72 48 99 23 27 18 92 84 44 73 60 78 53 29 68 99 19 39 61 40 69 6 77 12 47 29 15 4 8 45",
"output": "53"
},
{
"input": "100\n82 40 31 53 8 50 85 93 3 84 54 17 96 59 51 42 18 19 35 84 79 31 17 46 54 82 72 49 35 73 26 89 61 73 3 50 12 29 25 77 88 21 58 24 22 89 96 54 82 29 96 56 77 16 1 68 90 93 20 23 57 22 31 18 92 90 51 14 50 72 31 54 12 50 66 62 2 34 17 45 68 50 87 97 23 71 1 72 17 82 42 15 20 78 4 49 66 59 10 17",
"output": "54"
},
{
"input": "100\n32 82 82 24 39 53 48 5 29 24 9 37 91 37 91 95 1 97 84 52 12 56 93 47 22 20 14 17 40 22 79 34 24 2 69 30 69 29 3 89 21 46 60 92 39 29 18 24 49 18 40 22 60 13 77 50 39 64 50 70 99 8 66 31 90 38 20 54 7 21 5 56 41 68 69 20 54 89 69 62 9 53 43 89 81 97 15 2 52 78 89 65 16 61 59 42 56 25 32 52",
"output": "49"
},
{
"input": "100\n72 54 23 24 97 14 99 87 15 25 7 23 17 87 72 31 71 87 34 82 51 77 74 85 62 38 24 7 84 48 98 21 29 71 70 84 25 58 67 92 18 44 32 9 81 15 53 29 63 18 86 16 7 31 38 99 70 32 89 16 23 11 66 96 69 82 97 59 6 9 49 80 85 19 6 9 52 51 85 74 53 46 73 55 31 63 78 61 34 80 77 65 87 77 92 52 89 8 52 31",
"output": "44"
},
{
"input": "100\n56 88 8 19 7 15 11 54 35 50 19 57 63 72 51 43 50 19 57 90 40 100 8 92 11 96 30 32 59 65 93 47 62 3 50 41 30 50 72 83 61 46 83 60 20 46 33 1 5 18 83 22 34 16 41 95 63 63 7 59 55 95 91 29 64 60 64 81 45 45 10 9 88 37 69 85 21 82 41 76 42 34 47 78 51 83 65 100 13 22 59 76 63 1 26 86 36 94 99 74",
"output": "46"
},
{
"input": "100\n27 89 67 60 62 80 43 50 28 88 72 5 94 11 63 91 18 78 99 3 71 26 12 97 74 62 23 24 22 3 100 72 98 7 94 32 12 75 61 88 42 48 10 14 45 9 48 56 73 76 70 70 79 90 35 39 96 37 81 11 19 65 99 39 23 79 34 61 35 74 90 37 73 23 46 21 94 84 73 58 11 89 13 9 10 85 42 78 73 32 53 39 49 90 43 5 28 31 97 75",
"output": "53"
},
{
"input": "100\n33 24 97 96 1 14 99 51 13 65 67 20 46 88 42 44 20 49 5 89 98 83 15 40 74 83 58 3 10 79 34 2 69 28 37 100 55 52 14 8 44 94 97 89 6 42 11 28 30 33 55 56 20 57 52 25 75 1 87 42 62 41 37 12 54 85 95 80 42 36 94 96 28 76 54 36 4 17 26 24 62 15 17 79 84 36 92 78 74 91 96 77 54 92 81 91 62 98 37 37",
"output": "43"
},
{
"input": "100\n86 24 61 15 11 85 1 31 47 36 23 36 59 34 3 27 16 29 82 28 58 52 52 66 71 61 98 39 60 20 67 41 67 90 73 29 92 17 70 95 58 98 58 32 21 73 46 56 87 72 80 75 40 27 94 31 59 92 93 37 14 99 96 21 97 23 81 91 52 52 96 94 92 28 38 29 52 16 57 27 17 24 91 21 79 55 96 98 95 94 23 78 79 12 77 35 32 75 100 82",
"output": "51"
},
{
"input": "100\n88 85 41 37 69 21 7 69 36 5 92 26 64 75 22 46 67 20 70 22 62 66 38 24 47 49 68 30 90 31 67 86 86 82 9 51 43 45 48 42 73 44 31 94 45 60 54 66 20 87 11 94 34 32 87 66 56 28 75 39 37 90 72 93 55 72 31 42 30 71 87 61 4 12 12 81 23 61 56 98 71 32 30 33 96 63 92 16 8 78 47 91 47 54 49 3 81 82 41 5",
"output": "47"
},
{
"input": "1\n2",
"output": "1"
},
{
"input": "5\n1 1 3 2 2",
"output": "3"
}
] | 1,686,135,208
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 52
| 92
| 0
|
n = int(input())
arr = [int(i) for i in input().split()]
count = 0
for i in arr:
if (sum(arr) - i) % 2 == 0:
count += 1
print(count)
|
Title: Cookies
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Olga came to visit the twins Anna and Maria and saw that they have many cookies. The cookies are distributed into bags. As there are many cookies, Olga decided that it's no big deal if she steals a bag. However, she doesn't want the sisters to quarrel because of nothing when they divide the cookies. That's why Olga wants to steal a bag with cookies so that the number of cookies in the remaining bags was even, that is, so that Anna and Maria could evenly divide it into two (even 0 remaining cookies will do, just as any other even number). How many ways there are to steal exactly one cookie bag so that the total number of cookies in the remaining bags was even?
Input Specification:
The first line contains the only integer *n* (1<=≤<=*n*<=≤<=100) — the number of cookie bags Anna and Maria have. The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=100) — the number of cookies in the *i*-th bag.
Output Specification:
Print in the only line the only number — the sought number of ways. If there are no such ways print 0.
Demo Input:
['1\n1\n', '10\n1 2 2 3 4 4 4 2 2 2\n', '11\n2 2 2 2 2 2 2 2 2 2 99\n']
Demo Output:
['1\n', '8\n', '1\n']
Note:
In the first sample Olga should take the only bag so that the twins ended up with the even number of cookies.
In the second sample Olga can take any of five bags with two cookies or any of three bags with four cookies — 5 + 3 = 8 ways in total.
In the third sample, no matter which bag with two cookies Olga chooses, the twins are left with 2 * 9 + 99 = 117 cookies. Thus, Olga has only one option: to take the bag with 99 cookies.
|
```python
n = int(input())
arr = [int(i) for i in input().split()]
count = 0
for i in arr:
if (sum(arr) - i) % 2 == 0:
count += 1
print(count)
```
| 3
|
|
863
|
A
|
Quasi-palindrome
|
PROGRAMMING
| 900
|
[
"brute force",
"implementation"
] | null | null |
Let quasi-palindromic number be such number that adding some leading zeros (possible none) to it produces a palindromic string.
String *t* is called a palindrome, if it reads the same from left to right and from right to left.
For example, numbers 131 and 2010200 are quasi-palindromic, they can be transformed to strings "131" and "002010200", respectively, which are palindromes.
You are given some integer number *x*. Check if it's a quasi-palindromic number.
|
The first line contains one integer number *x* (1<=≤<=*x*<=≤<=109). This number is given without any leading zeroes.
|
Print "YES" if number *x* is quasi-palindromic. Otherwise, print "NO" (without quotes).
|
[
"131\n",
"320\n",
"2010200\n"
] |
[
"YES\n",
"NO\n",
"YES\n"
] |
none
| 0
|
[
{
"input": "131",
"output": "YES"
},
{
"input": "320",
"output": "NO"
},
{
"input": "2010200",
"output": "YES"
},
{
"input": "1",
"output": "YES"
},
{
"input": "1000000000",
"output": "YES"
},
{
"input": "999999999",
"output": "YES"
},
{
"input": "999999998",
"output": "NO"
},
{
"input": "102000",
"output": "NO"
},
{
"input": "210000000",
"output": "NO"
},
{
"input": "213443120",
"output": "YES"
},
{
"input": "99",
"output": "YES"
},
{
"input": "22002",
"output": "NO"
},
{
"input": "1010",
"output": "YES"
},
{
"input": "1201",
"output": "NO"
},
{
"input": "6460046",
"output": "NO"
},
{
"input": "503435",
"output": "NO"
},
{
"input": "21002",
"output": "NO"
},
{
"input": "101001",
"output": "NO"
},
{
"input": "200102",
"output": "NO"
},
{
"input": "20010002",
"output": "NO"
},
{
"input": "33003",
"output": "NO"
},
{
"input": "100101",
"output": "NO"
},
{
"input": "1021",
"output": "NO"
},
{
"input": "1101",
"output": "NO"
},
{
"input": "10101100",
"output": "NO"
},
{
"input": "101",
"output": "YES"
},
{
"input": "1011",
"output": "NO"
},
{
"input": "11010",
"output": "NO"
},
{
"input": "10110",
"output": "NO"
},
{
"input": "110000",
"output": "YES"
},
{
"input": "2011",
"output": "NO"
},
{
"input": "10020001",
"output": "NO"
},
{
"input": "12505021",
"output": "NO"
},
{
"input": "12310",
"output": "NO"
},
{
"input": "100501",
"output": "NO"
},
{
"input": "11001",
"output": "NO"
},
{
"input": "20020002",
"output": "NO"
},
{
"input": "202002",
"output": "NO"
},
{
"input": "1001",
"output": "YES"
},
{
"input": "1020021",
"output": "NO"
},
{
"input": "60660",
"output": "NO"
},
{
"input": "98809",
"output": "NO"
},
{
"input": "11000000",
"output": "YES"
},
{
"input": "807008",
"output": "NO"
}
] | 1,623,314,528
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 44
| 62
| 0
|
a=input()
while(a[-1]=='0'):a=a[:-1]
if a==a[::-1]:print("YES")
else: print("NO")
|
Title: Quasi-palindrome
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Let quasi-palindromic number be such number that adding some leading zeros (possible none) to it produces a palindromic string.
String *t* is called a palindrome, if it reads the same from left to right and from right to left.
For example, numbers 131 and 2010200 are quasi-palindromic, they can be transformed to strings "131" and "002010200", respectively, which are palindromes.
You are given some integer number *x*. Check if it's a quasi-palindromic number.
Input Specification:
The first line contains one integer number *x* (1<=≤<=*x*<=≤<=109). This number is given without any leading zeroes.
Output Specification:
Print "YES" if number *x* is quasi-palindromic. Otherwise, print "NO" (without quotes).
Demo Input:
['131\n', '320\n', '2010200\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n']
Note:
none
|
```python
a=input()
while(a[-1]=='0'):a=a[:-1]
if a==a[::-1]:print("YES")
else: print("NO")
```
| 3
|
|
658
|
A
|
Bear and Reverse Radewoosh
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Limak and Radewoosh are going to compete against each other in the upcoming algorithmic contest. They are equally skilled but they won't solve problems in the same order.
There will be *n* problems. The *i*-th problem has initial score *p**i* and it takes exactly *t**i* minutes to solve it. Problems are sorted by difficulty — it's guaranteed that *p**i*<=<<=*p**i*<=+<=1 and *t**i*<=<<=*t**i*<=+<=1.
A constant *c* is given too, representing the speed of loosing points. Then, submitting the *i*-th problem at time *x* (*x* minutes after the start of the contest) gives *max*(0,<= *p**i*<=-<=*c*·*x*) points.
Limak is going to solve problems in order 1,<=2,<=...,<=*n* (sorted increasingly by *p**i*). Radewoosh is going to solve them in order *n*,<=*n*<=-<=1,<=...,<=1 (sorted decreasingly by *p**i*). Your task is to predict the outcome — print the name of the winner (person who gets more points at the end) or a word "Tie" in case of a tie.
You may assume that the duration of the competition is greater or equal than the sum of all *t**i*. That means both Limak and Radewoosh will accept all *n* problems.
|
The first line contains two integers *n* and *c* (1<=≤<=*n*<=≤<=50,<=1<=≤<=*c*<=≤<=1000) — the number of problems and the constant representing the speed of loosing points.
The second line contains *n* integers *p*1,<=*p*2,<=...,<=*p**n* (1<=≤<=*p**i*<=≤<=1000,<=*p**i*<=<<=*p**i*<=+<=1) — initial scores.
The third line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=1000,<=*t**i*<=<<=*t**i*<=+<=1) where *t**i* denotes the number of minutes one needs to solve the *i*-th problem.
|
Print "Limak" (without quotes) if Limak will get more points in total. Print "Radewoosh" (without quotes) if Radewoosh will get more points in total. Print "Tie" (without quotes) if Limak and Radewoosh will get the same total number of points.
|
[
"3 2\n50 85 250\n10 15 25\n",
"3 6\n50 85 250\n10 15 25\n",
"8 1\n10 20 30 40 50 60 70 80\n8 10 58 63 71 72 75 76\n"
] |
[
"Limak\n",
"Radewoosh\n",
"Tie\n"
] |
In the first sample, there are 3 problems. Limak solves them as follows:
1. Limak spends 10 minutes on the 1-st problem and he gets 50 - *c*·10 = 50 - 2·10 = 30 points. 1. Limak spends 15 minutes on the 2-nd problem so he submits it 10 + 15 = 25 minutes after the start of the contest. For the 2-nd problem he gets 85 - 2·25 = 35 points. 1. He spends 25 minutes on the 3-rd problem so he submits it 10 + 15 + 25 = 50 minutes after the start. For this problem he gets 250 - 2·50 = 150 points.
So, Limak got 30 + 35 + 150 = 215 points.
Radewoosh solves problem in the reversed order:
1. Radewoosh solves 3-rd problem after 25 minutes so he gets 250 - 2·25 = 200 points. 1. He spends 15 minutes on the 2-nd problem so he submits it 25 + 15 = 40 minutes after the start. He gets 85 - 2·40 = 5 points for this problem. 1. He spends 10 minutes on the 1-st problem so he submits it 25 + 15 + 10 = 50 minutes after the start. He gets *max*(0, 50 - 2·50) = *max*(0, - 50) = 0 points.
Radewoosh got 200 + 5 + 0 = 205 points in total. Limak has 215 points so Limak wins.
In the second sample, Limak will get 0 points for each problem and Radewoosh will first solve the hardest problem and he will get 250 - 6·25 = 100 points for that. Radewoosh will get 0 points for other two problems but he is the winner anyway.
In the third sample, Limak will get 2 points for the 1-st problem and 2 points for the 2-nd problem. Radewoosh will get 4 points for the 8-th problem. They won't get points for other problems and thus there is a tie because 2 + 2 = 4.
| 500
|
[
{
"input": "3 2\n50 85 250\n10 15 25",
"output": "Limak"
},
{
"input": "3 6\n50 85 250\n10 15 25",
"output": "Radewoosh"
},
{
"input": "8 1\n10 20 30 40 50 60 70 80\n8 10 58 63 71 72 75 76",
"output": "Tie"
},
{
"input": "4 1\n3 5 6 9\n1 2 4 8",
"output": "Limak"
},
{
"input": "4 1\n1 3 6 10\n1 5 7 8",
"output": "Radewoosh"
},
{
"input": "4 1\n2 4 5 10\n2 3 9 10",
"output": "Tie"
},
{
"input": "18 4\n68 97 121 132 146 277 312 395 407 431 458 461 595 634 751 855 871 994\n1 2 3 4 9 10 13 21 22 29 31 34 37 38 39 41 48 49",
"output": "Radewoosh"
},
{
"input": "50 1\n5 14 18 73 137 187 195 197 212 226 235 251 262 278 287 304 310 322 342 379 393 420 442 444 448 472 483 485 508 515 517 523 559 585 618 627 636 646 666 682 703 707 780 853 937 951 959 989 991 992\n30 84 113 173 199 220 235 261 266 277 300 306 310 312 347 356 394 396 397 409 414 424 446 462 468 487 507 517 537 566 594 643 656 660 662 668 706 708 773 774 779 805 820 827 868 896 929 942 961 995",
"output": "Tie"
},
{
"input": "4 1\n4 6 9 10\n2 3 4 5",
"output": "Radewoosh"
},
{
"input": "4 1\n4 6 9 10\n3 4 5 7",
"output": "Radewoosh"
},
{
"input": "4 1\n1 6 7 10\n2 7 8 10",
"output": "Tie"
},
{
"input": "4 1\n4 5 7 9\n1 4 5 8",
"output": "Limak"
},
{
"input": "50 1\n6 17 44 82 94 127 134 156 187 211 212 252 256 292 294 303 352 355 379 380 398 409 424 434 480 524 584 594 631 714 745 756 777 778 789 793 799 821 841 849 859 878 879 895 925 932 944 952 958 990\n15 16 40 42 45 71 99 100 117 120 174 181 186 204 221 268 289 332 376 394 403 409 411 444 471 487 499 539 541 551 567 589 619 623 639 669 689 722 735 776 794 822 830 840 847 907 917 927 936 988",
"output": "Radewoosh"
},
{
"input": "50 10\n25 49 52 73 104 117 127 136 149 164 171 184 226 251 257 258 286 324 337 341 386 390 428 453 464 470 492 517 543 565 609 634 636 660 678 693 710 714 729 736 739 749 781 836 866 875 956 960 977 979\n2 4 7 10 11 22 24 26 27 28 31 35 37 38 42 44 45 46 52 53 55 56 57 59 60 61 64 66 67 68 69 71 75 76 77 78 79 81 83 85 86 87 89 90 92 93 94 98 99 100",
"output": "Limak"
},
{
"input": "50 10\n11 15 25 71 77 83 95 108 143 150 182 183 198 203 213 223 279 280 346 348 350 355 375 376 412 413 415 432 470 545 553 562 589 595 607 633 635 637 688 719 747 767 771 799 842 883 905 924 942 944\n1 3 5 6 7 10 11 12 13 14 15 16 19 20 21 23 25 32 35 36 37 38 40 41 42 43 47 50 51 54 55 56 57 58 59 60 62 63 64 65 66 68 69 70 71 72 73 75 78 80",
"output": "Radewoosh"
},
{
"input": "32 6\n25 77 141 148 157 159 192 196 198 244 245 255 332 392 414 457 466 524 575 603 629 700 738 782 838 841 845 847 870 945 984 985\n1 2 4 5 8 9 10 12 13 14 15 16 17 18 20 21 22 23 24 26 28 31 38 39 40 41 42 43 45 47 48 49",
"output": "Radewoosh"
},
{
"input": "5 1\n256 275 469 671 842\n7 9 14 17 26",
"output": "Limak"
},
{
"input": "2 1000\n1 2\n1 2",
"output": "Tie"
},
{
"input": "3 1\n1 50 809\n2 8 800",
"output": "Limak"
},
{
"input": "1 13\n866\n10",
"output": "Tie"
},
{
"input": "15 1\n9 11 66 128 199 323 376 386 393 555 585 718 935 960 971\n3 11 14 19 20 21 24 26 32 38 40 42 44 47 50",
"output": "Limak"
},
{
"input": "1 10\n546\n45",
"output": "Tie"
},
{
"input": "50 20\n21 43 51 99 117 119 158 167 175 190 196 244 250 316 335 375 391 403 423 428 451 457 460 480 487 522 539 559 566 584 598 602 604 616 626 666 675 730 771 787 828 841 861 867 886 889 898 970 986 991\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50",
"output": "Limak"
},
{
"input": "50 21\n13 20 22 38 62 84 118 135 141 152 170 175 194 218 227 229 232 253 260 263 278 313 329 357 396 402 422 452 454 533 575 576 580 594 624 644 653 671 676 759 789 811 816 823 831 833 856 924 933 987\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50",
"output": "Tie"
},
{
"input": "1 36\n312\n42",
"output": "Tie"
},
{
"input": "1 1000\n1\n1000",
"output": "Tie"
},
{
"input": "1 1\n1000\n1",
"output": "Tie"
},
{
"input": "50 35\n9 17 28 107 136 152 169 174 186 188 201 262 291 312 324 330 341 358 385 386 393 397 425 431 479 498 502 523 530 540 542 554 578 588 622 623 684 696 709 722 784 819 836 845 850 932 945 969 983 984\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50",
"output": "Tie"
},
{
"input": "50 20\n12 113 116 120 138 156 167 183 185 194 211 228 234 261 278 287 310 317 346 361 364 397 424 470 496 522 527 536 611 648 668 704 707 712 717 752 761 766 815 828 832 864 872 885 889 901 904 929 982 993\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50",
"output": "Limak"
}
] | 1,586,203,058
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 29
| 109
| 307,200
|
def ll():
return list(map(int,input().split()))
s =ll()
first_list = []
reverse_list = []
second_list = []
reverse_second_list = []
temp = 0
count = 0
inc = 0
inc2 = 0
for i in range(1):
n = ll()
first_list = first_list+n
reverse_list = first_list[::-1]
for t in range(1):
n = ll()
second_list = second_list+ n
reverse_second_list = second_list[::-1]
for j in range(len(first_list)):
inc +=second_list[j]
inc2+=reverse_second_list[j]
temp += max(0,first_list[j]-s[1]*inc)
count+=max(0,reverse_list[j]-s[1]*inc2)
if temp>count:
print("Limak")
elif count>temp:
print("Radewoosh")
else:
print("Tie")
|
Title: Bear and Reverse Radewoosh
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Limak and Radewoosh are going to compete against each other in the upcoming algorithmic contest. They are equally skilled but they won't solve problems in the same order.
There will be *n* problems. The *i*-th problem has initial score *p**i* and it takes exactly *t**i* minutes to solve it. Problems are sorted by difficulty — it's guaranteed that *p**i*<=<<=*p**i*<=+<=1 and *t**i*<=<<=*t**i*<=+<=1.
A constant *c* is given too, representing the speed of loosing points. Then, submitting the *i*-th problem at time *x* (*x* minutes after the start of the contest) gives *max*(0,<= *p**i*<=-<=*c*·*x*) points.
Limak is going to solve problems in order 1,<=2,<=...,<=*n* (sorted increasingly by *p**i*). Radewoosh is going to solve them in order *n*,<=*n*<=-<=1,<=...,<=1 (sorted decreasingly by *p**i*). Your task is to predict the outcome — print the name of the winner (person who gets more points at the end) or a word "Tie" in case of a tie.
You may assume that the duration of the competition is greater or equal than the sum of all *t**i*. That means both Limak and Radewoosh will accept all *n* problems.
Input Specification:
The first line contains two integers *n* and *c* (1<=≤<=*n*<=≤<=50,<=1<=≤<=*c*<=≤<=1000) — the number of problems and the constant representing the speed of loosing points.
The second line contains *n* integers *p*1,<=*p*2,<=...,<=*p**n* (1<=≤<=*p**i*<=≤<=1000,<=*p**i*<=<<=*p**i*<=+<=1) — initial scores.
The third line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=1000,<=*t**i*<=<<=*t**i*<=+<=1) where *t**i* denotes the number of minutes one needs to solve the *i*-th problem.
Output Specification:
Print "Limak" (without quotes) if Limak will get more points in total. Print "Radewoosh" (without quotes) if Radewoosh will get more points in total. Print "Tie" (without quotes) if Limak and Radewoosh will get the same total number of points.
Demo Input:
['3 2\n50 85 250\n10 15 25\n', '3 6\n50 85 250\n10 15 25\n', '8 1\n10 20 30 40 50 60 70 80\n8 10 58 63 71 72 75 76\n']
Demo Output:
['Limak\n', 'Radewoosh\n', 'Tie\n']
Note:
In the first sample, there are 3 problems. Limak solves them as follows:
1. Limak spends 10 minutes on the 1-st problem and he gets 50 - *c*·10 = 50 - 2·10 = 30 points. 1. Limak spends 15 minutes on the 2-nd problem so he submits it 10 + 15 = 25 minutes after the start of the contest. For the 2-nd problem he gets 85 - 2·25 = 35 points. 1. He spends 25 minutes on the 3-rd problem so he submits it 10 + 15 + 25 = 50 minutes after the start. For this problem he gets 250 - 2·50 = 150 points.
So, Limak got 30 + 35 + 150 = 215 points.
Radewoosh solves problem in the reversed order:
1. Radewoosh solves 3-rd problem after 25 minutes so he gets 250 - 2·25 = 200 points. 1. He spends 15 minutes on the 2-nd problem so he submits it 25 + 15 = 40 minutes after the start. He gets 85 - 2·40 = 5 points for this problem. 1. He spends 10 minutes on the 1-st problem so he submits it 25 + 15 + 10 = 50 minutes after the start. He gets *max*(0, 50 - 2·50) = *max*(0, - 50) = 0 points.
Radewoosh got 200 + 5 + 0 = 205 points in total. Limak has 215 points so Limak wins.
In the second sample, Limak will get 0 points for each problem and Radewoosh will first solve the hardest problem and he will get 250 - 6·25 = 100 points for that. Radewoosh will get 0 points for other two problems but he is the winner anyway.
In the third sample, Limak will get 2 points for the 1-st problem and 2 points for the 2-nd problem. Radewoosh will get 4 points for the 8-th problem. They won't get points for other problems and thus there is a tie because 2 + 2 = 4.
|
```python
def ll():
return list(map(int,input().split()))
s =ll()
first_list = []
reverse_list = []
second_list = []
reverse_second_list = []
temp = 0
count = 0
inc = 0
inc2 = 0
for i in range(1):
n = ll()
first_list = first_list+n
reverse_list = first_list[::-1]
for t in range(1):
n = ll()
second_list = second_list+ n
reverse_second_list = second_list[::-1]
for j in range(len(first_list)):
inc +=second_list[j]
inc2+=reverse_second_list[j]
temp += max(0,first_list[j]-s[1]*inc)
count+=max(0,reverse_list[j]-s[1]*inc2)
if temp>count:
print("Limak")
elif count>temp:
print("Radewoosh")
else:
print("Tie")
```
| 3
|
|
294
|
A
|
Shaass and Oskols
|
PROGRAMMING
| 800
|
[
"implementation",
"math"
] | null | null |
Shaass has decided to hunt some birds. There are *n* horizontal electricity wires aligned parallel to each other. Wires are numbered 1 to *n* from top to bottom. On each wire there are some oskols sitting next to each other. Oskol is the name of a delicious kind of birds in Shaass's territory. Supposed there are *a**i* oskols sitting on the *i*-th wire.
Sometimes Shaass shots one of the birds and the bird dies (suppose that this bird sat at the *i*-th wire). Consequently all the birds on the *i*-th wire to the left of the dead bird get scared and jump up on the wire number *i*<=-<=1, if there exists no upper wire they fly away. Also all the birds to the right of the dead bird jump down on wire number *i*<=+<=1, if there exists no such wire they fly away.
Shaass has shot *m* birds. You're given the initial number of birds on each wire, tell him how many birds are sitting on each wire after the shots.
|
The first line of the input contains an integer *n*, (1<=≤<=*n*<=≤<=100). The next line contains a list of space-separated integers *a*1,<=*a*2,<=...,<=*a**n*, (0<=≤<=*a**i*<=≤<=100).
The third line contains an integer *m*, (0<=≤<=*m*<=≤<=100). Each of the next *m* lines contains two integers *x**i* and *y**i*. The integers mean that for the *i*-th time Shaass shoot the *y**i*-th (from left) bird on the *x**i*-th wire, (1<=≤<=*x**i*<=≤<=*n*,<=1<=≤<=*y**i*). It's guaranteed there will be at least *y**i* birds on the *x**i*-th wire at that moment.
|
On the *i*-th line of the output print the number of birds on the *i*-th wire.
|
[
"5\n10 10 10 10 10\n5\n2 5\n3 13\n2 12\n1 13\n4 6\n",
"3\n2 4 1\n1\n2 2\n"
] |
[
"0\n12\n5\n0\n16\n",
"3\n0\n3\n"
] |
none
| 500
|
[
{
"input": "5\n10 10 10 10 10\n5\n2 5\n3 13\n2 12\n1 13\n4 6",
"output": "0\n12\n5\n0\n16"
},
{
"input": "3\n2 4 1\n1\n2 2",
"output": "3\n0\n3"
},
{
"input": "5\n58 51 45 27 48\n5\n4 9\n5 15\n4 5\n5 8\n1 43",
"output": "0\n66\n57\n7\n0"
},
{
"input": "10\n48 53 10 28 91 56 81 2 67 52\n2\n2 40\n6 51",
"output": "87\n0\n23\n28\n141\n0\n86\n2\n67\n52"
},
{
"input": "2\n72 45\n6\n1 69\n2 41\n1 19\n2 7\n1 5\n2 1",
"output": "0\n0"
},
{
"input": "10\n95 54 36 39 98 30 19 24 14 12\n3\n9 5\n8 15\n7 5",
"output": "95\n54\n36\n39\n98\n34\n0\n28\n13\n21"
},
{
"input": "100\n95 15 25 18 64 62 23 59 70 84 50 26 87 35 75 86 0 22 77 60 66 41 21 9 75 50 25 3 69 14 39 68 64 46 59 99 2 0 21 76 90 12 61 42 6 91 36 39 47 41 93 81 66 57 70 36 68 89 52 1 19 93 67 22 76 20 8 81 98 18 100 73 61 93 75 80 53 72 40 20 2 86 33 59 27 16 11 26 55 44 47 55 94 58 2 55 82 22 9 95\n0",
"output": "95\n15\n25\n18\n64\n62\n23\n59\n70\n84\n50\n26\n87\n35\n75\n86\n0\n22\n77\n60\n66\n41\n21\n9\n75\n50\n25\n3\n69\n14\n39\n68\n64\n46\n59\n99\n2\n0\n21\n76\n90\n12\n61\n42\n6\n91\n36\n39\n47\n41\n93\n81\n66\n57\n70\n36\n68\n89\n52\n1\n19\n93\n67\n22\n76\n20\n8\n81\n98\n18\n100\n73\n61\n93\n75\n80\n53\n72\n40\n20\n2\n86\n33\n59\n27\n16\n11\n26\n55\n44\n47\n55\n94\n58\n2\n55\n82\n22\n9\n95"
},
{
"input": "1\n100\n1\n1 100",
"output": "0"
},
{
"input": "1\n100\n1\n1 1",
"output": "0"
},
{
"input": "1\n50\n1\n1 25",
"output": "0"
},
{
"input": "2\n50 0\n1\n1 1",
"output": "0\n49"
},
{
"input": "1\n10\n0",
"output": "10"
}
] | 1,692,164,212
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 31
| 124
| 0
|
n = int(input())
a = input().split(" ")
a = [int(i) for i in a]
a.insert(0, 0)
a.append(0)
m = int(input())
for i in range(m):
x, y = input().split(" ")
x, y = int(x), int(y)
a[x-1] += y - 1
a[x+1] += a[x] - y
a[x] = 0
for i in range(1,len(a)-1):
print(a[i])
|
Title: Shaass and Oskols
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Shaass has decided to hunt some birds. There are *n* horizontal electricity wires aligned parallel to each other. Wires are numbered 1 to *n* from top to bottom. On each wire there are some oskols sitting next to each other. Oskol is the name of a delicious kind of birds in Shaass's territory. Supposed there are *a**i* oskols sitting on the *i*-th wire.
Sometimes Shaass shots one of the birds and the bird dies (suppose that this bird sat at the *i*-th wire). Consequently all the birds on the *i*-th wire to the left of the dead bird get scared and jump up on the wire number *i*<=-<=1, if there exists no upper wire they fly away. Also all the birds to the right of the dead bird jump down on wire number *i*<=+<=1, if there exists no such wire they fly away.
Shaass has shot *m* birds. You're given the initial number of birds on each wire, tell him how many birds are sitting on each wire after the shots.
Input Specification:
The first line of the input contains an integer *n*, (1<=≤<=*n*<=≤<=100). The next line contains a list of space-separated integers *a*1,<=*a*2,<=...,<=*a**n*, (0<=≤<=*a**i*<=≤<=100).
The third line contains an integer *m*, (0<=≤<=*m*<=≤<=100). Each of the next *m* lines contains two integers *x**i* and *y**i*. The integers mean that for the *i*-th time Shaass shoot the *y**i*-th (from left) bird on the *x**i*-th wire, (1<=≤<=*x**i*<=≤<=*n*,<=1<=≤<=*y**i*). It's guaranteed there will be at least *y**i* birds on the *x**i*-th wire at that moment.
Output Specification:
On the *i*-th line of the output print the number of birds on the *i*-th wire.
Demo Input:
['5\n10 10 10 10 10\n5\n2 5\n3 13\n2 12\n1 13\n4 6\n', '3\n2 4 1\n1\n2 2\n']
Demo Output:
['0\n12\n5\n0\n16\n', '3\n0\n3\n']
Note:
none
|
```python
n = int(input())
a = input().split(" ")
a = [int(i) for i in a]
a.insert(0, 0)
a.append(0)
m = int(input())
for i in range(m):
x, y = input().split(" ")
x, y = int(x), int(y)
a[x-1] += y - 1
a[x+1] += a[x] - y
a[x] = 0
for i in range(1,len(a)-1):
print(a[i])
```
| 3
|
|
961
|
A
|
Tetris
|
PROGRAMMING
| 900
|
[
"implementation"
] | null | null |
You are given a following process.
There is a platform with $n$ columns. $1 \times 1$ squares are appearing one after another in some columns on this platform. If there are no squares in the column, a square will occupy the bottom row. Otherwise a square will appear at the top of the highest square of this column.
When all of the $n$ columns have at least one square in them, the bottom row is being removed. You will receive $1$ point for this, and all the squares left will fall down one row.
You task is to calculate the amount of points you will receive.
|
The first line of input contain 2 integer numbers $n$ and $m$ ($1 \le n, m \le 1000$) — the length of the platform and the number of the squares.
The next line contain $m$ integer numbers $c_1, c_2, \dots, c_m$ ($1 \le c_i \le n$) — column in which $i$-th square will appear.
|
Print one integer — the amount of points you will receive.
|
[
"3 9\n1 1 2 2 2 3 1 2 3\n"
] |
[
"2\n"
] |
In the sample case the answer will be equal to $2$ because after the appearing of $6$-th square will be removed one row (counts of the squares on the platform will look like $[2~ 3~ 1]$, and after removing one row will be $[1~ 2~ 0]$).
After the appearing of $9$-th square counts will be $[2~ 3~ 1]$, and after removing one row it will look like $[1~ 2~ 0]$.
So the answer will be equal to $2$.
| 0
|
[
{
"input": "3 9\n1 1 2 2 2 3 1 2 3",
"output": "2"
},
{
"input": "1 7\n1 1 1 1 1 1 1",
"output": "7"
},
{
"input": "1 1\n1",
"output": "1"
},
{
"input": "3 5\n1 1 1 2 3",
"output": "1"
},
{
"input": "4 6\n4 4 4 4 4 4",
"output": "0"
},
{
"input": "4 6\n2 3 4 4 4 4",
"output": "0"
},
{
"input": "3 12\n1 1 1 1 2 2 2 2 3 3 3 3",
"output": "4"
},
{
"input": "8 8\n2 2 3 4 5 6 7 8",
"output": "0"
},
{
"input": "100 1\n50",
"output": "0"
},
{
"input": "2 1\n2",
"output": "0"
},
{
"input": "2 1\n1",
"output": "0"
},
{
"input": "2 4\n1 2 1 1",
"output": "1"
},
{
"input": "3 4\n3 2 2 2",
"output": "0"
},
{
"input": "2 2\n2 2",
"output": "0"
},
{
"input": "2 5\n2 1 1 2 1",
"output": "2"
},
{
"input": "15 3\n13 14 15",
"output": "0"
},
{
"input": "4 9\n1 2 3 1 2 3 1 2 3",
"output": "0"
},
{
"input": "100 3\n1 2 3",
"output": "0"
},
{
"input": "1000 10\n999 999 998 34 454 546 343 35 34 1000",
"output": "0"
},
{
"input": "4 2\n1 2",
"output": "0"
}
] | 1,660,932,895
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 29
| 46
| 0
|
class Tetris:
def __init__(self, length):
# Inicializar tabuleiro com zeros
self.board = [0] * length
# Inicializar pontuação
self.points = 0
def addBlock(self, column):
self.board[column] += 1
def checkBoard(self):
while not 0 in self.board:
self.points += 1
self.board = [n-1 for n in self.board]
return self.points
if __name__ == "__main__":
# Ler primeira linha da entrada
line1 = input().split()
length = int(line1[0])
# Instanciar classe
tetris = Tetris(length)
# Ler segunda linha da entrada
line2 = input().split()
for column in line2:
# Computar bloco
tetris.addBlock(int(column)-1)
# Computar pontuação
print(tetris.checkBoard())
|
Title: Tetris
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a following process.
There is a platform with $n$ columns. $1 \times 1$ squares are appearing one after another in some columns on this platform. If there are no squares in the column, a square will occupy the bottom row. Otherwise a square will appear at the top of the highest square of this column.
When all of the $n$ columns have at least one square in them, the bottom row is being removed. You will receive $1$ point for this, and all the squares left will fall down one row.
You task is to calculate the amount of points you will receive.
Input Specification:
The first line of input contain 2 integer numbers $n$ and $m$ ($1 \le n, m \le 1000$) — the length of the platform and the number of the squares.
The next line contain $m$ integer numbers $c_1, c_2, \dots, c_m$ ($1 \le c_i \le n$) — column in which $i$-th square will appear.
Output Specification:
Print one integer — the amount of points you will receive.
Demo Input:
['3 9\n1 1 2 2 2 3 1 2 3\n']
Demo Output:
['2\n']
Note:
In the sample case the answer will be equal to $2$ because after the appearing of $6$-th square will be removed one row (counts of the squares on the platform will look like $[2~ 3~ 1]$, and after removing one row will be $[1~ 2~ 0]$).
After the appearing of $9$-th square counts will be $[2~ 3~ 1]$, and after removing one row it will look like $[1~ 2~ 0]$.
So the answer will be equal to $2$.
|
```python
class Tetris:
def __init__(self, length):
# Inicializar tabuleiro com zeros
self.board = [0] * length
# Inicializar pontuação
self.points = 0
def addBlock(self, column):
self.board[column] += 1
def checkBoard(self):
while not 0 in self.board:
self.points += 1
self.board = [n-1 for n in self.board]
return self.points
if __name__ == "__main__":
# Ler primeira linha da entrada
line1 = input().split()
length = int(line1[0])
# Instanciar classe
tetris = Tetris(length)
# Ler segunda linha da entrada
line2 = input().split()
for column in line2:
# Computar bloco
tetris.addBlock(int(column)-1)
# Computar pontuação
print(tetris.checkBoard())
```
| 3
|
|
41
|
A
|
Translation
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] |
A. Translation
|
2
|
256
|
The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly.
|
The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols.
|
If the word *t* is a word *s*, written reversely, print YES, otherwise print NO.
|
[
"code\nedoc\n",
"abb\naba\n",
"code\ncode\n"
] |
[
"YES\n",
"NO\n",
"NO\n"
] |
none
| 500
|
[
{
"input": "code\nedoc",
"output": "YES"
},
{
"input": "abb\naba",
"output": "NO"
},
{
"input": "code\ncode",
"output": "NO"
},
{
"input": "abacaba\nabacaba",
"output": "YES"
},
{
"input": "q\nq",
"output": "YES"
},
{
"input": "asrgdfngfnmfgnhweratgjkk\nasrgdfngfnmfgnhweratgjkk",
"output": "NO"
},
{
"input": "z\na",
"output": "NO"
},
{
"input": "asd\ndsa",
"output": "YES"
},
{
"input": "abcdef\nfecdba",
"output": "NO"
},
{
"input": "ywjjbirapvskozubvxoemscfwl\ngnduubaogtfaiowjizlvjcu",
"output": "NO"
},
{
"input": "mfrmqxtzvgaeuleubcmcxcfqyruwzenguhgrmkuhdgnhgtgkdszwqyd\nmfxufheiperjnhyczclkmzyhcxntdfskzkzdwzzujdinf",
"output": "NO"
},
{
"input": "bnbnemvybqizywlnghlykniaxxxlkhftppbdeqpesrtgkcpoeqowjwhrylpsziiwcldodcoonpimudvrxejjo\ntiynnekmlalogyvrgptbinkoqdwzuiyjlrldxhzjmmp",
"output": "NO"
},
{
"input": "pwlpubwyhzqvcitemnhvvwkmwcaawjvdiwtoxyhbhbxerlypelevasmelpfqwjk\nstruuzebbcenziscuoecywugxncdwzyfozhljjyizpqcgkyonyetarcpwkqhuugsqjuixsxptmbnlfupdcfigacdhhrzb",
"output": "NO"
},
{
"input": "gdvqjoyxnkypfvdxssgrihnwxkeojmnpdeobpecytkbdwujqfjtxsqspxvxpqioyfagzjxupqqzpgnpnpxcuipweunqch\nkkqkiwwasbhezqcfeceyngcyuogrkhqecwsyerdniqiocjehrpkljiljophqhyaiefjpavoom",
"output": "NO"
},
{
"input": "umeszdawsvgkjhlqwzents\nhxqhdungbylhnikwviuh",
"output": "NO"
},
{
"input": "juotpscvyfmgntshcealgbsrwwksgrwnrrbyaqqsxdlzhkbugdyx\nibqvffmfktyipgiopznsqtrtxiijntdbgyy",
"output": "NO"
},
{
"input": "zbwueheveouatecaglziqmudxemhrsozmaujrwlqmppzoumxhamwugedikvkblvmxwuofmpafdprbcftew\nulczwrqhctbtbxrhhodwbcxwimncnexosksujlisgclllxokrsbnozthajnnlilyffmsyko",
"output": "NO"
},
{
"input": "nkgwuugukzcv\nqktnpxedwxpxkrxdvgmfgoxkdfpbzvwsduyiybynbkouonhvmzakeiruhfmvrktghadbfkmwxduoqv",
"output": "NO"
},
{
"input": "incenvizhqpcenhjhehvjvgbsnfixbatrrjstxjzhlmdmxijztphxbrldlqwdfimweepkggzcxsrwelodpnryntepioqpvk\ndhjbjjftlvnxibkklxquwmzhjfvnmwpapdrslioxisbyhhfymyiaqhlgecpxamqnocizwxniubrmpyubvpenoukhcobkdojlybxd",
"output": "NO"
},
{
"input": "w\nw",
"output": "YES"
},
{
"input": "vz\nzv",
"output": "YES"
},
{
"input": "ry\nyr",
"output": "YES"
},
{
"input": "xou\nuox",
"output": "YES"
},
{
"input": "axg\ngax",
"output": "NO"
},
{
"input": "zdsl\nlsdz",
"output": "YES"
},
{
"input": "kudl\nldku",
"output": "NO"
},
{
"input": "zzlzwnqlcl\nlclqnwzlzz",
"output": "YES"
},
{
"input": "vzzgicnzqooejpjzads\nsdazjpjeooqzncigzzv",
"output": "YES"
},
{
"input": "raqhmvmzuwaykjpyxsykr\nxkysrypjkyawuzmvmhqar",
"output": "NO"
},
{
"input": "ngedczubzdcqbxksnxuavdjaqtmdwncjnoaicvmodcqvhfezew\nwezefhvqcdomvciaonjcnwdmtqajdvauxnskxbqcdzbuzcdegn",
"output": "YES"
},
{
"input": "muooqttvrrljcxbroizkymuidvfmhhsjtumksdkcbwwpfqdyvxtrlymofendqvznzlmim\nmimlznzvqdnefomylrtxvydqfpwwbckdskmutjshhmfvdiumykziorbxcjlrrvttqooum",
"output": "YES"
},
{
"input": "vxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaivg\ngviayyikkitmuomcpiakhbxszgbnhvwyzkftwoagzixaearxpjacrnvpvbuzenvovehkmmxvblqyxvctroddksdsgebcmlluqpxv",
"output": "YES"
},
{
"input": "mnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfdc\ncdfmkdgrdptkpewbsqvszipgxvgvuiuzbkkwuowbafkikgvnqdkxnayzdjygvezmtsgywnupocdntipiyiorblqkrzjpzatxahnm",
"output": "NO"
},
{
"input": "dgxmzbqofstzcdgthbaewbwocowvhqpinehpjatnnbrijcolvsatbblsrxabzrpszoiecpwhfjmwuhqrapvtcgvikuxtzbftydkw\nwkdytfbztxukivgctvparqhuwmjfhwpceiozsprzbaxrslbbqasvlocjirbnntajphenipthvwocowbweabhtgdcztsfoqbzmxgd",
"output": "NO"
},
{
"input": "gxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwgeh\nhegwxvocotmzstqfbmpjvijgkcyodlxyjawrpkczpmdspsuhoiruavnnnuwvtwohglkdxjetshkboalvzqbgjgthoteceixioxg",
"output": "YES"
},
{
"input": "sihxuwvmaambplxvjfoskinghzicyfqebjtkysotattkahssumfcgrkheotdxwjckpvapbkaepqrxseyfrwtyaycmrzsrsngkh\nhkgnsrszrmcyaytwrfyesxrqpeakbpavpkcjwxdtoehkrgcfmusshakttatosyktjbeqfycizhgniksofjvxlpbmaamvwuxhis",
"output": "YES"
},
{
"input": "ycnahksbughnonldzrhkysujmylcgcfuludjvjiahtkyzqvkopzqcnwhltbzfugzojqkjjlggmvnultascmygelkiktmfieok\nkoeifmtkiklegkmcsatlunvmggkjjlqjozgufzbtlhwncqzpokvqzykthaijvjdulufcgclymjusyyhrzdlnonhgubskhancy",
"output": "NO"
},
{
"input": "wbqasaehtkfojruzyhrlgwmtyiovmzyfifslvlemhqheyaelzwnthrenjsbmntwaoryzwfbxmscmypvxlfmzpnkkjlvwvmtz\nztmvwvljkknpzmflxvpymcsmxbfwzyroawtnmbsjnerhtnwzleayehqhmelvlsfifyzmvoiytmwglrhyzurjofktheasaqbw",
"output": "YES"
},
{
"input": "imippqurprbhfugngtgifelytadegwrgaefnfhbjjnmzikvjaccotqzemufqieqldgnbmviisgkynzeldlhqxuqphjfmyij\njiymfjhpquxqhldleznykgsiivmbngdlqeiqfumezqtoccajvkizmnjjbhfnfeagrwgedatylefigtgngufhbrpruqppimi",
"output": "YES"
},
{
"input": "bikydffiuisckpvzqlteqfhegsagimodb\nbdomigasgehfqetlqzvpkcsiuiffdykib",
"output": "YES"
}
] | 1,620,025,112
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 40
| 124
| 0
|
s = input()
t = input()
reversed = (t[::-1])
if s == reversed:
print ("YES")
else:
print ("NO")
|
Title: Translation
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly.
Input Specification:
The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols.
Output Specification:
If the word *t* is a word *s*, written reversely, print YES, otherwise print NO.
Demo Input:
['code\nedoc\n', 'abb\naba\n', 'code\ncode\n']
Demo Output:
['YES\n', 'NO\n', 'NO\n']
Note:
none
|
```python
s = input()
t = input()
reversed = (t[::-1])
if s == reversed:
print ("YES")
else:
print ("NO")
```
| 3.969
|
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