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|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
218
|
B
|
Airport
|
PROGRAMMING
| 1,100
|
[
"implementation"
] | null | null |
Lolek and Bolek are about to travel abroad by plane. The local airport has a special "Choose Your Plane" offer. The offer's conditions are as follows:
- it is up to a passenger to choose a plane to fly on; - if the chosen plane has *x* (*x*<=><=0) empty seats at the given moment, then the ticket for such a plane costs *x* zlotys (units of Polish currency).
The only ticket office of the airport already has a queue of *n* passengers in front of it. Lolek and Bolek have not stood in the queue yet, but they are already wondering what is the maximum and the minimum number of zlotys the airport administration can earn if all *n* passengers buy tickets according to the conditions of this offer?
The passengers buy tickets in turn, the first person in the queue goes first, then goes the second one, and so on up to *n*-th person.
|
The first line contains two integers *n* and *m* (1<=β€<=*n*,<=*m*<=β€<=1000) β the number of passengers in the queue and the number of planes in the airport, correspondingly. The next line contains *m* integers *a*1,<=*a*2,<=...,<=*a**m* (1<=β€<=*a**i*<=β€<=1000) β *a**i* stands for the number of empty seats in the *i*-th plane before the ticket office starts selling tickets.
The numbers in the lines are separated by a space. It is guaranteed that there are at least *n* empty seats in total.
|
Print two integers β the maximum and the minimum number of zlotys that the airport administration can earn, correspondingly.
|
[
"4 3\n2 1 1\n",
"4 3\n2 2 2\n"
] |
[
"5 5\n",
"7 6\n"
] |
In the first test sample the number of passengers is equal to the number of empty seats, so regardless of the way the planes are chosen, the administration will earn the same sum.
In the second sample the sum is maximized if the 1-st person in the queue buys a ticket to the 1-st plane, the 2-nd person β to the 2-nd plane, the 3-rd person β to the 3-rd plane, the 4-th person β to the 1-st plane. The sum is minimized if the 1-st person in the queue buys a ticket to the 1-st plane, the 2-nd person β to the 1-st plane, the 3-rd person β to the 2-nd plane, the 4-th person β to the 2-nd plane.
| 500
|
[
{
"input": "4 3\n2 1 1",
"output": "5 5"
},
{
"input": "4 3\n2 2 2",
"output": "7 6"
},
{
"input": "10 5\n10 3 3 1 2",
"output": "58 26"
},
{
"input": "10 1\n10",
"output": "55 55"
},
{
"input": "10 1\n100",
"output": "955 955"
},
{
"input": "10 2\n4 7",
"output": "37 37"
},
{
"input": "40 10\n1 2 3 4 5 6 7 10 10 10",
"output": "223 158"
},
{
"input": "1 1\n6",
"output": "6 6"
},
{
"input": "1 2\n10 9",
"output": "10 9"
},
{
"input": "2 1\n7",
"output": "13 13"
},
{
"input": "2 2\n7 2",
"output": "13 3"
},
{
"input": "3 2\n4 7",
"output": "18 9"
},
{
"input": "3 3\n2 1 1",
"output": "4 4"
},
{
"input": "3 3\n2 1 1",
"output": "4 4"
},
{
"input": "10 10\n3 1 2 2 1 1 2 1 2 3",
"output": "20 13"
},
{
"input": "10 2\n7 3",
"output": "34 34"
},
{
"input": "10 1\n19",
"output": "145 145"
},
{
"input": "100 3\n29 36 35",
"output": "1731 1731"
},
{
"input": "100 5\n3 38 36 35 2",
"output": "2019 1941"
},
{
"input": "510 132\n50 76 77 69 94 30 47 65 14 62 18 121 26 35 49 17 105 93 47 16 78 3 7 74 7 37 30 36 30 83 71 113 7 58 86 10 65 57 34 102 55 44 43 47 106 44 115 75 109 70 47 45 16 57 62 55 20 88 74 40 45 84 41 1 9 53 65 25 67 31 115 2 63 51 123 70 65 65 18 14 75 14 103 26 117 105 36 104 81 37 35 61 44 90 71 70 88 89 26 21 64 77 89 16 87 99 13 79 27 3 46 120 116 11 14 17 32 70 113 94 108 57 29 100 53 48 44 29 70 30 32 62",
"output": "50279 5479"
},
{
"input": "510 123\n5 2 3 2 5 7 2 3 1 3 6 6 3 1 5 3 5 6 2 2 1 5 5 5 2 2 3 1 6 3 5 8 4 6 1 5 4 5 1 6 5 5 3 6 4 1 6 1 3 5 2 7 5 2 4 4 5 6 5 5 4 3 4 6 5 4 4 3 5 8 5 5 6 3 1 7 4 4 3 3 5 3 6 3 3 6 2 5 3 2 4 5 4 5 2 2 4 4 4 7 3 4 6 5 3 6 4 7 1 6 5 7 6 5 7 3 7 4 4 1 6 6 4",
"output": "1501 1501"
},
{
"input": "610 33\n15 44 8 8 17 11 39 39 38 25 17 36 17 25 21 37 10 11 34 30 29 50 29 50 4 20 32 13 41 14 2 11 2",
"output": "12204 8871"
}
] | 1,596,369,657
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 33
| 278
| 6,963,200
|
n, m = map(int,input().split())
l = list(map(int,input().split()))
def g(f):
A = l[:]; s = 0
for _ in range(n):
i = A.index(f(A)); s += A[i]
if A[i] > 1: A[i] -= 1;
else:
A.pop(i)
return s
print (g(max), g(min))
|
Title: Airport
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Lolek and Bolek are about to travel abroad by plane. The local airport has a special "Choose Your Plane" offer. The offer's conditions are as follows:
- it is up to a passenger to choose a plane to fly on; - if the chosen plane has *x* (*x*<=><=0) empty seats at the given moment, then the ticket for such a plane costs *x* zlotys (units of Polish currency).
The only ticket office of the airport already has a queue of *n* passengers in front of it. Lolek and Bolek have not stood in the queue yet, but they are already wondering what is the maximum and the minimum number of zlotys the airport administration can earn if all *n* passengers buy tickets according to the conditions of this offer?
The passengers buy tickets in turn, the first person in the queue goes first, then goes the second one, and so on up to *n*-th person.
Input Specification:
The first line contains two integers *n* and *m* (1<=β€<=*n*,<=*m*<=β€<=1000) β the number of passengers in the queue and the number of planes in the airport, correspondingly. The next line contains *m* integers *a*1,<=*a*2,<=...,<=*a**m* (1<=β€<=*a**i*<=β€<=1000) β *a**i* stands for the number of empty seats in the *i*-th plane before the ticket office starts selling tickets.
The numbers in the lines are separated by a space. It is guaranteed that there are at least *n* empty seats in total.
Output Specification:
Print two integers β the maximum and the minimum number of zlotys that the airport administration can earn, correspondingly.
Demo Input:
['4 3\n2 1 1\n', '4 3\n2 2 2\n']
Demo Output:
['5 5\n', '7 6\n']
Note:
In the first test sample the number of passengers is equal to the number of empty seats, so regardless of the way the planes are chosen, the administration will earn the same sum.
In the second sample the sum is maximized if the 1-st person in the queue buys a ticket to the 1-st plane, the 2-nd person β to the 2-nd plane, the 3-rd person β to the 3-rd plane, the 4-th person β to the 1-st plane. The sum is minimized if the 1-st person in the queue buys a ticket to the 1-st plane, the 2-nd person β to the 1-st plane, the 3-rd person β to the 2-nd plane, the 4-th person β to the 2-nd plane.
|
```python
n, m = map(int,input().split())
l = list(map(int,input().split()))
def g(f):
A = l[:]; s = 0
for _ in range(n):
i = A.index(f(A)); s += A[i]
if A[i] > 1: A[i] -= 1;
else:
A.pop(i)
return s
print (g(max), g(min))
```
| 3
|
|
431
|
A
|
Black Square
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Quite recently, a very smart student named Jury decided that lectures are boring, so he downloaded a game called "Black Square" on his super cool touchscreen phone.
In this game, the phone's screen is divided into four vertical strips. Each second, a black square appears on some of the strips. According to the rules of the game, Jury must use this second to touch the corresponding strip to make the square go away. As Jury is both smart and lazy, he counted that he wastes exactly *a**i* calories on touching the *i*-th strip.
You've got a string *s*, describing the process of the game and numbers *a*1,<=*a*2,<=*a*3,<=*a*4. Calculate how many calories Jury needs to destroy all the squares?
|
The first line contains four space-separated integers *a*1, *a*2, *a*3, *a*4 (0<=β€<=*a*1,<=*a*2,<=*a*3,<=*a*4<=β€<=104).
The second line contains string *s* (1<=β€<=|*s*|<=β€<=105), where the *Ρ*-th character of the string equals "1", if on the *i*-th second of the game the square appears on the first strip, "2", if it appears on the second strip, "3", if it appears on the third strip, "4", if it appears on the fourth strip.
|
Print a single integer β the total number of calories that Jury wastes.
|
[
"1 2 3 4\n123214\n",
"1 5 3 2\n11221\n"
] |
[
"13\n",
"13\n"
] |
none
| 500
|
[
{
"input": "1 2 3 4\n123214",
"output": "13"
},
{
"input": "1 5 3 2\n11221",
"output": "13"
},
{
"input": "5 5 5 1\n3422",
"output": "16"
},
{
"input": "4 3 2 1\n2",
"output": "3"
},
{
"input": "5651 6882 6954 4733\n2442313421",
"output": "60055"
},
{
"input": "0 0 0 0\n4132",
"output": "0"
},
{
"input": "3163 5778 83 7640\n11141442444",
"output": "64270"
},
{
"input": "1809 1302 7164 6122\n3144121413113111223311232232114144321414421243443243422322144324121433444342231344234443332241322442",
"output": "420780"
},
{
"input": "0 0 0 0\n1",
"output": "0"
},
{
"input": "1 2 3 4\n4",
"output": "4"
},
{
"input": "2343 7653 1242 5432\n1",
"output": "2343"
},
{
"input": "2343 7653 1242 5432\n2",
"output": "7653"
},
{
"input": "2343 7653 1242 5432\n3",
"output": "1242"
},
{
"input": "2343 7653 1242 5432\n4",
"output": "5432"
},
{
"input": "1 2 3 4\n123412",
"output": "13"
},
{
"input": "50 50 50 50\n11111111111111111111111111111111111111111111111111111",
"output": "2650"
},
{
"input": "1 2 3 4\n11111111111111111111111111111111111111111111111111",
"output": "50"
},
{
"input": "1 2 3 4\n23123231321231231231231231221232123121312321",
"output": "87"
},
{
"input": "1 2 3 4\n1111111111111222222222233333333333444444444444444",
"output": "126"
},
{
"input": "2 3 1 4\n121321232412342112312313213123123412131231231232",
"output": "105"
}
] | 1,674,062,410
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 49
| 78
| 0
|
n = list(map(int, input().strip().split()))
m = input()
s = 0
for i in range(len(m)):
s += n[int(m[i])-1]
print(s)
|
Title: Black Square
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Quite recently, a very smart student named Jury decided that lectures are boring, so he downloaded a game called "Black Square" on his super cool touchscreen phone.
In this game, the phone's screen is divided into four vertical strips. Each second, a black square appears on some of the strips. According to the rules of the game, Jury must use this second to touch the corresponding strip to make the square go away. As Jury is both smart and lazy, he counted that he wastes exactly *a**i* calories on touching the *i*-th strip.
You've got a string *s*, describing the process of the game and numbers *a*1,<=*a*2,<=*a*3,<=*a*4. Calculate how many calories Jury needs to destroy all the squares?
Input Specification:
The first line contains four space-separated integers *a*1, *a*2, *a*3, *a*4 (0<=β€<=*a*1,<=*a*2,<=*a*3,<=*a*4<=β€<=104).
The second line contains string *s* (1<=β€<=|*s*|<=β€<=105), where the *Ρ*-th character of the string equals "1", if on the *i*-th second of the game the square appears on the first strip, "2", if it appears on the second strip, "3", if it appears on the third strip, "4", if it appears on the fourth strip.
Output Specification:
Print a single integer β the total number of calories that Jury wastes.
Demo Input:
['1 2 3 4\n123214\n', '1 5 3 2\n11221\n']
Demo Output:
['13\n', '13\n']
Note:
none
|
```python
n = list(map(int, input().strip().split()))
m = input()
s = 0
for i in range(len(m)):
s += n[int(m[i])-1]
print(s)
```
| 3
|
|
719
|
A
|
Vitya in the Countryside
|
PROGRAMMING
| 1,100
|
[
"implementation"
] | null | null |
Every summer Vitya comes to visit his grandmother in the countryside. This summer, he got a huge wart. Every grandma knows that one should treat warts when the moon goes down. Thus, Vitya has to catch the moment when the moon is down.
Moon cycle lasts 30 days. The size of the visible part of the moon (in Vitya's units) for each day is 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, and then cycle repeats, thus after the second 1 again goes 0.
As there is no internet in the countryside, Vitya has been watching the moon for *n* consecutive days and for each of these days he wrote down the size of the visible part of the moon. Help him find out whether the moon will be up or down next day, or this cannot be determined by the data he has.
|
The first line of the input contains a single integer *n* (1<=β€<=*n*<=β€<=92)Β β the number of consecutive days Vitya was watching the size of the visible part of the moon.
The second line contains *n* integers *a**i* (0<=β€<=*a**i*<=β€<=15)Β β Vitya's records.
It's guaranteed that the input data is consistent.
|
If Vitya can be sure that the size of visible part of the moon on day *n*<=+<=1 will be less than the size of the visible part on day *n*, then print "DOWN" at the only line of the output. If he might be sure that the size of the visible part will increase, then print "UP". If it's impossible to determine what exactly will happen with the moon, print -1.
|
[
"5\n3 4 5 6 7\n",
"7\n12 13 14 15 14 13 12\n",
"1\n8\n"
] |
[
"UP\n",
"DOWN\n",
"-1\n"
] |
In the first sample, the size of the moon on the next day will be equal to 8, thus the answer is "UP".
In the second sample, the size of the moon on the next day will be 11, thus the answer is "DOWN".
In the third sample, there is no way to determine whether the size of the moon on the next day will be 7 or 9, thus the answer is -1.
| 500
|
[
{
"input": "5\n3 4 5 6 7",
"output": "UP"
},
{
"input": "7\n12 13 14 15 14 13 12",
"output": "DOWN"
},
{
"input": "1\n8",
"output": "-1"
},
{
"input": "44\n7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12 11 10",
"output": "DOWN"
},
{
"input": "92\n3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4",
"output": "UP"
},
{
"input": "6\n10 11 12 13 14 15",
"output": "DOWN"
},
{
"input": "27\n11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15",
"output": "DOWN"
},
{
"input": "6\n8 7 6 5 4 3",
"output": "DOWN"
},
{
"input": "27\n14 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10",
"output": "UP"
},
{
"input": "79\n7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7 6 5",
"output": "DOWN"
},
{
"input": "25\n1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7",
"output": "DOWN"
},
{
"input": "21\n3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7",
"output": "DOWN"
},
{
"input": "56\n1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7 6",
"output": "DOWN"
},
{
"input": "19\n4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14",
"output": "UP"
},
{
"input": "79\n5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13",
"output": "UP"
},
{
"input": "87\n14 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10",
"output": "UP"
},
{
"input": "13\n10 9 8 7 6 5 4 3 2 1 0 1 2",
"output": "UP"
},
{
"input": "2\n8 9",
"output": "UP"
},
{
"input": "3\n10 11 12",
"output": "UP"
},
{
"input": "1\n1",
"output": "-1"
},
{
"input": "1\n2",
"output": "-1"
},
{
"input": "1\n3",
"output": "-1"
},
{
"input": "1\n4",
"output": "-1"
},
{
"input": "1\n5",
"output": "-1"
},
{
"input": "1\n6",
"output": "-1"
},
{
"input": "1\n7",
"output": "-1"
},
{
"input": "1\n9",
"output": "-1"
},
{
"input": "1\n10",
"output": "-1"
},
{
"input": "1\n11",
"output": "-1"
},
{
"input": "1\n12",
"output": "-1"
},
{
"input": "1\n13",
"output": "-1"
},
{
"input": "1\n14",
"output": "-1"
},
{
"input": "1\n15",
"output": "DOWN"
},
{
"input": "1\n0",
"output": "UP"
},
{
"input": "3\n11 12 13",
"output": "UP"
},
{
"input": "2\n10 9",
"output": "DOWN"
},
{
"input": "92\n10 11 12 13 14 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11",
"output": "UP"
},
{
"input": "92\n7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7 6",
"output": "DOWN"
},
{
"input": "2\n14 15",
"output": "DOWN"
},
{
"input": "2\n1 0",
"output": "UP"
},
{
"input": "2\n15 14",
"output": "DOWN"
},
{
"input": "92\n7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8",
"output": "UP"
},
{
"input": "92\n13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12",
"output": "DOWN"
},
{
"input": "92\n4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7 6 5 4 3",
"output": "DOWN"
},
{
"input": "92\n14 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15",
"output": "DOWN"
},
{
"input": "92\n1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0",
"output": "UP"
},
{
"input": "2\n2 1",
"output": "DOWN"
},
{
"input": "3\n2 1 0",
"output": "UP"
},
{
"input": "5\n4 3 2 1 0",
"output": "UP"
},
{
"input": "2\n5 4",
"output": "DOWN"
},
{
"input": "4\n3 2 1 0",
"output": "UP"
},
{
"input": "3\n13 12 11",
"output": "DOWN"
},
{
"input": "2\n1 2",
"output": "UP"
},
{
"input": "2\n0 1",
"output": "UP"
},
{
"input": "2\n13 14",
"output": "UP"
},
{
"input": "14\n13 12 11 10 9 8 7 6 5 4 3 2 1 0",
"output": "UP"
}
] | 1,477,465,385
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 56
| 62
| 0
|
n=int(input())
l=list(map(int,input().split()))
if n==1 and l[0]!=15 and l[0]!=0 :
print(-1)
else :
if l[n-1]==15 :
print('DOWN')
if l[n-1]==0 :
print('UP')
if l[n-1]!=15 and l[n-1]!=0 and n!=1 :
if l[n-1]-l[n-2]>0 :
print('UP')
else :
print('DOWN')
|
Title: Vitya in the Countryside
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Every summer Vitya comes to visit his grandmother in the countryside. This summer, he got a huge wart. Every grandma knows that one should treat warts when the moon goes down. Thus, Vitya has to catch the moment when the moon is down.
Moon cycle lasts 30 days. The size of the visible part of the moon (in Vitya's units) for each day is 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, and then cycle repeats, thus after the second 1 again goes 0.
As there is no internet in the countryside, Vitya has been watching the moon for *n* consecutive days and for each of these days he wrote down the size of the visible part of the moon. Help him find out whether the moon will be up or down next day, or this cannot be determined by the data he has.
Input Specification:
The first line of the input contains a single integer *n* (1<=β€<=*n*<=β€<=92)Β β the number of consecutive days Vitya was watching the size of the visible part of the moon.
The second line contains *n* integers *a**i* (0<=β€<=*a**i*<=β€<=15)Β β Vitya's records.
It's guaranteed that the input data is consistent.
Output Specification:
If Vitya can be sure that the size of visible part of the moon on day *n*<=+<=1 will be less than the size of the visible part on day *n*, then print "DOWN" at the only line of the output. If he might be sure that the size of the visible part will increase, then print "UP". If it's impossible to determine what exactly will happen with the moon, print -1.
Demo Input:
['5\n3 4 5 6 7\n', '7\n12 13 14 15 14 13 12\n', '1\n8\n']
Demo Output:
['UP\n', 'DOWN\n', '-1\n']
Note:
In the first sample, the size of the moon on the next day will be equal to 8, thus the answer is "UP".
In the second sample, the size of the moon on the next day will be 11, thus the answer is "DOWN".
In the third sample, there is no way to determine whether the size of the moon on the next day will be 7 or 9, thus the answer is -1.
|
```python
n=int(input())
l=list(map(int,input().split()))
if n==1 and l[0]!=15 and l[0]!=0 :
print(-1)
else :
if l[n-1]==15 :
print('DOWN')
if l[n-1]==0 :
print('UP')
if l[n-1]!=15 and l[n-1]!=0 and n!=1 :
if l[n-1]-l[n-2]>0 :
print('UP')
else :
print('DOWN')
```
| 3
|
|
13
|
A
|
Numbers
|
PROGRAMMING
| 1,000
|
[
"implementation",
"math"
] |
A. Numbers
|
1
|
64
|
Little Petya likes numbers a lot. He found that number 123 in base 16 consists of two digits: the first is 7 and the second is 11. So the sum of digits of 123 in base 16 is equal to 18.
Now he wonders what is an average value of sum of digits of the number *A* written in all bases from 2 to *A*<=-<=1.
Note that all computations should be done in base 10. You should find the result as an irreducible fraction, written in base 10.
|
Input contains one integer number *A* (3<=β€<=*A*<=β€<=1000).
|
Output should contain required average value in format Β«X/YΒ», where X is the numerator and Y is the denominator.
|
[
"5\n",
"3\n"
] |
[
"7/3\n",
"2/1\n"
] |
In the first sample number 5 written in all bases from 2 to 4 looks so: 101, 12, 11. Sums of digits are 2, 3 and 2, respectively.
| 0
|
[
{
"input": "5",
"output": "7/3"
},
{
"input": "3",
"output": "2/1"
},
{
"input": "1000",
"output": "90132/499"
},
{
"input": "927",
"output": "155449/925"
},
{
"input": "260",
"output": "6265/129"
},
{
"input": "131",
"output": "3370/129"
},
{
"input": "386",
"output": "857/12"
},
{
"input": "277",
"output": "2864/55"
},
{
"input": "766",
"output": "53217/382"
},
{
"input": "28",
"output": "85/13"
},
{
"input": "406",
"output": "7560/101"
},
{
"input": "757",
"output": "103847/755"
},
{
"input": "6",
"output": "9/4"
},
{
"input": "239",
"output": "10885/237"
},
{
"input": "322",
"output": "2399/40"
},
{
"input": "98",
"output": "317/16"
},
{
"input": "208",
"output": "4063/103"
},
{
"input": "786",
"output": "55777/392"
},
{
"input": "879",
"output": "140290/877"
},
{
"input": "702",
"output": "89217/700"
},
{
"input": "948",
"output": "7369/43"
},
{
"input": "537",
"output": "52753/535"
},
{
"input": "984",
"output": "174589/982"
},
{
"input": "934",
"output": "157951/932"
},
{
"input": "726",
"output": "95491/724"
},
{
"input": "127",
"output": "3154/125"
},
{
"input": "504",
"output": "23086/251"
},
{
"input": "125",
"output": "3080/123"
},
{
"input": "604",
"output": "33178/301"
},
{
"input": "115",
"output": "2600/113"
},
{
"input": "27",
"output": "167/25"
},
{
"input": "687",
"output": "85854/685"
},
{
"input": "880",
"output": "69915/439"
},
{
"input": "173",
"output": "640/19"
},
{
"input": "264",
"output": "6438/131"
},
{
"input": "785",
"output": "111560/783"
},
{
"input": "399",
"output": "29399/397"
},
{
"input": "514",
"output": "6031/64"
},
{
"input": "381",
"output": "26717/379"
},
{
"input": "592",
"output": "63769/590"
},
{
"input": "417",
"output": "32002/415"
},
{
"input": "588",
"output": "62723/586"
},
{
"input": "852",
"output": "131069/850"
},
{
"input": "959",
"output": "5059/29"
},
{
"input": "841",
"output": "127737/839"
},
{
"input": "733",
"output": "97598/731"
},
{
"input": "692",
"output": "87017/690"
},
{
"input": "69",
"output": "983/67"
},
{
"input": "223",
"output": "556/13"
},
{
"input": "93",
"output": "246/13"
},
{
"input": "643",
"output": "75503/641"
},
{
"input": "119",
"output": "2833/117"
},
{
"input": "498",
"output": "1459/16"
},
{
"input": "155",
"output": "4637/153"
},
{
"input": "305",
"output": "17350/303"
},
{
"input": "454",
"output": "37893/452"
},
{
"input": "88",
"output": "1529/86"
},
{
"input": "850",
"output": "32645/212"
},
{
"input": "474",
"output": "20581/236"
},
{
"input": "309",
"output": "17731/307"
},
{
"input": "762",
"output": "105083/760"
},
{
"input": "591",
"output": "63761/589"
},
{
"input": "457",
"output": "38317/455"
},
{
"input": "141",
"output": "3832/139"
},
{
"input": "385",
"output": "27232/383"
},
{
"input": "387",
"output": "27628/385"
},
{
"input": "469",
"output": "40306/467"
},
{
"input": "624",
"output": "35285/311"
},
{
"input": "330",
"output": "487/8"
},
{
"input": "31",
"output": "222/29"
},
{
"input": "975",
"output": "171679/973"
},
{
"input": "584",
"output": "62183/582"
},
{
"input": "668",
"output": "81127/666"
},
{
"input": "331",
"output": "20297/329"
},
{
"input": "189",
"output": "6789/187"
},
{
"input": "251",
"output": "11939/249"
},
{
"input": "876",
"output": "69196/437"
},
{
"input": "615",
"output": "68987/613"
},
{
"input": "451",
"output": "37258/449"
},
{
"input": "499",
"output": "45727/497"
},
{
"input": "699",
"output": "89117/697"
},
{
"input": "619",
"output": "70019/617"
},
{
"input": "413",
"output": "10515/137"
},
{
"input": "197",
"output": "7399/195"
},
{
"input": "794",
"output": "14281/99"
},
{
"input": "659",
"output": "79403/657"
},
{
"input": "653",
"output": "77695/651"
},
{
"input": "23",
"output": "45/7"
},
{
"input": "430",
"output": "16985/214"
},
{
"input": "249",
"output": "11659/247"
},
{
"input": "837",
"output": "126869/835"
},
{
"input": "258",
"output": "12373/256"
},
{
"input": "995",
"output": "59665/331"
},
{
"input": "102",
"output": "504/25"
},
{
"input": "989",
"output": "177124/987"
},
{
"input": "376",
"output": "13008/187"
},
{
"input": "657",
"output": "15715/131"
},
{
"input": "746",
"output": "50509/372"
},
{
"input": "602",
"output": "13177/120"
}
] | 1,649,863,186
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 99
| 92
| 0
|
N = int(input())
s = 0
def gcd(a, b):
while(b):
a, b = b, a%b
return a
for i in range(2, N):
n = N
while(n>0):
s += n%i
n = n//i
t = gcd(s, N-2)
print(str(s//t)+"/"+str((N-2)//t))
|
Title: Numbers
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
Little Petya likes numbers a lot. He found that number 123 in base 16 consists of two digits: the first is 7 and the second is 11. So the sum of digits of 123 in base 16 is equal to 18.
Now he wonders what is an average value of sum of digits of the number *A* written in all bases from 2 to *A*<=-<=1.
Note that all computations should be done in base 10. You should find the result as an irreducible fraction, written in base 10.
Input Specification:
Input contains one integer number *A* (3<=β€<=*A*<=β€<=1000).
Output Specification:
Output should contain required average value in format Β«X/YΒ», where X is the numerator and Y is the denominator.
Demo Input:
['5\n', '3\n']
Demo Output:
['7/3\n', '2/1\n']
Note:
In the first sample number 5 written in all bases from 2 to 4 looks so: 101, 12, 11. Sums of digits are 2, 3 and 2, respectively.
|
```python
N = int(input())
s = 0
def gcd(a, b):
while(b):
a, b = b, a%b
return a
for i in range(2, N):
n = N
while(n>0):
s += n%i
n = n//i
t = gcd(s, N-2)
print(str(s//t)+"/"+str((N-2)//t))
```
| 3.954
|
58
|
A
|
Chat room
|
PROGRAMMING
| 1,000
|
[
"greedy",
"strings"
] |
A. Chat room
|
1
|
256
|
Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.
|
The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.
|
If Vasya managed to say hello, print "YES", otherwise print "NO".
|
[
"ahhellllloou\n",
"hlelo\n"
] |
[
"YES\n",
"NO\n"
] |
none
| 500
|
[
{
"input": "ahhellllloou",
"output": "YES"
},
{
"input": "hlelo",
"output": "NO"
},
{
"input": "helhcludoo",
"output": "YES"
},
{
"input": "hehwelloho",
"output": "YES"
},
{
"input": "pnnepelqomhhheollvlo",
"output": "YES"
},
{
"input": "tymbzjyqhymedasloqbq",
"output": "NO"
},
{
"input": "yehluhlkwo",
"output": "NO"
},
{
"input": "hatlevhhalrohairnolsvocafgueelrqmlqlleello",
"output": "YES"
},
{
"input": "hhhtehdbllnhwmbyhvelqqyoulretpbfokflhlhreeflxeftelziclrwllrpflflbdtotvlqgoaoqldlroovbfsq",
"output": "YES"
},
{
"input": "rzlvihhghnelqtwlexmvdjjrliqllolhyewgozkuovaiezgcilelqapuoeglnwmnlftxxiigzczlouooi",
"output": "YES"
},
{
"input": "pfhhwctyqdlkrwhebfqfelhyebwllhemtrmeblgrynmvyhioesqklclocxmlffuormljszllpoo",
"output": "YES"
},
{
"input": "lqllcolohwflhfhlnaow",
"output": "NO"
},
{
"input": "heheeellollvoo",
"output": "YES"
},
{
"input": "hellooo",
"output": "YES"
},
{
"input": "o",
"output": "NO"
},
{
"input": "hhqhzeclohlehljlhtesllylrolmomvuhcxsobtsckogdv",
"output": "YES"
},
{
"input": "yoegfuzhqsihygnhpnukluutocvvwuldiighpogsifealtgkfzqbwtmgghmythcxflebrkctlldlkzlagovwlstsghbouk",
"output": "YES"
},
{
"input": "uatqtgbvrnywfacwursctpagasnhydvmlinrcnqrry",
"output": "NO"
},
{
"input": "tndtbldbllnrwmbyhvqaqqyoudrstpbfokfoclnraefuxtftmgzicorwisrpfnfpbdtatvwqgyalqtdtrjqvbfsq",
"output": "NO"
},
{
"input": "rzlvirhgemelnzdawzpaoqtxmqucnahvqnwldklrmjiiyageraijfivigvozgwngiulttxxgzczptusoi",
"output": "YES"
},
{
"input": "kgyelmchocojsnaqdsyeqgnllytbqietpdlgknwwumqkxrexgdcnwoldicwzwofpmuesjuxzrasscvyuqwspm",
"output": "YES"
},
{
"input": "pnyvrcotjvgynbeldnxieghfltmexttuxzyac",
"output": "NO"
},
{
"input": "dtwhbqoumejligbenxvzhjlhosqojetcqsynlzyhfaevbdpekgbtjrbhlltbceobcok",
"output": "YES"
},
{
"input": "crrfpfftjwhhikwzeedrlwzblckkteseofjuxjrktcjfsylmlsvogvrcxbxtffujqshslemnixoeezivksouefeqlhhokwbqjz",
"output": "YES"
},
{
"input": "jhfbndhyzdvhbvhmhmefqllujdflwdpjbehedlsqfdsqlyelwjtyloxwsvasrbqosblzbowlqjmyeilcvotdlaouxhdpoeloaovb",
"output": "YES"
},
{
"input": "hwlghueoemiqtjhhpashjsouyegdlvoyzeunlroypoprnhlyiwiuxrghekaylndhrhllllwhbebezoglydcvykllotrlaqtvmlla",
"output": "YES"
},
{
"input": "wshiaunnqnqxodholbipwhhjmyeblhgpeleblklpzwhdunmpqkbuzloetmwwxmeltkrcomulxauzlwmlklldjodozxryghsnwgcz",
"output": "YES"
},
{
"input": "shvksednttggehroewuiptvvxtrzgidravtnjwuqrlnnkxbplctzkckinpkgjopjfoxdbojtcvsuvablcbkrzajrlhgobkcxeqti",
"output": "YES"
},
{
"input": "hyyhddqhxhekehkwfhlnlsihzefwchzerevcjtokefplholrbvxlltdlafjxrfhleglrvlolojoqaolagtbeyogxlbgfolllslli",
"output": "YES"
},
{
"input": "iaagrdhhelxpdegueiulflquvlzidoprzkehthkldaytrjwhyhprwjxlltinxvuilxohqgjqcvkkdcuoliekcnlilwgqlnlzouoo",
"output": "YES"
},
{
"input": "wfluaeseldgxyvxpwuhkptdmlflnlhktwxiabsvkolsquymrmhzczzoybvlilhmvsuunggvgxzgyyffk",
"output": "NO"
},
{
"input": "loee",
"output": "NO"
},
{
"input": "izdipwylefqmjbuoznfglgcfvedeouflkeehxbasaqmiooppfsqlhprospqgxvzrcpwlfdddngoqvpwvggsnvvxhmjoslwjudjlo",
"output": "NO"
},
{
"input": "pplynflbjlseflaegzqpdxswgunhnlglqhycrbcwdfhclmcrwekkzsoosmfblljeenwfjyifcvxihhhbyhmhlkvtskeglaovjnsi",
"output": "YES"
},
{
"input": "xqjqmenkodmlhzyzmmvofdngktygbbxbzpluzcohohmalkoeuwfikblltaaigv",
"output": "NO"
},
{
"input": "hello",
"output": "YES"
},
{
"input": "oohell",
"output": "NO"
},
{
"input": "hell",
"output": "NO"
},
{
"input": "eloellohoelo",
"output": "NO"
},
{
"input": "helllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllo",
"output": "YES"
}
] | 1,579,816,269
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 40
| 140
| 0
|
w=input()
a="hello"
s=""
c=0
for i in w:
if i == a[c]:
c+=1
s+=i
if c==5:
break
if s==a:
print("YES")
else:
print("NO")
|
Title: Chat room
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.
Input Specification:
The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.
Output Specification:
If Vasya managed to say hello, print "YES", otherwise print "NO".
Demo Input:
['ahhellllloou\n', 'hlelo\n']
Demo Output:
['YES\n', 'NO\n']
Note:
none
|
```python
w=input()
a="hello"
s=""
c=0
for i in w:
if i == a[c]:
c+=1
s+=i
if c==5:
break
if s==a:
print("YES")
else:
print("NO")
```
| 3.93
|
169
|
A
|
Chores
|
PROGRAMMING
| 800
|
[
"sortings"
] | null | null |
Petya and Vasya are brothers. Today is a special day for them as their parents left them home alone and commissioned them to do *n* chores. Each chore is characterized by a single parameter β its complexity. The complexity of the *i*-th chore equals *h**i*.
As Petya is older, he wants to take the chores with complexity larger than some value *x* (*h**i*<=><=*x*) to leave to Vasya the chores with complexity less than or equal to *x* (*h**i*<=β€<=*x*). The brothers have already decided that Petya will do exactly *a* chores and Vasya will do exactly *b* chores (*a*<=+<=*b*<==<=*n*).
In how many ways can they choose an integer *x* so that Petya got exactly *a* chores and Vasya got exactly *b* chores?
|
The first input line contains three integers *n*,<=*a* and *b* (2<=β€<=*n*<=β€<=2000; *a*,<=*b*<=β₯<=1; *a*<=+<=*b*<==<=*n*) β the total number of chores, the number of Petya's chores and the number of Vasya's chores.
The next line contains a sequence of integers *h*1,<=*h*2,<=...,<=*h**n* (1<=β€<=*h**i*<=β€<=109), *h**i* is the complexity of the *i*-th chore. The numbers in the given sequence are not necessarily different.
All numbers on the lines are separated by single spaces.
|
Print the required number of ways to choose an integer value of *x*. If there are no such ways, print 0.
|
[
"5 2 3\n6 2 3 100 1\n",
"7 3 4\n1 1 9 1 1 1 1\n"
] |
[
"3\n",
"0\n"
] |
In the first sample the possible values of *x* are 3, 4 or 5.
In the second sample it is impossible to find such *x*, that Petya got 3 chores and Vasya got 4.
| 500
|
[
{
"input": "5 2 3\n6 2 3 100 1",
"output": "3"
},
{
"input": "7 3 4\n1 1 9 1 1 1 1",
"output": "0"
},
{
"input": "2 1 1\n10 2",
"output": "8"
},
{
"input": "2 1 1\n7 7",
"output": "0"
},
{
"input": "2 1 1\n1 1000000000",
"output": "999999999"
},
{
"input": "3 1 2\n6 5 5",
"output": "1"
},
{
"input": "3 2 1\n10 10 8",
"output": "2"
},
{
"input": "8 3 5\n42 55 61 72 83 10 22 33",
"output": "6"
},
{
"input": "10 5 5\n1 2 3 4 5 999999999 999999998 999999997 999999996 999999995",
"output": "999999990"
},
{
"input": "4 1 3\n10 8 7 3",
"output": "2"
},
{
"input": "4 2 2\n402 10 10 402",
"output": "392"
},
{
"input": "4 1 3\n10 402 402 10",
"output": "0"
},
{
"input": "4 3 1\n100 100 200 200",
"output": "0"
},
{
"input": "150 10 140\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 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 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": "0"
},
{
"input": "102 101 1\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 1 1",
"output": "0"
}
] | 1,642,081,320
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 29
| 46
| 0
|
n,a,b= map(int,input().split())
x = list(map(int,input().split()))
x.sort()
print(x[b]-x[b-1])
|
Title: Chores
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya and Vasya are brothers. Today is a special day for them as their parents left them home alone and commissioned them to do *n* chores. Each chore is characterized by a single parameter β its complexity. The complexity of the *i*-th chore equals *h**i*.
As Petya is older, he wants to take the chores with complexity larger than some value *x* (*h**i*<=><=*x*) to leave to Vasya the chores with complexity less than or equal to *x* (*h**i*<=β€<=*x*). The brothers have already decided that Petya will do exactly *a* chores and Vasya will do exactly *b* chores (*a*<=+<=*b*<==<=*n*).
In how many ways can they choose an integer *x* so that Petya got exactly *a* chores and Vasya got exactly *b* chores?
Input Specification:
The first input line contains three integers *n*,<=*a* and *b* (2<=β€<=*n*<=β€<=2000; *a*,<=*b*<=β₯<=1; *a*<=+<=*b*<==<=*n*) β the total number of chores, the number of Petya's chores and the number of Vasya's chores.
The next line contains a sequence of integers *h*1,<=*h*2,<=...,<=*h**n* (1<=β€<=*h**i*<=β€<=109), *h**i* is the complexity of the *i*-th chore. The numbers in the given sequence are not necessarily different.
All numbers on the lines are separated by single spaces.
Output Specification:
Print the required number of ways to choose an integer value of *x*. If there are no such ways, print 0.
Demo Input:
['5 2 3\n6 2 3 100 1\n', '7 3 4\n1 1 9 1 1 1 1\n']
Demo Output:
['3\n', '0\n']
Note:
In the first sample the possible values of *x* are 3, 4 or 5.
In the second sample it is impossible to find such *x*, that Petya got 3 chores and Vasya got 4.
|
```python
n,a,b= map(int,input().split())
x = list(map(int,input().split()))
x.sort()
print(x[b]-x[b-1])
```
| 3
|
|
443
|
A
|
Anton and Letters
|
PROGRAMMING
| 800
|
[
"constructive algorithms",
"implementation"
] | null | null |
Recently, Anton has found a set. The set consists of small English letters. Anton carefully wrote out all the letters from the set in one line, separated by a comma. He also added an opening curved bracket at the beginning of the line and a closing curved bracket at the end of the line.
Unfortunately, from time to time Anton would forget writing some letter and write it again. He asks you to count the total number of distinct letters in his set.
|
The first and the single line contains the set of letters. The length of the line doesn't exceed 1000. It is guaranteed that the line starts from an opening curved bracket and ends with a closing curved bracket. Between them, small English letters are listed, separated by a comma. Each comma is followed by a space.
|
Print a single number β the number of distinct letters in Anton's set.
|
[
"{a, b, c}\n",
"{b, a, b, a}\n",
"{}\n"
] |
[
"3\n",
"2\n",
"0\n"
] |
none
| 500
|
[
{
"input": "{a, b, c}",
"output": "3"
},
{
"input": "{b, a, b, a}",
"output": "2"
},
{
"input": "{}",
"output": "0"
},
{
"input": "{a, a, c, b, b, b, c, c, c, c}",
"output": "3"
},
{
"input": "{a, c, b, b}",
"output": "3"
},
{
"input": "{a, b}",
"output": "2"
},
{
"input": "{a}",
"output": "1"
},
{
"input": "{b, a, b, a, b, c, c, b, c, b}",
"output": "3"
},
{
"input": "{e, g, c, e}",
"output": "3"
},
{
"input": "{a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a}",
"output": "1"
},
{
"input": "{a, a, b}",
"output": "2"
},
{
"input": "{a, b, b, b, a, b, a, a, a, a, a, a, b, a, b, a, a, a, a, a, b, a, b, a}",
"output": "2"
},
{
"input": "{j, u, a, c, f, w, e, w, x, t, h, p, v, n, i, l, x, n, i, b, u, c, a, a}",
"output": "16"
},
{
"input": "{x, i, w, c, p, e, h, z, k, i}",
"output": "9"
},
{
"input": "{t, k, o, x, r, d, q, j, k, e, z, w, y, r, z, s, s, e, s, b, k, i}",
"output": "15"
},
{
"input": "{y}",
"output": "1"
},
{
"input": "{x}",
"output": "1"
},
{
"input": "{b, z, a, z}",
"output": "3"
},
{
"input": "{z}",
"output": "1"
},
{
"input": "{a, z}",
"output": "2"
},
{
"input": "{a, b, z}",
"output": "3"
},
{
"input": "{s, q, z, r, t, a, b, h, j, i, o, z, r, q}",
"output": "11"
}
] | 1,697,813,734
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 27
| 92
| 0
|
s = input()
print(len(set(s.lstrip("{").rstrip("}").split(", "))) if s != "{}" else 0)
|
Title: Anton and Letters
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Recently, Anton has found a set. The set consists of small English letters. Anton carefully wrote out all the letters from the set in one line, separated by a comma. He also added an opening curved bracket at the beginning of the line and a closing curved bracket at the end of the line.
Unfortunately, from time to time Anton would forget writing some letter and write it again. He asks you to count the total number of distinct letters in his set.
Input Specification:
The first and the single line contains the set of letters. The length of the line doesn't exceed 1000. It is guaranteed that the line starts from an opening curved bracket and ends with a closing curved bracket. Between them, small English letters are listed, separated by a comma. Each comma is followed by a space.
Output Specification:
Print a single number β the number of distinct letters in Anton's set.
Demo Input:
['{a, b, c}\n', '{b, a, b, a}\n', '{}\n']
Demo Output:
['3\n', '2\n', '0\n']
Note:
none
|
```python
s = input()
print(len(set(s.lstrip("{").rstrip("}").split(", "))) if s != "{}" else 0)
```
| 3
|
|
431
|
A
|
Black Square
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Quite recently, a very smart student named Jury decided that lectures are boring, so he downloaded a game called "Black Square" on his super cool touchscreen phone.
In this game, the phone's screen is divided into four vertical strips. Each second, a black square appears on some of the strips. According to the rules of the game, Jury must use this second to touch the corresponding strip to make the square go away. As Jury is both smart and lazy, he counted that he wastes exactly *a**i* calories on touching the *i*-th strip.
You've got a string *s*, describing the process of the game and numbers *a*1,<=*a*2,<=*a*3,<=*a*4. Calculate how many calories Jury needs to destroy all the squares?
|
The first line contains four space-separated integers *a*1, *a*2, *a*3, *a*4 (0<=β€<=*a*1,<=*a*2,<=*a*3,<=*a*4<=β€<=104).
The second line contains string *s* (1<=β€<=|*s*|<=β€<=105), where the *Ρ*-th character of the string equals "1", if on the *i*-th second of the game the square appears on the first strip, "2", if it appears on the second strip, "3", if it appears on the third strip, "4", if it appears on the fourth strip.
|
Print a single integer β the total number of calories that Jury wastes.
|
[
"1 2 3 4\n123214\n",
"1 5 3 2\n11221\n"
] |
[
"13\n",
"13\n"
] |
none
| 500
|
[
{
"input": "1 2 3 4\n123214",
"output": "13"
},
{
"input": "1 5 3 2\n11221",
"output": "13"
},
{
"input": "5 5 5 1\n3422",
"output": "16"
},
{
"input": "4 3 2 1\n2",
"output": "3"
},
{
"input": "5651 6882 6954 4733\n2442313421",
"output": "60055"
},
{
"input": "0 0 0 0\n4132",
"output": "0"
},
{
"input": "3163 5778 83 7640\n11141442444",
"output": "64270"
},
{
"input": "1809 1302 7164 6122\n3144121413113111223311232232114144321414421243443243422322144324121433444342231344234443332241322442",
"output": "420780"
},
{
"input": "0 0 0 0\n1",
"output": "0"
},
{
"input": "1 2 3 4\n4",
"output": "4"
},
{
"input": "2343 7653 1242 5432\n1",
"output": "2343"
},
{
"input": "2343 7653 1242 5432\n2",
"output": "7653"
},
{
"input": "2343 7653 1242 5432\n3",
"output": "1242"
},
{
"input": "2343 7653 1242 5432\n4",
"output": "5432"
},
{
"input": "1 2 3 4\n123412",
"output": "13"
},
{
"input": "50 50 50 50\n11111111111111111111111111111111111111111111111111111",
"output": "2650"
},
{
"input": "1 2 3 4\n11111111111111111111111111111111111111111111111111",
"output": "50"
},
{
"input": "1 2 3 4\n23123231321231231231231231221232123121312321",
"output": "87"
},
{
"input": "1 2 3 4\n1111111111111222222222233333333333444444444444444",
"output": "126"
},
{
"input": "2 3 1 4\n121321232412342112312313213123123412131231231232",
"output": "105"
}
] | 1,675,894,587
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 49
| 92
| 0
|
a = []
a = [int(item) for item in input().split()] # fill the list with numbers
s = input() # take the string
x = len(s)
result = 0
for i in range(x):
if s[i]=='1':
result+=a[0]
elif s[i]== '2' :
result += a[1]
elif s[i] == '3' :
result += a[2]
elif s[i] == '4' :
result += a[3]
print (result)
# problem_link : https://codeforces.com/contest/431/problem/A
|
Title: Black Square
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Quite recently, a very smart student named Jury decided that lectures are boring, so he downloaded a game called "Black Square" on his super cool touchscreen phone.
In this game, the phone's screen is divided into four vertical strips. Each second, a black square appears on some of the strips. According to the rules of the game, Jury must use this second to touch the corresponding strip to make the square go away. As Jury is both smart and lazy, he counted that he wastes exactly *a**i* calories on touching the *i*-th strip.
You've got a string *s*, describing the process of the game and numbers *a*1,<=*a*2,<=*a*3,<=*a*4. Calculate how many calories Jury needs to destroy all the squares?
Input Specification:
The first line contains four space-separated integers *a*1, *a*2, *a*3, *a*4 (0<=β€<=*a*1,<=*a*2,<=*a*3,<=*a*4<=β€<=104).
The second line contains string *s* (1<=β€<=|*s*|<=β€<=105), where the *Ρ*-th character of the string equals "1", if on the *i*-th second of the game the square appears on the first strip, "2", if it appears on the second strip, "3", if it appears on the third strip, "4", if it appears on the fourth strip.
Output Specification:
Print a single integer β the total number of calories that Jury wastes.
Demo Input:
['1 2 3 4\n123214\n', '1 5 3 2\n11221\n']
Demo Output:
['13\n', '13\n']
Note:
none
|
```python
a = []
a = [int(item) for item in input().split()] # fill the list with numbers
s = input() # take the string
x = len(s)
result = 0
for i in range(x):
if s[i]=='1':
result+=a[0]
elif s[i]== '2' :
result += a[1]
elif s[i] == '3' :
result += a[2]
elif s[i] == '4' :
result += a[3]
print (result)
# problem_link : https://codeforces.com/contest/431/problem/A
```
| 3
|
|
712
|
A
|
Memory and Crow
|
PROGRAMMING
| 800
|
[
"implementation",
"math"
] | null | null |
There are *n* integers *b*1,<=*b*2,<=...,<=*b**n* written in a row. For all *i* from 1 to *n*, values *a**i* are defined by the crows performing the following procedure:
- The crow sets *a**i* initially 0. - The crow then adds *b**i* to *a**i*, subtracts *b**i*<=+<=1, adds the *b**i*<=+<=2 number, and so on until the *n*'th number. Thus, *a**i*<==<=*b**i*<=-<=*b**i*<=+<=1<=+<=*b**i*<=+<=2<=-<=*b**i*<=+<=3....
Memory gives you the values *a*1,<=*a*2,<=...,<=*a**n*, and he now wants you to find the initial numbers *b*1,<=*b*2,<=...,<=*b**n* written in the row? Can you do it?
|
The first line of the input contains a single integer *n* (2<=β€<=*n*<=β€<=100<=000)Β β the number of integers written in the row.
The next line contains *n*, the *i*'th of which is *a**i* (<=-<=109<=β€<=*a**i*<=β€<=109)Β β the value of the *i*'th number.
|
Print *n* integers corresponding to the sequence *b*1,<=*b*2,<=...,<=*b**n*. It's guaranteed that the answer is unique and fits in 32-bit integer type.
|
[
"5\n6 -4 8 -2 3\n",
"5\n3 -2 -1 5 6\n"
] |
[
"2 4 6 1 3 \n",
"1 -3 4 11 6 \n"
] |
In the first sample test, the crows report the numbers 6,β-β4, 8,β-β2, and 3 when he starts at indices 1, 2, 3, 4 and 5 respectively. It is easy to check that the sequence 2 4 6 1 3 satisfies the reports. For example, 6β=β2β-β4β+β6β-β1β+β3, and β-β4β=β4β-β6β+β1β-β3.
In the second sample test, the sequence 1, β-β3, 4, 11, 6 satisfies the reports. For example, 5β=β11β-β6 and 6β=β6.
| 500
|
[
{
"input": "5\n6 -4 8 -2 3",
"output": "2 4 6 1 3 "
},
{
"input": "5\n3 -2 -1 5 6",
"output": "1 -3 4 11 6 "
},
{
"input": "10\n13 -2 532 -63 -23 -63 -64 -23 12 10",
"output": "11 530 469 -86 -86 -127 -87 -11 22 10 "
},
{
"input": "10\n0 0 0 0 0 0 0 0 0 0",
"output": "0 0 0 0 0 0 0 0 0 0 "
},
{
"input": "10\n1 -1 1 -1 1 -1 1 -1 1 -1",
"output": "0 0 0 0 0 0 0 0 0 -1 "
},
{
"input": "10\n-1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000",
"output": "0 0 0 0 0 0 0 0 0 1000000000 "
},
{
"input": "10\n124 532 -642 6482 -124 952 -235 7594 34 -12",
"output": "656 -110 5840 6358 828 717 7359 7628 22 -12 "
},
{
"input": "10\n1294 35852 32749 537295 12048 53729 29357 58320 64739 1240",
"output": "37146 68601 570044 549343 65777 83086 87677 123059 65979 1240 "
}
] | 1,473,527,311
| 1,411
|
Python 3
|
OK
|
TESTS
| 49
| 405
| 8,396,800
|
n=int(input())
v=[int(i) for i in input().split()]
b=[v[i]+v[i+1] for i in range(n-1)]
b.append(v[-1])
for i in range(n):
print(b[i],end=" ")
|
Title: Memory and Crow
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* integers *b*1,<=*b*2,<=...,<=*b**n* written in a row. For all *i* from 1 to *n*, values *a**i* are defined by the crows performing the following procedure:
- The crow sets *a**i* initially 0. - The crow then adds *b**i* to *a**i*, subtracts *b**i*<=+<=1, adds the *b**i*<=+<=2 number, and so on until the *n*'th number. Thus, *a**i*<==<=*b**i*<=-<=*b**i*<=+<=1<=+<=*b**i*<=+<=2<=-<=*b**i*<=+<=3....
Memory gives you the values *a*1,<=*a*2,<=...,<=*a**n*, and he now wants you to find the initial numbers *b*1,<=*b*2,<=...,<=*b**n* written in the row? Can you do it?
Input Specification:
The first line of the input contains a single integer *n* (2<=β€<=*n*<=β€<=100<=000)Β β the number of integers written in the row.
The next line contains *n*, the *i*'th of which is *a**i* (<=-<=109<=β€<=*a**i*<=β€<=109)Β β the value of the *i*'th number.
Output Specification:
Print *n* integers corresponding to the sequence *b*1,<=*b*2,<=...,<=*b**n*. It's guaranteed that the answer is unique and fits in 32-bit integer type.
Demo Input:
['5\n6 -4 8 -2 3\n', '5\n3 -2 -1 5 6\n']
Demo Output:
['2 4 6 1 3 \n', '1 -3 4 11 6 \n']
Note:
In the first sample test, the crows report the numbers 6,β-β4, 8,β-β2, and 3 when he starts at indices 1, 2, 3, 4 and 5 respectively. It is easy to check that the sequence 2 4 6 1 3 satisfies the reports. For example, 6β=β2β-β4β+β6β-β1β+β3, and β-β4β=β4β-β6β+β1β-β3.
In the second sample test, the sequence 1, β-β3, 4, 11, 6 satisfies the reports. For example, 5β=β11β-β6 and 6β=β6.
|
```python
n=int(input())
v=[int(i) for i in input().split()]
b=[v[i]+v[i+1] for i in range(n-1)]
b.append(v[-1])
for i in range(n):
print(b[i],end=" ")
```
| 3
|
|
990
|
B
|
Micro-World
|
PROGRAMMING
| 1,200
|
[
"greedy",
"sortings"
] | null | null |
You have a Petri dish with bacteria and you are preparing to dive into the harsh micro-world. But, unfortunately, you don't have any microscope nearby, so you can't watch them.
You know that you have $n$ bacteria in the Petri dish and size of the $i$-th bacteria is $a_i$. Also you know intergalactic positive integer constant $K$.
The $i$-th bacteria can swallow the $j$-th bacteria if and only if $a_i > a_j$ and $a_i \le a_j + K$. The $j$-th bacteria disappear, but the $i$-th bacteria doesn't change its size. The bacteria can perform multiple swallows. On each swallow operation any bacteria $i$ can swallow any bacteria $j$ if $a_i > a_j$ and $a_i \le a_j + K$. The swallow operations go one after another.
For example, the sequence of bacteria sizes $a=[101, 53, 42, 102, 101, 55, 54]$ and $K=1$. The one of possible sequences of swallows is: $[101, 53, 42, 102, \underline{101}, 55, 54]$ $\to$ $[101, \underline{53}, 42, 102, 55, 54]$ $\to$ $[\underline{101}, 42, 102, 55, 54]$ $\to$ $[42, 102, 55, \underline{54}]$ $\to$ $[42, 102, 55]$. In total there are $3$ bacteria remained in the Petri dish.
Since you don't have a microscope, you can only guess, what the minimal possible number of bacteria can remain in your Petri dish when you finally will find any microscope.
|
The first line contains two space separated positive integers $n$ and $K$ ($1 \le n \le 2 \cdot 10^5$, $1 \le K \le 10^6$) β number of bacteria and intergalactic constant $K$.
The second line contains $n$ space separated integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^6$) β sizes of bacteria you have.
|
Print the only integer β minimal possible number of bacteria can remain.
|
[
"7 1\n101 53 42 102 101 55 54\n",
"6 5\n20 15 10 15 20 25\n",
"7 1000000\n1 1 1 1 1 1 1\n"
] |
[
"3\n",
"1\n",
"7\n"
] |
The first example is clarified in the problem statement.
In the second example an optimal possible sequence of swallows is: $[20, 15, 10, 15, \underline{20}, 25]$ $\to$ $[20, 15, 10, \underline{15}, 25]$ $\to$ $[20, 15, \underline{10}, 25]$ $\to$ $[20, \underline{15}, 25]$ $\to$ $[\underline{20}, 25]$ $\to$ $[25]$.
In the third example no bacteria can swallow any other bacteria.
| 0
|
[
{
"input": "7 1\n101 53 42 102 101 55 54",
"output": "3"
},
{
"input": "6 5\n20 15 10 15 20 25",
"output": "1"
},
{
"input": "7 1000000\n1 1 1 1 1 1 1",
"output": "7"
},
{
"input": "1 1\n1",
"output": "1"
},
{
"input": "1 4\n8",
"output": "1"
},
{
"input": "10 1\n1 2 3 5 6 8 10 11 9 4",
"output": "2"
},
{
"input": "9 2\n1 6 1 5 5 8 6 8 7",
"output": "4"
},
{
"input": "15 1\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "15"
},
{
"input": "2 1000000\n1 1000000",
"output": "1"
},
{
"input": "7 2\n1 5 5 8 9 8 8",
"output": "4"
},
{
"input": "10 1\n2 6 3 4 2 4 4 3 2 1",
"output": "4"
},
{
"input": "4 1\n2 2 1 1",
"output": "2"
},
{
"input": "10 1\n6 3 1 3 6 4 1 3 6 4",
"output": "7"
},
{
"input": "2 1\n1 1",
"output": "2"
},
{
"input": "2 1\n1 2",
"output": "1"
},
{
"input": "8 2\n3 13 9 8 3 13 9 14",
"output": "5"
},
{
"input": "8 1000000\n1 1 5 1000000 1000000 2 2 2",
"output": "2"
},
{
"input": "2 1\n999152 999153",
"output": "1"
}
] | 1,534,143,527
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 36
| 560
| 14,336,000
|
n ,k = map(int,input().split())
a = list(map(int,input().split()))
a.sort()
out = 0
i = 0
for j in range(n):
while(i+1 < n and a[j] == a[i]):
i += 1
if a[j] < a[i] and a[j] + k >= a[i]:
out += 1
print(n-out)
|
Title: Micro-World
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You have a Petri dish with bacteria and you are preparing to dive into the harsh micro-world. But, unfortunately, you don't have any microscope nearby, so you can't watch them.
You know that you have $n$ bacteria in the Petri dish and size of the $i$-th bacteria is $a_i$. Also you know intergalactic positive integer constant $K$.
The $i$-th bacteria can swallow the $j$-th bacteria if and only if $a_i > a_j$ and $a_i \le a_j + K$. The $j$-th bacteria disappear, but the $i$-th bacteria doesn't change its size. The bacteria can perform multiple swallows. On each swallow operation any bacteria $i$ can swallow any bacteria $j$ if $a_i > a_j$ and $a_i \le a_j + K$. The swallow operations go one after another.
For example, the sequence of bacteria sizes $a=[101, 53, 42, 102, 101, 55, 54]$ and $K=1$. The one of possible sequences of swallows is: $[101, 53, 42, 102, \underline{101}, 55, 54]$ $\to$ $[101, \underline{53}, 42, 102, 55, 54]$ $\to$ $[\underline{101}, 42, 102, 55, 54]$ $\to$ $[42, 102, 55, \underline{54}]$ $\to$ $[42, 102, 55]$. In total there are $3$ bacteria remained in the Petri dish.
Since you don't have a microscope, you can only guess, what the minimal possible number of bacteria can remain in your Petri dish when you finally will find any microscope.
Input Specification:
The first line contains two space separated positive integers $n$ and $K$ ($1 \le n \le 2 \cdot 10^5$, $1 \le K \le 10^6$) β number of bacteria and intergalactic constant $K$.
The second line contains $n$ space separated integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^6$) β sizes of bacteria you have.
Output Specification:
Print the only integer β minimal possible number of bacteria can remain.
Demo Input:
['7 1\n101 53 42 102 101 55 54\n', '6 5\n20 15 10 15 20 25\n', '7 1000000\n1 1 1 1 1 1 1\n']
Demo Output:
['3\n', '1\n', '7\n']
Note:
The first example is clarified in the problem statement.
In the second example an optimal possible sequence of swallows is: $[20, 15, 10, 15, \underline{20}, 25]$ $\to$ $[20, 15, 10, \underline{15}, 25]$ $\to$ $[20, 15, \underline{10}, 25]$ $\to$ $[20, \underline{15}, 25]$ $\to$ $[\underline{20}, 25]$ $\to$ $[25]$.
In the third example no bacteria can swallow any other bacteria.
|
```python
n ,k = map(int,input().split())
a = list(map(int,input().split()))
a.sort()
out = 0
i = 0
for j in range(n):
while(i+1 < n and a[j] == a[i]):
i += 1
if a[j] < a[i] and a[j] + k >= a[i]:
out += 1
print(n-out)
```
| 3
|
|
808
|
A
|
Lucky Year
|
PROGRAMMING
| 900
|
[
"implementation"
] | null | null |
Apart from having lots of holidays throughout the year, residents of Berland also have whole lucky years. Year is considered lucky if it has no more than 1 non-zero digit in its number. So years 100, 40000, 5 are lucky and 12, 3001 and 12345 are not.
You are given current year in Berland. Your task is to find how long will residents of Berland wait till the next lucky year.
|
The first line contains integer number *n* (1<=β€<=*n*<=β€<=109) β current year in Berland.
|
Output amount of years from the current year to the next lucky one.
|
[
"4\n",
"201\n",
"4000\n"
] |
[
"1\n",
"99\n",
"1000\n"
] |
In the first example next lucky year is 5. In the second one β 300. In the third β 5000.
| 0
|
[
{
"input": "4",
"output": "1"
},
{
"input": "201",
"output": "99"
},
{
"input": "4000",
"output": "1000"
},
{
"input": "9",
"output": "1"
},
{
"input": "10",
"output": "10"
},
{
"input": "1",
"output": "1"
},
{
"input": "100000000",
"output": "100000000"
},
{
"input": "900000000",
"output": "100000000"
},
{
"input": "999999999",
"output": "1"
},
{
"input": "1000000000",
"output": "1000000000"
},
{
"input": "9999999",
"output": "1"
},
{
"input": "100000001",
"output": "99999999"
},
{
"input": "3660",
"output": "340"
},
{
"input": "21",
"output": "9"
},
{
"input": "900000001",
"output": "99999999"
},
{
"input": "62911",
"output": "7089"
},
{
"input": "11",
"output": "9"
},
{
"input": "940302010",
"output": "59697990"
},
{
"input": "91",
"output": "9"
},
{
"input": "101",
"output": "99"
},
{
"input": "1090",
"output": "910"
},
{
"input": "987654321",
"output": "12345679"
},
{
"input": "703450474",
"output": "96549526"
},
{
"input": "1091",
"output": "909"
},
{
"input": "89",
"output": "1"
},
{
"input": "109",
"output": "91"
},
{
"input": "190",
"output": "10"
},
{
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"output": "1"
},
{
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"output": "1"
},
{
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"output": "18"
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{
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"output": "1"
},
{
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"output": "1"
},
{
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"output": "1"
},
{
"input": "4",
"output": "1"
},
{
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"output": "1"
},
{
"input": "6",
"output": "1"
},
{
"input": "7",
"output": "1"
},
{
"input": "8",
"output": "1"
},
{
"input": "9",
"output": "1"
},
{
"input": "10",
"output": "10"
},
{
"input": "11",
"output": "9"
},
{
"input": "12",
"output": "8"
},
{
"input": "13",
"output": "7"
},
{
"input": "14",
"output": "6"
},
{
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"output": "5"
},
{
"input": "16",
"output": "4"
},
{
"input": "17",
"output": "3"
},
{
"input": "18",
"output": "2"
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{
"input": "19",
"output": "1"
},
{
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{
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},
{
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},
{
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"output": "7"
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{
"input": "24",
"output": "6"
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{
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},
{
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{
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{
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{
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{
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{
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{
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{
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{
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{
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{
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{
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{
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{
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{
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{
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"output": "5"
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{
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{
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{
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{
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{
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{
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{
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{
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{
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{
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{
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{
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{
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{
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{
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{
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{
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{
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{
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{
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{
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},
{
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"output": "2"
},
{
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"output": "1"
},
{
"input": "100",
"output": "100"
},
{
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"output": "100"
},
{
"input": "100",
"output": "100"
},
{
"input": "1000",
"output": "1000"
},
{
"input": "1000",
"output": "1000"
},
{
"input": "1000",
"output": "1000"
},
{
"input": "10000",
"output": "10000"
},
{
"input": "10000",
"output": "10000"
},
{
"input": "101",
"output": "99"
},
{
"input": "110",
"output": "90"
},
{
"input": "1001",
"output": "999"
},
{
"input": "1100",
"output": "900"
},
{
"input": "1010",
"output": "990"
},
{
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"output": "9990"
},
{
"input": "10100",
"output": "9900"
},
{
"input": "102",
"output": "98"
},
{
"input": "120",
"output": "80"
},
{
"input": "1002",
"output": "998"
},
{
"input": "1200",
"output": "800"
},
{
"input": "1020",
"output": "980"
},
{
"input": "10020",
"output": "9980"
},
{
"input": "10200",
"output": "9800"
},
{
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},
{
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"output": "20"
},
{
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"output": "992"
},
{
"input": "1800",
"output": "200"
},
{
"input": "1080",
"output": "920"
},
{
"input": "10080",
"output": "9920"
},
{
"input": "10800",
"output": "9200"
},
{
"input": "109",
"output": "91"
},
{
"input": "190",
"output": "10"
},
{
"input": "1009",
"output": "991"
},
{
"input": "1900",
"output": "100"
},
{
"input": "1090",
"output": "910"
},
{
"input": "10090",
"output": "9910"
},
{
"input": "10900",
"output": "9100"
},
{
"input": "200",
"output": "100"
},
{
"input": "200",
"output": "100"
},
{
"input": "2000",
"output": "1000"
},
{
"input": "2000",
"output": "1000"
},
{
"input": "2000",
"output": "1000"
},
{
"input": "20000",
"output": "10000"
},
{
"input": "20000",
"output": "10000"
},
{
"input": "201",
"output": "99"
},
{
"input": "210",
"output": "90"
},
{
"input": "2001",
"output": "999"
},
{
"input": "2100",
"output": "900"
},
{
"input": "2010",
"output": "990"
},
{
"input": "20010",
"output": "9990"
},
{
"input": "20100",
"output": "9900"
},
{
"input": "202",
"output": "98"
},
{
"input": "220",
"output": "80"
},
{
"input": "2002",
"output": "998"
},
{
"input": "2200",
"output": "800"
},
{
"input": "2020",
"output": "980"
},
{
"input": "20020",
"output": "9980"
},
{
"input": "20200",
"output": "9800"
},
{
"input": "208",
"output": "92"
},
{
"input": "280",
"output": "20"
},
{
"input": "2008",
"output": "992"
},
{
"input": "2800",
"output": "200"
},
{
"input": "2080",
"output": "920"
},
{
"input": "20080",
"output": "9920"
},
{
"input": "20800",
"output": "9200"
},
{
"input": "209",
"output": "91"
},
{
"input": "290",
"output": "10"
},
{
"input": "2009",
"output": "991"
},
{
"input": "2900",
"output": "100"
},
{
"input": "2090",
"output": "910"
},
{
"input": "20090",
"output": "9910"
},
{
"input": "20900",
"output": "9100"
},
{
"input": "800",
"output": "100"
},
{
"input": "800",
"output": "100"
},
{
"input": "8000",
"output": "1000"
},
{
"input": "8000",
"output": "1000"
},
{
"input": "8000",
"output": "1000"
},
{
"input": "80000",
"output": "10000"
},
{
"input": "80000",
"output": "10000"
},
{
"input": "801",
"output": "99"
},
{
"input": "810",
"output": "90"
},
{
"input": "8001",
"output": "999"
},
{
"input": "8100",
"output": "900"
},
{
"input": "8010",
"output": "990"
},
{
"input": "80010",
"output": "9990"
},
{
"input": "80100",
"output": "9900"
},
{
"input": "802",
"output": "98"
},
{
"input": "820",
"output": "80"
},
{
"input": "8002",
"output": "998"
},
{
"input": "8200",
"output": "800"
},
{
"input": "8020",
"output": "980"
},
{
"input": "80020",
"output": "9980"
},
{
"input": "80200",
"output": "9800"
},
{
"input": "808",
"output": "92"
},
{
"input": "880",
"output": "20"
},
{
"input": "8008",
"output": "992"
},
{
"input": "8800",
"output": "200"
},
{
"input": "8080",
"output": "920"
},
{
"input": "80080",
"output": "9920"
},
{
"input": "80800",
"output": "9200"
},
{
"input": "809",
"output": "91"
},
{
"input": "890",
"output": "10"
},
{
"input": "8009",
"output": "991"
},
{
"input": "8900",
"output": "100"
},
{
"input": "8090",
"output": "910"
},
{
"input": "80090",
"output": "9910"
},
{
"input": "80900",
"output": "9100"
},
{
"input": "900",
"output": "100"
},
{
"input": "900",
"output": "100"
},
{
"input": "9000",
"output": "1000"
},
{
"input": "9000",
"output": "1000"
},
{
"input": "9000",
"output": "1000"
},
{
"input": "90000",
"output": "10000"
},
{
"input": "90000",
"output": "10000"
},
{
"input": "901",
"output": "99"
},
{
"input": "910",
"output": "90"
},
{
"input": "9001",
"output": "999"
},
{
"input": "9100",
"output": "900"
},
{
"input": "9010",
"output": "990"
},
{
"input": "90010",
"output": "9990"
},
{
"input": "90100",
"output": "9900"
},
{
"input": "902",
"output": "98"
},
{
"input": "920",
"output": "80"
},
{
"input": "9002",
"output": "998"
},
{
"input": "9200",
"output": "800"
},
{
"input": "9020",
"output": "980"
},
{
"input": "90020",
"output": "9980"
},
{
"input": "90200",
"output": "9800"
},
{
"input": "908",
"output": "92"
},
{
"input": "980",
"output": "20"
},
{
"input": "9008",
"output": "992"
},
{
"input": "9800",
"output": "200"
},
{
"input": "9080",
"output": "920"
},
{
"input": "90080",
"output": "9920"
},
{
"input": "90800",
"output": "9200"
},
{
"input": "909",
"output": "91"
},
{
"input": "990",
"output": "10"
},
{
"input": "9009",
"output": "991"
},
{
"input": "9900",
"output": "100"
},
{
"input": "9090",
"output": "910"
},
{
"input": "90090",
"output": "9910"
},
{
"input": "90900",
"output": "9100"
},
{
"input": "92651241",
"output": "7348759"
}
] | 1,587,786,889
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 271
| 155
| 0
|
n=int(input())
s=str(n)
r=int(s[0])+1
ss=str(r)+'0'*(len(s)-1)
ss=int(ss)
print(ss-n)
|
Title: Lucky Year
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Apart from having lots of holidays throughout the year, residents of Berland also have whole lucky years. Year is considered lucky if it has no more than 1 non-zero digit in its number. So years 100, 40000, 5 are lucky and 12, 3001 and 12345 are not.
You are given current year in Berland. Your task is to find how long will residents of Berland wait till the next lucky year.
Input Specification:
The first line contains integer number *n* (1<=β€<=*n*<=β€<=109) β current year in Berland.
Output Specification:
Output amount of years from the current year to the next lucky one.
Demo Input:
['4\n', '201\n', '4000\n']
Demo Output:
['1\n', '99\n', '1000\n']
Note:
In the first example next lucky year is 5. In the second one β 300. In the third β 5000.
|
```python
n=int(input())
s=str(n)
r=int(s[0])+1
ss=str(r)+'0'*(len(s)-1)
ss=int(ss)
print(ss-n)
```
| 3
|
|
761
|
A
|
Dasha and Stairs
|
PROGRAMMING
| 1,000
|
[
"brute force",
"constructive algorithms",
"implementation",
"math"
] | null | null |
On her way to programming school tiger Dasha faced her first test β a huge staircase!
The steps were numbered from one to infinity. As we know, tigers are very fond of all striped things, it is possible that it has something to do with their color. So on some interval of her way she calculated two values β the number of steps with even and odd numbers.
You need to check whether there is an interval of steps from the *l*-th to the *r*-th (1<=β€<=*l*<=β€<=*r*), for which values that Dasha has found are correct.
|
In the only line you are given two integers *a*, *b* (0<=β€<=*a*,<=*b*<=β€<=100) β the number of even and odd steps, accordingly.
|
In the only line print "YES", if the interval of steps described above exists, and "NO" otherwise.
|
[
"2 3\n",
"3 1\n"
] |
[
"YES\n",
"NO\n"
] |
In the first example one of suitable intervals is from 1 to 5. The interval contains two even stepsΒ β 2 and 4, and three odd: 1, 3 and 5.
| 500
|
[
{
"input": "2 3",
"output": "YES"
},
{
"input": "3 1",
"output": "NO"
},
{
"input": "5 4",
"output": "YES"
},
{
"input": "9 9",
"output": "YES"
},
{
"input": "85 95",
"output": "NO"
},
{
"input": "0 1",
"output": "YES"
},
{
"input": "89 25",
"output": "NO"
},
{
"input": "74 73",
"output": "YES"
},
{
"input": "62 39",
"output": "NO"
},
{
"input": "57 57",
"output": "YES"
},
{
"input": "100 99",
"output": "YES"
},
{
"input": "0 0",
"output": "NO"
},
{
"input": "98 100",
"output": "NO"
},
{
"input": "99 100",
"output": "YES"
},
{
"input": "1 0",
"output": "YES"
},
{
"input": "100 0",
"output": "NO"
},
{
"input": "0 100",
"output": "NO"
},
{
"input": "100 98",
"output": "NO"
},
{
"input": "100 100",
"output": "YES"
},
{
"input": "0 5",
"output": "NO"
},
{
"input": "2 2",
"output": "YES"
}
] | 1,582,462,490
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 21
| 109
| 0
|
n,m=map(int,input().split())
z=abs(n-m)
if((z==0 and n!=0) or z==1):
print("YES")
else:
print("NO")
|
Title: Dasha and Stairs
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
On her way to programming school tiger Dasha faced her first test β a huge staircase!
The steps were numbered from one to infinity. As we know, tigers are very fond of all striped things, it is possible that it has something to do with their color. So on some interval of her way she calculated two values β the number of steps with even and odd numbers.
You need to check whether there is an interval of steps from the *l*-th to the *r*-th (1<=β€<=*l*<=β€<=*r*), for which values that Dasha has found are correct.
Input Specification:
In the only line you are given two integers *a*, *b* (0<=β€<=*a*,<=*b*<=β€<=100) β the number of even and odd steps, accordingly.
Output Specification:
In the only line print "YES", if the interval of steps described above exists, and "NO" otherwise.
Demo Input:
['2 3\n', '3 1\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first example one of suitable intervals is from 1 to 5. The interval contains two even stepsΒ β 2 and 4, and three odd: 1, 3 and 5.
|
```python
n,m=map(int,input().split())
z=abs(n-m)
if((z==0 and n!=0) or z==1):
print("YES")
else:
print("NO")
```
| 3
|
|
612
|
A
|
The Text Splitting
|
PROGRAMMING
| 1,300
|
[
"brute force",
"implementation",
"strings"
] | null | null |
You are given the string *s* of length *n* and the numbers *p*,<=*q*. Split the string *s* to pieces of length *p* and *q*.
For example, the string "Hello" for *p*<==<=2, *q*<==<=3 can be split to the two strings "Hel" and "lo" or to the two strings "He" and "llo".
Note it is allowed to split the string *s* to the strings only of length *p* or to the strings only of length *q* (see the second sample test).
|
The first line contains three positive integers *n*,<=*p*,<=*q* (1<=β€<=*p*,<=*q*<=β€<=*n*<=β€<=100).
The second line contains the string *s* consists of lowercase and uppercase latin letters and digits.
|
If it's impossible to split the string *s* to the strings of length *p* and *q* print the only number "-1".
Otherwise in the first line print integer *k* β the number of strings in partition of *s*.
Each of the next *k* lines should contain the strings in partition. Each string should be of the length *p* or *q*. The string should be in order of their appearing in string *s* β from left to right.
If there are several solutions print any of them.
|
[
"5 2 3\nHello\n",
"10 9 5\nCodeforces\n",
"6 4 5\nPrivet\n",
"8 1 1\nabacabac\n"
] |
[
"2\nHe\nllo\n",
"2\nCodef\norces\n",
"-1\n",
"8\na\nb\na\nc\na\nb\na\nc\n"
] |
none
| 0
|
[
{
"input": "5 2 3\nHello",
"output": "2\nHe\nllo"
},
{
"input": "10 9 5\nCodeforces",
"output": "2\nCodef\norces"
},
{
"input": "6 4 5\nPrivet",
"output": "-1"
},
{
"input": "8 1 1\nabacabac",
"output": "8\na\nb\na\nc\na\nb\na\nc"
},
{
"input": "1 1 1\n1",
"output": "1\n1"
},
{
"input": "10 8 1\nuTl9w4lcdo",
"output": "10\nu\nT\nl\n9\nw\n4\nl\nc\nd\no"
},
{
"input": "20 6 4\nfmFRpk2NrzSvnQC9gB61",
"output": "5\nfmFR\npk2N\nrzSv\nnQC9\ngB61"
},
{
"input": "30 23 6\nWXDjl9kitaDTY673R5xyTlbL9gqeQ6",
"output": "5\nWXDjl9\nkitaDT\nY673R5\nxyTlbL\n9gqeQ6"
},
{
"input": "40 14 3\nSOHBIkWEv7ScrkHgMtFFxP9G7JQLYXFoH1sJDAde",
"output": "6\nSOHBIkWEv7Scrk\nHgMtFFxP9G7JQL\nYXF\noH1\nsJD\nAde"
},
{
"input": "50 16 3\nXCgVJUu4aMQ7HMxZjNxe3XARNiahK303g9y7NV8oN6tWdyXrlu",
"output": "8\nXCgVJUu4aMQ7HMxZ\njNxe3XARNiahK303\ng9y\n7NV\n8oN\n6tW\ndyX\nrlu"
},
{
"input": "60 52 8\nhae0PYwXcW2ziQCOSci5VaElHLZCZI81ULSHgpyG3fuZaP0fHjN4hCKogONj",
"output": "2\nhae0PYwXcW2ziQCOSci5VaElHLZCZI81ULSHgpyG3fuZaP0fHjN4\nhCKogONj"
},
{
"input": "70 50 5\n1BH1ECq7hjzooQOZdbiYHTAgATcP5mxI7kLI9rqA9AriWc9kE5KoLa1zmuTDFsd2ClAPPY",
"output": "14\n1BH1E\nCq7hj\nzooQO\nZdbiY\nHTAgA\nTcP5m\nxI7kL\nI9rqA\n9AriW\nc9kE5\nKoLa1\nzmuTD\nFsd2C\nlAPPY"
},
{
"input": "80 51 8\no2mpu1FCofuiLQb472qczCNHfVzz5TfJtVMrzgN3ff7FwlAY0fQ0ROhWmIX2bggodORNA76bHMjA5yyc",
"output": "10\no2mpu1FC\nofuiLQb4\n72qczCNH\nfVzz5TfJ\ntVMrzgN3\nff7FwlAY\n0fQ0ROhW\nmIX2bggo\ndORNA76b\nHMjA5yyc"
},
{
"input": "90 12 7\nclcImtsw176FFOA6OHGFxtEfEyhFh5bH4iktV0Y8onIcn0soTwiiHUFRWC6Ow36tT5bsQjgrVSTcB8fAVoe7dJIWkE",
"output": "10\nclcImtsw176F\nFOA6OHGFxtEf\nEyhFh5bH4ikt\nV0Y8onIcn0so\nTwiiHUF\nRWC6Ow3\n6tT5bsQ\njgrVSTc\nB8fAVoe\n7dJIWkE"
},
{
"input": "100 25 5\n2SRB9mRpXMRND5zQjeRxc4GhUBlEQSmLgnUtB9xTKoC5QM9uptc8dKwB88XRJy02r7edEtN2C6D60EjzK1EHPJcWNj6fbF8kECeB",
"output": "20\n2SRB9\nmRpXM\nRND5z\nQjeRx\nc4GhU\nBlEQS\nmLgnU\ntB9xT\nKoC5Q\nM9upt\nc8dKw\nB88XR\nJy02r\n7edEt\nN2C6D\n60Ejz\nK1EHP\nJcWNj\n6fbF8\nkECeB"
},
{
"input": "100 97 74\nxL8yd8lENYnXZs28xleyci4SxqsjZqkYzkEbQXfLQ4l4gKf9QQ9xjBjeZ0f9xQySf5psDUDkJEtPLsa62n4CLc6lF6E2yEqvt4EJ",
"output": "-1"
},
{
"input": "51 25 11\nwpk5wqrB6d3qE1slUrzJwMFafnnOu8aESlvTEb7Pp42FDG2iGQn",
"output": "-1"
},
{
"input": "70 13 37\nfzL91QIJvNoZRP4A9aNRT2GTksd8jEb1713pnWFaCGKHQ1oYvlTHXIl95lqyZRKJ1UPYvT",
"output": "-1"
},
{
"input": "10 3 1\nXQ2vXLPShy",
"output": "10\nX\nQ\n2\nv\nX\nL\nP\nS\nh\ny"
},
{
"input": "4 2 3\naaaa",
"output": "2\naa\naa"
},
{
"input": "100 1 1\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaabbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb",
"output": "100\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb"
},
{
"input": "99 2 4\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "-1"
},
{
"input": "11 2 3\nhavanahavan",
"output": "4\nha\nvan\naha\nvan"
},
{
"input": "100 2 2\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "50\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa"
},
{
"input": "17 3 5\ngopstopmipodoshli",
"output": "5\ngop\nsto\npmi\npod\noshli"
},
{
"input": "5 4 3\nfoyku",
"output": "-1"
},
{
"input": "99 2 2\n123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789",
"output": "-1"
},
{
"input": "99 2 2\nrecursionishellrecursionishellrecursionishellrecursionishellrecursionishellrecursionishelldontuseit",
"output": "-1"
},
{
"input": "11 2 3\nqibwnnvqqgo",
"output": "4\nqi\nbwn\nnvq\nqgo"
},
{
"input": "4 4 3\nhhhh",
"output": "1\nhhhh"
},
{
"input": "99 2 2\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "-1"
},
{
"input": "99 2 5\nhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh",
"output": "21\nhh\nhh\nhhhhh\nhhhhh\nhhhhh\nhhhhh\nhhhhh\nhhhhh\nhhhhh\nhhhhh\nhhhhh\nhhhhh\nhhhhh\nhhhhh\nhhhhh\nhhhhh\nhhhhh\nhhhhh\nhhhhh\nhhhhh\nhhhhh"
},
{
"input": "10 5 9\nCodeforces",
"output": "2\nCodef\norces"
},
{
"input": "10 5 9\naaaaaaaaaa",
"output": "2\naaaaa\naaaaa"
},
{
"input": "11 3 2\nmlmqpohwtsf",
"output": "5\nmlm\nqp\noh\nwt\nsf"
},
{
"input": "3 3 2\nzyx",
"output": "1\nzyx"
},
{
"input": "100 3 3\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "-1"
},
{
"input": "4 2 3\nzyxw",
"output": "2\nzy\nxw"
},
{
"input": "3 2 3\nejt",
"output": "1\nejt"
},
{
"input": "5 2 4\nzyxwv",
"output": "-1"
},
{
"input": "100 1 1\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "100\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na"
},
{
"input": "100 5 4\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "25\naaaa\naaaa\naaaa\naaaa\naaaa\naaaa\naaaa\naaaa\naaaa\naaaa\naaaa\naaaa\naaaa\naaaa\naaaa\naaaa\naaaa\naaaa\naaaa\naaaa\naaaa\naaaa\naaaa\naaaa\naaaa"
},
{
"input": "3 2 2\nzyx",
"output": "-1"
},
{
"input": "99 2 2\nhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh",
"output": "-1"
},
{
"input": "26 8 9\nabcabcabcabcabcabcabcabcab",
"output": "3\nabcabcab\ncabcabcab\ncabcabcab"
},
{
"input": "6 3 5\naaaaaa",
"output": "2\naaa\naaa"
},
{
"input": "3 2 3\nzyx",
"output": "1\nzyx"
},
{
"input": "5 5 2\naaaaa",
"output": "1\naaaaa"
},
{
"input": "4 3 2\nzyxw",
"output": "2\nzy\nxw"
},
{
"input": "5 4 3\nzyxwv",
"output": "-1"
},
{
"input": "95 3 29\nabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcab",
"output": "23\nabc\nabc\nabc\nabc\nabc\nabc\nabc\nabc\nabc\nabc\nabc\nabc\nabc\nabc\nabc\nabc\nabc\nabc\nabc\nabc\nabc\nabc\nabcabcabcabcabcabcabcabcabcab"
},
{
"input": "3 2 2\naaa",
"output": "-1"
},
{
"input": "91 62 3\nfjzhkfwzoabaauvbkuzaahkozofaophaafhfpuhobufawkzbavaazwavwppfwapkapaofbfjwaavajojgjguahphofj",
"output": "-1"
},
{
"input": "99 2 2\nabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabc",
"output": "-1"
},
{
"input": "56 13 5\nabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcab",
"output": "8\nabcabcabcabca\nbcabcabcabcab\ncabca\nbcabc\nabcab\ncabca\nbcabc\nabcab"
},
{
"input": "79 7 31\nabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabca",
"output": "-1"
},
{
"input": "92 79 6\nxlvplpckwnhmctoethhslkcyashqtsoeltriddglfwtgkfvkvgytygbcyohrvcxvosdioqvackxiuifmkgdngvbbudcb",
"output": "-1"
},
{
"input": "48 16 13\nibhfinipihcbsqnvtgsbkobepmwymlyfmlfgblvhlfhyojsy",
"output": "3\nibhfinipihcbsqnv\ntgsbkobepmwymlyf\nmlfgblvhlfhyojsy"
},
{
"input": "16 3 7\naaaaaaaaaaaaaaaa",
"output": "4\naaa\naaa\naaa\naaaaaaa"
},
{
"input": "11 10 3\naaaaaaaaaaa",
"output": "-1"
},
{
"input": "11 8 8\naaaaaaaaaaa",
"output": "-1"
},
{
"input": "11 7 3\naaaaaaaaaaa",
"output": "-1"
},
{
"input": "41 3 4\nabcabcabcabcabcabcabcabcabcabcabcabcabcab",
"output": "11\nabc\nabc\nabc\nabca\nbcab\ncabc\nabca\nbcab\ncabc\nabca\nbcab"
},
{
"input": "11 3 2\naaaaaaaaaaa",
"output": "5\naaa\naa\naa\naa\naa"
},
{
"input": "14 9 4\nabcdefghijklmn",
"output": "-1"
},
{
"input": "9 9 5\n123456789",
"output": "1\n123456789"
},
{
"input": "92 10 41\nmeotryyfneonmnrvfnhqlehlxtvpracifpadcofecvbikoitrlgeftiqofpvacgocrdiquhatlqosqvtduenaqkwrnnw",
"output": "3\nmeotryyfne\nonmnrvfnhqlehlxtvpracifpadcofecvbikoitrlg\neftiqofpvacgocrdiquhatlqosqvtduenaqkwrnnw"
},
{
"input": "17 16 3\ndxyhgtsxtuyljmclj",
"output": "-1"
},
{
"input": "82 13 30\nfmtwumakkejtolxqxrnydhqoufwtdwldfxcfjrndauqcarhbwmdwxsxfbqjsfspuxobywhcrvlndsdmkqd",
"output": "5\nfmtwumakkejto\nlxqxrnydhqouf\nwtdwldfxcfjrn\ndauqcarhbwmdw\nxsxfbqjsfspuxobywhcrvlndsdmkqd"
},
{
"input": "95 3 3\nihnfqcswushyoirjxlxxnwqtwtaowounkumxukwpacxwatimhhhoggqwkkspcplvyndfukbxickcixidgxkjtnpkoeiwlor",
"output": "-1"
},
{
"input": "7 5 3\nzyxwvut",
"output": "-1"
},
{
"input": "17 16 4\nctvfhkiakagcilrdw",
"output": "-1"
}
] | 1,613,387,060
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 72
| 93
| 1,331,200
|
n, p, q = [int(j) for j in input().split()]
s = str(input())
x, y = -1, -1
found = False
for j in range(101):
for k in range(101):
res = j * p + k * q
if res == n:
x, y = j, k
found = True
break
elif res > n:
break
if found == True or j * p > n:
break
if found == False:
print(-1)
else:
p1, p2 = 0, p
List = []
for j in range(x):
List.append(s[p1 : p2])
p1 += p
p2 += p
p2 = p1 + q
for j in range(y):
List.append(s[p1 : p2])
p1 += q
p2 += q
print(len(List))
for j in range(len(List)):
print(List[j])
|
Title: The Text Splitting
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given the string *s* of length *n* and the numbers *p*,<=*q*. Split the string *s* to pieces of length *p* and *q*.
For example, the string "Hello" for *p*<==<=2, *q*<==<=3 can be split to the two strings "Hel" and "lo" or to the two strings "He" and "llo".
Note it is allowed to split the string *s* to the strings only of length *p* or to the strings only of length *q* (see the second sample test).
Input Specification:
The first line contains three positive integers *n*,<=*p*,<=*q* (1<=β€<=*p*,<=*q*<=β€<=*n*<=β€<=100).
The second line contains the string *s* consists of lowercase and uppercase latin letters and digits.
Output Specification:
If it's impossible to split the string *s* to the strings of length *p* and *q* print the only number "-1".
Otherwise in the first line print integer *k* β the number of strings in partition of *s*.
Each of the next *k* lines should contain the strings in partition. Each string should be of the length *p* or *q*. The string should be in order of their appearing in string *s* β from left to right.
If there are several solutions print any of them.
Demo Input:
['5 2 3\nHello\n', '10 9 5\nCodeforces\n', '6 4 5\nPrivet\n', '8 1 1\nabacabac\n']
Demo Output:
['2\nHe\nllo\n', '2\nCodef\norces\n', '-1\n', '8\na\nb\na\nc\na\nb\na\nc\n']
Note:
none
|
```python
n, p, q = [int(j) for j in input().split()]
s = str(input())
x, y = -1, -1
found = False
for j in range(101):
for k in range(101):
res = j * p + k * q
if res == n:
x, y = j, k
found = True
break
elif res > n:
break
if found == True or j * p > n:
break
if found == False:
print(-1)
else:
p1, p2 = 0, p
List = []
for j in range(x):
List.append(s[p1 : p2])
p1 += p
p2 += p
p2 = p1 + q
for j in range(y):
List.append(s[p1 : p2])
p1 += q
p2 += q
print(len(List))
for j in range(len(List)):
print(List[j])
```
| 3
|
|
20
|
A
|
BerOS file system
|
PROGRAMMING
| 1,700
|
[
"implementation"
] |
A. BerOS file system
|
2
|
64
|
The new operating system BerOS has a nice feature. It is possible to use any number of characters '/' as a delimiter in path instead of one traditional '/'. For example, strings //usr///local//nginx/sbin// and /usr/local/nginx///sbin are equivalent. The character '/' (or some sequence of such characters) at the end of the path is required only in case of the path to the root directory, which can be represented as single character '/'.
A path called normalized if it contains the smallest possible number of characters '/'.
Your task is to transform a given path to the normalized form.
|
The first line of the input contains only lowercase Latin letters and character '/'Β β the path to some directory. All paths start with at least one character '/'. The length of the given line is no more than 100 characters, it is not empty.
|
The path in normalized form.
|
[
"//usr///local//nginx/sbin\n"
] |
[
"/usr/local/nginx/sbin\n"
] |
none
| 500
|
[
{
"input": "//usr///local//nginx/sbin",
"output": "/usr/local/nginx/sbin"
},
{
"input": "////a//b/////g",
"output": "/a/b/g"
},
{
"input": "/a/b/c",
"output": "/a/b/c"
},
{
"input": "/",
"output": "/"
},
{
"input": "////",
"output": "/"
},
{
"input": "/a//aa/a//",
"output": "/a/aa/a"
},
{
"input": "/aa//b/aa",
"output": "/aa/b/aa"
},
{
"input": "////////////////////////////////////////////////////////////////////////////////////////////////////",
"output": "/"
},
{
"input": "/opt///pokerband///srvb/opt///pokerband///srvb////pokerband///srvb/",
"output": "/opt/pokerband/srvb/opt/pokerband/srvb/pokerband/srvb"
},
{
"input": "/root/pokerband/trunk/server/game/code/src/main/java/com/uosipa/pokerband/gameserver/game/",
"output": "/root/pokerband/trunk/server/game/code/src/main/java/com/uosipa/pokerband/gameserver/game"
},
{
"input": "/root//pokerband/trunk/server/game/code/src///main/java/com/uosipa/pokerband//gameserver/game//",
"output": "/root/pokerband/trunk/server/game/code/src/main/java/com/uosipa/pokerband/gameserver/game"
},
{
"input": "/root//pokerband/trunk/server/game/code/src///main/java/com/uosipa/pokerband//gameserver/game//test/",
"output": "/root/pokerband/trunk/server/game/code/src/main/java/com/uosipa/pokerband/gameserver/game/test"
},
{
"input": "/root//pokerband/trunk/server/game/code/src///main/java/com/uosipa/pokerband//gameserver/game//testt",
"output": "/root/pokerband/trunk/server/game/code/src/main/java/com/uosipa/pokerband/gameserver/game/testt"
},
{
"input": "///a//a////a/a//a//a//a/////",
"output": "/a/a/a/a/a/a/a"
},
{
"input": "/zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz",
"output": "/zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz"
},
{
"input": "/zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz/",
"output": "/zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz"
},
{
"input": "/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a",
"output": "/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a"
},
{
"input": "///////////////////////////////////////////////////////////////////////////////////////////////////z",
"output": "/z"
},
{
"input": "/z//////////////////////////////////////////////////////////////////////////////////////////////////",
"output": "/z"
}
] | 1,674,109,233
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 19
| 186
| 0
|
print("/"+"/".join(filter(None, input().split("/"))))
|
Title: BerOS file system
Time Limit: 2 seconds
Memory Limit: 64 megabytes
Problem Description:
The new operating system BerOS has a nice feature. It is possible to use any number of characters '/' as a delimiter in path instead of one traditional '/'. For example, strings //usr///local//nginx/sbin// and /usr/local/nginx///sbin are equivalent. The character '/' (or some sequence of such characters) at the end of the path is required only in case of the path to the root directory, which can be represented as single character '/'.
A path called normalized if it contains the smallest possible number of characters '/'.
Your task is to transform a given path to the normalized form.
Input Specification:
The first line of the input contains only lowercase Latin letters and character '/'Β β the path to some directory. All paths start with at least one character '/'. The length of the given line is no more than 100 characters, it is not empty.
Output Specification:
The path in normalized form.
Demo Input:
['//usr///local//nginx/sbin\n']
Demo Output:
['/usr/local/nginx/sbin\n']
Note:
none
|
```python
print("/"+"/".join(filter(None, input().split("/"))))
```
| 3.9535
|
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<=β€<=*n*<=<<=109) β 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"
}
] | 1,623,375,612
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 54
| 154
| 0
|
n = int(input())
if n == 0:
print(0, 0, 0)
elif n == 1:
print(0, 0, 1)
elif n == 2:
print(0, 1, 1)
else:
arr = [0, 1]
i = 1
while True:
m = arr[i]+arr[i-1]
if m <= n:
arr.append(m)
i += 1
else:
break
print(arr[-2], arr[-4], arr[-5])
|
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<=β€<=*n*<=<<=109) β 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 == 0:
print(0, 0, 0)
elif n == 1:
print(0, 0, 1)
elif n == 2:
print(0, 1, 1)
else:
arr = [0, 1]
i = 1
while True:
m = arr[i]+arr[i-1]
if m <= n:
arr.append(m)
i += 1
else:
break
print(arr[-2], arr[-4], arr[-5])
```
| 3
|
|
776
|
A
|
A Serial Killer
|
PROGRAMMING
| 900
|
[
"brute force",
"implementation",
"strings"
] | null | null |
Our beloved detective, Sherlock is currently trying to catch a serial killer who kills a person each day. Using his powers of deduction, he came to know that the killer has a strategy for selecting his next victim.
The killer starts with two potential victims on his first day, selects one of these two, kills selected victim and replaces him with a new person. He repeats this procedure each day. This way, each day he has two potential victims to choose from. Sherlock knows the initial two potential victims. Also, he knows the murder that happened on a particular day and the new person who replaced this victim.
You need to help him get all the pairs of potential victims at each day so that Sherlock can observe some pattern.
|
First line of input contains two names (length of each of them doesn't exceed 10), the two initials potential victims. Next line contains integer *n* (1<=β€<=*n*<=β€<=1000), the number of days.
Next *n* lines contains two names (length of each of them doesn't exceed 10), first being the person murdered on this day and the second being the one who replaced that person.
The input format is consistent, that is, a person murdered is guaranteed to be from the two potential victims at that time. Also, all the names are guaranteed to be distinct and consists of lowercase English letters.
|
Output *n*<=+<=1 lines, the *i*-th line should contain the two persons from which the killer selects for the *i*-th murder. The (*n*<=+<=1)-th line should contain the two persons from which the next victim is selected. In each line, the two names can be printed in any order.
|
[
"ross rachel\n4\nross joey\nrachel phoebe\nphoebe monica\nmonica chandler\n",
"icm codeforces\n1\ncodeforces technex\n"
] |
[
"ross rachel\njoey rachel\njoey phoebe\njoey monica\njoey chandler\n",
"icm codeforces\nicm technex\n"
] |
In first example, the killer starts with ross and rachel.
- After day 1, ross is killed and joey appears. - After day 2, rachel is killed and phoebe appears. - After day 3, phoebe is killed and monica appears. - After day 4, monica is killed and chandler appears.
| 500
|
[
{
"input": "ross rachel\n4\nross joey\nrachel phoebe\nphoebe monica\nmonica chandler",
"output": "ross rachel\njoey rachel\njoey phoebe\njoey monica\njoey chandler"
},
{
"input": "icm codeforces\n1\ncodeforces technex",
"output": "icm codeforces\nicm technex"
},
{
"input": "a b\n3\na c\nb d\nd e",
"output": "a b\nc b\nc d\nc e"
},
{
"input": "ze udggmyop\n4\nze szhrbmft\nudggmyop mjorab\nszhrbmft ojdtfnzxj\nojdtfnzxj yjlkg",
"output": "ze udggmyop\nszhrbmft udggmyop\nszhrbmft mjorab\nojdtfnzxj mjorab\nyjlkg mjorab"
},
{
"input": "q s\n10\nq b\nb j\ns g\nj f\nf m\ng c\nc a\nm d\nd z\nz o",
"output": "q s\nb s\nj s\nj g\nf g\nm g\nm c\nm a\nd a\nz a\no a"
},
{
"input": "iii iiiiii\n7\niii iiiiiiiiii\niiiiiiiiii iiii\niiii i\niiiiii iiiiiiii\niiiiiiii iiiiiiiii\ni iiiii\niiiii ii",
"output": "iii iiiiii\niiiiiiiiii iiiiii\niiii iiiiii\ni iiiiii\ni iiiiiiii\ni iiiiiiiii\niiiii iiiiiiiii\nii iiiiiiiii"
},
{
"input": "bwyplnjn zkms\n26\nzkms nzmcsytxh\nnzmcsytxh yujsb\nbwyplnjn gtbzhudpb\ngtbzhudpb hpk\nyujsb xvy\nhpk wrwnfokml\nwrwnfokml ndouuikw\nndouuikw ucgrja\nucgrja tgfmpldz\nxvy nycrfphn\nnycrfphn quvs\nquvs htdy\nhtdy k\ntgfmpldz xtdpkxm\nxtdpkxm suwqxs\nk fv\nsuwqxs qckllwy\nqckllwy diun\nfv lefa\nlefa gdoqjysx\ndiun dhpz\ngdoqjysx bdmqdyt\ndhpz dgz\ndgz v\nbdmqdyt aswy\naswy ydkayhlrnm",
"output": "bwyplnjn zkms\nbwyplnjn nzmcsytxh\nbwyplnjn yujsb\ngtbzhudpb yujsb\nhpk yujsb\nhpk xvy\nwrwnfokml xvy\nndouuikw xvy\nucgrja xvy\ntgfmpldz xvy\ntgfmpldz nycrfphn\ntgfmpldz quvs\ntgfmpldz htdy\ntgfmpldz k\nxtdpkxm k\nsuwqxs k\nsuwqxs fv\nqckllwy fv\ndiun fv\ndiun lefa\ndiun gdoqjysx\ndhpz gdoqjysx\ndhpz bdmqdyt\ndgz bdmqdyt\nv bdmqdyt\nv aswy\nv ydkayhlrnm"
},
{
"input": "wxz hbeqwqp\n7\nhbeqwqp cpieghnszh\ncpieghnszh tlqrpd\ntlqrpd ttwrtio\nttwrtio xapvds\nxapvds zk\nwxz yryk\nzk b",
"output": "wxz hbeqwqp\nwxz cpieghnszh\nwxz tlqrpd\nwxz ttwrtio\nwxz xapvds\nwxz zk\nyryk zk\nyryk b"
},
{
"input": "wced gnsgv\n23\ngnsgv japawpaf\njapawpaf nnvpeu\nnnvpeu a\na ddupputljq\nddupputljq qyhnvbh\nqyhnvbh pqwijl\nwced khuvs\nkhuvs bjkh\npqwijl ysacmboc\nbjkh srf\nsrf jknoz\njknoz hodf\nysacmboc xqtkoyh\nhodf rfp\nxqtkoyh bivgnwqvoe\nbivgnwqvoe nknf\nnknf wuig\nrfp e\ne bqqknq\nwuig sznhhhu\nbqqknq dhrtdld\ndhrtdld n\nsznhhhu bguylf",
"output": "wced gnsgv\nwced japawpaf\nwced nnvpeu\nwced a\nwced ddupputljq\nwced qyhnvbh\nwced pqwijl\nkhuvs pqwijl\nbjkh pqwijl\nbjkh ysacmboc\nsrf ysacmboc\njknoz ysacmboc\nhodf ysacmboc\nhodf xqtkoyh\nrfp xqtkoyh\nrfp bivgnwqvoe\nrfp nknf\nrfp wuig\ne wuig\nbqqknq wuig\nbqqknq sznhhhu\ndhrtdld sznhhhu\nn sznhhhu\nn bguylf"
},
{
"input": "qqqqqqqqqq qqqqqqqq\n3\nqqqqqqqq qqqqqqqqq\nqqqqqqqqq qqqqq\nqqqqq q",
"output": "qqqqqqqqqq qqqqqqqq\nqqqqqqqqqq qqqqqqqqq\nqqqqqqqqqq qqqqq\nqqqqqqqqqq q"
},
{
"input": "wwwww w\n8\nwwwww wwwwwwww\nwwwwwwww wwwwwwwww\nwwwwwwwww wwwwwwwwww\nw www\nwwwwwwwwww wwww\nwwww ww\nwww wwwwww\nwwwwww wwwwwww",
"output": "wwwww w\nwwwwwwww w\nwwwwwwwww w\nwwwwwwwwww w\nwwwwwwwwww www\nwwww www\nww www\nww wwwwww\nww wwwwwww"
},
{
"input": "k d\n17\nk l\nd v\nv z\nl r\nz i\nr s\ns p\np w\nw j\nj h\ni c\nh m\nm q\nc o\no g\nq x\nx n",
"output": "k d\nl d\nl v\nl z\nr z\nr i\ns i\np i\nw i\nj i\nh i\nh c\nm c\nq c\nq o\nq g\nx g\nn g"
}
] | 1,647,694,132
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 57
| 46
| 0
|
x, y = input().split()
print(x, y)
for _ in range(int(input())):
a, b = input().split()
if x == a:
x = b
elif y == a:
y = b
print(x, y)
|
Title: A Serial Killer
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Our beloved detective, Sherlock is currently trying to catch a serial killer who kills a person each day. Using his powers of deduction, he came to know that the killer has a strategy for selecting his next victim.
The killer starts with two potential victims on his first day, selects one of these two, kills selected victim and replaces him with a new person. He repeats this procedure each day. This way, each day he has two potential victims to choose from. Sherlock knows the initial two potential victims. Also, he knows the murder that happened on a particular day and the new person who replaced this victim.
You need to help him get all the pairs of potential victims at each day so that Sherlock can observe some pattern.
Input Specification:
First line of input contains two names (length of each of them doesn't exceed 10), the two initials potential victims. Next line contains integer *n* (1<=β€<=*n*<=β€<=1000), the number of days.
Next *n* lines contains two names (length of each of them doesn't exceed 10), first being the person murdered on this day and the second being the one who replaced that person.
The input format is consistent, that is, a person murdered is guaranteed to be from the two potential victims at that time. Also, all the names are guaranteed to be distinct and consists of lowercase English letters.
Output Specification:
Output *n*<=+<=1 lines, the *i*-th line should contain the two persons from which the killer selects for the *i*-th murder. The (*n*<=+<=1)-th line should contain the two persons from which the next victim is selected. In each line, the two names can be printed in any order.
Demo Input:
['ross rachel\n4\nross joey\nrachel phoebe\nphoebe monica\nmonica chandler\n', 'icm codeforces\n1\ncodeforces technex\n']
Demo Output:
['ross rachel\njoey rachel\njoey phoebe\njoey monica\njoey chandler\n', 'icm codeforces\nicm technex\n']
Note:
In first example, the killer starts with ross and rachel.
- After day 1, ross is killed and joey appears. - After day 2, rachel is killed and phoebe appears. - After day 3, phoebe is killed and monica appears. - After day 4, monica is killed and chandler appears.
|
```python
x, y = input().split()
print(x, y)
for _ in range(int(input())):
a, b = input().split()
if x == a:
x = b
elif y == a:
y = b
print(x, y)
```
| 3
|
|
242
|
B
|
Big Segment
|
PROGRAMMING
| 1,100
|
[
"implementation",
"sortings"
] | null | null |
A coordinate line has *n* segments, the *i*-th segment starts at the position *l**i* and ends at the position *r**i*. We will denote such a segment as [*l**i*,<=*r**i*].
You have suggested that one of the defined segments covers all others. In other words, there is such segment in the given set, which contains all other ones. Now you want to test your assumption. Find in the given set the segment which covers all other segments, and print its number. If such a segment doesn't exist, print -1.
Formally we will assume that segment [*a*,<=*b*] covers segment [*c*,<=*d*], if they meet this condition *a*<=β€<=*c*<=β€<=*d*<=β€<=*b*.
|
The first line contains integer *n* (1<=β€<=*n*<=β€<=105) β the number of segments. Next *n* lines contain the descriptions of the segments. The *i*-th line contains two space-separated integers *l**i*,<=*r**i* (1<=β€<=*l**i*<=β€<=*r**i*<=β€<=109) β the borders of the *i*-th segment.
It is guaranteed that no two segments coincide.
|
Print a single integer β the number of the segment that covers all other segments in the set. If there's no solution, print -1.
The segments are numbered starting from 1 in the order in which they appear in the input.
|
[
"3\n1 1\n2 2\n3 3\n",
"6\n1 5\n2 3\n1 10\n7 10\n7 7\n10 10\n"
] |
[
"-1\n",
"3\n"
] |
none
| 1,000
|
[
{
"input": "3\n1 1\n2 2\n3 3",
"output": "-1"
},
{
"input": "6\n1 5\n2 3\n1 10\n7 10\n7 7\n10 10",
"output": "3"
},
{
"input": "4\n1 5\n2 2\n2 4\n2 5",
"output": "1"
},
{
"input": "5\n3 3\n1 3\n2 2\n2 3\n1 2",
"output": "2"
},
{
"input": "7\n7 7\n8 8\n3 7\n1 6\n1 7\n4 7\n2 8",
"output": "-1"
},
{
"input": "3\n2 5\n3 4\n2 3",
"output": "1"
},
{
"input": "16\n15 15\n8 12\n6 9\n15 16\n8 14\n3 12\n7 19\n9 13\n5 16\n9 17\n10 15\n9 14\n9 9\n18 19\n5 15\n6 19",
"output": "-1"
},
{
"input": "9\n1 10\n7 8\n6 7\n1 4\n5 9\n2 8\n3 10\n1 1\n2 3",
"output": "1"
},
{
"input": "1\n1 100000",
"output": "1"
},
{
"input": "6\n2 2\n3 3\n3 5\n4 5\n1 1\n1 5",
"output": "6"
},
{
"input": "33\n2 18\n4 14\n2 16\n10 12\n4 6\n9 17\n2 8\n4 12\n8 20\n1 10\n11 14\n11 17\n8 15\n3 16\n3 4\n6 9\n6 19\n4 17\n17 19\n6 16\n3 12\n1 7\n6 20\n8 16\n12 19\n1 3\n12 18\n6 11\n7 20\n16 18\n4 15\n3 15\n15 19",
"output": "-1"
},
{
"input": "34\n3 8\n5 9\n2 9\n1 4\n3 7\n3 3\n8 9\n6 10\n4 7\n6 7\n5 8\n5 10\n1 5\n8 8\n2 5\n3 5\n7 7\n2 8\n4 5\n1 1\n7 9\n5 6\n2 3\n1 2\n2 4\n8 10\n7 8\n1 3\n4 8\n9 10\n1 7\n10 10\n2 2\n1 8",
"output": "-1"
},
{
"input": "55\n3 4\n6 8\n9 10\n3 9\n9 9\n2 5\n4 8\n3 8\n8 10\n1 1\n4 9\n10 10\n6 6\n8 8\n1 8\n5 5\n4 5\n5 9\n2 2\n3 10\n4 6\n3 6\n1 6\n1 7\n6 10\n2 6\n3 7\n2 4\n4 4\n5 10\n1 4\n2 9\n1 3\n7 9\n7 8\n1 9\n1 10\n2 8\n8 9\n6 7\n1 2\n6 9\n7 7\n4 7\n3 3\n2 7\n4 10\n7 10\n2 3\n2 10\n5 7\n3 5\n5 8\n1 5\n5 6",
"output": "37"
},
{
"input": "1\n999999999 1000000000",
"output": "1"
},
{
"input": "3\n1 20\n2 22\n3 18",
"output": "-1"
},
{
"input": "1\n1000000000 1000000000",
"output": "1"
},
{
"input": "2\n100001 100008\n100005 100006",
"output": "1"
},
{
"input": "1\n1000000 10000000",
"output": "1"
},
{
"input": "3\n3 6\n2 4\n1 5",
"output": "-1"
},
{
"input": "2\n3 5\n1 2",
"output": "-1"
}
] | 1,662,149,152
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 45
| 530
| 135,270,400
|
from collections import defaultdict, deque
from functools import lru_cache
from heapq import heappush, heappop
from typing import Counter
import math
hpop = heappop
hpush = heappush
"""
num = int(input())
arr = list(map(int, input().split()))
a,b = map(int, input().split())
graph = defaultdict(list)
for i in range(#)
graph[a].append(b)
graph[b].append(a)
MOD = 10**9 + 7
"""
def solution():
n = int(input())
mn = float("inf")
mx = float("-inf")
res = [-1,mn,mx]
for i in range(n):
a,b = map(int, input().split())
mn = min(mn, a)
mx = max(mx, b)
if a == mn and b == mx:
res = [i,a,b]
if mn == float("inf") or mx == float("-inf"):
print(-1)
i,a,b = res
if mn == a and mx == b:
print(i+1)
else:
print(-1)
def main():
t = 1
#t = int(input())
# n + 1 diffrences
for _ in range(t):
solution()
import sys
import threading
sys.setrecursionlimit(1 << 30)
threading.stack_size(1 << 27)
thread = threading.Thread(target=main)
thread.start(); thread.join()
#main()
|
Title: Big Segment
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A coordinate line has *n* segments, the *i*-th segment starts at the position *l**i* and ends at the position *r**i*. We will denote such a segment as [*l**i*,<=*r**i*].
You have suggested that one of the defined segments covers all others. In other words, there is such segment in the given set, which contains all other ones. Now you want to test your assumption. Find in the given set the segment which covers all other segments, and print its number. If such a segment doesn't exist, print -1.
Formally we will assume that segment [*a*,<=*b*] covers segment [*c*,<=*d*], if they meet this condition *a*<=β€<=*c*<=β€<=*d*<=β€<=*b*.
Input Specification:
The first line contains integer *n* (1<=β€<=*n*<=β€<=105) β the number of segments. Next *n* lines contain the descriptions of the segments. The *i*-th line contains two space-separated integers *l**i*,<=*r**i* (1<=β€<=*l**i*<=β€<=*r**i*<=β€<=109) β the borders of the *i*-th segment.
It is guaranteed that no two segments coincide.
Output Specification:
Print a single integer β the number of the segment that covers all other segments in the set. If there's no solution, print -1.
The segments are numbered starting from 1 in the order in which they appear in the input.
Demo Input:
['3\n1 1\n2 2\n3 3\n', '6\n1 5\n2 3\n1 10\n7 10\n7 7\n10 10\n']
Demo Output:
['-1\n', '3\n']
Note:
none
|
```python
from collections import defaultdict, deque
from functools import lru_cache
from heapq import heappush, heappop
from typing import Counter
import math
hpop = heappop
hpush = heappush
"""
num = int(input())
arr = list(map(int, input().split()))
a,b = map(int, input().split())
graph = defaultdict(list)
for i in range(#)
graph[a].append(b)
graph[b].append(a)
MOD = 10**9 + 7
"""
def solution():
n = int(input())
mn = float("inf")
mx = float("-inf")
res = [-1,mn,mx]
for i in range(n):
a,b = map(int, input().split())
mn = min(mn, a)
mx = max(mx, b)
if a == mn and b == mx:
res = [i,a,b]
if mn == float("inf") or mx == float("-inf"):
print(-1)
i,a,b = res
if mn == a and mx == b:
print(i+1)
else:
print(-1)
def main():
t = 1
#t = int(input())
# n + 1 diffrences
for _ in range(t):
solution()
import sys
import threading
sys.setrecursionlimit(1 << 30)
threading.stack_size(1 << 27)
thread = threading.Thread(target=main)
thread.start(); thread.join()
#main()
```
| 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<=β€<=*n*<=β€<=50) β the length of the ticket number that needs to be checked. The second line contains an integer whose length equals exactly *n* β 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 (4ββ β7).
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"
}
] | 1,651,397,994
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 46
| 92
| 0
|
n=int(input())
b=input()
a=[]
da=True
for i in b:
a.append(int(i))
for i in a:
if i!=7 and i!=4:
da=False
break
if sum(a[0:n//2])==sum(a[n//2:]) and da:
print('YES')
else:
print('NO')
|
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<=β€<=*n*<=β€<=50) β the length of the ticket number that needs to be checked. The second line contains an integer whose length equals exactly *n* β 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 (4ββ β7).
In the second sample the ticket number is not the lucky number.
|
```python
n=int(input())
b=input()
a=[]
da=True
for i in b:
a.append(int(i))
for i in a:
if i!=7 and i!=4:
da=False
break
if sum(a[0:n//2])==sum(a[n//2:]) and da:
print('YES')
else:
print('NO')
```
| 3
|
|
845
|
A
|
Chess Tourney
|
PROGRAMMING
| 1,100
|
[
"implementation",
"sortings"
] | null | null |
Berland annual chess tournament is coming!
Organizers have gathered 2Β·*n* chess players who should be divided into two teams with *n* people each. The first team is sponsored by BerOil and the second team is sponsored by BerMobile. Obviously, organizers should guarantee the win for the team of BerOil.
Thus, organizers should divide all 2Β·*n* players into two teams with *n* people each in such a way that the first team always wins.
Every chess player has its rating *r**i*. It is known that chess player with the greater rating always wins the player with the lower rating. If their ratings are equal then any of the players can win.
After teams assignment there will come a drawing to form *n* pairs of opponents: in each pair there is a player from the first team and a player from the second team. Every chess player should be in exactly one pair. Every pair plays once. The drawing is totally random.
Is it possible to divide all 2Β·*n* players into two teams with *n* people each so that the player from the first team in every pair wins regardless of the results of the drawing?
|
The first line contains one integer *n* (1<=β€<=*n*<=β€<=100).
The second line contains 2Β·*n* integers *a*1,<=*a*2,<=... *a*2*n* (1<=β€<=*a**i*<=β€<=1000).
|
If it's possible to divide all 2Β·*n* players into two teams with *n* people each so that the player from the first team in every pair wins regardless of the results of the drawing, then print "YES". Otherwise print "NO".
|
[
"2\n1 3 2 4\n",
"1\n3 3\n"
] |
[
"YES\n",
"NO\n"
] |
none
| 0
|
[
{
"input": "2\n1 3 2 4",
"output": "YES"
},
{
"input": "1\n3 3",
"output": "NO"
},
{
"input": "5\n1 1 1 1 2 2 3 3 3 3",
"output": "NO"
},
{
"input": "5\n1 1 1 1 1 2 2 2 2 2",
"output": "YES"
},
{
"input": "10\n1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000",
"output": "NO"
},
{
"input": "1\n2 3",
"output": "YES"
},
{
"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 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 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": "NO"
},
{
"input": "35\n919 240 231 858 456 891 959 965 758 30 431 73 505 694 874 543 975 445 16 147 904 690 940 278 562 127 724 314 30 233 389 442 353 652 581 383 340 445 487 283 85 845 578 946 228 557 906 572 919 388 686 181 958 955 736 438 991 170 632 593 475 264 178 344 159 414 739 590 348 884",
"output": "YES"
},
{
"input": "5\n1 2 3 4 10 10 6 7 8 9",
"output": "YES"
},
{
"input": "2\n1 1 1 2",
"output": "NO"
},
{
"input": "2\n10 4 4 4",
"output": "NO"
},
{
"input": "2\n2 3 3 3",
"output": "NO"
},
{
"input": "4\n1 2 3 4 5 4 6 7",
"output": "NO"
},
{
"input": "4\n2 5 4 5 8 3 1 5",
"output": "YES"
},
{
"input": "4\n8 2 2 4 1 4 10 9",
"output": "NO"
},
{
"input": "2\n3 8 10 2",
"output": "YES"
},
{
"input": "3\n1 3 4 4 5 6",
"output": "NO"
},
{
"input": "2\n3 3 3 4",
"output": "NO"
},
{
"input": "2\n1 1 2 2",
"output": "YES"
},
{
"input": "2\n1 1 3 3",
"output": "YES"
},
{
"input": "2\n1 2 3 2",
"output": "NO"
},
{
"input": "10\n1 2 7 3 9 4 1 5 10 3 6 1 10 7 8 5 7 6 1 4",
"output": "NO"
},
{
"input": "3\n1 2 3 3 4 5",
"output": "NO"
},
{
"input": "2\n2 2 1 1",
"output": "YES"
},
{
"input": "7\n1 2 3 4 5 6 7 7 8 9 10 11 12 19",
"output": "NO"
},
{
"input": "5\n1 2 3 4 5 3 3 5 6 7",
"output": "YES"
},
{
"input": "4\n1 1 2 2 3 3 3 3",
"output": "YES"
},
{
"input": "51\n576 377 63 938 667 992 959 997 476 94 652 272 108 410 543 456 942 800 917 163 931 584 357 890 895 318 544 179 268 130 649 916 581 350 573 223 495 26 377 695 114 587 380 424 744 434 332 249 318 522 908 815 313 384 981 773 585 747 376 812 538 525 997 896 859 599 437 163 878 14 224 733 369 741 473 178 153 678 12 894 630 921 505 635 128 404 64 499 208 325 343 996 970 39 380 80 12 756 580 57 934 224",
"output": "YES"
},
{
"input": "3\n3 3 3 2 3 2",
"output": "NO"
},
{
"input": "2\n5 3 3 6",
"output": "YES"
},
{
"input": "2\n1 2 2 3",
"output": "NO"
},
{
"input": "2\n1 3 2 2",
"output": "NO"
},
{
"input": "2\n1 3 3 4",
"output": "NO"
},
{
"input": "2\n1 2 2 2",
"output": "NO"
},
{
"input": "3\n1 2 7 19 19 7",
"output": "NO"
},
{
"input": "3\n1 2 3 3 5 6",
"output": "NO"
},
{
"input": "2\n1 2 2 4",
"output": "NO"
},
{
"input": "2\n6 6 5 5",
"output": "YES"
},
{
"input": "2\n3 1 3 1",
"output": "YES"
},
{
"input": "3\n1 2 3 3 1 1",
"output": "YES"
},
{
"input": "3\n3 2 1 3 4 5",
"output": "NO"
},
{
"input": "3\n4 5 6 4 2 1",
"output": "NO"
},
{
"input": "3\n1 1 2 3 2 4",
"output": "NO"
},
{
"input": "3\n100 99 1 1 1 1",
"output": "NO"
},
{
"input": "3\n1 2 3 6 5 3",
"output": "NO"
},
{
"input": "2\n2 2 1 2",
"output": "NO"
},
{
"input": "4\n1 2 3 4 5 6 7 4",
"output": "NO"
},
{
"input": "3\n1 2 3 1 1 1",
"output": "NO"
},
{
"input": "3\n6 5 3 3 1 3",
"output": "NO"
},
{
"input": "2\n1 2 1 2",
"output": "YES"
},
{
"input": "3\n1 2 5 6 8 6",
"output": "YES"
},
{
"input": "5\n1 2 3 4 5 3 3 3 3 3",
"output": "NO"
},
{
"input": "2\n1 2 4 2",
"output": "NO"
},
{
"input": "3\n7 7 4 5 319 19",
"output": "NO"
},
{
"input": "3\n1 2 4 4 3 5",
"output": "YES"
},
{
"input": "3\n3 2 3 4 5 2",
"output": "NO"
},
{
"input": "5\n1 2 3 4 4 5 3 6 7 8",
"output": "NO"
},
{
"input": "3\n3 3 4 4 5 1",
"output": "YES"
},
{
"input": "2\n3 4 3 3",
"output": "NO"
},
{
"input": "2\n2 5 4 4",
"output": "NO"
},
{
"input": "5\n1 2 3 3 4 5 6 7 8 4",
"output": "NO"
},
{
"input": "3\n1 2 3 3 5 5",
"output": "NO"
},
{
"input": "2\n3 4 4 4",
"output": "NO"
},
{
"input": "2\n1 4 5 4",
"output": "NO"
},
{
"input": "2\n1 2 3 3",
"output": "YES"
},
{
"input": "2\n1 1 2 1",
"output": "NO"
},
{
"input": "4\n1 1 1 1 2 2 2 2",
"output": "YES"
},
{
"input": "4\n1 2 3 5 6 7 8 5",
"output": "NO"
},
{
"input": "2\n4 3 3 1",
"output": "NO"
},
{
"input": "3\n3 1 2 4 3 5",
"output": "NO"
},
{
"input": "3\n1 2 3 3 4 6",
"output": "NO"
},
{
"input": "4\n2 2 2 4 5 5 5 5",
"output": "YES"
},
{
"input": "2\n1 3 4 3",
"output": "NO"
},
{
"input": "2\n3 3 2 3",
"output": "NO"
},
{
"input": "2\n1 2 1 1",
"output": "NO"
},
{
"input": "3\n1 3 4 4 2 5",
"output": "YES"
},
{
"input": "4\n4 7 1 2 3 5 6 4",
"output": "NO"
},
{
"input": "2\n3 2 2 2",
"output": "NO"
},
{
"input": "1\n2 1",
"output": "YES"
},
{
"input": "2\n3 3 1 2",
"output": "YES"
},
{
"input": "1\n8 6",
"output": "YES"
},
{
"input": "7\n6 7 6 7 3 1 9 4 6 10 8 2 5 7",
"output": "NO"
},
{
"input": "2\n3 9 2 1",
"output": "YES"
},
{
"input": "2\n3 3 3 3",
"output": "NO"
}
] | 1,640,892,029
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 88
| 109
| 0
|
n=int(input())
a=sorted(list(map(int,input().split())))
print('YES' if a[n-1]<a[n] else 'NO')
|
Title: Chess Tourney
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Berland annual chess tournament is coming!
Organizers have gathered 2Β·*n* chess players who should be divided into two teams with *n* people each. The first team is sponsored by BerOil and the second team is sponsored by BerMobile. Obviously, organizers should guarantee the win for the team of BerOil.
Thus, organizers should divide all 2Β·*n* players into two teams with *n* people each in such a way that the first team always wins.
Every chess player has its rating *r**i*. It is known that chess player with the greater rating always wins the player with the lower rating. If their ratings are equal then any of the players can win.
After teams assignment there will come a drawing to form *n* pairs of opponents: in each pair there is a player from the first team and a player from the second team. Every chess player should be in exactly one pair. Every pair plays once. The drawing is totally random.
Is it possible to divide all 2Β·*n* players into two teams with *n* people each so that the player from the first team in every pair wins regardless of the results of the drawing?
Input Specification:
The first line contains one integer *n* (1<=β€<=*n*<=β€<=100).
The second line contains 2Β·*n* integers *a*1,<=*a*2,<=... *a*2*n* (1<=β€<=*a**i*<=β€<=1000).
Output Specification:
If it's possible to divide all 2Β·*n* players into two teams with *n* people each so that the player from the first team in every pair wins regardless of the results of the drawing, then print "YES". Otherwise print "NO".
Demo Input:
['2\n1 3 2 4\n', '1\n3 3\n']
Demo Output:
['YES\n', 'NO\n']
Note:
none
|
```python
n=int(input())
a=sorted(list(map(int,input().split())))
print('YES' if a[n-1]<a[n] else 'NO')
```
| 3
|
|
166
|
A
|
Rank List
|
PROGRAMMING
| 1,100
|
[
"binary search",
"implementation",
"sortings"
] | null | null |
Another programming contest is over. You got hold of the contest's final results table. The table has the following data. For each team we are shown two numbers: the number of problems and the total penalty time. However, for no team we are shown its final place.
You know the rules of comparing the results of two given teams very well. Let's say that team *a* solved *p**a* problems with total penalty time *t**a* and team *b* solved *p**b* problems with total penalty time *t**b*. Team *a* gets a higher place than team *b* in the end, if it either solved more problems on the contest, or solved the same number of problems but in less total time. In other words, team *a* gets a higher place than team *b* in the final results' table if either *p**a*<=><=*p**b*, or *p**a*<==<=*p**b* and *t**a*<=<<=*t**b*.
It is considered that the teams that solve the same number of problems with the same penalty time share all corresponding places. More formally, let's say there is a group of *x* teams that solved the same number of problems with the same penalty time. Let's also say that *y* teams performed better than the teams from this group. In this case all teams from the group share places *y*<=+<=1, *y*<=+<=2, ..., *y*<=+<=*x*. The teams that performed worse than the teams from this group, get their places in the results table starting from the *y*<=+<=*x*<=+<=1-th place.
Your task is to count what number of teams from the given list shared the *k*-th place.
|
The first line contains two integers *n* and *k* (1<=β€<=*k*<=β€<=*n*<=β€<=50). Then *n* lines contain the description of the teams: the *i*-th line contains two integers *p**i* and *t**i* (1<=β€<=*p**i*,<=*t**i*<=β€<=50) β the number of solved problems and the total penalty time of the *i*-th team, correspondingly. All numbers in the lines are separated by spaces.
|
In the only line print the sought number of teams that got the *k*-th place in the final results' table.
|
[
"7 2\n4 10\n4 10\n4 10\n3 20\n2 1\n2 1\n1 10\n",
"5 4\n3 1\n3 1\n5 3\n3 1\n3 1\n"
] |
[
"3\n",
"4\n"
] |
The final results' table for the first sample is:
- 1-3 places β 4 solved problems, the penalty time equals 10 - 4 place β 3 solved problems, the penalty time equals 20 - 5-6 places β 2 solved problems, the penalty time equals 1 - 7 place β 1 solved problem, the penalty time equals 10
The table shows that the second place is shared by the teams that solved 4 problems with penalty time 10. There are 3 such teams.
The final table for the second sample is:
- 1 place β 5 solved problems, the penalty time equals 3 - 2-5 places β 3 solved problems, the penalty time equals 1
The table shows that the fourth place is shared by the teams that solved 3 problems with penalty time 1. There are 4 such teams.
| 500
|
[
{
"input": "7 2\n4 10\n4 10\n4 10\n3 20\n2 1\n2 1\n1 10",
"output": "3"
},
{
"input": "5 4\n3 1\n3 1\n5 3\n3 1\n3 1",
"output": "4"
},
{
"input": "5 1\n2 2\n1 1\n1 1\n1 1\n2 2",
"output": "2"
},
{
"input": "6 3\n2 2\n3 1\n2 2\n4 5\n2 2\n4 5",
"output": "1"
},
{
"input": "5 5\n3 1\n10 2\n2 2\n1 10\n10 2",
"output": "1"
},
{
"input": "3 2\n3 3\n3 3\n3 3",
"output": "3"
},
{
"input": "4 3\n10 3\n6 10\n5 2\n5 2",
"output": "2"
},
{
"input": "5 3\n10 10\n10 10\n1 1\n10 10\n4 3",
"output": "3"
},
{
"input": "3 1\n2 1\n1 1\n1 2",
"output": "1"
},
{
"input": "1 1\n28 28",
"output": "1"
},
{
"input": "2 2\n1 2\n1 2",
"output": "2"
},
{
"input": "5 3\n2 3\n4 2\n5 3\n2 4\n3 5",
"output": "1"
},
{
"input": "50 22\n4 9\n8 1\n3 7\n1 2\n3 8\n9 8\n8 5\n2 10\n5 8\n1 3\n1 8\n2 3\n7 9\n10 2\n9 9\n7 3\n8 6\n10 6\n5 4\n8 1\n1 5\n6 8\n9 5\n9 5\n3 2\n3 3\n3 8\n7 5\n4 5\n8 10\n8 2\n3 5\n3 2\n1 1\n7 2\n2 7\n6 8\n10 4\n7 5\n1 7\n6 5\n3 1\n4 9\n2 3\n3 6\n5 8\n4 10\n10 7\n7 10\n9 8",
"output": "1"
},
{
"input": "50 6\n11 20\n18 13\n1 13\n3 11\n4 17\n15 10\n15 8\n9 16\n11 17\n16 3\n3 20\n14 13\n12 15\n9 10\n14 2\n12 12\n13 17\n6 10\n20 9\n2 8\n13 7\n7 20\n15 3\n1 20\n2 13\n2 5\n14 7\n10 13\n15 12\n15 5\n17 6\n9 11\n18 5\n10 1\n15 14\n3 16\n6 12\n4 1\n14 9\n7 14\n8 17\n17 13\n4 6\n19 16\n5 6\n3 15\n4 19\n15 20\n2 10\n20 10",
"output": "1"
},
{
"input": "50 12\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1",
"output": "50"
},
{
"input": "50 28\n2 2\n1 1\n2 1\n1 2\n1 1\n1 1\n1 1\n2 2\n2 2\n2 2\n2 1\n2 2\n2 1\n2 1\n1 2\n1 2\n1 2\n1 1\n2 2\n1 2\n2 2\n2 2\n2 1\n1 1\n1 2\n1 2\n1 1\n1 1\n1 1\n2 2\n2 1\n2 1\n2 2\n1 2\n1 2\n1 2\n1 1\n2 2\n1 2\n1 1\n2 2\n2 2\n1 1\n2 1\n2 1\n1 1\n2 2\n2 2\n2 2\n2 2",
"output": "13"
},
{
"input": "50 40\n2 3\n3 1\n2 1\n2 1\n2 1\n3 1\n1 1\n1 2\n2 3\n1 3\n1 3\n2 1\n3 1\n1 1\n3 1\n3 1\n2 2\n1 1\n3 3\n3 1\n3 2\n2 3\n3 3\n3 1\n1 3\n2 3\n2 1\n3 2\n3 3\n3 1\n2 1\n2 2\n1 3\n3 3\n1 1\n3 2\n1 2\n2 3\n2 1\n2 2\n3 2\n1 3\n3 1\n1 1\n3 3\n2 3\n2 1\n2 3\n2 3\n1 2",
"output": "5"
},
{
"input": "50 16\n2 1\n3 2\n5 2\n2 2\n3 4\n4 4\n3 3\n4 1\n2 3\n1 5\n4 1\n2 2\n1 5\n3 2\n2 1\n5 4\n5 2\n5 4\n1 1\n3 5\n2 1\n4 5\n5 1\n5 5\n5 4\n2 4\n1 2\n5 5\n4 4\n1 5\n4 2\n5 1\n2 4\n2 5\n2 2\n3 4\n3 1\n1 1\n5 5\n2 2\n3 4\n2 4\n5 2\n4 1\n3 1\n1 1\n4 1\n4 4\n1 4\n1 3",
"output": "1"
},
{
"input": "50 32\n6 6\n4 2\n5 5\n1 1\n2 4\n6 5\n2 3\n6 5\n2 3\n6 3\n1 4\n1 6\n3 3\n2 4\n3 2\n6 2\n4 1\n3 3\n3 1\n5 5\n1 2\n2 1\n5 4\n3 1\n4 4\n5 6\n4 1\n2 5\n3 1\n4 6\n2 3\n1 1\n6 5\n2 6\n3 3\n2 6\n2 3\n2 6\n3 4\n2 6\n4 5\n5 4\n1 6\n3 2\n5 1\n4 1\n4 6\n4 2\n1 2\n5 2",
"output": "1"
},
{
"input": "50 48\n5 1\n6 4\n3 2\n2 1\n4 7\n3 6\n7 1\n7 5\n6 5\n5 6\n4 7\n5 7\n5 7\n5 5\n7 3\n3 5\n4 3\n5 4\n6 2\n1 6\n6 3\n6 5\n5 2\n4 2\n3 1\n1 1\n5 6\n1 3\n6 5\n3 7\n1 5\n7 5\n6 5\n3 6\n2 7\n5 3\n5 3\n4 7\n5 2\n6 5\n5 7\n7 1\n2 3\n5 5\n2 6\n4 1\n6 2\n6 5\n3 3\n1 6",
"output": "1"
},
{
"input": "50 8\n5 3\n7 3\n4 3\n7 4\n2 2\n4 4\n5 4\n1 1\n7 7\n4 8\n1 1\n6 3\n1 5\n7 3\n6 5\n4 5\n8 6\n3 6\n2 1\n3 2\n2 5\n7 6\n5 8\n1 3\n5 5\n8 4\n4 5\n4 4\n8 8\n7 2\n7 2\n3 6\n2 8\n8 3\n3 2\n4 5\n8 1\n3 2\n8 7\n6 3\n2 3\n5 1\n3 4\n7 2\n6 3\n7 3\n3 3\n6 4\n2 2\n5 1",
"output": "3"
},
{
"input": "20 16\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1",
"output": "20"
},
{
"input": "20 20\n1 2\n2 2\n1 1\n2 1\n2 2\n1 1\n1 1\n2 1\n1 1\n1 2\n2 2\n1 2\n1 2\n2 2\n2 2\n1 2\n2 1\n2 1\n1 2\n2 2",
"output": "6"
},
{
"input": "30 16\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1",
"output": "30"
},
{
"input": "30 22\n2 1\n1 2\n2 1\n2 2\n2 1\n1 2\n2 2\n1 2\n2 2\n1 2\n2 2\n1 2\n1 2\n2 1\n1 2\n2 2\n2 2\n1 2\n2 1\n1 1\n1 2\n1 2\n1 1\n1 2\n1 2\n2 2\n1 2\n2 2\n2 1\n1 1",
"output": "13"
},
{
"input": "30 22\n1 1\n1 3\n2 3\n3 1\n2 3\n3 1\n1 2\n3 3\n2 1\n2 1\n2 2\n3 1\n3 2\n2 3\n3 1\n1 3\n2 3\n3 1\n1 2\n1 2\n2 3\n2 1\n3 3\n3 2\n1 3\n3 3\n3 3\n3 3\n3 3\n3 1",
"output": "5"
},
{
"input": "50 16\n2 1\n3 2\n5 2\n2 2\n3 4\n4 4\n3 3\n4 1\n2 3\n1 5\n4 1\n2 2\n1 5\n3 2\n2 1\n5 4\n5 2\n5 4\n1 1\n3 5\n2 1\n4 5\n5 1\n5 5\n5 4\n2 4\n1 2\n5 5\n4 4\n1 5\n4 2\n5 1\n2 4\n2 5\n2 2\n3 4\n3 1\n1 1\n5 5\n2 2\n3 4\n2 4\n5 2\n4 1\n3 1\n1 1\n4 1\n4 4\n1 4\n1 3",
"output": "1"
},
{
"input": "50 22\n4 9\n8 1\n3 7\n1 2\n3 8\n9 8\n8 5\n2 10\n5 8\n1 3\n1 8\n2 3\n7 9\n10 2\n9 9\n7 3\n8 6\n10 6\n5 4\n8 1\n1 5\n6 8\n9 5\n9 5\n3 2\n3 3\n3 8\n7 5\n4 5\n8 10\n8 2\n3 5\n3 2\n1 1\n7 2\n2 7\n6 8\n10 4\n7 5\n1 7\n6 5\n3 1\n4 9\n2 3\n3 6\n5 8\n4 10\n10 7\n7 10\n9 8",
"output": "1"
},
{
"input": "50 22\n29 15\n18 10\n6 23\n38 28\n34 40\n40 1\n16 26\n22 33\n14 30\n26 7\n15 16\n22 40\n14 15\n6 28\n32 27\n33 3\n38 22\n40 17\n16 27\n21 27\n34 26\n5 15\n34 9\n38 23\n7 36\n17 6\n19 37\n40 1\n10 28\n9 14\n8 31\n40 8\n14 2\n24 16\n38 33\n3 37\n2 9\n21 21\n40 26\n28 33\n24 31\n10 12\n27 27\n17 4\n38 5\n21 31\n5 12\n29 7\n39 12\n26 14",
"output": "1"
},
{
"input": "50 14\n4 20\n37 50\n46 19\n20 25\n47 10\n6 34\n12 41\n47 9\n22 28\n41 34\n47 40\n12 42\n9 4\n15 15\n27 8\n38 9\n4 17\n8 13\n47 7\n9 38\n30 48\n50 7\n41 34\n23 11\n16 37\n2 32\n18 46\n37 48\n47 41\n13 9\n24 50\n46 14\n33 49\n9 50\n35 30\n49 44\n42 49\n39 15\n33 42\n3 18\n44 15\n44 28\n9 17\n16 4\n10 36\n4 22\n47 17\n24 12\n2 31\n6 30",
"output": "2"
},
{
"input": "2 1\n50 50\n50 50",
"output": "2"
},
{
"input": "2 2\n50 50\n50 50",
"output": "2"
},
{
"input": "2 1\n50 50\n50 49",
"output": "1"
},
{
"input": "2 2\n50 50\n50 49",
"output": "1"
},
{
"input": "50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50",
"output": "50"
},
{
"input": "50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n49 50",
"output": "1"
},
{
"input": "50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 50\n50 49",
"output": "49"
},
{
"input": "10 1\n3 1\n3 2\n3 2\n2 3\n2 3\n2 3\n2 4\n2 4\n2 4\n2 4",
"output": "1"
},
{
"input": "4 3\n1 1\n1 2\n2 1\n2 2",
"output": "1"
},
{
"input": "4 2\n1 2\n1 3\n1 5\n1 2",
"output": "2"
},
{
"input": "6 1\n4 3\n4 2\n5 1\n5 2\n5 3\n5 3",
"output": "1"
},
{
"input": "5 3\n2 1\n1 3\n1 2\n1 1\n1 1",
"output": "2"
},
{
"input": "3 1\n1 1\n1 2\n1 3",
"output": "1"
},
{
"input": "10 10\n3 1\n3 2\n3 2\n2 3\n2 3\n2 3\n2 4\n2 4\n2 4\n2 4",
"output": "4"
},
{
"input": "5 1\n5 5\n5 6\n5 7\n5 8\n5 5",
"output": "2"
},
{
"input": "4 1\n1 1\n1 1\n1 2\n1 3",
"output": "2"
}
] | 1,619,338,614
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 46
| 124
| 307,200
|
n,k = map(int,input().split())
a,d = [],{}
for i in range(n):
l = tuple(map(int,input().split()))
if l in d:
d[l]+=1
else:
d[l]=1
a.append(l)
b = sorted(a, key=lambda x: (x[0], 10-x[1]), reverse=True)
print(d[b[k-1]])
|
Title: Rank List
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Another programming contest is over. You got hold of the contest's final results table. The table has the following data. For each team we are shown two numbers: the number of problems and the total penalty time. However, for no team we are shown its final place.
You know the rules of comparing the results of two given teams very well. Let's say that team *a* solved *p**a* problems with total penalty time *t**a* and team *b* solved *p**b* problems with total penalty time *t**b*. Team *a* gets a higher place than team *b* in the end, if it either solved more problems on the contest, or solved the same number of problems but in less total time. In other words, team *a* gets a higher place than team *b* in the final results' table if either *p**a*<=><=*p**b*, or *p**a*<==<=*p**b* and *t**a*<=<<=*t**b*.
It is considered that the teams that solve the same number of problems with the same penalty time share all corresponding places. More formally, let's say there is a group of *x* teams that solved the same number of problems with the same penalty time. Let's also say that *y* teams performed better than the teams from this group. In this case all teams from the group share places *y*<=+<=1, *y*<=+<=2, ..., *y*<=+<=*x*. The teams that performed worse than the teams from this group, get their places in the results table starting from the *y*<=+<=*x*<=+<=1-th place.
Your task is to count what number of teams from the given list shared the *k*-th place.
Input Specification:
The first line contains two integers *n* and *k* (1<=β€<=*k*<=β€<=*n*<=β€<=50). Then *n* lines contain the description of the teams: the *i*-th line contains two integers *p**i* and *t**i* (1<=β€<=*p**i*,<=*t**i*<=β€<=50) β the number of solved problems and the total penalty time of the *i*-th team, correspondingly. All numbers in the lines are separated by spaces.
Output Specification:
In the only line print the sought number of teams that got the *k*-th place in the final results' table.
Demo Input:
['7 2\n4 10\n4 10\n4 10\n3 20\n2 1\n2 1\n1 10\n', '5 4\n3 1\n3 1\n5 3\n3 1\n3 1\n']
Demo Output:
['3\n', '4\n']
Note:
The final results' table for the first sample is:
- 1-3 places β 4 solved problems, the penalty time equals 10 - 4 place β 3 solved problems, the penalty time equals 20 - 5-6 places β 2 solved problems, the penalty time equals 1 - 7 place β 1 solved problem, the penalty time equals 10
The table shows that the second place is shared by the teams that solved 4 problems with penalty time 10. There are 3 such teams.
The final table for the second sample is:
- 1 place β 5 solved problems, the penalty time equals 3 - 2-5 places β 3 solved problems, the penalty time equals 1
The table shows that the fourth place is shared by the teams that solved 3 problems with penalty time 1. There are 4 such teams.
|
```python
n,k = map(int,input().split())
a,d = [],{}
for i in range(n):
l = tuple(map(int,input().split()))
if l in d:
d[l]+=1
else:
d[l]=1
a.append(l)
b = sorted(a, key=lambda x: (x[0], 10-x[1]), reverse=True)
print(d[b[k-1]])
```
| 3
|
|
686
|
A
|
Free Ice Cream
|
PROGRAMMING
| 800
|
[
"constructive algorithms",
"implementation"
] | null | null |
After their adventure with the magic mirror Kay and Gerda have returned home and sometimes give free ice cream to kids in the summer.
At the start of the day they have *x* ice cream packs. Since the ice cream is free, people start standing in the queue before Kay and Gerda's house even in the night. Each person in the queue wants either to take several ice cream packs for himself and his friends or to give several ice cream packs to Kay and Gerda (carriers that bring ice cream have to stand in the same queue).
If a carrier with *d* ice cream packs comes to the house, then Kay and Gerda take all his packs. If a child who wants to take *d* ice cream packs comes to the house, then Kay and Gerda will give him *d* packs if they have enough ice cream, otherwise the child will get no ice cream at all and will leave in distress.
Kay wants to find the amount of ice cream they will have after all people will leave from the queue, and Gerda wants to find the number of distressed kids.
|
The first line contains two space-separated integers *n* and *x* (1<=β€<=*n*<=β€<=1000, 0<=β€<=*x*<=β€<=109).
Each of the next *n* lines contains a character '+' or '-', and an integer *d**i*, separated by a space (1<=β€<=*d**i*<=β€<=109). Record "+ *d**i*" in *i*-th line means that a carrier with *d**i* ice cream packs occupies *i*-th place from the start of the queue, and record "- *d**i*" means that a child who wants to take *d**i* packs stands in *i*-th place.
|
Print two space-separated integersΒ β number of ice cream packs left after all operations, and number of kids that left the house in distress.
|
[
"5 7\n+ 5\n- 10\n- 20\n+ 40\n- 20\n",
"5 17\n- 16\n- 2\n- 98\n+ 100\n- 98\n"
] |
[
"22 1\n",
"3 2\n"
] |
Consider the first sample.
1. Initially Kay and Gerda have 7 packs of ice cream. 1. Carrier brings 5 more, so now they have 12 packs. 1. A kid asks for 10 packs and receives them. There are only 2 packs remaining. 1. Another kid asks for 20 packs. Kay and Gerda do not have them, so the kid goes away distressed. 1. Carrier bring 40 packs, now Kay and Gerda have 42 packs. 1. Kid asks for 20 packs and receives them. There are 22 packs remaining.
| 500
|
[
{
"input": "5 7\n+ 5\n- 10\n- 20\n+ 40\n- 20",
"output": "22 1"
},
{
"input": "5 17\n- 16\n- 2\n- 98\n+ 100\n- 98",
"output": "3 2"
},
{
"input": "6 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000",
"output": "7000000000 0"
},
{
"input": "5 12\n- 12\n+ 7\n- 6\n- 1\n+ 46",
"output": "46 0"
},
{
"input": "11 1000\n- 100\n+ 100\n+ 100\n+ 100\n+ 100\n- 100\n- 100\n- 100\n- 100\n- 100\n- 100",
"output": "700 0"
},
{
"input": "1 0\n- 526403222",
"output": "0 1"
},
{
"input": "1 897986543\n- 371188251",
"output": "526798292 0"
},
{
"input": "1 0\n+ 1",
"output": "1 0"
},
{
"input": "1 0\n- 1",
"output": "0 1"
},
{
"input": "1 10\n+ 10",
"output": "20 0"
},
{
"input": "1 3\n- 5",
"output": "3 1"
},
{
"input": "1 0\n- 5",
"output": "0 1"
},
{
"input": "1 0\n+ 5",
"output": "5 0"
}
] | 1,633,450,302
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 34
| 140
| 21,606,400
|
a,b = map(int,input().split())
c = 0
for i in range(a):
n = input().split()
n1 = int(n[1])
if n[0] == "+":
b += n1
else:
if b < n1:
c += 1
else:
b -= n1
print(b,c)
|
Title: Free Ice Cream
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
After their adventure with the magic mirror Kay and Gerda have returned home and sometimes give free ice cream to kids in the summer.
At the start of the day they have *x* ice cream packs. Since the ice cream is free, people start standing in the queue before Kay and Gerda's house even in the night. Each person in the queue wants either to take several ice cream packs for himself and his friends or to give several ice cream packs to Kay and Gerda (carriers that bring ice cream have to stand in the same queue).
If a carrier with *d* ice cream packs comes to the house, then Kay and Gerda take all his packs. If a child who wants to take *d* ice cream packs comes to the house, then Kay and Gerda will give him *d* packs if they have enough ice cream, otherwise the child will get no ice cream at all and will leave in distress.
Kay wants to find the amount of ice cream they will have after all people will leave from the queue, and Gerda wants to find the number of distressed kids.
Input Specification:
The first line contains two space-separated integers *n* and *x* (1<=β€<=*n*<=β€<=1000, 0<=β€<=*x*<=β€<=109).
Each of the next *n* lines contains a character '+' or '-', and an integer *d**i*, separated by a space (1<=β€<=*d**i*<=β€<=109). Record "+ *d**i*" in *i*-th line means that a carrier with *d**i* ice cream packs occupies *i*-th place from the start of the queue, and record "- *d**i*" means that a child who wants to take *d**i* packs stands in *i*-th place.
Output Specification:
Print two space-separated integersΒ β number of ice cream packs left after all operations, and number of kids that left the house in distress.
Demo Input:
['5 7\n+ 5\n- 10\n- 20\n+ 40\n- 20\n', '5 17\n- 16\n- 2\n- 98\n+ 100\n- 98\n']
Demo Output:
['22 1\n', '3 2\n']
Note:
Consider the first sample.
1. Initially Kay and Gerda have 7 packs of ice cream. 1. Carrier brings 5 more, so now they have 12 packs. 1. A kid asks for 10 packs and receives them. There are only 2 packs remaining. 1. Another kid asks for 20 packs. Kay and Gerda do not have them, so the kid goes away distressed. 1. Carrier bring 40 packs, now Kay and Gerda have 42 packs. 1. Kid asks for 20 packs and receives them. There are 22 packs remaining.
|
```python
a,b = map(int,input().split())
c = 0
for i in range(a):
n = input().split()
n1 = int(n[1])
if n[0] == "+":
b += n1
else:
if b < n1:
c += 1
else:
b -= n1
print(b,c)
```
| 3
|
|
855
|
A
|
Tom Riddle's Diary
|
PROGRAMMING
| 800
|
[
"brute force",
"implementation",
"strings"
] | null | null |
Harry Potter is on a mission to destroy You-Know-Who's Horcruxes. The first Horcrux that he encountered in the Chamber of Secrets is Tom Riddle's diary. The diary was with Ginny and it forced her to open the Chamber of Secrets. Harry wants to know the different people who had ever possessed the diary to make sure they are not under its influence.
He has names of *n* people who possessed the diary in order. You need to tell, for each person, if he/she possessed the diary at some point before or not.
Formally, for a name *s**i* in the *i*-th line, output "YES" (without quotes) if there exists an index *j* such that *s**i*<==<=*s**j* and *j*<=<<=*i*, otherwise, output "NO" (without quotes).
|
First line of input contains an integer *n* (1<=β€<=*n*<=β€<=100)Β β the number of names in the list.
Next *n* lines each contain a string *s**i*, consisting of lowercase English letters. The length of each string is between 1 and 100.
|
Output *n* lines each containing either "YES" or "NO" (without quotes), depending on whether this string was already present in the stream or not.
You can print each letter in any case (upper or lower).
|
[
"6\ntom\nlucius\nginny\nharry\nginny\nharry\n",
"3\na\na\na\n"
] |
[
"NO\nNO\nNO\nNO\nYES\nYES\n",
"NO\nYES\nYES\n"
] |
In test case 1, for *i*β=β5 there exists *j*β=β3 such that *s*<sub class="lower-index">*i*</sub>β=β*s*<sub class="lower-index">*j*</sub> and *j*β<β*i*, which means that answer for *i*β=β5 is "YES".
| 500
|
[
{
"input": "6\ntom\nlucius\nginny\nharry\nginny\nharry",
"output": "NO\nNO\nNO\nNO\nYES\nYES"
},
{
"input": "3\na\na\na",
"output": "NO\nYES\nYES"
},
{
"input": "1\nzn",
"output": "NO"
},
{
"input": "9\nliyzmbjwnzryjokufuxcqtzwworjeoxkbaqrujrhdidqdvwdfzilwszgnzglnnbogaclckfnbqovtziuhwvyrqwmskx\nliyzmbjwnzryjokufuxcqtzwworjeoxkbaqrujrhdidqdvwdfzilwszgnzglnnbogaclckfnbqovtziuhwvyrqwmskx\nliyzmbjwnzryjokufuxcqtzwworjeoxkbaqrujrhdidqdvwdfzilwszgnzglnnbogaclckfnbqovtziuhwvyrqwmskx\nhrtm\nssjqvixduertmotgagizamvfucfwtxqnhuowbqbzctgznivehelpcyigwrbbdsxnewfqvcf\nhyrtxvozpbveexfkgalmguozzakitjiwsduqxonb\nwcyxteiwtcyuztaguilqpbiwcwjaiq\nwcyxteiwtcyuztaguilqpbiwcwjaiq\nbdbivqzvhggth",
"output": "NO\nYES\nYES\nNO\nNO\nNO\nNO\nYES\nNO"
},
{
"input": "10\nkkiubdktydpdcbbttwpfdplhhjhrpqmpg\nkkiubdktydpdcbbttwpfdplhhjhrpqmpg\nmvutw\nqooeqoxzxwetlpecqiwgdbogiqqulttysyohwhzxzphvsfmnplizxoebzcvvfyppqbhxjksuzepuezqqzxlfmdanoeaoqmor\nmvutw\nvchawxjoreboqzuklifv\nvchawxjoreboqzuklifv\nnivijte\nrflybruq\nvchawxjoreboqzuklifv",
"output": "NO\nYES\nNO\nNO\nYES\nNO\nYES\nNO\nNO\nYES"
},
{
"input": "1\nz",
"output": "NO"
},
{
"input": "9\nl\ny\nm\nj\nn\nr\nj\nk\nf",
"output": "NO\nNO\nNO\nNO\nNO\nNO\nYES\nNO\nNO"
},
{
"input": "14\nw\na\nh\np\nk\nw\ny\nv\ns\nf\nx\nd\nk\nr",
"output": "NO\nNO\nNO\nNO\nNO\nYES\nNO\nNO\nNO\nNO\nNO\nNO\nYES\nNO"
},
{
"input": "25\np\nk\nu\nl\nf\nt\nc\ns\nq\nd\nb\nq\no\ni\ni\nd\ni\nw\nn\ng\nw\nt\na\ne\ni",
"output": "NO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nYES\nNO\nNO\nYES\nYES\nYES\nNO\nNO\nNO\nYES\nYES\nNO\nNO\nYES"
},
{
"input": "20\nd\nh\ng\no\np\ne\nt\nj\nv\ni\nt\nh\ns\ni\nw\nf\nx\na\nl\ni",
"output": "NO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nYES\nYES\nNO\nYES\nNO\nNO\nNO\nNO\nNO\nYES"
},
{
"input": "3\nbbbbbbb\nbbbbbbbbb\nbbbbbbbbbbbbbbbbbbbbbbbbbbbb",
"output": "NO\nNO\nNO"
},
{
"input": "2\nab\nba",
"output": "NO\nNO"
},
{
"input": "6\ntom\nlucius\nginnys\nharpy\nginny\nharry",
"output": "NO\nNO\nNO\nNO\nNO\nNO"
},
{
"input": "2\nabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcde\nabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcde",
"output": "NO\nYES"
},
{
"input": "42\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na",
"output": "NO\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES"
}
] | 1,672,207,454
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 55
| 46
| 0
|
def tomRiddleDiary(s):
if s in a:
return "YES"
else:
return "NO"
a = set()
t = int(input())
while t > 0:
s = input()
print(tomRiddleDiary(s))
t = t - 1
a.add(s)
|
Title: Tom Riddle's Diary
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Harry Potter is on a mission to destroy You-Know-Who's Horcruxes. The first Horcrux that he encountered in the Chamber of Secrets is Tom Riddle's diary. The diary was with Ginny and it forced her to open the Chamber of Secrets. Harry wants to know the different people who had ever possessed the diary to make sure they are not under its influence.
He has names of *n* people who possessed the diary in order. You need to tell, for each person, if he/she possessed the diary at some point before or not.
Formally, for a name *s**i* in the *i*-th line, output "YES" (without quotes) if there exists an index *j* such that *s**i*<==<=*s**j* and *j*<=<<=*i*, otherwise, output "NO" (without quotes).
Input Specification:
First line of input contains an integer *n* (1<=β€<=*n*<=β€<=100)Β β the number of names in the list.
Next *n* lines each contain a string *s**i*, consisting of lowercase English letters. The length of each string is between 1 and 100.
Output Specification:
Output *n* lines each containing either "YES" or "NO" (without quotes), depending on whether this string was already present in the stream or not.
You can print each letter in any case (upper or lower).
Demo Input:
['6\ntom\nlucius\nginny\nharry\nginny\nharry\n', '3\na\na\na\n']
Demo Output:
['NO\nNO\nNO\nNO\nYES\nYES\n', 'NO\nYES\nYES\n']
Note:
In test case 1, for *i*β=β5 there exists *j*β=β3 such that *s*<sub class="lower-index">*i*</sub>β=β*s*<sub class="lower-index">*j*</sub> and *j*β<β*i*, which means that answer for *i*β=β5 is "YES".
|
```python
def tomRiddleDiary(s):
if s in a:
return "YES"
else:
return "NO"
a = set()
t = int(input())
while t > 0:
s = input()
print(tomRiddleDiary(s))
t = t - 1
a.add(s)
```
| 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<=β€<=*n*,<=*k*<=β€<=100) β the number of buckets and the length of the garden, respectively.
The second line of input contains *n* integer numbers *a**i* (1<=β€<=*a**i*<=β€<=100) β 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 β 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"
}
] | 1,671,364,069
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 83
| 46
| 0
|
def f():
num,k=map(int,input().split())
n=map(int, input().split())
maxi=0
ans=0
for i in n:
if(k%i==0 and i>maxi):
maxi=i
ans=int(k/i)
print(ans)
f()
|
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<=β€<=*n*,<=*k*<=β€<=100) β the number of buckets and the length of the garden, respectively.
The second line of input contains *n* integer numbers *a**i* (1<=β€<=*a**i*<=β€<=100) β 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 β 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
def f():
num,k=map(int,input().split())
n=map(int, input().split())
maxi=0
ans=0
for i in n:
if(k%i==0 and i>maxi):
maxi=i
ans=int(k/i)
print(ans)
f()
```
| 3
|
|
540
|
A
|
Combination Lock
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Scrooge McDuck keeps his most treasured savings in a home safe with a combination lock. Each time he wants to put there the treasures that he's earned fair and square, he has to open the lock.
The combination lock is represented by *n* rotating disks with digits from 0 to 9 written on them. Scrooge McDuck has to turn some disks so that the combination of digits on the disks forms a secret combination. In one move, he can rotate one disk one digit forwards or backwards. In particular, in one move he can go from digit 0 to digit 9 and vice versa. What minimum number of actions does he need for that?
|
The first line contains a single integer *n* (1<=β€<=*n*<=β€<=1000)Β β the number of disks on the combination lock.
The second line contains a string of *n* digitsΒ β the original state of the disks.
The third line contains a string of *n* digitsΒ β Scrooge McDuck's combination that opens the lock.
|
Print a single integerΒ β the minimum number of moves Scrooge McDuck needs to open the lock.
|
[
"5\n82195\n64723\n"
] |
[
"13\n"
] |
In the sample he needs 13 moves:
- 1 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/b8967f65a723782358b93eff9ce69f336817cf70.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 2 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/07fa58573ece0d32c4d555e498d2b24d2f70f36a.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 3 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/cc2275d9252aae35a6867c6a5b4ba7596e9a7626.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 4 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/b100aea470fcaaab4e9529b234ba0d7875943c10.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 5 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/eb2cbe4324cebca65b85816262a85e473cd65967.png" style="max-width: 100.0%;max-height: 100.0%;"/>
| 500
|
[
{
"input": "5\n82195\n64723",
"output": "13"
},
{
"input": "12\n102021090898\n010212908089",
"output": "16"
},
{
"input": "1\n8\n1",
"output": "3"
},
{
"input": "2\n83\n57",
"output": "7"
},
{
"input": "10\n0728592530\n1362615763",
"output": "27"
},
{
"input": "100\n4176196363694273682807653052945037727131821799902563705176501742060696655282954944720643131654235909\n3459912084922154505910287499879975659298239371519889866585472674423008837878123067103005344986554746",
"output": "245"
},
{
"input": "1\n8\n1",
"output": "3"
},
{
"input": "2\n83\n57",
"output": "7"
},
{
"input": "3\n607\n684",
"output": "5"
},
{
"input": "4\n0809\n0636",
"output": "8"
},
{
"input": "5\n84284\n08941",
"output": "16"
},
{
"input": "25\n8037856825987124762280548\n9519431339078678836940020",
"output": "72"
},
{
"input": "125\n23269567683904664184142384849516523616863461607751021071772615078579713054027902974007001544768640273491193035874486891541257\n47635110303703399505805044019026243695451609639556649012447370081552870340011971572363458960190590266459684717415349529509024",
"output": "305"
},
{
"input": "5\n84284\n08941",
"output": "16"
},
{
"input": "25\n8037856825987124762285484\n9519431339078678836940202",
"output": "74"
},
{
"input": "125\n23269567689466418414238845152168634610771021717726157879713054270294007001544768647391193035874486891412573389247025830678706\n47635110307339950580504010224954516093956649124473708152870340117152363458960190596659684717415349529090241694059599629136831",
"output": "357"
},
{
"input": "5\n84284\n08941",
"output": "16"
},
{
"input": "25\n8378525987476228048406972\n9194339078883694020217816",
"output": "55"
},
{
"input": "125\n23269576839046618414238484916523616863461607750210717761078579713054027902974007015447686027349193035874486891541257338624472\n47635103037033950580504401926243695451609639556490124437081552870340011971572363489601905026645984717415349529509024169604599",
"output": "274"
},
{
"input": "1\n0\n0",
"output": "0"
},
{
"input": "1\n7\n7",
"output": "0"
},
{
"input": "1\n0\n5",
"output": "5"
},
{
"input": "1\n2\n7",
"output": "5"
},
{
"input": "1\n7\n9",
"output": "2"
},
{
"input": "1\n9\n7",
"output": "2"
},
{
"input": "1\n2\n9",
"output": "3"
},
{
"input": "1\n9\n2",
"output": "3"
},
{
"input": "25\n3164978461316464614169874\n9413979197249127496597357",
"output": "66"
},
{
"input": "4\n9999\n9999",
"output": "0"
},
{
"input": "2\n11\n11",
"output": "0"
}
] | 1,664,536,124
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 43
| 77
| 1,433,600
|
n = int(input())
comb1 = list(map(int, ' '.join(input()).split()))
comb2 = list(map(int, ' '.join(input()).split()))
count = 0
for i in range(n):
a = abs(comb1[i] - comb2[i])
b = 10 - a
count += min(a, b)
print(count)
|
Title: Combination Lock
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Scrooge McDuck keeps his most treasured savings in a home safe with a combination lock. Each time he wants to put there the treasures that he's earned fair and square, he has to open the lock.
The combination lock is represented by *n* rotating disks with digits from 0 to 9 written on them. Scrooge McDuck has to turn some disks so that the combination of digits on the disks forms a secret combination. In one move, he can rotate one disk one digit forwards or backwards. In particular, in one move he can go from digit 0 to digit 9 and vice versa. What minimum number of actions does he need for that?
Input Specification:
The first line contains a single integer *n* (1<=β€<=*n*<=β€<=1000)Β β the number of disks on the combination lock.
The second line contains a string of *n* digitsΒ β the original state of the disks.
The third line contains a string of *n* digitsΒ β Scrooge McDuck's combination that opens the lock.
Output Specification:
Print a single integerΒ β the minimum number of moves Scrooge McDuck needs to open the lock.
Demo Input:
['5\n82195\n64723\n']
Demo Output:
['13\n']
Note:
In the sample he needs 13 moves:
- 1 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/b8967f65a723782358b93eff9ce69f336817cf70.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 2 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/07fa58573ece0d32c4d555e498d2b24d2f70f36a.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 3 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/cc2275d9252aae35a6867c6a5b4ba7596e9a7626.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 4 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/b100aea470fcaaab4e9529b234ba0d7875943c10.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 5 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/eb2cbe4324cebca65b85816262a85e473cd65967.png" style="max-width: 100.0%;max-height: 100.0%;"/>
|
```python
n = int(input())
comb1 = list(map(int, ' '.join(input()).split()))
comb2 = list(map(int, ' '.join(input()).split()))
count = 0
for i in range(n):
a = abs(comb1[i] - comb2[i])
b = 10 - a
count += min(a, b)
print(count)
```
| 3
|
|
669
|
A
|
Little Artem and Presents
|
PROGRAMMING
| 800
|
[
"math"
] | null | null |
Little Artem got *n* stones on his birthday and now wants to give some of them to Masha. He knows that Masha cares more about the fact of receiving the present, rather than the value of that present, so he wants to give her stones as many times as possible. However, Masha remembers the last present she received, so Artem can't give her the same number of stones twice in a row. For example, he can give her 3 stones, then 1 stone, then again 3 stones, but he can't give her 3 stones and then again 3 stones right after that.
How many times can Artem give presents to Masha?
|
The only line of the input contains a single integer *n* (1<=β€<=*n*<=β€<=109)Β β number of stones Artem received on his birthday.
|
Print the maximum possible number of times Artem can give presents to Masha.
|
[
"1\n",
"2\n",
"3\n",
"4\n"
] |
[
"1\n",
"1\n",
"2\n",
"3\n"
] |
In the first sample, Artem can only give 1 stone to Masha.
In the second sample, Atrem can give Masha 1 or 2 stones, though he can't give her 1 stone two times.
In the third sample, Atrem can first give Masha 2 stones, a then 1 more stone.
In the fourth sample, Atrem can first give Masha 1 stone, then 2 stones, and finally 1 stone again.
| 500
|
[
{
"input": "1",
"output": "1"
},
{
"input": "2",
"output": "1"
},
{
"input": "3",
"output": "2"
},
{
"input": "4",
"output": "3"
},
{
"input": "100",
"output": "67"
},
{
"input": "101",
"output": "67"
},
{
"input": "102",
"output": "68"
},
{
"input": "1000000000",
"output": "666666667"
},
{
"input": "5",
"output": "3"
},
{
"input": "6",
"output": "4"
},
{
"input": "999999999",
"output": "666666666"
},
{
"input": "999999998",
"output": "666666665"
},
{
"input": "999999997",
"output": "666666665"
},
{
"input": "999999996",
"output": "666666664"
},
{
"input": "999999995",
"output": "666666663"
},
{
"input": "999999994",
"output": "666666663"
},
{
"input": "999999993",
"output": "666666662"
},
{
"input": "999999992",
"output": "666666661"
},
{
"input": "999999991",
"output": "666666661"
},
{
"input": "1000",
"output": "667"
},
{
"input": "10000",
"output": "6667"
},
{
"input": "100000",
"output": "66667"
},
{
"input": "1000000",
"output": "666667"
},
{
"input": "10000000",
"output": "6666667"
},
{
"input": "100000000",
"output": "66666667"
},
{
"input": "7",
"output": "5"
}
] | 1,632,339,994
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 26
| 109
| 20,172,800
|
a = int(input())
print(a // 3 * 2 + (a % 3 > 0))
|
Title: Little Artem and Presents
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Artem got *n* stones on his birthday and now wants to give some of them to Masha. He knows that Masha cares more about the fact of receiving the present, rather than the value of that present, so he wants to give her stones as many times as possible. However, Masha remembers the last present she received, so Artem can't give her the same number of stones twice in a row. For example, he can give her 3 stones, then 1 stone, then again 3 stones, but he can't give her 3 stones and then again 3 stones right after that.
How many times can Artem give presents to Masha?
Input Specification:
The only line of the input contains a single integer *n* (1<=β€<=*n*<=β€<=109)Β β number of stones Artem received on his birthday.
Output Specification:
Print the maximum possible number of times Artem can give presents to Masha.
Demo Input:
['1\n', '2\n', '3\n', '4\n']
Demo Output:
['1\n', '1\n', '2\n', '3\n']
Note:
In the first sample, Artem can only give 1 stone to Masha.
In the second sample, Atrem can give Masha 1 or 2 stones, though he can't give her 1 stone two times.
In the third sample, Atrem can first give Masha 2 stones, a then 1 more stone.
In the fourth sample, Atrem can first give Masha 1 stone, then 2 stones, and finally 1 stone again.
|
```python
a = int(input())
print(a // 3 * 2 + (a % 3 > 0))
```
| 3
|
|
192
|
A
|
Funky Numbers
|
PROGRAMMING
| 1,300
|
[
"binary search",
"brute force",
"implementation"
] | null | null |
As you very well know, this year's funkiest numbers are so called triangular numbers (that is, integers that are representable as , where *k* is some positive integer), and the coolest numbers are those that are representable as a sum of two triangular numbers.
A well-known hipster Andrew adores everything funky and cool but unfortunately, he isn't good at maths. Given number *n*, help him define whether this number can be represented by a sum of two triangular numbers (not necessarily different)!
|
The first input line contains an integer *n* (1<=β€<=*n*<=β€<=109).
|
Print "YES" (without the quotes), if *n* can be represented as a sum of two triangular numbers, otherwise print "NO" (without the quotes).
|
[
"256\n",
"512\n"
] |
[
"YES\n",
"NO\n"
] |
In the first sample number <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/92095692c6ea93e9e3b837a0408ba7543549d5b2.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second sample number 512 can not be represented as a sum of two triangular numbers.
| 500
|
[
{
"input": "256",
"output": "YES"
},
{
"input": "512",
"output": "NO"
},
{
"input": "80",
"output": "NO"
},
{
"input": "828",
"output": "YES"
},
{
"input": "6035",
"output": "NO"
},
{
"input": "39210",
"output": "YES"
},
{
"input": "79712",
"output": "NO"
},
{
"input": "190492",
"output": "YES"
},
{
"input": "5722367",
"output": "NO"
},
{
"input": "816761542",
"output": "YES"
},
{
"input": "1",
"output": "NO"
},
{
"input": "2",
"output": "YES"
},
{
"input": "3",
"output": "NO"
},
{
"input": "4",
"output": "YES"
},
{
"input": "5",
"output": "NO"
},
{
"input": "6",
"output": "YES"
},
{
"input": "7",
"output": "YES"
},
{
"input": "8",
"output": "NO"
},
{
"input": "9",
"output": "YES"
},
{
"input": "10",
"output": "NO"
},
{
"input": "12",
"output": "YES"
},
{
"input": "13",
"output": "YES"
},
{
"input": "14",
"output": "NO"
},
{
"input": "15",
"output": "NO"
},
{
"input": "16",
"output": "YES"
},
{
"input": "17",
"output": "NO"
},
{
"input": "18",
"output": "YES"
},
{
"input": "19",
"output": "NO"
},
{
"input": "20",
"output": "YES"
},
{
"input": "41",
"output": "NO"
},
{
"input": "11",
"output": "YES"
},
{
"input": "69",
"output": "YES"
},
{
"input": "82",
"output": "NO"
},
{
"input": "85",
"output": "NO"
},
{
"input": "736",
"output": "NO"
},
{
"input": "895",
"output": "YES"
},
{
"input": "934",
"output": "YES"
},
{
"input": "6213",
"output": "YES"
},
{
"input": "7405",
"output": "NO"
},
{
"input": "9919",
"output": "NO"
},
{
"input": "40942",
"output": "YES"
},
{
"input": "41992",
"output": "NO"
},
{
"input": "68535",
"output": "NO"
},
{
"input": "405718",
"output": "NO"
},
{
"input": "1046146",
"output": "YES"
},
{
"input": "3761248",
"output": "YES"
},
{
"input": "6195181",
"output": "YES"
},
{
"input": "35354345",
"output": "NO"
},
{
"input": "81282830",
"output": "NO"
},
{
"input": "187719774",
"output": "NO"
},
{
"input": "296798673",
"output": "NO"
},
{
"input": "938938476",
"output": "NO"
},
{
"input": "1000000000",
"output": "NO"
},
{
"input": "999887464",
"output": "YES"
},
{
"input": "999111944",
"output": "NO"
},
{
"input": "999966520",
"output": "YES"
},
{
"input": "999912080",
"output": "NO"
},
{
"input": "999992017",
"output": "YES"
},
{
"input": "999990474",
"output": "NO"
},
{
"input": "999999190",
"output": "YES"
},
{
"input": "999999125",
"output": "NO"
},
{
"input": "999999940",
"output": "YES"
},
{
"input": "999999995",
"output": "NO"
},
{
"input": "1000000000",
"output": "NO"
},
{
"input": "1",
"output": "NO"
},
{
"input": "999999999",
"output": "YES"
},
{
"input": "83495494",
"output": "NO"
},
{
"input": "968022000",
"output": "YES"
},
{
"input": "399980000",
"output": "YES"
},
{
"input": "4",
"output": "YES"
},
{
"input": "999999998",
"output": "NO"
}
] | 1,600,323,557
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 71
| 498
| 3,481,600
|
import sys, os.path
from collections import*
from copy import*
import math
mod=10**9+7
if(os.path.exists('input.txt')):
sys.stdin = open("input.txt","r")
sys.stdout = open("output.txt","w")
def bs(left,right,x):
while(left<=right):
mid=left+(right-left)//2
b=(mid*(mid+1))//2
if(b==x):
return 1
elif(b>x):
right=mid-1
else:
left=mid+1
return 0
n=int(input())
n1=10**5
flag=0
for i in range(1,n1+1):
a=(i*(i+1))//2
if(bs(1,n1,n-a)):
flag=1
break
if(flag==0):
print('NO')
else:
print('YES')
|
Title: Funky Numbers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
As you very well know, this year's funkiest numbers are so called triangular numbers (that is, integers that are representable as , where *k* is some positive integer), and the coolest numbers are those that are representable as a sum of two triangular numbers.
A well-known hipster Andrew adores everything funky and cool but unfortunately, he isn't good at maths. Given number *n*, help him define whether this number can be represented by a sum of two triangular numbers (not necessarily different)!
Input Specification:
The first input line contains an integer *n* (1<=β€<=*n*<=β€<=109).
Output Specification:
Print "YES" (without the quotes), if *n* can be represented as a sum of two triangular numbers, otherwise print "NO" (without the quotes).
Demo Input:
['256\n', '512\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first sample number <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/92095692c6ea93e9e3b837a0408ba7543549d5b2.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second sample number 512 can not be represented as a sum of two triangular numbers.
|
```python
import sys, os.path
from collections import*
from copy import*
import math
mod=10**9+7
if(os.path.exists('input.txt')):
sys.stdin = open("input.txt","r")
sys.stdout = open("output.txt","w")
def bs(left,right,x):
while(left<=right):
mid=left+(right-left)//2
b=(mid*(mid+1))//2
if(b==x):
return 1
elif(b>x):
right=mid-1
else:
left=mid+1
return 0
n=int(input())
n1=10**5
flag=0
for i in range(1,n1+1):
a=(i*(i+1))//2
if(bs(1,n1,n-a)):
flag=1
break
if(flag==0):
print('NO')
else:
print('YES')
```
| 3
|
|
214
|
A
|
System of Equations
|
PROGRAMMING
| 800
|
[
"brute force"
] | null | null |
Furik loves math lessons very much, so he doesn't attend them, unlike Rubik. But now Furik wants to get a good mark for math. For that Ms. Ivanova, his math teacher, gave him a new task. Furik solved the task immediately. Can you?
You are given a system of equations:
You should count, how many there are pairs of integers (*a*,<=*b*) (0<=β€<=*a*,<=*b*) which satisfy the system.
|
A single line contains two integers *n*,<=*m* (1<=β€<=*n*,<=*m*<=β€<=1000) β the parameters of the system. The numbers on the line are separated by a space.
|
On a single line print the answer to the problem.
|
[
"9 3\n",
"14 28\n",
"4 20\n"
] |
[
"1\n",
"1\n",
"0\n"
] |
In the first sample the suitable pair is integers (3,β0). In the second sample the suitable pair is integers (3,β5). In the third sample there is no suitable pair.
| 500
|
[
{
"input": "9 3",
"output": "1"
},
{
"input": "14 28",
"output": "1"
},
{
"input": "4 20",
"output": "0"
},
{
"input": "18 198",
"output": "1"
},
{
"input": "22 326",
"output": "1"
},
{
"input": "26 104",
"output": "1"
},
{
"input": "14 10",
"output": "0"
},
{
"input": "8 20",
"output": "0"
},
{
"input": "2 8",
"output": "0"
},
{
"input": "20 11",
"output": "0"
},
{
"input": "57 447",
"output": "1"
},
{
"input": "1 1",
"output": "2"
},
{
"input": "66 296",
"output": "1"
},
{
"input": "75 683",
"output": "1"
},
{
"input": "227 975",
"output": "1"
},
{
"input": "247 499",
"output": "1"
},
{
"input": "266 116",
"output": "1"
},
{
"input": "286 916",
"output": "1"
},
{
"input": "307 341",
"output": "1"
},
{
"input": "451 121",
"output": "1"
},
{
"input": "471 921",
"output": "1"
},
{
"input": "502 346",
"output": "1"
},
{
"input": "535 59",
"output": "1"
},
{
"input": "555 699",
"output": "1"
},
{
"input": "747 351",
"output": "1"
},
{
"input": "790 64",
"output": "1"
},
{
"input": "810 704",
"output": "1"
},
{
"input": "855 225",
"output": "1"
},
{
"input": "902 34",
"output": "1"
},
{
"input": "922 514",
"output": "1"
},
{
"input": "971 131",
"output": "1"
},
{
"input": "991 931",
"output": "1"
},
{
"input": "840 780",
"output": "0"
},
{
"input": "102 595",
"output": "0"
},
{
"input": "139 433",
"output": "0"
},
{
"input": "968 288",
"output": "0"
},
{
"input": "563 354",
"output": "0"
},
{
"input": "994 975",
"output": "0"
},
{
"input": "456 221",
"output": "0"
},
{
"input": "205 210",
"output": "0"
},
{
"input": "1 11",
"output": "0"
},
{
"input": "1000 1000",
"output": "0"
},
{
"input": "3 3",
"output": "0"
},
{
"input": "11 99",
"output": "0"
},
{
"input": "2 2",
"output": "1"
},
{
"input": "11 1",
"output": "0"
},
{
"input": "6 6",
"output": "1"
},
{
"input": "100 452",
"output": "0"
},
{
"input": "420 380",
"output": "0"
},
{
"input": "31 31",
"output": "0"
},
{
"input": "2 6",
"output": "0"
},
{
"input": "112 134",
"output": "0"
},
{
"input": "13 13",
"output": "0"
},
{
"input": "1 571",
"output": "0"
}
] | 1,685,299,983
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 54
| 560
| 0
|
import sys
import math
def function():
n, m = map(int, input().split())
ans = 0
for a in range(1001):
for b in range(1001):
if (a**2) + b == n and a + (b**2) == m:
ans+=1
print(ans)
return
if __name__ == '__main__':
function()
|
Title: System of Equations
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Furik loves math lessons very much, so he doesn't attend them, unlike Rubik. But now Furik wants to get a good mark for math. For that Ms. Ivanova, his math teacher, gave him a new task. Furik solved the task immediately. Can you?
You are given a system of equations:
You should count, how many there are pairs of integers (*a*,<=*b*) (0<=β€<=*a*,<=*b*) which satisfy the system.
Input Specification:
A single line contains two integers *n*,<=*m* (1<=β€<=*n*,<=*m*<=β€<=1000) β the parameters of the system. The numbers on the line are separated by a space.
Output Specification:
On a single line print the answer to the problem.
Demo Input:
['9 3\n', '14 28\n', '4 20\n']
Demo Output:
['1\n', '1\n', '0\n']
Note:
In the first sample the suitable pair is integers (3,β0). In the second sample the suitable pair is integers (3,β5). In the third sample there is no suitable pair.
|
```python
import sys
import math
def function():
n, m = map(int, input().split())
ans = 0
for a in range(1001):
for b in range(1001):
if (a**2) + b == n and a + (b**2) == m:
ans+=1
print(ans)
return
if __name__ == '__main__':
function()
```
| 3
|
|
43
|
A
|
Football
|
PROGRAMMING
| 1,000
|
[
"strings"
] |
A. Football
|
2
|
256
|
One day Vasya decided to have a look at the results of Berland 1910 Football Championshipβs finals. Unfortunately he didn't find the overall score of the match; however, he got hold of a profound description of the match's process. On the whole there are *n* lines in that description each of which described one goal. Every goal was marked with the name of the team that had scored it. Help Vasya, learn the name of the team that won the finals. It is guaranteed that the match did not end in a tie.
|
The first line contains an integer *n* (1<=β€<=*n*<=β€<=100) β the number of lines in the description. Then follow *n* lines β for each goal the names of the teams that scored it. The names are non-empty lines consisting of uppercase Latin letters whose lengths do not exceed 10 symbols. It is guaranteed that the match did not end in a tie and the description contains no more than two different teams.
|
Print the name of the winning team. We remind you that in football the team that scores more goals is considered the winner.
|
[
"1\nABC\n",
"5\nA\nABA\nABA\nA\nA\n"
] |
[
"ABC\n",
"A\n"
] |
none
| 500
|
[
{
"input": "1\nABC",
"output": "ABC"
},
{
"input": "5\nA\nABA\nABA\nA\nA",
"output": "A"
},
{
"input": "2\nXTSJEP\nXTSJEP",
"output": "XTSJEP"
},
{
"input": "3\nXZYDJAEDZ\nXZYDJAEDZ\nXZYDJAEDZ",
"output": "XZYDJAEDZ"
},
{
"input": "3\nQCCYXL\nQCCYXL\nAXGLFQDD",
"output": "QCCYXL"
},
{
"input": "3\nAZID\nEERWBC\nEERWBC",
"output": "EERWBC"
},
{
"input": "3\nHNCGYL\nHNCGYL\nHNCGYL",
"output": "HNCGYL"
},
{
"input": "4\nZZWZTG\nZZWZTG\nZZWZTG\nZZWZTG",
"output": "ZZWZTG"
},
{
"input": "4\nA\nA\nKUDLJMXCSE\nA",
"output": "A"
},
{
"input": "5\nPHBTW\nPHBTW\nPHBTW\nPHBTW\nPHBTW",
"output": "PHBTW"
},
{
"input": "5\nPKUZYTFYWN\nPKUZYTFYWN\nSTC\nPKUZYTFYWN\nPKUZYTFYWN",
"output": "PKUZYTFYWN"
},
{
"input": "5\nHH\nHH\nNTQWPA\nNTQWPA\nHH",
"output": "HH"
},
{
"input": "10\nW\nW\nW\nW\nW\nD\nW\nD\nD\nW",
"output": "W"
},
{
"input": "19\nXBCP\nTGACNIH\nXBCP\nXBCP\nXBCP\nXBCP\nXBCP\nTGACNIH\nXBCP\nXBCP\nXBCP\nXBCP\nXBCP\nTGACNIH\nXBCP\nXBCP\nTGACNIH\nTGACNIH\nXBCP",
"output": "XBCP"
},
{
"input": "33\nOWQWCKLLF\nOWQWCKLLF\nOWQWCKLLF\nPYPAS\nPYPAS\nPYPAS\nOWQWCKLLF\nPYPAS\nOWQWCKLLF\nPYPAS\nPYPAS\nOWQWCKLLF\nOWQWCKLLF\nOWQWCKLLF\nPYPAS\nOWQWCKLLF\nPYPAS\nPYPAS\nPYPAS\nPYPAS\nOWQWCKLLF\nPYPAS\nPYPAS\nOWQWCKLLF\nOWQWCKLLF\nPYPAS\nOWQWCKLLF\nOWQWCKLLF\nPYPAS\nPYPAS\nOWQWCKLLF\nPYPAS\nPYPAS",
"output": "PYPAS"
},
{
"input": "51\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC",
"output": "NC"
},
{
"input": "89\nH\nVOCI\nVOCI\nH\nVOCI\nH\nH\nVOCI\nVOCI\nVOCI\nH\nH\nH\nVOCI\nVOCI\nVOCI\nH\nVOCI\nVOCI\nH\nVOCI\nVOCI\nVOCI\nH\nVOCI\nH\nVOCI\nH\nVOCI\nH\nVOCI\nVOCI\nH\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nH\nVOCI\nVOCI\nVOCI\nVOCI\nH\nVOCI\nH\nH\nVOCI\nH\nVOCI\nH\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nH\nH\nVOCI\nH\nH\nVOCI\nH\nVOCI\nH\nVOCI\nVOCI\nH\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nH\nH\nH\nH\nH\nVOCI\nH\nVOCI\nH\nVOCI\nVOCI",
"output": "VOCI"
},
{
"input": "100\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nM\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA",
"output": "HA"
},
{
"input": "100\nG\nG\nS\nS\nG\nG\nS\nS\nG\nS\nS\nS\nG\nS\nG\nG\nS\nG\nS\nS\nG\nS\nS\nS\nS\nS\nG\nS\nG\nS\nS\nG\nG\nG\nS\nS\nS\nS\nG\nS\nS\nG\nG\nG\nG\nG\nS\nG\nG\nS\nS\nS\nS\nS\nG\nG\nS\nG\nG\nG\nG\nG\nS\nS\nG\nS\nS\nS\nS\nG\nS\nS\nG\nS\nG\nG\nG\nG\nG\nG\nG\nG\nG\nG\nG\nS\nS\nG\nS\nS\nS\nS\nG\nG\nG\nS\nG\nG\nG\nS",
"output": "G"
},
{
"input": "100\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nOBH\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL",
"output": "WL"
}
] | 1,640,018,288
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 34
| 92
| 0
|
#football 43A
n_l = int(input())
team = []
for i in range(n_l):
team.append(input())
print(max(team, key = team.count))
|
Title: Football
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
One day Vasya decided to have a look at the results of Berland 1910 Football Championshipβs finals. Unfortunately he didn't find the overall score of the match; however, he got hold of a profound description of the match's process. On the whole there are *n* lines in that description each of which described one goal. Every goal was marked with the name of the team that had scored it. Help Vasya, learn the name of the team that won the finals. It is guaranteed that the match did not end in a tie.
Input Specification:
The first line contains an integer *n* (1<=β€<=*n*<=β€<=100) β the number of lines in the description. Then follow *n* lines β for each goal the names of the teams that scored it. The names are non-empty lines consisting of uppercase Latin letters whose lengths do not exceed 10 symbols. It is guaranteed that the match did not end in a tie and the description contains no more than two different teams.
Output Specification:
Print the name of the winning team. We remind you that in football the team that scores more goals is considered the winner.
Demo Input:
['1\nABC\n', '5\nA\nABA\nABA\nA\nA\n']
Demo Output:
['ABC\n', 'A\n']
Note:
none
|
```python
#football 43A
n_l = int(input())
team = []
for i in range(n_l):
team.append(input())
print(max(team, key = team.count))
```
| 3.977
|
805
|
B
|
3-palindrome
|
PROGRAMMING
| 1,000
|
[
"constructive algorithms"
] | null | null |
In the beginning of the new year Keivan decided to reverse his name. He doesn't like palindromes, so he changed Naviek to Navick.
He is too selfish, so for a given *n* he wants to obtain a string of *n* characters, each of which is either 'a', 'b' or 'c', with no palindromes of length 3 appearing in the string as a substring. For example, the strings "abc" and "abca" suit him, while the string "aba" doesn't. He also want the number of letters 'c' in his string to be as little as possible.
|
The first line contains single integer *n* (1<=β€<=*n*<=β€<=2Β·105)Β β the length of the string.
|
Print the string that satisfies all the constraints.
If there are multiple answers, print any of them.
|
[
"2\n",
"3\n"
] |
[
"aa\n",
"bba\n"
] |
A palindrome is a sequence of characters which reads the same backward and forward.
| 1,000
|
[
{
"input": "2",
"output": "aa"
},
{
"input": "3",
"output": "aab"
},
{
"input": "38",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaa"
},
{
"input": "47",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab"
},
{
"input": "59",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab"
},
{
"input": "67",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab"
},
{
"input": "77",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabba"
},
{
"input": "89",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabba"
},
{
"input": "98",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaa"
},
{
"input": "109",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabba"
},
{
"input": "117",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabba"
},
{
"input": "1",
"output": "a"
},
{
"input": "200000",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab..."
},
{
"input": "100000",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab..."
},
{
"input": "143670",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab..."
},
{
"input": "104217",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab..."
},
{
"input": "17879",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab..."
},
{
"input": "131809",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab..."
},
{
"input": "140873",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab..."
},
{
"input": "77859",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab..."
},
{
"input": "153022",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab..."
},
{
"input": "179227",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab..."
},
{
"input": "182801",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab..."
},
{
"input": "5188",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab..."
},
{
"input": "86539",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab..."
},
{
"input": "12802",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab..."
},
{
"input": "120289",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab..."
},
{
"input": "132866",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab..."
},
{
"input": "133377",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab..."
},
{
"input": "31775",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab..."
},
{
"input": "160397",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab..."
},
{
"input": "161415",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab..."
},
{
"input": "163623",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab..."
},
{
"input": "22942",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab..."
},
{
"input": "134767",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab..."
},
{
"input": "52257",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab..."
},
{
"input": "64905",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab..."
},
{
"input": "183758",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab..."
},
{
"input": "60131",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab..."
},
{
"input": "146883",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab..."
},
{
"input": "6",
"output": "aabbaa"
},
{
"input": "4",
"output": "aabb"
}
] | 1,607,960,318
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 42
| 108
| 409,600
|
n = int(input())
k="aabb"
print(k*(n//4)+k[:n%4])
|
Title: 3-palindrome
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In the beginning of the new year Keivan decided to reverse his name. He doesn't like palindromes, so he changed Naviek to Navick.
He is too selfish, so for a given *n* he wants to obtain a string of *n* characters, each of which is either 'a', 'b' or 'c', with no palindromes of length 3 appearing in the string as a substring. For example, the strings "abc" and "abca" suit him, while the string "aba" doesn't. He also want the number of letters 'c' in his string to be as little as possible.
Input Specification:
The first line contains single integer *n* (1<=β€<=*n*<=β€<=2Β·105)Β β the length of the string.
Output Specification:
Print the string that satisfies all the constraints.
If there are multiple answers, print any of them.
Demo Input:
['2\n', '3\n']
Demo Output:
['aa\n', 'bba\n']
Note:
A palindrome is a sequence of characters which reads the same backward and forward.
|
```python
n = int(input())
k="aabb"
print(k*(n//4)+k[:n%4])
```
| 3
|
|
709
|
A
|
Juicer
|
PROGRAMMING
| 900
|
[
"implementation"
] | null | null |
Kolya is going to make fresh orange juice. He has *n* oranges of sizes *a*1,<=*a*2,<=...,<=*a**n*. Kolya will put them in the juicer in the fixed order, starting with orange of size *a*1, then orange of size *a*2 and so on. To be put in the juicer the orange must have size not exceeding *b*, so if Kolya sees an orange that is strictly greater he throws it away and continues with the next one.
The juicer has a special section to collect waste. It overflows if Kolya squeezes oranges of the total size strictly greater than *d*. When it happens Kolya empties the waste section (even if there are no more oranges) and continues to squeeze the juice. How many times will he have to empty the waste section?
|
The first line of the input contains three integers *n*, *b* and *d* (1<=β€<=*n*<=β€<=100<=000, 1<=β€<=*b*<=β€<=*d*<=β€<=1<=000<=000)Β β the number of oranges, the maximum size of the orange that fits in the juicer and the value *d*, which determines the condition when the waste section should be emptied.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=β€<=*a**i*<=β€<=1<=000<=000)Β β sizes of the oranges listed in the order Kolya is going to try to put them in the juicer.
|
Print one integerΒ β the number of times Kolya will have to empty the waste section.
|
[
"2 7 10\n5 6\n",
"1 5 10\n7\n",
"3 10 10\n5 7 7\n",
"1 1 1\n1\n"
] |
[
"1\n",
"0\n",
"1\n",
"0\n"
] |
In the first sample, Kolya will squeeze the juice from two oranges and empty the waste section afterwards.
In the second sample, the orange won't fit in the juicer so Kolya will have no juice at all.
| 500
|
[
{
"input": "2 7 10\n5 6",
"output": "1"
},
{
"input": "1 5 10\n7",
"output": "0"
},
{
"input": "3 10 10\n5 7 7",
"output": "1"
},
{
"input": "1 1 1\n1",
"output": "0"
},
{
"input": "2 951637 951638\n44069 951637",
"output": "1"
},
{
"input": "50 100 129\n55 130 91 19 116 3 63 52 104 76 75 27 151 99 149 147 39 148 84 9 132 49 40 112 124 141 144 93 36 32 146 74 48 38 150 55 94 32 107 69 77 81 33 57 62 98 78 127 154 126",
"output": "12"
},
{
"input": "100 1000 1083\n992 616 818 359 609 783 263 989 501 929 362 394 919 1081 870 830 1097 975 62 346 531 367 323 457 707 360 949 334 867 116 478 417 961 963 1029 114 867 1008 988 916 983 1077 959 942 572 961 579 318 721 337 488 717 111 70 416 685 987 130 353 107 61 191 827 849 106 815 211 953 111 398 889 860 801 71 375 320 395 1059 116 222 931 444 582 74 677 655 88 173 686 491 661 186 114 832 615 814 791 464 517 850",
"output": "36"
},
{
"input": "2 6 8\n2 1",
"output": "0"
},
{
"input": "5 15 16\n7 11 5 12 8",
"output": "2"
},
{
"input": "15 759966 759967\n890397 182209 878577 548548 759966 812923 759966 860479 200595 381358 299175 339368 759966 907668 69574",
"output": "4"
},
{
"input": "5 234613 716125\n642626 494941 234613 234613 234613",
"output": "0"
},
{
"input": "50 48547 567054\n529808 597004 242355 559114 78865 537318 631455 733020 655072 645093 309010 855034 306058 625046 524574 834944 27330 664392 443637 821584 338013 490702 289520 675471 885846 258814 134220 571301 84875 94132 200425 928833 375166 521232 317961 175315 947093 89971 322071 174033 48547 998535 954205 704114 943163 438900 48547 538422 48547 48547",
"output": "0"
},
{
"input": "5 10 20\n10 10 10 10 1",
"output": "1"
},
{
"input": "5 10 11\n10 10 10 10 1",
"output": "2"
},
{
"input": "3 10 10\n4 3 3",
"output": "0"
},
{
"input": "3 5 5\n5 5 5",
"output": "1"
},
{
"input": "3 4 14\n5 5 5",
"output": "0"
},
{
"input": "2 7 10\n1234 1234",
"output": "0"
},
{
"input": "1 5 6\n10",
"output": "0"
},
{
"input": "3 4 6\n1 2 3",
"output": "0"
},
{
"input": "5 10 12\n13 13 13 13 13",
"output": "0"
},
{
"input": "3 4 5\n5 7 9",
"output": "0"
},
{
"input": "3 10 100\n5 5 5",
"output": "0"
},
{
"input": "5 1 2\n2 2 2 2 2",
"output": "0"
},
{
"input": "5 5 5\n5 5 5 5 5",
"output": "2"
},
{
"input": "2 2 5\n5 5",
"output": "0"
},
{
"input": "3 1 4\n2 2 2",
"output": "0"
}
] | 1,612,469,279
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 58
| 109
| 6,144,000
|
def getNumOfTimes(b, d, oranges):
num = 0
waste = 0
for orange in oranges:
if orange > b:
continue
if waste + orange > d:
num += 1
waste = 0
else:
waste += orange
return num
if __name__ == "__main__":
_, b, d = map(int, input().split())
oranges = map(int, input().split())
print(getNumOfTimes(b, d, oranges))
|
Title: Juicer
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kolya is going to make fresh orange juice. He has *n* oranges of sizes *a*1,<=*a*2,<=...,<=*a**n*. Kolya will put them in the juicer in the fixed order, starting with orange of size *a*1, then orange of size *a*2 and so on. To be put in the juicer the orange must have size not exceeding *b*, so if Kolya sees an orange that is strictly greater he throws it away and continues with the next one.
The juicer has a special section to collect waste. It overflows if Kolya squeezes oranges of the total size strictly greater than *d*. When it happens Kolya empties the waste section (even if there are no more oranges) and continues to squeeze the juice. How many times will he have to empty the waste section?
Input Specification:
The first line of the input contains three integers *n*, *b* and *d* (1<=β€<=*n*<=β€<=100<=000, 1<=β€<=*b*<=β€<=*d*<=β€<=1<=000<=000)Β β the number of oranges, the maximum size of the orange that fits in the juicer and the value *d*, which determines the condition when the waste section should be emptied.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=β€<=*a**i*<=β€<=1<=000<=000)Β β sizes of the oranges listed in the order Kolya is going to try to put them in the juicer.
Output Specification:
Print one integerΒ β the number of times Kolya will have to empty the waste section.
Demo Input:
['2 7 10\n5 6\n', '1 5 10\n7\n', '3 10 10\n5 7 7\n', '1 1 1\n1\n']
Demo Output:
['1\n', '0\n', '1\n', '0\n']
Note:
In the first sample, Kolya will squeeze the juice from two oranges and empty the waste section afterwards.
In the second sample, the orange won't fit in the juicer so Kolya will have no juice at all.
|
```python
def getNumOfTimes(b, d, oranges):
num = 0
waste = 0
for orange in oranges:
if orange > b:
continue
if waste + orange > d:
num += 1
waste = 0
else:
waste += orange
return num
if __name__ == "__main__":
_, b, d = map(int, input().split())
oranges = map(int, input().split())
print(getNumOfTimes(b, d, oranges))
```
| 3
|
|
546
|
A
|
Soldier and Bananas
|
PROGRAMMING
| 800
|
[
"brute force",
"implementation",
"math"
] | null | null |
A soldier wants to buy *w* bananas in the shop. He has to pay *k* dollars for the first banana, 2*k* dollars for the second one and so on (in other words, he has to pay *i*Β·*k* dollars for the *i*-th banana).
He has *n* dollars. How many dollars does he have to borrow from his friend soldier to buy *w* bananas?
|
The first line contains three positive integers *k*,<=*n*,<=*w* (1<=<=β€<=<=*k*,<=*w*<=<=β€<=<=1000, 0<=β€<=*n*<=β€<=109), the cost of the first banana, initial number of dollars the soldier has and number of bananas he wants.
|
Output one integer β the amount of dollars that the soldier must borrow from his friend. If he doesn't have to borrow money, output 0.
|
[
"3 17 4\n"
] |
[
"13"
] |
none
| 500
|
[
{
"input": "3 17 4",
"output": "13"
},
{
"input": "1 2 1",
"output": "0"
},
{
"input": "1 1 1",
"output": "0"
},
{
"input": "1 5 6",
"output": "16"
},
{
"input": "1 1000000000 1",
"output": "0"
},
{
"input": "1000 0 1000",
"output": "500500000"
},
{
"input": "859 453892 543",
"output": "126416972"
},
{
"input": "1000 1000000000 1000",
"output": "0"
},
{
"input": "1000 500500000 1000",
"output": "0"
},
{
"input": "1000 500500001 1000",
"output": "0"
},
{
"input": "1000 500499999 1000",
"output": "1"
},
{
"input": "634 87973 214",
"output": "14497197"
},
{
"input": "432 10000 241",
"output": "12587552"
},
{
"input": "111 111111111 111",
"output": "0"
},
{
"input": "20 43 3",
"output": "77"
}
] | 1,694,085,800
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 15
| 46
| 0
|
a,b,n=map(int,input().split())
borrow=0
tcost=(n*(n+1))//2 *a
borrow=tcost-b
if borrow<=0:
print(0)
else:
print(borrow)
|
Title: Soldier and Bananas
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A soldier wants to buy *w* bananas in the shop. He has to pay *k* dollars for the first banana, 2*k* dollars for the second one and so on (in other words, he has to pay *i*Β·*k* dollars for the *i*-th banana).
He has *n* dollars. How many dollars does he have to borrow from his friend soldier to buy *w* bananas?
Input Specification:
The first line contains three positive integers *k*,<=*n*,<=*w* (1<=<=β€<=<=*k*,<=*w*<=<=β€<=<=1000, 0<=β€<=*n*<=β€<=109), the cost of the first banana, initial number of dollars the soldier has and number of bananas he wants.
Output Specification:
Output one integer β the amount of dollars that the soldier must borrow from his friend. If he doesn't have to borrow money, output 0.
Demo Input:
['3 17 4\n']
Demo Output:
['13']
Note:
none
|
```python
a,b,n=map(int,input().split())
borrow=0
tcost=(n*(n+1))//2 *a
borrow=tcost-b
if borrow<=0:
print(0)
else:
print(borrow)
```
| 3
|
|
268
|
B
|
Buttons
|
PROGRAMMING
| 1,000
|
[
"implementation",
"math"
] | null | null |
Manao is trying to open a rather challenging lock. The lock has *n* buttons on it and to open it, you should press the buttons in a certain order to open the lock. When you push some button, it either stays pressed into the lock (that means that you've guessed correctly and pushed the button that goes next in the sequence), or all pressed buttons return to the initial position. When all buttons are pressed into the lock at once, the lock opens.
Consider an example with three buttons. Let's say that the opening sequence is: {2, 3, 1}. If you first press buttons 1 or 3, the buttons unpress immediately. If you first press button 2, it stays pressed. If you press 1 after 2, all buttons unpress. If you press 3 after 2, buttons 3 and 2 stay pressed. As soon as you've got two pressed buttons, you only need to press button 1 to open the lock.
Manao doesn't know the opening sequence. But he is really smart and he is going to act in the optimal way. Calculate the number of times he's got to push a button in order to open the lock in the worst-case scenario.
|
A single line contains integer *n* (1<=β€<=*n*<=β€<=2000) β the number of buttons the lock has.
|
In a single line print the number of times Manao has to push a button in the worst-case scenario.
|
[
"2\n",
"3\n"
] |
[
"3\n",
"7\n"
] |
Consider the first test sample. Manao can fail his first push and push the wrong button. In this case he will already be able to guess the right one with his second push. And his third push will push the second right button. Thus, in the worst-case scenario he will only need 3 pushes.
| 1,000
|
[
{
"input": "2",
"output": "3"
},
{
"input": "3",
"output": "7"
},
{
"input": "4",
"output": "14"
},
{
"input": "1",
"output": "1"
},
{
"input": "10",
"output": "175"
},
{
"input": "2000",
"output": "1333335000"
},
{
"input": "1747",
"output": "888644743"
},
{
"input": "889",
"output": "117099969"
},
{
"input": "1999",
"output": "1331335999"
},
{
"input": "914",
"output": "127259419"
},
{
"input": "996",
"output": "164675486"
},
{
"input": "17",
"output": "833"
},
{
"input": "50",
"output": "20875"
},
{
"input": "91",
"output": "125671"
},
{
"input": "92",
"output": "129858"
},
{
"input": "256",
"output": "2796416"
},
{
"input": "512",
"output": "22370048"
},
{
"input": "666",
"output": "49235271"
},
{
"input": "667",
"output": "49457383"
},
{
"input": "314",
"output": "5160119"
},
{
"input": "1241",
"output": "318541121"
},
{
"input": "1500",
"output": "562501250"
},
{
"input": "1837",
"output": "1033182073"
},
{
"input": "1000",
"output": "166667500"
}
] | 1,691,774,191
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 24
| 92
| 0
|
n = int(input())
result = 0
for i in range(1, n + 1):
result += i * (n - i) + 1
print(result)
|
Title: Buttons
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Manao is trying to open a rather challenging lock. The lock has *n* buttons on it and to open it, you should press the buttons in a certain order to open the lock. When you push some button, it either stays pressed into the lock (that means that you've guessed correctly and pushed the button that goes next in the sequence), or all pressed buttons return to the initial position. When all buttons are pressed into the lock at once, the lock opens.
Consider an example with three buttons. Let's say that the opening sequence is: {2, 3, 1}. If you first press buttons 1 or 3, the buttons unpress immediately. If you first press button 2, it stays pressed. If you press 1 after 2, all buttons unpress. If you press 3 after 2, buttons 3 and 2 stay pressed. As soon as you've got two pressed buttons, you only need to press button 1 to open the lock.
Manao doesn't know the opening sequence. But he is really smart and he is going to act in the optimal way. Calculate the number of times he's got to push a button in order to open the lock in the worst-case scenario.
Input Specification:
A single line contains integer *n* (1<=β€<=*n*<=β€<=2000) β the number of buttons the lock has.
Output Specification:
In a single line print the number of times Manao has to push a button in the worst-case scenario.
Demo Input:
['2\n', '3\n']
Demo Output:
['3\n', '7\n']
Note:
Consider the first test sample. Manao can fail his first push and push the wrong button. In this case he will already be able to guess the right one with his second push. And his third push will push the second right button. Thus, in the worst-case scenario he will only need 3 pushes.
|
```python
n = int(input())
result = 0
for i in range(1, n + 1):
result += i * (n - i) + 1
print(result)
```
| 3
|
|
275
|
A
|
Lights Out
|
PROGRAMMING
| 900
|
[
"implementation"
] | null | null |
Lenny is playing a game on a 3<=Γ<=3 grid of lights. In the beginning of the game all lights are switched on. Pressing any of the lights will toggle it and all side-adjacent lights. The goal of the game is to switch all the lights off. We consider the toggling as follows: if the light was switched on then it will be switched off, if it was switched off then it will be switched on.
Lenny has spent some time playing with the grid and by now he has pressed each light a certain number of times. Given the number of times each light is pressed, you have to print the current state of each light.
|
The input consists of three rows. Each row contains three integers each between 0 to 100 inclusive. The *j*-th number in the *i*-th row is the number of times the *j*-th light of the *i*-th row of the grid is pressed.
|
Print three lines, each containing three characters. The *j*-th character of the *i*-th line is "1" if and only if the corresponding light is switched on, otherwise it's "0".
|
[
"1 0 0\n0 0 0\n0 0 1\n",
"1 0 1\n8 8 8\n2 0 3\n"
] |
[
"001\n010\n100\n",
"010\n011\n100\n"
] |
none
| 500
|
[
{
"input": "1 0 0\n0 0 0\n0 0 1",
"output": "001\n010\n100"
},
{
"input": "1 0 1\n8 8 8\n2 0 3",
"output": "010\n011\n100"
},
{
"input": "13 85 77\n25 50 45\n65 79 9",
"output": "000\n010\n000"
},
{
"input": "96 95 5\n8 84 74\n67 31 61",
"output": "011\n011\n101"
},
{
"input": "24 54 37\n60 63 6\n1 84 26",
"output": "110\n101\n011"
},
{
"input": "23 10 40\n15 6 40\n92 80 77",
"output": "101\n100\n000"
},
{
"input": "62 74 80\n95 74 93\n2 47 95",
"output": "010\n001\n110"
},
{
"input": "80 83 48\n26 0 66\n47 76 37",
"output": "000\n000\n010"
},
{
"input": "32 15 65\n7 54 36\n5 51 3",
"output": "111\n101\n001"
},
{
"input": "22 97 12\n71 8 24\n100 21 64",
"output": "100\n001\n100"
},
{
"input": "46 37 13\n87 0 50\n90 8 55",
"output": "111\n011\n000"
},
{
"input": "57 43 58\n20 82 83\n66 16 52",
"output": "111\n010\n110"
},
{
"input": "45 56 93\n47 51 59\n18 51 63",
"output": "101\n011\n100"
},
{
"input": "47 66 67\n14 1 37\n27 81 69",
"output": "001\n001\n110"
},
{
"input": "26 69 69\n85 18 23\n14 22 74",
"output": "110\n001\n010"
},
{
"input": "10 70 65\n94 27 25\n74 66 30",
"output": "111\n010\n100"
},
{
"input": "97 1 74\n15 99 1\n88 68 86",
"output": "001\n011\n000"
},
{
"input": "36 48 42\n45 41 66\n26 64 1",
"output": "001\n111\n010"
},
{
"input": "52 81 97\n29 77 71\n66 11 2",
"output": "100\n100\n111"
},
{
"input": "18 66 33\n19 49 49\n48 46 26",
"output": "011\n100\n000"
},
{
"input": "68 79 52\n51 39 100\n29 14 26",
"output": "110\n000\n111"
},
{
"input": "91 69 77\n91 26 64\n91 88 57",
"output": "001\n011\n110"
},
{
"input": "16 69 64\n48 21 80\n81 51 51",
"output": "010\n101\n111"
},
{
"input": "96 14 2\n100 18 12\n65 34 89",
"output": "111\n010\n010"
},
{
"input": "93 95 90\n8 59 42\n53 13 19",
"output": "100\n001\n111"
},
{
"input": "71 84 18\n100 19 67\n9 76 15",
"output": "010\n010\n001"
},
{
"input": "38 93 85\n21 88 64\n4 96 25",
"output": "111\n011\n000"
},
{
"input": "75 20 20\n60 5 78\n77 4 69",
"output": "011\n001\n000"
},
{
"input": "65 70 96\n19 6 83\n33 37 82",
"output": "100\n000\n011"
},
{
"input": "11 13 60\n17 13 46\n42 21 39",
"output": "000\n011\n101"
},
{
"input": "0 0 0\n0 0 0\n0 0 0",
"output": "111\n111\n111"
},
{
"input": "0 0 0\n0 1 0\n0 0 0",
"output": "101\n000\n101"
},
{
"input": "0 0 0\n0 0 0\n0 0 1",
"output": "111\n110\n100"
}
] | 1,661,157,611
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 33
| 77
| 0
|
a=input()
a=a.split()
a=list(map(int,a))
b=input()
b=b.split()
b=list(map(int,b))
c=input()
c=c.split()
c=list(map(int,c))
aa=''
bb=''
cc=''
if (a[0]+b[0]+a[1])%2==0:
aa+='1'
else:
aa+='0'
if (a[0]+b[1]+a[1]+a[2])%2==0:
aa+='1'
else:
aa+='0'
if (a[2]+b[2]+a[1])%2==0:
aa+='1'
else:
aa+='0'
if (a[0]+b[0]+c[0]+b[1])%2==0:
bb+='1'
else:
bb+='0'
if (a[1]+b[0]+c[1]+b[1]+b[2])%2==0:
bb+='1'
else:
bb+='0'
if (a[2]+b[2]+c[2]+b[1])%2==0:
bb+='1'
else:
bb+='0'
if (c[0]+b[0]+c[1])%2==0:
cc+='1'
else:
cc+='0'
if (c[0]+b[1]+c[1]+c[2])%2==0:
cc+='1'
else:
cc+='0'
if (c[2]+b[2]+c[1])%2==0:
cc+='1'
else:
cc+='0'
print(aa)
print(bb)
print(cc)
|
Title: Lights Out
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Lenny is playing a game on a 3<=Γ<=3 grid of lights. In the beginning of the game all lights are switched on. Pressing any of the lights will toggle it and all side-adjacent lights. The goal of the game is to switch all the lights off. We consider the toggling as follows: if the light was switched on then it will be switched off, if it was switched off then it will be switched on.
Lenny has spent some time playing with the grid and by now he has pressed each light a certain number of times. Given the number of times each light is pressed, you have to print the current state of each light.
Input Specification:
The input consists of three rows. Each row contains three integers each between 0 to 100 inclusive. The *j*-th number in the *i*-th row is the number of times the *j*-th light of the *i*-th row of the grid is pressed.
Output Specification:
Print three lines, each containing three characters. The *j*-th character of the *i*-th line is "1" if and only if the corresponding light is switched on, otherwise it's "0".
Demo Input:
['1 0 0\n0 0 0\n0 0 1\n', '1 0 1\n8 8 8\n2 0 3\n']
Demo Output:
['001\n010\n100\n', '010\n011\n100\n']
Note:
none
|
```python
a=input()
a=a.split()
a=list(map(int,a))
b=input()
b=b.split()
b=list(map(int,b))
c=input()
c=c.split()
c=list(map(int,c))
aa=''
bb=''
cc=''
if (a[0]+b[0]+a[1])%2==0:
aa+='1'
else:
aa+='0'
if (a[0]+b[1]+a[1]+a[2])%2==0:
aa+='1'
else:
aa+='0'
if (a[2]+b[2]+a[1])%2==0:
aa+='1'
else:
aa+='0'
if (a[0]+b[0]+c[0]+b[1])%2==0:
bb+='1'
else:
bb+='0'
if (a[1]+b[0]+c[1]+b[1]+b[2])%2==0:
bb+='1'
else:
bb+='0'
if (a[2]+b[2]+c[2]+b[1])%2==0:
bb+='1'
else:
bb+='0'
if (c[0]+b[0]+c[1])%2==0:
cc+='1'
else:
cc+='0'
if (c[0]+b[1]+c[1]+c[2])%2==0:
cc+='1'
else:
cc+='0'
if (c[2]+b[2]+c[1])%2==0:
cc+='1'
else:
cc+='0'
print(aa)
print(bb)
print(cc)
```
| 3
|
|
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<=β€<=*n*<=β€<=1000) β 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"
}
] | 1,689,054,805
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 34
| 46
| 307,200
|
from collections import deque
n = int(input())
a = deque([int(i) for i in input().split()])
sereza = 0
dima = 0
turns = 1
while len(a) > 0:
if turns % 2 == 1:
if a[0] > a[-1]:
sereza += a[0]
a.popleft()
else:
sereza += a[-1]
a.pop()
else:
if a[0] > a[-1]:
dima += a[0]
a.popleft()
else:
dima += a[-1]
a.pop()
turns += 1
print(sereza, dima)
|
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<=β€<=*n*<=β€<=1000) β 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
from collections import deque
n = int(input())
a = deque([int(i) for i in input().split()])
sereza = 0
dima = 0
turns = 1
while len(a) > 0:
if turns % 2 == 1:
if a[0] > a[-1]:
sereza += a[0]
a.popleft()
else:
sereza += a[-1]
a.pop()
else:
if a[0] > a[-1]:
dima += a[0]
a.popleft()
else:
dima += a[-1]
a.pop()
turns += 1
print(sereza, dima)
```
| 3
|
|
50
|
A
|
Domino piling
|
PROGRAMMING
| 800
|
[
"greedy",
"math"
] |
A. Domino piling
|
2
|
256
|
You are given a rectangular board of *M*<=Γ<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=Γ<=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* β board sizes in squares (1<=β€<=*M*<=β€<=*N*<=β€<=16).
|
Output one number β 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"
}
] | 1,647,238,852
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 35
| 92
| 0
|
first_multiple_input = input().rstrip().split()
n = int(first_multiple_input[0])
k = int(first_multiple_input[1])
print(n*k//2)
|
Title: Domino piling
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
You are given a rectangular board of *M*<=Γ<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=Γ<=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* β board sizes in squares (1<=β€<=*M*<=β€<=*N*<=β€<=16).
Output Specification:
Output one number β 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
first_multiple_input = input().rstrip().split()
n = int(first_multiple_input[0])
k = int(first_multiple_input[1])
print(n*k//2)
```
| 3.977
|
63
|
A
|
Sinking Ship
|
PROGRAMMING
| 900
|
[
"implementation",
"sortings",
"strings"
] |
A. Sinking Ship
|
2
|
256
|
The ship crashed into a reef and is sinking. Now the entire crew must be evacuated. All *n* crew members have already lined up in a row (for convenience let's label them all from left to right with positive integers from 1 to *n*) and await further instructions. However, one should evacuate the crew properly, in a strict order. Specifically:
The first crew members to leave the ship are rats. Then women and children (both groups have the same priority) leave the ship. After that all men are evacuated from the ship. The captain leaves the sinking ship last.
If we cannot determine exactly who should leave the ship first for any two members of the crew by the rules from the previous paragraph, then the one who stands to the left in the line leaves the ship first (or in other words, the one whose number in the line is less).
For each crew member we know his status as a crew member, and also his name. All crew members have different names. Determine the order in which to evacuate the crew.
|
The first line contains an integer *n*, which is the number of people in the crew (1<=β€<=*n*<=β€<=100). Then follow *n* lines. The *i*-th of those lines contains two words β the name of the crew member who is *i*-th in line, and his status on the ship. The words are separated by exactly one space. There are no other spaces in the line. The names consist of Latin letters, the first letter is uppercase, the rest are lowercase. The length of any name is from 1 to 10 characters. The status can have the following values: rat for a rat, woman for a woman, child for a child, man for a man, captain for the captain. The crew contains exactly one captain.
|
Print *n* lines. The *i*-th of them should contain the name of the crew member who must be the *i*-th one to leave the ship.
|
[
"6\nJack captain\nAlice woman\nCharlie man\nTeddy rat\nBob child\nJulia woman\n"
] |
[
"Teddy\nAlice\nBob\nJulia\nCharlie\nJack\n"
] |
none
| 500
|
[
{
"input": "6\nJack captain\nAlice woman\nCharlie man\nTeddy rat\nBob child\nJulia woman",
"output": "Teddy\nAlice\nBob\nJulia\nCharlie\nJack"
},
{
"input": "1\nA captain",
"output": "A"
},
{
"input": "1\nAbcdefjhij captain",
"output": "Abcdefjhij"
},
{
"input": "5\nA captain\nB man\nD woman\nC child\nE rat",
"output": "E\nD\nC\nB\nA"
},
{
"input": "10\nCap captain\nD child\nC woman\nA woman\nE child\nMan man\nB child\nF woman\nRat rat\nRatt rat",
"output": "Rat\nRatt\nD\nC\nA\nE\nB\nF\nMan\nCap"
},
{
"input": "5\nJoyxnkypf captain\nDxssgr woman\nKeojmnpd rat\nGdv man\nHnw man",
"output": "Keojmnpd\nDxssgr\nGdv\nHnw\nJoyxnkypf"
},
{
"input": "11\nJue rat\nWyglbyphk rat\nGjlgu child\nGi man\nAttx rat\nTheorpkgx man\nYm rat\nX child\nB captain\nEnualf rat\nKktsgyuyv woman",
"output": "Jue\nWyglbyphk\nAttx\nYm\nEnualf\nGjlgu\nX\nKktsgyuyv\nGi\nTheorpkgx\nB"
},
{
"input": "22\nWswwcvvm woman\nBtmfats rat\nI rat\nOcmtsnwx man\nUrcqv rat\nYghnogt woman\nWtyfc man\nWqle child\nUjfrelpu rat\nDstixj man\nAhksnio woman\nKhkvaap woman\nSjppvwm rat\nEgdmsv rat\nDank rat\nNquicjnw rat\nLh captain\nTdyaqaqln rat\nQtj rat\nTfgwijvq rat\nNbiso child\nNqthvbf woman",
"output": "Btmfats\nI\nUrcqv\nUjfrelpu\nSjppvwm\nEgdmsv\nDank\nNquicjnw\nTdyaqaqln\nQtj\nTfgwijvq\nWswwcvvm\nYghnogt\nWqle\nAhksnio\nKhkvaap\nNbiso\nNqthvbf\nOcmtsnwx\nWtyfc\nDstixj\nLh"
},
{
"input": "36\nKqxmtwmsf child\nIze woman\nDlpr child\nK woman\nF captain\nRjwfeuhba rat\nBbv rat\nS rat\nMnmg woman\nSmzyx woman\nSr man\nQmhroracn rat\nSoqpuqock rat\nPibdq man\nIlrkrptx rat\nZaecfyqka man\nMmersfs child\nVvvocqi man\nHjeqxvq rat\nMpmb woman\nWmgu woman\nCerelmhoxi child\nA man\nDylv man\nXrdgmmtcpq woman\nXj woman\nCeh child\nOfccguo man\nNevr man\nAknkwdkq child\nBmo man\nLqo woman\nVej child\nT child\nRlijc man\nNsdtpdg woman",
"output": "Rjwfeuhba\nBbv\nS\nQmhroracn\nSoqpuqock\nIlrkrptx\nHjeqxvq\nKqxmtwmsf\nIze\nDlpr\nK\nMnmg\nSmzyx\nMmersfs\nMpmb\nWmgu\nCerelmhoxi\nXrdgmmtcpq\nXj\nCeh\nAknkwdkq\nLqo\nVej\nT\nNsdtpdg\nSr\nPibdq\nZaecfyqka\nVvvocqi\nA\nDylv\nOfccguo\nNevr\nBmo\nRlijc\nF"
},
{
"input": "2\nA captain\nB man",
"output": "B\nA"
},
{
"input": "2\nB woman\nA captain",
"output": "B\nA"
},
{
"input": "2\nX child\nY captain",
"output": "X\nY"
},
{
"input": "2\nX captain\nY rat",
"output": "Y\nX"
},
{
"input": "6\nAbdelrhman captain\nAlice woman\nCharlie man\nTeddy rat\nBob child\nJulia woman",
"output": "Teddy\nAlice\nBob\nJulia\nCharlie\nAbdelrhman"
},
{
"input": "4\nAratos captain\nMind child\nAman woman\nCarrot man",
"output": "Mind\nAman\nCarrot\nAratos"
}
] | 1,699,681,780
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 26
| 216
| 1,843,200
|
from functools import cmp_to_key
class Info:
def __init__(self, _id: int, _name: str, _status: str):
self.id = _id
self.name = _name
self.status = _status
def __str__(self):
return f'id: {self.id}, name: {self.name}, status: {self.status}'
def customCompare(obj1:Info, obj2:Info) -> int:
if obj1.status != "rat" and obj2.status == "rat":
return 1
if obj1.status == "rat" and obj2.status != "rat":
return -1
if (obj1.status != "woman" and obj1.status != "child") and (obj2.status == "woman" or obj2.status == "child"):
return 1
if (obj1.status == "woman" or obj1.status == "child") and (obj2.status != "woman" and obj2.status != "child"):
return -1
if obj1.status != "man" and obj2.status == "man":
return 1
if obj1.status == "man" and obj2.status != "man":
return -1
return obj1.id - obj2.id
def output(cont:list):
for obj in cont:
print(obj)
n = int(input())
cont = []
for i in range(n):
_name, _status = input().split()
obj = Info(i + 1, _name, _status)
cont.append(obj)
cont.sort(key=cmp_to_key(customCompare))
for obj in cont:
print(obj.name)
|
Title: Sinking Ship
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
The ship crashed into a reef and is sinking. Now the entire crew must be evacuated. All *n* crew members have already lined up in a row (for convenience let's label them all from left to right with positive integers from 1 to *n*) and await further instructions. However, one should evacuate the crew properly, in a strict order. Specifically:
The first crew members to leave the ship are rats. Then women and children (both groups have the same priority) leave the ship. After that all men are evacuated from the ship. The captain leaves the sinking ship last.
If we cannot determine exactly who should leave the ship first for any two members of the crew by the rules from the previous paragraph, then the one who stands to the left in the line leaves the ship first (or in other words, the one whose number in the line is less).
For each crew member we know his status as a crew member, and also his name. All crew members have different names. Determine the order in which to evacuate the crew.
Input Specification:
The first line contains an integer *n*, which is the number of people in the crew (1<=β€<=*n*<=β€<=100). Then follow *n* lines. The *i*-th of those lines contains two words β the name of the crew member who is *i*-th in line, and his status on the ship. The words are separated by exactly one space. There are no other spaces in the line. The names consist of Latin letters, the first letter is uppercase, the rest are lowercase. The length of any name is from 1 to 10 characters. The status can have the following values: rat for a rat, woman for a woman, child for a child, man for a man, captain for the captain. The crew contains exactly one captain.
Output Specification:
Print *n* lines. The *i*-th of them should contain the name of the crew member who must be the *i*-th one to leave the ship.
Demo Input:
['6\nJack captain\nAlice woman\nCharlie man\nTeddy rat\nBob child\nJulia woman\n']
Demo Output:
['Teddy\nAlice\nBob\nJulia\nCharlie\nJack\n']
Note:
none
|
```python
from functools import cmp_to_key
class Info:
def __init__(self, _id: int, _name: str, _status: str):
self.id = _id
self.name = _name
self.status = _status
def __str__(self):
return f'id: {self.id}, name: {self.name}, status: {self.status}'
def customCompare(obj1:Info, obj2:Info) -> int:
if obj1.status != "rat" and obj2.status == "rat":
return 1
if obj1.status == "rat" and obj2.status != "rat":
return -1
if (obj1.status != "woman" and obj1.status != "child") and (obj2.status == "woman" or obj2.status == "child"):
return 1
if (obj1.status == "woman" or obj1.status == "child") and (obj2.status != "woman" and obj2.status != "child"):
return -1
if obj1.status != "man" and obj2.status == "man":
return 1
if obj1.status == "man" and obj2.status != "man":
return -1
return obj1.id - obj2.id
def output(cont:list):
for obj in cont:
print(obj)
n = int(input())
cont = []
for i in range(n):
_name, _status = input().split()
obj = Info(i + 1, _name, _status)
cont.append(obj)
cont.sort(key=cmp_to_key(customCompare))
for obj in cont:
print(obj.name)
```
| 3.942567
|
638
|
A
|
Home Numbers
|
PROGRAMMING
| 1,100
|
[
"*special",
"constructive algorithms",
"math"
] | null | null |
The main street of Berland is a straight line with *n* houses built along it (*n* is an even number). The houses are located at both sides of the street. The houses with odd numbers are at one side of the street and are numbered from 1 to *n*<=-<=1 in the order from the beginning of the street to the end (in the picture: from left to right). The houses with even numbers are at the other side of the street and are numbered from 2 to *n* in the order from the end of the street to its beginning (in the picture: from right to left). The corresponding houses with even and odd numbers are strictly opposite each other, that is, house 1 is opposite house *n*, house 3 is opposite house *n*<=-<=2, house 5 is opposite house *n*<=-<=4 and so on.
Vasya needs to get to house number *a* as quickly as possible. He starts driving from the beginning of the street and drives his car to house *a*. To get from the beginning of the street to houses number 1 and *n*, he spends exactly 1 second. He also spends exactly one second to drive the distance between two neighbouring houses. Vasya can park at any side of the road, so the distance between the beginning of the street at the houses that stand opposite one another should be considered the same.
Your task is: find the minimum time Vasya needs to reach house *a*.
|
The first line of the input contains two integers, *n* and *a* (1<=β€<=*a*<=β€<=*n*<=β€<=100<=000)Β β the number of houses on the street and the number of the house that Vasya needs to reach, correspondingly. It is guaranteed that number *n* is even.
|
Print a single integer β the minimum time Vasya needs to get from the beginning of the street to house *a*.
|
[
"4 2\n",
"8 5\n"
] |
[
"2\n",
"3\n"
] |
In the first sample there are only four houses on the street, two houses at each side. House 2 will be the last at Vasya's right.
The second sample corresponds to picture with *n*β=β8. House 5 is the one before last at Vasya's left.
| 500
|
[
{
"input": "4 2",
"output": "2"
},
{
"input": "8 5",
"output": "3"
},
{
"input": "2 1",
"output": "1"
},
{
"input": "2 2",
"output": "1"
},
{
"input": "10 1",
"output": "1"
},
{
"input": "10 10",
"output": "1"
},
{
"input": "100000 100000",
"output": "1"
},
{
"input": "100000 2",
"output": "50000"
},
{
"input": "100000 3",
"output": "2"
},
{
"input": "100000 99999",
"output": "50000"
},
{
"input": "100 100",
"output": "1"
},
{
"input": "3000 34",
"output": "1484"
},
{
"input": "2000 1",
"output": "1"
},
{
"input": "100000 1",
"output": "1"
},
{
"input": "24842 1038",
"output": "11903"
},
{
"input": "1628 274",
"output": "678"
},
{
"input": "16186 337",
"output": "169"
},
{
"input": "24562 2009",
"output": "1005"
},
{
"input": "9456 3443",
"output": "1722"
},
{
"input": "5610 332",
"output": "2640"
},
{
"input": "1764 1288",
"output": "239"
},
{
"input": "28588 13902",
"output": "7344"
},
{
"input": "92480 43074",
"output": "24704"
},
{
"input": "40022 26492",
"output": "6766"
},
{
"input": "85766 64050",
"output": "10859"
},
{
"input": "67808 61809",
"output": "30905"
},
{
"input": "80124 68695",
"output": "34348"
},
{
"input": "95522 91716",
"output": "1904"
},
{
"input": "7752 2915",
"output": "1458"
},
{
"input": "5094 5058",
"output": "19"
},
{
"input": "6144 4792",
"output": "677"
},
{
"input": "34334 20793",
"output": "10397"
},
{
"input": "23538 10243",
"output": "5122"
},
{
"input": "9328 7933",
"output": "3967"
},
{
"input": "11110 9885",
"output": "4943"
},
{
"input": "26096 2778",
"output": "11660"
},
{
"input": "75062 5323",
"output": "2662"
},
{
"input": "94790 7722",
"output": "43535"
},
{
"input": "90616 32240",
"output": "29189"
},
{
"input": "96998 8992",
"output": "44004"
},
{
"input": "95130 19219",
"output": "9610"
},
{
"input": "92586 8812",
"output": "41888"
},
{
"input": "3266 3044",
"output": "112"
},
{
"input": "5026 4697",
"output": "2349"
},
{
"input": "3044 2904",
"output": "71"
},
{
"input": "6022 5396",
"output": "314"
},
{
"input": "31270 25522",
"output": "2875"
},
{
"input": "82156 75519",
"output": "37760"
},
{
"input": "34614 27913",
"output": "13957"
},
{
"input": "88024 61143",
"output": "30572"
},
{
"input": "91870 55672",
"output": "18100"
},
{
"input": "95718 4868",
"output": "45426"
},
{
"input": "99564 358",
"output": "49604"
},
{
"input": "89266 13047",
"output": "6524"
},
{
"input": "90904 16455",
"output": "8228"
},
{
"input": "94750 13761",
"output": "6881"
},
{
"input": "100000 23458",
"output": "38272"
},
{
"input": "100000 23457",
"output": "11729"
},
{
"input": "59140 24272",
"output": "17435"
},
{
"input": "9860 8516",
"output": "673"
},
{
"input": "25988 2733",
"output": "1367"
},
{
"input": "9412 5309",
"output": "2655"
},
{
"input": "25540 23601",
"output": "11801"
},
{
"input": "76260 6050",
"output": "35106"
},
{
"input": "92388 39118",
"output": "26636"
},
{
"input": "8516 5495",
"output": "2748"
},
{
"input": "91940 37847",
"output": "18924"
},
{
"input": "30518 286",
"output": "15117"
},
{
"input": "46646 19345",
"output": "9673"
}
] | 1,458,487,878
| 12,678
|
PyPy 3
|
OK
|
TESTS
| 69
| 109
| 23,244,800
|
n, a = [int(x) for x in input().split()]
ans = 1
if a % 2 == 0:
while n != a:
n -= 2
ans += 1
else:
t = 1
while t != a:
t += 2
ans += 1
print(ans)
|
Title: Home Numbers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The main street of Berland is a straight line with *n* houses built along it (*n* is an even number). The houses are located at both sides of the street. The houses with odd numbers are at one side of the street and are numbered from 1 to *n*<=-<=1 in the order from the beginning of the street to the end (in the picture: from left to right). The houses with even numbers are at the other side of the street and are numbered from 2 to *n* in the order from the end of the street to its beginning (in the picture: from right to left). The corresponding houses with even and odd numbers are strictly opposite each other, that is, house 1 is opposite house *n*, house 3 is opposite house *n*<=-<=2, house 5 is opposite house *n*<=-<=4 and so on.
Vasya needs to get to house number *a* as quickly as possible. He starts driving from the beginning of the street and drives his car to house *a*. To get from the beginning of the street to houses number 1 and *n*, he spends exactly 1 second. He also spends exactly one second to drive the distance between two neighbouring houses. Vasya can park at any side of the road, so the distance between the beginning of the street at the houses that stand opposite one another should be considered the same.
Your task is: find the minimum time Vasya needs to reach house *a*.
Input Specification:
The first line of the input contains two integers, *n* and *a* (1<=β€<=*a*<=β€<=*n*<=β€<=100<=000)Β β the number of houses on the street and the number of the house that Vasya needs to reach, correspondingly. It is guaranteed that number *n* is even.
Output Specification:
Print a single integer β the minimum time Vasya needs to get from the beginning of the street to house *a*.
Demo Input:
['4 2\n', '8 5\n']
Demo Output:
['2\n', '3\n']
Note:
In the first sample there are only four houses on the street, two houses at each side. House 2 will be the last at Vasya's right.
The second sample corresponds to picture with *n*β=β8. House 5 is the one before last at Vasya's left.
|
```python
n, a = [int(x) for x in input().split()]
ans = 1
if a % 2 == 0:
while n != a:
n -= 2
ans += 1
else:
t = 1
while t != a:
t += 2
ans += 1
print(ans)
```
| 3
|
|
581
|
A
|
Vasya the Hipster
|
PROGRAMMING
| 800
|
[
"implementation",
"math"
] | null | null |
One day Vasya the Hipster decided to count how many socks he had. It turned out that he had *a* red socks and *b* blue socks.
According to the latest fashion, hipsters should wear the socks of different colors: a red one on the left foot, a blue one on the right foot.
Every day Vasya puts on new socks in the morning and throws them away before going to bed as he doesn't want to wash them.
Vasya wonders, what is the maximum number of days when he can dress fashionable and wear different socks, and after that, for how many days he can then wear the same socks until he either runs out of socks or cannot make a single pair from the socks he's got.
Can you help him?
|
The single line of the input contains two positive integers *a* and *b* (1<=β€<=*a*,<=*b*<=β€<=100) β the number of red and blue socks that Vasya's got.
|
Print two space-separated integers β the maximum number of days when Vasya can wear different socks and the number of days when he can wear the same socks until he either runs out of socks or cannot make a single pair from the socks he's got.
Keep in mind that at the end of the day Vasya throws away the socks that he's been wearing on that day.
|
[
"3 1\n",
"2 3\n",
"7 3\n"
] |
[
"1 1\n",
"2 0\n",
"3 2\n"
] |
In the first sample Vasya can first put on one pair of different socks, after that he has two red socks left to wear on the second day.
| 500
|
[
{
"input": "3 1",
"output": "1 1"
},
{
"input": "2 3",
"output": "2 0"
},
{
"input": "7 3",
"output": "3 2"
},
{
"input": "100 100",
"output": "100 0"
},
{
"input": "4 10",
"output": "4 3"
},
{
"input": "6 10",
"output": "6 2"
},
{
"input": "6 11",
"output": "6 2"
},
{
"input": "10 40",
"output": "10 15"
},
{
"input": "11 56",
"output": "11 22"
},
{
"input": "34 30",
"output": "30 2"
},
{
"input": "33 33",
"output": "33 0"
},
{
"input": "100 45",
"output": "45 27"
},
{
"input": "100 23",
"output": "23 38"
},
{
"input": "45 12",
"output": "12 16"
},
{
"input": "1 1",
"output": "1 0"
},
{
"input": "1 100",
"output": "1 49"
},
{
"input": "100 1",
"output": "1 49"
},
{
"input": "68 59",
"output": "59 4"
},
{
"input": "45 99",
"output": "45 27"
},
{
"input": "99 100",
"output": "99 0"
},
{
"input": "100 98",
"output": "98 1"
},
{
"input": "59 12",
"output": "12 23"
},
{
"input": "86 4",
"output": "4 41"
},
{
"input": "68 21",
"output": "21 23"
},
{
"input": "100 11",
"output": "11 44"
},
{
"input": "100 10",
"output": "10 45"
},
{
"input": "15 45",
"output": "15 15"
},
{
"input": "11 32",
"output": "11 10"
},
{
"input": "34 96",
"output": "34 31"
},
{
"input": "89 89",
"output": "89 0"
}
] | 1,687,523,565
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 30
| 46
| 0
|
socks = list(map(int, input().split(' ')))
first = 0
second = 0
while socks[0] != 0 and socks[1] != 0:
socks[0] -= 1
socks[1] -= 1
first += 1
if socks[0] == 0: socks = socks[1]
elif socks[1] == 0: socks = socks[0]
while socks != 1 and socks != 0:
socks -= 2
second += 1
print(first, second)
|
Title: Vasya the Hipster
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Vasya the Hipster decided to count how many socks he had. It turned out that he had *a* red socks and *b* blue socks.
According to the latest fashion, hipsters should wear the socks of different colors: a red one on the left foot, a blue one on the right foot.
Every day Vasya puts on new socks in the morning and throws them away before going to bed as he doesn't want to wash them.
Vasya wonders, what is the maximum number of days when he can dress fashionable and wear different socks, and after that, for how many days he can then wear the same socks until he either runs out of socks or cannot make a single pair from the socks he's got.
Can you help him?
Input Specification:
The single line of the input contains two positive integers *a* and *b* (1<=β€<=*a*,<=*b*<=β€<=100) β the number of red and blue socks that Vasya's got.
Output Specification:
Print two space-separated integers β the maximum number of days when Vasya can wear different socks and the number of days when he can wear the same socks until he either runs out of socks or cannot make a single pair from the socks he's got.
Keep in mind that at the end of the day Vasya throws away the socks that he's been wearing on that day.
Demo Input:
['3 1\n', '2 3\n', '7 3\n']
Demo Output:
['1 1\n', '2 0\n', '3 2\n']
Note:
In the first sample Vasya can first put on one pair of different socks, after that he has two red socks left to wear on the second day.
|
```python
socks = list(map(int, input().split(' ')))
first = 0
second = 0
while socks[0] != 0 and socks[1] != 0:
socks[0] -= 1
socks[1] -= 1
first += 1
if socks[0] == 0: socks = socks[1]
elif socks[1] == 0: socks = socks[0]
while socks != 1 and socks != 0:
socks -= 2
second += 1
print(first, second)
```
| 3
|
|
888
|
A
|
Local Extrema
|
PROGRAMMING
| 800
|
[
"brute force",
"implementation"
] | null | null |
You are given an array *a*. Some element of this array *a**i* is a local minimum iff it is strictly less than both of its neighbours (that is, *a**i*<=<<=*a**i*<=-<=1 and *a**i*<=<<=*a**i*<=+<=1). Also the element can be called local maximum iff it is strictly greater than its neighbours (that is, *a**i*<=><=*a**i*<=-<=1 and *a**i*<=><=*a**i*<=+<=1). Since *a*1 and *a**n* have only one neighbour each, they are neither local minima nor local maxima.
An element is called a local extremum iff it is either local maximum or local minimum. Your task is to calculate the number of local extrema in the given array.
|
The first line contains one integer *n* (1<=β€<=*n*<=β€<=1000) β the number of elements in array *a*.
The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=β€<=*a**i*<=β€<=1000) β the elements of array *a*.
|
Print the number of local extrema in the given array.
|
[
"3\n1 2 3\n",
"4\n1 5 2 5\n"
] |
[
"0\n",
"2\n"
] |
none
| 0
|
[
{
"input": "3\n1 2 3",
"output": "0"
},
{
"input": "4\n1 5 2 5",
"output": "2"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "1\n548",
"output": "0"
},
{
"input": "2\n1 1",
"output": "0"
},
{
"input": "3\n3 2 3",
"output": "1"
},
{
"input": "2\n1 1000",
"output": "0"
},
{
"input": "3\n1 2 1",
"output": "1"
}
] | 1,656,443,702
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 14
| 46
| 0
|
a = int(input())
local = list(map(int, input().split()))
ans = 0
for i in range(1, a - 1):
if local[i] > local[i - 1] and local[i] > local[i + 1]:
ans += 1
if local[i] < local[i - 1] and local[i] < local[i + 1]:
ans += 1
print(ans)
|
Title: Local Extrema
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given an array *a*. Some element of this array *a**i* is a local minimum iff it is strictly less than both of its neighbours (that is, *a**i*<=<<=*a**i*<=-<=1 and *a**i*<=<<=*a**i*<=+<=1). Also the element can be called local maximum iff it is strictly greater than its neighbours (that is, *a**i*<=><=*a**i*<=-<=1 and *a**i*<=><=*a**i*<=+<=1). Since *a*1 and *a**n* have only one neighbour each, they are neither local minima nor local maxima.
An element is called a local extremum iff it is either local maximum or local minimum. Your task is to calculate the number of local extrema in the given array.
Input Specification:
The first line contains one integer *n* (1<=β€<=*n*<=β€<=1000) β the number of elements in array *a*.
The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=β€<=*a**i*<=β€<=1000) β the elements of array *a*.
Output Specification:
Print the number of local extrema in the given array.
Demo Input:
['3\n1 2 3\n', '4\n1 5 2 5\n']
Demo Output:
['0\n', '2\n']
Note:
none
|
```python
a = int(input())
local = list(map(int, input().split()))
ans = 0
for i in range(1, a - 1):
if local[i] > local[i - 1] and local[i] > local[i + 1]:
ans += 1
if local[i] < local[i - 1] and local[i] < local[i + 1]:
ans += 1
print(ans)
```
| 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<=β€<=*n*<=β€<=100,<=1<=β€<=*a*<=β€<=*n*,<=<=-<=100<=β€<=*b*<=β€<=100)Β β 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<=β€<=*k*<=β€<=*n*)Β β 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"
}
] | 1,503,859,206
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 50
| 62
| 0
|
n,a,b=map(int,input().split(' '))
z=(((a+b)%n)+n)%n
if(z==0):
print(n)
else:
print(z)
|
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<=β€<=*n*<=β€<=100,<=1<=β€<=*a*<=β€<=*n*,<=<=-<=100<=β€<=*b*<=β€<=100)Β β 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<=β€<=*k*<=β€<=*n*)Β β 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(' '))
z=(((a+b)%n)+n)%n
if(z==0):
print(n)
else:
print(z)
```
| 3
|
|
732
|
A
|
Buy a Shovel
|
PROGRAMMING
| 800
|
[
"brute force",
"constructive algorithms",
"implementation",
"math"
] | null | null |
Polycarp urgently needs a shovel! He comes to the shop and chooses an appropriate one. The shovel that Policarp chooses is sold for *k* burles. Assume that there is an unlimited number of such shovels in the shop.
In his pocket Polycarp has an unlimited number of "10-burle coins" and exactly one coin of *r* burles (1<=β€<=*r*<=β€<=9).
What is the minimum number of shovels Polycarp has to buy so that he can pay for the purchase without any change? It is obvious that he can pay for 10 shovels without any change (by paying the requied amount of 10-burle coins and not using the coin of *r* burles). But perhaps he can buy fewer shovels and pay without any change. Note that Polycarp should buy at least one shovel.
|
The single line of input contains two integers *k* and *r* (1<=β€<=*k*<=β€<=1000, 1<=β€<=*r*<=β€<=9)Β β the price of one shovel and the denomination of the coin in Polycarp's pocket that is different from "10-burle coins".
Remember that he has an unlimited number of coins in the denomination of 10, that is, Polycarp has enough money to buy any number of shovels.
|
Print the required minimum number of shovels Polycarp has to buy so that he can pay for them without any change.
|
[
"117 3\n",
"237 7\n",
"15 2\n"
] |
[
"9\n",
"1\n",
"2\n"
] |
In the first example Polycarp can buy 9 shovels and pay 9Β·117β=β1053 burles. Indeed, he can pay this sum by using 10-burle coins and one 3-burle coin. He can't buy fewer shovels without any change.
In the second example it is enough for Polycarp to buy one shovel.
In the third example Polycarp should buy two shovels and pay 2Β·15β=β30 burles. It is obvious that he can pay this sum without any change.
| 500
|
[
{
"input": "117 3",
"output": "9"
},
{
"input": "237 7",
"output": "1"
},
{
"input": "15 2",
"output": "2"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "1 9",
"output": "9"
},
{
"input": "1000 3",
"output": "1"
},
{
"input": "1000 1",
"output": "1"
},
{
"input": "1000 9",
"output": "1"
},
{
"input": "1 2",
"output": "2"
},
{
"input": "999 9",
"output": "1"
},
{
"input": "999 8",
"output": "2"
},
{
"input": "105 6",
"output": "2"
},
{
"input": "403 9",
"output": "3"
},
{
"input": "546 4",
"output": "4"
},
{
"input": "228 9",
"output": "5"
},
{
"input": "57 2",
"output": "6"
},
{
"input": "437 9",
"output": "7"
},
{
"input": "997 6",
"output": "8"
},
{
"input": "109 1",
"output": "9"
},
{
"input": "998 9",
"output": "5"
},
{
"input": "4 2",
"output": "3"
},
{
"input": "9 3",
"output": "7"
},
{
"input": "8 2",
"output": "4"
},
{
"input": "1 3",
"output": "3"
},
{
"input": "1 4",
"output": "4"
},
{
"input": "1 5",
"output": "5"
},
{
"input": "1 6",
"output": "6"
},
{
"input": "1 7",
"output": "7"
},
{
"input": "1 8",
"output": "8"
},
{
"input": "100 3",
"output": "1"
},
{
"input": "1000 2",
"output": "1"
},
{
"input": "1000 4",
"output": "1"
},
{
"input": "1000 5",
"output": "1"
},
{
"input": "1000 6",
"output": "1"
},
{
"input": "1000 7",
"output": "1"
},
{
"input": "1000 8",
"output": "1"
},
{
"input": "23 4",
"output": "8"
},
{
"input": "33 1",
"output": "7"
},
{
"input": "33 2",
"output": "4"
},
{
"input": "666 5",
"output": "5"
},
{
"input": "2 3",
"output": "5"
},
{
"input": "5 5",
"output": "1"
},
{
"input": "3 6",
"output": "2"
},
{
"input": "12 4",
"output": "2"
},
{
"input": "15 5",
"output": "1"
},
{
"input": "2 5",
"output": "5"
},
{
"input": "25 5",
"output": "1"
},
{
"input": "2 9",
"output": "5"
},
{
"input": "6 7",
"output": "5"
},
{
"input": "8 9",
"output": "5"
},
{
"input": "2 7",
"output": "5"
},
{
"input": "4 7",
"output": "5"
},
{
"input": "2 1",
"output": "5"
},
{
"input": "261 1",
"output": "1"
}
] | 1,691,372,387
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 54
| 62
| 0
|
k,r=map(int,input().split())
for i in range(1,11):
if (k*i)%10 in [0,r]:
print(i)
quit()
|
Title: Buy a Shovel
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp urgently needs a shovel! He comes to the shop and chooses an appropriate one. The shovel that Policarp chooses is sold for *k* burles. Assume that there is an unlimited number of such shovels in the shop.
In his pocket Polycarp has an unlimited number of "10-burle coins" and exactly one coin of *r* burles (1<=β€<=*r*<=β€<=9).
What is the minimum number of shovels Polycarp has to buy so that he can pay for the purchase without any change? It is obvious that he can pay for 10 shovels without any change (by paying the requied amount of 10-burle coins and not using the coin of *r* burles). But perhaps he can buy fewer shovels and pay without any change. Note that Polycarp should buy at least one shovel.
Input Specification:
The single line of input contains two integers *k* and *r* (1<=β€<=*k*<=β€<=1000, 1<=β€<=*r*<=β€<=9)Β β the price of one shovel and the denomination of the coin in Polycarp's pocket that is different from "10-burle coins".
Remember that he has an unlimited number of coins in the denomination of 10, that is, Polycarp has enough money to buy any number of shovels.
Output Specification:
Print the required minimum number of shovels Polycarp has to buy so that he can pay for them without any change.
Demo Input:
['117 3\n', '237 7\n', '15 2\n']
Demo Output:
['9\n', '1\n', '2\n']
Note:
In the first example Polycarp can buy 9 shovels and pay 9Β·117β=β1053 burles. Indeed, he can pay this sum by using 10-burle coins and one 3-burle coin. He can't buy fewer shovels without any change.
In the second example it is enough for Polycarp to buy one shovel.
In the third example Polycarp should buy two shovels and pay 2Β·15β=β30 burles. It is obvious that he can pay this sum without any change.
|
```python
k,r=map(int,input().split())
for i in range(1,11):
if (k*i)%10 in [0,r]:
print(i)
quit()
```
| 3
|
|
732
|
A
|
Buy a Shovel
|
PROGRAMMING
| 800
|
[
"brute force",
"constructive algorithms",
"implementation",
"math"
] | null | null |
Polycarp urgently needs a shovel! He comes to the shop and chooses an appropriate one. The shovel that Policarp chooses is sold for *k* burles. Assume that there is an unlimited number of such shovels in the shop.
In his pocket Polycarp has an unlimited number of "10-burle coins" and exactly one coin of *r* burles (1<=β€<=*r*<=β€<=9).
What is the minimum number of shovels Polycarp has to buy so that he can pay for the purchase without any change? It is obvious that he can pay for 10 shovels without any change (by paying the requied amount of 10-burle coins and not using the coin of *r* burles). But perhaps he can buy fewer shovels and pay without any change. Note that Polycarp should buy at least one shovel.
|
The single line of input contains two integers *k* and *r* (1<=β€<=*k*<=β€<=1000, 1<=β€<=*r*<=β€<=9)Β β the price of one shovel and the denomination of the coin in Polycarp's pocket that is different from "10-burle coins".
Remember that he has an unlimited number of coins in the denomination of 10, that is, Polycarp has enough money to buy any number of shovels.
|
Print the required minimum number of shovels Polycarp has to buy so that he can pay for them without any change.
|
[
"117 3\n",
"237 7\n",
"15 2\n"
] |
[
"9\n",
"1\n",
"2\n"
] |
In the first example Polycarp can buy 9 shovels and pay 9Β·117β=β1053 burles. Indeed, he can pay this sum by using 10-burle coins and one 3-burle coin. He can't buy fewer shovels without any change.
In the second example it is enough for Polycarp to buy one shovel.
In the third example Polycarp should buy two shovels and pay 2Β·15β=β30 burles. It is obvious that he can pay this sum without any change.
| 500
|
[
{
"input": "117 3",
"output": "9"
},
{
"input": "237 7",
"output": "1"
},
{
"input": "15 2",
"output": "2"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "1 9",
"output": "9"
},
{
"input": "1000 3",
"output": "1"
},
{
"input": "1000 1",
"output": "1"
},
{
"input": "1000 9",
"output": "1"
},
{
"input": "1 2",
"output": "2"
},
{
"input": "999 9",
"output": "1"
},
{
"input": "999 8",
"output": "2"
},
{
"input": "105 6",
"output": "2"
},
{
"input": "403 9",
"output": "3"
},
{
"input": "546 4",
"output": "4"
},
{
"input": "228 9",
"output": "5"
},
{
"input": "57 2",
"output": "6"
},
{
"input": "437 9",
"output": "7"
},
{
"input": "997 6",
"output": "8"
},
{
"input": "109 1",
"output": "9"
},
{
"input": "998 9",
"output": "5"
},
{
"input": "4 2",
"output": "3"
},
{
"input": "9 3",
"output": "7"
},
{
"input": "8 2",
"output": "4"
},
{
"input": "1 3",
"output": "3"
},
{
"input": "1 4",
"output": "4"
},
{
"input": "1 5",
"output": "5"
},
{
"input": "1 6",
"output": "6"
},
{
"input": "1 7",
"output": "7"
},
{
"input": "1 8",
"output": "8"
},
{
"input": "100 3",
"output": "1"
},
{
"input": "1000 2",
"output": "1"
},
{
"input": "1000 4",
"output": "1"
},
{
"input": "1000 5",
"output": "1"
},
{
"input": "1000 6",
"output": "1"
},
{
"input": "1000 7",
"output": "1"
},
{
"input": "1000 8",
"output": "1"
},
{
"input": "23 4",
"output": "8"
},
{
"input": "33 1",
"output": "7"
},
{
"input": "33 2",
"output": "4"
},
{
"input": "666 5",
"output": "5"
},
{
"input": "2 3",
"output": "5"
},
{
"input": "5 5",
"output": "1"
},
{
"input": "3 6",
"output": "2"
},
{
"input": "12 4",
"output": "2"
},
{
"input": "15 5",
"output": "1"
},
{
"input": "2 5",
"output": "5"
},
{
"input": "25 5",
"output": "1"
},
{
"input": "2 9",
"output": "5"
},
{
"input": "6 7",
"output": "5"
},
{
"input": "8 9",
"output": "5"
},
{
"input": "2 7",
"output": "5"
},
{
"input": "4 7",
"output": "5"
},
{
"input": "2 1",
"output": "5"
},
{
"input": "261 1",
"output": "1"
}
] | 1,684,145,748
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 54
| 77
| 0
|
price, coin = map(int, input().split())
for i in range(1, 11):
if (price * i) % 10 == 0 or coin == (price * i) % 10:
print(i)
break
|
Title: Buy a Shovel
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp urgently needs a shovel! He comes to the shop and chooses an appropriate one. The shovel that Policarp chooses is sold for *k* burles. Assume that there is an unlimited number of such shovels in the shop.
In his pocket Polycarp has an unlimited number of "10-burle coins" and exactly one coin of *r* burles (1<=β€<=*r*<=β€<=9).
What is the minimum number of shovels Polycarp has to buy so that he can pay for the purchase without any change? It is obvious that he can pay for 10 shovels without any change (by paying the requied amount of 10-burle coins and not using the coin of *r* burles). But perhaps he can buy fewer shovels and pay without any change. Note that Polycarp should buy at least one shovel.
Input Specification:
The single line of input contains two integers *k* and *r* (1<=β€<=*k*<=β€<=1000, 1<=β€<=*r*<=β€<=9)Β β the price of one shovel and the denomination of the coin in Polycarp's pocket that is different from "10-burle coins".
Remember that he has an unlimited number of coins in the denomination of 10, that is, Polycarp has enough money to buy any number of shovels.
Output Specification:
Print the required minimum number of shovels Polycarp has to buy so that he can pay for them without any change.
Demo Input:
['117 3\n', '237 7\n', '15 2\n']
Demo Output:
['9\n', '1\n', '2\n']
Note:
In the first example Polycarp can buy 9 shovels and pay 9Β·117β=β1053 burles. Indeed, he can pay this sum by using 10-burle coins and one 3-burle coin. He can't buy fewer shovels without any change.
In the second example it is enough for Polycarp to buy one shovel.
In the third example Polycarp should buy two shovels and pay 2Β·15β=β30 burles. It is obvious that he can pay this sum without any change.
|
```python
price, coin = map(int, input().split())
for i in range(1, 11):
if (price * i) % 10 == 0 or coin == (price * i) % 10:
print(i)
break
```
| 3
|
|
59
|
A
|
Word
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] |
A. Word
|
2
|
256
|
Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP β with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word.
|
The first line contains a word *s* β it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100.
|
Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one.
|
[
"HoUse\n",
"ViP\n",
"maTRIx\n"
] |
[
"house\n",
"VIP\n",
"matrix\n"
] |
none
| 500
|
[
{
"input": "HoUse",
"output": "house"
},
{
"input": "ViP",
"output": "VIP"
},
{
"input": "maTRIx",
"output": "matrix"
},
{
"input": "BNHWpnpawg",
"output": "bnhwpnpawg"
},
{
"input": "VTYGP",
"output": "VTYGP"
},
{
"input": "CHNenu",
"output": "chnenu"
},
{
"input": "ERPZGrodyu",
"output": "erpzgrodyu"
},
{
"input": "KSXBXWpebh",
"output": "KSXBXWPEBH"
},
{
"input": "qvxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaiv",
"output": "qvxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaiv"
},
{
"input": "Amnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfd",
"output": "amnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfd"
},
{
"input": "ISAGFJFARYFBLOPQDSHWGMCNKMFTLVFUGNJEWGWNBLXUIATXEkqiettmmjgydwcpafqrppdsrrrtguinqbgmzzfqwonkpgpcwenv",
"output": "isagfjfaryfblopqdshwgmcnkmftlvfugnjewgwnblxuiatxekqiettmmjgydwcpafqrppdsrrrtguinqbgmzzfqwonkpgpcwenv"
},
{
"input": "XHRPXZEGHSOCJPICUIXSKFUZUPYTSGJSDIYBCMNMNBPNDBXLXBzhbfnqvwcffvrdhtickyqhupmcehlsyvncqmfhautvxudqdhgg",
"output": "xhrpxzeghsocjpicuixskfuzupytsgjsdiybcmnmnbpndbxlxbzhbfnqvwcffvrdhtickyqhupmcehlsyvncqmfhautvxudqdhgg"
},
{
"input": "RJIQZMJCIMSNDBOHBRAWIENODSALETAKGKPYUFGVEFGCBRENZGAdkcetqjljtmttlonpekcovdzebzdkzggwfsxhapmjkdbuceak",
"output": "RJIQZMJCIMSNDBOHBRAWIENODSALETAKGKPYUFGVEFGCBRENZGADKCETQJLJTMTTLONPEKCOVDZEBZDKZGGWFSXHAPMJKDBUCEAK"
},
{
"input": "DWLWOBHNMMGTFOLFAECKBRNNGLYLYDXTGTVRLMEESZOIUATZZZXUFUZDLSJXMEVRTESSFBWLNZZCLCQWEVNNUCXYVHNGNXHCBDFw",
"output": "DWLWOBHNMMGTFOLFAECKBRNNGLYLYDXTGTVRLMEESZOIUATZZZXUFUZDLSJXMEVRTESSFBWLNZZCLCQWEVNNUCXYVHNGNXHCBDFW"
},
{
"input": "NYCNHJWGBOCOTSPETKKHVWFGAQYNHOVJWJHCIEFOUQZXOYUIEQDZALFKTEHTVDBVJMEUBJUBCMNVPWGDPNCHQHZJRCHYRFPVIGUB",
"output": "NYCNHJWGBOCOTSPETKKHVWFGAQYNHOVJWJHCIEFOUQZXOYUIEQDZALFKTEHTVDBVJMEUBJUBCMNVPWGDPNCHQHZJRCHYRFPVIGUB"
},
{
"input": "igxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwge",
"output": "igxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwge"
},
{
"input": "Ykkekrsqolzryiwsmdlnbmfautxxxauoojrddvwklgnlyrfcvhorrzbmtcrvpaypqhcffdqhwziipyyskcmztjprjqvmzzqhqnw",
"output": "ykkekrsqolzryiwsmdlnbmfautxxxauoojrddvwklgnlyrfcvhorrzbmtcrvpaypqhcffdqhwziipyyskcmztjprjqvmzzqhqnw"
},
{
"input": "YQOMLKYAORUQQUCQZCDYMIVDHGWZFFRMUVTAWCHERFPMNRYRIkgqrciokgajamehmcxgerpudvsqyonjonsxgbnefftzmygncks",
"output": "yqomlkyaoruqqucqzcdymivdhgwzffrmuvtawcherfpmnryrikgqrciokgajamehmcxgerpudvsqyonjonsxgbnefftzmygncks"
},
{
"input": "CDOZDPBVVVHNBJVBYHEOXWFLJKRWJCAJMIFCOZWWYFKVWOGTVJcuusigdqfkumewjtdyitveeiaybwrhomrwmpdipjwiuxfnwuz",
"output": "CDOZDPBVVVHNBJVBYHEOXWFLJKRWJCAJMIFCOZWWYFKVWOGTVJCUUSIGDQFKUMEWJTDYITVEEIAYBWRHOMRWMPDIPJWIUXFNWUZ"
},
{
"input": "WHIUVEXHVOOIJIDVJVPQUBJMEVPMPDKQWJKFBZSGSKUXMIPPMJWuckzcpxosodcjaaakvlxpbiigsiauviilylnnqlyucziihqg",
"output": "WHIUVEXHVOOIJIDVJVPQUBJMEVPMPDKQWJKFBZSGSKUXMIPPMJWUCKZCPXOSODCJAAAKVLXPBIIGSIAUVIILYLNNQLYUCZIIHQG"
},
{
"input": "VGHUNFOXKETUYMZDJNGTAOIOANYXSGYNFOGOFFLDAWEUKYFOZXCJTCAFXZYLQZERYZLRSQXYQGAPCSUDPMEYTNCTTTMFAGVDWBO",
"output": "VGHUNFOXKETUYMZDJNGTAOIOANYXSGYNFOGOFFLDAWEUKYFOZXCJTCAFXZYLQZERYZLRSQXYQGAPCSUDPMEYTNCTTTMFAGVDWBO"
},
{
"input": "EDUANHCQDOCJHFONTTSWBUJSTTIXBIXMAIUFSGFLXWAYIURTVAVZPYQDLAWIGCLMPHXCEFCJPFAAHXVNGQUFNHADAIUAZIDMHDd",
"output": "EDUANHCQDOCJHFONTTSWBUJSTTIXBIXMAIUFSGFLXWAYIURTVAVZPYQDLAWIGCLMPHXCEFCJPFAAHXVNGQUFNHADAIUAZIDMHDD"
},
{
"input": "EBWOVSNLYTWWXrnovgugogtjahnmatomubebyxakas",
"output": "ebwovsnlytwwxrnovgugogtjahnmatomubebyxakas"
},
{
"input": "AORNNDKTRLRVGDPXJKXFTPFpopuzrlqumrxssskvbm",
"output": "AORNNDKTRLRVGDPXJKXFTPFPOPUZRLQUMRXSSSKVBM"
},
{
"input": "DBBZJCWQEVGYHEEUHJAWRHBDPsnplijbwtdibqrrheebfxqcvn",
"output": "dbbzjcwqevgyheeuhjawrhbdpsnplijbwtdibqrrheebfxqcvn"
},
{
"input": "t",
"output": "t"
},
{
"input": "N",
"output": "N"
},
{
"input": "kv",
"output": "kv"
},
{
"input": "Ur",
"output": "ur"
},
{
"input": "CN",
"output": "CN"
}
] | 1,696,065,925
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 30
| 92
| 0
|
s=input()
u=0
l=0
o=""
for i in range(0,len(s)):
a=s[i]
if(a.isupper()):
u=u+1
else:
l=l+1
if u>l:
o=s.upper()
elif u<l:
o=s.lower()
else:
o=s.lower()
print(o)
|
Title: Word
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP β with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word.
Input Specification:
The first line contains a word *s* β it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100.
Output Specification:
Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one.
Demo Input:
['HoUse\n', 'ViP\n', 'maTRIx\n']
Demo Output:
['house\n', 'VIP\n', 'matrix\n']
Note:
none
|
```python
s=input()
u=0
l=0
o=""
for i in range(0,len(s)):
a=s[i]
if(a.isupper()):
u=u+1
else:
l=l+1
if u>l:
o=s.upper()
elif u<l:
o=s.lower()
else:
o=s.lower()
print(o)
```
| 3.977
|
735
|
D
|
Taxes
|
PROGRAMMING
| 1,600
|
[
"math",
"number theory"
] | null | null |
Mr. Funt now lives in a country with a very specific tax laws. The total income of mr. Funt during this year is equal to *n* (*n*<=β₯<=2) burles and the amount of tax he has to pay is calculated as the maximum divisor of *n* (not equal to *n*, of course). For example, if *n*<==<=6 then Funt has to pay 3 burles, while for *n*<==<=25 he needs to pay 5 and if *n*<==<=2 he pays only 1 burle.
As mr. Funt is a very opportunistic person he wants to cheat a bit. In particular, he wants to split the initial *n* in several parts *n*1<=+<=*n*2<=+<=...<=+<=*n**k*<==<=*n* (here *k* is arbitrary, even *k*<==<=1 is allowed) and pay the taxes for each part separately. He can't make some part equal to 1 because it will reveal him. So, the condition *n**i*<=β₯<=2 should hold for all *i* from 1 to *k*.
Ostap Bender wonders, how many money Funt has to pay (i.e. minimal) if he chooses and optimal way to split *n* in parts.
|
The first line of the input contains a single integer *n* (2<=β€<=*n*<=β€<=2Β·109)Β β the total year income of mr. Funt.
|
Print one integerΒ β minimum possible number of burles that mr. Funt has to pay as a tax.
|
[
"4\n",
"27\n"
] |
[
"2\n",
"3\n"
] |
none
| 1,750
|
[
{
"input": "4",
"output": "2"
},
{
"input": "27",
"output": "3"
},
{
"input": "3",
"output": "1"
},
{
"input": "5",
"output": "1"
},
{
"input": "10",
"output": "2"
},
{
"input": "2000000000",
"output": "2"
},
{
"input": "26",
"output": "2"
},
{
"input": "7",
"output": "1"
},
{
"input": "2",
"output": "1"
},
{
"input": "11",
"output": "1"
},
{
"input": "1000000007",
"output": "1"
},
{
"input": "1000000009",
"output": "1"
},
{
"input": "1999999999",
"output": "3"
},
{
"input": "1000000011",
"output": "2"
},
{
"input": "101",
"output": "1"
},
{
"input": "103",
"output": "1"
},
{
"input": "1001",
"output": "3"
},
{
"input": "1003",
"output": "3"
},
{
"input": "10001",
"output": "3"
},
{
"input": "10003",
"output": "3"
},
{
"input": "129401294",
"output": "2"
},
{
"input": "234911024",
"output": "2"
},
{
"input": "192483501",
"output": "3"
},
{
"input": "1234567890",
"output": "2"
},
{
"input": "719241201",
"output": "3"
},
{
"input": "9",
"output": "2"
},
{
"input": "33",
"output": "2"
},
{
"input": "25",
"output": "2"
},
{
"input": "15",
"output": "2"
},
{
"input": "147",
"output": "3"
},
{
"input": "60119912",
"output": "2"
},
{
"input": "45",
"output": "2"
},
{
"input": "21",
"output": "2"
},
{
"input": "9975",
"output": "2"
},
{
"input": "17",
"output": "1"
},
{
"input": "99",
"output": "2"
},
{
"input": "49",
"output": "2"
},
{
"input": "243",
"output": "2"
},
{
"input": "43",
"output": "1"
},
{
"input": "39",
"output": "2"
},
{
"input": "6",
"output": "2"
},
{
"input": "8",
"output": "2"
},
{
"input": "12",
"output": "2"
},
{
"input": "13",
"output": "1"
},
{
"input": "14",
"output": "2"
},
{
"input": "16",
"output": "2"
},
{
"input": "18",
"output": "2"
},
{
"input": "19",
"output": "1"
},
{
"input": "20",
"output": "2"
},
{
"input": "22",
"output": "2"
},
{
"input": "23",
"output": "1"
},
{
"input": "24",
"output": "2"
},
{
"input": "962",
"output": "2"
},
{
"input": "29",
"output": "1"
},
{
"input": "55",
"output": "2"
},
{
"input": "125",
"output": "3"
},
{
"input": "1999999929",
"output": "2"
},
{
"input": "493",
"output": "2"
},
{
"input": "10000021",
"output": "2"
},
{
"input": "541",
"output": "1"
},
{
"input": "187",
"output": "3"
},
{
"input": "95",
"output": "3"
},
{
"input": "999991817",
"output": "3"
},
{
"input": "37998938",
"output": "2"
},
{
"input": "1847133842",
"output": "2"
},
{
"input": "1000000005",
"output": "3"
},
{
"input": "19828",
"output": "2"
},
{
"input": "998321704",
"output": "2"
},
{
"input": "370359",
"output": "3"
},
{
"input": "115",
"output": "2"
},
{
"input": "200000015",
"output": "3"
},
{
"input": "479001600",
"output": "2"
},
{
"input": "536870912",
"output": "2"
},
{
"input": "10759922",
"output": "2"
},
{
"input": "1999999927",
"output": "1"
},
{
"input": "123",
"output": "3"
},
{
"input": "200743933",
"output": "3"
},
{
"input": "949575615",
"output": "3"
},
{
"input": "99990001",
"output": "1"
},
{
"input": "715827883",
"output": "1"
},
{
"input": "5592406",
"output": "2"
},
{
"input": "8388609",
"output": "3"
},
{
"input": "1908903481",
"output": "3"
},
{
"input": "1076153021",
"output": "3"
},
{
"input": "344472101",
"output": "3"
}
] | 1,561,148,769
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 85
| 155
| 1,228,800
|
from math import sqrt
def prime(n):
for d in range(3, int(sqrt(n)) + 1, 2):
if n % d == 0:
return 0
return 1
n = int(input())
if n == 2:
print(1)
elif n % 2 == 0:
print(2)
elif prime(n):
print(1)
elif prime(n - 2):
print(2)
else:
print(3)
|
Title: Taxes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mr. Funt now lives in a country with a very specific tax laws. The total income of mr. Funt during this year is equal to *n* (*n*<=β₯<=2) burles and the amount of tax he has to pay is calculated as the maximum divisor of *n* (not equal to *n*, of course). For example, if *n*<==<=6 then Funt has to pay 3 burles, while for *n*<==<=25 he needs to pay 5 and if *n*<==<=2 he pays only 1 burle.
As mr. Funt is a very opportunistic person he wants to cheat a bit. In particular, he wants to split the initial *n* in several parts *n*1<=+<=*n*2<=+<=...<=+<=*n**k*<==<=*n* (here *k* is arbitrary, even *k*<==<=1 is allowed) and pay the taxes for each part separately. He can't make some part equal to 1 because it will reveal him. So, the condition *n**i*<=β₯<=2 should hold for all *i* from 1 to *k*.
Ostap Bender wonders, how many money Funt has to pay (i.e. minimal) if he chooses and optimal way to split *n* in parts.
Input Specification:
The first line of the input contains a single integer *n* (2<=β€<=*n*<=β€<=2Β·109)Β β the total year income of mr. Funt.
Output Specification:
Print one integerΒ β minimum possible number of burles that mr. Funt has to pay as a tax.
Demo Input:
['4\n', '27\n']
Demo Output:
['2\n', '3\n']
Note:
none
|
```python
from math import sqrt
def prime(n):
for d in range(3, int(sqrt(n)) + 1, 2):
if n % d == 0:
return 0
return 1
n = int(input())
if n == 2:
print(1)
elif n % 2 == 0:
print(2)
elif prime(n):
print(1)
elif prime(n - 2):
print(2)
else:
print(3)
```
| 3
|
|
233
|
A
|
Perfect Permutation
|
PROGRAMMING
| 800
|
[
"implementation",
"math"
] | null | null |
A permutation is a sequence of integers *p*1,<=*p*2,<=...,<=*p**n*, consisting of *n* distinct positive integers, each of them doesn't exceed *n*. Let's denote the *i*-th element of permutation *p* as *p**i*. We'll call number *n* the size of permutation *p*1,<=*p*2,<=...,<=*p**n*.
Nickolas adores permutations. He likes some permutations more than the others. He calls such permutations perfect. A perfect permutation is such permutation *p* that for any *i* (1<=β€<=*i*<=β€<=*n*) (*n* is the permutation size) the following equations hold *p**p**i*<==<=*i* and *p**i*<=β <=*i*. Nickolas asks you to print any perfect permutation of size *n* for the given *n*.
|
A single line contains a single integer *n* (1<=β€<=*n*<=β€<=100) β the permutation size.
|
If a perfect permutation of size *n* doesn't exist, print a single integer -1. Otherwise print *n* distinct integers from 1 to *n*, *p*1,<=*p*2,<=...,<=*p**n* β permutation *p*, that is perfect. Separate printed numbers by whitespaces.
|
[
"1\n",
"2\n",
"4\n"
] |
[
"-1\n",
"2 1 \n",
"2 1 4 3 \n"
] |
none
| 500
|
[
{
"input": "1",
"output": "-1"
},
{
"input": "2",
"output": "2 1 "
},
{
"input": "4",
"output": "2 1 4 3 "
},
{
"input": "3",
"output": "-1"
},
{
"input": "5",
"output": "-1"
},
{
"input": "6",
"output": "2 1 4 3 6 5 "
},
{
"input": "7",
"output": "-1"
},
{
"input": "20",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 "
},
{
"input": "8",
"output": "2 1 4 3 6 5 8 7 "
},
{
"input": "9",
"output": "-1"
},
{
"input": "10",
"output": "2 1 4 3 6 5 8 7 10 9 "
},
{
"input": "11",
"output": "-1"
},
{
"input": "21",
"output": "-1"
},
{
"input": "50",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 42 41 44 43 46 45 48 47 50 49 "
},
{
"input": "51",
"output": "-1"
},
{
"input": "52",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 42 41 44 43 46 45 48 47 50 49 52 51 "
},
{
"input": "84",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 42 41 44 43 46 45 48 47 50 49 52 51 54 53 56 55 58 57 60 59 62 61 64 63 66 65 68 67 70 69 72 71 74 73 76 75 78 77 80 79 82 81 84 83 "
},
{
"input": "86",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 42 41 44 43 46 45 48 47 50 49 52 51 54 53 56 55 58 57 60 59 62 61 64 63 66 65 68 67 70 69 72 71 74 73 76 75 78 77 80 79 82 81 84 83 86 85 "
},
{
"input": "100",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 42 41 44 43 46 45 48 47 50 49 52 51 54 53 56 55 58 57 60 59 62 61 64 63 66 65 68 67 70 69 72 71 74 73 76 75 78 77 80 79 82 81 84 83 86 85 88 87 90 89 92 91 94 93 96 95 98 97 100 99 "
},
{
"input": "98",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 42 41 44 43 46 45 48 47 50 49 52 51 54 53 56 55 58 57 60 59 62 61 64 63 66 65 68 67 70 69 72 71 74 73 76 75 78 77 80 79 82 81 84 83 86 85 88 87 90 89 92 91 94 93 96 95 98 97 "
},
{
"input": "96",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 42 41 44 43 46 45 48 47 50 49 52 51 54 53 56 55 58 57 60 59 62 61 64 63 66 65 68 67 70 69 72 71 74 73 76 75 78 77 80 79 82 81 84 83 86 85 88 87 90 89 92 91 94 93 96 95 "
},
{
"input": "33",
"output": "-1"
},
{
"input": "34",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 "
},
{
"input": "36",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 "
},
{
"input": "38",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 "
},
{
"input": "40",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 "
},
{
"input": "42",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 42 41 "
},
{
"input": "44",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 42 41 44 43 "
},
{
"input": "46",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 42 41 44 43 46 45 "
},
{
"input": "48",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 42 41 44 43 46 45 48 47 "
}
] | 1,649,937,084
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 30
| 92
| 0
|
# -*- coding: utf-8 -*-
"""
Created on Thu Apr 14 12:45:15 2022
@author: dehon
"""
n = int(input())
if(n%2 == 1):
print(-1)
else:
for i in range(1,n+1):
if(i%2 == 1):
print(i+1, end=' ')
else:
print(i-1, end=' ')
|
Title: Perfect Permutation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A permutation is a sequence of integers *p*1,<=*p*2,<=...,<=*p**n*, consisting of *n* distinct positive integers, each of them doesn't exceed *n*. Let's denote the *i*-th element of permutation *p* as *p**i*. We'll call number *n* the size of permutation *p*1,<=*p*2,<=...,<=*p**n*.
Nickolas adores permutations. He likes some permutations more than the others. He calls such permutations perfect. A perfect permutation is such permutation *p* that for any *i* (1<=β€<=*i*<=β€<=*n*) (*n* is the permutation size) the following equations hold *p**p**i*<==<=*i* and *p**i*<=β <=*i*. Nickolas asks you to print any perfect permutation of size *n* for the given *n*.
Input Specification:
A single line contains a single integer *n* (1<=β€<=*n*<=β€<=100) β the permutation size.
Output Specification:
If a perfect permutation of size *n* doesn't exist, print a single integer -1. Otherwise print *n* distinct integers from 1 to *n*, *p*1,<=*p*2,<=...,<=*p**n* β permutation *p*, that is perfect. Separate printed numbers by whitespaces.
Demo Input:
['1\n', '2\n', '4\n']
Demo Output:
['-1\n', '2 1 \n', '2 1 4 3 \n']
Note:
none
|
```python
# -*- coding: utf-8 -*-
"""
Created on Thu Apr 14 12:45:15 2022
@author: dehon
"""
n = int(input())
if(n%2 == 1):
print(-1)
else:
for i in range(1,n+1):
if(i%2 == 1):
print(i+1, end=' ')
else:
print(i-1, end=' ')
```
| 3
|
|
665
|
C
|
Simple Strings
|
PROGRAMMING
| 1,300
|
[
"dp",
"greedy",
"strings"
] | null | null |
zscoder loves simple strings! A string *t* is called simple if every pair of adjacent characters are distinct. For example ab, aba, zscoder are simple whereas aa, add are not simple.
zscoder is given a string *s*. He wants to change a minimum number of characters so that the string *s* becomes simple. Help him with this task!
|
The only line contains the string *s* (1<=β€<=|*s*|<=β€<=2Β·105) β the string given to zscoder. The string *s* consists of only lowercase English letters.
|
Print the simple string *s*' β the string *s* after the minimal number of changes. If there are multiple solutions, you may output any of them.
Note that the string *s*' should also consist of only lowercase English letters.
|
[
"aab\n",
"caaab\n",
"zscoder\n"
] |
[
"bab\n",
"cabab\n",
"zscoder\n"
] |
none
| 0
|
[
{
"input": "aab",
"output": "bab"
},
{
"input": "caaab",
"output": "cabab"
},
{
"input": "zscoder",
"output": "zscoder"
},
{
"input": "u",
"output": "u"
},
{
"input": "h",
"output": "h"
},
{
"input": "dtottttotd",
"output": "dtotataotd"
},
{
"input": "rxxxrrxrxxxxxrrrrrxxxxrrrrxrxxrxxrxrxrrrxrrxrrxrxxxrxrrxrrxrxrxxxxxrxxxxrrrxrxxrxxrxxxrrrrrxrrxrrxrr",
"output": "rxaxraxrxaxaxrararxaxararaxrxarxarxrxrarxraxraxrxaxrxraxraxrxrxaxaxrxaxararxrxarxarxaxrararxraxraxra"
},
{
"input": "aazz",
"output": "baza"
},
{
"input": "zz",
"output": "za"
},
{
"input": "gg",
"output": "ga"
},
{
"input": "qasdasd",
"output": "qasdasd"
},
{
"input": "aa",
"output": "ba"
},
{
"input": "ab",
"output": "ab"
},
{
"input": "zza",
"output": "zba"
},
{
"input": "g",
"output": "g"
},
{
"input": "nnop",
"output": "naop"
},
{
"input": "xx",
"output": "xa"
}
] | 1,664,936,705
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 27
| 109
| 15,769,600
|
def add(a, b):
for j in range(26):
c = chr(j + ord('a'))
if(c != a and c != b):
return c
s = list(input())
n = len(s)
for i in range(1, n - 1):
if(s[i] == s[i - 1]):
s[i] = add(s[i - 1], s[i + 1])
if(n > 1):
if(s[-1] == s[-2]):
for j in range(26):
c = chr(j + ord('a'))
if(c != s[-2]):
s[-1] = c
break
print(''.join(s))
|
Title: Simple Strings
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
zscoder loves simple strings! A string *t* is called simple if every pair of adjacent characters are distinct. For example ab, aba, zscoder are simple whereas aa, add are not simple.
zscoder is given a string *s*. He wants to change a minimum number of characters so that the string *s* becomes simple. Help him with this task!
Input Specification:
The only line contains the string *s* (1<=β€<=|*s*|<=β€<=2Β·105) β the string given to zscoder. The string *s* consists of only lowercase English letters.
Output Specification:
Print the simple string *s*' β the string *s* after the minimal number of changes. If there are multiple solutions, you may output any of them.
Note that the string *s*' should also consist of only lowercase English letters.
Demo Input:
['aab\n', 'caaab\n', 'zscoder\n']
Demo Output:
['bab\n', 'cabab\n', 'zscoder\n']
Note:
none
|
```python
def add(a, b):
for j in range(26):
c = chr(j + ord('a'))
if(c != a and c != b):
return c
s = list(input())
n = len(s)
for i in range(1, n - 1):
if(s[i] == s[i - 1]):
s[i] = add(s[i - 1], s[i + 1])
if(n > 1):
if(s[-1] == s[-2]):
for j in range(26):
c = chr(j + ord('a'))
if(c != s[-2]):
s[-1] = c
break
print(''.join(s))
```
| 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<=β€<=*x*<=β€<=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"
}
] | 1,674,300,241
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 44
| 46
| 0
|
inputStr = input()
tempRev = inputStr[::-1]
ok = False
if(inputStr == tempRev):
ok = True
else:
zerosAdded = 0
temp = list(inputStr)
for i in reversed(temp):
if(i != '0'):
break
else:
zerosAdded += 1
inputStr =inputStr.rjust(len(inputStr) + zerosAdded, "0")
tempRev = inputStr[::-1]
if(inputStr == tempRev):
ok = True
print("YES" if ok else "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<=β€<=*x*<=β€<=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
inputStr = input()
tempRev = inputStr[::-1]
ok = False
if(inputStr == tempRev):
ok = True
else:
zerosAdded = 0
temp = list(inputStr)
for i in reversed(temp):
if(i != '0'):
break
else:
zerosAdded += 1
inputStr =inputStr.rjust(len(inputStr) + zerosAdded, "0")
tempRev = inputStr[::-1]
if(inputStr == tempRev):
ok = True
print("YES" if ok else "NO")
```
| 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 β 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<=β€<=*n*<=β€<=100) β 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"
}
] | 1,532,606,696
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 32
| 218
| 0
|
n = int(input())
m = list(map(int, input().split()))
k=0
z=0
for i in range(n):
if m[i] % 2 !=0:
k+=1
b=i+1
elif m[i] % 2 == 0:
z+=1
j=i+1
if k > z:
print(j)
else:
print(b)
|
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 β 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<=β€<=*n*<=β€<=100) β 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())
m = list(map(int, input().split()))
k=0
z=0
for i in range(n):
if m[i] % 2 !=0:
k+=1
b=i+1
elif m[i] % 2 == 0:
z+=1
j=i+1
if k > z:
print(j)
else:
print(b)
```
| 3.9455
|
1,006
|
C
|
Three Parts of the Array
|
PROGRAMMING
| 1,200
|
[
"binary search",
"data structures",
"two pointers"
] | null | null |
You are given an array $d_1, d_2, \dots, d_n$ consisting of $n$ integer numbers.
Your task is to split this array into three parts (some of which may be empty) in such a way that each element of the array belongs to exactly one of the three parts, and each of the parts forms a consecutive contiguous subsegment (possibly, empty) of the original array.
Let the sum of elements of the first part be $sum_1$, the sum of elements of the second part be $sum_2$ and the sum of elements of the third part be $sum_3$. Among all possible ways to split the array you have to choose a way such that $sum_1 = sum_3$ and $sum_1$ is maximum possible.
More formally, if the first part of the array contains $a$ elements, the second part of the array contains $b$ elements and the third part contains $c$ elements, then:
$$sum_1 = \sum\limits_{1 \le i \le a}d_i,$$ $$sum_2 = \sum\limits_{a + 1 \le i \le a + b}d_i,$$ $$sum_3 = \sum\limits_{a + b + 1 \le i \le a + b + c}d_i.$$
The sum of an empty array is $0$.
Your task is to find a way to split the array such that $sum_1 = sum_3$ and $sum_1$ is maximum possible.
|
The first line of the input contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) β the number of elements in the array $d$.
The second line of the input contains $n$ integers $d_1, d_2, \dots, d_n$ ($1 \le d_i \le 10^9$) β the elements of the array $d$.
|
Print a single integer β the maximum possible value of $sum_1$, considering that the condition $sum_1 = sum_3$ must be met.
Obviously, at least one valid way to split the array exists (use $a=c=0$ and $b=n$).
|
[
"5\n1 3 1 1 4\n",
"5\n1 3 2 1 4\n",
"3\n4 1 2\n"
] |
[
"5\n",
"4\n",
"0\n"
] |
In the first example there is only one possible splitting which maximizes $sum_1$: $[1, 3, 1], [~], [1, 4]$.
In the second example the only way to have $sum_1=4$ is: $[1, 3], [2, 1], [4]$.
In the third example there is only one way to split the array: $[~], [4, 1, 2], [~]$.
| 0
|
[
{
"input": "5\n1 3 1 1 4",
"output": "5"
},
{
"input": "5\n1 3 2 1 4",
"output": "4"
},
{
"input": "3\n4 1 2",
"output": "0"
},
{
"input": "1\n1000000000",
"output": "0"
},
{
"input": "2\n1 1",
"output": "1"
},
{
"input": "5\n1 3 5 4 5",
"output": "9"
}
] | 1,691,890,894
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 27
| 109
| 16,998,400
|
n = int(input())
arr = [int(x) for x in input().split(" ")]
soma1 = 0
soma2 = 0
l = 0
r = n-1
ans = 0
while l<=r:
if soma1 < soma2:
soma1+=arr[l]
l += 1
else:
soma2 += arr[r]
r -= 1
if soma1 == soma2:
ans = soma1
print(ans)
|
Title: Three Parts of the Array
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given an array $d_1, d_2, \dots, d_n$ consisting of $n$ integer numbers.
Your task is to split this array into three parts (some of which may be empty) in such a way that each element of the array belongs to exactly one of the three parts, and each of the parts forms a consecutive contiguous subsegment (possibly, empty) of the original array.
Let the sum of elements of the first part be $sum_1$, the sum of elements of the second part be $sum_2$ and the sum of elements of the third part be $sum_3$. Among all possible ways to split the array you have to choose a way such that $sum_1 = sum_3$ and $sum_1$ is maximum possible.
More formally, if the first part of the array contains $a$ elements, the second part of the array contains $b$ elements and the third part contains $c$ elements, then:
$$sum_1 = \sum\limits_{1 \le i \le a}d_i,$$ $$sum_2 = \sum\limits_{a + 1 \le i \le a + b}d_i,$$ $$sum_3 = \sum\limits_{a + b + 1 \le i \le a + b + c}d_i.$$
The sum of an empty array is $0$.
Your task is to find a way to split the array such that $sum_1 = sum_3$ and $sum_1$ is maximum possible.
Input Specification:
The first line of the input contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) β the number of elements in the array $d$.
The second line of the input contains $n$ integers $d_1, d_2, \dots, d_n$ ($1 \le d_i \le 10^9$) β the elements of the array $d$.
Output Specification:
Print a single integer β the maximum possible value of $sum_1$, considering that the condition $sum_1 = sum_3$ must be met.
Obviously, at least one valid way to split the array exists (use $a=c=0$ and $b=n$).
Demo Input:
['5\n1 3 1 1 4\n', '5\n1 3 2 1 4\n', '3\n4 1 2\n']
Demo Output:
['5\n', '4\n', '0\n']
Note:
In the first example there is only one possible splitting which maximizes $sum_1$: $[1, 3, 1], [~], [1, 4]$.
In the second example the only way to have $sum_1=4$ is: $[1, 3], [2, 1], [4]$.
In the third example there is only one way to split the array: $[~], [4, 1, 2], [~]$.
|
```python
n = int(input())
arr = [int(x) for x in input().split(" ")]
soma1 = 0
soma2 = 0
l = 0
r = n-1
ans = 0
while l<=r:
if soma1 < soma2:
soma1+=arr[l]
l += 1
else:
soma2 += arr[r]
r -= 1
if soma1 == soma2:
ans = soma1
print(ans)
```
| 3
|
|
38
|
A
|
Army
|
PROGRAMMING
| 800
|
[
"implementation"
] |
A. Army
|
2
|
256
|
The Berland Armed Forces System consists of *n* ranks that are numbered using natural numbers from 1 to *n*, where 1 is the lowest rank and *n* is the highest rank.
One needs exactly *d**i* years to rise from rank *i* to rank *i*<=+<=1. Reaching a certain rank *i* having not reached all the previous *i*<=-<=1 ranks is impossible.
Vasya has just reached a new rank of *a*, but he dreams of holding the rank of *b*. Find for how many more years Vasya should serve in the army until he can finally realize his dream.
|
The first input line contains an integer *n* (2<=β€<=*n*<=β€<=100). The second line contains *n*<=-<=1 integers *d**i* (1<=β€<=*d**i*<=β€<=100). The third input line contains two integers *a* and *b* (1<=β€<=*a*<=<<=*b*<=β€<=*n*). The numbers on the lines are space-separated.
|
Print the single number which is the number of years that Vasya needs to rise from rank *a* to rank *b*.
|
[
"3\n5 6\n1 2\n",
"3\n5 6\n1 3\n"
] |
[
"5\n",
"11\n"
] |
none
| 0
|
[
{
"input": "3\n5 6\n1 2",
"output": "5"
},
{
"input": "3\n5 6\n1 3",
"output": "11"
},
{
"input": "2\n55\n1 2",
"output": "55"
},
{
"input": "3\n85 78\n1 3",
"output": "163"
},
{
"input": "4\n63 4 49\n2 3",
"output": "4"
},
{
"input": "5\n93 83 42 56\n2 5",
"output": "181"
},
{
"input": "6\n22 9 87 89 57\n1 6",
"output": "264"
},
{
"input": "7\n52 36 31 23 74 78\n2 7",
"output": "242"
},
{
"input": "8\n82 14 24 5 91 49 94\n3 8",
"output": "263"
},
{
"input": "9\n12 40 69 39 59 21 59 5\n4 6",
"output": "98"
},
{
"input": "10\n95 81 32 59 71 30 50 61 100\n1 6",
"output": "338"
},
{
"input": "15\n89 55 94 4 15 69 19 60 91 77 3 94 91 62\n3 14",
"output": "617"
},
{
"input": "20\n91 1 41 51 95 67 92 35 23 70 44 91 57 50 21 8 9 71 40\n8 17",
"output": "399"
},
{
"input": "25\n70 95 21 84 97 39 12 98 53 24 78 29 84 65 70 22 100 17 69 27 62 48 35 80\n8 23",
"output": "846"
},
{
"input": "30\n35 69 50 44 19 56 86 56 98 24 21 2 61 24 85 30 2 22 57 35 59 84 12 77 92 53 50 92 9\n1 16",
"output": "730"
},
{
"input": "35\n2 34 47 15 27 61 6 88 67 20 53 65 29 68 77 5 78 86 44 98 32 81 91 79 54 84 95 23 65 97 22 33 42 87\n8 35",
"output": "1663"
},
{
"input": "40\n32 88 59 36 95 45 28 78 73 30 97 13 13 47 48 100 43 21 22 45 88 25 15 13 63 25 72 92 29 5 25 11 50 5 54 51 48 84 23\n7 26",
"output": "862"
},
{
"input": "45\n83 74 73 95 10 31 100 26 29 15 80 100 22 70 31 88 9 56 19 70 2 62 48 30 27 47 52 50 94 44 21 94 23 85 15 3 95 72 43 62 94 89 68 88\n17 40",
"output": "1061"
},
{
"input": "50\n28 8 16 29 19 82 70 51 96 84 74 72 17 69 12 21 37 21 39 3 18 66 19 49 86 96 94 93 2 90 96 84 59 88 58 15 61 33 55 22 35 54 51 29 64 68 29 38 40\n23 28",
"output": "344"
},
{
"input": "60\n24 28 25 21 43 71 64 73 71 90 51 83 69 43 75 43 78 72 56 61 99 7 23 86 9 16 16 94 23 74 18 56 20 72 13 31 75 34 35 86 61 49 4 72 84 7 65 70 66 52 21 38 6 43 69 40 73 46 5\n28 60",
"output": "1502"
},
{
"input": "70\n69 95 34 14 67 61 6 95 94 44 28 94 73 66 39 13 19 71 73 71 28 48 26 22 32 88 38 95 43 59 88 77 80 55 17 95 40 83 67 1 38 95 58 63 56 98 49 2 41 4 73 8 78 41 64 71 60 71 41 61 67 4 4 19 97 14 39 20 27\n9 41",
"output": "1767"
},
{
"input": "80\n65 15 43 6 43 98 100 16 69 98 4 54 25 40 2 35 12 23 38 29 10 89 30 6 4 8 7 96 64 43 11 49 89 38 20 59 54 85 46 16 16 89 60 54 28 37 32 34 67 9 78 30 50 87 58 53 99 48 77 3 5 6 19 99 16 20 31 10 80 76 82 56 56 83 72 81 84 60 28\n18 24",
"output": "219"
},
{
"input": "90\n61 35 100 99 67 87 42 90 44 4 81 65 29 63 66 56 53 22 55 87 39 30 34 42 27 80 29 97 85 28 81 22 50 22 24 75 67 86 78 79 94 35 13 97 48 76 68 66 94 13 82 1 22 85 5 36 86 73 65 97 43 56 35 26 87 25 74 47 81 67 73 75 99 75 53 38 70 21 66 78 38 17 57 40 93 57 68 55 1\n12 44",
"output": "1713"
},
{
"input": "95\n37 74 53 96 65 84 65 72 95 45 6 77 91 35 58 50 51 51 97 30 51 20 79 81 92 10 89 34 40 76 71 54 26 34 73 72 72 28 53 19 95 64 97 10 44 15 12 38 5 63 96 95 86 8 36 96 45 53 81 5 18 18 47 97 65 9 33 53 41 86 37 53 5 40 15 76 83 45 33 18 26 5 19 90 46 40 100 42 10 90 13 81 40 53\n6 15",
"output": "570"
},
{
"input": "96\n51 32 95 75 23 54 70 89 67 3 1 51 4 100 97 30 9 35 56 38 54 77 56 98 43 17 60 43 72 46 87 61 100 65 81 22 74 38 16 96 5 10 54 22 23 22 10 91 9 54 49 82 29 73 33 98 75 8 4 26 24 90 71 42 90 24 94 74 94 10 41 98 56 63 18 43 56 21 26 64 74 33 22 38 67 66 38 60 64 76 53 10 4 65 76\n21 26",
"output": "328"
},
{
"input": "97\n18 90 84 7 33 24 75 55 86 10 96 72 16 64 37 9 19 71 62 97 5 34 85 15 46 72 82 51 52 16 55 68 27 97 42 72 76 97 32 73 14 56 11 86 2 81 59 95 60 93 1 22 71 37 77 100 6 16 78 47 78 62 94 86 16 91 56 46 47 35 93 44 7 86 70 10 29 45 67 62 71 61 74 39 36 92 24 26 65 14 93 92 15 28 79 59\n6 68",
"output": "3385"
},
{
"input": "98\n32 47 26 86 43 42 79 72 6 68 40 46 29 80 24 89 29 7 21 56 8 92 13 33 50 79 5 7 84 85 24 23 1 80 51 21 26 55 96 51 24 2 68 98 81 88 57 100 64 84 54 10 14 2 74 1 89 71 1 20 84 85 17 31 42 58 69 67 48 60 97 90 58 10 21 29 2 21 60 61 68 89 77 39 57 18 61 44 67 100 33 74 27 40 83 29 6\n8 77",
"output": "3319"
},
{
"input": "99\n46 5 16 66 53 12 84 89 26 27 35 68 41 44 63 17 88 43 80 15 59 1 42 50 53 34 75 16 16 55 92 30 28 11 12 71 27 65 11 28 86 47 24 10 60 47 7 53 16 75 6 49 56 66 70 3 20 78 75 41 38 57 89 23 16 74 30 39 1 32 49 84 9 33 25 95 75 45 54 59 17 17 29 40 79 96 47 11 69 86 73 56 91 4 87 47 31 24\n23 36",
"output": "514"
},
{
"input": "100\n63 65 21 41 95 23 3 4 12 23 95 50 75 63 58 34 71 27 75 31 23 94 96 74 69 34 43 25 25 55 44 19 43 86 68 17 52 65 36 29 72 96 84 25 84 23 71 54 6 7 71 7 21 100 99 58 93 35 62 47 36 70 68 9 75 13 35 70 76 36 62 22 52 51 2 87 66 41 54 35 78 62 30 35 65 44 74 93 78 37 96 70 26 32 71 27 85 85 63\n43 92",
"output": "2599"
},
{
"input": "51\n85 38 22 38 42 36 55 24 36 80 49 15 66 91 88 61 46 82 1 61 89 92 6 56 28 8 46 80 56 90 91 38 38 17 69 64 57 68 13 44 45 38 8 72 61 39 87 2 73 88\n15 27",
"output": "618"
},
{
"input": "2\n3\n1 2",
"output": "3"
},
{
"input": "5\n6 8 22 22\n2 3",
"output": "8"
},
{
"input": "6\n3 12 27 28 28\n3 4",
"output": "27"
},
{
"input": "9\n1 2 2 2 2 3 3 5\n3 7",
"output": "9"
},
{
"input": "10\n1 1 1 1 1 1 1 1 1\n6 8",
"output": "2"
},
{
"input": "20\n1 1 1 1 1 1 1 1 2 2 2 2 2 3 3 3 3 3 3\n5 17",
"output": "23"
},
{
"input": "25\n1 1 1 4 5 6 8 11 11 11 11 12 13 14 14 14 15 16 16 17 17 17 19 19\n4 8",
"output": "23"
},
{
"input": "35\n1 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\n30 31",
"output": "2"
},
{
"input": "45\n1 1 1 1 2 2 2 2 2 2 2 3 3 3 3 3 3 4 5 5 5 5 6 6 6 6 6 6 6 7 7 7 7 8 8 8 9 9 9 9 9 10 10 10\n42 45",
"output": "30"
},
{
"input": "50\n1 8 8 13 14 15 15 16 19 21 22 24 26 31 32 37 45 47 47 47 50 50 51 54 55 56 58 61 61 61 63 63 64 66 66 67 67 70 71 80 83 84 85 92 92 94 95 95 100\n4 17",
"output": "285"
},
{
"input": "60\n1 2 4 4 4 6 6 8 9 10 10 13 14 18 20 20 21 22 23 23 26 29 30 32 33 34 35 38 40 42 44 44 46 48 52 54 56 56 60 60 66 67 68 68 69 73 73 74 80 80 81 81 82 84 86 86 87 89 89\n56 58",
"output": "173"
},
{
"input": "70\n1 2 3 3 4 5 5 7 7 7 8 8 8 8 9 9 10 12 12 12 12 13 16 16 16 16 16 16 17 17 18 18 20 20 21 23 24 25 25 26 29 29 29 29 31 32 32 34 35 36 36 37 37 38 39 39 40 40 40 40 41 41 42 43 44 44 44 45 45\n62 65",
"output": "126"
},
{
"input": "80\n1 1 1 1 1 1 1 1 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 4 4 4 4 5 5 5 5 5 5 5 6 7 7 7 7 7 7 8 8 8 8 9 9 9 9 9 9 9 9 9 10 10 10 10 10 10 10 10 10 11 11 11 11 11 11 11 12 12 12 12 12 12 12 12\n17 65",
"output": "326"
},
{
"input": "90\n1 1 3 5 8 9 10 11 11 11 11 12 13 14 15 15 15 16 16 19 19 20 22 23 24 25 25 28 29 29 30 31 33 34 35 37 37 38 41 43 43 44 45 47 51 54 55 56 58 58 59 59 60 62 66 67 67 67 68 68 69 70 71 72 73 73 76 77 77 78 78 78 79 79 79 82 83 84 85 85 87 87 89 93 93 93 95 99 99\n28 48",
"output": "784"
},
{
"input": "95\n2 2 3 3 4 6 6 7 7 7 9 10 12 12 12 12 13 14 15 16 17 18 20 20 20 20 21 21 21 21 22 22 22 22 22 23 23 23 25 26 26 27 27 27 28 29 29 30 30 31 32 33 34 36 37 37 38 39 39 39 42 43 43 43 45 47 48 50 50 51 52 53 54 54 54 55 55 55 58 59 60 61 61 61 61 62 62 63 64 65 66 67 67 67\n64 93",
"output": "1636"
},
{
"input": "96\n1 1 2 3 3 5 8 9 9 10 10 10 11 11 11 11 11 12 13 13 13 14 15 15 16 16 17 17 17 17 18 18 20 20 20 21 21 21 23 24 24 25 25 26 27 27 27 27 29 29 29 30 30 30 32 32 32 32 32 32 33 33 34 34 34 35 35 35 36 36 37 37 37 38 39 40 41 41 41 41 42 42 43 43 45 45 45 46 46 47 47 49 50 52 52\n76 96",
"output": "898"
},
{
"input": "98\n2 3 4 4 5 7 8 10 10 10 11 11 12 12 12 12 13 14 15 15 16 16 18 19 19 20 21 21 21 21 22 23 24 25 26 26 27 27 27 27 29 29 30 30 31 31 37 40 40 40 41 41 41 42 43 44 44 44 46 46 47 49 49 50 50 50 51 53 55 55 56 56 56 56 56 57 57 58 59 60 60 60 62 62 63 64 64 64 65 66 66 67 68 70 70 71 71\n8 90",
"output": "3016"
},
{
"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\n66 95",
"output": "29"
},
{
"input": "100\n1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 3 3 3 4 4 4 4 4 4 4 4 4 4 5 5 5 5 5 5 6 6 6 6 6 6 6 6 6 6 6 6 7 7 7 7 7 7 8 8 8 8 9 9 9 9 10 10 10 10 11 11 11 11 12 12 12 13 13 13 13 13 13 13 13 13 13 14 14 14 14 14 14 15 15 15 15 15 15 16 16 16 17 17\n39 52",
"output": "82"
},
{
"input": "51\n5 7 9 16 19 25 26 29 29 30 31 32 32 41 43 44 47 49 50 50 51 52 54 55 56 63 64 66 67 70 74 74 77 78 79 80 80 85 86 87 89 89 90 92 93 94 94 95 95 97\n3 44",
"output": "2268"
}
] | 1,585,222,557
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 50
| 278
| 0
|
### A. Army
n=int(input())
arr=[int(x) for x in input().split()]
a,b=map(int,input().split())
total=0
for i in range(a,b):
total+=arr[i-1]
print(total)
|
Title: Army
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
The Berland Armed Forces System consists of *n* ranks that are numbered using natural numbers from 1 to *n*, where 1 is the lowest rank and *n* is the highest rank.
One needs exactly *d**i* years to rise from rank *i* to rank *i*<=+<=1. Reaching a certain rank *i* having not reached all the previous *i*<=-<=1 ranks is impossible.
Vasya has just reached a new rank of *a*, but he dreams of holding the rank of *b*. Find for how many more years Vasya should serve in the army until he can finally realize his dream.
Input Specification:
The first input line contains an integer *n* (2<=β€<=*n*<=β€<=100). The second line contains *n*<=-<=1 integers *d**i* (1<=β€<=*d**i*<=β€<=100). The third input line contains two integers *a* and *b* (1<=β€<=*a*<=<<=*b*<=β€<=*n*). The numbers on the lines are space-separated.
Output Specification:
Print the single number which is the number of years that Vasya needs to rise from rank *a* to rank *b*.
Demo Input:
['3\n5 6\n1 2\n', '3\n5 6\n1 3\n']
Demo Output:
['5\n', '11\n']
Note:
none
|
```python
### A. Army
n=int(input())
arr=[int(x) for x in input().split()]
a,b=map(int,input().split())
total=0
for i in range(a,b):
total+=arr[i-1]
print(total)
```
| 3.9305
|
714
|
A
|
Meeting of Old Friends
|
PROGRAMMING
| 1,100
|
[
"implementation",
"math"
] | null | null |
Today an outstanding event is going to happen in the forestΒ β hedgehog Filya will come to his old fried Sonya!
Sonya is an owl and she sleeps during the day and stay awake from minute *l*1 to minute *r*1 inclusive. Also, during the minute *k* she prinks and is unavailable for Filya.
Filya works a lot and he plans to visit Sonya from minute *l*2 to minute *r*2 inclusive.
Calculate the number of minutes they will be able to spend together.
|
The only line of the input contains integers *l*1, *r*1, *l*2, *r*2 and *k* (1<=β€<=*l*1,<=*r*1,<=*l*2,<=*r*2,<=*k*<=β€<=1018, *l*1<=β€<=*r*1, *l*2<=β€<=*r*2), providing the segments of time for Sonya and Filya and the moment of time when Sonya prinks.
|
Print one integerΒ β the number of minutes Sonya and Filya will be able to spend together.
|
[
"1 10 9 20 1\n",
"1 100 50 200 75\n"
] |
[
"2\n",
"50\n"
] |
In the first sample, they will be together during minutes 9 and 10.
In the second sample, they will be together from minute 50 to minute 74 and from minute 76 to minute 100.
| 500
|
[
{
"input": "1 10 9 20 1",
"output": "2"
},
{
"input": "1 100 50 200 75",
"output": "50"
},
{
"input": "6 6 5 8 9",
"output": "1"
},
{
"input": "1 1000000000 1 1000000000 1",
"output": "999999999"
},
{
"input": "5 100 8 8 8",
"output": "0"
},
{
"input": "1 1000000000000000000 2 99999999999999999 1000000000",
"output": "99999999999999997"
},
{
"input": "1 1 1 1 1",
"output": "0"
},
{
"input": "1 2 3 4 5",
"output": "0"
},
{
"input": "1 1000000000 2 999999999 3141592",
"output": "999999997"
},
{
"input": "24648817341102 41165114064236 88046848035 13602161452932 10000831349205",
"output": "0"
},
{
"input": "1080184299348 34666828555290 6878390132365 39891656267344 15395310291636",
"output": "27788438422925"
},
{
"input": "11814 27385 22309 28354 23595",
"output": "5076"
},
{
"input": "4722316546398 36672578279675 796716437180 33840047334985 13411035401708",
"output": "29117730788587"
},
{
"input": "14300093617438 14381698008501 6957847034861 32510754974307 66056597033082",
"output": "81604391064"
},
{
"input": "700062402405871919 762322967106512617 297732773882447821 747309903322652819 805776739998108178",
"output": "47247500916780901"
},
{
"input": "59861796371397621 194872039092923459 668110259718450585 841148673332698972 928360292123223779",
"output": "0"
},
{
"input": "298248781360904821 346420922793050061 237084570581741798 726877079564549183 389611850470532358",
"output": "48172141432145241"
},
{
"input": "420745791717606818 864206437350900994 764928840030524015 966634105370748487 793326512080703489",
"output": "99277597320376979"
},
{
"input": "519325240668210886 776112702001665034 360568516809443669 875594219634943179 994594983925273138",
"output": "256787461333454149"
},
{
"input": "170331212821058551 891149660635282032 125964175621755330 208256491683509799 526532153531983174",
"output": "37925278862451249"
},
{
"input": "1 3 3 5 3",
"output": "0"
},
{
"input": "1 5 8 10 9",
"output": "0"
},
{
"input": "1 2 4 5 10",
"output": "0"
},
{
"input": "1 2 2 3 5",
"output": "1"
},
{
"input": "2 4 3 7 3",
"output": "1"
},
{
"input": "1 2 9 10 1",
"output": "0"
},
{
"input": "5 15 1 10 5",
"output": "5"
},
{
"input": "1 4 9 20 25",
"output": "0"
},
{
"input": "2 4 1 2 5",
"output": "1"
},
{
"input": "10 1000 1 100 2",
"output": "91"
},
{
"input": "1 3 3 8 10",
"output": "1"
},
{
"input": "4 6 6 8 9",
"output": "1"
},
{
"input": "2 3 1 4 3",
"output": "1"
},
{
"input": "1 2 2 3 100",
"output": "1"
},
{
"input": "1 2 100 120 2",
"output": "0"
},
{
"input": "1 3 5 7 4",
"output": "0"
},
{
"input": "1 3 5 7 5",
"output": "0"
},
{
"input": "1 4 8 10 6",
"output": "0"
},
{
"input": "1 2 5 6 100",
"output": "0"
},
{
"input": "1 2 5 10 20",
"output": "0"
},
{
"input": "1 2 5 6 7",
"output": "0"
},
{
"input": "2 5 7 12 6",
"output": "0"
},
{
"input": "10 20 50 100 80",
"output": "0"
},
{
"input": "1 2 5 10 2",
"output": "0"
},
{
"input": "1 2 5 6 4",
"output": "0"
},
{
"input": "5 9 1 2 3",
"output": "0"
},
{
"input": "50 100 1 20 3",
"output": "0"
},
{
"input": "10 20 3 7 30",
"output": "0"
},
{
"input": "1 5 10 10 100",
"output": "0"
},
{
"input": "100 101 1 2 3",
"output": "0"
},
{
"input": "1 5 10 20 6",
"output": "0"
},
{
"input": "1 10 15 25 5",
"output": "0"
},
{
"input": "1 2 5 10 3",
"output": "0"
},
{
"input": "2 3 5 6 100",
"output": "0"
},
{
"input": "1 2 4 5 6",
"output": "0"
},
{
"input": "6 10 1 2 40",
"output": "0"
},
{
"input": "20 30 1 5 1",
"output": "0"
},
{
"input": "20 40 50 100 50",
"output": "0"
},
{
"input": "1 1 4 9 2",
"output": "0"
},
{
"input": "1 2 5 6 1",
"output": "0"
},
{
"input": "1 100 400 500 450",
"output": "0"
},
{
"input": "5 6 1 2 5",
"output": "0"
},
{
"input": "1 10 21 30 50",
"output": "0"
},
{
"input": "100 200 300 400 101",
"output": "0"
},
{
"input": "2 8 12 16 9",
"output": "0"
},
{
"input": "1 5 7 9 6",
"output": "0"
},
{
"input": "300 400 100 200 101",
"output": "0"
},
{
"input": "1 2 2 3 10",
"output": "1"
},
{
"input": "1 10 100 200 5",
"output": "0"
},
{
"input": "1 3 3 4 4",
"output": "1"
},
{
"input": "10 20 30 40 25",
"output": "0"
},
{
"input": "1 2 5 10 1",
"output": "0"
},
{
"input": "2 4 8 10 1",
"output": "0"
},
{
"input": "2 5 10 15 7",
"output": "0"
},
{
"input": "100 200 5 10 1",
"output": "0"
},
{
"input": "1 2 100 200 300",
"output": "0"
},
{
"input": "30 100 10 20 25",
"output": "0"
},
{
"input": "10 20 1 5 6",
"output": "0"
},
{
"input": "4 5 1 2 4",
"output": "0"
},
{
"input": "11 100 1 9 1000",
"output": "0"
},
{
"input": "1 1 10 10 228",
"output": "0"
},
{
"input": "5 7 10 20 15",
"output": "0"
},
{
"input": "1 3 8 9 7",
"output": "0"
},
{
"input": "1 10 2 8 8",
"output": "6"
},
{
"input": "1 5 9 15 1",
"output": "0"
},
{
"input": "1 3 5 6 12",
"output": "0"
},
{
"input": "1 100 500 1000 3",
"output": "0"
},
{
"input": "1 1 1 1 2",
"output": "1"
},
{
"input": "1 1000 100 1000 200",
"output": "900"
},
{
"input": "4 5 1 4 1",
"output": "1"
},
{
"input": "1 5 5 7 3",
"output": "1"
},
{
"input": "1 4 4 10 11",
"output": "1"
},
{
"input": "1 1 3 4 100",
"output": "0"
},
{
"input": "1 4 3 5 6",
"output": "2"
},
{
"input": "10 100 20 30 40",
"output": "11"
},
{
"input": "5 9 1 11 7",
"output": "4"
}
] | 1,592,417,055
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 96
| 389
| 10,444,800
|
import operator as op
import re
import sys
from bisect import bisect, bisect_left, insort, insort_left
from collections import Counter, defaultdict, deque
from copy import deepcopy
from decimal import Decimal
from functools import reduce
from itertools import (
accumulate, combinations, combinations_with_replacement, groupby,
permutations, product)
from math import (acos, asin, atan, ceil, cos, degrees, factorial, gcd, hypot,
log2, pi, radians, sin, sqrt, tan)
from operator import itemgetter, mul
from string import ascii_lowercase, ascii_uppercase, digits
def inp():
return(int(input()))
def inlist():
return(list(map(int, input().split())))
def instr():
s = input()
return(list(s[:len(s)]))
def invr():
return(map(int, input().split()))
def def_value():
return 0
# For getting input from input.txt file
#sys.stdin = open('input.txt', 'r')
# Printing the Output to output.txt file
#sys.stdout = open('output.txt', 'w')
l1, r1, l2, r2, k = invr()
if r2 < l1 or l2 > r1:
print(0)
sys.exit()
L = max(l1, l2)
R = min(r1, r2)
res = R - L + 1
if k >= L and k <= R:
res -= 1
if res < 0:
print(0)
else:
print(res)
|
Title: Meeting of Old Friends
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Today an outstanding event is going to happen in the forestΒ β hedgehog Filya will come to his old fried Sonya!
Sonya is an owl and she sleeps during the day and stay awake from minute *l*1 to minute *r*1 inclusive. Also, during the minute *k* she prinks and is unavailable for Filya.
Filya works a lot and he plans to visit Sonya from minute *l*2 to minute *r*2 inclusive.
Calculate the number of minutes they will be able to spend together.
Input Specification:
The only line of the input contains integers *l*1, *r*1, *l*2, *r*2 and *k* (1<=β€<=*l*1,<=*r*1,<=*l*2,<=*r*2,<=*k*<=β€<=1018, *l*1<=β€<=*r*1, *l*2<=β€<=*r*2), providing the segments of time for Sonya and Filya and the moment of time when Sonya prinks.
Output Specification:
Print one integerΒ β the number of minutes Sonya and Filya will be able to spend together.
Demo Input:
['1 10 9 20 1\n', '1 100 50 200 75\n']
Demo Output:
['2\n', '50\n']
Note:
In the first sample, they will be together during minutes 9 and 10.
In the second sample, they will be together from minute 50 to minute 74 and from minute 76 to minute 100.
|
```python
import operator as op
import re
import sys
from bisect import bisect, bisect_left, insort, insort_left
from collections import Counter, defaultdict, deque
from copy import deepcopy
from decimal import Decimal
from functools import reduce
from itertools import (
accumulate, combinations, combinations_with_replacement, groupby,
permutations, product)
from math import (acos, asin, atan, ceil, cos, degrees, factorial, gcd, hypot,
log2, pi, radians, sin, sqrt, tan)
from operator import itemgetter, mul
from string import ascii_lowercase, ascii_uppercase, digits
def inp():
return(int(input()))
def inlist():
return(list(map(int, input().split())))
def instr():
s = input()
return(list(s[:len(s)]))
def invr():
return(map(int, input().split()))
def def_value():
return 0
# For getting input from input.txt file
#sys.stdin = open('input.txt', 'r')
# Printing the Output to output.txt file
#sys.stdout = open('output.txt', 'w')
l1, r1, l2, r2, k = invr()
if r2 < l1 or l2 > r1:
print(0)
sys.exit()
L = max(l1, l2)
R = min(r1, r2)
res = R - L + 1
if k >= L and k <= R:
res -= 1
if res < 0:
print(0)
else:
print(res)
```
| 3
|
|
50
|
A
|
Domino piling
|
PROGRAMMING
| 800
|
[
"greedy",
"math"
] |
A. Domino piling
|
2
|
256
|
You are given a rectangular board of *M*<=Γ<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=Γ<=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* β board sizes in squares (1<=β€<=*M*<=β€<=*N*<=β€<=16).
|
Output one number β 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"
}
] | 1,694,771,897
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 35
| 122
| 0
|
val = [int(x) for x in input().split()]
print((val[0] * val[1])//2 if val[0]*val[1] % 2 == 0 else ((val[0]*val[1]) - 1)//2)
|
Title: Domino piling
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
You are given a rectangular board of *M*<=Γ<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=Γ<=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* β board sizes in squares (1<=β€<=*M*<=β€<=*N*<=β€<=16).
Output Specification:
Output one number β 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
val = [int(x) for x in input().split()]
print((val[0] * val[1])//2 if val[0]*val[1] % 2 == 0 else ((val[0]*val[1]) - 1)//2)
```
| 3.9695
|
43
|
A
|
Football
|
PROGRAMMING
| 1,000
|
[
"strings"
] |
A. Football
|
2
|
256
|
One day Vasya decided to have a look at the results of Berland 1910 Football Championshipβs finals. Unfortunately he didn't find the overall score of the match; however, he got hold of a profound description of the match's process. On the whole there are *n* lines in that description each of which described one goal. Every goal was marked with the name of the team that had scored it. Help Vasya, learn the name of the team that won the finals. It is guaranteed that the match did not end in a tie.
|
The first line contains an integer *n* (1<=β€<=*n*<=β€<=100) β the number of lines in the description. Then follow *n* lines β for each goal the names of the teams that scored it. The names are non-empty lines consisting of uppercase Latin letters whose lengths do not exceed 10 symbols. It is guaranteed that the match did not end in a tie and the description contains no more than two different teams.
|
Print the name of the winning team. We remind you that in football the team that scores more goals is considered the winner.
|
[
"1\nABC\n",
"5\nA\nABA\nABA\nA\nA\n"
] |
[
"ABC\n",
"A\n"
] |
none
| 500
|
[
{
"input": "1\nABC",
"output": "ABC"
},
{
"input": "5\nA\nABA\nABA\nA\nA",
"output": "A"
},
{
"input": "2\nXTSJEP\nXTSJEP",
"output": "XTSJEP"
},
{
"input": "3\nXZYDJAEDZ\nXZYDJAEDZ\nXZYDJAEDZ",
"output": "XZYDJAEDZ"
},
{
"input": "3\nQCCYXL\nQCCYXL\nAXGLFQDD",
"output": "QCCYXL"
},
{
"input": "3\nAZID\nEERWBC\nEERWBC",
"output": "EERWBC"
},
{
"input": "3\nHNCGYL\nHNCGYL\nHNCGYL",
"output": "HNCGYL"
},
{
"input": "4\nZZWZTG\nZZWZTG\nZZWZTG\nZZWZTG",
"output": "ZZWZTG"
},
{
"input": "4\nA\nA\nKUDLJMXCSE\nA",
"output": "A"
},
{
"input": "5\nPHBTW\nPHBTW\nPHBTW\nPHBTW\nPHBTW",
"output": "PHBTW"
},
{
"input": "5\nPKUZYTFYWN\nPKUZYTFYWN\nSTC\nPKUZYTFYWN\nPKUZYTFYWN",
"output": "PKUZYTFYWN"
},
{
"input": "5\nHH\nHH\nNTQWPA\nNTQWPA\nHH",
"output": "HH"
},
{
"input": "10\nW\nW\nW\nW\nW\nD\nW\nD\nD\nW",
"output": "W"
},
{
"input": "19\nXBCP\nTGACNIH\nXBCP\nXBCP\nXBCP\nXBCP\nXBCP\nTGACNIH\nXBCP\nXBCP\nXBCP\nXBCP\nXBCP\nTGACNIH\nXBCP\nXBCP\nTGACNIH\nTGACNIH\nXBCP",
"output": "XBCP"
},
{
"input": "33\nOWQWCKLLF\nOWQWCKLLF\nOWQWCKLLF\nPYPAS\nPYPAS\nPYPAS\nOWQWCKLLF\nPYPAS\nOWQWCKLLF\nPYPAS\nPYPAS\nOWQWCKLLF\nOWQWCKLLF\nOWQWCKLLF\nPYPAS\nOWQWCKLLF\nPYPAS\nPYPAS\nPYPAS\nPYPAS\nOWQWCKLLF\nPYPAS\nPYPAS\nOWQWCKLLF\nOWQWCKLLF\nPYPAS\nOWQWCKLLF\nOWQWCKLLF\nPYPAS\nPYPAS\nOWQWCKLLF\nPYPAS\nPYPAS",
"output": "PYPAS"
},
{
"input": "51\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC",
"output": "NC"
},
{
"input": "89\nH\nVOCI\nVOCI\nH\nVOCI\nH\nH\nVOCI\nVOCI\nVOCI\nH\nH\nH\nVOCI\nVOCI\nVOCI\nH\nVOCI\nVOCI\nH\nVOCI\nVOCI\nVOCI\nH\nVOCI\nH\nVOCI\nH\nVOCI\nH\nVOCI\nVOCI\nH\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nH\nVOCI\nVOCI\nVOCI\nVOCI\nH\nVOCI\nH\nH\nVOCI\nH\nVOCI\nH\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nH\nH\nVOCI\nH\nH\nVOCI\nH\nVOCI\nH\nVOCI\nVOCI\nH\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nH\nH\nH\nH\nH\nVOCI\nH\nVOCI\nH\nVOCI\nVOCI",
"output": "VOCI"
},
{
"input": "100\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nM\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA",
"output": "HA"
},
{
"input": "100\nG\nG\nS\nS\nG\nG\nS\nS\nG\nS\nS\nS\nG\nS\nG\nG\nS\nG\nS\nS\nG\nS\nS\nS\nS\nS\nG\nS\nG\nS\nS\nG\nG\nG\nS\nS\nS\nS\nG\nS\nS\nG\nG\nG\nG\nG\nS\nG\nG\nS\nS\nS\nS\nS\nG\nG\nS\nG\nG\nG\nG\nG\nS\nS\nG\nS\nS\nS\nS\nG\nS\nS\nG\nS\nG\nG\nG\nG\nG\nG\nG\nG\nG\nG\nG\nS\nS\nG\nS\nS\nS\nS\nG\nG\nG\nS\nG\nG\nG\nS",
"output": "G"
},
{
"input": "100\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nOBH\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL",
"output": "WL"
}
] | 1,596,554,131
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 34
| 280
| 20,172,800
|
n = int(input())
cnt1, cnt2 = 1, 0
first, second = str(input()), ''
for i in range(0,n-1):
ip = str(input())
if ip==first:
cnt1+=1
else:
second = ip
cnt2+=1
out = first if cnt1>cnt2 else second
print(out)
|
Title: Football
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
One day Vasya decided to have a look at the results of Berland 1910 Football Championshipβs finals. Unfortunately he didn't find the overall score of the match; however, he got hold of a profound description of the match's process. On the whole there are *n* lines in that description each of which described one goal. Every goal was marked with the name of the team that had scored it. Help Vasya, learn the name of the team that won the finals. It is guaranteed that the match did not end in a tie.
Input Specification:
The first line contains an integer *n* (1<=β€<=*n*<=β€<=100) β the number of lines in the description. Then follow *n* lines β for each goal the names of the teams that scored it. The names are non-empty lines consisting of uppercase Latin letters whose lengths do not exceed 10 symbols. It is guaranteed that the match did not end in a tie and the description contains no more than two different teams.
Output Specification:
Print the name of the winning team. We remind you that in football the team that scores more goals is considered the winner.
Demo Input:
['1\nABC\n', '5\nA\nABA\nABA\nA\nA\n']
Demo Output:
['ABC\n', 'A\n']
Note:
none
|
```python
n = int(input())
cnt1, cnt2 = 1, 0
first, second = str(input()), ''
for i in range(0,n-1):
ip = str(input())
if ip==first:
cnt1+=1
else:
second = ip
cnt2+=1
out = first if cnt1>cnt2 else second
print(out)
```
| 3.892425
|
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<=β€<=*y*<=β€<=9000) β the year number.
|
Print a single integer β 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"
}
] | 1,695,644,624
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 27
| 92
| 0
|
y = input()
y = str(int(y) + 1)
while y[0] == y[1] or y[0] == y[2] or y[0] == y[3] or y[1] == y[2] or y[1] == y[3] or y[2] == y[3]:
y = str(int(y)+1)
else:
print(y)
|
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<=β€<=*y*<=β€<=9000) β the year number.
Output Specification:
Print a single integer β 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
y = input()
y = str(int(y) + 1)
while y[0] == y[1] or y[0] == y[2] or y[0] == y[3] or y[1] == y[2] or y[1] == y[3] or y[2] == y[3]:
y = str(int(y)+1)
else:
print(y)
```
| 3
|
|
723
|
A
|
The New Year: Meeting Friends
|
PROGRAMMING
| 800
|
[
"implementation",
"math",
"sortings"
] | null | null |
There are three friend living on the straight line *Ox* in Lineland. The first friend lives at the point *x*1, the second friend lives at the point *x*2, and the third friend lives at the point *x*3. They plan to celebrate the New Year together, so they need to meet at one point. What is the minimum total distance they have to travel in order to meet at some point and celebrate the New Year?
It's guaranteed that the optimal answer is always integer.
|
The first line of the input contains three distinct integers *x*1, *x*2 and *x*3 (1<=β€<=*x*1,<=*x*2,<=*x*3<=β€<=100)Β β the coordinates of the houses of the first, the second and the third friends respectively.
|
Print one integerΒ β the minimum total distance the friends need to travel in order to meet together.
|
[
"7 1 4\n",
"30 20 10\n"
] |
[
"6\n",
"20\n"
] |
In the first sample, friends should meet at the point 4. Thus, the first friend has to travel the distance of 3 (from the point 7 to the point 4), the second friend also has to travel the distance of 3 (from the point 1 to the point 4), while the third friend should not go anywhere because he lives at the point 4.
| 500
|
[
{
"input": "7 1 4",
"output": "6"
},
{
"input": "30 20 10",
"output": "20"
},
{
"input": "1 4 100",
"output": "99"
},
{
"input": "100 1 91",
"output": "99"
},
{
"input": "1 45 100",
"output": "99"
},
{
"input": "1 2 3",
"output": "2"
},
{
"input": "71 85 88",
"output": "17"
},
{
"input": "30 38 99",
"output": "69"
},
{
"input": "23 82 95",
"output": "72"
},
{
"input": "22 41 47",
"output": "25"
},
{
"input": "9 94 77",
"output": "85"
},
{
"input": "1 53 51",
"output": "52"
},
{
"input": "25 97 93",
"output": "72"
},
{
"input": "42 53 51",
"output": "11"
},
{
"input": "81 96 94",
"output": "15"
},
{
"input": "21 5 93",
"output": "88"
},
{
"input": "50 13 75",
"output": "62"
},
{
"input": "41 28 98",
"output": "70"
},
{
"input": "69 46 82",
"output": "36"
},
{
"input": "87 28 89",
"output": "61"
},
{
"input": "44 45 40",
"output": "5"
},
{
"input": "86 97 68",
"output": "29"
},
{
"input": "43 92 30",
"output": "62"
},
{
"input": "16 70 1",
"output": "69"
},
{
"input": "40 46 19",
"output": "27"
},
{
"input": "71 38 56",
"output": "33"
},
{
"input": "82 21 80",
"output": "61"
},
{
"input": "75 8 35",
"output": "67"
},
{
"input": "75 24 28",
"output": "51"
},
{
"input": "78 23 56",
"output": "55"
},
{
"input": "85 31 10",
"output": "75"
},
{
"input": "76 50 9",
"output": "67"
},
{
"input": "95 37 34",
"output": "61"
},
{
"input": "84 61 35",
"output": "49"
},
{
"input": "87 85 37",
"output": "50"
},
{
"input": "1 3 2",
"output": "2"
},
{
"input": "4 2 6",
"output": "4"
},
{
"input": "6 9 3",
"output": "6"
},
{
"input": "12 4 8",
"output": "8"
},
{
"input": "15 10 5",
"output": "10"
},
{
"input": "1 50 17",
"output": "49"
},
{
"input": "10 5 15",
"output": "10"
},
{
"input": "8 1 9",
"output": "8"
},
{
"input": "3 5 4",
"output": "2"
},
{
"input": "2 1 3",
"output": "2"
},
{
"input": "1 8 2",
"output": "7"
},
{
"input": "1 100 2",
"output": "99"
},
{
"input": "1 4 6",
"output": "5"
}
] | 1,690,548,503
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 48
| 46
| 0
|
x1,x2,x3=map(int,input().split())
y=max(x1,x2,x3)
z=min(x1,x2,x3)
print(y-z)
|
Title: The New Year: Meeting Friends
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are three friend living on the straight line *Ox* in Lineland. The first friend lives at the point *x*1, the second friend lives at the point *x*2, and the third friend lives at the point *x*3. They plan to celebrate the New Year together, so they need to meet at one point. What is the minimum total distance they have to travel in order to meet at some point and celebrate the New Year?
It's guaranteed that the optimal answer is always integer.
Input Specification:
The first line of the input contains three distinct integers *x*1, *x*2 and *x*3 (1<=β€<=*x*1,<=*x*2,<=*x*3<=β€<=100)Β β the coordinates of the houses of the first, the second and the third friends respectively.
Output Specification:
Print one integerΒ β the minimum total distance the friends need to travel in order to meet together.
Demo Input:
['7 1 4\n', '30 20 10\n']
Demo Output:
['6\n', '20\n']
Note:
In the first sample, friends should meet at the point 4. Thus, the first friend has to travel the distance of 3 (from the point 7 to the point 4), the second friend also has to travel the distance of 3 (from the point 1 to the point 4), while the third friend should not go anywhere because he lives at the point 4.
|
```python
x1,x2,x3=map(int,input().split())
y=max(x1,x2,x3)
z=min(x1,x2,x3)
print(y-z)
```
| 3
|
|
652
|
B
|
z-sort
|
PROGRAMMING
| 1,000
|
[
"sortings"
] | null | null |
A student of *z*-school found a kind of sorting called *z*-sort. The array *a* with *n* elements are *z*-sorted if two conditions hold:
1. *a**i*<=β₯<=*a**i*<=-<=1 for all even *i*, 1. *a**i*<=β€<=*a**i*<=-<=1 for all odd *i*<=><=1.
For example the arrays [1,2,1,2] and [1,1,1,1] are *z*-sorted while the array [1,2,3,4] isnβt *z*-sorted.
Can you make the array *z*-sorted?
|
The first line contains a single integer *n* (1<=β€<=*n*<=β€<=1000) β the number of elements in the array *a*.
The second line contains *n* integers *a**i* (1<=β€<=*a**i*<=β€<=109) β the elements of the array *a*.
|
If it's possible to make the array *a* *z*-sorted print *n* space separated integers *a**i* β the elements after *z*-sort. Otherwise print the only word "Impossible".
|
[
"4\n1 2 2 1\n",
"5\n1 3 2 2 5\n"
] |
[
"1 2 1 2\n",
"1 5 2 3 2\n"
] |
none
| 0
|
[
{
"input": "4\n1 2 2 1",
"output": "1 2 1 2"
},
{
"input": "5\n1 3 2 2 5",
"output": "1 5 2 3 2"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "10\n1 1 1 1 1 1 1 1 1 1",
"output": "1 1 1 1 1 1 1 1 1 1"
},
{
"input": "10\n1 9 7 6 2 4 7 8 1 3",
"output": "1 9 1 8 2 7 3 7 4 6"
},
{
"input": "100\n82 51 81 14 37 17 78 92 64 15 8 86 89 8 87 77 66 10 15 12 100 25 92 47 21 78 20 63 13 49 41 36 41 79 16 87 87 69 3 76 80 60 100 49 70 59 72 8 38 71 45 97 71 14 76 54 81 4 59 46 39 29 92 3 49 22 53 99 59 52 74 31 92 43 42 23 44 9 82 47 7 40 12 9 3 55 37 85 46 22 84 52 98 41 21 77 63 17 62 91",
"output": "3 100 3 100 3 99 4 98 7 97 8 92 8 92 8 92 9 92 9 91 10 89 12 87 12 87 13 87 14 86 14 85 15 84 15 82 16 82 17 81 17 81 20 80 21 79 21 78 22 78 22 77 23 77 25 76 29 76 31 74 36 72 37 71 37 71 38 70 39 69 40 66 41 64 41 63 41 63 42 62 43 60 44 59 45 59 46 59 46 55 47 54 47 53 49 52 49 52 49 51"
},
{
"input": "3\n1 2 6",
"output": "1 6 2"
},
{
"input": "136\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 1 1 1 1 1 1 1 1 1 1 1 1 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": "1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 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": "3\n1 2 3",
"output": "1 3 2"
},
{
"input": "7\n999999998 999999999 999999999 999999999 999999999 999999999 1000000000",
"output": "999999998 1000000000 999999999 999999999 999999999 999999999 999999999"
},
{
"input": "3\n100 1 2",
"output": "1 100 2"
}
] | 1,619,300,629
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 16
| 62
| 0
|
n = int(input())
a = [int(x) for x in input().split()]
a = sorted(a)
saida = []
for i in range(n):
if i%2 == 0:
saida.append(a[i//2])
else:
saida.append(a[n-1-i//2])
print(*saida)
|
Title: z-sort
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A student of *z*-school found a kind of sorting called *z*-sort. The array *a* with *n* elements are *z*-sorted if two conditions hold:
1. *a**i*<=β₯<=*a**i*<=-<=1 for all even *i*, 1. *a**i*<=β€<=*a**i*<=-<=1 for all odd *i*<=><=1.
For example the arrays [1,2,1,2] and [1,1,1,1] are *z*-sorted while the array [1,2,3,4] isnβt *z*-sorted.
Can you make the array *z*-sorted?
Input Specification:
The first line contains a single integer *n* (1<=β€<=*n*<=β€<=1000) β the number of elements in the array *a*.
The second line contains *n* integers *a**i* (1<=β€<=*a**i*<=β€<=109) β the elements of the array *a*.
Output Specification:
If it's possible to make the array *a* *z*-sorted print *n* space separated integers *a**i* β the elements after *z*-sort. Otherwise print the only word "Impossible".
Demo Input:
['4\n1 2 2 1\n', '5\n1 3 2 2 5\n']
Demo Output:
['1 2 1 2\n', '1 5 2 3 2\n']
Note:
none
|
```python
n = int(input())
a = [int(x) for x in input().split()]
a = sorted(a)
saida = []
for i in range(n):
if i%2 == 0:
saida.append(a[i//2])
else:
saida.append(a[n-1-i//2])
print(*saida)
```
| 3
|
|
757
|
A
|
Gotta Catch Em' All!
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
Bash wants to become a Pokemon master one day. Although he liked a lot of Pokemon, he has always been fascinated by Bulbasaur the most. Soon, things started getting serious and his fascination turned into an obsession. Since he is too young to go out and catch Bulbasaur, he came up with his own way of catching a Bulbasaur.
Each day, he takes the front page of the newspaper. He cuts out the letters one at a time, from anywhere on the front page of the newspaper to form the word "Bulbasaur" (without quotes) and sticks it on his wall. Bash is very particular about caseΒ β the first letter of "Bulbasaur" must be upper case and the rest must be lower case. By doing this he thinks he has caught one Bulbasaur. He then repeats this step on the left over part of the newspaper. He keeps doing this until it is not possible to form the word "Bulbasaur" from the newspaper.
Given the text on the front page of the newspaper, can you tell how many Bulbasaurs he will catch today?
Note: uppercase and lowercase letters are considered different.
|
Input contains a single line containing a string *s* (1<=<=β€<=<=|*s*|<=<=β€<=<=105)Β β the text on the front page of the newspaper without spaces and punctuation marks. |*s*| is the length of the string *s*.
The string *s* contains lowercase and uppercase English letters, i.e. .
|
Output a single integer, the answer to the problem.
|
[
"Bulbbasaur\n",
"F\n",
"aBddulbasaurrgndgbualdBdsagaurrgndbb\n"
] |
[
"1\n",
"0\n",
"2\n"
] |
In the first case, you could pick: Bulbbasaur.
In the second case, there is no way to pick even a single Bulbasaur.
In the third case, you can rearrange the string to BulbasaurBulbasauraddrgndgddgargndbb to get two words "Bulbasaur".
| 500
|
[
{
"input": "Bulbbasaur",
"output": "1"
},
{
"input": "F",
"output": "0"
},
{
"input": "aBddulbasaurrgndgbualdBdsagaurrgndbb",
"output": "2"
},
{
"input": "BBBBBBBBBBbbbbbbbbbbuuuuuuuuuullllllllllssssssssssaaaaaaaaaarrrrrrrrrr",
"output": "5"
},
{
"input": "BBBBBBBBBBbbbbbbbbbbbbbbbbbbbbuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuussssssssssssssssssssaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "0"
},
{
"input": "BBBBBBBBBBssssssssssssssssssssaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaarrrrrrrrrr",
"output": "0"
},
{
"input": "BBBBBBBBBBbbbbbbbbbbbbbbbbbbbbuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuullllllllllllllllllllssssssssssssssssssssaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaarrrrrrrrrrrrrrrrrrrr",
"output": "10"
},
{
"input": "BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBbbbbbbbbbbbbbbbbbbbbuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuullllllllllllllllllllssssssssssssssssssssaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaarrrrrrrrrrrrrrrrrrrrrrrrrrrrrr",
"output": "20"
},
{
"input": "CeSlSwec",
"output": "0"
},
{
"input": "PnMrWPBGzVcmRcO",
"output": "0"
},
{
"input": "hHPWBQeEmCuhdCnzrqYtuFtwxokGhdGkFtsFICVqYfJeUrSBtSxEbzMCblOgqOvjXURhSKivPcseqgiNuUgIboEYMvVeRBbpzCGCfVydDvZNFGSFidwUtNbmPSfSYdMNmHgchIsiVswzFsGQewlMVEzicOagpWMdCWrCdPmexfnM",
"output": "0"
},
{
"input": "BBBBBBBBBBbbbbbbbbbbbbuuuuuuuuuuuullllllllllllssssssssssssaaaaaaaaaaaarrrrrrrrrrrrZBphUC",
"output": "6"
},
{
"input": "bulsar",
"output": "0"
},
{
"input": "Bblsar",
"output": "0"
},
{
"input": "Bbusar",
"output": "0"
},
{
"input": "Bbular",
"output": "0"
},
{
"input": "Bbulsr",
"output": "0"
},
{
"input": "Bbulsa",
"output": "0"
},
{
"input": "Bbulsar",
"output": "0"
},
{
"input": "Bbulsar",
"output": "0"
},
{
"input": "CaQprCjTiQACZjUJjSmMHVTDorSUugvTtksEjptVzNLhClWaVVWszIixBlqFkvjDmbRjarQoUWhXHoCgYNNjvEgRTgKpbdEMFsmqcTyvJzupKgYiYMtrZWXIAGVhmDURtddbBZIMgIgXqQUmXpssLSaVCDGZDHimNthwiAWabjtcraAQugMCpBPQZbBGZyqUZmzDVSvJZmDWfZEUHGJVtiJANAIbvjTxtvvTbjWRpNQZlxAqpLCLRVwYWqLaHOTvzgeNGdxiBwsAVKKsewXMTwZUUfxYwrwsiaRBwEdvDDoPsQUtinvajBoRzLBUuQekhjsfDAOQzIABSVPitRuhvvqeAahsSELTGbCPh",
"output": "2"
},
{
"input": "Bulbasaur",
"output": "1"
},
{
"input": "BulbasaurBulbasaur",
"output": "2"
},
{
"input": "Bulbbasar",
"output": "0"
},
{
"input": "Bulbasur",
"output": "0"
},
{
"input": "Bulbsaur",
"output": "0"
},
{
"input": "BulbsurBulbsurBulbsurBulbsur",
"output": "0"
},
{
"input": "Blbbasar",
"output": "0"
},
{
"input": "Bulbasar",
"output": "0"
},
{
"input": "BBullllbbaassaauurr",
"output": "1"
},
{
"input": "BulbasaurBulbasar",
"output": "1"
},
{
"input": "BulbasaurBulbsaur",
"output": "1"
},
{
"input": "Bubasaur",
"output": "0"
},
{
"input": "ulbasaurulbasaur",
"output": "0"
},
{
"input": "Bulbasr",
"output": "0"
},
{
"input": "BBBuuulllbbbaaasssaaauuurrr",
"output": "3"
},
{
"input": "BBuuuullbbaaaassrr",
"output": "2"
},
{
"input": "BBBBBBBuuuuuuuullllllllllllbbbbaaaaaassssssssssssssssaaaaauuuuuuuuuuuuurrrrrrrrrrrrrrrr",
"output": "4"
},
{
"input": "BBuullbbaassaarr",
"output": "1"
},
{
"input": "Bulbasau",
"output": "0"
},
{
"input": "BBuullbbaassaauurr",
"output": "2"
},
{
"input": "BulbasauBulbasauBulbasauBulbasauBulbasauBulbasauBulbasauBulbasau",
"output": "0"
},
{
"input": "Blbasaur",
"output": "0"
},
{
"input": "BulbasaurBulbasaurd",
"output": "2"
},
{
"input": "ulbasaur",
"output": "0"
},
{
"input": "Bulbaaur",
"output": "0"
},
{
"input": "BBuuuullbbbbbbbbbbbbbbbaassrr",
"output": "1"
},
{
"input": "Bulbasua",
"output": "0"
},
{
"input": "Bubbasaur",
"output": "0"
},
{
"input": "BulbasauBulbasauBulbasauBulbasauBulbasauBulbasaurrr",
"output": "3"
},
{
"input": "BulbasaurBubasaur",
"output": "1"
},
{
"input": "Baab",
"output": "0"
},
{
"input": "BulbasaurBulbasau",
"output": "1"
},
{
"input": "Bulbasauu",
"output": "0"
},
{
"input": "BulbasauBulbasau",
"output": "0"
},
{
"input": "BBBBBBBBBBB",
"output": "0"
},
{
"input": "Bulbbasau",
"output": "0"
},
{
"input": "BulbbasaurBulbbasar",
"output": "1"
},
{
"input": "Bulaaaasaur",
"output": "0"
},
{
"input": "BulbasaurBulbasauBulbasauBulbasau",
"output": "1"
}
] | 1,566,253,197
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 107
| 171
| 4,403,200
|
#391_A
st = list(input())
#B = 1, u = 2, l = 1, b = 1, a = 2, s = 1, r = 1
dct = {"B": 0, "u": 0, "l": 0, "b": 0, "a": 0, "s": 0, "r": 0}
for i in range(0, len(st)):
if st[i] in dct:
dct[st[i]] += 1
n = min(dct["B"], dct["u"] // 2, dct["l"], dct["b"], dct["a"] // 2, dct["s"], dct["r"])
print(n)
|
Title: Gotta Catch Em' All!
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bash wants to become a Pokemon master one day. Although he liked a lot of Pokemon, he has always been fascinated by Bulbasaur the most. Soon, things started getting serious and his fascination turned into an obsession. Since he is too young to go out and catch Bulbasaur, he came up with his own way of catching a Bulbasaur.
Each day, he takes the front page of the newspaper. He cuts out the letters one at a time, from anywhere on the front page of the newspaper to form the word "Bulbasaur" (without quotes) and sticks it on his wall. Bash is very particular about caseΒ β the first letter of "Bulbasaur" must be upper case and the rest must be lower case. By doing this he thinks he has caught one Bulbasaur. He then repeats this step on the left over part of the newspaper. He keeps doing this until it is not possible to form the word "Bulbasaur" from the newspaper.
Given the text on the front page of the newspaper, can you tell how many Bulbasaurs he will catch today?
Note: uppercase and lowercase letters are considered different.
Input Specification:
Input contains a single line containing a string *s* (1<=<=β€<=<=|*s*|<=<=β€<=<=105)Β β the text on the front page of the newspaper without spaces and punctuation marks. |*s*| is the length of the string *s*.
The string *s* contains lowercase and uppercase English letters, i.e. .
Output Specification:
Output a single integer, the answer to the problem.
Demo Input:
['Bulbbasaur\n', 'F\n', 'aBddulbasaurrgndgbualdBdsagaurrgndbb\n']
Demo Output:
['1\n', '0\n', '2\n']
Note:
In the first case, you could pick: Bulbbasaur.
In the second case, there is no way to pick even a single Bulbasaur.
In the third case, you can rearrange the string to BulbasaurBulbasauraddrgndgddgargndbb to get two words "Bulbasaur".
|
```python
#391_A
st = list(input())
#B = 1, u = 2, l = 1, b = 1, a = 2, s = 1, r = 1
dct = {"B": 0, "u": 0, "l": 0, "b": 0, "a": 0, "s": 0, "r": 0}
for i in range(0, len(st)):
if st[i] in dct:
dct[st[i]] += 1
n = min(dct["B"], dct["u"] // 2, dct["l"], dct["b"], dct["a"] // 2, dct["s"], dct["r"])
print(n)
```
| 3
|
|
584
|
B
|
Kolya and Tanya
|
PROGRAMMING
| 1,500
|
[
"combinatorics"
] | null | null |
Kolya loves putting gnomes at the circle table and giving them coins, and Tanya loves studying triplets of gnomes, sitting in the vertexes of an equilateral triangle.
More formally, there are 3*n* gnomes sitting in a circle. Each gnome can have from 1 to 3 coins. Let's number the places in the order they occur in the circle by numbers from 0 to 3*n*<=-<=1, let the gnome sitting on the *i*-th place have *a**i* coins. If there is an integer *i* (0<=β€<=*i*<=<<=*n*) such that *a**i*<=+<=*a**i*<=+<=*n*<=+<=*a**i*<=+<=2*n*<=β <=6, then Tanya is satisfied.
Count the number of ways to choose *a**i* so that Tanya is satisfied. As there can be many ways of distributing coins, print the remainder of this number modulo 109<=+<=7. Two ways, *a* and *b*, are considered distinct if there is index *i* (0<=β€<=*i*<=<<=3*n*), such that *a**i*<=β <=*b**i* (that is, some gnome got different number of coins in these two ways).
|
A single line contains number *n* (1<=β€<=*n*<=β€<=105) β the number of the gnomes divided by three.
|
Print a single number β the remainder of the number of variants of distributing coins that satisfy Tanya modulo 109<=+<=7.
|
[
"1\n",
"2\n"
] |
[
"20",
"680"
] |
20 ways for *n*β=β1 (gnome with index 0 sits on the top of the triangle, gnome 1 on the right vertex, gnome 2 on the left vertex): <img class="tex-graphics" src="https://espresso.codeforces.com/64df38b85ccb482cf88d02dc52e348e33313f9da.png" style="max-width: 100.0%;max-height: 100.0%;"/>
| 1,000
|
[
{
"input": "1",
"output": "20"
},
{
"input": "2",
"output": "680"
},
{
"input": "3",
"output": "19340"
},
{
"input": "4",
"output": "529040"
},
{
"input": "5",
"output": "14332100"
},
{
"input": "7",
"output": "459529590"
},
{
"input": "10",
"output": "848178170"
},
{
"input": "14",
"output": "356372551"
},
{
"input": "58000",
"output": "66667472"
},
{
"input": "6",
"output": "387302840"
},
{
"input": "8",
"output": "423769706"
},
{
"input": "9",
"output": "557078005"
},
{
"input": "15",
"output": "83420839"
},
{
"input": "20",
"output": "208683407"
},
{
"input": "22",
"output": "695178405"
},
{
"input": "48",
"output": "974152357"
},
{
"input": "74",
"output": "972015869"
},
{
"input": "100",
"output": "364994225"
},
{
"input": "866",
"output": "363347020"
},
{
"input": "1673",
"output": "112069243"
},
{
"input": "555",
"output": "506150275"
},
{
"input": "2000",
"output": "394190362"
},
{
"input": "9877",
"output": "915063121"
},
{
"input": "5984",
"output": "546302050"
},
{
"input": "2091",
"output": "982683445"
},
{
"input": "10000",
"output": "246305722"
},
{
"input": "10868",
"output": "118127713"
},
{
"input": "66115",
"output": "413345995"
},
{
"input": "62212",
"output": "587210092"
},
{
"input": "100000",
"output": "192156836"
},
{
"input": "2441",
"output": "149858187"
},
{
"input": "18569",
"output": "54080463"
},
{
"input": "69289",
"output": "582363404"
},
{
"input": "85417",
"output": "796810422"
},
{
"input": "68841",
"output": "371243141"
},
{
"input": "10000",
"output": "246305722"
},
{
"input": "4343",
"output": "682818914"
},
{
"input": "12312",
"output": "837337190"
},
{
"input": "5462",
"output": "719513799"
},
{
"input": "100",
"output": "364994225"
},
{
"input": "11",
"output": "550315374"
},
{
"input": "100000",
"output": "192156836"
}
] | 1,664,464,355
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 42
| 77
| 2,457,600
|
n = int(input())
print((3**(3*n)-7**n) % (int(1e9)+7))
|
Title: Kolya and Tanya
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kolya loves putting gnomes at the circle table and giving them coins, and Tanya loves studying triplets of gnomes, sitting in the vertexes of an equilateral triangle.
More formally, there are 3*n* gnomes sitting in a circle. Each gnome can have from 1 to 3 coins. Let's number the places in the order they occur in the circle by numbers from 0 to 3*n*<=-<=1, let the gnome sitting on the *i*-th place have *a**i* coins. If there is an integer *i* (0<=β€<=*i*<=<<=*n*) such that *a**i*<=+<=*a**i*<=+<=*n*<=+<=*a**i*<=+<=2*n*<=β <=6, then Tanya is satisfied.
Count the number of ways to choose *a**i* so that Tanya is satisfied. As there can be many ways of distributing coins, print the remainder of this number modulo 109<=+<=7. Two ways, *a* and *b*, are considered distinct if there is index *i* (0<=β€<=*i*<=<<=3*n*), such that *a**i*<=β <=*b**i* (that is, some gnome got different number of coins in these two ways).
Input Specification:
A single line contains number *n* (1<=β€<=*n*<=β€<=105) β the number of the gnomes divided by three.
Output Specification:
Print a single number β the remainder of the number of variants of distributing coins that satisfy Tanya modulo 109<=+<=7.
Demo Input:
['1\n', '2\n']
Demo Output:
['20', '680']
Note:
20 ways for *n*β=β1 (gnome with index 0 sits on the top of the triangle, gnome 1 on the right vertex, gnome 2 on the left vertex): <img class="tex-graphics" src="https://espresso.codeforces.com/64df38b85ccb482cf88d02dc52e348e33313f9da.png" style="max-width: 100.0%;max-height: 100.0%;"/>
|
```python
n = int(input())
print((3**(3*n)-7**n) % (int(1e9)+7))
```
| 3
|
|
424
|
A
|
Squats
|
PROGRAMMING
| 900
|
[
"implementation"
] | null | null |
Pasha has many hamsters and he makes them work out. Today, *n* hamsters (*n* is even) came to work out. The hamsters lined up and each hamster either sat down or stood up.
For another exercise, Pasha needs exactly hamsters to stand up and the other hamsters to sit down. In one minute, Pasha can make some hamster ether sit down or stand up. How many minutes will he need to get what he wants if he acts optimally well?
|
The first line contains integer *n* (2<=β€<=*n*<=β€<=200; *n* is even). The next line contains *n* characters without spaces. These characters describe the hamsters' position: the *i*-th character equals 'X', if the *i*-th hamster in the row is standing, and 'x', if he is sitting.
|
In the first line, print a single integer β the minimum required number of minutes. In the second line, print a string that describes the hamsters' position after Pasha makes the required changes. If there are multiple optimal positions, print any of them.
|
[
"4\nxxXx\n",
"2\nXX\n",
"6\nxXXxXx\n"
] |
[
"1\nXxXx\n",
"1\nxX\n",
"0\nxXXxXx\n"
] |
none
| 500
|
[
{
"input": "4\nxxXx",
"output": "1\nXxXx"
},
{
"input": "2\nXX",
"output": "1\nxX"
},
{
"input": "6\nxXXxXx",
"output": "0\nxXXxXx"
},
{
"input": "4\nxXXX",
"output": "1\nxxXX"
},
{
"input": "2\nXx",
"output": "0\nXx"
},
{
"input": "22\nXXxXXxxXxXxXXXXXXXXXxx",
"output": "4\nxxxxxxxXxXxXXXXXXXXXxx"
},
{
"input": "30\nXXxXxxXXXXxxXXxxXXxxxxXxxXXXxx",
"output": "0\nXXxXxxXXXXxxXXxxXXxxxxXxxXXXxx"
},
{
"input": "104\nxxXxXxxXXXxxXxXxxXXXxxxXxxXXXxxXXXxXxXxXXxxXxxxxxXXXXxXXXXxXXXxxxXxxxxxxxXxxXxXXxxXXXXxXXXxxXXXXXXXXXxXX",
"output": "4\nxxxxxxxxxXxxXxXxxXXXxxxXxxXXXxxXXXxXxXxXXxxXxxxxxXXXXxXXXXxXXXxxxXxxxxxxxXxxXxXXxxXXXXxXXXxxXXXXXXXXXxXX"
},
{
"input": "78\nxxxXxxXxXxxXxxxxxXxXXXxXXXXxxxxxXxXXXxxXxXXXxxxxXxxXXXxxxxxxxxXXXXxXxXXxXXXxXX",
"output": "3\nXXXXxxXxXxxXxxxxxXxXXXxXXXXxxxxxXxXXXxxXxXXXxxxxXxxXXXxxxxxxxxXXXXxXxXXxXXXxXX"
},
{
"input": "200\nxxXXxxXXxXxxXxxXxXxxXxXxXxXxxxxxXXxXXxxXXXXxXXXxXXxXxXxxxxXxxXXXxxxXxXxxxXxxXXxXxXxxxxxxxXxxXxXxxXxXXXxxXxXXXXxxXxxxXxXXXXXXxXxXXxxxxXxxxXxxxXxXXXxXxXXXXxXXxxxXxXXxxXXxxxXxXxXXxXXXxXxXxxxXXxxxxXXxXXXX",
"output": "4\nXXXXXXXXxXxxXxxXxXxxXxXxXxXxxxxxXXxXXxxXXXXxXXXxXXxXxXxxxxXxxXXXxxxXxXxxxXxxXXxXxXxxxxxxxXxxXxXxxXxXXXxxXxXXXXxxXxxxXxXXXXXXxXxXXxxxxXxxxXxxxXxXXXxXxXXXXxXXxxxXxXXxxXXxxxXxXxXXxXXXxXxXxxxXXxxxxXXxXXXX"
},
{
"input": "198\nxXxxXxxXxxXXxXxXxXxxXXXxxXxxxxXXXXxxXxxxxXXXXxXxXXxxxXXXXXXXxXXXxxxxXXxXXxXxXXxxxxXxXXXXXXxXxxXxXxxxXxXXXXxxXXxxXxxxXXxXxXXxXxXXxXXXXxxxxxXxXXxxxXxXXXXxXxXXxxXxXXxXxXXxxxXxXXXXxXxxXxXXXxxxxXxXXXXxXx",
"output": "5\nxxxxxxxxxxxxxXxXxXxxXXXxxXxxxxXXXXxxXxxxxXXXXxXxXXxxxXXXXXXXxXXXxxxxXXxXXxXxXXxxxxXxXXXXXXxXxxXxXxxxXxXXXXxxXXxxXxxxXXxXxXXxXxXXxXXXXxxxxxXxXXxxxXxXXXXxXxXXxxXxXXxXxXXxxxXxXXXXxXxxXxXXXxxxxXxXXXXxXx"
},
{
"input": "200\nxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx",
"output": "100\nXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx"
},
{
"input": "198\nxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx",
"output": "99\nXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx"
},
{
"input": "200\nXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX",
"output": "100\nxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX"
},
{
"input": "198\nXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX",
"output": "99\nxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX"
},
{
"input": "2\nxx",
"output": "1\nXx"
},
{
"input": "2\nXx",
"output": "0\nXx"
},
{
"input": "2\nxX",
"output": "0\nxX"
},
{
"input": "4\nXXXX",
"output": "2\nxxXX"
},
{
"input": "4\nxxxx",
"output": "2\nXXxx"
},
{
"input": "4\nxxXX",
"output": "0\nxxXX"
},
{
"input": "4\nXXxx",
"output": "0\nXXxx"
},
{
"input": "4\nxXxx",
"output": "1\nXXxx"
},
{
"input": "4\nXxxx",
"output": "1\nXXxx"
},
{
"input": "4\nxxxX",
"output": "1\nXxxX"
},
{
"input": "4\nxxXx",
"output": "1\nXxXx"
},
{
"input": "4\nXXXx",
"output": "1\nxXXx"
},
{
"input": "4\nxXXX",
"output": "1\nxxXX"
},
{
"input": "4\nXxXX",
"output": "1\nxxXX"
},
{
"input": "4\nXXxX",
"output": "1\nxXxX"
},
{
"input": "4\nXxXx",
"output": "0\nXxXx"
},
{
"input": "6\nxXXxXX",
"output": "1\nxxXxXX"
}
] | 1,586,084,583
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 32
| 93
| 307,200
|
# -*- coding: utf-8 -*-
"""
Created on Sun Apr 5 05:45:34 2020
@author: alexi
"""
#https://codeforces.com/problemset/problem/424/A --- Alexis Galvan
def hamster_squat():
total = int(input())
hamsters = input()
dic = {'x':0,'X':0}
for i in range(len(hamsters)):
dic[hamsters[i]] += 1
if dic['x'] == dic['X']:
print(0)
print(hamsters)
return
sat = dic['x']
stood = dic['X']
add_lower = False
if sat < stood:
add_lower = True
maximum = max(sat, stood)
add = maximum - (int((sat+stood)/2))
print(add)
if add_lower:
output = ''
hamsters = [i for i in hamsters]
for i in range(len(hamsters)):
if add > 0:
if hamsters[i] == 'X':
hamsters[i] = 'x'
add -= 1
output = output + hamsters[i]
else:
output = ''
hamsters = [i for i in hamsters]
for i in range(len(hamsters)):
if add > 0:
if hamsters[i] == 'x':
hamsters[i] = 'X'
add -= 1
output = output + hamsters[i]
print(output)
hamster_squat()
|
Title: Squats
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Pasha has many hamsters and he makes them work out. Today, *n* hamsters (*n* is even) came to work out. The hamsters lined up and each hamster either sat down or stood up.
For another exercise, Pasha needs exactly hamsters to stand up and the other hamsters to sit down. In one minute, Pasha can make some hamster ether sit down or stand up. How many minutes will he need to get what he wants if he acts optimally well?
Input Specification:
The first line contains integer *n* (2<=β€<=*n*<=β€<=200; *n* is even). The next line contains *n* characters without spaces. These characters describe the hamsters' position: the *i*-th character equals 'X', if the *i*-th hamster in the row is standing, and 'x', if he is sitting.
Output Specification:
In the first line, print a single integer β the minimum required number of minutes. In the second line, print a string that describes the hamsters' position after Pasha makes the required changes. If there are multiple optimal positions, print any of them.
Demo Input:
['4\nxxXx\n', '2\nXX\n', '6\nxXXxXx\n']
Demo Output:
['1\nXxXx\n', '1\nxX\n', '0\nxXXxXx\n']
Note:
none
|
```python
# -*- coding: utf-8 -*-
"""
Created on Sun Apr 5 05:45:34 2020
@author: alexi
"""
#https://codeforces.com/problemset/problem/424/A --- Alexis Galvan
def hamster_squat():
total = int(input())
hamsters = input()
dic = {'x':0,'X':0}
for i in range(len(hamsters)):
dic[hamsters[i]] += 1
if dic['x'] == dic['X']:
print(0)
print(hamsters)
return
sat = dic['x']
stood = dic['X']
add_lower = False
if sat < stood:
add_lower = True
maximum = max(sat, stood)
add = maximum - (int((sat+stood)/2))
print(add)
if add_lower:
output = ''
hamsters = [i for i in hamsters]
for i in range(len(hamsters)):
if add > 0:
if hamsters[i] == 'X':
hamsters[i] = 'x'
add -= 1
output = output + hamsters[i]
else:
output = ''
hamsters = [i for i in hamsters]
for i in range(len(hamsters)):
if add > 0:
if hamsters[i] == 'x':
hamsters[i] = 'X'
add -= 1
output = output + hamsters[i]
print(output)
hamster_squat()
```
| 3
|
|
224
|
A
|
Parallelepiped
|
PROGRAMMING
| 1,100
|
[
"brute force",
"geometry",
"math"
] | null | null |
You've got a rectangular parallelepiped with integer edge lengths. You know the areas of its three faces that have a common vertex. Your task is to find the sum of lengths of all 12 edges of this parallelepiped.
|
The first and the single line contains three space-separated integers β the areas of the parallelepiped's faces. The area's values are positive (<=><=0) and do not exceed 104. It is guaranteed that there exists at least one parallelepiped that satisfies the problem statement.
|
Print a single number β the sum of all edges of the parallelepiped.
|
[
"1 1 1\n",
"4 6 6\n"
] |
[
"12\n",
"28\n"
] |
In the first sample the parallelepiped has sizes 1βΓβ1βΓβ1, in the second oneΒ β 2βΓβ2βΓβ3.
| 500
|
[
{
"input": "1 1 1",
"output": "12"
},
{
"input": "4 6 6",
"output": "28"
},
{
"input": "20 10 50",
"output": "68"
},
{
"input": "9 4 36",
"output": "56"
},
{
"input": "324 9 36",
"output": "184"
},
{
"input": "1333 93 129",
"output": "308"
},
{
"input": "1022 584 112",
"output": "380"
},
{
"input": "66 174 319",
"output": "184"
},
{
"input": "912 276 1748",
"output": "444"
},
{
"input": "65 156 60",
"output": "120"
},
{
"input": "1 10000 10000",
"output": "40008"
},
{
"input": "1485 55 27",
"output": "332"
},
{
"input": "152 108 4104",
"output": "528"
},
{
"input": "1656 6900 1350",
"output": "740"
},
{
"input": "12 14 42",
"output": "60"
},
{
"input": "615 18 1230",
"output": "856"
},
{
"input": "680 60 408",
"output": "336"
},
{
"input": "644 966 6",
"output": "1308"
},
{
"input": "1 432 432",
"output": "1736"
},
{
"input": "2239 2239 1",
"output": "8964"
},
{
"input": "4106 8212 2",
"output": "16436"
},
{
"input": "10000 10000 10000",
"output": "1200"
},
{
"input": "3623 3623 1",
"output": "14500"
},
{
"input": "9801 9801 9801",
"output": "1188"
},
{
"input": "10000 1 10000",
"output": "40008"
},
{
"input": "9 9 9",
"output": "36"
},
{
"input": "9801 9702 9702",
"output": "1184"
}
] | 1,692,183,711
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 27
| 92
| 0
|
import math
z = []
z[0:] = map(int, input().split())
z.sort()
lst1 = []
lst2 = []
lst3 = []
lst = [1]
for i in range(1,z[1]):
if i not in lst and z[0] % i == 0 and z[2] % i == 0: #and i not in lst:# and z[2] % i == 0:
lst.append(i)
#break
if i not in lst and z[0] % i == 0 and z[1] % i == 0: #and i not in lst:# and z[2] % i == 0:
lst.append(i)
#break
#print(lst)
count = sum(z)
if len(lst1) == 0:
lst1.append(1)
lst1.append(z[1] // lst1[0])
lst1.append(z[2] // lst1[1])
else:
lst1.append(z[1] // lst1[0])
lst1.append(z[2] // lst1[1])
for i in range(len(lst)):
if (lst1[0] * lst1[1] + lst1[0] * lst1[2] + lst1[1] * lst1[2]) != count:
lst1.clear()
lst1.append(lst[len(lst1) - 1 -i])
lst1.append(z[1] // lst1[0])
lst1.append(z[2] // lst1[1])
else:
break
print(sum(lst1) * 4)
|
Title: Parallelepiped
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You've got a rectangular parallelepiped with integer edge lengths. You know the areas of its three faces that have a common vertex. Your task is to find the sum of lengths of all 12 edges of this parallelepiped.
Input Specification:
The first and the single line contains three space-separated integers β the areas of the parallelepiped's faces. The area's values are positive (<=><=0) and do not exceed 104. It is guaranteed that there exists at least one parallelepiped that satisfies the problem statement.
Output Specification:
Print a single number β the sum of all edges of the parallelepiped.
Demo Input:
['1 1 1\n', '4 6 6\n']
Demo Output:
['12\n', '28\n']
Note:
In the first sample the parallelepiped has sizes 1βΓβ1βΓβ1, in the second oneΒ β 2βΓβ2βΓβ3.
|
```python
import math
z = []
z[0:] = map(int, input().split())
z.sort()
lst1 = []
lst2 = []
lst3 = []
lst = [1]
for i in range(1,z[1]):
if i not in lst and z[0] % i == 0 and z[2] % i == 0: #and i not in lst:# and z[2] % i == 0:
lst.append(i)
#break
if i not in lst and z[0] % i == 0 and z[1] % i == 0: #and i not in lst:# and z[2] % i == 0:
lst.append(i)
#break
#print(lst)
count = sum(z)
if len(lst1) == 0:
lst1.append(1)
lst1.append(z[1] // lst1[0])
lst1.append(z[2] // lst1[1])
else:
lst1.append(z[1] // lst1[0])
lst1.append(z[2] // lst1[1])
for i in range(len(lst)):
if (lst1[0] * lst1[1] + lst1[0] * lst1[2] + lst1[1] * lst1[2]) != count:
lst1.clear()
lst1.append(lst[len(lst1) - 1 -i])
lst1.append(z[1] // lst1[0])
lst1.append(z[2] // lst1[1])
else:
break
print(sum(lst1) * 4)
```
| 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<=β€<=*n*<=β€<=100) β the number of orange-containing drinks in Vasya's fridge. The second line contains *n* integers *p**i* (0<=β€<=*p**i*<=β€<=100) β 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 3Β·*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"
}
] | 1,692,529,668
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 31
| 124
| 0
|
n = int(input())
def sum(l,n):
sum1 = 0
for i in range(n):
sum1 += l[i]
return sum1
l = [int(x) for x in input().split()]
sum(l, n)
print(sum(l,n)/n)
|
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<=β€<=*n*<=β€<=100) β the number of orange-containing drinks in Vasya's fridge. The second line contains *n* integers *p**i* (0<=β€<=*p**i*<=β€<=100) β 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 3Β·*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
n = int(input())
def sum(l,n):
sum1 = 0
for i in range(n):
sum1 += l[i]
return sum1
l = [int(x) for x in input().split()]
sum(l, n)
print(sum(l,n)/n)
```
| 3
|
|
242
|
C
|
King's Path
|
PROGRAMMING
| 1,800
|
[
"dfs and similar",
"graphs",
"hashing",
"shortest paths"
] | null | null |
The black king is standing on a chess field consisting of 109 rows and 109 columns. We will consider the rows of the field numbered with integers from 1 to 109 from top to bottom. The columns are similarly numbered with integers from 1 to 109 from left to right. We will denote a cell of the field that is located in the *i*-th row and *j*-th column as (*i*,<=*j*).
You know that some squares of the given chess field are allowed. All allowed cells of the chess field are given as *n* segments. Each segment is described by three integers *r**i*,<=*a**i*,<=*b**i* (*a**i*<=β€<=*b**i*), denoting that cells in columns from number *a**i* to number *b**i* inclusive in the *r**i*-th row are allowed.
Your task is to find the minimum number of moves the king needs to get from square (*x*0,<=*y*0) to square (*x*1,<=*y*1), provided that he only moves along the allowed cells. In other words, the king can be located only on allowed cells on his way.
Let us remind you that a chess king can move to any of the neighboring cells in one move. Two cells of a chess field are considered neighboring if they share at least one point.
|
The first line contains four space-separated integers *x*0,<=*y*0,<=*x*1,<=*y*1 (1<=β€<=*x*0,<=*y*0,<=*x*1,<=*y*1<=β€<=109), denoting the initial and the final positions of the king.
The second line contains a single integer *n* (1<=β€<=*n*<=β€<=105), denoting the number of segments of allowed cells. Next *n* lines contain the descriptions of these segments. The *i*-th line contains three space-separated integers *r**i*,<=*a**i*,<=*b**i* (1<=β€<=*r**i*,<=*a**i*,<=*b**i*<=β€<=109,<=*a**i*<=β€<=*b**i*), denoting that cells in columns from number *a**i* to number *b**i* inclusive in the *r**i*-th row are allowed. Note that the segments of the allowed cells can intersect and embed arbitrarily.
It is guaranteed that the king's initial and final position are allowed cells. It is guaranteed that the king's initial and the final positions do not coincide. It is guaranteed that the total length of all given segments doesn't exceed 105.
|
If there is no path between the initial and final position along allowed cells, print -1.
Otherwise print a single integer β the minimum number of moves the king needs to get from the initial position to the final one.
|
[
"5 7 6 11\n3\n5 3 8\n6 7 11\n5 2 5\n",
"3 4 3 10\n3\n3 1 4\n4 5 9\n3 10 10\n",
"1 1 2 10\n2\n1 1 3\n2 6 10\n"
] |
[
"4\n",
"6\n",
"-1\n"
] |
none
| 1,500
|
[
{
"input": "5 7 6 11\n3\n5 3 8\n6 7 11\n5 2 5",
"output": "4"
},
{
"input": "3 4 3 10\n3\n3 1 4\n4 5 9\n3 10 10",
"output": "6"
},
{
"input": "1 1 2 10\n2\n1 1 3\n2 6 10",
"output": "-1"
},
{
"input": "9 8 7 8\n9\n10 6 6\n10 6 6\n7 7 8\n9 5 6\n8 9 9\n9 5 5\n9 8 8\n8 5 6\n9 10 10",
"output": "2"
},
{
"input": "6 15 7 15\n9\n6 15 15\n7 14 14\n6 15 15\n9 14 14\n7 14 16\n6 15 15\n6 15 15\n7 14 14\n8 15 15",
"output": "1"
},
{
"input": "13 16 20 10\n18\n13 16 16\n20 10 10\n19 10 10\n12 15 15\n20 10 10\n18 11 11\n19 10 10\n19 10 10\n20 10 10\n19 10 10\n20 10 10\n20 10 10\n19 10 10\n18 11 11\n13 16 16\n12 15 15\n19 10 10\n19 10 10",
"output": "-1"
},
{
"input": "89 29 88 30\n16\n87 31 31\n14 95 95\n98 88 89\n96 88 88\n14 97 97\n13 97 98\n100 88 88\n88 32 32\n99 88 89\n90 29 29\n87 31 31\n15 94 96\n89 29 29\n88 32 32\n97 89 89\n88 29 30",
"output": "1"
},
{
"input": "30 14 39 19\n31\n35 7 11\n37 11 12\n32 13 13\n37 5 6\n46 13 13\n37 14 14\n31 13 13\n43 13 19\n45 15 19\n46 13 13\n32 17 17\n41 14 19\n30 14 14\n43 13 17\n34 16 18\n44 11 19\n38 13 13\n40 12 20\n37 16 18\n46 16 18\n34 10 14\n36 9 10\n36 15 19\n38 15 19\n42 13 19\n33 14 15\n35 15 19\n33 17 18\n39 12 20\n36 5 7\n45 12 12",
"output": "9"
},
{
"input": "2 1 1 1\n2\n1 1 2\n2 1 2",
"output": "1"
},
{
"input": "1 1 1 2\n5\n1000000000 1 10000\n19920401 1188 5566\n1000000000 1 10000\n1 1 10000\n5 100 200",
"output": "1"
},
{
"input": "1 1 1000000000 2\n5\n1000000000 1 10000\n19920401 1188 5566\n1000000000 1 10000\n1 1 10000\n5 100 200",
"output": "-1"
}
] | 1,638,729,291
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 43
| 686
| 11,980,800
|
import sys
import math
from sys import stdin, stdout
# TAKE INPUT
def get_ints_in_variables():
return map(int, sys.stdin.readline().strip().split())
def get_int(): return int(input())
def get_ints_in_list(): return list(
map(int, sys.stdin.readline().strip().split()))
def get_list_of_list(n): return [list(
map(int, sys.stdin.readline().strip().split())) for _ in range(n)]
def get_string(): return sys.stdin.readline().strip()
def main():
# Write Your Code Here
x0,y0,x1,y1 = get_ints_in_variables()
n = int(input())
g = {}
for i in range(0, n):
r,a,b = get_ints_in_variables()
for i in range(a, b+1):
g[(r, i)] = -1
g[(x0, y0)] = 0
# {[i-1][0], [i-1][j-1], [i-1][j+1], [i][j-1], [i][j+1], [i+1][j], [i+1][j-1], [i+1][j+1]}
dx = [-1, -1, -1, 0, 0, 1, 1, 1]
dy = [0, -1, 1, -1, 1, 0, -1, 1]
queue = [(x0, y0)]
while len(queue):
node = queue.pop(0)
for j in range(8):
v = (node[0]+dx[j], node[1]+dy[j])
# print(v, "v")
if not v in g:
continue
if v in g and g[v] != -1:
continue
g[v] = g[node]+1
queue.append(v)
print(g[(x1, y1)])
# calling main Function
if __name__ == "__main__":
main()
|
Title: King's Path
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The black king is standing on a chess field consisting of 109 rows and 109 columns. We will consider the rows of the field numbered with integers from 1 to 109 from top to bottom. The columns are similarly numbered with integers from 1 to 109 from left to right. We will denote a cell of the field that is located in the *i*-th row and *j*-th column as (*i*,<=*j*).
You know that some squares of the given chess field are allowed. All allowed cells of the chess field are given as *n* segments. Each segment is described by three integers *r**i*,<=*a**i*,<=*b**i* (*a**i*<=β€<=*b**i*), denoting that cells in columns from number *a**i* to number *b**i* inclusive in the *r**i*-th row are allowed.
Your task is to find the minimum number of moves the king needs to get from square (*x*0,<=*y*0) to square (*x*1,<=*y*1), provided that he only moves along the allowed cells. In other words, the king can be located only on allowed cells on his way.
Let us remind you that a chess king can move to any of the neighboring cells in one move. Two cells of a chess field are considered neighboring if they share at least one point.
Input Specification:
The first line contains four space-separated integers *x*0,<=*y*0,<=*x*1,<=*y*1 (1<=β€<=*x*0,<=*y*0,<=*x*1,<=*y*1<=β€<=109), denoting the initial and the final positions of the king.
The second line contains a single integer *n* (1<=β€<=*n*<=β€<=105), denoting the number of segments of allowed cells. Next *n* lines contain the descriptions of these segments. The *i*-th line contains three space-separated integers *r**i*,<=*a**i*,<=*b**i* (1<=β€<=*r**i*,<=*a**i*,<=*b**i*<=β€<=109,<=*a**i*<=β€<=*b**i*), denoting that cells in columns from number *a**i* to number *b**i* inclusive in the *r**i*-th row are allowed. Note that the segments of the allowed cells can intersect and embed arbitrarily.
It is guaranteed that the king's initial and final position are allowed cells. It is guaranteed that the king's initial and the final positions do not coincide. It is guaranteed that the total length of all given segments doesn't exceed 105.
Output Specification:
If there is no path between the initial and final position along allowed cells, print -1.
Otherwise print a single integer β the minimum number of moves the king needs to get from the initial position to the final one.
Demo Input:
['5 7 6 11\n3\n5 3 8\n6 7 11\n5 2 5\n', '3 4 3 10\n3\n3 1 4\n4 5 9\n3 10 10\n', '1 1 2 10\n2\n1 1 3\n2 6 10\n']
Demo Output:
['4\n', '6\n', '-1\n']
Note:
none
|
```python
import sys
import math
from sys import stdin, stdout
# TAKE INPUT
def get_ints_in_variables():
return map(int, sys.stdin.readline().strip().split())
def get_int(): return int(input())
def get_ints_in_list(): return list(
map(int, sys.stdin.readline().strip().split()))
def get_list_of_list(n): return [list(
map(int, sys.stdin.readline().strip().split())) for _ in range(n)]
def get_string(): return sys.stdin.readline().strip()
def main():
# Write Your Code Here
x0,y0,x1,y1 = get_ints_in_variables()
n = int(input())
g = {}
for i in range(0, n):
r,a,b = get_ints_in_variables()
for i in range(a, b+1):
g[(r, i)] = -1
g[(x0, y0)] = 0
# {[i-1][0], [i-1][j-1], [i-1][j+1], [i][j-1], [i][j+1], [i+1][j], [i+1][j-1], [i+1][j+1]}
dx = [-1, -1, -1, 0, 0, 1, 1, 1]
dy = [0, -1, 1, -1, 1, 0, -1, 1]
queue = [(x0, y0)]
while len(queue):
node = queue.pop(0)
for j in range(8):
v = (node[0]+dx[j], node[1]+dy[j])
# print(v, "v")
if not v in g:
continue
if v in g and g[v] != -1:
continue
g[v] = g[node]+1
queue.append(v)
print(g[(x1, y1)])
# calling main Function
if __name__ == "__main__":
main()
```
| 3
|
|
910
|
B
|
Door Frames
|
PROGRAMMING
| 1,600
|
[
"greedy",
"implementation"
] | null | null |
Petya has equal wooden bars of length *n*. He wants to make a frame for two equal doors. Each frame has two vertical (left and right) sides of length *a* and one top side of length *b*. A solid (i.e. continuous without breaks) piece of bar is needed for each side.
Determine a minimal number of wooden bars which are needed to make the frames for two doors. Petya can cut the wooden bars into any parts, but each side of each door should be a solid piece of a wooden bar (or a whole wooden bar).
|
The first line contains a single integer *n* (1<=β€<=*n*<=β€<=1<=000) β the length of each wooden bar.
The second line contains a single integer *a* (1<=β€<=*a*<=β€<=*n*) β the length of the vertical (left and right) sides of a door frame.
The third line contains a single integer *b* (1<=β€<=*b*<=β€<=*n*) β the length of the upper side of a door frame.
|
Print the minimal number of wooden bars with length *n* which are needed to make the frames for two doors.
|
[
"8\n1\n2\n",
"5\n3\n4\n",
"6\n4\n2\n",
"20\n5\n6\n"
] |
[
"1\n",
"6\n",
"4\n",
"2\n"
] |
In the first example one wooden bar is enough, since the total length of all six sides of the frames for two doors is 8.
In the second example 6 wooden bars is enough, because for each side of the frames the new wooden bar is needed.
| 1,000
|
[
{
"input": "8\n1\n2",
"output": "1"
},
{
"input": "5\n3\n4",
"output": "6"
},
{
"input": "6\n4\n2",
"output": "4"
},
{
"input": "20\n5\n6",
"output": "2"
},
{
"input": "1\n1\n1",
"output": "6"
},
{
"input": "3\n1\n2",
"output": "3"
},
{
"input": "3\n2\n1",
"output": "4"
},
{
"input": "1000\n1\n1",
"output": "1"
},
{
"input": "1000\n1000\n1000",
"output": "6"
},
{
"input": "1000\n1\n999",
"output": "3"
},
{
"input": "1000\n1\n498",
"output": "1"
},
{
"input": "1000\n1\n998",
"output": "2"
},
{
"input": "31\n5\n6",
"output": "2"
},
{
"input": "400\n100\n2",
"output": "2"
},
{
"input": "399\n100\n2",
"output": "2"
},
{
"input": "800\n401\n400",
"output": "5"
},
{
"input": "141\n26\n11",
"output": "1"
},
{
"input": "717\n40\n489",
"output": "2"
},
{
"input": "293\n47\n30",
"output": "1"
},
{
"input": "165\n59\n40",
"output": "2"
},
{
"input": "404\n5\n183",
"output": "1"
},
{
"input": "828\n468\n726",
"output": "6"
},
{
"input": "956\n153\n941",
"output": "3"
},
{
"input": "676\n175\n514",
"output": "4"
},
{
"input": "296\n1\n10",
"output": "1"
},
{
"input": "872\n3\n182",
"output": "1"
},
{
"input": "448\n15\n126",
"output": "1"
},
{
"input": "24\n2\n5",
"output": "1"
},
{
"input": "289\n56\n26",
"output": "1"
},
{
"input": "713\n150\n591",
"output": "3"
},
{
"input": "841\n62\n704",
"output": "2"
},
{
"input": "266\n38\n164",
"output": "2"
},
{
"input": "156\n34\n7",
"output": "1"
},
{
"input": "28\n14\n9",
"output": "3"
},
{
"input": "604\n356\n239",
"output": "4"
},
{
"input": "180\n18\n76",
"output": "2"
},
{
"input": "879\n545\n607",
"output": "6"
},
{
"input": "599\n160\n520",
"output": "4"
},
{
"input": "727\n147\n693",
"output": "3"
},
{
"input": "151\n27\n135",
"output": "3"
},
{
"input": "504\n71\n73",
"output": "1"
},
{
"input": "80\n57\n31",
"output": "5"
},
{
"input": "951\n225\n352",
"output": "2"
},
{
"input": "823\n168\n141",
"output": "2"
},
{
"input": "956\n582\n931",
"output": "6"
},
{
"input": "380\n108\n356",
"output": "4"
},
{
"input": "804\n166\n472",
"output": "2"
},
{
"input": "228\n12\n159",
"output": "2"
},
{
"input": "380\n126\n82",
"output": "2"
},
{
"input": "252\n52\n178",
"output": "3"
},
{
"input": "828\n363\n56",
"output": "2"
},
{
"input": "404\n122\n36",
"output": "2"
},
{
"input": "314\n4\n237",
"output": "2"
},
{
"input": "34\n5\n17",
"output": "2"
},
{
"input": "162\n105\n160",
"output": "6"
},
{
"input": "586\n22\n272",
"output": "2"
},
{
"input": "32\n9\n2",
"output": "2"
},
{
"input": "904\n409\n228",
"output": "3"
},
{
"input": "480\n283\n191",
"output": "4"
},
{
"input": "56\n37\n10",
"output": "4"
},
{
"input": "429\n223\n170",
"output": "4"
},
{
"input": "149\n124\n129",
"output": "6"
},
{
"input": "277\n173\n241",
"output": "6"
},
{
"input": "701\n211\n501",
"output": "4"
},
{
"input": "172\n144\n42",
"output": "5"
},
{
"input": "748\n549\n256",
"output": "5"
},
{
"input": "324\n284\n26",
"output": "4"
},
{
"input": "900\n527\n298",
"output": "4"
},
{
"input": "648\n624\n384",
"output": "6"
},
{
"input": "72\n48\n54",
"output": "6"
},
{
"input": "200\n194\n87",
"output": "5"
},
{
"input": "624\n510\n555",
"output": "6"
},
{
"input": "17\n16\n2",
"output": "5"
},
{
"input": "593\n442\n112",
"output": "4"
},
{
"input": "169\n158\n11",
"output": "4"
},
{
"input": "41\n38\n17",
"output": "5"
},
{
"input": "762\n609\n442",
"output": "6"
},
{
"input": "186\n98\n104",
"output": "6"
},
{
"input": "314\n304\n294",
"output": "6"
},
{
"input": "35\n35\n33",
"output": "6"
},
{
"input": "8\n3\n5",
"output": "3"
},
{
"input": "11\n3\n5",
"output": "2"
},
{
"input": "5\n4\n2",
"output": "5"
},
{
"input": "41\n5\n36",
"output": "3"
},
{
"input": "7\n4\n1",
"output": "4"
},
{
"input": "6\n1\n4",
"output": "2"
},
{
"input": "597\n142\n484",
"output": "3"
},
{
"input": "6\n6\n1",
"output": "5"
},
{
"input": "8\n4\n2",
"output": "3"
},
{
"input": "4\n1\n4",
"output": "3"
},
{
"input": "7\n2\n3",
"output": "2"
},
{
"input": "100\n100\n50",
"output": "5"
},
{
"input": "5\n1\n3",
"output": "2"
},
{
"input": "10\n4\n6",
"output": "3"
},
{
"input": "8\n8\n2",
"output": "5"
},
{
"input": "5\n2\n4",
"output": "4"
},
{
"input": "11\n5\n3",
"output": "3"
},
{
"input": "668\n248\n336",
"output": "3"
},
{
"input": "2\n2\n1",
"output": "5"
},
{
"input": "465\n126\n246",
"output": "3"
},
{
"input": "5\n1\n5",
"output": "3"
},
{
"input": "132\n34\n64",
"output": "2"
},
{
"input": "11\n1\n6",
"output": "2"
},
{
"input": "8\n4\n5",
"output": "4"
},
{
"input": "4\n2\n4",
"output": "4"
},
{
"input": "576\n238\n350",
"output": "4"
},
{
"input": "6\n1\n5",
"output": "3"
},
{
"input": "5\n1\n4",
"output": "3"
},
{
"input": "9\n2\n8",
"output": "3"
},
{
"input": "7\n3\n4",
"output": "3"
},
{
"input": "9\n4\n5",
"output": "3"
},
{
"input": "10\n3\n4",
"output": "2"
},
{
"input": "18\n5\n8",
"output": "2"
},
{
"input": "2\n1\n1",
"output": "3"
},
{
"input": "100\n40\n60",
"output": "3"
},
{
"input": "6\n4\n4",
"output": "6"
},
{
"input": "3\n1\n1",
"output": "2"
},
{
"input": "10\n3\n7",
"output": "3"
},
{
"input": "9\n2\n5",
"output": "2"
},
{
"input": "6\n2\n3",
"output": "3"
}
] | 1,535,312,179
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 120
| 124
| 0
|
import math
[n,a,b],r,i,j=[int(input())for x in range(3)],6,4,5
while i>=0:
l,c,o=[b if x in[i,j]else a for x in range(6)],0,n
for k in l:
if o<k:
o,c=n-k,c+1
else:o-=k
r=min(r,c if o==n else c+1)
j-=1
if i==j:i,j=i-1,5
print(r)
# Made By Mostafa_Khaled
|
Title: Door Frames
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya has equal wooden bars of length *n*. He wants to make a frame for two equal doors. Each frame has two vertical (left and right) sides of length *a* and one top side of length *b*. A solid (i.e. continuous without breaks) piece of bar is needed for each side.
Determine a minimal number of wooden bars which are needed to make the frames for two doors. Petya can cut the wooden bars into any parts, but each side of each door should be a solid piece of a wooden bar (or a whole wooden bar).
Input Specification:
The first line contains a single integer *n* (1<=β€<=*n*<=β€<=1<=000) β the length of each wooden bar.
The second line contains a single integer *a* (1<=β€<=*a*<=β€<=*n*) β the length of the vertical (left and right) sides of a door frame.
The third line contains a single integer *b* (1<=β€<=*b*<=β€<=*n*) β the length of the upper side of a door frame.
Output Specification:
Print the minimal number of wooden bars with length *n* which are needed to make the frames for two doors.
Demo Input:
['8\n1\n2\n', '5\n3\n4\n', '6\n4\n2\n', '20\n5\n6\n']
Demo Output:
['1\n', '6\n', '4\n', '2\n']
Note:
In the first example one wooden bar is enough, since the total length of all six sides of the frames for two doors is 8.
In the second example 6 wooden bars is enough, because for each side of the frames the new wooden bar is needed.
|
```python
import math
[n,a,b],r,i,j=[int(input())for x in range(3)],6,4,5
while i>=0:
l,c,o=[b if x in[i,j]else a for x in range(6)],0,n
for k in l:
if o<k:
o,c=n-k,c+1
else:o-=k
r=min(r,c if o==n else c+1)
j-=1
if i==j:i,j=i-1,5
print(r)
# Made By Mostafa_Khaled
```
| 3
|
|
214
|
A
|
System of Equations
|
PROGRAMMING
| 800
|
[
"brute force"
] | null | null |
Furik loves math lessons very much, so he doesn't attend them, unlike Rubik. But now Furik wants to get a good mark for math. For that Ms. Ivanova, his math teacher, gave him a new task. Furik solved the task immediately. Can you?
You are given a system of equations:
You should count, how many there are pairs of integers (*a*,<=*b*) (0<=β€<=*a*,<=*b*) which satisfy the system.
|
A single line contains two integers *n*,<=*m* (1<=β€<=*n*,<=*m*<=β€<=1000) β the parameters of the system. The numbers on the line are separated by a space.
|
On a single line print the answer to the problem.
|
[
"9 3\n",
"14 28\n",
"4 20\n"
] |
[
"1\n",
"1\n",
"0\n"
] |
In the first sample the suitable pair is integers (3,β0). In the second sample the suitable pair is integers (3,β5). In the third sample there is no suitable pair.
| 500
|
[
{
"input": "9 3",
"output": "1"
},
{
"input": "14 28",
"output": "1"
},
{
"input": "4 20",
"output": "0"
},
{
"input": "18 198",
"output": "1"
},
{
"input": "22 326",
"output": "1"
},
{
"input": "26 104",
"output": "1"
},
{
"input": "14 10",
"output": "0"
},
{
"input": "8 20",
"output": "0"
},
{
"input": "2 8",
"output": "0"
},
{
"input": "20 11",
"output": "0"
},
{
"input": "57 447",
"output": "1"
},
{
"input": "1 1",
"output": "2"
},
{
"input": "66 296",
"output": "1"
},
{
"input": "75 683",
"output": "1"
},
{
"input": "227 975",
"output": "1"
},
{
"input": "247 499",
"output": "1"
},
{
"input": "266 116",
"output": "1"
},
{
"input": "286 916",
"output": "1"
},
{
"input": "307 341",
"output": "1"
},
{
"input": "451 121",
"output": "1"
},
{
"input": "471 921",
"output": "1"
},
{
"input": "502 346",
"output": "1"
},
{
"input": "535 59",
"output": "1"
},
{
"input": "555 699",
"output": "1"
},
{
"input": "747 351",
"output": "1"
},
{
"input": "790 64",
"output": "1"
},
{
"input": "810 704",
"output": "1"
},
{
"input": "855 225",
"output": "1"
},
{
"input": "902 34",
"output": "1"
},
{
"input": "922 514",
"output": "1"
},
{
"input": "971 131",
"output": "1"
},
{
"input": "991 931",
"output": "1"
},
{
"input": "840 780",
"output": "0"
},
{
"input": "102 595",
"output": "0"
},
{
"input": "139 433",
"output": "0"
},
{
"input": "968 288",
"output": "0"
},
{
"input": "563 354",
"output": "0"
},
{
"input": "994 975",
"output": "0"
},
{
"input": "456 221",
"output": "0"
},
{
"input": "205 210",
"output": "0"
},
{
"input": "1 11",
"output": "0"
},
{
"input": "1000 1000",
"output": "0"
},
{
"input": "3 3",
"output": "0"
},
{
"input": "11 99",
"output": "0"
},
{
"input": "2 2",
"output": "1"
},
{
"input": "11 1",
"output": "0"
},
{
"input": "6 6",
"output": "1"
},
{
"input": "100 452",
"output": "0"
},
{
"input": "420 380",
"output": "0"
},
{
"input": "31 31",
"output": "0"
},
{
"input": "2 6",
"output": "0"
},
{
"input": "112 134",
"output": "0"
},
{
"input": "13 13",
"output": "0"
},
{
"input": "1 571",
"output": "0"
}
] | 1,656,565,182
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 54
| 218
| 2,048,000
|
n,m=map(int,input().split())
c=0
for a in range(300):
for b in range(300):
if a**2+b==n and a+b**2==m:
c+=1
print(c)
|
Title: System of Equations
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Furik loves math lessons very much, so he doesn't attend them, unlike Rubik. But now Furik wants to get a good mark for math. For that Ms. Ivanova, his math teacher, gave him a new task. Furik solved the task immediately. Can you?
You are given a system of equations:
You should count, how many there are pairs of integers (*a*,<=*b*) (0<=β€<=*a*,<=*b*) which satisfy the system.
Input Specification:
A single line contains two integers *n*,<=*m* (1<=β€<=*n*,<=*m*<=β€<=1000) β the parameters of the system. The numbers on the line are separated by a space.
Output Specification:
On a single line print the answer to the problem.
Demo Input:
['9 3\n', '14 28\n', '4 20\n']
Demo Output:
['1\n', '1\n', '0\n']
Note:
In the first sample the suitable pair is integers (3,β0). In the second sample the suitable pair is integers (3,β5). In the third sample there is no suitable pair.
|
```python
n,m=map(int,input().split())
c=0
for a in range(300):
for b in range(300):
if a**2+b==n and a+b**2==m:
c+=1
print(c)
```
| 3
|
|
884
|
B
|
Japanese Crosswords Strike Back
|
PROGRAMMING
| 1,100
|
[
"implementation"
] | null | null |
A one-dimensional Japanese crossword can be represented as a binary string of length *x*. An encoding of this crossword is an array *a* of size *n*, where *n* is the number of segments formed completely of 1's, and *a**i* is the length of *i*-th segment. No two segments touch or intersect.
For example:
- If *x*<==<=6 and the crossword is 111011, then its encoding is an array {3,<=2}; - If *x*<==<=8 and the crossword is 01101010, then its encoding is an array {2,<=1,<=1}; - If *x*<==<=5 and the crossword is 11111, then its encoding is an array {5}; - If *x*<==<=5 and the crossword is 00000, then its encoding is an empty array.
Mishka wants to create a new one-dimensional Japanese crossword. He has already picked the length and the encoding for this crossword. And now he needs to check if there is exactly one crossword such that its length and encoding are equal to the length and encoding he picked. Help him to check it!
|
The first line contains two integer numbers *n* and *x* (1<=β€<=*n*<=β€<=100000, 1<=β€<=*x*<=β€<=109) β the number of elements in the encoding and the length of the crossword Mishka picked.
The second line contains *n* integer numbers *a*1, *a*2, ..., *a**n* (1<=β€<=*a**i*<=β€<=10000) β the encoding.
|
Print YES if there exists exaclty one crossword with chosen length and encoding. Otherwise, print NO.
|
[
"2 4\n1 3\n",
"3 10\n3 3 2\n",
"2 10\n1 3\n"
] |
[
"NO\n",
"YES\n",
"NO\n"
] |
none
| 0
|
[
{
"input": "2 4\n1 3",
"output": "NO"
},
{
"input": "3 10\n3 3 2",
"output": "YES"
},
{
"input": "2 10\n1 3",
"output": "NO"
},
{
"input": "1 1\n1",
"output": "YES"
},
{
"input": "1 10\n10",
"output": "YES"
},
{
"input": "1 10000\n10000",
"output": "YES"
},
{
"input": "10 1\n5 78 3 87 4 9 5 8 9 1235",
"output": "NO"
},
{
"input": "3 12\n3 3 3",
"output": "NO"
},
{
"input": "3 9\n2 2 2",
"output": "NO"
},
{
"input": "2 5\n1 1",
"output": "NO"
},
{
"input": "1 2\n1",
"output": "NO"
},
{
"input": "3 13\n3 3 3",
"output": "NO"
},
{
"input": "3 6\n1 1 1",
"output": "NO"
},
{
"input": "1 6\n5",
"output": "NO"
},
{
"input": "3 11\n3 3 2",
"output": "NO"
},
{
"input": "2 6\n1 3",
"output": "NO"
},
{
"input": "3 10\n2 2 2",
"output": "NO"
},
{
"input": "3 8\n2 1 1",
"output": "NO"
},
{
"input": "1 5\n2",
"output": "NO"
},
{
"input": "1 3\n1",
"output": "NO"
},
{
"input": "5 5\n1 1 1 1 1",
"output": "NO"
},
{
"input": "2 10\n4 4",
"output": "NO"
},
{
"input": "2 8\n2 3",
"output": "NO"
},
{
"input": "2 4\n1 1",
"output": "NO"
},
{
"input": "3 10\n1 2 4",
"output": "NO"
},
{
"input": "3 10\n2 1 3",
"output": "NO"
},
{
"input": "2 6\n1 2",
"output": "NO"
},
{
"input": "3 4\n1 1 1",
"output": "NO"
},
{
"input": "3 11\n1 2 4",
"output": "NO"
},
{
"input": "3 12\n3 3 2",
"output": "NO"
},
{
"input": "4 9\n1 1 1 1",
"output": "NO"
},
{
"input": "1 10\n9",
"output": "NO"
},
{
"input": "1 7\n5",
"output": "NO"
},
{
"input": "2 5\n1 2",
"output": "NO"
},
{
"input": "3 8\n1 1 2",
"output": "NO"
},
{
"input": "3 7\n1 1 1",
"output": "NO"
},
{
"input": "1 10\n1",
"output": "NO"
},
{
"input": "3 10\n2 3 4",
"output": "NO"
},
{
"input": "3 9\n1 2 3",
"output": "NO"
},
{
"input": "3 9\n3 3 2",
"output": "NO"
},
{
"input": "3 6\n3 3 2",
"output": "NO"
},
{
"input": "1 1\n3",
"output": "NO"
},
{
"input": "1 3\n2",
"output": "NO"
},
{
"input": "3 10\n3 3 3",
"output": "NO"
},
{
"input": "3 5\n1 1 1",
"output": "YES"
},
{
"input": "2 1\n100 100",
"output": "NO"
},
{
"input": "3 3\n3 3 3",
"output": "NO"
},
{
"input": "4 17\n3 3 9 1",
"output": "NO"
},
{
"input": "4 1660\n1505 13 37 100",
"output": "NO"
},
{
"input": "3 5\n3 3 2",
"output": "NO"
},
{
"input": "4 10\n3 3 2 5",
"output": "NO"
},
{
"input": "5 5\n5 5 5 5 5",
"output": "NO"
},
{
"input": "1 1\n2",
"output": "NO"
},
{
"input": "5 10\n1 2 2 4 5",
"output": "NO"
},
{
"input": "2 1\n1 1",
"output": "NO"
}
] | 1,584,995,645
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 66
| 139
| 6,963,200
|
n,x=map(int,input().split(' '))
val=[int(i) for i in input().split(' ') ]
s=len(val)-1
for i in val:
s+=i
if s==x:
print('YES')
else:
print('NO')
|
Title: Japanese Crosswords Strike Back
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A one-dimensional Japanese crossword can be represented as a binary string of length *x*. An encoding of this crossword is an array *a* of size *n*, where *n* is the number of segments formed completely of 1's, and *a**i* is the length of *i*-th segment. No two segments touch or intersect.
For example:
- If *x*<==<=6 and the crossword is 111011, then its encoding is an array {3,<=2}; - If *x*<==<=8 and the crossword is 01101010, then its encoding is an array {2,<=1,<=1}; - If *x*<==<=5 and the crossword is 11111, then its encoding is an array {5}; - If *x*<==<=5 and the crossword is 00000, then its encoding is an empty array.
Mishka wants to create a new one-dimensional Japanese crossword. He has already picked the length and the encoding for this crossword. And now he needs to check if there is exactly one crossword such that its length and encoding are equal to the length and encoding he picked. Help him to check it!
Input Specification:
The first line contains two integer numbers *n* and *x* (1<=β€<=*n*<=β€<=100000, 1<=β€<=*x*<=β€<=109) β the number of elements in the encoding and the length of the crossword Mishka picked.
The second line contains *n* integer numbers *a*1, *a*2, ..., *a**n* (1<=β€<=*a**i*<=β€<=10000) β the encoding.
Output Specification:
Print YES if there exists exaclty one crossword with chosen length and encoding. Otherwise, print NO.
Demo Input:
['2 4\n1 3\n', '3 10\n3 3 2\n', '2 10\n1 3\n']
Demo Output:
['NO\n', 'YES\n', 'NO\n']
Note:
none
|
```python
n,x=map(int,input().split(' '))
val=[int(i) for i in input().split(' ') ]
s=len(val)-1
for i in val:
s+=i
if s==x:
print('YES')
else:
print('NO')
```
| 3
|
|
996
|
A
|
Hit the Lottery
|
PROGRAMMING
| 800
|
[
"dp",
"greedy"
] | null | null |
Allen has a LOT of money. He has $n$ dollars in the bank. For security reasons, he wants to withdraw it in cash (we will not disclose the reasons here). The denominations for dollar bills are $1$, $5$, $10$, $20$, $100$. What is the minimum number of bills Allen could receive after withdrawing his entire balance?
|
The first and only line of input contains a single integer $n$ ($1 \le n \le 10^9$).
|
Output the minimum number of bills that Allen could receive.
|
[
"125\n",
"43\n",
"1000000000\n"
] |
[
"3\n",
"5\n",
"10000000\n"
] |
In the first sample case, Allen can withdraw this with a $100$ dollar bill, a $20$ dollar bill, and a $5$ dollar bill. There is no way for Allen to receive $125$ dollars in one or two bills.
In the second sample case, Allen can withdraw two $20$ dollar bills and three $1$ dollar bills.
In the third sample case, Allen can withdraw $100000000$ (ten million!) $100$ dollar bills.
| 500
|
[
{
"input": "125",
"output": "3"
},
{
"input": "43",
"output": "5"
},
{
"input": "1000000000",
"output": "10000000"
},
{
"input": "4",
"output": "4"
},
{
"input": "5",
"output": "1"
},
{
"input": "1",
"output": "1"
},
{
"input": "74",
"output": "8"
},
{
"input": "31",
"output": "3"
},
{
"input": "59",
"output": "8"
},
{
"input": "79",
"output": "9"
},
{
"input": "7",
"output": "3"
},
{
"input": "55",
"output": "4"
},
{
"input": "40",
"output": "2"
},
{
"input": "719",
"output": "13"
},
{
"input": "847",
"output": "13"
},
{
"input": "225",
"output": "4"
},
{
"input": "4704",
"output": "51"
},
{
"input": "1132",
"output": "15"
},
{
"input": "7811",
"output": "80"
},
{
"input": "7981",
"output": "84"
},
{
"input": "82655",
"output": "830"
},
{
"input": "6364",
"output": "70"
},
{
"input": "74611",
"output": "748"
},
{
"input": "45391",
"output": "459"
},
{
"input": "620448",
"output": "6210"
},
{
"input": "265145",
"output": "2654"
},
{
"input": "671704",
"output": "6721"
},
{
"input": "365173",
"output": "3658"
},
{
"input": "7130872",
"output": "71314"
},
{
"input": "9628747",
"output": "96292"
},
{
"input": "8898399",
"output": "88993"
},
{
"input": "9497953",
"output": "94985"
},
{
"input": "19070947",
"output": "190714"
},
{
"input": "20185520",
"output": "201856"
},
{
"input": "91402248",
"output": "914028"
},
{
"input": "27035533",
"output": "270360"
},
{
"input": "22717528",
"output": "227180"
},
{
"input": "24403439",
"output": "244041"
},
{
"input": "163565555",
"output": "1635659"
},
{
"input": "152541908",
"output": "1525423"
},
{
"input": "668439837",
"output": "6684403"
},
{
"input": "20",
"output": "1"
},
{
"input": "999999999",
"output": "10000009"
},
{
"input": "10",
"output": "1"
},
{
"input": "7",
"output": "3"
},
{
"input": "6",
"output": "2"
},
{
"input": "99999999",
"output": "1000009"
},
{
"input": "1000001",
"output": "10001"
},
{
"input": "2521",
"output": "27"
}
] | 1,698,825,161
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 49
| 46
| 0
|
x = int(input(""))
bills = 0
bills = bills + x//100
x = x%100
bills = bills + x//20
x = x%20
bills = bills + x//10
x = x%10
bills = bills + x//5
x = x%5
bills = bills + x//1
x = x%1
print(bills)
|
Title: Hit the Lottery
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Allen has a LOT of money. He has $n$ dollars in the bank. For security reasons, he wants to withdraw it in cash (we will not disclose the reasons here). The denominations for dollar bills are $1$, $5$, $10$, $20$, $100$. What is the minimum number of bills Allen could receive after withdrawing his entire balance?
Input Specification:
The first and only line of input contains a single integer $n$ ($1 \le n \le 10^9$).
Output Specification:
Output the minimum number of bills that Allen could receive.
Demo Input:
['125\n', '43\n', '1000000000\n']
Demo Output:
['3\n', '5\n', '10000000\n']
Note:
In the first sample case, Allen can withdraw this with a $100$ dollar bill, a $20$ dollar bill, and a $5$ dollar bill. There is no way for Allen to receive $125$ dollars in one or two bills.
In the second sample case, Allen can withdraw two $20$ dollar bills and three $1$ dollar bills.
In the third sample case, Allen can withdraw $100000000$ (ten million!) $100$ dollar bills.
|
```python
x = int(input(""))
bills = 0
bills = bills + x//100
x = x%100
bills = bills + x//20
x = x%20
bills = bills + x//10
x = x%10
bills = bills + x//5
x = x%5
bills = bills + x//1
x = x%1
print(bills)
```
| 3
|
|
61
|
B
|
Hard Work
|
PROGRAMMING
| 1,300
|
[
"strings"
] |
B. Hard Work
|
2
|
256
|
After the contest in comparing numbers, Shapur's teacher found out that he is a real genius and that no one could possibly do the calculations faster than him even using a super computer!
Some days before the contest, the teacher took a very simple-looking exam and all his *n* students took part in the exam. The teacher gave them 3 strings and asked them to concatenate them. Concatenating strings means to put them in some arbitrary order one after the other. For example from concatenating Alireza and Amir we can get to AlirezaAmir or AmirAlireza depending on the order of concatenation.
Unfortunately enough, the teacher forgot to ask students to concatenate their strings in a pre-defined order so each student did it the way he/she liked.
Now the teacher knows that Shapur is such a fast-calculating genius boy and asks him to correct the students' papers.
Shapur is not good at doing such a time-taking task. He rather likes to finish up with it as soon as possible and take his time to solve 3-SAT in polynomial time. Moreover, the teacher has given some advice that Shapur has to follow. Here's what the teacher said:
- As I expect you know, the strings I gave to my students (including you) contained only lowercase and uppercase Persian Mikhi-Script letters. These letters are too much like Latin letters, so to make your task much harder I converted all the initial strings and all of the students' answers to Latin. - As latin alphabet has much less characters than Mikhi-Script, I added three odd-looking characters to the answers, these include "-", ";" and "_". These characters are my own invention of course! And I call them Signs. - The length of all initial strings was less than or equal to 100 and the lengths of my students' answers are less than or equal to 600 - My son, not all students are genius as you are. It is quite possible that they make minor mistakes changing case of some characters. For example they may write ALiReZaAmIR instead of AlirezaAmir. Don't be picky and ignore these mistakes. - Those signs which I previously talked to you about are not important. You can ignore them, since many students are in the mood for adding extra signs or forgetting about a sign. So something like Iran;;-- is the same as --;IRAN - You should indicate for any of my students if his answer was right or wrong. Do this by writing "WA" for Wrong answer or "ACC" for a correct answer. - I should remind you that none of the strings (initial strings or answers) are empty. - Finally, do these as soon as possible. You have less than 2 hours to complete this.
|
The first three lines contain a string each. These are the initial strings. They consists only of lowercase and uppercase Latin letters and signs ("-", ";" and "_"). All the initial strings have length from 1 to 100, inclusively.
In the fourth line there is a single integer *n* (0<=β€<=*n*<=β€<=1000), the number of students.
Next *n* lines contain a student's answer each. It is guaranteed that the answer meets what the teacher said. Each answer iconsists only of lowercase and uppercase Latin letters and signs ("-", ";" and "_"). Length is from 1 to 600, inclusively.
|
For each student write in a different line. Print "WA" if his answer is wrong or "ACC" if his answer is OK.
|
[
"Iran_\nPersian;\nW_o;n;d;e;r;f;u;l;\n7\nWonderfulPersianIran\nwonderful_PersIAN_IRAN;;_\nWONDERFUL___IRAN__PERSIAN__;;\nIra__Persiann__Wonderful\nWonder;;fulPersian___;I;r;a;n;\n__________IranPersianWonderful__________\nPersianIran_is_Wonderful\n",
"Shapur;;\nis___\na_genius\n3\nShapur__a_is___geniUs\nis___shapur___a__Genius;\nShapur;;is;;a;;geni;;us;;\n"
] |
[
"ACC\nACC\nACC\nWA\nACC\nACC\nWA\n",
"WA\nACC\nACC\n"
] |
none
| 1,000
|
[
{
"input": "Iran_\nPersian;\nW_o;n;d;e;r;f;u;l;\n7\nWonderfulPersianIran\nwonderful_PersIAN_IRAN;;_\nWONDERFUL___IRAN__PERSIAN__;;\nIra__Persiann__Wonderful\nWonder;;fulPersian___;I;r;a;n;\n__________IranPersianWonderful__________\nPersianIran_is_Wonderful",
"output": "ACC\nACC\nACC\nWA\nACC\nACC\nWA"
},
{
"input": "Shapur;;\nis___\na_genius\n3\nShapur__a_is___geniUs\nis___shapur___a__Genius;\nShapur;;is;;a;;geni;;us;;",
"output": "WA\nACC\nACC"
},
{
"input": "rr\nrrx\nab\n1\nrabrrrx",
"output": "WA"
},
{
"input": "AB\nBC\nCD\n1\nABCDZZ",
"output": "WA"
},
{
"input": "aa\naaa\nz\n1\naazaaa",
"output": "ACC"
},
{
"input": "aa\naaa\nz\n1\naaazaa",
"output": "ACC"
},
{
"input": "as\nav\nax\n1\n-------",
"output": "WA"
},
{
"input": "a\nab\nb\n1\nabcd",
"output": "WA"
},
{
"input": "c\naba\ncc\n2\nccabac\nabcacc",
"output": "ACC\nWA"
},
{
"input": "ab\na\nb\n1\nabcd",
"output": "WA"
},
{
"input": "ACB\nTB\nAC\n1\nATBACBC",
"output": "WA"
},
{
"input": "cc\naba\ncc\n1\nccabaxx",
"output": "WA"
}
] | 1,646,915,967
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 43
| 93
| 5,734,400
|
import math,sys;input=sys.stdin.readline;S=lambda:input().rstrip();I=lambda:int(S());M=lambda:map(int,S().split());L=lambda:list(M());mod1=1000000007;mod2=998244353
strings = []
for i in range(3):
s = S()
letter = ""
for j in range(len(s)):
if 97<=ord(s[j])<=122:
letter += chr(ord(s[j])-32)
elif 90<ord(s[j])<97 or ord(s[j])>122 or ord(s[j])<65:
continue
else:
letter+= s[j]
strings.append(letter)
possible = []
for i in range(3):
for j in range(3):
for k in range(3):
if i!=j and i!=k and j!=k:
m = strings[i] + strings[j] + strings[k]
possible.append(m)
# print(strings)
n = I()
for i in range(n):
t = S()
given = ""
for j in range(len(t)):
if 97<=ord(t[j])<=122:
given += chr(ord(t[j])-32)
elif 90<ord(t[j])<97 or ord(t[j])>122 or ord(t[j])<65:
continue
else:
given+= t[j]
if given in possible:
print("ACC")
else:
print("WA")
|
Title: Hard Work
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
After the contest in comparing numbers, Shapur's teacher found out that he is a real genius and that no one could possibly do the calculations faster than him even using a super computer!
Some days before the contest, the teacher took a very simple-looking exam and all his *n* students took part in the exam. The teacher gave them 3 strings and asked them to concatenate them. Concatenating strings means to put them in some arbitrary order one after the other. For example from concatenating Alireza and Amir we can get to AlirezaAmir or AmirAlireza depending on the order of concatenation.
Unfortunately enough, the teacher forgot to ask students to concatenate their strings in a pre-defined order so each student did it the way he/she liked.
Now the teacher knows that Shapur is such a fast-calculating genius boy and asks him to correct the students' papers.
Shapur is not good at doing such a time-taking task. He rather likes to finish up with it as soon as possible and take his time to solve 3-SAT in polynomial time. Moreover, the teacher has given some advice that Shapur has to follow. Here's what the teacher said:
- As I expect you know, the strings I gave to my students (including you) contained only lowercase and uppercase Persian Mikhi-Script letters. These letters are too much like Latin letters, so to make your task much harder I converted all the initial strings and all of the students' answers to Latin. - As latin alphabet has much less characters than Mikhi-Script, I added three odd-looking characters to the answers, these include "-", ";" and "_". These characters are my own invention of course! And I call them Signs. - The length of all initial strings was less than or equal to 100 and the lengths of my students' answers are less than or equal to 600 - My son, not all students are genius as you are. It is quite possible that they make minor mistakes changing case of some characters. For example they may write ALiReZaAmIR instead of AlirezaAmir. Don't be picky and ignore these mistakes. - Those signs which I previously talked to you about are not important. You can ignore them, since many students are in the mood for adding extra signs or forgetting about a sign. So something like Iran;;-- is the same as --;IRAN - You should indicate for any of my students if his answer was right or wrong. Do this by writing "WA" for Wrong answer or "ACC" for a correct answer. - I should remind you that none of the strings (initial strings or answers) are empty. - Finally, do these as soon as possible. You have less than 2 hours to complete this.
Input Specification:
The first three lines contain a string each. These are the initial strings. They consists only of lowercase and uppercase Latin letters and signs ("-", ";" and "_"). All the initial strings have length from 1 to 100, inclusively.
In the fourth line there is a single integer *n* (0<=β€<=*n*<=β€<=1000), the number of students.
Next *n* lines contain a student's answer each. It is guaranteed that the answer meets what the teacher said. Each answer iconsists only of lowercase and uppercase Latin letters and signs ("-", ";" and "_"). Length is from 1 to 600, inclusively.
Output Specification:
For each student write in a different line. Print "WA" if his answer is wrong or "ACC" if his answer is OK.
Demo Input:
['Iran_\nPersian;\nW_o;n;d;e;r;f;u;l;\n7\nWonderfulPersianIran\nwonderful_PersIAN_IRAN;;_\nWONDERFUL___IRAN__PERSIAN__;;\nIra__Persiann__Wonderful\nWonder;;fulPersian___;I;r;a;n;\n__________IranPersianWonderful__________\nPersianIran_is_Wonderful\n', 'Shapur;;\nis___\na_genius\n3\nShapur__a_is___geniUs\nis___shapur___a__Genius;\nShapur;;is;;a;;geni;;us;;\n']
Demo Output:
['ACC\nACC\nACC\nWA\nACC\nACC\nWA\n', 'WA\nACC\nACC\n']
Note:
none
|
```python
import math,sys;input=sys.stdin.readline;S=lambda:input().rstrip();I=lambda:int(S());M=lambda:map(int,S().split());L=lambda:list(M());mod1=1000000007;mod2=998244353
strings = []
for i in range(3):
s = S()
letter = ""
for j in range(len(s)):
if 97<=ord(s[j])<=122:
letter += chr(ord(s[j])-32)
elif 90<ord(s[j])<97 or ord(s[j])>122 or ord(s[j])<65:
continue
else:
letter+= s[j]
strings.append(letter)
possible = []
for i in range(3):
for j in range(3):
for k in range(3):
if i!=j and i!=k and j!=k:
m = strings[i] + strings[j] + strings[k]
possible.append(m)
# print(strings)
n = I()
for i in range(n):
t = S()
given = ""
for j in range(len(t)):
if 97<=ord(t[j])<=122:
given += chr(ord(t[j])-32)
elif 90<ord(t[j])<97 or ord(t[j])>122 or ord(t[j])<65:
continue
else:
given+= t[j]
if given in possible:
print("ACC")
else:
print("WA")
```
| 3.966069
|
427
|
A
|
Police Recruits
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
The police department of your city has just started its journey. Initially, they donβt have any manpower. So, they started hiring new recruits in groups.
Meanwhile, crimes keeps occurring within the city. One member of the police force can investigate only one crime during his/her lifetime.
If there is no police officer free (isn't busy with crime) during the occurrence of a crime, it will go untreated.
Given the chronological order of crime occurrences and recruit hirings, find the number of crimes which will go untreated.
|
The first line of input will contain an integer *n*Β (1<=β€<=*n*<=β€<=105), the number of events. The next line will contain *n* space-separated integers.
If the integer is -1 then it means a crime has occurred. Otherwise, the integer will be positive, the number of officers recruited together at that time. No more than 10 officers will be recruited at a time.
|
Print a single integer, the number of crimes which will go untreated.
|
[
"3\n-1 -1 1\n",
"8\n1 -1 1 -1 -1 1 1 1\n",
"11\n-1 -1 2 -1 -1 -1 -1 -1 -1 -1 -1\n"
] |
[
"2\n",
"1\n",
"8\n"
] |
Lets consider the second example:
1. Firstly one person is hired. 1. Then crime appears, the last hired person will investigate this crime. 1. One more person is hired. 1. One more crime appears, the last hired person will investigate this crime. 1. Crime appears. There is no free policeman at the time, so this crime will go untreated. 1. One more person is hired. 1. One more person is hired. 1. One more person is hired.
The answer is one, as one crime (on step 5) will go untreated.
| 500
|
[
{
"input": "3\n-1 -1 1",
"output": "2"
},
{
"input": "8\n1 -1 1 -1 -1 1 1 1",
"output": "1"
},
{
"input": "11\n-1 -1 2 -1 -1 -1 -1 -1 -1 -1 -1",
"output": "8"
},
{
"input": "7\n-1 -1 1 1 -1 -1 1",
"output": "2"
},
{
"input": "21\n-1 -1 -1 -1 -1 3 2 -1 6 -1 -1 2 1 -1 2 2 1 6 5 -1 5",
"output": "5"
},
{
"input": "98\n-1 -1 1 -1 -1 -1 -1 1 -1 -1 1 -1 -1 1 -1 1 1 1 -1 1 1 1 1 1 -1 1 -1 -1 -1 -1 1 -1 -1 1 1 -1 1 1 1 -1 -1 -1 -1 -1 -1 1 -1 -1 -1 1 -1 1 -1 1 -1 1 1 1 1 1 1 1 -1 -1 1 1 -1 -1 -1 -1 -1 -1 -1 -1 -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": "13"
},
{
"input": "3\n-1 5 4",
"output": "1"
},
{
"input": "146\n4 -1 -1 -1 -1 -1 -1 -1 -1 -1 4 -1 3 -1 3 -1 -1 1 4 -1 2 -1 -1 3 -1 -1 -1 4 1 -1 -1 -1 4 -1 -1 -1 -1 -1 -1 3 2 -1 3 2 4 5 2 4 1 5 -1 -1 2 -1 -1 1 -1 5 3 -1 1 2 2 3 5 3 -1 -1 3 -1 -1 3 5 5 -1 -1 5 -1 4 4 1 -1 -1 -1 2 1 -1 -1 -1 2 5 3 -1 -1 -1 3 -1 5 4 -1 1 -1 -1 3 -1 -1 3 1 1 2 -1 -1 -1 1 3 1 -1 2 -1 -1 5 5 -1 -1 3 4 5 1 -1 2 -1 -1 -1 3 -1 5 3 2 -1 2 -1 -1 5 -1 3 -1",
"output": "5"
},
{
"input": "1\n2",
"output": "0"
},
{
"input": "1\n-1",
"output": "1"
},
{
"input": "2\n1 -1",
"output": "0"
},
{
"input": "2\n-1 1",
"output": "1"
},
{
"input": "2\n1 1",
"output": "0"
},
{
"input": "2\n-1 -1",
"output": "2"
},
{
"input": "4\n10 -1 -1 -1",
"output": "0"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "4\n-1 -1 1 1",
"output": "2"
}
] | 1,681,128,954
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 63
| 93
| 5,324,800
|
_ = input()
events = input().split(" ")
crime_counter = police_counter = 0
for event in events:
event = int(event)
if event > 0 or police_counter > 0:
police_counter += event
else:
crime_counter += 1
print(crime_counter)
|
Title: Police Recruits
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The police department of your city has just started its journey. Initially, they donβt have any manpower. So, they started hiring new recruits in groups.
Meanwhile, crimes keeps occurring within the city. One member of the police force can investigate only one crime during his/her lifetime.
If there is no police officer free (isn't busy with crime) during the occurrence of a crime, it will go untreated.
Given the chronological order of crime occurrences and recruit hirings, find the number of crimes which will go untreated.
Input Specification:
The first line of input will contain an integer *n*Β (1<=β€<=*n*<=β€<=105), the number of events. The next line will contain *n* space-separated integers.
If the integer is -1 then it means a crime has occurred. Otherwise, the integer will be positive, the number of officers recruited together at that time. No more than 10 officers will be recruited at a time.
Output Specification:
Print a single integer, the number of crimes which will go untreated.
Demo Input:
['3\n-1 -1 1\n', '8\n1 -1 1 -1 -1 1 1 1\n', '11\n-1 -1 2 -1 -1 -1 -1 -1 -1 -1 -1\n']
Demo Output:
['2\n', '1\n', '8\n']
Note:
Lets consider the second example:
1. Firstly one person is hired. 1. Then crime appears, the last hired person will investigate this crime. 1. One more person is hired. 1. One more crime appears, the last hired person will investigate this crime. 1. Crime appears. There is no free policeman at the time, so this crime will go untreated. 1. One more person is hired. 1. One more person is hired. 1. One more person is hired.
The answer is one, as one crime (on step 5) will go untreated.
|
```python
_ = input()
events = input().split(" ")
crime_counter = police_counter = 0
for event in events:
event = int(event)
if event > 0 or police_counter > 0:
police_counter += event
else:
crime_counter += 1
print(crime_counter)
```
| 3
|
|
995
|
B
|
Suit and Tie
|
PROGRAMMING
| 1,400
|
[
"greedy",
"implementation",
"math"
] | null | null |
Allen is hosting a formal dinner party. $2n$ people come to the event in $n$ pairs (couples). After a night of fun, Allen wants to line everyone up for a final picture. The $2n$ people line up, but Allen doesn't like the ordering. Allen prefers if each pair occupies adjacent positions in the line, as this makes the picture more aesthetic.
Help Allen find the minimum number of swaps of adjacent positions he must perform to make it so that each couple occupies adjacent positions in the line.
|
The first line contains a single integer $n$ ($1 \le n \le 100$), the number of pairs of people.
The second line contains $2n$ integers $a_1, a_2, \dots, a_{2n}$. For each $i$ with $1 \le i \le n$, $i$ appears exactly twice. If $a_j = a_k = i$, that means that the $j$-th and $k$-th people in the line form a couple.
|
Output a single integer, representing the minimum number of adjacent swaps needed to line the people up so that each pair occupies adjacent positions.
|
[
"4\n1 1 2 3 3 2 4 4\n",
"3\n1 1 2 2 3 3\n",
"3\n3 1 2 3 1 2\n"
] |
[
"2\n",
"0\n",
"3\n"
] |
In the first sample case, we can transform $1 1 2 3 3 2 4 4 \rightarrow 1 1 2 3 2 3 4 4 \rightarrow 1 1 2 2 3 3 4 4$ in two steps. Note that the sequence $1 1 2 3 3 2 4 4 \rightarrow 1 1 3 2 3 2 4 4 \rightarrow 1 1 3 3 2 2 4 4$ also works in the same number of steps.
The second sample case already satisfies the constraints; therefore we need $0$ swaps.
| 750
|
[
{
"input": "4\n1 1 2 3 3 2 4 4",
"output": "2"
},
{
"input": "3\n1 1 2 2 3 3",
"output": "0"
},
{
"input": "3\n3 1 2 3 1 2",
"output": "3"
},
{
"input": "8\n7 6 2 1 4 3 3 7 2 6 5 1 8 5 8 4",
"output": "27"
},
{
"input": "2\n1 2 1 2",
"output": "1"
},
{
"input": "3\n1 2 3 3 1 2",
"output": "5"
},
{
"input": "38\n26 28 23 34 33 14 38 15 35 36 30 1 19 17 18 28 22 15 9 27 11 16 17 32 7 21 6 8 32 26 33 23 18 4 2 25 29 3 35 8 38 37 31 37 12 25 3 27 16 24 5 20 12 13 29 11 30 22 9 19 2 24 7 10 34 4 36 21 14 31 13 6 20 10 5 1",
"output": "744"
},
{
"input": "24\n21 21 22 5 8 5 15 11 13 16 17 9 3 18 15 1 12 12 7 2 22 19 20 19 23 14 8 24 4 23 16 17 9 10 1 6 4 2 7 3 18 11 24 10 13 6 20 14",
"output": "259"
},
{
"input": "1\n1 1",
"output": "0"
},
{
"input": "19\n15 19 18 8 12 2 11 7 5 2 1 1 9 9 3 3 16 6 15 17 13 18 4 14 5 8 10 12 6 11 17 13 14 16 19 7 4 10",
"output": "181"
},
{
"input": "8\n3 1 5 2 1 6 3 5 6 2 4 8 8 4 7 7",
"output": "13"
},
{
"input": "2\n2 1 1 2",
"output": "2"
},
{
"input": "81\n48 22 31 24 73 77 79 75 37 78 43 56 20 33 70 34 6 50 51 21 39 29 20 11 73 53 39 61 28 17 55 52 28 57 52 74 35 13 55 2 57 9 46 81 60 47 21 68 1 53 31 64 42 9 79 80 69 30 32 24 15 2 69 10 22 3 71 19 67 66 17 50 62 36 32 65 58 18 25 59 38 10 14 51 23 16 29 81 45 40 18 54 47 12 45 74 41 34 75 44 19 77 71 67 7 16 35 49 15 3 38 4 7 25 76 66 5 65 27 6 1 72 37 42 26 60 12 64 44 41 80 13 49 68 76 48 11 78 40 61 30 43 62 58 5 4 33 26 54 27 36 72 63 63 59 70 23 8 56 8 46 14",
"output": "3186"
},
{
"input": "84\n10 29 12 22 55 3 81 33 64 78 46 44 69 41 34 71 24 12 22 54 63 9 65 40 36 81 32 37 83 50 28 84 53 25 72 77 41 35 50 8 29 78 72 53 21 63 16 1 79 20 66 23 38 18 44 5 27 77 32 52 42 60 67 62 64 52 14 80 4 19 15 45 40 47 42 46 68 18 70 8 3 36 65 38 73 43 59 20 66 6 51 10 58 55 51 13 4 5 43 82 71 21 9 33 47 11 61 30 76 27 24 48 75 15 48 75 2 31 83 67 59 74 56 11 39 13 45 76 26 30 39 17 61 57 68 7 70 62 49 57 49 84 31 26 56 54 74 16 60 1 80 35 82 28 79 73 14 69 6 19 25 34 23 2 58 37 7 17",
"output": "3279"
},
{
"input": "4\n3 4 2 4 1 2 1 3",
"output": "8"
},
{
"input": "75\n28 28 42 3 39 39 73 73 75 75 30 30 21 9 57 41 26 70 15 15 65 65 24 24 4 4 62 62 17 17 29 29 37 37 18 18 1 1 8 8 63 63 49 49 5 5 59 59 19 19 34 34 48 48 10 10 14 42 22 22 38 38 50 50 60 60 64 35 47 31 72 72 41 52 46 46 20 20 21 9 7 7 36 36 2 2 6 6 70 26 69 69 16 16 61 61 66 66 33 33 44 44 11 11 23 23 40 40 12 12 64 35 56 56 27 27 53 53 3 14 43 43 31 47 68 68 13 13 74 74 67 67 71 71 45 45 57 52 32 32 25 25 58 58 55 55 51 51 54 54",
"output": "870"
},
{
"input": "35\n6 32 4 19 9 34 20 29 22 26 19 14 33 11 17 31 30 13 7 12 8 16 5 5 21 15 18 28 34 3 2 10 23 24 35 6 32 4 25 9 1 11 24 20 26 25 2 13 22 17 31 30 33 7 12 8 16 27 27 21 15 18 28 1 3 14 10 23 29 35",
"output": "673"
},
{
"input": "86\n33 6 22 8 54 43 57 85 70 41 20 17 35 12 66 25 45 78 67 55 50 19 31 75 77 29 58 78 34 15 40 48 14 82 6 37 44 53 62 23 56 22 34 18 71 83 21 80 47 38 3 42 60 9 73 49 84 7 76 30 5 4 11 28 69 16 26 10 59 48 64 46 32 68 24 63 79 36 13 1 27 61 39 74 2 51 51 2 74 39 61 27 1 13 36 79 86 24 68 32 46 64 63 59 10 26 16 69 28 11 4 5 30 76 7 84 49 73 9 60 42 3 38 47 80 21 83 72 18 52 65 56 23 62 53 44 37 81 82 14 86 40 15 52 72 58 29 77 85 31 19 50 55 67 71 45 25 66 12 35 17 20 41 70 75 57 43 54 8 65 81 33",
"output": "6194"
}
] | 1,622,719,459
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 22
| 108
| 0
|
n=int(input())
lst = list(map(int, input().strip().split(' ')))
c=0
while(len(lst)!=0):
p=lst[0]
del lst[0]
i=lst.index(p)
c+=i
del lst[i]
print(c)
|
Title: Suit and Tie
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Allen is hosting a formal dinner party. $2n$ people come to the event in $n$ pairs (couples). After a night of fun, Allen wants to line everyone up for a final picture. The $2n$ people line up, but Allen doesn't like the ordering. Allen prefers if each pair occupies adjacent positions in the line, as this makes the picture more aesthetic.
Help Allen find the minimum number of swaps of adjacent positions he must perform to make it so that each couple occupies adjacent positions in the line.
Input Specification:
The first line contains a single integer $n$ ($1 \le n \le 100$), the number of pairs of people.
The second line contains $2n$ integers $a_1, a_2, \dots, a_{2n}$. For each $i$ with $1 \le i \le n$, $i$ appears exactly twice. If $a_j = a_k = i$, that means that the $j$-th and $k$-th people in the line form a couple.
Output Specification:
Output a single integer, representing the minimum number of adjacent swaps needed to line the people up so that each pair occupies adjacent positions.
Demo Input:
['4\n1 1 2 3 3 2 4 4\n', '3\n1 1 2 2 3 3\n', '3\n3 1 2 3 1 2\n']
Demo Output:
['2\n', '0\n', '3\n']
Note:
In the first sample case, we can transform $1 1 2 3 3 2 4 4 \rightarrow 1 1 2 3 2 3 4 4 \rightarrow 1 1 2 2 3 3 4 4$ in two steps. Note that the sequence $1 1 2 3 3 2 4 4 \rightarrow 1 1 3 2 3 2 4 4 \rightarrow 1 1 3 3 2 2 4 4$ also works in the same number of steps.
The second sample case already satisfies the constraints; therefore we need $0$ swaps.
|
```python
n=int(input())
lst = list(map(int, input().strip().split(' ')))
c=0
while(len(lst)!=0):
p=lst[0]
del lst[0]
i=lst.index(p)
c+=i
del lst[i]
print(c)
```
| 3
|
|
712
|
B
|
Memory and Trident
|
PROGRAMMING
| 1,100
|
[
"implementation",
"strings"
] | null | null |
Memory is performing a walk on the two-dimensional plane, starting at the origin. He is given a string *s* with his directions for motion:
- An 'L' indicates he should move one unit left. - An 'R' indicates he should move one unit right. - A 'U' indicates he should move one unit up. - A 'D' indicates he should move one unit down.
But now Memory wants to end at the origin. To do this, he has a special trident. This trident can replace any character in *s* with any of 'L', 'R', 'U', or 'D'. However, because he doesn't want to wear out the trident, he wants to make the minimum number of edits possible. Please tell Memory what is the minimum number of changes he needs to make to produce a string that, when walked, will end at the origin, or if there is no such string.
|
The first and only line contains the string *s* (1<=β€<=|*s*|<=β€<=100<=000)Β β the instructions Memory is given.
|
If there is a string satisfying the conditions, output a single integerΒ β the minimum number of edits required. In case it's not possible to change the sequence in such a way that it will bring Memory to to the origin, output -1.
|
[
"RRU\n",
"UDUR\n",
"RUUR\n"
] |
[
"-1\n",
"1\n",
"2\n"
] |
In the first sample test, Memory is told to walk right, then right, then up. It is easy to see that it is impossible to edit these instructions to form a valid walk.
In the second sample test, Memory is told to walk up, then down, then up, then right. One possible solution is to change *s* to "LDUR". This string uses 1 edit, which is the minimum possible. It also ends at the origin.
| 1,000
|
[
{
"input": "RRU",
"output": "-1"
},
{
"input": "UDUR",
"output": "1"
},
{
"input": "RUUR",
"output": "2"
},
{
"input": "DDDD",
"output": "2"
},
{
"input": "RRRR",
"output": "2"
},
{
"input": "RRRUUD",
"output": "2"
},
{
"input": "UDURLRDURLRD",
"output": "1"
},
{
"input": "RLRU",
"output": "1"
},
{
"input": "RDDLLDLUUUDDRDRURLUUURLLDDLRLUURRLLRRLDRLLUDRLRULLDLRRLRLRLRUDUUDLULURLLDUURULURLLRRRURRRDRUUDLDRLRDRLRRDDLDLDLLUDRUDRLLLLDRDUULRUURRDLULLULDUDULRURRDDDLLUDRLUDDLDDDRRDDDULLLLDLDRLRRLRRDDRLULURRUDRDUUUULDURUDRDLDDUDUDRRURDULRRUDRLRRDLUURURDLDRLRDUDDDLDDDURURLUULRDUUULRURUDUDRRUDULLLUUUDRLLDRRDDLRUDRDRDLLRURURRRULURURRRLUUULRRRUURUUDURUDDLLDLDRLRDLDRLLDLDRDRRLRRRURUUUDRDLRLRUDRLULUUULUDDLULDLRLLRDUULLRLRURLRURULLLUDUDDLRULRDUURURLDLUURRRDURRLLDRUUDRDLLDUUDLURUDDUUUULRLLURLUDDRLRRDRURLRUDRLDDRLLL",
"output": "-1"
},
{
"input": "LDLDLDLDLDRULD",
"output": "5"
},
{
"input": "LULULURULLLU",
"output": "5"
},
{
"input": "DRDRDDRR",
"output": "4"
},
{
"input": "LR",
"output": "0"
},
{
"input": "UL",
"output": "1"
},
{
"input": "UD",
"output": "0"
},
{
"input": "R",
"output": "-1"
},
{
"input": "LU",
"output": "1"
},
{
"input": "RR",
"output": "1"
},
{
"input": "UDLR",
"output": "0"
},
{
"input": "RRRRRLLUUUUDD",
"output": "-1"
},
{
"input": "UUUUDLLLLR",
"output": "3"
},
{
"input": "LLRUD",
"output": "-1"
},
{
"input": "LLRDDU",
"output": "1"
}
] | 1,474,129,156
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 75
| 77
| 307,200
|
def main():
s = input()
if len(s) % 2 == 1:
print('-1')
return
cnt = {'L':0,'R':0,'U':0,'D':0}
for c in s:
cnt[c] += 1
reduce1 = abs(cnt['L'] - cnt['R'])
reduce2 = abs(cnt['U'] - cnt['D'])
print((reduce1 + reduce2) // 2)
if __name__ == '__main__':
main()
|
Title: Memory and Trident
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Memory is performing a walk on the two-dimensional plane, starting at the origin. He is given a string *s* with his directions for motion:
- An 'L' indicates he should move one unit left. - An 'R' indicates he should move one unit right. - A 'U' indicates he should move one unit up. - A 'D' indicates he should move one unit down.
But now Memory wants to end at the origin. To do this, he has a special trident. This trident can replace any character in *s* with any of 'L', 'R', 'U', or 'D'. However, because he doesn't want to wear out the trident, he wants to make the minimum number of edits possible. Please tell Memory what is the minimum number of changes he needs to make to produce a string that, when walked, will end at the origin, or if there is no such string.
Input Specification:
The first and only line contains the string *s* (1<=β€<=|*s*|<=β€<=100<=000)Β β the instructions Memory is given.
Output Specification:
If there is a string satisfying the conditions, output a single integerΒ β the minimum number of edits required. In case it's not possible to change the sequence in such a way that it will bring Memory to to the origin, output -1.
Demo Input:
['RRU\n', 'UDUR\n', 'RUUR\n']
Demo Output:
['-1\n', '1\n', '2\n']
Note:
In the first sample test, Memory is told to walk right, then right, then up. It is easy to see that it is impossible to edit these instructions to form a valid walk.
In the second sample test, Memory is told to walk up, then down, then up, then right. One possible solution is to change *s* to "LDUR". This string uses 1 edit, which is the minimum possible. It also ends at the origin.
|
```python
def main():
s = input()
if len(s) % 2 == 1:
print('-1')
return
cnt = {'L':0,'R':0,'U':0,'D':0}
for c in s:
cnt[c] += 1
reduce1 = abs(cnt['L'] - cnt['R'])
reduce2 = abs(cnt['U'] - cnt['D'])
print((reduce1 + reduce2) // 2)
if __name__ == '__main__':
main()
```
| 3
|
|
116
|
A
|
Tram
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Linear Kingdom has exactly one tram line. It has *n* stops, numbered from 1 to *n* in the order of tram's movement. At the *i*-th stop *a**i* passengers exit the tram, while *b**i* passengers enter it. The tram is empty before it arrives at the first stop. Also, when the tram arrives at the last stop, all passengers exit so that it becomes empty.
Your task is to calculate the tram's minimum capacity such that the number of people inside the tram at any time never exceeds this capacity. Note that at each stop all exiting passengers exit before any entering passenger enters the tram.
|
The first line contains a single number *n* (2<=β€<=*n*<=β€<=1000) β the number of the tram's stops.
Then *n* lines follow, each contains two integers *a**i* and *b**i* (0<=β€<=*a**i*,<=*b**i*<=β€<=1000) β the number of passengers that exits the tram at the *i*-th stop, and the number of passengers that enter the tram at the *i*-th stop. The stops are given from the first to the last stop in the order of tram's movement.
- The number of people who exit at a given stop does not exceed the total number of people in the tram immediately before it arrives at the stop. More formally, . This particularly means that *a*1<==<=0. - At the last stop, all the passengers exit the tram and it becomes empty. More formally, . - No passenger will enter the train at the last stop. That is, *b**n*<==<=0.
|
Print a single integer denoting the minimum possible capacity of the tram (0 is allowed).
|
[
"4\n0 3\n2 5\n4 2\n4 0\n"
] |
[
"6\n"
] |
For the first example, a capacity of 6 is sufficient:
- At the first stop, the number of passengers inside the tram before arriving is 0. Then, 3 passengers enter the tram, and the number of passengers inside the tram becomes 3. - At the second stop, 2 passengers exit the tram (1 passenger remains inside). Then, 5 passengers enter the tram. There are 6 passengers inside the tram now. - At the third stop, 4 passengers exit the tram (2 passengers remain inside). Then, 2 passengers enter the tram. There are 4 passengers inside the tram now. - Finally, all the remaining passengers inside the tram exit the tram at the last stop. There are no passenger inside the tram now, which is in line with the constraints.
Since the number of passengers inside the tram never exceeds 6, a capacity of 6 is sufficient. Furthermore it is not possible for the tram to have a capacity less than 6. Hence, 6 is the correct answer.
| 500
|
[
{
"input": "4\n0 3\n2 5\n4 2\n4 0",
"output": "6"
},
{
"input": "5\n0 4\n4 6\n6 5\n5 4\n4 0",
"output": "6"
},
{
"input": "10\n0 5\n1 7\n10 8\n5 3\n0 5\n3 3\n8 8\n0 6\n10 1\n9 0",
"output": "18"
},
{
"input": "3\n0 1\n1 1\n1 0",
"output": "1"
},
{
"input": "4\n0 1\n0 1\n1 0\n1 0",
"output": "2"
},
{
"input": "3\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "3\n0 1000\n1000 1000\n1000 0",
"output": "1000"
},
{
"input": "5\n0 73\n73 189\n189 766\n766 0\n0 0",
"output": "766"
},
{
"input": "5\n0 0\n0 0\n0 0\n0 1\n1 0",
"output": "1"
},
{
"input": "5\n0 917\n917 923\n904 992\n1000 0\n11 0",
"output": "1011"
},
{
"input": "5\n0 1\n1 2\n2 1\n1 2\n2 0",
"output": "2"
},
{
"input": "5\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "20\n0 7\n2 1\n2 2\n5 7\n2 6\n6 10\n2 4\n0 4\n7 4\n8 0\n10 6\n2 1\n6 1\n1 7\n0 3\n8 7\n6 3\n6 3\n1 1\n3 0",
"output": "22"
},
{
"input": "5\n0 1000\n1000 1000\n1000 1000\n1000 1000\n1000 0",
"output": "1000"
},
{
"input": "10\n0 592\n258 598\n389 203\n249 836\n196 635\n478 482\n994 987\n1000 0\n769 0\n0 0",
"output": "1776"
},
{
"input": "10\n0 1\n1 0\n0 0\n0 0\n0 0\n0 1\n1 1\n0 1\n1 0\n1 0",
"output": "2"
},
{
"input": "10\n0 926\n926 938\n938 931\n931 964\n937 989\n983 936\n908 949\n997 932\n945 988\n988 0",
"output": "1016"
},
{
"input": "10\n0 1\n1 2\n1 2\n2 2\n2 2\n2 2\n1 1\n1 1\n2 1\n2 0",
"output": "3"
},
{
"input": "10\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "10\n0 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 0",
"output": "1000"
},
{
"input": "50\n0 332\n332 268\n268 56\n56 711\n420 180\n160 834\n149 341\n373 777\n763 93\n994 407\n86 803\n700 132\n471 608\n429 467\n75 5\n638 305\n405 853\n316 478\n643 163\n18 131\n648 241\n241 766\n316 847\n640 380\n923 759\n789 41\n125 421\n421 9\n9 388\n388 829\n408 108\n462 856\n816 411\n518 688\n290 7\n405 912\n397 772\n396 652\n394 146\n27 648\n462 617\n514 433\n780 35\n710 705\n460 390\n194 508\n643 56\n172 469\n1000 0\n194 0",
"output": "2071"
},
{
"input": "50\n0 0\n0 1\n1 1\n0 1\n0 0\n1 0\n0 0\n1 0\n0 0\n0 0\n0 0\n0 0\n0 1\n0 0\n0 0\n0 1\n1 0\n0 1\n0 0\n1 1\n1 0\n0 1\n0 0\n1 1\n0 1\n1 0\n1 1\n1 0\n0 0\n1 1\n1 0\n0 1\n0 0\n0 1\n1 1\n1 1\n1 1\n1 0\n1 1\n1 0\n0 1\n1 0\n0 0\n0 1\n1 1\n1 1\n0 1\n0 0\n1 0\n1 0",
"output": "3"
},
{
"input": "50\n0 926\n926 971\n915 980\n920 965\n954 944\n928 952\n955 980\n916 980\n906 935\n944 913\n905 923\n912 922\n965 934\n912 900\n946 930\n931 983\n979 905\n925 969\n924 926\n910 914\n921 977\n934 979\n962 986\n942 909\n976 903\n982 982\n991 941\n954 929\n902 980\n947 983\n919 924\n917 943\n916 905\n907 913\n964 977\n984 904\n905 999\n950 970\n986 906\n993 970\n960 994\n963 983\n918 986\n980 900\n931 986\n993 997\n941 909\n907 909\n1000 0\n278 0",
"output": "1329"
},
{
"input": "2\n0 863\n863 0",
"output": "863"
},
{
"input": "50\n0 1\n1 2\n2 2\n1 1\n1 1\n1 2\n1 2\n1 1\n1 2\n1 1\n1 1\n1 2\n1 2\n1 1\n2 1\n2 2\n1 2\n2 2\n1 2\n2 1\n2 1\n2 2\n2 1\n1 2\n1 2\n2 1\n1 1\n2 2\n1 1\n2 1\n2 2\n2 1\n1 2\n2 2\n1 2\n1 1\n1 1\n2 1\n2 1\n2 2\n2 1\n2 1\n1 2\n1 2\n1 2\n1 2\n2 0\n2 0\n2 0\n0 0",
"output": "8"
},
{
"input": "50\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "100\n0 1\n0 0\n0 0\n1 0\n0 0\n0 1\n0 1\n1 1\n0 0\n0 0\n1 1\n0 0\n1 1\n0 1\n1 1\n0 1\n1 1\n1 0\n1 0\n0 0\n1 0\n0 1\n1 0\n0 0\n0 0\n1 1\n1 1\n0 1\n0 0\n1 0\n1 1\n0 1\n1 0\n1 1\n0 1\n1 1\n1 0\n0 0\n0 0\n0 1\n0 0\n0 1\n1 1\n0 0\n1 1\n1 1\n0 0\n0 1\n1 0\n0 1\n0 0\n0 1\n0 1\n1 1\n1 1\n1 1\n0 0\n0 0\n1 1\n0 1\n0 1\n1 0\n0 0\n0 0\n1 1\n0 1\n0 1\n1 1\n1 1\n0 1\n1 1\n1 1\n0 0\n1 0\n0 1\n0 0\n0 0\n1 1\n1 1\n1 1\n1 1\n0 1\n1 0\n1 0\n1 0\n1 0\n1 0\n0 0\n1 0\n1 0\n0 0\n1 0\n0 0\n0 1\n1 0\n0 1\n1 0\n1 0\n1 0\n1 0",
"output": "11"
},
{
"input": "100\n0 2\n1 2\n2 1\n1 2\n1 2\n2 1\n2 2\n1 1\n1 1\n2 1\n1 2\n2 1\n1 2\n2 2\n2 2\n2 2\n1 2\n2 2\n2 1\n1 1\n1 1\n1 1\n2 2\n1 2\n2 2\n1 1\n1 1\n1 1\n1 1\n2 2\n1 2\n2 1\n1 1\n2 2\n1 1\n2 1\n1 1\n2 2\n2 1\n1 2\n1 1\n1 2\n2 1\n2 2\n1 1\n2 1\n1 1\n2 1\n1 1\n1 2\n2 2\n2 2\n1 1\n2 2\n1 2\n2 1\n2 1\n1 1\n1 1\n1 2\n1 2\n1 1\n1 1\n2 1\n1 2\n1 2\n2 1\n2 2\n2 2\n2 2\n2 1\n2 2\n1 1\n1 2\n1 2\n1 1\n2 2\n2 2\n1 1\n2 1\n1 1\n1 2\n1 2\n1 2\n1 1\n1 1\n2 2\n1 2\n2 1\n2 1\n2 1\n1 2\n1 2\n1 1\n2 2\n1 2\n2 0\n2 0\n2 0\n1 0",
"output": "7"
},
{
"input": "100\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "2\n0 1\n1 0",
"output": "1"
},
{
"input": "2\n0 900\n900 0",
"output": "900"
},
{
"input": "2\n0 1\n1 0",
"output": "1"
},
{
"input": "2\n0 0\n0 0",
"output": "0"
},
{
"input": "2\n0 1000\n1000 0",
"output": "1000"
},
{
"input": "3\n0 802\n175 188\n815 0",
"output": "815"
},
{
"input": "3\n0 910\n910 976\n976 0",
"output": "976"
},
{
"input": "3\n0 2\n2 1\n1 0",
"output": "2"
}
] | 1,694,351,536
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 71
| 186
| 1,843,200
|
total_stop = int(input())
max = 0
current = 0
for i in range(total_stop):
a,b = map(int,input().split())
if i == 0:
max = b
current = b
else:
current = (current - a) + b
if current > max:
max = current
print(max)
|
Title: Tram
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Linear Kingdom has exactly one tram line. It has *n* stops, numbered from 1 to *n* in the order of tram's movement. At the *i*-th stop *a**i* passengers exit the tram, while *b**i* passengers enter it. The tram is empty before it arrives at the first stop. Also, when the tram arrives at the last stop, all passengers exit so that it becomes empty.
Your task is to calculate the tram's minimum capacity such that the number of people inside the tram at any time never exceeds this capacity. Note that at each stop all exiting passengers exit before any entering passenger enters the tram.
Input Specification:
The first line contains a single number *n* (2<=β€<=*n*<=β€<=1000) β the number of the tram's stops.
Then *n* lines follow, each contains two integers *a**i* and *b**i* (0<=β€<=*a**i*,<=*b**i*<=β€<=1000) β the number of passengers that exits the tram at the *i*-th stop, and the number of passengers that enter the tram at the *i*-th stop. The stops are given from the first to the last stop in the order of tram's movement.
- The number of people who exit at a given stop does not exceed the total number of people in the tram immediately before it arrives at the stop. More formally, . This particularly means that *a*1<==<=0. - At the last stop, all the passengers exit the tram and it becomes empty. More formally, . - No passenger will enter the train at the last stop. That is, *b**n*<==<=0.
Output Specification:
Print a single integer denoting the minimum possible capacity of the tram (0 is allowed).
Demo Input:
['4\n0 3\n2 5\n4 2\n4 0\n']
Demo Output:
['6\n']
Note:
For the first example, a capacity of 6 is sufficient:
- At the first stop, the number of passengers inside the tram before arriving is 0. Then, 3 passengers enter the tram, and the number of passengers inside the tram becomes 3. - At the second stop, 2 passengers exit the tram (1 passenger remains inside). Then, 5 passengers enter the tram. There are 6 passengers inside the tram now. - At the third stop, 4 passengers exit the tram (2 passengers remain inside). Then, 2 passengers enter the tram. There are 4 passengers inside the tram now. - Finally, all the remaining passengers inside the tram exit the tram at the last stop. There are no passenger inside the tram now, which is in line with the constraints.
Since the number of passengers inside the tram never exceeds 6, a capacity of 6 is sufficient. Furthermore it is not possible for the tram to have a capacity less than 6. Hence, 6 is the correct answer.
|
```python
total_stop = int(input())
max = 0
current = 0
for i in range(total_stop):
a,b = map(int,input().split())
if i == 0:
max = b
current = b
else:
current = (current - a) + b
if current > max:
max = current
print(max)
```
| 3
|
|
559
|
B
|
Equivalent Strings
|
PROGRAMMING
| 1,700
|
[
"divide and conquer",
"hashing",
"sortings",
"strings"
] | null | null |
Today on a lecture about strings Gerald learned a new definition of string equivalency. Two strings *a* and *b* of equal length are called equivalent in one of the two cases:
1. They are equal. 1. If we split string *a* into two halves of the same size *a*1 and *a*2, and string *b* into two halves of the same size *b*1 and *b*2, then one of the following is correct: *a*1 is equivalent to *b*1, and *a*2 is equivalent to *b*2 1. *a*1 is equivalent to *b*2, and *a*2 is equivalent to *b*1
As a home task, the teacher gave two strings to his students and asked to determine if they are equivalent.
Gerald has already completed this home task. Now it's your turn!
|
The first two lines of the input contain two strings given by the teacher. Each of them has the length from 1 to 200<=000 and consists of lowercase English letters. The strings have the same length.
|
Print "YES" (without the quotes), if these two strings are equivalent, and "NO" (without the quotes) otherwise.
|
[
"aaba\nabaa\n",
"aabb\nabab\n"
] |
[
"YES\n",
"NO\n"
] |
In the first sample you should split the first string into strings "aa" and "ba", the second one β into strings "ab" and "aa". "aa" is equivalent to "aa"; "ab" is equivalent to "ba" as "ab" = "a" + "b", "ba" = "b" + "a".
In the second sample the first string can be splitted into strings "aa" and "bb", that are equivalent only to themselves. That's why string "aabb" is equivalent only to itself and to string "bbaa".
| 1,000
|
[
{
"input": "aaba\nabaa",
"output": "YES"
},
{
"input": "aabb\nabab",
"output": "NO"
},
{
"input": "a\na",
"output": "YES"
},
{
"input": "a\nb",
"output": "NO"
},
{
"input": "ab\nab",
"output": "YES"
},
{
"input": "ab\nba",
"output": "YES"
},
{
"input": "ab\nbb",
"output": "NO"
},
{
"input": "zzaa\naazz",
"output": "YES"
},
{
"input": "azza\nzaaz",
"output": "YES"
},
{
"input": "abc\nabc",
"output": "YES"
},
{
"input": "abc\nacb",
"output": "NO"
},
{
"input": "azzz\nzzaz",
"output": "YES"
},
{
"input": "abcd\ndcab",
"output": "YES"
},
{
"input": "abcd\ncdab",
"output": "YES"
},
{
"input": "abcd\ndcba",
"output": "YES"
},
{
"input": "abcd\nacbd",
"output": "NO"
},
{
"input": "oloaxgddgujq\noloaxgujqddg",
"output": "YES"
},
{
"input": "uwzwdxfmosmqatyv\ndxfmzwwusomqvyta",
"output": "YES"
},
{
"input": "hagnzomowtledfdotnll\nledfdotnllomowthagnz",
"output": "YES"
},
{
"input": "snyaydaeobufdg\nsnyaydaeobufdg",
"output": "YES"
},
{
"input": "baaaaa\nabaaaa",
"output": "NO"
},
{
"input": "hhiisug\nmzdjwju",
"output": "NO"
},
{
"input": "bbbabbabaaab\naaaabbabbbbb",
"output": "NO"
},
{
"input": "bbaaab\naababb",
"output": "NO"
},
{
"input": "aabbaaaa\naaaaabab",
"output": "NO"
},
{
"input": "aab\naba",
"output": "NO"
},
{
"input": "abcddd\nbacddd",
"output": "NO"
},
{
"input": "qgiufelsfhanx\naaaaaaaaaaaaa",
"output": "NO"
},
{
"input": "aabaababaaba\naababaaababa",
"output": "NO"
},
{
"input": "nocdqzdriyyil\naaaaaaaaaaaaa",
"output": "NO"
},
{
"input": "zdmctxl\nkojqhgw",
"output": "NO"
},
{
"input": "yhwepqwyhwepqwyhwepqweahnqtueahnqtueahnqtuyhwepqwyhwepqwyhwepqwyhwepqweahnqtueahnqtuyhwepqweahnqtueahnqtueahnqtueahnqtueahnqtueahnqtu\neahnqtueahnqtueahnqtuyhwepqweahnqtuyhwepqwyhwepqweahnqtuyhwepqweahnqtuyhwepqweahnqtueahnqtuyhwepqweahnqtueahnqtuyhwepqwyhwepqwyhwepqw",
"output": "NO"
},
{
"input": "abc\nbac",
"output": "NO"
},
{
"input": "ottceez\npcstdvz",
"output": "NO"
}
] | 1,699,960,116
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 104
| 967
| 113,664,000
|
memo = {}
def solve(a, b):
if (a,b) in memo:
return memo[a,b]
if len(a)%2 == 1:
return a == b
half = len(a)//2
a1, a2 = a[:half], a[half:]
b1, b2 = b[:half], b[half:]
memo[a,b] = (solve(a1, b1) and solve(a2, b2)) or (solve(a1, b2) and solve(a2, b1))
return memo[a,b]
s1 = input()
s2 = input()
ans = solve(s1, s2)
print("YES" if ans else "NO")
|
Title: Equivalent Strings
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Today on a lecture about strings Gerald learned a new definition of string equivalency. Two strings *a* and *b* of equal length are called equivalent in one of the two cases:
1. They are equal. 1. If we split string *a* into two halves of the same size *a*1 and *a*2, and string *b* into two halves of the same size *b*1 and *b*2, then one of the following is correct: *a*1 is equivalent to *b*1, and *a*2 is equivalent to *b*2 1. *a*1 is equivalent to *b*2, and *a*2 is equivalent to *b*1
As a home task, the teacher gave two strings to his students and asked to determine if they are equivalent.
Gerald has already completed this home task. Now it's your turn!
Input Specification:
The first two lines of the input contain two strings given by the teacher. Each of them has the length from 1 to 200<=000 and consists of lowercase English letters. The strings have the same length.
Output Specification:
Print "YES" (without the quotes), if these two strings are equivalent, and "NO" (without the quotes) otherwise.
Demo Input:
['aaba\nabaa\n', 'aabb\nabab\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first sample you should split the first string into strings "aa" and "ba", the second one β into strings "ab" and "aa". "aa" is equivalent to "aa"; "ab" is equivalent to "ba" as "ab" = "a" + "b", "ba" = "b" + "a".
In the second sample the first string can be splitted into strings "aa" and "bb", that are equivalent only to themselves. That's why string "aabb" is equivalent only to itself and to string "bbaa".
|
```python
memo = {}
def solve(a, b):
if (a,b) in memo:
return memo[a,b]
if len(a)%2 == 1:
return a == b
half = len(a)//2
a1, a2 = a[:half], a[half:]
b1, b2 = b[:half], b[half:]
memo[a,b] = (solve(a1, b1) and solve(a2, b2)) or (solve(a1, b2) and solve(a2, b1))
return memo[a,b]
s1 = input()
s2 = input()
ans = solve(s1, s2)
print("YES" if ans else "NO")
```
| 3
|
|
50
|
A
|
Domino piling
|
PROGRAMMING
| 800
|
[
"greedy",
"math"
] |
A. Domino piling
|
2
|
256
|
You are given a rectangular board of *M*<=Γ<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=Γ<=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* β board sizes in squares (1<=β€<=*M*<=β€<=*N*<=β€<=16).
|
Output one number β 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"
}
] | 1,622,833,485
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 35
| 340
| 0
|
if __name__ == '__main__':
s = input()
l = list(map(int, s.split()))
print(l[0] * l[1] // 2)
|
Title: Domino piling
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
You are given a rectangular board of *M*<=Γ<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=Γ<=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* β board sizes in squares (1<=β€<=*M*<=β€<=*N*<=β€<=16).
Output Specification:
Output one number β 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
if __name__ == '__main__':
s = input()
l = list(map(int, s.split()))
print(l[0] * l[1] // 2)
```
| 3.915
|
895
|
B
|
XK Segments
|
PROGRAMMING
| 1,700
|
[
"binary search",
"math",
"sortings",
"two pointers"
] | null | null |
While Vasya finished eating his piece of pizza, the lesson has already started. For being late for the lesson, the teacher suggested Vasya to solve one interesting problem. Vasya has an array *a* and integer *x*. He should find the number of different ordered pairs of indexes (*i*,<=*j*) such that *a**i*<=β€<=*a**j* and there are exactly *k* integers *y* such that *a**i*<=β€<=*y*<=β€<=*a**j* and *y* is divisible by *x*.
In this problem it is meant that pair (*i*,<=*j*) is equal to (*j*,<=*i*) only if *i* is equal to *j*. For example pair (1,<=2) is not the same as (2,<=1).
|
The first line contains 3 integers *n*,<=*x*,<=*k* (1<=β€<=*n*<=β€<=105,<=1<=β€<=*x*<=β€<=109,<=0<=β€<=*k*<=β€<=109), where *n* is the size of the array *a* and *x* and *k* are numbers from the statement.
The second line contains *n* integers *a**i* (1<=β€<=*a**i*<=β€<=109)Β β the elements of the array *a*.
|
Print one integerΒ β the answer to the problem.
|
[
"4 2 1\n1 3 5 7\n",
"4 2 0\n5 3 1 7\n",
"5 3 1\n3 3 3 3 3\n"
] |
[
"3\n",
"4\n",
"25\n"
] |
In first sample there are only three suitable pairs of indexesΒ β (1,β2),β(2,β3),β(3,β4).
In second sample there are four suitable pairs of indexes(1,β1),β(2,β2),β(3,β3),β(4,β4).
In third sample every pair (*i*,β*j*) is suitable, so the answer is 5β*β5β=β25.
| 1,000
|
[
{
"input": "4 2 1\n1 3 5 7",
"output": "3"
},
{
"input": "4 2 0\n5 3 1 7",
"output": "4"
},
{
"input": "5 3 1\n3 3 3 3 3",
"output": "25"
},
{
"input": "5 3 4\n24 13 1 24 24",
"output": "4"
},
{
"input": "4 2 2\n1 3 5 7",
"output": "2"
},
{
"input": "5 1 0\n10 12 14 20 30",
"output": "0"
},
{
"input": "7 3 1\n3 3 6 9 12 15 18",
"output": "9"
},
{
"input": "2 5 0\n3 4",
"output": "3"
},
{
"input": "3 5 0\n4 4 4",
"output": "9"
},
{
"input": "3 6 0\n3 4 5",
"output": "6"
},
{
"input": "10 2 1\n2 2 2 2 2 2 2 1 2 1",
"output": "80"
},
{
"input": "100 2 1\n1 2 2 2 1 2 1 2 2 2 2 1 2 1 1 2 1 2 2 1 1 2 1 1 2 2 1 1 2 1 2 2 2 2 2 1 1 1 2 1 2 1 2 1 2 1 2 2 1 2 2 2 1 1 1 2 2 1 1 2 2 1 1 1 1 1 1 2 2 1 1 2 2 1 2 2 1 2 1 2 2 1 2 1 1 1 2 2 1 1 2 1 2 2 1 1 2 2 2 1",
"output": "5200"
},
{
"input": "1 13 1\n13",
"output": "1"
},
{
"input": "2 1 1000000000\n1 1000000000",
"output": "1"
},
{
"input": "2 1000000000 1\n1 1000000000",
"output": "2"
},
{
"input": "1 1 1\n1",
"output": "1"
},
{
"input": "5 1000000000 1\n1000000000 1 1 1 1",
"output": "5"
},
{
"input": "2 1000 1000000\n1 1000000000",
"output": "1"
},
{
"input": "6 2 10\n1 1 1 20 20 20",
"output": "9"
},
{
"input": "1 1000000000 1\n1000000000",
"output": "1"
},
{
"input": "4 65536 65536\n1 2 3 4",
"output": "0"
},
{
"input": "1 600000000 1\n500000000",
"output": "0"
},
{
"input": "5 3 2\n2 3 4 5 8",
"output": "2"
}
] | 1,512,220,345
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 69
| 389
| 36,761,600
|
from bisect import bisect_left as b
f = lambda: map(int, input().split())
n, x, k = f()
s, t = 0, sorted(f())
p = [(q, ((q - 1) // x + k) * x) for q in t]
for q, d in p: s += b(t, d + x) - b(t, max(q, d))
print(s)
|
Title: XK Segments
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
While Vasya finished eating his piece of pizza, the lesson has already started. For being late for the lesson, the teacher suggested Vasya to solve one interesting problem. Vasya has an array *a* and integer *x*. He should find the number of different ordered pairs of indexes (*i*,<=*j*) such that *a**i*<=β€<=*a**j* and there are exactly *k* integers *y* such that *a**i*<=β€<=*y*<=β€<=*a**j* and *y* is divisible by *x*.
In this problem it is meant that pair (*i*,<=*j*) is equal to (*j*,<=*i*) only if *i* is equal to *j*. For example pair (1,<=2) is not the same as (2,<=1).
Input Specification:
The first line contains 3 integers *n*,<=*x*,<=*k* (1<=β€<=*n*<=β€<=105,<=1<=β€<=*x*<=β€<=109,<=0<=β€<=*k*<=β€<=109), where *n* is the size of the array *a* and *x* and *k* are numbers from the statement.
The second line contains *n* integers *a**i* (1<=β€<=*a**i*<=β€<=109)Β β the elements of the array *a*.
Output Specification:
Print one integerΒ β the answer to the problem.
Demo Input:
['4 2 1\n1 3 5 7\n', '4 2 0\n5 3 1 7\n', '5 3 1\n3 3 3 3 3\n']
Demo Output:
['3\n', '4\n', '25\n']
Note:
In first sample there are only three suitable pairs of indexesΒ β (1,β2),β(2,β3),β(3,β4).
In second sample there are four suitable pairs of indexes(1,β1),β(2,β2),β(3,β3),β(4,β4).
In third sample every pair (*i*,β*j*) is suitable, so the answer is 5β*β5β=β25.
|
```python
from bisect import bisect_left as b
f = lambda: map(int, input().split())
n, x, k = f()
s, t = 0, sorted(f())
p = [(q, ((q - 1) // x + k) * x) for q in t]
for q, d in p: s += b(t, d + x) - b(t, max(q, d))
print(s)
```
| 3
|
|
579
|
A
|
Raising Bacteria
|
PROGRAMMING
| 1,000
|
[
"bitmasks"
] | null | null |
You are a lover of bacteria. You want to raise some bacteria in a box.
Initially, the box is empty. Each morning, you can put any number of bacteria into the box. And each night, every bacterium in the box will split into two bacteria. You hope to see exactly *x* bacteria in the box at some moment.
What is the minimum number of bacteria you need to put into the box across those days?
|
The only line containing one integer *x* (1<=β€<=*x*<=β€<=109).
|
The only line containing one integer: the answer.
|
[
"5\n",
"8\n"
] |
[
"2\n",
"1\n"
] |
For the first sample, we can add one bacterium in the box in the first day morning and at the third morning there will be 4 bacteria in the box. Now we put one more resulting 5 in the box. We added 2 bacteria in the process so the answer is 2.
For the second sample, we can put one in the first morning and in the 4-th morning there will be 8 in the box. So the answer is 1.
| 250
|
[
{
"input": "5",
"output": "2"
},
{
"input": "8",
"output": "1"
},
{
"input": "536870911",
"output": "29"
},
{
"input": "1",
"output": "1"
},
{
"input": "343000816",
"output": "14"
},
{
"input": "559980448",
"output": "12"
},
{
"input": "697681824",
"output": "14"
},
{
"input": "41313494",
"output": "14"
},
{
"input": "673935585",
"output": "13"
},
{
"input": "117422204",
"output": "19"
},
{
"input": "954746654",
"output": "15"
},
{
"input": "536838144",
"output": "14"
},
{
"input": "1000000000",
"output": "13"
},
{
"input": "2",
"output": "1"
},
{
"input": "3",
"output": "2"
},
{
"input": "4",
"output": "1"
},
{
"input": "6",
"output": "2"
},
{
"input": "7",
"output": "3"
},
{
"input": "9",
"output": "2"
},
{
"input": "10",
"output": "2"
},
{
"input": "999999999",
"output": "21"
},
{
"input": "999999998",
"output": "20"
},
{
"input": "999999997",
"output": "20"
},
{
"input": "999999996",
"output": "19"
},
{
"input": "999999995",
"output": "20"
},
{
"input": "999999994",
"output": "19"
},
{
"input": "999999993",
"output": "19"
},
{
"input": "999999992",
"output": "18"
},
{
"input": "999999991",
"output": "20"
},
{
"input": "999999990",
"output": "19"
},
{
"input": "536870910",
"output": "28"
},
{
"input": "536870912",
"output": "1"
},
{
"input": "536870913",
"output": "2"
},
{
"input": "536870914",
"output": "2"
},
{
"input": "22",
"output": "3"
},
{
"input": "14",
"output": "3"
}
] | 1,688,390,445
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 36
| 46
| 0
|
w2=int(input())
w=bin(w2)
w1=w[2:]
c=0
for i in w1:
if int(i)%2!=0:
c+=1
print(c)
|
Title: Raising Bacteria
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are a lover of bacteria. You want to raise some bacteria in a box.
Initially, the box is empty. Each morning, you can put any number of bacteria into the box. And each night, every bacterium in the box will split into two bacteria. You hope to see exactly *x* bacteria in the box at some moment.
What is the minimum number of bacteria you need to put into the box across those days?
Input Specification:
The only line containing one integer *x* (1<=β€<=*x*<=β€<=109).
Output Specification:
The only line containing one integer: the answer.
Demo Input:
['5\n', '8\n']
Demo Output:
['2\n', '1\n']
Note:
For the first sample, we can add one bacterium in the box in the first day morning and at the third morning there will be 4 bacteria in the box. Now we put one more resulting 5 in the box. We added 2 bacteria in the process so the answer is 2.
For the second sample, we can put one in the first morning and in the 4-th morning there will be 8 in the box. So the answer is 1.
|
```python
w2=int(input())
w=bin(w2)
w1=w[2:]
c=0
for i in w1:
if int(i)%2!=0:
c+=1
print(c)
```
| 3
|
|
260
|
A
|
Adding Digits
|
PROGRAMMING
| 1,400
|
[
"implementation",
"math"
] | null | null |
Vasya has got two number: *a* and *b*. However, Vasya finds number *a* too short. So he decided to repeat the operation of lengthening number *a* *n* times.
One operation of lengthening a number means adding exactly one digit to the number (in the decimal notation) to the right provided that the resulting number is divisible by Vasya's number *b*. If it is impossible to obtain the number which is divisible by *b*, then the lengthening operation cannot be performed.
Your task is to help Vasya and print the number he can get after applying the lengthening operation to number *a* *n* times.
|
The first line contains three integers: *a*,<=*b*,<=*n* (1<=β€<=*a*,<=*b*,<=*n*<=β€<=105).
|
In a single line print the integer without leading zeros, which Vasya can get when he applies the lengthening operations to number *a* *n* times. If no such number exists, then print number -1. If there are multiple possible answers, print any of them.
|
[
"5 4 5\n",
"12 11 1\n",
"260 150 10\n"
] |
[
"524848\n",
"121\n",
"-1\n"
] |
none
| 500
|
[
{
"input": "5 4 5",
"output": "524848"
},
{
"input": "12 11 1",
"output": "121"
},
{
"input": "260 150 10",
"output": "-1"
},
{
"input": "78843 5684 42717",
"output": "-1"
},
{
"input": "93248 91435 1133",
"output": "-1"
},
{
"input": "100000 10 64479",
"output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "99999 21 73839",
"output": "9999990000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "99991 623 36438",
"output": "9999150000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "99999 334 94854",
"output": "9999960000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "99252 9827 84849",
"output": "9925270000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "99313 9833 10561",
"output": "9931330000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "94885 55815 11417",
"output": "9488550000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "99492 58525 53481",
"output": "9949250000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "99858 28531 79193",
"output": "9985850000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "99136 47208 42607",
"output": "9913680000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "63270 19953 5555",
"output": "-1"
},
{
"input": "10240 128 100000",
"output": "1024000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "12 11 3",
"output": "12100"
},
{
"input": "14 12 99998",
"output": "1440000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "1 11 3",
"output": "1100"
},
{
"input": "3 40 1",
"output": "-1"
},
{
"input": "150 100 10",
"output": "1500000000000"
},
{
"input": "5 10 1",
"output": "50"
},
{
"input": "1 15 10",
"output": "15000000000"
},
{
"input": "3 13 2",
"output": "390"
}
] | 1,608,641,756
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 25
| 483
| 8,704,000
|
a,b,n=map(int,input().split())
N=n
c=0
if(a%b==0):
print(int(str(a)+'0'*n))
else:
for i in range(n):
for j in range(0, 10):
s = str(a) + str(j)
num = int(s)
if (num % b == 0):
ans = num
break
else:
c += 1
if (c == n * 10):
print(-1)
else:
if(len(str(num))==N+len(str(a))):
print(num)
else:
l=len(str(a))+n-len(str(num))
number=int(str(num)+'0'*l)
print(number)
|
Title: Adding Digits
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya has got two number: *a* and *b*. However, Vasya finds number *a* too short. So he decided to repeat the operation of lengthening number *a* *n* times.
One operation of lengthening a number means adding exactly one digit to the number (in the decimal notation) to the right provided that the resulting number is divisible by Vasya's number *b*. If it is impossible to obtain the number which is divisible by *b*, then the lengthening operation cannot be performed.
Your task is to help Vasya and print the number he can get after applying the lengthening operation to number *a* *n* times.
Input Specification:
The first line contains three integers: *a*,<=*b*,<=*n* (1<=β€<=*a*,<=*b*,<=*n*<=β€<=105).
Output Specification:
In a single line print the integer without leading zeros, which Vasya can get when he applies the lengthening operations to number *a* *n* times. If no such number exists, then print number -1. If there are multiple possible answers, print any of them.
Demo Input:
['5 4 5\n', '12 11 1\n', '260 150 10\n']
Demo Output:
['524848\n', '121\n', '-1\n']
Note:
none
|
```python
a,b,n=map(int,input().split())
N=n
c=0
if(a%b==0):
print(int(str(a)+'0'*n))
else:
for i in range(n):
for j in range(0, 10):
s = str(a) + str(j)
num = int(s)
if (num % b == 0):
ans = num
break
else:
c += 1
if (c == n * 10):
print(-1)
else:
if(len(str(num))==N+len(str(a))):
print(num)
else:
l=len(str(a))+n-len(str(num))
number=int(str(num)+'0'*l)
print(number)
```
| 3
|
|
544
|
A
|
Set of Strings
|
PROGRAMMING
| 1,100
|
[
"implementation",
"strings"
] | null | null |
You are given a string *q*. A sequence of *k* strings *s*1,<=*s*2,<=...,<=*s**k* is called beautiful, if the concatenation of these strings is string *q* (formally, *s*1<=+<=*s*2<=+<=...<=+<=*s**k*<==<=*q*) and the first characters of these strings are distinct.
Find any beautiful sequence of strings or determine that the beautiful sequence doesn't exist.
|
The first line contains a positive integer *k* (1<=β€<=*k*<=β€<=26) β the number of strings that should be in a beautiful sequence.
The second line contains string *q*, consisting of lowercase Latin letters. The length of the string is within range from 1 to 100, inclusive.
|
If such sequence doesn't exist, then print in a single line "NO" (without the quotes). Otherwise, print in the first line "YES" (without the quotes) and in the next *k* lines print the beautiful sequence of strings *s*1,<=*s*2,<=...,<=*s**k*.
If there are multiple possible answers, print any of them.
|
[
"1\nabca\n",
"2\naaacas\n",
"4\nabc\n"
] |
[
"YES\nabca\n",
"YES\naaa\ncas\n",
"NO\n"
] |
In the second sample there are two possible answers: {"*aaaca*",β"*s*"} and {"*aaa*",β"*cas*"}.
| 500
|
[
{
"input": "1\nabca",
"output": "YES\nabca"
},
{
"input": "2\naaacas",
"output": "YES\naaa\ncas"
},
{
"input": "4\nabc",
"output": "NO"
},
{
"input": "3\nnddkhkhkdndknndkhrnhddkrdhrnrrnkkdnnndndrdhnknknhnrnnkrrdhrkhkrkhnkhkhhrhdnrndnknrrhdrdrkhdrkkhkrnkk",
"output": "YES\nn\ndd\nkhkhkdndknndkhrnhddkrdhrnrrnkkdnnndndrdhnknknhnrnnkrrdhrkhkrkhnkhkhhrhdnrndnknrrhdrdrkhdrkkhkrnkk"
},
{
"input": "26\nbiibfmmfifmffbmmfmbmbmiimbmiffmffibibfbiffibibiiimbffbbfbifmiibffbmbbbfmfibmibfffibfbffmfmimbmmmfmfm",
"output": "NO"
},
{
"input": "3\nkydoybxlfeugtrbvqnrjtzshorrsrwsxkvlwyolbaadtzpmyyfllxuciia",
"output": "YES\nk\ny\ndoybxlfeugtrbvqnrjtzshorrsrwsxkvlwyolbaadtzpmyyfllxuciia"
},
{
"input": "3\nssussususskkskkskuusksuuussksukkskuksukukusssususuususkkuukssuksskusukkssuksskskuskusussusskskksksus",
"output": "YES\nss\nussususs\nkkskkskuusksuuussksukkskuksukukusssususuususkkuukssuksskusukkssuksskskuskusussusskskksksus"
},
{
"input": "5\naaaaabcdef",
"output": "YES\naaaaa\nb\nc\nd\nef"
},
{
"input": "3\niiiiiiimiriiriwmimtmwrhhxmbmhwgghhgbqhywebrblyhlxjrthoooltehrmdhqhuodjmsjwcgrfnttiitpmqvbhlafwtzyikc",
"output": "YES\niiiiiii\nmi\nriiriwmimtmwrhhxmbmhwgghhgbqhywebrblyhlxjrthoooltehrmdhqhuodjmsjwcgrfnttiitpmqvbhlafwtzyikc"
},
{
"input": "20\ngggggllglgllltgtlglttstsgtttsslhhlssghgagtlsaghhoggtfgsaahtotdodthfltdxggxislnttlanxonhnkddtigppitdh",
"output": "NO"
},
{
"input": "16\nkkkkkkyykkynkknkkonyokdndkyonokdywkwykdkdotknnwzkoywiooinkcyzyntcdnitnppnpziomyzdspomoqmomcyrrospppn",
"output": "NO"
},
{
"input": "15\nwwwgggowgwwhoohwgwghwyohhggywhyyodgwydwgggkhgyydqyggkgkpokgthqghidhworprodtcogqkwgtfiodwdurcctkmrfmh",
"output": "YES\nwww\nggg\nowgww\nhoohwgwghw\nyohhggywhyyo\ndgwydwggg\nkhgyyd\nqyggkgk\npokg\nthqgh\nidhwo\nrprodt\ncogqkwgt\nfiodwd\nurcctkmrfmh"
},
{
"input": "15\nnnnnnntnttttttqqnqqynnqqwwnnnwneenhwtyhhoqeyeqyeuthwtnhtpnphhwetjhouhwnpojvvovoswwjryrwerbwwpbvrwvjj",
"output": "YES\nnnnnnn\ntntttttt\nqqnqq\nynnqq\nwwnnnwn\neen\nhwtyhh\noqeyeqye\nuthwtnht\npnphhwet\njhouhwnpoj\nvvovo\nswwj\nryrwer\nbwwpbvrwvjj"
},
{
"input": "15\nvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv",
"output": "NO"
},
{
"input": "1\niiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiaaaaaiiiiaiaiiiiaaiaiiiaiiaiaaiaiiaiiiiiaiiiaiiiaiaiaai",
"output": "YES\niiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiaaaaaiiiiaiaiiiiaaiaiiiaiiaiaaiaiiaiiiiiaiiiaiiiaiaiaai"
},
{
"input": "26\nvvvnnsnnnpsnnswwspncvshtncwphaphmwnwkhvvhuvctvnehemowkmtzissswjaxuuvphzrmfzihamdqmmyhhijbitlipgltyy",
"output": "YES\nvvv\nnn\nsnnn\npsnns\nwwspn\ncvs\nh\ntncwph\naph\nmwnw\nkhvvh\nuvctvn\nehem\nowkmt\nz\nisssw\nja\nxuuvphz\nrm\nfziham\nd\nqmm\nyhhij\nbit\nlip\ngltyy"
},
{
"input": "26\njexzsbwaih",
"output": "NO"
},
{
"input": "1\nk",
"output": "YES\nk"
},
{
"input": "1\nzz",
"output": "YES\nzz"
},
{
"input": "3\nziw",
"output": "YES\nz\ni\nw"
},
{
"input": "26\ntjmbyqwuahlixegopkzrfndcsv",
"output": "YES\nt\nj\nm\nb\ny\nq\nw\nu\na\nh\nl\ni\nx\ne\ng\no\np\nk\nz\nr\nf\nn\nd\nc\ns\nv"
},
{
"input": "25\nvobekscyadzqwnjxruplifmthg",
"output": "YES\nv\no\nb\ne\nk\ns\nc\ny\na\nd\nz\nq\nw\nn\nj\nx\nr\nu\np\nl\ni\nf\nm\nt\nhg"
},
{
"input": "26\nlllplzkkzflzflffzznnnnfgflqlttlmtnkzlztskngyymitqagattkdllyutzimsrskpapcmuupjdopxqlnhqcscwvdtxbflefy",
"output": "YES\nlll\npl\nz\nkkz\nflzflffzz\nnnnnf\ngfl\nql\nttl\nmtnkzlzt\nskng\nyym\nitq\nagattk\ndlly\nutzims\nrskpap\ncmuup\njd\nop\nxqln\nhqcsc\nw\nvdtx\nbfl\nefy"
},
{
"input": "25\nkkrrkrkrkrsrskpskbrppdsdbgbkrbllkbswdwcchgskmkhwiidicczlscsodtjglxbmeotzxnmbjmoqgkquglaoxgcykxvbhdi",
"output": "YES\nkk\nrrkrkrkr\nsrsk\npsk\nbrpp\ndsdb\ngbkrb\nllkbs\nwdw\ncc\nhgsk\nmkhw\niidicc\nzlscs\nod\nt\njgl\nxbm\neotzx\nnmbjmo\nqgkq\nugl\naoxgc\nykx\nvbhdi"
},
{
"input": "25\nuuuuuccpucubccbupxubcbpujiliwbpqbpyiweuywaxwqasbsllwehceruytjvphytraawgbjmerfeymoayujqranlvkpkiypadr",
"output": "YES\nuuuuu\ncc\npucu\nbccbup\nxubcbpu\nj\ni\nli\nwbp\nqbp\nyiw\neuyw\naxwqa\nsbsllwe\nhce\nruy\ntj\nvphytraaw\ngbj\nmer\nfeym\noayujqra\nnlv\nkpkiypa\ndr"
},
{
"input": "26\nxxjxodrogovufvohrodliretxxyjqnrbzmicorptkjafiwmsbwml",
"output": "YES\nxx\njx\no\nd\nro\ngo\nv\nu\nfvo\nhrod\nl\nir\ne\ntxx\nyj\nq\nnr\nb\nz\nmi\ncor\npt\nkj\nafi\nwm\nsbwml"
},
{
"input": "26\npjhsxjbvkqntwmsdnrguecaofylzti",
"output": "YES\np\nj\nh\ns\nxj\nb\nv\nk\nq\nn\nt\nw\nms\ndn\nr\ng\nu\ne\nc\na\no\nf\ny\nl\nzt\ni"
},
{
"input": "25\nrrrrqqwrlqrwglrlylwhrrwyvrhvzgvqahrhgsvavtggyduayivxzgeicinlnrkapoepbsfyjjrt",
"output": "YES\nrrrr\nqq\nwr\nlqrw\nglrl\nylw\nhrrwy\nvrhv\nzgvq\nahrhg\nsvav\ntggy\nd\nuay\niv\nxzg\nei\nci\nnlnr\nka\np\noep\nbs\nfy\njjrt"
},
{
"input": "26\ncccccccaacwwaxxaacczacnnnqqwnaggzqrwagcnabxnrcvgjqjamqzgdntzanaxvjfwqlvdttuzjoxiwtkqvrydospmpeirhg",
"output": "YES\nccccccc\naac\nwwa\nxxaacc\nzac\nnnn\nqqwna\nggzq\nrwagcna\nbxnrc\nvg\njqja\nmqzg\ndn\ntzanaxvj\nfwq\nlvdtt\nuzj\nox\niwt\nkqvr\nydo\ns\npmp\neir\nhg"
},
{
"input": "4\nsssssssssssssssssssssssssslsslslsllsslssslslssllaaslsaalsasaassllasasalrlrslarlaarlsrsaslasarlr",
"output": "YES\nssssssssssssssssssssssssss\nlsslslsllsslssslslssll\naaslsaalsasaassllasasal\nrlrslarlaarlsrsaslasarlr"
},
{
"input": "26\na",
"output": "NO"
},
{
"input": "26\nab",
"output": "NO"
},
{
"input": "2\nab",
"output": "YES\na\nb"
}
] | 1,537,999,835
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 33
| 140
| 0
|
k = int(input())
s = input()
used = [False for _ in range(26)]
starts = []
for i, c in enumerate(s):
x = ord(c) - ord('a')
if not used[x]:
used[x] = True
starts.append(i)
if len(starts) < k:
print('NO')
else:
print('YES')
while len(starts) > k:
starts.pop()
for i in range(k - 1):
print(s[starts[i]:starts[i + 1]])
print(s[starts[-1]:])
|
Title: Set of Strings
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a string *q*. A sequence of *k* strings *s*1,<=*s*2,<=...,<=*s**k* is called beautiful, if the concatenation of these strings is string *q* (formally, *s*1<=+<=*s*2<=+<=...<=+<=*s**k*<==<=*q*) and the first characters of these strings are distinct.
Find any beautiful sequence of strings or determine that the beautiful sequence doesn't exist.
Input Specification:
The first line contains a positive integer *k* (1<=β€<=*k*<=β€<=26) β the number of strings that should be in a beautiful sequence.
The second line contains string *q*, consisting of lowercase Latin letters. The length of the string is within range from 1 to 100, inclusive.
Output Specification:
If such sequence doesn't exist, then print in a single line "NO" (without the quotes). Otherwise, print in the first line "YES" (without the quotes) and in the next *k* lines print the beautiful sequence of strings *s*1,<=*s*2,<=...,<=*s**k*.
If there are multiple possible answers, print any of them.
Demo Input:
['1\nabca\n', '2\naaacas\n', '4\nabc\n']
Demo Output:
['YES\nabca\n', 'YES\naaa\ncas\n', 'NO\n']
Note:
In the second sample there are two possible answers: {"*aaaca*",β"*s*"} and {"*aaa*",β"*cas*"}.
|
```python
k = int(input())
s = input()
used = [False for _ in range(26)]
starts = []
for i, c in enumerate(s):
x = ord(c) - ord('a')
if not used[x]:
used[x] = True
starts.append(i)
if len(starts) < k:
print('NO')
else:
print('YES')
while len(starts) > k:
starts.pop()
for i in range(k - 1):
print(s[starts[i]:starts[i + 1]])
print(s[starts[-1]:])
```
| 3
|
|
988
|
A
|
Diverse Team
|
PROGRAMMING
| 800
|
[
"brute force",
"implementation"
] | null | null |
There are $n$ students in a school class, the rating of the $i$-th student on Codehorses is $a_i$. You have to form a team consisting of $k$ students ($1 \le k \le n$) such that the ratings of all team members are distinct.
If it is impossible to form a suitable team, print "NO" (without quotes). Otherwise print "YES", and then print $k$ distinct numbers which should be the indices of students in the team you form. If there are multiple answers, print any of them.
|
The first line contains two integers $n$ and $k$ ($1 \le k \le n \le 100$) β the number of students and the size of the team you have to form.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 100$), where $a_i$ is the rating of $i$-th student.
|
If it is impossible to form a suitable team, print "NO" (without quotes). Otherwise print "YES", and then print $k$ distinct integers from $1$ to $n$ which should be the indices of students in the team you form. All the ratings of the students in the team should be distinct. You may print the indices in any order. If there are multiple answers, print any of them.
Assume that the students are numbered from $1$ to $n$.
|
[
"5 3\n15 13 15 15 12\n",
"5 4\n15 13 15 15 12\n",
"4 4\n20 10 40 30\n"
] |
[
"YES\n1 2 5 \n",
"NO\n",
"YES\n1 2 3 4 \n"
] |
All possible answers for the first example:
- {1 2 5} - {2 3 5} - {2 4 5}
Note that the order does not matter.
| 0
|
[
{
"input": "5 3\n15 13 15 15 12",
"output": "YES\n1 2 5 "
},
{
"input": "5 4\n15 13 15 15 12",
"output": "NO"
},
{
"input": "4 4\n20 10 40 30",
"output": "YES\n1 2 3 4 "
},
{
"input": "1 1\n1",
"output": "YES\n1 "
},
{
"input": "100 53\n16 17 1 2 27 5 9 9 53 24 17 33 35 24 20 48 56 73 12 14 39 55 58 13 59 73 29 26 40 33 22 29 34 22 55 38 63 66 36 13 60 42 10 15 21 9 11 5 23 37 79 47 26 3 79 53 44 8 71 75 42 11 34 39 79 33 10 26 23 23 17 14 54 41 60 31 83 5 45 4 14 35 6 60 28 48 23 18 60 36 21 28 7 34 9 25 52 43 54 19",
"output": "YES\n1 2 3 4 5 6 7 9 10 12 13 15 16 17 18 19 20 21 22 23 24 25 27 28 29 31 33 36 37 38 39 41 42 43 44 45 47 49 50 51 52 54 57 58 59 60 73 74 76 77 79 80 83 "
},
{
"input": "2 2\n100 100",
"output": "NO"
},
{
"input": "2 2\n100 99",
"output": "YES\n1 2 "
},
{
"input": "100 100\n63 100 75 32 53 24 73 98 76 15 70 48 8 81 88 58 95 78 27 92 14 16 72 43 46 39 66 38 64 42 59 9 22 51 4 6 10 94 28 99 68 80 35 50 45 20 47 7 30 26 49 91 77 19 96 57 65 1 11 13 31 12 82 87 93 34 62 3 21 79 56 41 89 18 44 23 74 86 2 33 69 36 61 67 25 83 5 84 90 37 40 29 97 60 52 55 54 71 17 85",
"output": "YES\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 99 100 "
},
{
"input": "100 41\n54 16 42 3 45 6 9 72 100 13 24 57 35 5 89 13 97 27 43 9 73 89 48 16 48 55 18 15 55 28 30 6 18 41 100 61 9 42 35 54 57 25 73 15 42 54 49 5 72 48 30 55 4 43 94 5 60 92 93 23 89 75 53 92 74 93 89 28 69 6 3 49 15 28 49 57 54 55 30 57 69 18 89 6 25 23 93 74 30 13 87 53 6 42 4 54 60 30 4 35",
"output": "NO"
},
{
"input": "100 2\n70 64 70 32 70 64 32 70 64 32 32 64 70 64 64 32 64 64 64 70 70 64 64 64 64 70 32 64 70 64 32 70 70 70 64 70 64 70 64 32 70 32 70 64 64 64 32 70 64 70 70 32 70 32 32 32 70 32 70 32 64 64 70 32 32 64 70 64 32 32 64 64 32 32 70 70 32 70 32 64 32 70 64 64 32 64 32 64 70 32 70 32 70 64 64 64 70 70 64 70",
"output": "YES\n1 2 "
}
] | 1,637,503,302
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 10
| 46
| 0
|
n,k=map(int,(input()).split(" "))
z=0
m=input().split(" ")
x=[]
x.extend(set(m))
if len(x)>= k:
print("YES")
for i in range(k):
print(m.index (x[i])+1,end=" ")
z+=1
else:
print("NO")
|
Title: Diverse Team
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are $n$ students in a school class, the rating of the $i$-th student on Codehorses is $a_i$. You have to form a team consisting of $k$ students ($1 \le k \le n$) such that the ratings of all team members are distinct.
If it is impossible to form a suitable team, print "NO" (without quotes). Otherwise print "YES", and then print $k$ distinct numbers which should be the indices of students in the team you form. If there are multiple answers, print any of them.
Input Specification:
The first line contains two integers $n$ and $k$ ($1 \le k \le n \le 100$) β the number of students and the size of the team you have to form.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 100$), where $a_i$ is the rating of $i$-th student.
Output Specification:
If it is impossible to form a suitable team, print "NO" (without quotes). Otherwise print "YES", and then print $k$ distinct integers from $1$ to $n$ which should be the indices of students in the team you form. All the ratings of the students in the team should be distinct. You may print the indices in any order. If there are multiple answers, print any of them.
Assume that the students are numbered from $1$ to $n$.
Demo Input:
['5 3\n15 13 15 15 12\n', '5 4\n15 13 15 15 12\n', '4 4\n20 10 40 30\n']
Demo Output:
['YES\n1 2 5 \n', 'NO\n', 'YES\n1 2 3 4 \n']
Note:
All possible answers for the first example:
- {1 2 5} - {2 3 5} - {2 4 5}
Note that the order does not matter.
|
```python
n,k=map(int,(input()).split(" "))
z=0
m=input().split(" ")
x=[]
x.extend(set(m))
if len(x)>= k:
print("YES")
for i in range(k):
print(m.index (x[i])+1,end=" ")
z+=1
else:
print("NO")
```
| 3
|
|
808
|
C
|
Tea Party
|
PROGRAMMING
| 1,400
|
[
"constructive algorithms",
"greedy",
"sortings"
] | null | null |
Polycarp invited all his friends to the tea party to celebrate the holiday. He has *n* cups, one for each of his *n* friends, with volumes *a*1,<=*a*2,<=...,<=*a**n*. His teapot stores *w* milliliters of tea (*w*<=β€<=*a*1<=+<=*a*2<=+<=...<=+<=*a**n*). Polycarp wants to pour tea in cups in such a way that:
- Every cup will contain tea for at least half of its volume - Every cup will contain integer number of milliliters of tea - All the tea from the teapot will be poured into cups - All friends will be satisfied.
Friend with cup *i* won't be satisfied, if there exists such cup *j* that cup *i* contains less tea than cup *j* but *a**i*<=><=*a**j*.
For each cup output how many milliliters of tea should be poured in it. If it's impossible to pour all the tea and satisfy all conditions then output -1.
|
The first line contains two integer numbers *n* and *w* (1<=β€<=*n*<=β€<=100, ).
The second line contains *n* numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=β€<=*a**i*<=β€<=100).
|
Output how many milliliters of tea every cup should contain. If there are multiple answers, print any of them.
If it's impossible to pour all the tea and satisfy all conditions then output -1.
|
[
"2 10\n8 7\n",
"4 4\n1 1 1 1\n",
"3 10\n9 8 10\n"
] |
[
"6 4 \n",
"1 1 1 1 \n",
"-1\n"
] |
In the third example you should pour to the first cup at least 5 milliliters, to the second one at least 4, to the third one at least 5. It sums up to 14, which is greater than 10 milliliters available.
| 0
|
[
{
"input": "2 10\n8 7",
"output": "6 4 "
},
{
"input": "4 4\n1 1 1 1",
"output": "1 1 1 1 "
},
{
"input": "3 10\n9 8 10",
"output": "-1"
},
{
"input": "1 1\n1",
"output": "1 "
},
{
"input": "1 1\n2",
"output": "1 "
},
{
"input": "1 10\n20",
"output": "10 "
},
{
"input": "3 10\n8 4 8",
"output": "4 2 4 "
},
{
"input": "3 100\n37 26 37",
"output": "37 26 37 "
},
{
"input": "3 60\n43 23 24",
"output": "36 12 12 "
},
{
"input": "20 14\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "-1"
},
{
"input": "20 8\n1 2 1 2 1 1 1 2 1 1 1 2 1 1 2 1 1 1 2 2",
"output": "-1"
},
{
"input": "50 1113\n25 21 23 37 28 23 19 25 5 12 3 11 46 50 13 50 7 1 8 40 4 6 34 27 11 39 45 31 10 12 48 2 19 37 47 45 30 24 21 42 36 14 31 30 31 50 6 3 33 49",
"output": "13 11 12 37 28 12 10 18 3 6 2 6 46 50 7 50 4 1 4 40 2 3 34 27 6 39 45 31 5 6 48 1 10 37 47 45 30 12 11 42 36 7 31 30 31 50 3 2 33 49 "
},
{
"input": "50 440\n14 69 33 38 83 65 21 66 89 3 93 60 31 16 61 20 42 64 13 1 50 50 74 58 67 61 52 22 69 68 18 33 28 59 4 8 96 32 84 85 87 87 61 89 2 47 15 64 88 18",
"output": "-1"
},
{
"input": "100 640\n82 51 81 14 37 17 78 92 64 15 8 86 89 8 87 77 66 10 15 12 100 25 92 47 21 78 20 63 13 49 41 36 41 79 16 87 87 69 3 76 80 60 100 49 70 59 72 8 38 71 45 97 71 14 76 54 81 4 59 46 39 29 92 3 49 22 53 99 59 52 74 31 92 43 42 23 44 9 82 47 7 40 12 9 3 55 37 85 46 22 84 52 98 41 21 77 63 17 62 91",
"output": "-1"
},
{
"input": "100 82\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": "-1"
},
{
"input": "100 55\n1 1 1 1 2 1 1 1 1 1 2 2 1 1 2 1 2 1 1 1 2 1 1 2 1 2 1 1 2 2 2 1 1 2 1 1 1 2 2 2 1 1 1 2 1 2 2 1 2 1 1 2 2 1 2 1 2 1 2 2 1 1 1 2 1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 1 1 1 1 2 2 2 2 2 2 2 1 1 1 2 1 2 1",
"output": "-1"
},
{
"input": "30 50\n3 1 2 4 1 2 2 4 3 4 4 3 3 3 3 5 3 2 5 4 3 3 5 3 3 5 4 5 3 5",
"output": "-1"
},
{
"input": "40 100\n3 3 3 3 4 1 1 1 1 1 2 2 1 3 1 2 3 2 1 2 2 2 1 4 2 2 3 3 3 2 4 6 4 4 3 2 2 2 4 5",
"output": "3 3 3 3 4 1 1 1 1 1 2 2 1 3 1 2 3 2 1 2 2 2 1 4 2 2 3 3 3 2 4 6 4 4 3 2 2 2 4 5 "
},
{
"input": "100 10000\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100",
"output": "100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 "
},
{
"input": "2 5\n3 4",
"output": "2 3 "
},
{
"input": "2 6\n2 6",
"output": "1 5 "
},
{
"input": "23 855\n5 63 94 57 38 84 77 79 83 36 47 31 60 79 75 48 88 17 46 33 23 15 27",
"output": "3 32 94 29 19 84 39 72 83 18 24 16 30 79 38 24 88 9 23 17 12 8 14 "
},
{
"input": "52 2615\n73 78 70 92 94 74 46 19 55 20 70 3 1 42 68 10 66 80 1 31 65 19 73 74 56 35 53 38 92 35 65 81 6 98 74 51 27 49 76 19 86 76 5 60 14 75 64 99 43 7 36 79",
"output": "73 78 70 92 94 74 46 10 55 10 70 2 1 42 68 5 66 80 1 16 65 10 73 74 56 18 53 38 92 30 65 81 3 98 74 51 14 49 76 10 86 76 3 60 7 75 64 99 43 4 36 79 "
},
{
"input": "11 287\n34 30 69 86 22 53 11 91 62 44 5",
"output": "17 15 35 43 11 27 6 77 31 22 3 "
},
{
"input": "55 1645\n60 53 21 20 87 48 10 21 76 35 52 41 82 86 93 11 93 86 34 15 37 63 57 3 57 57 32 8 55 25 29 38 46 22 13 87 27 35 40 83 5 7 6 18 88 25 4 59 95 62 31 93 98 50 62",
"output": "30 27 11 10 82 24 5 11 38 18 26 21 41 43 93 6 93 43 17 8 19 32 29 2 29 29 16 4 28 13 15 19 23 11 7 87 14 18 20 42 3 4 3 9 88 13 2 30 95 31 16 93 98 25 31 "
},
{
"input": "71 3512\n97 46 76 95 81 96 99 83 10 50 19 18 73 5 41 60 12 73 60 31 21 64 88 61 43 57 61 19 75 35 41 85 12 59 32 47 37 43 35 92 90 47 3 98 21 18 61 79 39 86 74 8 52 33 39 27 93 54 35 38 96 36 83 51 97 10 8 66 75 87 68",
"output": "97 46 76 95 81 96 99 83 5 50 10 9 73 3 41 60 6 73 60 16 11 64 88 61 43 57 61 10 75 18 41 85 6 59 16 47 19 43 18 92 90 47 2 98 11 9 61 79 20 86 74 4 52 17 21 14 93 54 18 19 96 18 83 51 97 5 4 66 75 87 68 "
},
{
"input": "100 2633\n99 50 64 81 75 73 26 31 31 36 95 12 100 2 70 72 78 56 76 23 94 8 91 1 39 82 97 67 64 25 71 90 48 34 31 46 64 37 46 50 99 93 14 56 1 89 95 89 50 52 12 58 43 65 45 88 90 14 38 19 6 15 91 67 43 48 82 20 11 48 33 20 39 52 73 5 25 84 26 54 42 56 10 28 9 63 60 98 30 1 25 74 86 56 85 9 12 94 80 95",
"output": "50 25 32 41 38 37 13 16 16 18 48 6 61 1 35 36 39 28 38 12 47 4 46 1 20 41 49 34 32 13 36 45 24 17 16 23 32 19 23 25 50 47 7 28 1 45 48 45 25 26 6 29 22 33 23 44 45 7 19 10 3 8 46 34 22 24 41 10 6 24 17 10 20 26 37 3 13 42 13 27 21 28 5 14 5 32 30 49 15 1 13 37 43 28 43 5 6 47 40 48 "
},
{
"input": "71 1899\n23 55 58 87 69 85 100 21 19 72 81 68 20 25 29 92 18 74 89 70 53 7 78 57 41 79 64 87 63 76 95 84 1 28 32 1 79 34 77 17 71 61 35 31 62 92 69 99 60 26 2 18 61 9 27 77 82 6 30 65 52 3 51 43 13 77 41 59 19 29 86",
"output": "12 28 29 44 35 43 95 11 10 36 41 34 10 13 15 46 9 37 45 35 27 4 39 29 21 40 32 44 32 38 48 42 1 14 16 1 40 17 39 9 36 31 18 16 31 46 35 50 30 13 1 9 31 5 14 39 41 3 15 33 26 2 26 22 7 39 21 30 10 15 43 "
},
{
"input": "10 21\n3 3 3 3 4 3 3 3 3 3",
"output": "2 2 2 2 3 2 2 2 2 2 "
}
] | 1,564,643,019
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 29
| 140
| 0
|
n, w = (int(i) for i in input().split())
mas = [int(i) for i in input().split()]
def qer():
global w
for i, size in enumerate(mas):
q = round(size / 2 + 0.1)
rer[size,i] = q
w -= q
if w < 0:
return {1:-1}
for size,i in sorted(rer, reverse = True):
o = size - rer[size,i]
rer[size,i] += min(w,o)
w -= min(o,w)
if w == 0:
break
return rer
rer = {}
print(*qer().values())
|
Title: Tea Party
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp invited all his friends to the tea party to celebrate the holiday. He has *n* cups, one for each of his *n* friends, with volumes *a*1,<=*a*2,<=...,<=*a**n*. His teapot stores *w* milliliters of tea (*w*<=β€<=*a*1<=+<=*a*2<=+<=...<=+<=*a**n*). Polycarp wants to pour tea in cups in such a way that:
- Every cup will contain tea for at least half of its volume - Every cup will contain integer number of milliliters of tea - All the tea from the teapot will be poured into cups - All friends will be satisfied.
Friend with cup *i* won't be satisfied, if there exists such cup *j* that cup *i* contains less tea than cup *j* but *a**i*<=><=*a**j*.
For each cup output how many milliliters of tea should be poured in it. If it's impossible to pour all the tea and satisfy all conditions then output -1.
Input Specification:
The first line contains two integer numbers *n* and *w* (1<=β€<=*n*<=β€<=100, ).
The second line contains *n* numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=β€<=*a**i*<=β€<=100).
Output Specification:
Output how many milliliters of tea every cup should contain. If there are multiple answers, print any of them.
If it's impossible to pour all the tea and satisfy all conditions then output -1.
Demo Input:
['2 10\n8 7\n', '4 4\n1 1 1 1\n', '3 10\n9 8 10\n']
Demo Output:
['6 4 \n', '1 1 1 1 \n', '-1\n']
Note:
In the third example you should pour to the first cup at least 5 milliliters, to the second one at least 4, to the third one at least 5. It sums up to 14, which is greater than 10 milliliters available.
|
```python
n, w = (int(i) for i in input().split())
mas = [int(i) for i in input().split()]
def qer():
global w
for i, size in enumerate(mas):
q = round(size / 2 + 0.1)
rer[size,i] = q
w -= q
if w < 0:
return {1:-1}
for size,i in sorted(rer, reverse = True):
o = size - rer[size,i]
rer[size,i] += min(w,o)
w -= min(o,w)
if w == 0:
break
return rer
rer = {}
print(*qer().values())
```
| 3
|
|
710
|
A
|
King Moves
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
The only king stands on the standard chess board. You are given his position in format "cd", where *c* is the column from 'a' to 'h' and *d* is the row from '1' to '8'. Find the number of moves permitted for the king.
Check the king's moves here [https://en.wikipedia.org/wiki/King_(chess)](https://en.wikipedia.org/wiki/King_(chess)).
|
The only line contains the king's position in the format "cd", where 'c' is the column from 'a' to 'h' and 'd' is the row from '1' to '8'.
|
Print the only integer *x* β the number of moves permitted for the king.
|
[
"e4\n"
] |
[
"8\n"
] |
none
| 0
|
[
{
"input": "e4",
"output": "8"
},
{
"input": "a1",
"output": "3"
},
{
"input": "h8",
"output": "3"
},
{
"input": "a4",
"output": "5"
},
{
"input": "g7",
"output": "8"
},
{
"input": "e1",
"output": "5"
},
{
"input": "b2",
"output": "8"
},
{
"input": "c7",
"output": "8"
},
{
"input": "h6",
"output": "5"
},
{
"input": "c8",
"output": "5"
},
{
"input": "h2",
"output": "5"
},
{
"input": "h5",
"output": "5"
},
{
"input": "a8",
"output": "3"
},
{
"input": "f8",
"output": "5"
},
{
"input": "h1",
"output": "3"
},
{
"input": "f2",
"output": "8"
},
{
"input": "e8",
"output": "5"
},
{
"input": "h3",
"output": "5"
},
{
"input": "b8",
"output": "5"
},
{
"input": "g8",
"output": "5"
},
{
"input": "d8",
"output": "5"
},
{
"input": "h4",
"output": "5"
},
{
"input": "b1",
"output": "5"
},
{
"input": "a2",
"output": "5"
}
] | 1,474,236,384
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 24
| 62
| 0
|
s=input()
if s in {"a8", "a1", "h1", "h8"}:
print(3)
elif s[0] in {'a', 'h'} or s[1] in {'1', '8'}:
print(5)
else:
print(8)
|
Title: King Moves
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The only king stands on the standard chess board. You are given his position in format "cd", where *c* is the column from 'a' to 'h' and *d* is the row from '1' to '8'. Find the number of moves permitted for the king.
Check the king's moves here [https://en.wikipedia.org/wiki/King_(chess)](https://en.wikipedia.org/wiki/King_(chess)).
Input Specification:
The only line contains the king's position in the format "cd", where 'c' is the column from 'a' to 'h' and 'd' is the row from '1' to '8'.
Output Specification:
Print the only integer *x* β the number of moves permitted for the king.
Demo Input:
['e4\n']
Demo Output:
['8\n']
Note:
none
|
```python
s=input()
if s in {"a8", "a1", "h1", "h8"}:
print(3)
elif s[0] in {'a', 'h'} or s[1] in {'1', '8'}:
print(5)
else:
print(8)
```
| 3
|
|
1,004
|
A
|
Sonya and Hotels
|
PROGRAMMING
| 900
|
[
"implementation"
] | null | null |
Sonya decided that having her own hotel business is the best way of earning money because she can profit and rest wherever she wants.
The country where Sonya lives is an endless line. There is a city in each integer coordinate on this line. She has $n$ hotels, where the $i$-th hotel is located in the city with coordinate $x_i$. Sonya is a smart girl, so she does not open two or more hotels in the same city.
Sonya understands that her business needs to be expanded by opening new hotels, so she decides to build one more. She wants to make the minimum distance from this hotel to all others to be equal to $d$. The girl understands that there are many possible locations to construct such a hotel. Thus she wants to know the number of possible coordinates of the cities where she can build a new hotel.
Because Sonya is lounging in a jacuzzi in one of her hotels, she is asking you to find the number of cities where she can build a new hotel so that the minimum distance from the original $n$ hotels to the new one is equal to $d$.
|
The first line contains two integers $n$ and $d$ ($1\leq n\leq 100$, $1\leq d\leq 10^9$)Β β the number of Sonya's hotels and the needed minimum distance from a new hotel to all others.
The second line contains $n$ different integers in strictly increasing order $x_1, x_2, \ldots, x_n$ ($-10^9\leq x_i\leq 10^9$)Β β coordinates of Sonya's hotels.
|
Print the number of cities where Sonya can build a new hotel so that the minimum distance from this hotel to all others is equal to $d$.
|
[
"4 3\n-3 2 9 16\n",
"5 2\n4 8 11 18 19\n"
] |
[
"6\n",
"5\n"
] |
In the first example, there are $6$ possible cities where Sonya can build a hotel. These cities have coordinates $-6$, $5$, $6$, $12$, $13$, and $19$.
In the second example, there are $5$ possible cities where Sonya can build a hotel. These cities have coordinates $2$, $6$, $13$, $16$, and $21$.
| 500
|
[
{
"input": "4 3\n-3 2 9 16",
"output": "6"
},
{
"input": "5 2\n4 8 11 18 19",
"output": "5"
},
{
"input": "10 10\n-67 -59 -49 -38 -8 20 41 59 74 83",
"output": "8"
},
{
"input": "10 10\n0 20 48 58 81 95 111 137 147 159",
"output": "9"
},
{
"input": "100 1\n0 1 2 3 4 5 7 8 10 11 12 13 14 15 16 17 19 21 22 23 24 25 26 27 28 30 32 33 36 39 40 41 42 46 48 53 54 55 59 60 61 63 65 68 70 71 74 75 76 79 80 81 82 84 88 89 90 91 93 94 96 97 98 100 101 102 105 106 107 108 109 110 111 113 114 115 116 117 118 120 121 122 125 126 128 131 132 133 134 135 137 138 139 140 143 144 146 147 148 149",
"output": "47"
},
{
"input": "1 1000000000\n-1000000000",
"output": "2"
},
{
"input": "2 1000000000\n-1000000000 1000000000",
"output": "3"
},
{
"input": "100 2\n1 3 5 6 8 9 12 13 14 17 18 21 22 23 24 25 26 27 29 30 34 35 36 39 41 44 46 48 52 53 55 56 57 59 61 63 64 66 68 69 70 71 72 73 75 76 77 79 80 81 82 87 88 91 92 93 94 95 96 97 99 100 102 103 104 106 109 110 111 112 113 114 115 117 118 119 120 122 124 125 127 128 129 130 131 132 133 134 136 137 139 140 141 142 143 145 146 148 149 150",
"output": "6"
},
{
"input": "100 3\n0 1 3 6 7 8 9 10 13 14 16 17 18 20 21 22 24 26 27 30 33 34 35 36 37 39 42 43 44 45 46 48 53 54 55 56 57 58 61 63 64 65 67 69 70 72 73 76 77 78 79 81 82 83 85 86 87 88 90 92 93 95 96 97 98 99 100 101 104 105 108 109 110 113 114 115 116 118 120 121 123 124 125 128 130 131 132 133 134 135 136 137 139 140 141 142 146 147 148 150",
"output": "2"
},
{
"input": "1 1000000000\n1000000000",
"output": "2"
},
{
"input": "10 2\n-93 -62 -53 -42 -38 11 57 58 87 94",
"output": "17"
},
{
"input": "2 500000000\n-1000000000 1000000000",
"output": "4"
},
{
"input": "100 10\n-489 -476 -445 -432 -430 -421 -420 -418 -412 -411 -404 -383 -356 -300 -295 -293 -287 -276 -265 -263 -258 -251 -249 -246 -220 -219 -205 -186 -166 -157 -143 -137 -136 -130 -103 -86 -80 -69 -67 -55 -43 -41 -40 -26 -19 -9 16 29 41 42 54 76 84 97 98 99 101 115 134 151 157 167 169 185 197 204 208 226 227 232 234 249 259 266 281 282 293 298 300 306 308 313 319 328 331 340 341 344 356 362 366 380 390 399 409 411 419 444 455 498",
"output": "23"
},
{
"input": "1 1000000000\n999999999",
"output": "2"
},
{
"input": "1 1\n-5",
"output": "2"
},
{
"input": "2 1\n-1000000000 1000000000",
"output": "4"
},
{
"input": "1 2\n1",
"output": "2"
},
{
"input": "4 5\n0 20 40 60",
"output": "8"
},
{
"input": "1 1\n-10",
"output": "2"
},
{
"input": "1 1\n-1000000000",
"output": "2"
},
{
"input": "1 1000000000\n0",
"output": "2"
},
{
"input": "1 2\n-10",
"output": "2"
},
{
"input": "1 1\n2",
"output": "2"
},
{
"input": "1 5\n-100",
"output": "2"
},
{
"input": "2 1000000000\n999999999 1000000000",
"output": "2"
},
{
"input": "2 2\n0 5",
"output": "4"
},
{
"input": "1 2\n-5",
"output": "2"
},
{
"input": "1 1\n1",
"output": "2"
},
{
"input": "1 5\n-20",
"output": "2"
},
{
"input": "2 100000000\n-1000000000 1000000000",
"output": "4"
},
{
"input": "1 1000000000\n147483641",
"output": "2"
},
{
"input": "3 3\n1 8 9",
"output": "4"
},
{
"input": "2 1\n1 1000000000",
"output": "4"
},
{
"input": "1 2\n-4",
"output": "2"
},
{
"input": "2 1\n-1000000000 -1",
"output": "4"
}
] | 1,594,065,824
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 45
| 109
| 6,656,000
|
n,k = map(int, input().split())
k *= 2
l = input().split()
sumi = 2
for x in range(len(l)-1):
y = int(l[x+1])-int(l[x])
if y == k:
sumi += 1
elif y > k:
sumi += 2
print(sumi)
|
Title: Sonya and Hotels
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Sonya decided that having her own hotel business is the best way of earning money because she can profit and rest wherever she wants.
The country where Sonya lives is an endless line. There is a city in each integer coordinate on this line. She has $n$ hotels, where the $i$-th hotel is located in the city with coordinate $x_i$. Sonya is a smart girl, so she does not open two or more hotels in the same city.
Sonya understands that her business needs to be expanded by opening new hotels, so she decides to build one more. She wants to make the minimum distance from this hotel to all others to be equal to $d$. The girl understands that there are many possible locations to construct such a hotel. Thus she wants to know the number of possible coordinates of the cities where she can build a new hotel.
Because Sonya is lounging in a jacuzzi in one of her hotels, she is asking you to find the number of cities where she can build a new hotel so that the minimum distance from the original $n$ hotels to the new one is equal to $d$.
Input Specification:
The first line contains two integers $n$ and $d$ ($1\leq n\leq 100$, $1\leq d\leq 10^9$)Β β the number of Sonya's hotels and the needed minimum distance from a new hotel to all others.
The second line contains $n$ different integers in strictly increasing order $x_1, x_2, \ldots, x_n$ ($-10^9\leq x_i\leq 10^9$)Β β coordinates of Sonya's hotels.
Output Specification:
Print the number of cities where Sonya can build a new hotel so that the minimum distance from this hotel to all others is equal to $d$.
Demo Input:
['4 3\n-3 2 9 16\n', '5 2\n4 8 11 18 19\n']
Demo Output:
['6\n', '5\n']
Note:
In the first example, there are $6$ possible cities where Sonya can build a hotel. These cities have coordinates $-6$, $5$, $6$, $12$, $13$, and $19$.
In the second example, there are $5$ possible cities where Sonya can build a hotel. These cities have coordinates $2$, $6$, $13$, $16$, and $21$.
|
```python
n,k = map(int, input().split())
k *= 2
l = input().split()
sumi = 2
for x in range(len(l)-1):
y = int(l[x+1])-int(l[x])
if y == k:
sumi += 1
elif y > k:
sumi += 2
print(sumi)
```
| 3
|
|
379
|
A
|
New Year Candles
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
Vasily the Programmer loves romance, so this year he decided to illuminate his room with candles.
Vasily has *a* candles.When Vasily lights up a new candle, it first burns for an hour and then it goes out. Vasily is smart, so he can make *b* went out candles into a new candle. As a result, this new candle can be used like any other new candle.
Now Vasily wonders: for how many hours can his candles light up the room if he acts optimally well? Help him find this number.
|
The single line contains two integers, *a* and *b* (1<=β€<=*a*<=β€<=1000;Β 2<=β€<=*b*<=β€<=1000).
|
Print a single integer β the number of hours Vasily can light up the room for.
|
[
"4 2\n",
"6 3\n"
] |
[
"7\n",
"8\n"
] |
Consider the first sample. For the first four hours Vasily lights up new candles, then he uses four burned out candles to make two new ones and lights them up. When these candles go out (stop burning), Vasily can make another candle. Overall, Vasily can light up the room for 7 hours.
| 500
|
[
{
"input": "4 2",
"output": "7"
},
{
"input": "6 3",
"output": "8"
},
{
"input": "1000 1000",
"output": "1001"
},
{
"input": "123 5",
"output": "153"
},
{
"input": "1000 2",
"output": "1999"
},
{
"input": "1 2",
"output": "1"
},
{
"input": "1 3",
"output": "1"
},
{
"input": "1 4",
"output": "1"
},
{
"input": "2 2",
"output": "3"
},
{
"input": "3 2",
"output": "5"
},
{
"input": "3 3",
"output": "4"
},
{
"input": "999 2",
"output": "1997"
},
{
"input": "1000 3",
"output": "1499"
},
{
"input": "1000 4",
"output": "1333"
},
{
"input": "1 1000",
"output": "1"
},
{
"input": "80 970",
"output": "80"
},
{
"input": "80 970",
"output": "80"
},
{
"input": "80 970",
"output": "80"
},
{
"input": "80 970",
"output": "80"
},
{
"input": "80 970",
"output": "80"
},
{
"input": "80 970",
"output": "80"
},
{
"input": "10 4",
"output": "13"
},
{
"input": "4 3",
"output": "5"
},
{
"input": "91 5",
"output": "113"
},
{
"input": "777 17",
"output": "825"
},
{
"input": "100 4",
"output": "133"
},
{
"input": "5 3",
"output": "7"
},
{
"input": "6 4",
"output": "7"
},
{
"input": "26 8",
"output": "29"
},
{
"input": "9 4",
"output": "11"
},
{
"input": "20 3",
"output": "29"
},
{
"input": "17 3",
"output": "25"
}
] | 1,693,897,207
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 32
| 62
| 0
|
a, b = map(int, input().split())
res = 0
while a >= b:
a += 1 - b
res += b
res += a
print(res)
|
Title: New Year Candles
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasily the Programmer loves romance, so this year he decided to illuminate his room with candles.
Vasily has *a* candles.When Vasily lights up a new candle, it first burns for an hour and then it goes out. Vasily is smart, so he can make *b* went out candles into a new candle. As a result, this new candle can be used like any other new candle.
Now Vasily wonders: for how many hours can his candles light up the room if he acts optimally well? Help him find this number.
Input Specification:
The single line contains two integers, *a* and *b* (1<=β€<=*a*<=β€<=1000;Β 2<=β€<=*b*<=β€<=1000).
Output Specification:
Print a single integer β the number of hours Vasily can light up the room for.
Demo Input:
['4 2\n', '6 3\n']
Demo Output:
['7\n', '8\n']
Note:
Consider the first sample. For the first four hours Vasily lights up new candles, then he uses four burned out candles to make two new ones and lights them up. When these candles go out (stop burning), Vasily can make another candle. Overall, Vasily can light up the room for 7 hours.
|
```python
a, b = map(int, input().split())
res = 0
while a >= b:
a += 1 - b
res += b
res += a
print(res)
```
| 3
|
|
38
|
A
|
Army
|
PROGRAMMING
| 800
|
[
"implementation"
] |
A. Army
|
2
|
256
|
The Berland Armed Forces System consists of *n* ranks that are numbered using natural numbers from 1 to *n*, where 1 is the lowest rank and *n* is the highest rank.
One needs exactly *d**i* years to rise from rank *i* to rank *i*<=+<=1. Reaching a certain rank *i* having not reached all the previous *i*<=-<=1 ranks is impossible.
Vasya has just reached a new rank of *a*, but he dreams of holding the rank of *b*. Find for how many more years Vasya should serve in the army until he can finally realize his dream.
|
The first input line contains an integer *n* (2<=β€<=*n*<=β€<=100). The second line contains *n*<=-<=1 integers *d**i* (1<=β€<=*d**i*<=β€<=100). The third input line contains two integers *a* and *b* (1<=β€<=*a*<=<<=*b*<=β€<=*n*). The numbers on the lines are space-separated.
|
Print the single number which is the number of years that Vasya needs to rise from rank *a* to rank *b*.
|
[
"3\n5 6\n1 2\n",
"3\n5 6\n1 3\n"
] |
[
"5\n",
"11\n"
] |
none
| 0
|
[
{
"input": "3\n5 6\n1 2",
"output": "5"
},
{
"input": "3\n5 6\n1 3",
"output": "11"
},
{
"input": "2\n55\n1 2",
"output": "55"
},
{
"input": "3\n85 78\n1 3",
"output": "163"
},
{
"input": "4\n63 4 49\n2 3",
"output": "4"
},
{
"input": "5\n93 83 42 56\n2 5",
"output": "181"
},
{
"input": "6\n22 9 87 89 57\n1 6",
"output": "264"
},
{
"input": "7\n52 36 31 23 74 78\n2 7",
"output": "242"
},
{
"input": "8\n82 14 24 5 91 49 94\n3 8",
"output": "263"
},
{
"input": "9\n12 40 69 39 59 21 59 5\n4 6",
"output": "98"
},
{
"input": "10\n95 81 32 59 71 30 50 61 100\n1 6",
"output": "338"
},
{
"input": "15\n89 55 94 4 15 69 19 60 91 77 3 94 91 62\n3 14",
"output": "617"
},
{
"input": "20\n91 1 41 51 95 67 92 35 23 70 44 91 57 50 21 8 9 71 40\n8 17",
"output": "399"
},
{
"input": "25\n70 95 21 84 97 39 12 98 53 24 78 29 84 65 70 22 100 17 69 27 62 48 35 80\n8 23",
"output": "846"
},
{
"input": "30\n35 69 50 44 19 56 86 56 98 24 21 2 61 24 85 30 2 22 57 35 59 84 12 77 92 53 50 92 9\n1 16",
"output": "730"
},
{
"input": "35\n2 34 47 15 27 61 6 88 67 20 53 65 29 68 77 5 78 86 44 98 32 81 91 79 54 84 95 23 65 97 22 33 42 87\n8 35",
"output": "1663"
},
{
"input": "40\n32 88 59 36 95 45 28 78 73 30 97 13 13 47 48 100 43 21 22 45 88 25 15 13 63 25 72 92 29 5 25 11 50 5 54 51 48 84 23\n7 26",
"output": "862"
},
{
"input": "45\n83 74 73 95 10 31 100 26 29 15 80 100 22 70 31 88 9 56 19 70 2 62 48 30 27 47 52 50 94 44 21 94 23 85 15 3 95 72 43 62 94 89 68 88\n17 40",
"output": "1061"
},
{
"input": "50\n28 8 16 29 19 82 70 51 96 84 74 72 17 69 12 21 37 21 39 3 18 66 19 49 86 96 94 93 2 90 96 84 59 88 58 15 61 33 55 22 35 54 51 29 64 68 29 38 40\n23 28",
"output": "344"
},
{
"input": "60\n24 28 25 21 43 71 64 73 71 90 51 83 69 43 75 43 78 72 56 61 99 7 23 86 9 16 16 94 23 74 18 56 20 72 13 31 75 34 35 86 61 49 4 72 84 7 65 70 66 52 21 38 6 43 69 40 73 46 5\n28 60",
"output": "1502"
},
{
"input": "70\n69 95 34 14 67 61 6 95 94 44 28 94 73 66 39 13 19 71 73 71 28 48 26 22 32 88 38 95 43 59 88 77 80 55 17 95 40 83 67 1 38 95 58 63 56 98 49 2 41 4 73 8 78 41 64 71 60 71 41 61 67 4 4 19 97 14 39 20 27\n9 41",
"output": "1767"
},
{
"input": "80\n65 15 43 6 43 98 100 16 69 98 4 54 25 40 2 35 12 23 38 29 10 89 30 6 4 8 7 96 64 43 11 49 89 38 20 59 54 85 46 16 16 89 60 54 28 37 32 34 67 9 78 30 50 87 58 53 99 48 77 3 5 6 19 99 16 20 31 10 80 76 82 56 56 83 72 81 84 60 28\n18 24",
"output": "219"
},
{
"input": "90\n61 35 100 99 67 87 42 90 44 4 81 65 29 63 66 56 53 22 55 87 39 30 34 42 27 80 29 97 85 28 81 22 50 22 24 75 67 86 78 79 94 35 13 97 48 76 68 66 94 13 82 1 22 85 5 36 86 73 65 97 43 56 35 26 87 25 74 47 81 67 73 75 99 75 53 38 70 21 66 78 38 17 57 40 93 57 68 55 1\n12 44",
"output": "1713"
},
{
"input": "95\n37 74 53 96 65 84 65 72 95 45 6 77 91 35 58 50 51 51 97 30 51 20 79 81 92 10 89 34 40 76 71 54 26 34 73 72 72 28 53 19 95 64 97 10 44 15 12 38 5 63 96 95 86 8 36 96 45 53 81 5 18 18 47 97 65 9 33 53 41 86 37 53 5 40 15 76 83 45 33 18 26 5 19 90 46 40 100 42 10 90 13 81 40 53\n6 15",
"output": "570"
},
{
"input": "96\n51 32 95 75 23 54 70 89 67 3 1 51 4 100 97 30 9 35 56 38 54 77 56 98 43 17 60 43 72 46 87 61 100 65 81 22 74 38 16 96 5 10 54 22 23 22 10 91 9 54 49 82 29 73 33 98 75 8 4 26 24 90 71 42 90 24 94 74 94 10 41 98 56 63 18 43 56 21 26 64 74 33 22 38 67 66 38 60 64 76 53 10 4 65 76\n21 26",
"output": "328"
},
{
"input": "97\n18 90 84 7 33 24 75 55 86 10 96 72 16 64 37 9 19 71 62 97 5 34 85 15 46 72 82 51 52 16 55 68 27 97 42 72 76 97 32 73 14 56 11 86 2 81 59 95 60 93 1 22 71 37 77 100 6 16 78 47 78 62 94 86 16 91 56 46 47 35 93 44 7 86 70 10 29 45 67 62 71 61 74 39 36 92 24 26 65 14 93 92 15 28 79 59\n6 68",
"output": "3385"
},
{
"input": "98\n32 47 26 86 43 42 79 72 6 68 40 46 29 80 24 89 29 7 21 56 8 92 13 33 50 79 5 7 84 85 24 23 1 80 51 21 26 55 96 51 24 2 68 98 81 88 57 100 64 84 54 10 14 2 74 1 89 71 1 20 84 85 17 31 42 58 69 67 48 60 97 90 58 10 21 29 2 21 60 61 68 89 77 39 57 18 61 44 67 100 33 74 27 40 83 29 6\n8 77",
"output": "3319"
},
{
"input": "99\n46 5 16 66 53 12 84 89 26 27 35 68 41 44 63 17 88 43 80 15 59 1 42 50 53 34 75 16 16 55 92 30 28 11 12 71 27 65 11 28 86 47 24 10 60 47 7 53 16 75 6 49 56 66 70 3 20 78 75 41 38 57 89 23 16 74 30 39 1 32 49 84 9 33 25 95 75 45 54 59 17 17 29 40 79 96 47 11 69 86 73 56 91 4 87 47 31 24\n23 36",
"output": "514"
},
{
"input": "100\n63 65 21 41 95 23 3 4 12 23 95 50 75 63 58 34 71 27 75 31 23 94 96 74 69 34 43 25 25 55 44 19 43 86 68 17 52 65 36 29 72 96 84 25 84 23 71 54 6 7 71 7 21 100 99 58 93 35 62 47 36 70 68 9 75 13 35 70 76 36 62 22 52 51 2 87 66 41 54 35 78 62 30 35 65 44 74 93 78 37 96 70 26 32 71 27 85 85 63\n43 92",
"output": "2599"
},
{
"input": "51\n85 38 22 38 42 36 55 24 36 80 49 15 66 91 88 61 46 82 1 61 89 92 6 56 28 8 46 80 56 90 91 38 38 17 69 64 57 68 13 44 45 38 8 72 61 39 87 2 73 88\n15 27",
"output": "618"
},
{
"input": "2\n3\n1 2",
"output": "3"
},
{
"input": "5\n6 8 22 22\n2 3",
"output": "8"
},
{
"input": "6\n3 12 27 28 28\n3 4",
"output": "27"
},
{
"input": "9\n1 2 2 2 2 3 3 5\n3 7",
"output": "9"
},
{
"input": "10\n1 1 1 1 1 1 1 1 1\n6 8",
"output": "2"
},
{
"input": "20\n1 1 1 1 1 1 1 1 2 2 2 2 2 3 3 3 3 3 3\n5 17",
"output": "23"
},
{
"input": "25\n1 1 1 4 5 6 8 11 11 11 11 12 13 14 14 14 15 16 16 17 17 17 19 19\n4 8",
"output": "23"
},
{
"input": "35\n1 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\n30 31",
"output": "2"
},
{
"input": "45\n1 1 1 1 2 2 2 2 2 2 2 3 3 3 3 3 3 4 5 5 5 5 6 6 6 6 6 6 6 7 7 7 7 8 8 8 9 9 9 9 9 10 10 10\n42 45",
"output": "30"
},
{
"input": "50\n1 8 8 13 14 15 15 16 19 21 22 24 26 31 32 37 45 47 47 47 50 50 51 54 55 56 58 61 61 61 63 63 64 66 66 67 67 70 71 80 83 84 85 92 92 94 95 95 100\n4 17",
"output": "285"
},
{
"input": "60\n1 2 4 4 4 6 6 8 9 10 10 13 14 18 20 20 21 22 23 23 26 29 30 32 33 34 35 38 40 42 44 44 46 48 52 54 56 56 60 60 66 67 68 68 69 73 73 74 80 80 81 81 82 84 86 86 87 89 89\n56 58",
"output": "173"
},
{
"input": "70\n1 2 3 3 4 5 5 7 7 7 8 8 8 8 9 9 10 12 12 12 12 13 16 16 16 16 16 16 17 17 18 18 20 20 21 23 24 25 25 26 29 29 29 29 31 32 32 34 35 36 36 37 37 38 39 39 40 40 40 40 41 41 42 43 44 44 44 45 45\n62 65",
"output": "126"
},
{
"input": "80\n1 1 1 1 1 1 1 1 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 4 4 4 4 5 5 5 5 5 5 5 6 7 7 7 7 7 7 8 8 8 8 9 9 9 9 9 9 9 9 9 10 10 10 10 10 10 10 10 10 11 11 11 11 11 11 11 12 12 12 12 12 12 12 12\n17 65",
"output": "326"
},
{
"input": "90\n1 1 3 5 8 9 10 11 11 11 11 12 13 14 15 15 15 16 16 19 19 20 22 23 24 25 25 28 29 29 30 31 33 34 35 37 37 38 41 43 43 44 45 47 51 54 55 56 58 58 59 59 60 62 66 67 67 67 68 68 69 70 71 72 73 73 76 77 77 78 78 78 79 79 79 82 83 84 85 85 87 87 89 93 93 93 95 99 99\n28 48",
"output": "784"
},
{
"input": "95\n2 2 3 3 4 6 6 7 7 7 9 10 12 12 12 12 13 14 15 16 17 18 20 20 20 20 21 21 21 21 22 22 22 22 22 23 23 23 25 26 26 27 27 27 28 29 29 30 30 31 32 33 34 36 37 37 38 39 39 39 42 43 43 43 45 47 48 50 50 51 52 53 54 54 54 55 55 55 58 59 60 61 61 61 61 62 62 63 64 65 66 67 67 67\n64 93",
"output": "1636"
},
{
"input": "96\n1 1 2 3 3 5 8 9 9 10 10 10 11 11 11 11 11 12 13 13 13 14 15 15 16 16 17 17 17 17 18 18 20 20 20 21 21 21 23 24 24 25 25 26 27 27 27 27 29 29 29 30 30 30 32 32 32 32 32 32 33 33 34 34 34 35 35 35 36 36 37 37 37 38 39 40 41 41 41 41 42 42 43 43 45 45 45 46 46 47 47 49 50 52 52\n76 96",
"output": "898"
},
{
"input": "98\n2 3 4 4 5 7 8 10 10 10 11 11 12 12 12 12 13 14 15 15 16 16 18 19 19 20 21 21 21 21 22 23 24 25 26 26 27 27 27 27 29 29 30 30 31 31 37 40 40 40 41 41 41 42 43 44 44 44 46 46 47 49 49 50 50 50 51 53 55 55 56 56 56 56 56 57 57 58 59 60 60 60 62 62 63 64 64 64 65 66 66 67 68 70 70 71 71\n8 90",
"output": "3016"
},
{
"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\n66 95",
"output": "29"
},
{
"input": "100\n1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 3 3 3 4 4 4 4 4 4 4 4 4 4 5 5 5 5 5 5 6 6 6 6 6 6 6 6 6 6 6 6 7 7 7 7 7 7 8 8 8 8 9 9 9 9 10 10 10 10 11 11 11 11 12 12 12 13 13 13 13 13 13 13 13 13 13 14 14 14 14 14 14 15 15 15 15 15 15 16 16 16 17 17\n39 52",
"output": "82"
},
{
"input": "51\n5 7 9 16 19 25 26 29 29 30 31 32 32 41 43 44 47 49 50 50 51 52 54 55 56 63 64 66 67 70 74 74 77 78 79 80 80 85 86 87 89 89 90 92 93 94 94 95 95 97\n3 44",
"output": "2268"
}
] | 1,632,665,224
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 50
| 124
| 6,963,200
|
# -*- coding: utf-8 -*-
"""
Created on Sun Sep 26 21:56:04 2021
@author: cheehong
"""
n=int(input())
l=list(map(int, input().split()))
k=list(map(int, input().split()))
x=k[1]-k[0]
total=0
while k[0]<k[1]:
total+=l[k[0]-1]
k[0]+=1
print(total)
|
Title: Army
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
The Berland Armed Forces System consists of *n* ranks that are numbered using natural numbers from 1 to *n*, where 1 is the lowest rank and *n* is the highest rank.
One needs exactly *d**i* years to rise from rank *i* to rank *i*<=+<=1. Reaching a certain rank *i* having not reached all the previous *i*<=-<=1 ranks is impossible.
Vasya has just reached a new rank of *a*, but he dreams of holding the rank of *b*. Find for how many more years Vasya should serve in the army until he can finally realize his dream.
Input Specification:
The first input line contains an integer *n* (2<=β€<=*n*<=β€<=100). The second line contains *n*<=-<=1 integers *d**i* (1<=β€<=*d**i*<=β€<=100). The third input line contains two integers *a* and *b* (1<=β€<=*a*<=<<=*b*<=β€<=*n*). The numbers on the lines are space-separated.
Output Specification:
Print the single number which is the number of years that Vasya needs to rise from rank *a* to rank *b*.
Demo Input:
['3\n5 6\n1 2\n', '3\n5 6\n1 3\n']
Demo Output:
['5\n', '11\n']
Note:
none
|
```python
# -*- coding: utf-8 -*-
"""
Created on Sun Sep 26 21:56:04 2021
@author: cheehong
"""
n=int(input())
l=list(map(int, input().split()))
k=list(map(int, input().split()))
x=k[1]-k[0]
total=0
while k[0]<k[1]:
total+=l[k[0]-1]
k[0]+=1
print(total)
```
| 3.95603
|
330
|
A
|
Cakeminator
|
PROGRAMMING
| 800
|
[
"brute force",
"implementation"
] | null | null |
You are given a rectangular cake, represented as an *r*<=Γ<=*c* grid. Each cell either has an evil strawberry, or is empty. For example, a 3<=Γ<=4 cake may look as follows:
The cakeminator is going to eat the cake! Each time he eats, he chooses a row or a column that does not contain any evil strawberries and contains at least one cake cell that has not been eaten before, and eats all the cake cells there. He may decide to eat any number of times.
Please output the maximum number of cake cells that the cakeminator can eat.
|
The first line contains two integers *r* and *c* (2<=β€<=*r*,<=*c*<=β€<=10), denoting the number of rows and the number of columns of the cake. The next *r* lines each contains *c* characters β the *j*-th character of the *i*-th line denotes the content of the cell at row *i* and column *j*, and is either one of these:
- '.' character denotes a cake cell with no evil strawberry; - 'S' character denotes a cake cell with an evil strawberry.
|
Output the maximum number of cake cells that the cakeminator can eat.
|
[
"3 4\nS...\n....\n..S.\n"
] |
[
"8\n"
] |
For the first example, one possible way to eat the maximum number of cake cells is as follows (perform 3 eats).
| 500
|
[
{
"input": "3 4\nS...\n....\n..S.",
"output": "8"
},
{
"input": "2 2\n..\n..",
"output": "4"
},
{
"input": "2 2\nSS\nSS",
"output": "0"
},
{
"input": "7 3\nS..\nS..\nS..\nS..\nS..\nS..\nS..",
"output": "14"
},
{
"input": "3 5\n..S..\nSSSSS\n..S..",
"output": "0"
},
{
"input": "10 10\nSSSSSSSSSS\nSSSSSSSSSS\nSSSSSSSSSS\nSSSSSSSSSS\nSSSSSSSSSS\nSSSSSSSSSS\nSSSSSSSSSS\nSSSSSSSSSS\nSSSSSSSSSS\nSSSSSSSSSS",
"output": "0"
},
{
"input": "10 10\nS...SSSSSS\nS...SSSSSS\nS...SSSSSS\nS...SSSSSS\nS...SSSSSS\nS...SSSSSS\nS...SSSSSS\nS...SSSSSS\nS...SSSSSS\nS...SSSSSS",
"output": "30"
},
{
"input": "10 10\n....S..S..\n....S..S..\n....S..S..\n....S..S..\n....S..S..\n....S..S..\n....S..S..\n....S..S..\n....S..S..\n....S..S..",
"output": "80"
},
{
"input": "9 5\nSSSSS\nSSSSS\nSSSSS\nSSSSS\nSSSSS\nSSSSS\nSSSSS\nSSSSS\nSSSSS",
"output": "0"
},
{
"input": "9 9\n...S.....\nS.S.....S\n.S....S..\n.S.....SS\n.........\n..S.S..S.\n.SS......\n....S....\n..S...S..",
"output": "17"
},
{
"input": "5 6\nSSSSSS\nSSSSSS\nSSSSSS\nSS.S..\nS.S.SS",
"output": "0"
},
{
"input": "9 8\n........\n.......S\n........\nS.......\n........\n........\nS.......\n........\n.......S",
"output": "64"
},
{
"input": "9 7\n......S\n......S\nS.S.S..\n.......\n.......\n.S.....\n.S....S\n..S....\n.S....S",
"output": "28"
},
{
"input": "10 10\n.....S....\n....SS..S.\n.S...S....\n........SS\n.S.......S\nSS..S.....\n.SS.....SS\nS..S......\n.......SSS\nSSSSS....S",
"output": "10"
},
{
"input": "6 7\n..S.SS.\n......S\n....S.S\nSS..S..\nS..SS.S\n.....S.",
"output": "0"
},
{
"input": "10 6\n.SSSSS\nSSS.SS\nSSSSSS\nS.SSSS\nSSSSS.\nS.SSSS\nSS.SSS\n.SSS.S\n.SSS..\nSS..SS",
"output": "0"
},
{
"input": "2 2\n..\n..",
"output": "4"
},
{
"input": "3 2\nS.\n.S\nS.",
"output": "0"
},
{
"input": "3 2\nS.\n.S\nS.",
"output": "0"
},
{
"input": "4 3\n.S.\nS.S\n.S.\nS.S",
"output": "0"
},
{
"input": "2 3\n...\nSSS",
"output": "3"
},
{
"input": "2 4\nS.SS\nS.SS",
"output": "2"
},
{
"input": "2 2\n..\n.S",
"output": "3"
},
{
"input": "3 2\n.S\n.S\nSS",
"output": "0"
},
{
"input": "2 4\nSS.S\n..S.",
"output": "0"
},
{
"input": "2 3\n...\nS..",
"output": "5"
}
] | 1,665,665,945
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 42
| 92
| 0
|
arr = input()
arr = arr.split(" ")
r = int(arr[0])
c = int(arr[1])
row = []
col = []
for i in range(r):
text = input()
for j in range(c):
if text[j] == "S":
if j not in col:
col.append(j)
if i not in row:
row.append(i)
print((r-len(row))*c + (c-len(col))*r - ((r-len(row))*(c-len(col))))
|
Title: Cakeminator
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a rectangular cake, represented as an *r*<=Γ<=*c* grid. Each cell either has an evil strawberry, or is empty. For example, a 3<=Γ<=4 cake may look as follows:
The cakeminator is going to eat the cake! Each time he eats, he chooses a row or a column that does not contain any evil strawberries and contains at least one cake cell that has not been eaten before, and eats all the cake cells there. He may decide to eat any number of times.
Please output the maximum number of cake cells that the cakeminator can eat.
Input Specification:
The first line contains two integers *r* and *c* (2<=β€<=*r*,<=*c*<=β€<=10), denoting the number of rows and the number of columns of the cake. The next *r* lines each contains *c* characters β the *j*-th character of the *i*-th line denotes the content of the cell at row *i* and column *j*, and is either one of these:
- '.' character denotes a cake cell with no evil strawberry; - 'S' character denotes a cake cell with an evil strawberry.
Output Specification:
Output the maximum number of cake cells that the cakeminator can eat.
Demo Input:
['3 4\nS...\n....\n..S.\n']
Demo Output:
['8\n']
Note:
For the first example, one possible way to eat the maximum number of cake cells is as follows (perform 3 eats).
|
```python
arr = input()
arr = arr.split(" ")
r = int(arr[0])
c = int(arr[1])
row = []
col = []
for i in range(r):
text = input()
for j in range(c):
if text[j] == "S":
if j not in col:
col.append(j)
if i not in row:
row.append(i)
print((r-len(row))*c + (c-len(col))*r - ((r-len(row))*(c-len(col))))
```
| 3
|
|
111
|
B
|
Petya and Divisors
|
PROGRAMMING
| 1,900
|
[
"binary search",
"data structures",
"number theory"
] |
B. Petya and Divisors
|
5
|
256
|
Little Petya loves looking for numbers' divisors. One day Petya came across the following problem:
You are given *n* queries in the form "*x**i* *y**i*". For each query Petya should count how many divisors of number *x**i* divide none of the numbers *x**i*<=-<=*y**i*,<=*x**i*<=-<=*y**i*<=+<=1,<=...,<=*x**i*<=-<=1. Help him.
|
The first line contains an integer *n* (1<=β€<=*n*<=β€<=105). Each of the following *n* lines contain two space-separated integers *x**i* and *y**i* (1<=β€<=*x**i*<=β€<=105, 0<=β€<=*y**i*<=β€<=*i*<=-<=1, where *i* is the query's ordinal number; the numeration starts with 1).
If *y**i*<==<=0 for the query, then the answer to the query will be the number of divisors of the number *x**i*. In this case you do not need to take the previous numbers *x* into consideration.
|
For each query print the answer on a single line: the number of positive integers *k* such that
|
[
"6\n4 0\n3 1\n5 2\n6 2\n18 4\n10000 3\n"
] |
[
"3\n1\n1\n2\n2\n22\n"
] |
Let's write out the divisors that give answers for the first 5 queries:
1) 1, 2, 4
2) 3
3) 5
4) 2, 6
5) 9, 18
| 1,000
|
[
{
"input": "6\n4 0\n3 1\n5 2\n6 2\n18 4\n10000 3",
"output": "3\n1\n1\n2\n2\n22"
},
{
"input": "5\n10 0\n10 0\n10 0\n10 0\n10 0",
"output": "4\n4\n4\n4\n4"
},
{
"input": "12\n41684 0\n95210 1\n60053 1\n32438 3\n97956 1\n21785 2\n14594 6\n17170 4\n93937 6\n70764 5\n13695 4\n14552 6",
"output": "12\n6\n7\n9\n22\n3\n2\n13\n1\n6\n13\n11"
},
{
"input": "10\n54972 0\n48015 1\n7114 1\n68273 2\n53650 4\n1716 1\n16165 2\n96062 5\n57750 1\n21071 5",
"output": "24\n21\n3\n3\n21\n22\n6\n6\n62\n3"
},
{
"input": "20\n68260 0\n819 1\n54174 1\n20460 1\n25696 2\n81647 4\n17736 4\n91307 5\n5210 4\n87730 2\n4653 8\n11044 6\n15776 4\n17068 7\n73738 7\n36004 12\n83183 7\n75700 12\n84270 14\n16120 5",
"output": "12\n11\n6\n44\n18\n1\n9\n7\n6\n12\n8\n8\n21\n3\n14\n3\n3\n13\n18\n26"
},
{
"input": "17\n81548 0\n69975 1\n1234 0\n72647 0\n81389 4\n77930 1\n19308 0\n86551 6\n69023 8\n38037 1\n133 9\n59290 8\n1106 11\n95012 10\n57693 11\n8467 6\n93732 13",
"output": "24\n17\n4\n2\n11\n7\n12\n3\n3\n7\n2\n27\n4\n3\n2\n1\n18"
},
{
"input": "15\n94836 0\n22780 1\n48294 0\n24834 3\n37083 2\n57862 0\n37231 1\n81795 7\n32835 2\n4696 8\n95612 0\n7536 6\n70084 5\n72956 10\n41647 7",
"output": "24\n21\n12\n4\n6\n8\n3\n27\n12\n5\n24\n15\n8\n21\n1"
},
{
"input": "12\n91771 0\n75584 1\n95355 1\n60669 1\n92776 0\n37793 3\n38802 4\n60688 0\n80296 5\n55003 8\n91092 3\n55782 8",
"output": "2\n13\n23\n17\n8\n2\n13\n10\n4\n2\n9\n10"
},
{
"input": "11\n5059 0\n28388 1\n42415 2\n12856 0\n48470 3\n34076 2\n40374 6\n55932 1\n44108 2\n5310 5\n86571 4",
"output": "2\n11\n7\n8\n13\n9\n10\n20\n3\n12\n3"
},
{
"input": "10\n18347 0\n81193 1\n89475 2\n65043 3\n4164 0\n14007 5\n41945 0\n51177 1\n91569 5\n71969 4",
"output": "4\n4\n11\n18\n12\n13\n4\n7\n6\n3"
}
] | 1,582,692,467
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 44
| 3,806
| 12,595,200
|
def R(): return map(int, input().split())
def I(): return int(input())
def S(): return str(input())
def L(): return list(R())
from collections import Counter
import math
import sys
from itertools import permutations
import bisect
div=[[] for i in range(10**5+1)]
for i in range(1,10**5+1):
for j in range(1,10**5//i+1):
div[i*j].append(i)
n=I()
last=[-1]*(10**5+1)
for j in range(n):
x,y=R()
ans=0
for i in div[x]:
if last[i]<j-y:
ans+=1
last[i]=j
print(ans)
|
Title: Petya and Divisors
Time Limit: 5 seconds
Memory Limit: 256 megabytes
Problem Description:
Little Petya loves looking for numbers' divisors. One day Petya came across the following problem:
You are given *n* queries in the form "*x**i* *y**i*". For each query Petya should count how many divisors of number *x**i* divide none of the numbers *x**i*<=-<=*y**i*,<=*x**i*<=-<=*y**i*<=+<=1,<=...,<=*x**i*<=-<=1. Help him.
Input Specification:
The first line contains an integer *n* (1<=β€<=*n*<=β€<=105). Each of the following *n* lines contain two space-separated integers *x**i* and *y**i* (1<=β€<=*x**i*<=β€<=105, 0<=β€<=*y**i*<=β€<=*i*<=-<=1, where *i* is the query's ordinal number; the numeration starts with 1).
If *y**i*<==<=0 for the query, then the answer to the query will be the number of divisors of the number *x**i*. In this case you do not need to take the previous numbers *x* into consideration.
Output Specification:
For each query print the answer on a single line: the number of positive integers *k* such that
Demo Input:
['6\n4 0\n3 1\n5 2\n6 2\n18 4\n10000 3\n']
Demo Output:
['3\n1\n1\n2\n2\n22\n']
Note:
Let's write out the divisors that give answers for the first 5 queries:
1) 1, 2, 4
2) 3
3) 5
4) 2, 6
5) 9, 18
|
```python
def R(): return map(int, input().split())
def I(): return int(input())
def S(): return str(input())
def L(): return list(R())
from collections import Counter
import math
import sys
from itertools import permutations
import bisect
div=[[] for i in range(10**5+1)]
for i in range(1,10**5+1):
for j in range(1,10**5//i+1):
div[i*j].append(i)
n=I()
last=[-1]*(10**5+1)
for j in range(n):
x,y=R()
ans=0
for i in div[x]:
if last[i]<j-y:
ans+=1
last[i]=j
print(ans)
```
| 3.59594
|
197
|
A
|
Plate Game
|
PROGRAMMING
| 1,600
|
[
"constructive algorithms",
"games",
"math"
] | null | null |
You've got a rectangular table with length *a* and width *b* and the infinite number of plates of radius *r*. Two players play the following game: they take turns to put the plates on the table so that the plates don't lie on each other (but they can touch each other), and so that any point on any plate is located within the table's border. During the game one cannot move the plates that already lie on the table. The player who cannot make another move loses. Determine which player wins, the one who moves first or the one who moves second, provided that both players play optimally well.
|
A single line contains three space-separated integers *a*, *b*, *r* (1<=β€<=*a*,<=*b*,<=*r*<=β€<=100) β the table sides and the plates' radius, correspondingly.
|
If wins the player who moves first, print "First" (without the quotes). Otherwise print "Second" (without the quotes).
|
[
"5 5 2\n",
"6 7 4\n"
] |
[
"First\n",
"Second\n"
] |
In the first sample the table has place for only one plate. The first player puts a plate on the table, the second player can't do that and loses.
In the second sample the table is so small that it doesn't have enough place even for one plate. So the first player loses without making a single move.
| 1,000
|
[
{
"input": "5 5 2",
"output": "First"
},
{
"input": "6 7 4",
"output": "Second"
},
{
"input": "100 100 1",
"output": "First"
},
{
"input": "1 1 100",
"output": "Second"
},
{
"input": "13 7 3",
"output": "First"
},
{
"input": "23 7 3",
"output": "First"
},
{
"input": "9 9 2",
"output": "First"
},
{
"input": "13 13 2",
"output": "First"
},
{
"input": "21 21 10",
"output": "First"
},
{
"input": "20 21 10",
"output": "First"
},
{
"input": "20 20 10",
"output": "First"
},
{
"input": "9 13 2",
"output": "First"
},
{
"input": "19 7 3",
"output": "First"
},
{
"input": "19 19 10",
"output": "Second"
},
{
"input": "19 20 10",
"output": "Second"
},
{
"input": "19 21 10",
"output": "Second"
},
{
"input": "1 100 1",
"output": "Second"
},
{
"input": "2 100 1",
"output": "First"
},
{
"input": "3 100 1",
"output": "First"
},
{
"input": "100 100 49",
"output": "First"
},
{
"input": "100 100 50",
"output": "First"
},
{
"input": "100 100 51",
"output": "Second"
},
{
"input": "100 99 50",
"output": "Second"
},
{
"input": "4 10 5",
"output": "Second"
},
{
"input": "8 11 2",
"output": "First"
},
{
"input": "3 12 5",
"output": "Second"
},
{
"input": "14 15 5",
"output": "First"
},
{
"input": "61 2 3",
"output": "Second"
},
{
"input": "82 20 5",
"output": "First"
},
{
"input": "16 80 10",
"output": "Second"
},
{
"input": "2 1 20",
"output": "Second"
},
{
"input": "78 82 5",
"output": "First"
},
{
"input": "8 55 7",
"output": "Second"
},
{
"input": "75 55 43",
"output": "Second"
},
{
"input": "34 43 70",
"output": "Second"
},
{
"input": "86 74 36",
"output": "First"
},
{
"input": "86 74 37",
"output": "First"
},
{
"input": "86 74 38",
"output": "Second"
},
{
"input": "24 70 11",
"output": "First"
},
{
"input": "24 70 12",
"output": "First"
},
{
"input": "24 70 13",
"output": "Second"
},
{
"input": "78 95 38",
"output": "First"
},
{
"input": "78 95 39",
"output": "First"
},
{
"input": "78 95 40",
"output": "Second"
},
{
"input": "88 43 21",
"output": "First"
},
{
"input": "88 43 22",
"output": "Second"
},
{
"input": "88 43 23",
"output": "Second"
},
{
"input": "30 40 14",
"output": "First"
},
{
"input": "30 40 15",
"output": "First"
},
{
"input": "30 40 16",
"output": "Second"
},
{
"input": "2 5 2",
"output": "Second"
},
{
"input": "5 100 3",
"output": "Second"
},
{
"input": "44 58 5",
"output": "First"
},
{
"input": "4 4 6",
"output": "Second"
},
{
"input": "10 20 6",
"output": "Second"
},
{
"input": "100 1 1",
"output": "Second"
},
{
"input": "60 60 1",
"output": "First"
},
{
"input": "100 1 2",
"output": "Second"
},
{
"input": "2 4 2",
"output": "Second"
},
{
"input": "10 90 11",
"output": "Second"
},
{
"input": "20 5 6",
"output": "Second"
},
{
"input": "1 44 2",
"output": "Second"
},
{
"input": "10 5 5",
"output": "Second"
},
{
"input": "5 100 4",
"output": "Second"
},
{
"input": "99 99 50",
"output": "Second"
},
{
"input": "1 100 2",
"output": "Second"
},
{
"input": "100 20 12",
"output": "Second"
},
{
"input": "10 2 4",
"output": "Second"
},
{
"input": "1 50 2",
"output": "Second"
},
{
"input": "10 4 3",
"output": "Second"
},
{
"input": "74 1 1",
"output": "Second"
},
{
"input": "6 6 1",
"output": "First"
},
{
"input": "10 10 1",
"output": "First"
},
{
"input": "21 41 5",
"output": "First"
},
{
"input": "13 1 2",
"output": "Second"
},
{
"input": "1 100 3",
"output": "Second"
},
{
"input": "1 64 2",
"output": "Second"
},
{
"input": "3 4 1",
"output": "First"
},
{
"input": "15 15 1",
"output": "First"
},
{
"input": "15 16 1",
"output": "First"
},
{
"input": "16 15 1",
"output": "First"
},
{
"input": "16 16 1",
"output": "First"
},
{
"input": "15 15 2",
"output": "First"
},
{
"input": "15 16 2",
"output": "First"
},
{
"input": "16 15 2",
"output": "First"
},
{
"input": "16 16 2",
"output": "First"
},
{
"input": "15 15 3",
"output": "First"
},
{
"input": "15 16 3",
"output": "First"
},
{
"input": "16 15 3",
"output": "First"
},
{
"input": "16 16 3",
"output": "First"
},
{
"input": "15 17 3",
"output": "First"
},
{
"input": "16 17 3",
"output": "First"
},
{
"input": "17 17 3",
"output": "First"
},
{
"input": "17 15 3",
"output": "First"
},
{
"input": "17 16 3",
"output": "First"
}
] | 1,652,775,968
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 95
| 92
| 4,300,800
|
A = list(map(int,input().split()))
a = A[0];b = A[1];r = A[2]
if 2*r <= min(a,b):
print('First')
else:
print('Second')
|
Title: Plate Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You've got a rectangular table with length *a* and width *b* and the infinite number of plates of radius *r*. Two players play the following game: they take turns to put the plates on the table so that the plates don't lie on each other (but they can touch each other), and so that any point on any plate is located within the table's border. During the game one cannot move the plates that already lie on the table. The player who cannot make another move loses. Determine which player wins, the one who moves first or the one who moves second, provided that both players play optimally well.
Input Specification:
A single line contains three space-separated integers *a*, *b*, *r* (1<=β€<=*a*,<=*b*,<=*r*<=β€<=100) β the table sides and the plates' radius, correspondingly.
Output Specification:
If wins the player who moves first, print "First" (without the quotes). Otherwise print "Second" (without the quotes).
Demo Input:
['5 5 2\n', '6 7 4\n']
Demo Output:
['First\n', 'Second\n']
Note:
In the first sample the table has place for only one plate. The first player puts a plate on the table, the second player can't do that and loses.
In the second sample the table is so small that it doesn't have enough place even for one plate. So the first player loses without making a single move.
|
```python
A = list(map(int,input().split()))
a = A[0];b = A[1];r = A[2]
if 2*r <= min(a,b):
print('First')
else:
print('Second')
```
| 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<=β€<=*a*<=β€<=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"
}
] | 1,658,231,599
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 4
| 92
| 0
|
f = open("input.txt", mode="r")
entered, holding = f.read().split()
f.close()
result = {"front": "1", "back": "2"}
f = open("output.txt", mode="w")
if result[entered] == holding:
f.write("L")
else:
f.write("R")
f.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<=β€<=*a*<=β€<=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
f = open("input.txt", mode="r")
entered, holding = f.read().split()
f.close()
result = {"front": "1", "back": "2"}
f = open("output.txt", mode="w")
if result[entered] == holding:
f.write("L")
else:
f.write("R")
f.close()
```
| 3
|
|
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<=β€<=*n*<=<<=109) β 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"
}
] | 1,593,151,347
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 54
| 218
| 6,758,400
|
n=int(input())
'''l=list(map(int, input().split()))
if(l.count(min(l))>1):
print("Still Rozdil")
else:
print(l.index(min(l))+1)'''
print('0 0',str(n))
|
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<=β€<=*n*<=<<=109) β 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())
'''l=list(map(int, input().split()))
if(l.count(min(l))>1):
print("Still Rozdil")
else:
print(l.index(min(l))+1)'''
print('0 0',str(n))
```
| 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 β 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<=β€<=*n*<=β€<=100) β 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"
}
] | 1,658,478,678
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 32
| 124
| 0
|
n=int(input())
a=list(map(int, input().split()))
c=""
for i in a:
if i%2==0:
c+="0"
else:
c+="1"
zhup=c.count("0")
tak=c.count("1")
if zhup> tak:
print(c.index("1")+1)
else:
print(c.index("0")+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 β 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<=β€<=*n*<=β€<=100) β 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()))
c=""
for i in a:
if i%2==0:
c+="0"
else:
c+="1"
zhup=c.count("0")
tak=c.count("1")
if zhup> tak:
print(c.index("1")+1)
else:
print(c.index("0")+1)
```
| 3.969
|
152
|
A
|
Marks
|
PROGRAMMING
| 900
|
[
"implementation"
] | null | null |
Vasya, or Mr. Vasily Petrov is a dean of a department in a local university. After the winter exams he got his hands on a group's gradebook.
Overall the group has *n* students. They received marks for *m* subjects. Each student got a mark from 1 to 9 (inclusive) for each subject.
Let's consider a student the best at some subject, if there is no student who got a higher mark for this subject. Let's consider a student successful, if there exists a subject he is the best at.
Your task is to find the number of successful students in the group.
|
The first input line contains two integers *n* and *m* (1<=β€<=*n*,<=*m*<=β€<=100) β the number of students and the number of subjects, correspondingly. Next *n* lines each containing *m* characters describe the gradebook. Each character in the gradebook is a number from 1 to 9. Note that the marks in a rows are not sepatated by spaces.
|
Print the single number β the number of successful students in the given group.
|
[
"3 3\n223\n232\n112\n",
"3 5\n91728\n11828\n11111\n"
] |
[
"2\n",
"3\n"
] |
In the first sample test the student number 1 is the best at subjects 1 and 3, student 2 is the best at subjects 1 and 2, but student 3 isn't the best at any subject.
In the second sample test each student is the best at at least one subject.
| 500
|
[
{
"input": "3 3\n223\n232\n112",
"output": "2"
},
{
"input": "3 5\n91728\n11828\n11111",
"output": "3"
},
{
"input": "2 2\n48\n27",
"output": "1"
},
{
"input": "2 1\n4\n6",
"output": "1"
},
{
"input": "1 2\n57",
"output": "1"
},
{
"input": "1 1\n5",
"output": "1"
},
{
"input": "3 4\n2553\n6856\n5133",
"output": "2"
},
{
"input": "8 7\n6264676\n7854895\n3244128\n2465944\n8958761\n1378945\n3859353\n6615285",
"output": "6"
},
{
"input": "9 8\n61531121\n43529859\n18841327\n88683622\n98995641\n62741632\n57441743\n49396792\n63381994",
"output": "4"
},
{
"input": "10 20\n26855662887514171367\n48525577498621511535\n47683778377545341138\n47331616748732562762\n44876938191354974293\n24577238399664382695\n42724955594463126746\n79187344479926159359\n48349683283914388185\n82157191115518781898",
"output": "9"
},
{
"input": "20 15\n471187383859588\n652657222494199\n245695867594992\n726154672861295\n614617827782772\n862889444974692\n373977167653235\n645434268565473\n785993468314573\n722176861496755\n518276853323939\n723712762593348\n728935312568886\n373898548522463\n769777587165681\n247592995114377\n182375946483965\n497496542536127\n988239919677856\n859844339819143",
"output": "18"
},
{
"input": "13 9\n514562255\n322655246\n135162979\n733845982\n473117129\n513967187\n965649829\n799122777\n661249521\n298618978\n659352422\n747778378\n723261619",
"output": "11"
},
{
"input": "75 1\n2\n3\n8\n3\n2\n1\n3\n1\n5\n1\n5\n4\n8\n8\n4\n2\n5\n1\n7\n6\n3\n2\n2\n3\n5\n5\n2\n4\n7\n7\n9\n2\n9\n5\n1\n4\n9\n5\n2\n4\n6\n6\n3\n3\n9\n3\n3\n2\n3\n4\n2\n6\n9\n1\n1\n1\n1\n7\n2\n3\n2\n9\n7\n4\n9\n1\n7\n5\n6\n8\n3\n4\n3\n4\n6",
"output": "7"
},
{
"input": "92 3\n418\n665\n861\n766\n529\n416\n476\n676\n561\n995\n415\n185\n291\n176\n776\n631\n556\n488\n118\n188\n437\n496\n466\n131\n914\n118\n766\n365\n113\n897\n386\n639\n276\n946\n759\n169\n494\n837\n338\n351\n783\n311\n261\n862\n598\n132\n246\n982\n575\n364\n615\n347\n374\n368\n523\n132\n774\n161\n552\n492\n598\n474\n639\n681\n635\n342\n516\n483\n141\n197\n571\n336\n175\n596\n481\n327\n841\n133\n142\n146\n246\n396\n287\n582\n556\n996\n479\n814\n497\n363\n963\n162",
"output": "23"
},
{
"input": "100 1\n1\n6\n9\n1\n1\n5\n5\n4\n6\n9\n6\n1\n7\n8\n7\n3\n8\n8\n7\n6\n2\n1\n5\n8\n7\n3\n5\n4\n9\n7\n1\n2\n4\n1\n6\n5\n1\n3\n9\n4\n5\n8\n1\n2\n1\n9\n7\n3\n7\n1\n2\n2\n2\n2\n3\n9\n7\n2\n4\n7\n1\n6\n8\n1\n5\n6\n1\n1\n2\n9\n7\n4\n9\n1\n9\n4\n1\n3\n5\n2\n4\n4\n6\n5\n1\n4\n5\n8\n4\n7\n6\n5\n6\n9\n5\n8\n1\n5\n1\n6",
"output": "10"
},
{
"input": "100 2\n71\n87\n99\n47\n22\n87\n49\n73\n21\n12\n77\n43\n18\n41\n78\n62\n61\n16\n64\n89\n81\n54\n53\n92\n93\n94\n68\n93\n15\n68\n42\n93\n28\n19\n86\n16\n97\n17\n11\n43\n72\n76\n54\n95\n58\n53\n48\n45\n85\n85\n74\n21\n44\n51\n89\n75\n76\n17\n38\n62\n81\n22\n66\n59\n89\n85\n91\n87\n12\n97\n52\n87\n43\n89\n51\n58\n57\n98\n78\n68\n82\n41\n87\n29\n75\n72\n48\n14\n35\n71\n74\n91\n66\n67\n42\n98\n52\n54\n22\n41",
"output": "21"
},
{
"input": "5 20\n11111111111111111111\n11111111111111111111\n11111111111111111111\n99999999999999999999\n11111111111111111119",
"output": "2"
},
{
"input": "3 3\n111\n111\n999",
"output": "1"
},
{
"input": "3 3\n119\n181\n711",
"output": "3"
},
{
"input": "15 5\n91728\n11828\n11111\n91728\n11828\n11111\n91728\n11828\n11111\n91728\n11828\n11111\n91728\n11828\n11111",
"output": "15"
},
{
"input": "2 20\n22222222222222222222\n11111111111111111111",
"output": "1"
},
{
"input": "3 3\n233\n222\n111",
"output": "2"
},
{
"input": "4 15\n222222222222222\n111111111111119\n111111111111119\n111111111111111",
"output": "3"
},
{
"input": "4 1\n1\n9\n9\n9",
"output": "3"
},
{
"input": "3 3\n123\n321\n132",
"output": "3"
},
{
"input": "3 3\n113\n332\n322",
"output": "3"
},
{
"input": "2 100\n2222222222222222222222222222222222222222222222222222222222222222222222221222222222222222222222222222\n1111111111111111111111111111111111111111111111111111111111111111111111119111111111111111111111111111",
"output": "2"
},
{
"input": "3 3\n321\n231\n123",
"output": "3"
},
{
"input": "2 100\n1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111\n2222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222",
"output": "1"
},
{
"input": "3 3\n221\n111\n111",
"output": "3"
}
] | 1,673,281,281
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 44
| 218
| 8,908,800
|
import sys,math
def get_ints(): return map(int, sys.stdin.readline().strip().split())
def get_list(): return list(map(int, sys.stdin.readline().strip().split()))
def get_string(): return sys.stdin.readline().strip()
n,m = get_ints()
ar = [0]*n
L = []
for i in range(n):
s = get_string()
L.append(s)
for j in range(m):
mx = 0
for i in range(n):
mx = max(mx,eval(L[i][j]))
for i in range(n):
if mx==eval(L[i][j]):
ar[i]=1
print(sum(ar))
|
Title: Marks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya, or Mr. Vasily Petrov is a dean of a department in a local university. After the winter exams he got his hands on a group's gradebook.
Overall the group has *n* students. They received marks for *m* subjects. Each student got a mark from 1 to 9 (inclusive) for each subject.
Let's consider a student the best at some subject, if there is no student who got a higher mark for this subject. Let's consider a student successful, if there exists a subject he is the best at.
Your task is to find the number of successful students in the group.
Input Specification:
The first input line contains two integers *n* and *m* (1<=β€<=*n*,<=*m*<=β€<=100) β the number of students and the number of subjects, correspondingly. Next *n* lines each containing *m* characters describe the gradebook. Each character in the gradebook is a number from 1 to 9. Note that the marks in a rows are not sepatated by spaces.
Output Specification:
Print the single number β the number of successful students in the given group.
Demo Input:
['3 3\n223\n232\n112\n', '3 5\n91728\n11828\n11111\n']
Demo Output:
['2\n', '3\n']
Note:
In the first sample test the student number 1 is the best at subjects 1 and 3, student 2 is the best at subjects 1 and 2, but student 3 isn't the best at any subject.
In the second sample test each student is the best at at least one subject.
|
```python
import sys,math
def get_ints(): return map(int, sys.stdin.readline().strip().split())
def get_list(): return list(map(int, sys.stdin.readline().strip().split()))
def get_string(): return sys.stdin.readline().strip()
n,m = get_ints()
ar = [0]*n
L = []
for i in range(n):
s = get_string()
L.append(s)
for j in range(m):
mx = 0
for i in range(n):
mx = max(mx,eval(L[i][j]))
for i in range(n):
if mx==eval(L[i][j]):
ar[i]=1
print(sum(ar))
```
| 3
|
|
818
|
A
|
Diplomas and Certificates
|
PROGRAMMING
| 800
|
[
"implementation",
"math"
] | null | null |
There are *n* students who have taken part in an olympiad. Now it's time to award the students.
Some of them will receive diplomas, some wiil get certificates, and others won't receive anything. Students with diplomas and certificates are called winners. But there are some rules of counting the number of diplomas and certificates. The number of certificates must be exactly *k* times greater than the number of diplomas. The number of winners must not be greater than half of the number of all students (i.e. not be greater than half of *n*). It's possible that there are no winners.
You have to identify the maximum possible number of winners, according to these rules. Also for this case you have to calculate the number of students with diplomas, the number of students with certificates and the number of students who are not winners.
|
The first (and the only) line of input contains two integers *n* and *k* (1<=β€<=*n*,<=*k*<=β€<=1012), where *n* is the number of students and *k* is the ratio between the number of certificates and the number of diplomas.
|
Output three numbers: the number of students with diplomas, the number of students with certificates and the number of students who are not winners in case when the number of winners is maximum possible.
It's possible that there are no winners.
|
[
"18 2\n",
"9 10\n",
"1000000000000 5\n",
"1000000000000 499999999999\n"
] |
[
"3 6 9\n",
"0 0 9\n",
"83333333333 416666666665 500000000002\n",
"1 499999999999 500000000000\n"
] |
none
| 0
|
[
{
"input": "18 2",
"output": "3 6 9"
},
{
"input": "9 10",
"output": "0 0 9"
},
{
"input": "1000000000000 5",
"output": "83333333333 416666666665 500000000002"
},
{
"input": "1000000000000 499999999999",
"output": "1 499999999999 500000000000"
},
{
"input": "1 1",
"output": "0 0 1"
},
{
"input": "5 3",
"output": "0 0 5"
},
{
"input": "42 6",
"output": "3 18 21"
},
{
"input": "1000000000000 1000",
"output": "499500499 499500499000 500000000501"
},
{
"input": "999999999999 999999",
"output": "499999 499998500001 500000999999"
},
{
"input": "732577309725 132613",
"output": "2762066 366285858458 366288689201"
},
{
"input": "152326362626 15",
"output": "4760198832 71402982480 76163181314"
},
{
"input": "2 1",
"output": "0 0 2"
},
{
"input": "1000000000000 500000000000",
"output": "0 0 1000000000000"
},
{
"input": "100000000000 50000000011",
"output": "0 0 100000000000"
},
{
"input": "1000000000000 32416187567",
"output": "15 486242813505 513757186480"
},
{
"input": "1000000000000 7777777777",
"output": "64 497777777728 502222222208"
},
{
"input": "1000000000000 77777777777",
"output": "6 466666666662 533333333332"
},
{
"input": "100000000000 578485652",
"output": "86 49749766072 50250233842"
},
{
"input": "999999999999 10000000000",
"output": "49 490000000000 509999999950"
},
{
"input": "7 2",
"output": "1 2 4"
},
{
"input": "420506530901 752346673804",
"output": "0 0 420506530901"
},
{
"input": "960375521135 321688347872",
"output": "1 321688347872 638687173262"
},
{
"input": "1000000000000 1000000000000",
"output": "0 0 1000000000000"
},
{
"input": "99999999999 15253636363",
"output": "3 45760909089 54239090907"
},
{
"input": "19 2",
"output": "3 6 10"
},
{
"input": "999999999999 1000000000000",
"output": "0 0 999999999999"
},
{
"input": "1000000000000 5915587276",
"output": "84 496909331184 503090668732"
},
{
"input": "1000000000000 1000000006",
"output": "499 499000002994 500999996507"
},
{
"input": "549755813888 134217728",
"output": "2047 274743689216 275012122625"
},
{
"input": "99999999999 3333333",
"output": "14999 49996661667 50003323333"
},
{
"input": "9 1",
"output": "2 2 5"
},
{
"input": "1000000000000 250000000001",
"output": "1 250000000001 749999999998"
},
{
"input": "5 1",
"output": "1 1 3"
},
{
"input": "3107038133 596040207",
"output": "2 1192080414 1914957717"
},
{
"input": "1000000000000 73786977",
"output": "6776 499980556152 500019437072"
},
{
"input": "1000000000000 73786976",
"output": "6776 499980549376 500019443848"
},
{
"input": "1000000000000 25000000000",
"output": "19 475000000000 524999999981"
},
{
"input": "216929598879 768233755932",
"output": "0 0 216929598879"
},
{
"input": "1000000000000 250000000000",
"output": "1 250000000000 749999999999"
},
{
"input": "1000000000000 100000000001",
"output": "4 400000000004 599999999992"
},
{
"input": "100000000000 100000000001",
"output": "0 0 100000000000"
},
{
"input": "900000000000 100281800001",
"output": "4 401127200004 498872799992"
},
{
"input": "906028900004 109123020071",
"output": "4 436492080284 469536819716"
},
{
"input": "1000000000000 1",
"output": "250000000000 250000000000 500000000000"
}
] | 1,518,203,982
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 44
| 61
| 5,632,000
|
n,k=map(int,input().split())
d=(n//2)//(k+1)
c=k*d
l=n-c-d
print(str(d)+" "+str(c)+" "+str(l))
|
Title: Diplomas and Certificates
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* students who have taken part in an olympiad. Now it's time to award the students.
Some of them will receive diplomas, some wiil get certificates, and others won't receive anything. Students with diplomas and certificates are called winners. But there are some rules of counting the number of diplomas and certificates. The number of certificates must be exactly *k* times greater than the number of diplomas. The number of winners must not be greater than half of the number of all students (i.e. not be greater than half of *n*). It's possible that there are no winners.
You have to identify the maximum possible number of winners, according to these rules. Also for this case you have to calculate the number of students with diplomas, the number of students with certificates and the number of students who are not winners.
Input Specification:
The first (and the only) line of input contains two integers *n* and *k* (1<=β€<=*n*,<=*k*<=β€<=1012), where *n* is the number of students and *k* is the ratio between the number of certificates and the number of diplomas.
Output Specification:
Output three numbers: the number of students with diplomas, the number of students with certificates and the number of students who are not winners in case when the number of winners is maximum possible.
It's possible that there are no winners.
Demo Input:
['18 2\n', '9 10\n', '1000000000000 5\n', '1000000000000 499999999999\n']
Demo Output:
['3 6 9\n', '0 0 9\n', '83333333333 416666666665 500000000002\n', '1 499999999999 500000000000\n']
Note:
none
|
```python
n,k=map(int,input().split())
d=(n//2)//(k+1)
c=k*d
l=n-c-d
print(str(d)+" "+str(c)+" "+str(l))
```
| 3
|
|
599
|
A
|
Patrick and Shopping
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Today Patrick waits for a visit from his friend Spongebob. To prepare for the visit, Patrick needs to buy some goodies in two stores located near his house. There is a *d*1 meter long road between his house and the first shop and a *d*2 meter long road between his house and the second shop. Also, there is a road of length *d*3 directly connecting these two shops to each other. Help Patrick calculate the minimum distance that he needs to walk in order to go to both shops and return to his house.
Patrick always starts at his house. He should visit both shops moving only along the three existing roads and return back to his house. He doesn't mind visiting the same shop or passing the same road multiple times. The only goal is to minimize the total distance traveled.
|
The first line of the input contains three integers *d*1, *d*2, *d*3 (1<=β€<=*d*1,<=*d*2,<=*d*3<=β€<=108)Β β the lengths of the paths.
- *d*1 is the length of the path connecting Patrick's house and the first shop; - *d*2 is the length of the path connecting Patrick's house and the second shop; - *d*3 is the length of the path connecting both shops.
|
Print the minimum distance that Patrick will have to walk in order to visit both shops and return to his house.
|
[
"10 20 30\n",
"1 1 5\n"
] |
[
"60\n",
"4\n"
] |
The first sample is shown on the picture in the problem statement. One of the optimal routes is: house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> first shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> second shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house.
In the second sample one of the optimal routes is: house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> first shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> second shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house.
| 500
|
[
{
"input": "10 20 30",
"output": "60"
},
{
"input": "1 1 5",
"output": "4"
},
{
"input": "100 33 34",
"output": "134"
},
{
"input": "777 777 777",
"output": "2331"
},
{
"input": "2 2 8",
"output": "8"
},
{
"input": "12 34 56",
"output": "92"
},
{
"input": "789 101112 131415",
"output": "203802"
},
{
"input": "27485716 99999999 35182",
"output": "55041796"
},
{
"input": "1 293548 5",
"output": "12"
},
{
"input": "12059 259855 5874875",
"output": "543828"
},
{
"input": "46981 105809 585858",
"output": "305580"
},
{
"input": "9889 1221 2442",
"output": "7326"
},
{
"input": "100500 200600 300700",
"output": "601800"
},
{
"input": "318476 318476 318476",
"output": "955428"
},
{
"input": "23985 3353 75633",
"output": "54676"
},
{
"input": "120 1298 2222",
"output": "2836"
},
{
"input": "98437 23487 666672",
"output": "243848"
},
{
"input": "100000000 100000000 100000000",
"output": "300000000"
},
{
"input": "2 5 2",
"output": "8"
},
{
"input": "1 1000 1",
"output": "4"
},
{
"input": "1 100000000 1",
"output": "4"
}
] | 1,606,459,997
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 21
| 124
| 0
|
def good(a,b,c):
return min(a,b+c)+min(c,a+b)+min(b,a+c)
a,b,c=map(int,input().split())
print(good(a, b, c))
|
Title: Patrick and Shopping
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Today Patrick waits for a visit from his friend Spongebob. To prepare for the visit, Patrick needs to buy some goodies in two stores located near his house. There is a *d*1 meter long road between his house and the first shop and a *d*2 meter long road between his house and the second shop. Also, there is a road of length *d*3 directly connecting these two shops to each other. Help Patrick calculate the minimum distance that he needs to walk in order to go to both shops and return to his house.
Patrick always starts at his house. He should visit both shops moving only along the three existing roads and return back to his house. He doesn't mind visiting the same shop or passing the same road multiple times. The only goal is to minimize the total distance traveled.
Input Specification:
The first line of the input contains three integers *d*1, *d*2, *d*3 (1<=β€<=*d*1,<=*d*2,<=*d*3<=β€<=108)Β β the lengths of the paths.
- *d*1 is the length of the path connecting Patrick's house and the first shop; - *d*2 is the length of the path connecting Patrick's house and the second shop; - *d*3 is the length of the path connecting both shops.
Output Specification:
Print the minimum distance that Patrick will have to walk in order to visit both shops and return to his house.
Demo Input:
['10 20 30\n', '1 1 5\n']
Demo Output:
['60\n', '4\n']
Note:
The first sample is shown on the picture in the problem statement. One of the optimal routes is: house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> first shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> second shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house.
In the second sample one of the optimal routes is: house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> first shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> second shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house.
|
```python
def good(a,b,c):
return min(a,b+c)+min(c,a+b)+min(b,a+c)
a,b,c=map(int,input().split())
print(good(a, b, c))
```
| 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"
}
] | 1,695,654,993
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 25
| 124
| 0
|
def solve():
s = input()
print(s[0].capitalize()+s[1:])
# t = int(input())
t = 1
while t:
solve()
t -= 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
def solve():
s = input()
print(s[0].capitalize()+s[1:])
# t = int(input())
t = 1
while t:
solve()
t -= 1
```
| 3
|
|
420
|
A
|
Start Up
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
Recently, a start up by two students of a state university of city F gained incredible popularity. Now it's time to start a new company. But what do we call it?
The market analysts came up with a very smart plan: the name of the company should be identical to its reflection in a mirror! In other words, if we write out the name of the company on a piece of paper in a line (horizontally, from left to right) with large English letters, then put this piece of paper in front of the mirror, then the reflection of the name in the mirror should perfectly match the line written on the piece of paper.
There are many suggestions for the company name, so coming up to the mirror with a piece of paper for each name wouldn't be sensible. The founders of the company decided to automatize this process. They asked you to write a program that can, given a word, determine whether the word is a 'mirror' word or not.
|
The first line contains a non-empty name that needs to be checked. The name contains at most 105 large English letters. The name will be written with the next sans serif font:
|
Print 'YES' (without the quotes), if the given name matches its mirror reflection. Otherwise, print 'NO' (without the quotes).
|
[
"AHA\n",
"Z\n",
"XO\n"
] |
[
"YES\n",
"NO\n",
"NO\n"
] |
none
| 500
|
[
{
"input": "AHA",
"output": "YES"
},
{
"input": "Z",
"output": "NO"
},
{
"input": "XO",
"output": "NO"
},
{
"input": "AAA",
"output": "YES"
},
{
"input": "AHHA",
"output": "YES"
},
{
"input": "BAB",
"output": "NO"
},
{
"input": "OMMMAAMMMO",
"output": "YES"
},
{
"input": "YYHUIUGYI",
"output": "NO"
},
{
"input": "TT",
"output": "YES"
},
{
"input": "UUU",
"output": "YES"
},
{
"input": "WYYW",
"output": "YES"
},
{
"input": "MITIM",
"output": "YES"
},
{
"input": "VO",
"output": "NO"
},
{
"input": "WWS",
"output": "NO"
},
{
"input": "VIYMAXXAVM",
"output": "NO"
},
{
"input": "OVWIHIWVYXMVAAAATOXWOIUUHYXHIHHVUIOOXWHOXTUUMUUVHVWWYUTIAUAITAOMHXWMTTOIVMIVOTHOVOIOHYHAOXWAUVWAVIVM",
"output": "NO"
},
{
"input": "CC",
"output": "NO"
},
{
"input": "QOQ",
"output": "NO"
},
{
"input": "AEEA",
"output": "NO"
},
{
"input": "OQQQO",
"output": "NO"
},
{
"input": "HNCMEEMCNH",
"output": "NO"
},
{
"input": "QDPINBMCRFWXPDBFGOZVVOCEMJRUCTOADEWEGTVBVBFWWRPGYEEYGPRWWFBVBVTGEWEDAOTCURJMECOVVZOGFBDPXWFRCMBNIPDQ",
"output": "NO"
},
{
"input": "A",
"output": "YES"
},
{
"input": "B",
"output": "NO"
},
{
"input": "C",
"output": "NO"
},
{
"input": "D",
"output": "NO"
},
{
"input": "E",
"output": "NO"
},
{
"input": "F",
"output": "NO"
},
{
"input": "G",
"output": "NO"
},
{
"input": "H",
"output": "YES"
},
{
"input": "I",
"output": "YES"
},
{
"input": "J",
"output": "NO"
},
{
"input": "K",
"output": "NO"
},
{
"input": "L",
"output": "NO"
},
{
"input": "M",
"output": "YES"
},
{
"input": "N",
"output": "NO"
},
{
"input": "O",
"output": "YES"
},
{
"input": "P",
"output": "NO"
},
{
"input": "Q",
"output": "NO"
},
{
"input": "R",
"output": "NO"
},
{
"input": "S",
"output": "NO"
},
{
"input": "T",
"output": "YES"
},
{
"input": "U",
"output": "YES"
},
{
"input": "V",
"output": "YES"
},
{
"input": "W",
"output": "YES"
},
{
"input": "X",
"output": "YES"
},
{
"input": "Y",
"output": "YES"
},
{
"input": "JL",
"output": "NO"
},
{
"input": "AAAKTAAA",
"output": "NO"
},
{
"input": "AKA",
"output": "NO"
},
{
"input": "AAJAA",
"output": "NO"
},
{
"input": "ABA",
"output": "NO"
},
{
"input": "AAAAAABAAAAAA",
"output": "NO"
},
{
"input": "ZZ",
"output": "NO"
},
{
"input": "ADA",
"output": "NO"
},
{
"input": "N",
"output": "NO"
},
{
"input": "P",
"output": "NO"
},
{
"input": "LAL",
"output": "NO"
},
{
"input": "AABAA",
"output": "NO"
},
{
"input": "AZA",
"output": "NO"
},
{
"input": "V",
"output": "YES"
},
{
"input": "SSS",
"output": "NO"
},
{
"input": "NNN",
"output": "NO"
},
{
"input": "S",
"output": "NO"
},
{
"input": "I",
"output": "YES"
},
{
"input": "SS",
"output": "NO"
},
{
"input": "E",
"output": "NO"
}
] | 1,398,500,809
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 80
| 77
| 512,000
|
import sys
def check(text):
legal = 'AHIMOTUVWXY'
for i in text:
if i not in legal:
return False
rev = list(text)
rev.reverse()
if text != ''.join(rev):
return False
return True
if __name__ == '__main__':
text = sys.stdin.readline().split()[0]
if check(text):
print('YES')
else:
print('NO')
|
Title: Start Up
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Recently, a start up by two students of a state university of city F gained incredible popularity. Now it's time to start a new company. But what do we call it?
The market analysts came up with a very smart plan: the name of the company should be identical to its reflection in a mirror! In other words, if we write out the name of the company on a piece of paper in a line (horizontally, from left to right) with large English letters, then put this piece of paper in front of the mirror, then the reflection of the name in the mirror should perfectly match the line written on the piece of paper.
There are many suggestions for the company name, so coming up to the mirror with a piece of paper for each name wouldn't be sensible. The founders of the company decided to automatize this process. They asked you to write a program that can, given a word, determine whether the word is a 'mirror' word or not.
Input Specification:
The first line contains a non-empty name that needs to be checked. The name contains at most 105 large English letters. The name will be written with the next sans serif font:
Output Specification:
Print 'YES' (without the quotes), if the given name matches its mirror reflection. Otherwise, print 'NO' (without the quotes).
Demo Input:
['AHA\n', 'Z\n', 'XO\n']
Demo Output:
['YES\n', 'NO\n', 'NO\n']
Note:
none
|
```python
import sys
def check(text):
legal = 'AHIMOTUVWXY'
for i in text:
if i not in legal:
return False
rev = list(text)
rev.reverse()
if text != ''.join(rev):
return False
return True
if __name__ == '__main__':
text = sys.stdin.readline().split()[0]
if check(text):
print('YES')
else:
print('NO')
```
| 3
|
|
978
|
A
|
Remove Duplicates
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Petya has an array $a$ consisting of $n$ integers. He wants to remove duplicate (equal) elements.
Petya wants to leave only the rightmost entry (occurrence) for each element of the array. The relative order of the remaining unique elements should not be changed.
|
The first line contains a single integer $n$ ($1 \le n \le 50$) β the number of elements in Petya's array.
The following line contains a sequence $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1\,000$) β the Petya's array.
|
In the first line print integer $x$ β the number of elements which will be left in Petya's array after he removed the duplicates.
In the second line print $x$ integers separated with a space β Petya's array after he removed the duplicates. For each unique element only the rightmost entry should be left.
|
[
"6\n1 5 5 1 6 1\n",
"5\n2 4 2 4 4\n",
"5\n6 6 6 6 6\n"
] |
[
"3\n5 6 1 \n",
"2\n2 4 \n",
"1\n6 \n"
] |
In the first example you should remove two integers $1$, which are in the positions $1$ and $4$. Also you should remove the integer $5$, which is in the position $2$.
In the second example you should remove integer $2$, which is in the position $1$, and two integers $4$, which are in the positions $2$ and $4$.
In the third example you should remove four integers $6$, which are in the positions $1$, $2$, $3$ and $4$.
| 0
|
[
{
"input": "6\n1 5 5 1 6 1",
"output": "3\n5 6 1 "
},
{
"input": "5\n2 4 2 4 4",
"output": "2\n2 4 "
},
{
"input": "5\n6 6 6 6 6",
"output": "1\n6 "
},
{
"input": "7\n1 2 3 4 2 2 3",
"output": "4\n1 4 2 3 "
},
{
"input": "9\n100 100 100 99 99 99 100 100 100",
"output": "2\n99 100 "
},
{
"input": "27\n489 489 487 488 750 230 43 645 42 42 489 42 973 42 973 750 645 355 868 112 868 489 750 489 887 489 868",
"output": "13\n487 488 230 43 42 973 645 355 112 750 887 489 868 "
},
{
"input": "40\n151 421 421 909 117 222 909 954 227 421 227 954 954 222 421 227 421 421 421 151 421 227 222 222 222 222 421 183 421 227 421 954 222 421 954 421 222 421 909 421",
"output": "8\n117 151 183 227 954 222 909 421 "
},
{
"input": "48\n2 2 2 903 903 2 726 2 2 2 2 2 2 2 2 2 2 726 2 2 2 2 2 2 2 726 2 2 2 2 62 2 2 2 2 2 2 2 2 726 62 726 2 2 2 903 903 2",
"output": "4\n62 726 903 2 "
},
{
"input": "1\n1",
"output": "1\n1 "
},
{
"input": "13\n5 37 375 5 37 33 37 375 37 2 3 3 2",
"output": "6\n5 33 375 37 3 2 "
},
{
"input": "50\n1 2 3 4 5 4 3 2 1 2 3 2 1 4 5 5 4 3 2 1 1 2 3 4 5 4 3 2 1 2 3 2 1 4 5 5 4 3 2 1 4 3 2 5 1 6 6 6 6 6",
"output": "6\n4 3 2 5 1 6 "
},
{
"input": "47\n233 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 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": "2\n233 1 "
},
{
"input": "47\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",
"output": "1\n1 "
},
{
"input": "2\n964 964",
"output": "1\n964 "
},
{
"input": "2\n1000 1000",
"output": "1\n1000 "
},
{
"input": "1\n1000",
"output": "1\n1000 "
},
{
"input": "45\n991 991 996 996 992 992 999 1000 998 1000 992 999 996 999 991 991 999 993 992 999 1000 997 992 999 996 991 994 996 991 999 1000 993 999 997 999 992 991 997 991 998 998 995 998 994 993",
"output": "10\n996 1000 999 992 997 991 995 998 994 993 "
},
{
"input": "6\n994 993 1000 998 991 994",
"output": "5\n993 1000 998 991 994 "
},
{
"input": "48\n992 995 992 991 994 992 995 999 996 993 999 995 993 992 1000 992 997 996 991 993 992 998 998 998 999 995 992 992 993 992 992 995 996 995 997 991 997 991 999 994 994 997 1000 998 1000 992 1000 999",
"output": "10\n993 996 995 991 994 997 998 992 1000 999 "
},
{
"input": "3\n6 6 3",
"output": "2\n6 3 "
},
{
"input": "36\n999 1000 993 993 1000 999 996 997 998 995 995 997 999 995 1000 998 998 994 993 998 994 999 1000 995 996 994 991 991 999 996 993 999 996 998 991 997",
"output": "9\n1000 995 994 993 999 996 998 991 997 "
},
{
"input": "49\n991 997 995 991 995 998 994 996 997 991 998 993 992 999 996 995 991 991 999 997 995 993 997 1000 997 993 993 994 999 994 992 991 1000 996 994 996 996 999 992 992 1000 991 997 993 991 994 997 991 996",
"output": "10\n998 995 999 992 1000 993 994 997 991 996 "
},
{
"input": "2\n1000 999",
"output": "2\n1000 999 "
},
{
"input": "28\n997 994 991 994 994 995 1000 992 995 994 994 995 991 996 991 996 991 999 999 993 994 997 995 992 991 992 998 1000",
"output": "10\n996 999 993 994 997 995 991 992 998 1000 "
},
{
"input": "15\n991 995 995 1000 991 993 997 998 1000 994 1000 992 991 995 992",
"output": "8\n993 997 998 994 1000 991 995 992 "
},
{
"input": "6\n1000 1 1000 2 1000 1",
"output": "3\n2 1000 1 "
},
{
"input": "6\n1000 1 1000 2 1 1000",
"output": "3\n2 1 1000 "
},
{
"input": "7\n1 1000 1 999 1000 1 2",
"output": "4\n999 1000 1 2 "
}
] | 1,662,422,449
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 28
| 46
| 0
|
no_inputs = int(input())
nums = list(map(int, input().split()))
unique_nums = []
for num in nums[::-1]:
if num not in unique_nums:
unique_nums.append(num)
print(len(unique_nums))
print(" ".join(str(item) for item in unique_nums[::-1]))
|
Title: Remove Duplicates
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya has an array $a$ consisting of $n$ integers. He wants to remove duplicate (equal) elements.
Petya wants to leave only the rightmost entry (occurrence) for each element of the array. The relative order of the remaining unique elements should not be changed.
Input Specification:
The first line contains a single integer $n$ ($1 \le n \le 50$) β the number of elements in Petya's array.
The following line contains a sequence $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1\,000$) β the Petya's array.
Output Specification:
In the first line print integer $x$ β the number of elements which will be left in Petya's array after he removed the duplicates.
In the second line print $x$ integers separated with a space β Petya's array after he removed the duplicates. For each unique element only the rightmost entry should be left.
Demo Input:
['6\n1 5 5 1 6 1\n', '5\n2 4 2 4 4\n', '5\n6 6 6 6 6\n']
Demo Output:
['3\n5 6 1 \n', '2\n2 4 \n', '1\n6 \n']
Note:
In the first example you should remove two integers $1$, which are in the positions $1$ and $4$. Also you should remove the integer $5$, which is in the position $2$.
In the second example you should remove integer $2$, which is in the position $1$, and two integers $4$, which are in the positions $2$ and $4$.
In the third example you should remove four integers $6$, which are in the positions $1$, $2$, $3$ and $4$.
|
```python
no_inputs = int(input())
nums = list(map(int, input().split()))
unique_nums = []
for num in nums[::-1]:
if num not in unique_nums:
unique_nums.append(num)
print(len(unique_nums))
print(" ".join(str(item) for item in unique_nums[::-1]))
```
| 3
|
|
740
|
A
|
Alyona and copybooks
|
PROGRAMMING
| 1,300
|
[
"brute force",
"implementation"
] | null | null |
Little girl Alyona is in a shop to buy some copybooks for school. She study four subjects so she wants to have equal number of copybooks for each of the subjects. There are three types of copybook's packs in the shop: it is possible to buy one copybook for *a* rubles, a pack of two copybooks for *b* rubles, and a pack of three copybooks for *c* rubles. Alyona already has *n* copybooks.
What is the minimum amount of rubles she should pay to buy such number of copybooks *k* that *n*<=+<=*k* is divisible by 4? There are infinitely many packs of any type in the shop. Alyona can buy packs of different type in the same purchase.
|
The only line contains 4 integers *n*, *a*, *b*, *c* (1<=β€<=*n*,<=*a*,<=*b*,<=*c*<=β€<=109).
|
Print the minimum amount of rubles she should pay to buy such number of copybooks *k* that *n*<=+<=*k* is divisible by 4.
|
[
"1 1 3 4\n",
"6 2 1 1\n",
"4 4 4 4\n",
"999999999 1000000000 1000000000 1000000000\n"
] |
[
"3\n",
"1\n",
"0\n",
"1000000000\n"
] |
In the first example Alyona can buy 3 packs of 1 copybook for 3*a*β=β3 rubles in total. After that she will have 4 copybooks which she can split between the subjects equally.
In the second example Alyuna can buy a pack of 2 copybooks for *b*β=β1 ruble. She will have 8 copybooks in total.
In the third example Alyona can split the copybooks she already has between the 4 subject equally, so she doesn't need to buy anything.
In the fourth example Alyona should buy one pack of one copybook.
| 500
|
[
{
"input": "1 1 3 4",
"output": "3"
},
{
"input": "6 2 1 1",
"output": "1"
},
{
"input": "4 4 4 4",
"output": "0"
},
{
"input": "999999999 1000000000 1000000000 1000000000",
"output": "1000000000"
},
{
"input": "1016 3 2 1",
"output": "0"
},
{
"input": "17 100 100 1",
"output": "1"
},
{
"input": "17 2 3 100",
"output": "5"
},
{
"input": "18 1 3 3",
"output": "2"
},
{
"input": "19 1 1 1",
"output": "1"
},
{
"input": "999999997 999999990 1000000000 1000000000",
"output": "1000000000"
},
{
"input": "999999998 1000000000 999999990 1000000000",
"output": "999999990"
},
{
"input": "634074578 336470888 481199252 167959139",
"output": "335918278"
},
{
"input": "999999999 1000000000 1000000000 999999990",
"output": "1000000000"
},
{
"input": "804928248 75475634 54748096 641009859",
"output": "0"
},
{
"input": "535590429 374288891 923264237 524125987",
"output": "524125987"
},
{
"input": "561219907 673102149 496813081 702209411",
"output": "673102149"
},
{
"input": "291882089 412106895 365329221 585325539",
"output": "585325539"
},
{
"input": "757703054 5887448 643910770 58376259",
"output": "11774896"
},
{
"input": "783332532 449924898 72235422 941492387",
"output": "0"
},
{
"input": "513994713 43705451 940751563 824608515",
"output": "131116353"
},
{
"input": "539624191 782710197 514300407 2691939",
"output": "8075817"
},
{
"input": "983359971 640274071 598196518 802030518",
"output": "640274071"
},
{
"input": "8989449 379278816 26521171 685146646",
"output": "405799987"
},
{
"input": "34618927 678092074 895037311 863230070",
"output": "678092074"
},
{
"input": "205472596 417096820 468586155 41313494",
"output": "0"
},
{
"input": "19 5 1 2",
"output": "3"
},
{
"input": "17 1 2 2",
"output": "2"
},
{
"input": "18 3 3 1",
"output": "2"
},
{
"input": "19 4 3 1",
"output": "3"
},
{
"input": "936134778 715910077 747167704 219396918",
"output": "438793836"
},
{
"input": "961764255 454914823 615683844 102513046",
"output": "307539138"
},
{
"input": "692426437 48695377 189232688 985629174",
"output": "146086131"
},
{
"input": "863280107 347508634 912524637 458679894",
"output": "347508634"
},
{
"input": "593942288 86513380 486073481 341796022",
"output": "0"
},
{
"input": "914539062 680293934 764655030 519879446",
"output": "764655030"
},
{
"input": "552472140 509061481 586588704 452405440",
"output": "0"
},
{
"input": "723325809 807874739 160137548 335521569",
"output": "335521569"
},
{
"input": "748955287 546879484 733686393 808572289",
"output": "546879484"
},
{
"input": "774584765 845692742 162011045 691688417",
"output": "691688417"
},
{
"input": "505246946 439473295 30527185 869771841",
"output": "30527185"
},
{
"input": "676100616 178478041 604076030 752887969",
"output": "0"
},
{
"input": "701730093 477291299 177624874 930971393",
"output": "654916173"
},
{
"input": "432392275 216296044 751173719 109054817",
"output": "216296044"
},
{
"input": "458021753 810076598 324722563 992170945",
"output": "992170945"
},
{
"input": "188683934 254114048 48014511 170254369",
"output": "48014511"
},
{
"input": "561775796 937657403 280013594 248004555",
"output": "0"
},
{
"input": "1000000000 1000000000 1000000000 1000000000",
"output": "0"
},
{
"input": "3 10000 10000 3",
"output": "9"
},
{
"input": "3 12 3 4",
"output": "7"
},
{
"input": "3 10000 10000 1",
"output": "3"
},
{
"input": "3 1000 1000 1",
"output": "3"
},
{
"input": "3 10 10 1",
"output": "3"
},
{
"input": "3 100 100 1",
"output": "3"
},
{
"input": "3 100000 10000 1",
"output": "3"
},
{
"input": "7 10 2 3",
"output": "5"
},
{
"input": "3 1000 1000 2",
"output": "6"
},
{
"input": "1 100000 1 100000",
"output": "100000"
},
{
"input": "7 4 3 1",
"output": "3"
},
{
"input": "3 1000 1000 3",
"output": "9"
},
{
"input": "3 1000 1 1",
"output": "2"
},
{
"input": "3 10 1 1",
"output": "2"
},
{
"input": "3 100000 1 1",
"output": "2"
},
{
"input": "3 100 1 1",
"output": "2"
},
{
"input": "3 100000 100000 1",
"output": "3"
},
{
"input": "3 1000 1 100",
"output": "101"
},
{
"input": "3 1000000000 1 1000000000",
"output": "1000000000"
},
{
"input": "3 1000 1 10",
"output": "11"
},
{
"input": "3 200 1 100",
"output": "101"
},
{
"input": "7 4 1 1",
"output": "2"
},
{
"input": "7 4 12 1",
"output": "3"
},
{
"input": "3 9 1 1",
"output": "2"
},
{
"input": "3 10000000 1000000 1",
"output": "3"
},
{
"input": "7 1000 1000 1",
"output": "3"
},
{
"input": "3 10000 1 30",
"output": "31"
},
{
"input": "3 1000 1 2",
"output": "3"
},
{
"input": "7 12 6 1",
"output": "3"
},
{
"input": "3 100000 1 1000",
"output": "1001"
},
{
"input": "7 1000 1000 3",
"output": "9"
},
{
"input": "3 4 3 1",
"output": "3"
},
{
"input": "3 3000000 1 100000",
"output": "100001"
},
{
"input": "3 3 1 1",
"output": "2"
},
{
"input": "3 10 1 5",
"output": "6"
},
{
"input": "3 2000 2000 1",
"output": "3"
},
{
"input": "3 10000000 10000000 1",
"output": "3"
},
{
"input": "3 5 1 1",
"output": "2"
},
{
"input": "3 100 1 33",
"output": "34"
},
{
"input": "7 9 2 7",
"output": "9"
},
{
"input": "4448 2 3 6",
"output": "0"
},
{
"input": "2228 1 6 3",
"output": "0"
}
] | 1,484,659,844
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 89
| 77
| 4,608,000
|
n,a,b,c=map(int,input().split())
if n%4==0:
print(0)
if (n+1)%4==0:
print(min(a,b+c,3*c))
if (n+2)%4==0:
print(min(2*a,b,c*2))
if (n+3)%4==0:
print(min(3*a,b+a,c))
|
Title: Alyona and copybooks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little girl Alyona is in a shop to buy some copybooks for school. She study four subjects so she wants to have equal number of copybooks for each of the subjects. There are three types of copybook's packs in the shop: it is possible to buy one copybook for *a* rubles, a pack of two copybooks for *b* rubles, and a pack of three copybooks for *c* rubles. Alyona already has *n* copybooks.
What is the minimum amount of rubles she should pay to buy such number of copybooks *k* that *n*<=+<=*k* is divisible by 4? There are infinitely many packs of any type in the shop. Alyona can buy packs of different type in the same purchase.
Input Specification:
The only line contains 4 integers *n*, *a*, *b*, *c* (1<=β€<=*n*,<=*a*,<=*b*,<=*c*<=β€<=109).
Output Specification:
Print the minimum amount of rubles she should pay to buy such number of copybooks *k* that *n*<=+<=*k* is divisible by 4.
Demo Input:
['1 1 3 4\n', '6 2 1 1\n', '4 4 4 4\n', '999999999 1000000000 1000000000 1000000000\n']
Demo Output:
['3\n', '1\n', '0\n', '1000000000\n']
Note:
In the first example Alyona can buy 3 packs of 1 copybook for 3*a*β=β3 rubles in total. After that she will have 4 copybooks which she can split between the subjects equally.
In the second example Alyuna can buy a pack of 2 copybooks for *b*β=β1 ruble. She will have 8 copybooks in total.
In the third example Alyona can split the copybooks she already has between the 4 subject equally, so she doesn't need to buy anything.
In the fourth example Alyona should buy one pack of one copybook.
|
```python
n,a,b,c=map(int,input().split())
if n%4==0:
print(0)
if (n+1)%4==0:
print(min(a,b+c,3*c))
if (n+2)%4==0:
print(min(2*a,b,c*2))
if (n+3)%4==0:
print(min(3*a,b+a,c))
```
| 3
|
|
75
|
A
|
Life Without Zeros
|
PROGRAMMING
| 1,000
|
[
"implementation"
] |
A. Life Without Zeros
|
2
|
256
|
Can you imagine our life if we removed all zeros from it? For sure we will have many problems.
In this problem we will have a simple example if we removed all zeros from our life, it's the addition operation. Let's assume you are given this equation *a*<=+<=*b*<==<=*c*, where *a* and *b* are positive integers, and *c* is the sum of *a* and *b*. Now let's remove all zeros from this equation. Will the equation remain correct after removing all zeros?
For example if the equation is 101<=+<=102<==<=203, if we removed all zeros it will be 11<=+<=12<==<=23 which is still a correct equation.
But if the equation is 105<=+<=106<==<=211, if we removed all zeros it will be 15<=+<=16<==<=211 which is not a correct equation.
|
The input will consist of two lines, the first line will contain the integer *a*, and the second line will contain the integer *b* which are in the equation as described above (1<=β€<=*a*,<=*b*<=β€<=109). There won't be any leading zeros in both. The value of *c* should be calculated as *c*<==<=*a*<=+<=*b*.
|
The output will be just one line, you should print "YES" if the equation will remain correct after removing all zeros, and print "NO" otherwise.
|
[
"101\n102\n",
"105\n106\n"
] |
[
"YES\n",
"NO\n"
] |
none
| 500
|
[
{
"input": "101\n102",
"output": "YES"
},
{
"input": "105\n106",
"output": "NO"
},
{
"input": "544\n397",
"output": "YES"
},
{
"input": "822\n280",
"output": "NO"
},
{
"input": "101\n413",
"output": "NO"
},
{
"input": "309\n139",
"output": "NO"
},
{
"input": "693\n970",
"output": "NO"
},
{
"input": "981\n1",
"output": "YES"
},
{
"input": "352\n276",
"output": "YES"
},
{
"input": "164\n691",
"output": "YES"
},
{
"input": "110036\n43",
"output": "YES"
},
{
"input": "100\n1000",
"output": "NO"
},
{
"input": "1000000000\n1000000000",
"output": "YES"
},
{
"input": "999999999\n999999999",
"output": "YES"
},
{
"input": "6\n4",
"output": "NO"
},
{
"input": "123456\n876543",
"output": "YES"
},
{
"input": "1234567\n9876543",
"output": "NO"
},
{
"input": "1111111\n1119111",
"output": "NO"
},
{
"input": "10502405\n40403403",
"output": "YES"
},
{
"input": "1\n1",
"output": "YES"
},
{
"input": "5\n4",
"output": "YES"
},
{
"input": "1\n6",
"output": "YES"
},
{
"input": "40\n17",
"output": "NO"
},
{
"input": "35\n81",
"output": "YES"
},
{
"input": "347\n908",
"output": "NO"
},
{
"input": "2081\n7450",
"output": "NO"
},
{
"input": "84355694\n373629503",
"output": "NO"
},
{
"input": "885992048\n510468672",
"output": "NO"
},
{
"input": "540144752\n647307841",
"output": "NO"
},
{
"input": "341781108\n784147010",
"output": "NO"
},
{
"input": "960440938\n572334653",
"output": "NO"
},
{
"input": "657414510\n768771903",
"output": "NO"
},
{
"input": "501871728\n725074574",
"output": "NO"
},
{
"input": "968\n851",
"output": "YES"
},
{
"input": "943\n939",
"output": "YES"
},
{
"input": "347\n908",
"output": "NO"
},
{
"input": "611\n41",
"output": "YES"
},
{
"input": "737\n289",
"output": "NO"
},
{
"input": "168\n571",
"output": "YES"
},
{
"input": "868\n463",
"output": "YES"
},
{
"input": "815732007\n913882008",
"output": "NO"
},
{
"input": "459039901\n453033850",
"output": "NO"
},
{
"input": "975305404\n484601256",
"output": "NO"
}
] | 1,500,357,997
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 43
| 124
| 4,608,000
|
n = int(input().strip())
m = int(input().strip())
c = str(n+m)
n = str(n)
m = str(m)
cc = mm = nn = []
jn = ''
jm = ''
jc = ''
for i in c:
if i != '0':
jc = jc + i
for i in m:
if i != '0':
jm = jm + i
for i in n :
if i != '0':
jn = jn + i
if int(jn) + int(jm) == int(jc):
print('YES')
else:
print('NO')
|
Title: Life Without Zeros
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Can you imagine our life if we removed all zeros from it? For sure we will have many problems.
In this problem we will have a simple example if we removed all zeros from our life, it's the addition operation. Let's assume you are given this equation *a*<=+<=*b*<==<=*c*, where *a* and *b* are positive integers, and *c* is the sum of *a* and *b*. Now let's remove all zeros from this equation. Will the equation remain correct after removing all zeros?
For example if the equation is 101<=+<=102<==<=203, if we removed all zeros it will be 11<=+<=12<==<=23 which is still a correct equation.
But if the equation is 105<=+<=106<==<=211, if we removed all zeros it will be 15<=+<=16<==<=211 which is not a correct equation.
Input Specification:
The input will consist of two lines, the first line will contain the integer *a*, and the second line will contain the integer *b* which are in the equation as described above (1<=β€<=*a*,<=*b*<=β€<=109). There won't be any leading zeros in both. The value of *c* should be calculated as *c*<==<=*a*<=+<=*b*.
Output Specification:
The output will be just one line, you should print "YES" if the equation will remain correct after removing all zeros, and print "NO" otherwise.
Demo Input:
['101\n102\n', '105\n106\n']
Demo Output:
['YES\n', 'NO\n']
Note:
none
|
```python
n = int(input().strip())
m = int(input().strip())
c = str(n+m)
n = str(n)
m = str(m)
cc = mm = nn = []
jn = ''
jm = ''
jc = ''
for i in c:
if i != '0':
jc = jc + i
for i in m:
if i != '0':
jm = jm + i
for i in n :
if i != '0':
jn = jn + i
if int(jn) + int(jm) == int(jc):
print('YES')
else:
print('NO')
```
| 3.960417
|
525
|
A
|
Vitaliy and Pie
|
PROGRAMMING
| 1,100
|
[
"greedy",
"hashing",
"strings"
] | null | null |
After a hard day Vitaly got very hungry and he wants to eat his favorite potato pie. But it's not that simple. Vitaly is in the first room of the house with *n* room located in a line and numbered starting from one from left to right. You can go from the first room to the second room, from the second room to the third room and so on β you can go from the (*n*<=-<=1)-th room to the *n*-th room. Thus, you can go to room *x* only from room *x*<=-<=1.
The potato pie is located in the *n*-th room and Vitaly needs to go there.
Each pair of consecutive rooms has a door between them. In order to go to room *x* from room *x*<=-<=1, you need to open the door between the rooms with the corresponding key.
In total the house has several types of doors (represented by uppercase Latin letters) and several types of keys (represented by lowercase Latin letters). The key of type *t* can open the door of type *T* if and only if *t* and *T* are the same letter, written in different cases. For example, key f can open door F.
Each of the first *n*<=-<=1 rooms contains exactly one key of some type that Vitaly can use to get to next rooms. Once the door is open with some key, Vitaly won't get the key from the keyhole but he will immediately run into the next room. In other words, each key can open no more than one door.
Vitaly realizes that he may end up in some room without the key that opens the door to the next room. Before the start his run for the potato pie Vitaly can buy any number of keys of any type that is guaranteed to get to room *n*.
Given the plan of the house, Vitaly wants to know what is the minimum number of keys he needs to buy to surely get to the room *n*, which has a delicious potato pie. Write a program that will help Vitaly find out this number.
|
The first line of the input contains a positive integer *n* (2<=β€<=*n*<=β€<=105)Β βΒ the number of rooms in the house.
The second line of the input contains string *s* of length 2Β·*n*<=-<=2. Let's number the elements of the string from left to right, starting from one.
The odd positions in the given string *s* contain lowercase Latin lettersΒ βΒ the types of the keys that lie in the corresponding rooms. Thus, each odd position *i* of the given string *s* contains a lowercase Latin letter β the type of the key that lies in room number (*i*<=+<=1)<=/<=2.
The even positions in the given string contain uppercase Latin letters β the types of doors between the rooms. Thus, each even position *i* of the given string *s* contains an uppercase letter β the type of the door that leads from room *i*<=/<=2 to room *i*<=/<=2<=+<=1.
|
Print the only integer β the minimum number of keys that Vitaly needs to buy to surely get from room one to room *n*.
|
[
"3\naAbB\n",
"4\naBaCaB\n",
"5\nxYyXzZaZ\n"
] |
[
"0\n",
"3\n",
"2\n"
] |
none
| 250
|
[
{
"input": "3\naAbB",
"output": "0"
},
{
"input": "4\naBaCaB",
"output": "3"
},
{
"input": "5\nxYyXzZaZ",
"output": "2"
},
{
"input": "26\naAbBcCdDeEfFgGhHiIjJkKlLmMnNoOpPqQrRsStTuUvVwWxXyY",
"output": "0"
},
{
"input": "26\nzAyBxCwDvEuFtGsHrIqJpKoLnMmNlOkPjQiRhSgTfUeVdWcXbY",
"output": "13"
},
{
"input": "5\naArRaRaR",
"output": "2"
},
{
"input": "2\ndA",
"output": "1"
},
{
"input": "2\ncB",
"output": "1"
},
{
"input": "10\nhNcMeXsSlHsUwYeMcA",
"output": "7"
},
{
"input": "100\nqDpInBmCrFwXpDbFgOzVvOcEmJrUcToAdEwEgTvBvBfWwRpGyEaXgDdRwVlQnYgWmWhMrHaIzPyXvGaFlRsVzHhZrOuVpXrKxFzAmWwPlFtNfPtJxVmLuHjKfYyArHrEnSwSzOvDpQhCgCqLlAcNpGhXrEeFuCmAqIkXyYtSsQwIxJzNiIuTgEbVuWrMwPrAlLyKaZ",
"output": "42"
},
{
"input": "2\ndD",
"output": "0"
},
{
"input": "2\ndE",
"output": "1"
},
{
"input": "3\ndDdD",
"output": "0"
},
{
"input": "3\ndEdD",
"output": "1"
},
{
"input": "3\ndEeD",
"output": "1"
},
{
"input": "3\ndEeF",
"output": "2"
}
] | 1,586,711,173
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 46
| 343
| 1,126,400
|
n = int(input())
s = input()
D = dict()
D_mins = dict()
for i in range(len(s)):
if i % 2 == 0:
if s[i] in D:
D[s[i]] += 1
else:
D[s[i]] = 1
if s[i] in D_mins:
D_mins[s[i]] = min(D_mins[s[i]], D[s[i]])
else:
D_mins[s[i]] = 1
else:
key = s[i].lower()
if key in D:
D[key] -= 1
else:
D[key] = -1
if key in D_mins:
D_mins[key] = min(D_mins[key], D[key])
else:
D_mins[key] = -1
key_to_buy = 0
for elem in D_mins.values():
if elem < 0:
key_to_buy += elem
print(-key_to_buy)
|
Title: Vitaliy and Pie
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
After a hard day Vitaly got very hungry and he wants to eat his favorite potato pie. But it's not that simple. Vitaly is in the first room of the house with *n* room located in a line and numbered starting from one from left to right. You can go from the first room to the second room, from the second room to the third room and so on β you can go from the (*n*<=-<=1)-th room to the *n*-th room. Thus, you can go to room *x* only from room *x*<=-<=1.
The potato pie is located in the *n*-th room and Vitaly needs to go there.
Each pair of consecutive rooms has a door between them. In order to go to room *x* from room *x*<=-<=1, you need to open the door between the rooms with the corresponding key.
In total the house has several types of doors (represented by uppercase Latin letters) and several types of keys (represented by lowercase Latin letters). The key of type *t* can open the door of type *T* if and only if *t* and *T* are the same letter, written in different cases. For example, key f can open door F.
Each of the first *n*<=-<=1 rooms contains exactly one key of some type that Vitaly can use to get to next rooms. Once the door is open with some key, Vitaly won't get the key from the keyhole but he will immediately run into the next room. In other words, each key can open no more than one door.
Vitaly realizes that he may end up in some room without the key that opens the door to the next room. Before the start his run for the potato pie Vitaly can buy any number of keys of any type that is guaranteed to get to room *n*.
Given the plan of the house, Vitaly wants to know what is the minimum number of keys he needs to buy to surely get to the room *n*, which has a delicious potato pie. Write a program that will help Vitaly find out this number.
Input Specification:
The first line of the input contains a positive integer *n* (2<=β€<=*n*<=β€<=105)Β βΒ the number of rooms in the house.
The second line of the input contains string *s* of length 2Β·*n*<=-<=2. Let's number the elements of the string from left to right, starting from one.
The odd positions in the given string *s* contain lowercase Latin lettersΒ βΒ the types of the keys that lie in the corresponding rooms. Thus, each odd position *i* of the given string *s* contains a lowercase Latin letter β the type of the key that lies in room number (*i*<=+<=1)<=/<=2.
The even positions in the given string contain uppercase Latin letters β the types of doors between the rooms. Thus, each even position *i* of the given string *s* contains an uppercase letter β the type of the door that leads from room *i*<=/<=2 to room *i*<=/<=2<=+<=1.
Output Specification:
Print the only integer β the minimum number of keys that Vitaly needs to buy to surely get from room one to room *n*.
Demo Input:
['3\naAbB\n', '4\naBaCaB\n', '5\nxYyXzZaZ\n']
Demo Output:
['0\n', '3\n', '2\n']
Note:
none
|
```python
n = int(input())
s = input()
D = dict()
D_mins = dict()
for i in range(len(s)):
if i % 2 == 0:
if s[i] in D:
D[s[i]] += 1
else:
D[s[i]] = 1
if s[i] in D_mins:
D_mins[s[i]] = min(D_mins[s[i]], D[s[i]])
else:
D_mins[s[i]] = 1
else:
key = s[i].lower()
if key in D:
D[key] -= 1
else:
D[key] = -1
if key in D_mins:
D_mins[key] = min(D_mins[key], D[key])
else:
D_mins[key] = -1
key_to_buy = 0
for elem in D_mins.values():
if elem < 0:
key_to_buy += elem
print(-key_to_buy)
```
| 3
|
|
701
|
A
|
Cards
|
PROGRAMMING
| 800
|
[
"greedy",
"implementation"
] | null | null |
There are *n* cards (*n* is even) in the deck. Each card has a positive integer written on it. *n*<=/<=2 people will play new card game. At the beginning of the game each player gets two cards, each card is given to exactly one player.
Find the way to distribute cards such that the sum of values written of the cards will be equal for each player. It is guaranteed that it is always possible.
|
The first line of the input contains integer *n* (2<=β€<=*n*<=β€<=100)Β β the number of cards in the deck. It is guaranteed that *n* is even.
The second line contains the sequence of *n* positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=β€<=*a**i*<=β€<=100), where *a**i* is equal to the number written on the *i*-th card.
|
Print *n*<=/<=2 pairs of integers, the *i*-th pair denote the cards that should be given to the *i*-th player. Each card should be given to exactly one player. Cards are numbered in the order they appear in the input.
It is guaranteed that solution exists. If there are several correct answers, you are allowed to print any of them.
|
[
"6\n1 5 7 4 4 3\n",
"4\n10 10 10 10\n"
] |
[
"1 3\n6 2\n4 5\n",
"1 2\n3 4\n"
] |
In the first sample, cards are distributed in such a way that each player has the sum of numbers written on his cards equal to 8.
In the second sample, all values *a*<sub class="lower-index">*i*</sub> are equal. Thus, any distribution is acceptable.
| 500
|
[
{
"input": "6\n1 5 7 4 4 3",
"output": "1 3\n6 2\n4 5"
},
{
"input": "4\n10 10 10 10",
"output": "1 4\n2 3"
},
{
"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": "1 100\n2 99\n3 98\n4 97\n5 96\n6 95\n7 94\n8 93\n9 92\n10 91\n11 90\n12 89\n13 88\n14 87\n15 86\n16 85\n17 84\n18 83\n19 82\n20 81\n21 80\n22 79\n23 78\n24 77\n25 76\n26 75\n27 74\n28 73\n29 72\n30 71\n31 70\n32 69\n33 68\n34 67\n35 66\n36 65\n37 64\n38 63\n39 62\n40 61\n41 60\n42 59\n43 58\n44 57\n45 56\n46 55\n47 54\n48 53\n49 52\n50 51"
},
{
"input": "4\n82 46 8 44",
"output": "3 1\n4 2"
},
{
"input": "2\n35 50",
"output": "1 2"
},
{
"input": "8\n24 39 49 38 44 64 44 50",
"output": "1 6\n4 8\n2 3\n5 7"
},
{
"input": "100\n23 44 35 88 10 78 8 84 46 19 69 36 81 60 46 12 53 22 83 73 6 18 80 14 54 39 74 42 34 20 91 70 32 11 80 53 70 21 24 12 87 68 35 39 8 84 81 70 8 54 73 2 60 71 4 33 65 48 69 58 55 57 78 61 45 50 55 72 86 37 5 11 12 81 32 19 22 11 22 82 23 56 61 84 47 59 31 38 31 90 57 1 24 38 68 27 80 9 37 14",
"output": "92 31\n52 90\n55 4\n71 41\n21 69\n7 84\n45 46\n49 8\n98 19\n5 80\n34 74\n72 47\n78 13\n16 97\n40 35\n73 23\n24 63\n100 6\n22 27\n10 51\n76 20\n30 68\n38 54\n18 48\n77 37\n79 32\n1 59\n81 11\n39 95\n93 42\n96 57\n87 83\n89 64\n33 53\n75 14\n56 86\n29 60\n3 91\n43 62\n12 82\n70 67\n99 61\n88 50\n94 25\n26 36\n44 17\n28 66\n2 58\n65 85\n9 15"
},
{
"input": "12\n22 83 2 67 55 12 40 93 83 73 12 28",
"output": "3 8\n6 9\n11 2\n1 10\n12 4\n7 5"
},
{
"input": "16\n10 33 36 32 48 25 31 27 45 13 37 26 22 21 15 43",
"output": "1 5\n10 9\n15 16\n14 11\n13 3\n6 2\n12 4\n8 7"
},
{
"input": "20\n18 13 71 60 28 10 20 65 65 12 13 14 64 68 6 50 72 7 66 58",
"output": "15 17\n18 3\n6 14\n10 19\n2 9\n11 8\n12 13\n1 4\n7 20\n5 16"
},
{
"input": "24\n59 39 25 22 46 21 24 70 60 11 46 42 44 37 13 37 41 58 72 23 25 61 58 62",
"output": "10 19\n15 8\n6 24\n4 22\n20 9\n7 1\n3 23\n21 18\n14 11\n16 5\n2 13\n17 12"
},
{
"input": "28\n22 1 51 31 83 35 3 64 59 10 61 25 19 53 55 80 78 8 82 22 67 4 27 64 33 6 85 76",
"output": "2 27\n7 5\n22 19\n26 16\n18 17\n10 28\n13 21\n1 24\n20 8\n12 11\n23 9\n4 15\n25 14\n6 3"
},
{
"input": "32\n41 42 22 68 40 52 66 16 73 25 41 21 36 60 46 30 24 55 35 10 54 52 70 24 20 56 3 34 35 6 51 8",
"output": "27 9\n30 23\n32 4\n20 7\n8 14\n25 26\n12 18\n3 21\n17 22\n24 6\n10 31\n16 15\n28 2\n19 11\n29 1\n13 5"
},
{
"input": "36\n1 10 61 43 27 49 55 33 7 30 45 78 69 34 38 19 36 49 55 11 30 63 46 24 16 68 71 18 11 52 72 24 60 68 8 41",
"output": "1 12\n9 31\n35 27\n2 13\n20 34\n29 26\n25 22\n28 3\n16 33\n24 19\n32 7\n5 30\n10 18\n21 6\n8 23\n14 11\n17 4\n15 36"
},
{
"input": "40\n7 30 13 37 37 56 45 28 61 28 23 33 44 63 58 52 21 2 42 19 10 32 9 7 61 15 58 20 45 4 46 24 35 17 50 4 20 48 41 55",
"output": "18 14\n30 25\n36 9\n1 27\n24 15\n23 6\n21 40\n3 16\n26 35\n34 38\n20 31\n28 29\n37 7\n17 13\n11 19\n32 39\n8 5\n10 4\n2 33\n22 12"
},
{
"input": "44\n7 12 46 78 24 68 86 22 71 79 85 14 58 72 26 46 54 39 35 13 31 45 81 21 15 8 47 64 69 87 57 6 18 80 47 29 36 62 34 67 59 48 75 25",
"output": "32 30\n1 7\n26 11\n2 23\n20 34\n12 10\n25 4\n33 43\n24 14\n8 9\n5 29\n44 6\n15 40\n36 28\n21 38\n39 41\n19 13\n37 31\n18 17\n22 42\n3 35\n16 27"
},
{
"input": "48\n57 38 16 25 34 57 29 38 60 51 72 78 22 39 10 33 20 16 12 3 51 74 9 88 4 70 56 65 86 18 33 12 77 78 52 87 68 85 81 5 61 2 52 39 80 13 74 30",
"output": "42 24\n20 36\n25 29\n40 38\n23 39\n15 45\n19 34\n32 12\n46 33\n3 47\n18 22\n30 11\n17 26\n13 37\n4 28\n7 41\n48 9\n16 6\n31 1\n5 27\n2 43\n8 35\n14 21\n44 10"
},
{
"input": "52\n57 12 13 40 68 31 18 4 31 18 65 3 62 32 6 3 49 48 51 33 53 40 9 32 47 53 58 19 14 23 32 38 39 69 19 20 62 52 68 17 39 22 54 59 3 2 52 9 67 68 24 39",
"output": "46 34\n12 50\n16 39\n45 5\n8 49\n15 11\n23 37\n48 13\n2 44\n3 27\n29 1\n40 43\n7 26\n10 21\n28 47\n35 38\n36 19\n42 17\n30 18\n51 25\n6 22\n9 4\n14 52\n24 41\n31 33\n20 32"
},
{
"input": "56\n53 59 66 68 71 25 48 32 12 61 72 69 30 6 56 55 25 49 60 47 46 46 66 19 31 9 23 15 10 12 71 53 51 32 39 31 66 66 17 52 12 7 7 22 49 12 71 29 63 7 47 29 18 39 27 26",
"output": "14 11\n42 47\n43 31\n50 5\n26 12\n29 4\n9 38\n30 37\n41 23\n46 3\n28 49\n39 10\n53 19\n24 2\n44 15\n27 16\n6 32\n17 1\n56 40\n55 33\n48 45\n52 18\n13 7\n25 51\n36 20\n8 22\n34 21\n35 54"
},
{
"input": "60\n47 63 20 68 46 12 45 44 14 38 28 73 60 5 20 18 70 64 37 47 26 47 37 61 29 61 23 28 30 68 55 22 25 60 38 7 63 12 38 15 14 30 11 5 70 15 53 52 7 57 49 45 55 37 45 28 50 2 31 30",
"output": "58 12\n14 45\n44 17\n36 30\n49 4\n43 18\n6 37\n38 2\n9 26\n41 24\n40 34\n46 13\n16 50\n3 53\n15 31\n32 47\n27 48\n33 57\n21 51\n11 22\n28 20\n56 1\n25 5\n29 55\n42 52\n60 7\n59 8\n19 39\n23 35\n54 10"
},
{
"input": "64\n63 39 19 5 48 56 49 45 29 68 25 59 37 69 62 26 60 44 60 6 67 68 2 40 56 6 19 12 17 70 23 11 59 37 41 55 30 68 72 14 38 34 3 71 2 4 55 15 31 66 15 51 36 72 18 7 6 14 43 33 8 35 57 18",
"output": "23 54\n45 39\n43 44\n46 30\n4 14\n20 38\n26 22\n57 10\n56 21\n61 50\n32 1\n28 15\n40 19\n58 17\n48 33\n51 12\n29 63\n55 25\n64 6\n3 47\n27 36\n31 52\n11 7\n16 5\n9 8\n37 18\n49 59\n60 35\n42 24\n62 2\n53 41\n13 34"
},
{
"input": "68\n58 68 40 55 62 15 10 54 19 18 69 27 15 53 8 18 8 33 15 49 20 9 70 8 18 64 14 59 9 64 3 35 46 11 5 65 58 55 28 58 4 55 64 5 68 24 4 58 23 45 58 50 38 68 5 15 20 9 5 53 20 63 69 68 15 53 65 65",
"output": "31 23\n41 63\n47 11\n35 64\n44 54\n55 45\n59 2\n15 68\n17 67\n24 36\n22 43\n29 30\n58 26\n7 62\n34 5\n27 28\n6 51\n13 48\n19 40\n56 37\n65 1\n10 42\n16 38\n25 4\n9 8\n21 66\n57 60\n61 14\n49 52\n46 20\n12 33\n39 50\n18 3\n32 53"
},
{
"input": "72\n61 13 55 23 24 55 44 33 59 19 14 17 66 40 27 33 29 37 28 74 50 56 59 65 64 17 42 56 73 51 64 23 22 26 38 22 36 47 60 14 52 28 14 12 6 41 73 5 64 67 61 74 54 34 45 34 44 4 34 49 18 72 44 47 31 19 11 31 5 4 45 50",
"output": "58 52\n70 20\n48 47\n69 29\n45 62\n67 50\n44 13\n2 24\n11 49\n40 31\n43 25\n12 51\n26 1\n61 39\n10 23\n66 9\n33 28\n36 22\n4 6\n32 3\n5 53\n34 41\n15 30\n19 72\n42 21\n17 60\n65 64\n68 38\n8 71\n16 55\n54 63\n56 57\n59 7\n37 27\n18 46\n35 14"
},
{
"input": "76\n73 37 73 67 26 45 43 74 47 31 43 81 4 3 39 79 48 81 67 39 67 66 43 67 80 51 34 79 5 58 45 10 39 50 9 78 6 18 75 17 45 17 51 71 34 53 33 11 17 15 11 69 50 41 13 74 10 33 77 41 11 64 36 74 17 32 3 10 27 20 5 73 52 41 7 57",
"output": "14 18\n67 12\n13 25\n29 28\n71 16\n37 36\n75 59\n35 39\n32 64\n57 56\n68 8\n48 72\n51 3\n61 1\n55 44\n50 52\n40 24\n42 21\n49 19\n65 4\n38 22\n70 62\n5 30\n69 76\n10 46\n66 73\n47 43\n58 26\n27 53\n45 34\n63 17\n2 9\n15 41\n20 31\n33 6\n54 23\n60 11\n74 7"
},
{
"input": "80\n18 38 65 1 20 9 57 2 36 26 15 17 33 61 65 27 10 35 49 42 40 32 19 33 12 36 56 31 10 41 8 54 56 60 5 47 61 43 23 19 20 30 7 6 38 60 29 58 35 64 30 51 6 17 30 24 47 1 37 47 34 36 48 28 5 25 47 19 30 39 36 23 31 28 46 46 59 43 19 49",
"output": "4 15\n58 3\n8 50\n35 37\n65 14\n44 46\n53 34\n43 77\n31 48\n6 7\n17 33\n29 27\n25 32\n11 52\n12 80\n54 19\n1 63\n23 67\n40 60\n68 57\n79 36\n5 76\n41 75\n39 78\n72 38\n56 20\n66 30\n10 21\n16 70\n64 45\n74 2\n47 59\n42 71\n51 62\n55 26\n69 9\n28 49\n73 18\n22 61\n13 24"
},
{
"input": "84\n59 41 54 14 42 55 29 28 41 73 40 15 1 1 66 49 76 59 68 60 42 81 19 23 33 12 80 81 42 22 54 54 2 22 22 28 27 60 36 57 17 76 38 20 40 65 23 9 81 50 25 13 46 36 59 53 6 35 47 40 59 19 67 46 63 49 12 33 23 49 33 23 32 62 60 70 44 1 6 63 28 16 70 69",
"output": "13 49\n14 28\n78 22\n33 27\n57 42\n79 17\n48 10\n26 83\n67 76\n52 84\n4 19\n12 63\n82 15\n41 46\n23 80\n62 65\n44 74\n30 75\n34 38\n35 20\n24 61\n47 55\n69 18\n72 1\n51 40\n37 6\n8 32\n36 31\n81 3\n7 56\n73 50\n25 70\n68 66\n71 16\n58 59\n39 64\n54 53\n43 77\n11 29\n45 21\n60 5\n2 9"
},
{
"input": "88\n10 28 71 6 58 66 45 52 13 71 39 1 10 29 30 70 14 17 15 38 4 60 5 46 66 41 40 58 2 57 32 44 21 26 13 40 64 63 56 33 46 8 30 43 67 55 44 28 32 62 14 58 42 67 45 59 32 68 10 31 51 6 42 34 9 12 51 27 20 14 62 42 16 5 1 14 30 62 40 59 58 26 25 15 27 47 21 57",
"output": "12 10\n75 3\n29 16\n21 58\n23 54\n74 45\n4 25\n62 6\n42 37\n65 38\n1 78\n13 71\n59 50\n66 22\n9 80\n35 56\n17 81\n51 52\n70 28\n76 5\n19 88\n84 30\n73 39\n18 46\n69 8\n33 67\n87 61\n83 86\n34 41\n82 24\n68 55\n85 7\n2 47\n48 32\n14 44\n15 72\n43 63\n77 53\n60 26\n31 79\n49 36\n57 27\n40 11\n64 20"
},
{
"input": "92\n17 37 81 15 29 70 73 42 49 23 44 77 27 44 74 11 43 66 15 41 60 36 33 11 2 76 16 51 45 21 46 16 85 29 76 79 16 6 60 13 25 44 62 28 43 35 63 24 76 71 62 15 57 72 45 10 71 59 74 14 53 13 58 72 14 72 73 11 25 1 57 42 86 63 50 30 64 38 10 77 75 24 58 8 54 12 43 30 27 71 52 34",
"output": "70 73\n25 33\n38 3\n84 36\n56 80\n79 12\n16 49\n24 35\n68 26\n86 81\n40 59\n62 15\n60 67\n65 7\n4 66\n19 64\n52 54\n27 90\n32 57\n37 50\n1 6\n30 18\n10 77\n48 74\n82 47\n41 51\n69 43\n13 39\n89 21\n44 58\n5 83\n34 63\n76 71\n88 53\n23 85\n92 61\n46 91\n22 28\n2 75\n78 9\n20 31\n8 55\n72 29\n17 42\n45 14\n87 11"
},
{
"input": "96\n77 7 47 19 73 31 46 13 89 69 52 9 26 77 6 87 55 45 71 2 79 1 80 20 4 82 64 20 75 86 84 24 77 56 16 54 53 35 74 73 40 29 63 20 83 39 58 16 31 41 40 16 11 90 30 48 62 39 55 8 50 3 77 73 75 66 14 90 18 54 38 10 53 22 67 38 27 91 62 37 85 13 92 7 18 83 10 3 86 54 80 59 34 16 39 43",
"output": "22 83\n20 78\n62 68\n88 54\n25 9\n15 16\n2 89\n84 30\n60 81\n12 31\n72 86\n87 45\n53 26\n8 91\n82 23\n67 21\n35 63\n48 33\n52 14\n94 1\n69 65\n85 29\n4 39\n24 64\n28 40\n44 5\n74 19\n32 10\n13 75\n77 66\n42 27\n55 43\n6 79\n49 57\n93 92\n38 47\n80 34\n71 59\n76 17\n46 90\n58 70\n95 36\n41 73\n51 37\n50 11\n96 61\n18 56\n7 3"
},
{
"input": "4\n100 100 1 1",
"output": "3 2\n4 1"
},
{
"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": "1 100\n2 99\n3 98\n4 97\n5 96\n6 95\n7 94\n8 93\n9 92\n10 91\n11 90\n12 89\n13 88\n14 87\n15 86\n16 85\n17 84\n18 83\n19 82\n20 81\n21 80\n22 79\n23 78\n24 77\n25 76\n26 75\n27 74\n28 73\n29 72\n30 71\n31 70\n32 69\n33 68\n34 67\n35 66\n36 65\n37 64\n38 63\n39 62\n40 61\n41 60\n42 59\n43 58\n44 57\n45 56\n46 55\n47 54\n48 53\n49 52\n50 51"
},
{
"input": "100\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100",
"output": "1 100\n2 99\n3 98\n4 97\n5 96\n6 95\n7 94\n8 93\n9 92\n10 91\n11 90\n12 89\n13 88\n14 87\n15 86\n16 85\n17 84\n18 83\n19 82\n20 81\n21 80\n22 79\n23 78\n24 77\n25 76\n26 75\n27 74\n28 73\n29 72\n30 71\n31 70\n32 69\n33 68\n34 67\n35 66\n36 65\n37 64\n38 63\n39 62\n40 61\n41 60\n42 59\n43 58\n44 57\n45 56\n46 55\n47 54\n48 53\n49 52\n50 51"
},
{
"input": "4\n3 4 4 5",
"output": "1 4\n2 3"
},
{
"input": "4\n1 1 2 2",
"output": "1 4\n2 3"
},
{
"input": "4\n1 2 3 4",
"output": "1 4\n2 3"
}
] | 1,544,757,035
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 35
| 124
| 0
|
n = int(input())
string = input().split(' ')
for i in range(n):
string[i] = int(string[i])
target = int(sum(string)/(n/2))
result = []
for i in range(n):
if not i in result:
result.append(i)
remaining = target - string[i]
for j in range(n):
if string[j] == remaining:
if not j in result:
result.append(j)
break
for i in range(n//2):
stringToPrint = str(result[2*i]+1) + " "
stringToPrint += str(result[2*i + 1]+1)
print(stringToPrint)
|
Title: Cards
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* cards (*n* is even) in the deck. Each card has a positive integer written on it. *n*<=/<=2 people will play new card game. At the beginning of the game each player gets two cards, each card is given to exactly one player.
Find the way to distribute cards such that the sum of values written of the cards will be equal for each player. It is guaranteed that it is always possible.
Input Specification:
The first line of the input contains integer *n* (2<=β€<=*n*<=β€<=100)Β β the number of cards in the deck. It is guaranteed that *n* is even.
The second line contains the sequence of *n* positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=β€<=*a**i*<=β€<=100), where *a**i* is equal to the number written on the *i*-th card.
Output Specification:
Print *n*<=/<=2 pairs of integers, the *i*-th pair denote the cards that should be given to the *i*-th player. Each card should be given to exactly one player. Cards are numbered in the order they appear in the input.
It is guaranteed that solution exists. If there are several correct answers, you are allowed to print any of them.
Demo Input:
['6\n1 5 7 4 4 3\n', '4\n10 10 10 10\n']
Demo Output:
['1 3\n6 2\n4 5\n', '1 2\n3 4\n']
Note:
In the first sample, cards are distributed in such a way that each player has the sum of numbers written on his cards equal to 8.
In the second sample, all values *a*<sub class="lower-index">*i*</sub> are equal. Thus, any distribution is acceptable.
|
```python
n = int(input())
string = input().split(' ')
for i in range(n):
string[i] = int(string[i])
target = int(sum(string)/(n/2))
result = []
for i in range(n):
if not i in result:
result.append(i)
remaining = target - string[i]
for j in range(n):
if string[j] == remaining:
if not j in result:
result.append(j)
break
for i in range(n//2):
stringToPrint = str(result[2*i]+1) + " "
stringToPrint += str(result[2*i + 1]+1)
print(stringToPrint)
```
| 3
|
|
266
|
B
|
Queue at the School
|
PROGRAMMING
| 800
|
[
"constructive algorithms",
"graph matchings",
"implementation",
"shortest paths"
] | null | null |
During the break the schoolchildren, boys and girls, formed a queue of *n* people in the canteen. Initially the children stood in the order they entered the canteen. However, after a while the boys started feeling awkward for standing in front of the girls in the queue and they started letting the girls move forward each second.
Let's describe the process more precisely. Let's say that the positions in the queue are sequentially numbered by integers from 1 to *n*, at that the person in the position number 1 is served first. Then, if at time *x* a boy stands on the *i*-th position and a girl stands on the (*i*<=+<=1)-th position, then at time *x*<=+<=1 the *i*-th position will have a girl and the (*i*<=+<=1)-th position will have a boy. The time is given in seconds.
You've got the initial position of the children, at the initial moment of time. Determine the way the queue is going to look after *t* seconds.
|
The first line contains two integers *n* and *t* (1<=β€<=*n*,<=*t*<=β€<=50), which represent the number of children in the queue and the time after which the queue will transform into the arrangement you need to find.
The next line contains string *s*, which represents the schoolchildren's initial arrangement. If the *i*-th position in the queue contains a boy, then the *i*-th character of string *s* equals "B", otherwise the *i*-th character equals "G".
|
Print string *a*, which describes the arrangement after *t* seconds. If the *i*-th position has a boy after the needed time, then the *i*-th character *a* must equal "B", otherwise it must equal "G".
|
[
"5 1\nBGGBG\n",
"5 2\nBGGBG\n",
"4 1\nGGGB\n"
] |
[
"GBGGB\n",
"GGBGB\n",
"GGGB\n"
] |
none
| 500
|
[
{
"input": "5 1\nBGGBG",
"output": "GBGGB"
},
{
"input": "5 2\nBGGBG",
"output": "GGBGB"
},
{
"input": "4 1\nGGGB",
"output": "GGGB"
},
{
"input": "2 1\nBB",
"output": "BB"
},
{
"input": "2 1\nBG",
"output": "GB"
},
{
"input": "6 2\nBBGBBG",
"output": "GBBGBB"
},
{
"input": "8 3\nBBGBGBGB",
"output": "GGBGBBBB"
},
{
"input": "10 3\nBBGBBBBBBG",
"output": "GBBBBBGBBB"
},
{
"input": "22 7\nGBGGBGGGGGBBBGGBGBGBBB",
"output": "GGGGGGGGBGGBGGBBBBBBBB"
},
{
"input": "50 4\nGBBGBBBGGGGGBBGGBBBBGGGBBBGBBBGGBGGBGBBBGGBGGBGGBG",
"output": "GGBGBGBGBGBGGGBBGBGBGBGBBBGBGBGBGBGBGBGBGBGBGGBGBB"
},
{
"input": "50 8\nGGGGBGGBGGGBGBBBGGGGGGGGBBGBGBGBBGGBGGBGGGGGGGGBBG",
"output": "GGGGGGGGGGGGBGGBGBGBGBGBGGGGGGBGBGBGBGBGBGGBGGBGBB"
},
{
"input": "50 30\nBGGGGGGBGGBGBGGGGBGBBGBBBGGBBBGBGBGGGGGBGBBGBGBGGG",
"output": "GGGGGGGGGGGGGGGGGGGGGGGGGGGGGGBBBBBBBBBBBBBBBBBBBB"
},
{
"input": "20 20\nBBGGBGGGGBBBGBBGGGBB",
"output": "GGGGGGGGGGBBBBBBBBBB"
},
{
"input": "27 6\nGBGBGBGGGGGGBGGBGGBBGBBBGBB",
"output": "GGGGGGGBGBGBGGGGGBGBBBBBBBB"
},
{
"input": "46 11\nBGGGGGBGBGGBGGGBBGBBGBBGGBBGBBGBGGGGGGGBGBGBGB",
"output": "GGGGGGGGGGGBGGGGGBBGBGBGBGBGBGBGBGBGBGBGBBBBBB"
},
{
"input": "50 6\nBGGBBBBGGBBBBBBGGBGBGBBBBGBBBBBBGBBBBBBBBBBBBBBBBB",
"output": "GGGGBBBBBGBGBGBGBBBGBBBBBBGBBBBBBBBBBBBBBBBBBBBBBB"
},
{
"input": "50 10\nBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB",
"output": "BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB"
},
{
"input": "50 8\nGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG",
"output": "GGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG"
},
{
"input": "50 10\nBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBGB",
"output": "BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBGBBBBBBBBBBB"
},
{
"input": "50 13\nGGGBGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG",
"output": "GGGGGGGGGGGGGGGGBGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG"
},
{
"input": "1 1\nB",
"output": "B"
},
{
"input": "1 1\nG",
"output": "G"
},
{
"input": "1 50\nB",
"output": "B"
},
{
"input": "1 50\nG",
"output": "G"
},
{
"input": "50 50\nBBBBBBBBGGBBBBBBGBBBBBBBBBBBGBBBBBBBBBBBBBBGBBBBBB",
"output": "GGGGGBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB"
},
{
"input": "50 50\nGGBBGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGBBGGGGGGBG",
"output": "GGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGBBBBB"
},
{
"input": "6 3\nGGBBBG",
"output": "GGGBBB"
},
{
"input": "26 3\nGBBGBBBBBGGGBGBGGGBGBGGBBG",
"output": "GGBBBBGBGBGBGGGBGBGGGBGBBB"
},
{
"input": "46 3\nGGBBGGGGBBGBGBBBBBGGGBGGGBBGGGBBBGGBGGBBBGBGBB",
"output": "GGGGBGBGGGBBBBBGBGBGBGGGBGGBGBGBGBGBGBGBGBBBBB"
},
{
"input": "44 8\nBGBBBBBBBBBGGBBGBGBGGBBBBBGBBGBBBBBBBBBGBBGB",
"output": "GBBGBGBGBGBGBGBBBBGBBGBBBBBBBBBGBBGBBBBBBBBB"
},
{
"input": "20 20\nBBGGBGGGGBBBGBBGGGBB",
"output": "GGGGGGGGGGBBBBBBBBBB"
},
{
"input": "30 25\nBGGBBGBGGBGBGBBGBGGGGBGBGGBBBB",
"output": "GGGGGGGGGGGGGGGBBBBBBBBBBBBBBB"
},
{
"input": "17 42\nBBGBGBGGGGGGBBGGG",
"output": "GGGGGGGGGGGBBBBBB"
},
{
"input": "30 50\nBGGBBGGGGGGGGBBGGGBBGGBBBGBBGG",
"output": "GGGGGGGGGGGGGGGGGGBBBBBBBBBBBB"
},
{
"input": "31 33\nBBGGBBGGBGBBBBGGBGBBBGBGGBBGBBB",
"output": "GGGGGGGGGGGGBBBBBBBBBBBBBBBBBBB"
},
{
"input": "2 2\nBG",
"output": "GB"
},
{
"input": "8 8\nGGGBGGBB",
"output": "GGGGGBBB"
},
{
"input": "34 34\nBGGBBBBGGBBGBGGGGGGBBGGGGGBGGBGGGB",
"output": "GGGGGGGGGGGGGGGGGGGGGBBBBBBBBBBBBB"
},
{
"input": "34 20\nBBBBGBGGGBGGGBGBGGBBBBGGBBGGGBGBBG",
"output": "GGGGGGGGGGGGGGGGGBBBBBBBBBBBBBBBBB"
},
{
"input": "50 50\nBBGBBBBBBBBBBBBGBBBGBBBBBBBGBBBBBBGBBBGBBGBBGBBBGG",
"output": "GGGGGGGGGGBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB"
},
{
"input": "10 10\nGGGGGGGGGG",
"output": "GGGGGGGGGG"
},
{
"input": "10 10\nBBBBBBBBBB",
"output": "BBBBBBBBBB"
},
{
"input": "10 10\nBGBGBGBGBG",
"output": "GGGGGBBBBB"
},
{
"input": "1 1\nB",
"output": "B"
}
] | 1,698,325,863
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 44
| 154
| 1,536,000
|
n, t = map(int, input().split())
q = list(input())
for _ in range(t):
i = 0
while i < len(q) - 1:
if q[i] == 'B' and q[i + 1] == 'G':
q[i], q[i + 1] = 'G', 'B'
i += 2
else:
i += 1
for _ in q:
print(_, end = '')
|
Title: Queue at the School
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
During the break the schoolchildren, boys and girls, formed a queue of *n* people in the canteen. Initially the children stood in the order they entered the canteen. However, after a while the boys started feeling awkward for standing in front of the girls in the queue and they started letting the girls move forward each second.
Let's describe the process more precisely. Let's say that the positions in the queue are sequentially numbered by integers from 1 to *n*, at that the person in the position number 1 is served first. Then, if at time *x* a boy stands on the *i*-th position and a girl stands on the (*i*<=+<=1)-th position, then at time *x*<=+<=1 the *i*-th position will have a girl and the (*i*<=+<=1)-th position will have a boy. The time is given in seconds.
You've got the initial position of the children, at the initial moment of time. Determine the way the queue is going to look after *t* seconds.
Input Specification:
The first line contains two integers *n* and *t* (1<=β€<=*n*,<=*t*<=β€<=50), which represent the number of children in the queue and the time after which the queue will transform into the arrangement you need to find.
The next line contains string *s*, which represents the schoolchildren's initial arrangement. If the *i*-th position in the queue contains a boy, then the *i*-th character of string *s* equals "B", otherwise the *i*-th character equals "G".
Output Specification:
Print string *a*, which describes the arrangement after *t* seconds. If the *i*-th position has a boy after the needed time, then the *i*-th character *a* must equal "B", otherwise it must equal "G".
Demo Input:
['5 1\nBGGBG\n', '5 2\nBGGBG\n', '4 1\nGGGB\n']
Demo Output:
['GBGGB\n', 'GGBGB\n', 'GGGB\n']
Note:
none
|
```python
n, t = map(int, input().split())
q = list(input())
for _ in range(t):
i = 0
while i < len(q) - 1:
if q[i] == 'B' and q[i + 1] == 'G':
q[i], q[i + 1] = 'G', 'B'
i += 2
else:
i += 1
for _ in q:
print(_, end = '')
```
| 3
|
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