contestId
int64 0
1.01k
| index
stringclasses 57
values | name
stringlengths 2
58
| type
stringclasses 2
values | rating
int64 0
3.5k
| tags
listlengths 0
11
| title
stringclasses 522
values | time-limit
stringclasses 8
values | memory-limit
stringclasses 8
values | problem-description
stringlengths 0
7.15k
| input-specification
stringlengths 0
2.05k
| output-specification
stringlengths 0
1.5k
| demo-input
listlengths 0
7
| demo-output
listlengths 0
7
| note
stringlengths 0
5.24k
| points
float64 0
425k
| test_cases
listlengths 0
402
| creationTimeSeconds
int64 1.37B
1.7B
| relativeTimeSeconds
int64 8
2.15B
| programmingLanguage
stringclasses 3
values | verdict
stringclasses 14
values | testset
stringclasses 12
values | passedTestCount
int64 0
1k
| timeConsumedMillis
int64 0
15k
| memoryConsumedBytes
int64 0
805M
| code
stringlengths 3
65.5k
| prompt
stringlengths 262
8.2k
| response
stringlengths 17
65.5k
| score
float64 -1
3.99
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
552
|
E
|
Vanya and Brackets
|
PROGRAMMING
| 2,100
|
[
"brute force",
"dp",
"expression parsing",
"greedy",
"implementation",
"strings"
] | null | null |
Vanya is doing his maths homework. He has an expression of form , where *x*1,<=*x*2,<=...,<=*x**n* are digits from 1 to 9, and sign represents either a plus '+' or the multiplication sign '*'. Vanya needs to add one pair of brackets in this expression so that to maximize the value of the resulting expression.
|
The first line contains expression *s* (1<=≤<=|*s*|<=≤<=5001, |*s*| is odd), its odd positions only contain digits from 1 to 9, and even positions only contain signs <=+<= and <=*<=.
The number of signs <=*<= doesn't exceed 15.
|
In the first line print the maximum possible value of an expression.
|
[
"3+5*7+8*4\n",
"2+3*5\n",
"3*4*5\n"
] |
[
"303\n",
"25\n",
"60\n"
] |
Note to the first sample test. 3 + 5 * (7 + 8) * 4 = 303.
Note to the second sample test. (2 + 3) * 5 = 25.
Note to the third sample test. (3 * 4) * 5 = 60 (also many other variants are valid, for instance, (3) * 4 * 5 = 60).
| 2,500
|
[
{
"input": "3+5*7+8*4",
"output": "303"
},
{
"input": "2+3*5",
"output": "25"
},
{
"input": "3*4*5",
"output": "60"
},
{
"input": "5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5",
"output": "152587890625"
},
{
"input": "2*2+2*2",
"output": "16"
},
{
"input": "1+1+1+1+1+1+1",
"output": "7"
},
{
"input": "1+5*6+7*8",
"output": "521"
},
{
"input": "9*8+7*6+5*4+3*2+1",
"output": "1987"
},
{
"input": "3*3*9+4+6+8*4+5+1*4*6",
"output": "12312"
},
{
"input": "4*9+4+5+8*4+6+9+8+2+5+2+5*7+6+8",
"output": "2450"
},
{
"input": "9+9+9*9*9*9+9+9",
"output": "19701"
},
{
"input": "9+9+9+9+9*9*9*9",
"output": "32805"
},
{
"input": "1*1*1*1*1*1*1*1+1*1*1*1*1*1*1*1",
"output": "2"
},
{
"input": "4+2*7+8+9*6+6*9+8+7*2+4",
"output": "1380"
},
{
"input": "5",
"output": "5"
},
{
"input": "4+6*7+4",
"output": "74"
},
{
"input": "2+7+3+5+4+2+3+9+9+6+9+2+3+6+5*3+4+5+6+5+8",
"output": "253"
},
{
"input": "3+2+2+3+7+1+9+1+6+8+3+2+2+6+7+2+8+8+1+4+9",
"output": "94"
},
{
"input": "3+9+3+1+6+4+7+9+5+8+2+6+1+4+4+5+1+7+5+4+6+4+3+1*9+7+7+4+5+2+3+2+6+5+5+8+7+8+2+3*3+8+3+4+9+8*5+9+2+8+2+8+6+6+9+6+4+2+5+3+1+2+6+6+2+4+6+4+2+7+2+7+6+9+9+3+6+7+8+3+3+2+3+7+9+7+8+5+5+5*1+3+2+5+8+5*6+5+4*6+2+5+5+4+9+8+3+5+1+3+1+6+2+2+1+3+2+3+3+3+2+8+3+2+8+8+5+2+6+6+3+1+1+5+5*1+5+7+5+8+4+1*7+5+9+5+8+1*8+5+9+3+1+8+6+7+8+3+5+1+5+6+9*9+6+1+9+8+9+1+5+9+9+6+3+8+8+6*9*3+9+7+7+4+3+8+8+6+7+1+8+6+3+1+7+7+1+1+3+9+8+5+5+6+8+2+4+1+5+7+2+3+7+1+5+1+6+1+7+3*5+5+9+2+1+3+9+4+8+6+5+5+2+3+7+9+5+6+8+3*3+2+4+4+6+3+2+4+1+4+8",
"output": "162353"
},
{
"input": "1*5*1+8*2*6*5*3*9+3+8+2+9*5+7+2+9+5+1*3+2*2*3*4*2*3",
"output": "19699205"
},
{
"input": "4+4+6+2+5+9+9+5+5+9+4+1*5+3+6+9+6+2+4+3+2+8+9*6+5+4+3+8+7+3+2*3+1+6+8+3+8+1+8+2+1+1+1+6+9+6+4+6+7+8+3+1+5+4+8+8+6+5+8+7+7+1+7+6+3+3+9+6+3+5+4+4+1+4+1+8+6+2+9+8+7+2+3+1+4+3+9+9+2*1+3+8+2+4+1+8+9+3*7+3+7+5+3+7+5+5+3+2+9+8+4+7+5+3+7+7+3+8+9+4+9+6*6+3+8+8*7+7+9+1+3+5+1+1+1+9+8+2+1+1+5+5+5+1+6+7+3+6+1+4+1+7+1+7+1+1+9+9*4+1+3+9+3+5+5+5+5+2+9+6+7+3+5+9+3+5+3+9+3+9+9+2+7+2+1*4+6*2+5+7+6+1+1+2+8+9+5+8+3+9+9+1+1+4+9+7+5+8*9+5+2+6+5+6*2+4+2+5+2+3+9+6+9+5+5+5*6+8+2+3+1+2+8+3+1+6+5+9+7+4+2+8+9+1+5+8+5+3+2+7+1",
"output": "82140"
},
{
"input": "6*9+9*5*5+1*2*9*9*1+4*8+8+9+5+6*5*6+4+2+2+1+5*5*7*8",
"output": "11294919"
},
{
"input": "5+3+5+9+3+9+1+3+1*7+7+1+9+3+7+7+6+6+3+7+4+3+6+4+5+1+2*3+6*5+5+6+2+8+3+3+9+9+1+1+2+8+4+8+9+3*7+3+2*8+9+8+1*9+9+7+4+8+6+7+3+5+6+4+4+9+2+2+8+6+7+1+5+4+4+6+6+6+9+8+7+2+3+5+4+6+1+8+8+9+1+9+6+3+8+5*7+3+1+6+7+9+1+6+2+2+8+8+9+3+7+7+2+5+8+6+7+9+7+2+4+9+8+3+7+4+5+7+6+5*6+4+6+4+6+2+2*6+2+5+5+1+8+7+7+6+6+8+2+8+8+6+7+1+1+1+2+5+1+1+8+9+9+6+5+8+7+5+8+4+8+8+1+4+6+7+3+2*1+1+3+5+3+3+3+9+8+7*2+4+7+5+8+3+3+9+3+7+2+1+1+7+6+2+5+5+2+1+8+8+2+9+9+2+4+6+6+4+8+9+3+7+1+3*9+8+7+4+9+4+6+2+9+8+8+5+8+8+2+5+6+6+4+7+9+4+7+2+3+1+7",
"output": "58437"
},
{
"input": "2+7+8*8*7+1+3+6*5*3*7*3*2+8+5*1+5*5+9*6+6*5+1*3+8+5",
"output": "1473847"
},
{
"input": "1+2+4+8+6+5+3+8+2+9+9+5+8+7+7+7+6+1+7+2+8+3+2+5+1+6+1+3+8+2+5+4+3+5+7+8+5+7+7+3+8+1+7+1+1+1+5+9+5+9+1+6+7+6+8+9+2+7+9+2+9+9+7+3+2+8+4+4+5+9+6+2+6+8+1+3+5+3+9+4+7+4+3+9+8+2+6+3+5+1*3+1+6+8+5+3+9+2+9+9+3+4+8*6+3+9+7+1+1+4+6+4+5*6*1+1*9+6+5+4+3+7+3+8+6+2+3+7+4+1+5+8+6+1+6+9+1+2+7+2+2+1+7+9+4+3+1+4+3+3+1+1+2+1+8+9+8+6+9+9+6+3+7*1+1+3+7+9+3+6+5+2*9+8+1+9+8+7+5+3+6+9+3+5+3+5+5+7+5+2*9+9+2+4+2+3+7+1+7+1+3+8+6+4+5+9+3*2+8+6+8+2*6+8+1+4+2+7+7+6+8+3+2+5+8+1+8+5+6+1+6+4+6+8+6+6+4+3+5+2+1+5+9+9+4+4*9+7+8+4+4",
"output": "178016"
},
{
"input": "8+3*6*9*6+5*1*8*2+1+9+2+1*3*2+9+5+4+3+1+3*9*6*8+4+1",
"output": "9027949"
},
{
"input": "1*1*1*1*1*1*1*1*1*1*1*1",
"output": "1"
},
{
"input": "5+5*5+5*5+5*5+5",
"output": "885"
},
{
"input": "8+7+3+6+3*8+8+9+8+4+2",
"output": "247"
},
{
"input": "7+8*4+9+5+3+2+3+3+2+9",
"output": "327"
},
{
"input": "1+1+7+1+7+7*7+5+3*9+3",
"output": "965"
},
{
"input": "9+6+9+7+8*2*9+8+6+7+5",
"output": "728"
},
{
"input": "8+8*3*8+1+9*4+9+2+8+4",
"output": "1759"
},
{
"input": "3+5+5+2+2+9*7+7+7*2*2",
"output": "773"
},
{
"input": "6+8+5+9*2+7*9*3+2*2+8",
"output": "3501"
},
{
"input": "2*3+9+6*5*8+2+9*6+3+9",
"output": "3447"
},
{
"input": "7+7*6+7+6*1+8+8*1*2*4",
"output": "1967"
},
{
"input": "3+2*5+9+5*2+5*5*7+9*2",
"output": "2051"
},
{
"input": "3+4*5+6",
"output": "47"
}
] | 1,475,622,266
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 7
| 77
| 0
|
s=input()
r=[-1]
t=s.find('*')
while-1!=t:r+=[t];t=s.find('*',t+1)
r+=[len(s)]
print(max(eval(s[:r[i-1]+1]+'('+s[r[i-1]+1:r[i]]+')'+s[r[i]:])for i in range(1,len(r))))
|
Title: Vanya and Brackets
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vanya is doing his maths homework. He has an expression of form , where *x*1,<=*x*2,<=...,<=*x**n* are digits from 1 to 9, and sign represents either a plus '+' or the multiplication sign '*'. Vanya needs to add one pair of brackets in this expression so that to maximize the value of the resulting expression.
Input Specification:
The first line contains expression *s* (1<=≤<=|*s*|<=≤<=5001, |*s*| is odd), its odd positions only contain digits from 1 to 9, and even positions only contain signs <=+<= and <=*<=.
The number of signs <=*<= doesn't exceed 15.
Output Specification:
In the first line print the maximum possible value of an expression.
Demo Input:
['3+5*7+8*4\n', '2+3*5\n', '3*4*5\n']
Demo Output:
['303\n', '25\n', '60\n']
Note:
Note to the first sample test. 3 + 5 * (7 + 8) * 4 = 303.
Note to the second sample test. (2 + 3) * 5 = 25.
Note to the third sample test. (3 * 4) * 5 = 60 (also many other variants are valid, for instance, (3) * 4 * 5 = 60).
|
```python
s=input()
r=[-1]
t=s.find('*')
while-1!=t:r+=[t];t=s.find('*',t+1)
r+=[len(s)]
print(max(eval(s[:r[i-1]+1]+'('+s[r[i-1]+1:r[i]]+')'+s[r[i]:])for i in range(1,len(r))))
```
| 0
|
|
4
|
A
|
Watermelon
|
PROGRAMMING
| 800
|
[
"brute force",
"math"
] |
A. Watermelon
|
1
|
64
|
One hot summer day Pete and his friend Billy decided to buy a watermelon. They chose the biggest and the ripest one, in their opinion. After that the watermelon was weighed, and the scales showed *w* kilos. They rushed home, dying of thirst, and decided to divide the berry, however they faced a hard problem.
Pete and Billy are great fans of even numbers, that's why they want to divide the watermelon in such a way that each of the two parts weighs even number of kilos, at the same time it is not obligatory that the parts are equal. The boys are extremely tired and want to start their meal as soon as possible, that's why you should help them and find out, if they can divide the watermelon in the way they want. For sure, each of them should get a part of positive weight.
|
The first (and the only) input line contains integer number *w* (1<=≤<=*w*<=≤<=100) — the weight of the watermelon bought by the boys.
|
Print YES, if the boys can divide the watermelon into two parts, each of them weighing even number of kilos; and NO in the opposite case.
|
[
"8\n"
] |
[
"YES\n"
] |
For example, the boys can divide the watermelon into two parts of 2 and 6 kilos respectively (another variant — two parts of 4 and 4 kilos).
| 0
|
[
{
"input": "8",
"output": "YES"
},
{
"input": "5",
"output": "NO"
},
{
"input": "4",
"output": "YES"
},
{
"input": "3",
"output": "NO"
},
{
"input": "2",
"output": "NO"
},
{
"input": "1",
"output": "NO"
},
{
"input": "7",
"output": "NO"
},
{
"input": "6",
"output": "YES"
},
{
"input": "10",
"output": "YES"
},
{
"input": "9",
"output": "NO"
},
{
"input": "53",
"output": "NO"
},
{
"input": "77",
"output": "NO"
},
{
"input": "32",
"output": "YES"
},
{
"input": "44",
"output": "YES"
},
{
"input": "98",
"output": "YES"
},
{
"input": "99",
"output": "NO"
},
{
"input": "90",
"output": "YES"
},
{
"input": "67",
"output": "NO"
},
{
"input": "100",
"output": "YES"
},
{
"input": "88",
"output": "YES"
}
] | 1,696,182,852
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 30
| 0
|
w=int(input())
while w<=100 and w>=100:
if w%2==0 and w!=2:
print ("YES")
elif w==2:
print ("NO")
else:
print ("NO")
|
Title: Watermelon
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
One hot summer day Pete and his friend Billy decided to buy a watermelon. They chose the biggest and the ripest one, in their opinion. After that the watermelon was weighed, and the scales showed *w* kilos. They rushed home, dying of thirst, and decided to divide the berry, however they faced a hard problem.
Pete and Billy are great fans of even numbers, that's why they want to divide the watermelon in such a way that each of the two parts weighs even number of kilos, at the same time it is not obligatory that the parts are equal. The boys are extremely tired and want to start their meal as soon as possible, that's why you should help them and find out, if they can divide the watermelon in the way they want. For sure, each of them should get a part of positive weight.
Input Specification:
The first (and the only) input line contains integer number *w* (1<=≤<=*w*<=≤<=100) — the weight of the watermelon bought by the boys.
Output Specification:
Print YES, if the boys can divide the watermelon into two parts, each of them weighing even number of kilos; and NO in the opposite case.
Demo Input:
['8\n']
Demo Output:
['YES\n']
Note:
For example, the boys can divide the watermelon into two parts of 2 and 6 kilos respectively (another variant — two parts of 4 and 4 kilos).
|
```python
w=int(input())
while w<=100 and w>=100:
if w%2==0 and w!=2:
print ("YES")
elif w==2:
print ("NO")
else:
print ("NO")
```
| 0
|
813
|
A
|
The Contest
|
PROGRAMMING
| 1,100
|
[
"implementation"
] | null | null |
Pasha is participating in a contest on one well-known website. This time he wants to win the contest and will do anything to get to the first place!
This contest consists of *n* problems, and Pasha solves *i*th problem in *a**i* time units (his solutions are always correct). At any moment of time he can be thinking about a solution to only one of the problems (that is, he cannot be solving two problems at the same time). The time Pasha spends to send his solutions is negligible. Pasha can send any number of solutions at the same moment.
Unfortunately, there are too many participants, and the website is not always working. Pasha received the information that the website will be working only during *m* time periods, *j*th period is represented by its starting moment *l**j* and ending moment *r**j*. Of course, Pasha can send his solution only when the website is working. In other words, Pasha can send his solution at some moment *T* iff there exists a period *x* such that *l**x*<=≤<=*T*<=≤<=*r**x*.
Pasha wants to know his best possible result. We need to tell him the minimal moment of time by which he is able to have solutions to all problems submitted, if he acts optimally, or say that it's impossible no matter how Pasha solves the problems.
|
The first line contains one integer *n* (1<=≤<=*n*<=≤<=1000) — the number of problems. The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=105) — the time Pasha needs to solve *i*th problem.
The third line contains one integer *m* (0<=≤<=*m*<=≤<=1000) — the number of periods of time when the website is working. Next *m* lines represent these periods. *j*th line contains two numbers *l**j* and *r**j* (1<=≤<=*l**j*<=<<=*r**j*<=≤<=105) — the starting and the ending moment of *j*th period.
It is guaranteed that the periods are not intersecting and are given in chronological order, so for every *j*<=><=1 the condition *l**j*<=><=*r**j*<=-<=1 is met.
|
If Pasha can solve and submit all the problems before the end of the contest, print the minimal moment of time by which he can have all the solutions submitted.
Otherwise print "-1" (without brackets).
|
[
"2\n3 4\n2\n1 4\n7 9\n",
"1\n5\n1\n1 4\n",
"1\n5\n1\n1 5\n"
] |
[
"7\n",
"-1\n",
"5\n"
] |
In the first example Pasha can act like this: he solves the second problem in 4 units of time and sends it immediately. Then he spends 3 time units to solve the first problem and sends it 7 time units after the contest starts, because at this moment the website starts working again.
In the second example Pasha invents the solution only after the website stops working for the last time.
In the third example Pasha sends the solution exactly at the end of the first period.
| 0
|
[
{
"input": "2\n3 4\n2\n1 4\n7 9",
"output": "7"
},
{
"input": "1\n5\n1\n1 4",
"output": "-1"
},
{
"input": "1\n5\n1\n1 5",
"output": "5"
},
{
"input": "5\n100000 100000 100000 100000 100000\n0",
"output": "-1"
},
{
"input": "5\n886 524 128 4068 298\n3\n416 3755\n4496 11945\n17198 18039",
"output": "5904"
},
{
"input": "10\n575 3526 1144 1161 889 1038 790 19 765 357\n2\n4475 10787\n16364 21678",
"output": "10264"
},
{
"input": "1\n4\n1\n5 9",
"output": "5"
},
{
"input": "1\n200\n4\n1 10\n20 40\n50 55\n190 210",
"output": "200"
},
{
"input": "4\n643 70 173 745\n14\n990 995\n1256 1259\n1494 1499\n1797 1804\n2443 2450\n2854 2859\n3164 3167\n4084 4092\n4615 4622\n5555 5563\n6412 6421\n7173 7180\n7566 7571\n8407 8415",
"output": "1797"
},
{
"input": "42\n749 516 256 497 37 315 184 518 103 726 80 983 474 884 209 706 10 543 587 371 199 315 967 707 948 736 590 734 715 184 230 513 199 898 287 468 250 600 352 29 408 22\n2\n312 314\n1293 1302",
"output": "-1"
},
{
"input": "1\n10000\n2\n1 10\n9998 10000",
"output": "10000"
},
{
"input": "1\n547\n15\n774 779\n1598 1605\n2458 2464\n3138 3140\n3372 3378\n4268 4272\n4730 4733\n5064 5067\n5074 5075\n5483 5490\n5894 5901\n5931 5938\n6750 6756\n7487 7491\n8328 8332",
"output": "774"
},
{
"input": "1\n10\n2\n1 2\n11 12",
"output": "11"
},
{
"input": "2\n4 6\n2\n5 10\n15 20",
"output": "10"
},
{
"input": "2\n16 5\n3\n5 10\n15 20\n25 30",
"output": "25"
},
{
"input": "1\n16\n2\n5 10\n15 20",
"output": "16"
},
{
"input": "44\n750 672 846 969 981 698 380 968 813 587 156 28 446 917 849 449 173 764 226 958 335 622 236 782 416 689 113 728 452 265 585 217 707 50 520 712 946 275 423 123 175 268 583 528\n4\n869 870\n1353 1354\n1683 1685\n2532 2540",
"output": "-1"
},
{
"input": "1\n1\n0",
"output": "-1"
},
{
"input": "3\n1 2 5\n3\n5 6\n7 8\n9 13",
"output": "8"
},
{
"input": "1\n2\n0",
"output": "-1"
},
{
"input": "1\n5\n3\n1 2\n3 4\n10 11",
"output": "10"
},
{
"input": "1\n4\n0",
"output": "-1"
},
{
"input": "1\n5\n0",
"output": "-1"
},
{
"input": "1\n239\n0",
"output": "-1"
}
] | 1,614,249,912
| 312
|
PyPy 3
|
OK
|
TESTS
| 29
| 108
| 102,400
|
import sys
#import re
#sys.stdin=open('.in','r')
#sys.stdout=open('.out','w')
#import math
#import random
#import time
#sys.setrecursionlimit(int(1e6))
input = sys.stdin.readline
############ ---- USER DEFINED INPUT FUNCTIONS ---- ############
def inp():
return(int(input()))
def inara():
return(list(map(int,input().split())))
def insr():
s = input()
return(list(s[:len(s) - 1]))
def invr():
return(map(int,input().split()))
################################################################
############ ---- THE ACTUAL CODE STARTS BELOW ---- ############
n=inp()
a=inara()
tot=sum(a)
m=inp()
for i in range(m):
l,r=invr()
if l<=tot and r>=tot:
print(tot)
exit(0)
elif l>tot:
print(l)
exit(0)
print(-1)
|
Title: The Contest
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Pasha is participating in a contest on one well-known website. This time he wants to win the contest and will do anything to get to the first place!
This contest consists of *n* problems, and Pasha solves *i*th problem in *a**i* time units (his solutions are always correct). At any moment of time he can be thinking about a solution to only one of the problems (that is, he cannot be solving two problems at the same time). The time Pasha spends to send his solutions is negligible. Pasha can send any number of solutions at the same moment.
Unfortunately, there are too many participants, and the website is not always working. Pasha received the information that the website will be working only during *m* time periods, *j*th period is represented by its starting moment *l**j* and ending moment *r**j*. Of course, Pasha can send his solution only when the website is working. In other words, Pasha can send his solution at some moment *T* iff there exists a period *x* such that *l**x*<=≤<=*T*<=≤<=*r**x*.
Pasha wants to know his best possible result. We need to tell him the minimal moment of time by which he is able to have solutions to all problems submitted, if he acts optimally, or say that it's impossible no matter how Pasha solves the problems.
Input Specification:
The first line contains one integer *n* (1<=≤<=*n*<=≤<=1000) — the number of problems. The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=105) — the time Pasha needs to solve *i*th problem.
The third line contains one integer *m* (0<=≤<=*m*<=≤<=1000) — the number of periods of time when the website is working. Next *m* lines represent these periods. *j*th line contains two numbers *l**j* and *r**j* (1<=≤<=*l**j*<=<<=*r**j*<=≤<=105) — the starting and the ending moment of *j*th period.
It is guaranteed that the periods are not intersecting and are given in chronological order, so for every *j*<=><=1 the condition *l**j*<=><=*r**j*<=-<=1 is met.
Output Specification:
If Pasha can solve and submit all the problems before the end of the contest, print the minimal moment of time by which he can have all the solutions submitted.
Otherwise print "-1" (without brackets).
Demo Input:
['2\n3 4\n2\n1 4\n7 9\n', '1\n5\n1\n1 4\n', '1\n5\n1\n1 5\n']
Demo Output:
['7\n', '-1\n', '5\n']
Note:
In the first example Pasha can act like this: he solves the second problem in 4 units of time and sends it immediately. Then he spends 3 time units to solve the first problem and sends it 7 time units after the contest starts, because at this moment the website starts working again.
In the second example Pasha invents the solution only after the website stops working for the last time.
In the third example Pasha sends the solution exactly at the end of the first period.
|
```python
import sys
#import re
#sys.stdin=open('.in','r')
#sys.stdout=open('.out','w')
#import math
#import random
#import time
#sys.setrecursionlimit(int(1e6))
input = sys.stdin.readline
############ ---- USER DEFINED INPUT FUNCTIONS ---- ############
def inp():
return(int(input()))
def inara():
return(list(map(int,input().split())))
def insr():
s = input()
return(list(s[:len(s) - 1]))
def invr():
return(map(int,input().split()))
################################################################
############ ---- THE ACTUAL CODE STARTS BELOW ---- ############
n=inp()
a=inara()
tot=sum(a)
m=inp()
for i in range(m):
l,r=invr()
if l<=tot and r>=tot:
print(tot)
exit(0)
elif l>tot:
print(l)
exit(0)
print(-1)
```
| 3
|
|
409
|
H
|
A + B Strikes Back
|
PROGRAMMING
| 1,500
|
[
"*special",
"brute force",
"constructive algorithms",
"dsu",
"implementation"
] | null | null |
A + B is often used as an example of the easiest problem possible to show some contest platform. However, some scientists have observed that sometimes this problem is not so easy to get accepted. Want to try?
|
The input contains two integers *a* and *b* (0<=≤<=*a*,<=*b*<=≤<=103), separated by a single space.
|
Output the sum of the given integers.
|
[
"5 14\n",
"381 492\n"
] |
[
"19\n",
"873\n"
] |
none
| 0
|
[
{
"input": "5 14",
"output": "19"
},
{
"input": "381 492",
"output": "873"
},
{
"input": "536 298",
"output": "834"
},
{
"input": "143 522",
"output": "665"
},
{
"input": "433 126",
"output": "559"
},
{
"input": "723 350",
"output": "1073"
},
{
"input": "632 264",
"output": "896"
},
{
"input": "923 488",
"output": "1411"
},
{
"input": "522 92",
"output": "614"
},
{
"input": "953 553",
"output": "1506"
},
{
"input": "242 86",
"output": "328"
},
{
"input": "151 691",
"output": "842"
},
{
"input": "441 915",
"output": "1356"
},
{
"input": "740 519",
"output": "1259"
},
{
"input": "339 433",
"output": "772"
},
{
"input": "629 657",
"output": "1286"
},
{
"input": "538 261",
"output": "799"
},
{
"input": "828 485",
"output": "1313"
},
{
"input": "117 399",
"output": "516"
},
{
"input": "399 280",
"output": "679"
}
] | 1,664,614,329
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 0
| 0
| 0
|
n, m = [int(i) for i in input().split()]
print(n + m)
|
Title: A + B Strikes Back
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A + B is often used as an example of the easiest problem possible to show some contest platform. However, some scientists have observed that sometimes this problem is not so easy to get accepted. Want to try?
Input Specification:
The input contains two integers *a* and *b* (0<=≤<=*a*,<=*b*<=≤<=103), separated by a single space.
Output Specification:
Output the sum of the given integers.
Demo Input:
['5 14\n', '381 492\n']
Demo Output:
['19\n', '873\n']
Note:
none
|
```python
n, m = [int(i) for i in input().split()]
print(n + m)
```
| 0
|
|
294
|
C
|
Shaass and Lights
|
PROGRAMMING
| 1,900
|
[
"combinatorics",
"number theory"
] | null | null |
There are *n* lights aligned in a row. These lights are numbered 1 to *n* from left to right. Initially some of the lights are switched on. Shaass wants to switch all the lights on. At each step he can switch a light on (this light should be switched off at that moment) if there's at least one adjacent light which is already switched on.
He knows the initial state of lights and he's wondering how many different ways there exist to switch all the lights on. Please find the required number of ways modulo 1000000007 (109<=+<=7).
|
The first line of the input contains two integers *n* and *m* where *n* is the number of lights in the sequence and *m* is the number of lights which are initially switched on, (1<=≤<=*n*<=≤<=1000,<=1<=≤<=*m*<=≤<=*n*). The second line contains *m* distinct integers, each between 1 to *n* inclusive, denoting the indices of lights which are initially switched on.
|
In the only line of the output print the number of different possible ways to switch on all the lights modulo 1000000007 (109<=+<=7).
|
[
"3 1\n1\n",
"4 2\n1 4\n",
"11 2\n4 8\n"
] |
[
"1\n",
"2\n",
"6720\n"
] |
none
| 1,500
|
[
{
"input": "3 1\n1",
"output": "1"
},
{
"input": "4 2\n1 4",
"output": "2"
},
{
"input": "11 2\n4 8",
"output": "6720"
},
{
"input": "4 2\n1 3",
"output": "2"
},
{
"input": "4 4\n1 2 3 4",
"output": "1"
},
{
"input": "4 2\n1 3",
"output": "2"
},
{
"input": "4 4\n1 2 3 4",
"output": "1"
},
{
"input": "1000 3\n100 900 10",
"output": "727202008"
},
{
"input": "74 13\n6 14 19 20 21 24 30 43 58 61 69 70 73",
"output": "16623551"
},
{
"input": "74 13\n6 14 19 20 21 24 30 43 58 61 69 70 73",
"output": "16623551"
},
{
"input": "74 13\n6 14 19 20 21 24 30 43 58 61 69 70 73",
"output": "16623551"
},
{
"input": "74 13\n6 14 19 20 21 24 30 43 58 61 69 70 73",
"output": "16623551"
},
{
"input": "74 13\n6 14 19 20 21 24 30 43 58 61 69 70 73",
"output": "16623551"
},
{
"input": "74 13\n6 14 19 20 21 24 30 43 58 61 69 70 73",
"output": "16623551"
},
{
"input": "68 37\n1 2 3 6 7 8 10 11 12 14 16 18 22 23 24 26 30 31 32 35 37 39 41 42 45 47 50 51 52 54 58 59 61 62 63 64 68",
"output": "867201120"
},
{
"input": "132 48\n6 7 8 12 15 17 18 19 22 24 25 26 30 33 35 38 40 43 46 49 50 51 52 54 59 60 66 70 76 79 87 89 91 92 94 98 99 101 102 105 106 109 113 115 116 118 120 129",
"output": "376947760"
},
{
"input": "36 24\n1 7 8 10 11 12 13 14 15 16 17 19 21 22 25 26 27 28 29 30 31 32 35 36",
"output": "63866880"
},
{
"input": "100 100\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",
"output": "1"
},
{
"input": "100 2\n11 64",
"output": "910895596"
},
{
"input": "100 90\n1 2 3 4 5 7 8 9 10 11 12 13 15 16 17 18 19 20 21 22 23 24 25 27 28 29 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 58 59 60 61 62 63 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 82 83 84 86 87 88 89 90 91 92 94 95 96 98 99 100",
"output": "3628800"
},
{
"input": "1000 1\n35",
"output": "253560421"
},
{
"input": "1000 2\n747 798",
"output": "474746180"
},
{
"input": "1000 3\n804 811 984",
"output": "600324842"
},
{
"input": "1 1\n1",
"output": "1"
}
] | 1,578,315,475
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 4
| 109
| 307,200
|
def fac(x):
pr = 1
for i in range(2, x + 1):
pr *= i
return pr
n, m = map(int, input().split())
a = list(map(int, input().split()))
x = a[0] - 1
y = n - a[-1]
mn1 = 1
mn2 = 1
for i in range(1, m):
mn1 *= fac(a[i] - a[i - 1] - 1)
mn2 *= (2 ** (a[i] - a[i - 1] - 2))
ans = fac(n - m) // (fac(x) * fac(y) * mn1) * mn2
print(ans % 1000000007)
|
Title: Shaass and Lights
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* lights aligned in a row. These lights are numbered 1 to *n* from left to right. Initially some of the lights are switched on. Shaass wants to switch all the lights on. At each step he can switch a light on (this light should be switched off at that moment) if there's at least one adjacent light which is already switched on.
He knows the initial state of lights and he's wondering how many different ways there exist to switch all the lights on. Please find the required number of ways modulo 1000000007 (109<=+<=7).
Input Specification:
The first line of the input contains two integers *n* and *m* where *n* is the number of lights in the sequence and *m* is the number of lights which are initially switched on, (1<=≤<=*n*<=≤<=1000,<=1<=≤<=*m*<=≤<=*n*). The second line contains *m* distinct integers, each between 1 to *n* inclusive, denoting the indices of lights which are initially switched on.
Output Specification:
In the only line of the output print the number of different possible ways to switch on all the lights modulo 1000000007 (109<=+<=7).
Demo Input:
['3 1\n1\n', '4 2\n1 4\n', '11 2\n4 8\n']
Demo Output:
['1\n', '2\n', '6720\n']
Note:
none
|
```python
def fac(x):
pr = 1
for i in range(2, x + 1):
pr *= i
return pr
n, m = map(int, input().split())
a = list(map(int, input().split()))
x = a[0] - 1
y = n - a[-1]
mn1 = 1
mn2 = 1
for i in range(1, m):
mn1 *= fac(a[i] - a[i - 1] - 1)
mn2 *= (2 ** (a[i] - a[i - 1] - 2))
ans = fac(n - m) // (fac(x) * fac(y) * mn1) * mn2
print(ans % 1000000007)
```
| 0
|
|
2
|
A
|
Winner
|
PROGRAMMING
| 1,500
|
[
"hashing",
"implementation"
] |
A. Winner
|
1
|
64
|
The winner of the card game popular in Berland "Berlogging" is determined according to the following rules. If at the end of the game there is only one player with the maximum number of points, he is the winner. The situation becomes more difficult if the number of such players is more than one. During each round a player gains or loses a particular number of points. In the course of the game the number of points is registered in the line "name score", where name is a player's name, and score is the number of points gained in this round, which is an integer number. If score is negative, this means that the player has lost in the round. So, if two or more players have the maximum number of points (say, it equals to *m*) at the end of the game, than wins the one of them who scored at least *m* points first. Initially each player has 0 points. It's guaranteed that at the end of the game at least one player has a positive number of points.
|
The first line contains an integer number *n* (1<=<=≤<=<=*n*<=<=≤<=<=1000), *n* is the number of rounds played. Then follow *n* lines, containing the information about the rounds in "name score" format in chronological order, where name is a string of lower-case Latin letters with the length from 1 to 32, and score is an integer number between -1000 and 1000, inclusive.
|
Print the name of the winner.
|
[
"3\nmike 3\nandrew 5\nmike 2\n",
"3\nandrew 3\nandrew 2\nmike 5\n"
] |
[
"andrew\n",
"andrew\n"
] |
none
| 0
|
[
{
"input": "3\nmike 3\nandrew 5\nmike 2",
"output": "andrew"
},
{
"input": "3\nandrew 3\nandrew 2\nmike 5",
"output": "andrew"
},
{
"input": "5\nkaxqybeultn -352\nmgochgrmeyieyskhuourfg -910\nkaxqybeultn 691\nmgochgrmeyieyskhuourfg -76\nkaxqybeultn -303",
"output": "kaxqybeultn"
},
{
"input": "7\nksjuuerbnlklcfdjeyq 312\ndthjlkrvvbyahttifpdewvyslsh -983\nksjuuerbnlklcfdjeyq 268\ndthjlkrvvbyahttifpdewvyslsh 788\nksjuuerbnlklcfdjeyq -79\nksjuuerbnlklcfdjeyq -593\nksjuuerbnlklcfdjeyq 734",
"output": "ksjuuerbnlklcfdjeyq"
},
{
"input": "12\natrtthfpcvishmqbakprquvnejr 185\natrtthfpcvishmqbakprquvnejr -699\natrtthfpcvishmqbakprquvnejr -911\natrtthfpcvishmqbakprquvnejr -220\nfcgslzkicjrpbqaifgweyzreajjfdo 132\nfcgslzkicjrpbqaifgweyzreajjfdo -242\nm 177\nm -549\natrtthfpcvishmqbakprquvnejr -242\nm 38\natrtthfpcvishmqbakprquvnejr -761\nfcgslzkicjrpbqaifgweyzreajjfdo 879",
"output": "fcgslzkicjrpbqaifgweyzreajjfdo"
},
{
"input": "15\naawtvezfntstrcpgbzjbf 681\nzhahpvqiptvksnbjkdvmknb -74\naawtvezfntstrcpgbzjbf 661\njpdwmyke 474\naawtvezfntstrcpgbzjbf -547\naawtvezfntstrcpgbzjbf 600\nzhahpvqiptvksnbjkdvmknb -11\njpdwmyke 711\nbjmj 652\naawtvezfntstrcpgbzjbf -1000\naawtvezfntstrcpgbzjbf -171\nbjmj -302\naawtvezfntstrcpgbzjbf 961\nzhahpvqiptvksnbjkdvmknb 848\nbjmj -735",
"output": "aawtvezfntstrcpgbzjbf"
},
{
"input": "17\nqdplghhx -649\nivhgbxiv 424\nivhgbxiv -72\nivhgbxiv 479\nugowfouebkelargxylsxn 83\nugowfouebkelargxylsxn -840\nivhgbxiv 382\nqdplghhx -904\nivhgbxiv -710\nugowfouebkelargxylsxn -838\nqdplghhx -974\nqdplghhx 571\nivhgbxiv -145\nugowfouebkelargxylsxn -788\nivhgbxiv 209\nivhgbxiv 420\nqdplghhx -158",
"output": "ivhgbxiv"
}
] | 1,674,635,958
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 1
| 62
| 0
|
import math
count_iter = int(input())
counter = {}
history = []
while count_iter:
name, score = input().split()
if name not in counter:
counter[name] = int(score)
else:
counter[name] += int(score)
history.append((int(score), name))
count_iter -= 1
max_number = -math.inf
for s in counter.values():
if s > max_number:
max_number = s
top_players = []
for n, s in counter.items():
if s == max_number:
top_players.append(n)
for i in history:
if i[0] >= max_number and i[1] in top_players:
print(i[1])
break
|
Title: Winner
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
The winner of the card game popular in Berland "Berlogging" is determined according to the following rules. If at the end of the game there is only one player with the maximum number of points, he is the winner. The situation becomes more difficult if the number of such players is more than one. During each round a player gains or loses a particular number of points. In the course of the game the number of points is registered in the line "name score", where name is a player's name, and score is the number of points gained in this round, which is an integer number. If score is negative, this means that the player has lost in the round. So, if two or more players have the maximum number of points (say, it equals to *m*) at the end of the game, than wins the one of them who scored at least *m* points first. Initially each player has 0 points. It's guaranteed that at the end of the game at least one player has a positive number of points.
Input Specification:
The first line contains an integer number *n* (1<=<=≤<=<=*n*<=<=≤<=<=1000), *n* is the number of rounds played. Then follow *n* lines, containing the information about the rounds in "name score" format in chronological order, where name is a string of lower-case Latin letters with the length from 1 to 32, and score is an integer number between -1000 and 1000, inclusive.
Output Specification:
Print the name of the winner.
Demo Input:
['3\nmike 3\nandrew 5\nmike 2\n', '3\nandrew 3\nandrew 2\nmike 5\n']
Demo Output:
['andrew\n', 'andrew\n']
Note:
none
|
```python
import math
count_iter = int(input())
counter = {}
history = []
while count_iter:
name, score = input().split()
if name not in counter:
counter[name] = int(score)
else:
counter[name] += int(score)
history.append((int(score), name))
count_iter -= 1
max_number = -math.inf
for s in counter.values():
if s > max_number:
max_number = s
top_players = []
for n, s in counter.items():
if s == max_number:
top_players.append(n)
for i in history:
if i[0] >= max_number and i[1] in top_players:
print(i[1])
break
```
| 0
|
29
|
A
|
Spit Problem
|
PROGRAMMING
| 1,000
|
[
"brute force"
] |
A. Spit Problem
|
2
|
256
|
In a Berland's zoo there is an enclosure with camels. It is known that camels like to spit. Bob watched these interesting animals for the whole day and registered in his notepad where each animal spitted. Now he wants to know if in the zoo there are two camels, which spitted at each other. Help him to solve this task.
The trajectory of a camel's spit is an arc, i.e. if the camel in position *x* spits *d* meters right, he can hit only the camel in position *x*<=+<=*d*, if such a camel exists.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the amount of camels in the zoo. Each of the following *n* lines contains two integers *x**i* and *d**i* (<=-<=104<=≤<=*x**i*<=≤<=104,<=1<=≤<=|*d**i*|<=≤<=2·104) — records in Bob's notepad. *x**i* is a position of the *i*-th camel, and *d**i* is a distance at which the *i*-th camel spitted. Positive values of *d**i* correspond to the spits right, negative values correspond to the spits left. No two camels may stand in the same position.
|
If there are two camels, which spitted at each other, output YES. Otherwise, output NO.
|
[
"2\n0 1\n1 -1\n",
"3\n0 1\n1 1\n2 -2\n",
"5\n2 -10\n3 10\n0 5\n5 -5\n10 1\n"
] |
[
"YES\n",
"NO\n",
"YES\n"
] |
none
| 500
|
[
{
"input": "2\n0 1\n1 -1",
"output": "YES"
},
{
"input": "3\n0 1\n1 1\n2 -2",
"output": "NO"
},
{
"input": "5\n2 -10\n3 10\n0 5\n5 -5\n10 1",
"output": "YES"
},
{
"input": "10\n-9897 -1144\n-4230 -6350\n2116 -3551\n-3635 4993\n3907 -9071\n-2362 4120\n-6542 984\n5807 3745\n7594 7675\n-5412 -6872",
"output": "NO"
},
{
"input": "11\n-1536 3809\n-2406 -8438\n-1866 395\n5636 -490\n-6867 -7030\n7525 3575\n-6796 2908\n3884 4629\n-2862 -6122\n-8984 6122\n7137 -326",
"output": "YES"
},
{
"input": "12\n-9765 1132\n-1382 -215\n-9405 7284\n-2040 3947\n-9360 3150\n6425 9386\n806 -2278\n-2121 -7284\n5663 -1608\n-8377 9297\n6245 708\n8470 6024",
"output": "YES"
},
{
"input": "15\n8122 -9991\n-4068 -3386\n8971 3731\n3458 5161\n-8700 7562\n2691 8735\n-1510 -3892\n5183 -3753\n-7018 6637\n-7454 3386\n-818 -6377\n6771 -8647\n-7357 -1246\n-6186 1922\n9889 -3627",
"output": "YES"
},
{
"input": "20\n-5264 6424\n-3664 -7459\n-2780 -9859\n-3317 6842\n5681 -8092\n1555 1904\n-6684 1414\n6593 -1253\n-5708 -1202\n335 1733\n-926 7579\n3459 -1904\n-4486 4006\n6201 3616\n2847 -5255\n8438 7057\n8171 6042\n-9102 3545\n7731 -233\n6264 6563",
"output": "YES"
},
{
"input": "30\n-398 -1774\n313 -6974\n2346 -4657\n8552 -9647\n-5265 1538\n8195 4864\n-5641 -5219\n-1394 8563\n-1190 1992\n-4669 -1156\n7574 256\n9206 -2414\n4140 -549\n-294 2169\n7029 -2871\n3808 -9799\n3141 5690\n4648 -2680\n-5990 9800\n-2299 1697\n6077 -7177\n-400 -9724\n-4644 -2392\n-2198 -9531\n-2105 9386\n-8165 -4201\n-1589 -7916\n2518 -7840\n4173 -6949\n-3368 -9943",
"output": "NO"
}
] | 1,584,542,998
| 298
|
PyPy 3
|
OK
|
TESTS
| 30
| 310
| 1,228,800
|
import sys
n=int(input())
a=[0]*n
b=[0]*n
for i in range(n):
a[i],b[i]=map(int,input().split())
b[i]=a[i]+b[i]
for i in range(n):
for j in range(n):
if a[i]==b[j] and b[i]==a[j] and i!=j:
print("YES")
sys.exit()
print("NO")
|
Title: Spit Problem
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
In a Berland's zoo there is an enclosure with camels. It is known that camels like to spit. Bob watched these interesting animals for the whole day and registered in his notepad where each animal spitted. Now he wants to know if in the zoo there are two camels, which spitted at each other. Help him to solve this task.
The trajectory of a camel's spit is an arc, i.e. if the camel in position *x* spits *d* meters right, he can hit only the camel in position *x*<=+<=*d*, if such a camel exists.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the amount of camels in the zoo. Each of the following *n* lines contains two integers *x**i* and *d**i* (<=-<=104<=≤<=*x**i*<=≤<=104,<=1<=≤<=|*d**i*|<=≤<=2·104) — records in Bob's notepad. *x**i* is a position of the *i*-th camel, and *d**i* is a distance at which the *i*-th camel spitted. Positive values of *d**i* correspond to the spits right, negative values correspond to the spits left. No two camels may stand in the same position.
Output Specification:
If there are two camels, which spitted at each other, output YES. Otherwise, output NO.
Demo Input:
['2\n0 1\n1 -1\n', '3\n0 1\n1 1\n2 -2\n', '5\n2 -10\n3 10\n0 5\n5 -5\n10 1\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n']
Note:
none
|
```python
import sys
n=int(input())
a=[0]*n
b=[0]*n
for i in range(n):
a[i],b[i]=map(int,input().split())
b[i]=a[i]+b[i]
for i in range(n):
for j in range(n):
if a[i]==b[j] and b[i]==a[j] and i!=j:
print("YES")
sys.exit()
print("NO")
```
| 3.920211
|
265
|
B
|
Roadside Trees (Simplified Edition)
|
PROGRAMMING
| 1,000
|
[
"greedy",
"implementation"
] | null | null |
Squirrel Liss loves nuts. There are *n* trees (numbered 1 to *n* from west to east) along a street and there is a delicious nut on the top of each tree. The height of the tree *i* is *h**i*. Liss wants to eat all nuts.
Now Liss is on the root of the tree with the number 1. In one second Liss can perform one of the following actions:
- Walk up or down one unit on a tree. - Eat a nut on the top of the current tree. - Jump to the next tree. In this action the height of Liss doesn't change. More formally, when Liss is at height *h* of the tree *i* (1<=≤<=*i*<=≤<=*n*<=-<=1), she jumps to height *h* of the tree *i*<=+<=1. This action can't be performed if *h*<=><=*h**i*<=+<=1.
Compute the minimal time (in seconds) required to eat all nuts.
|
The first line contains an integer *n* (1<=<=≤<=<=*n*<=≤<=105) — the number of trees.
Next *n* lines contains the height of trees: *i*-th line contains an integer *h**i* (1<=≤<=*h**i*<=≤<=104) — the height of the tree with the number *i*.
|
Print a single integer — the minimal time required to eat all nuts in seconds.
|
[
"2\n1\n2\n",
"5\n2\n1\n2\n1\n1\n"
] |
[
"5\n",
"14\n"
] |
none
| 1,000
|
[
{
"input": "2\n1\n2",
"output": "5"
},
{
"input": "5\n2\n1\n2\n1\n1",
"output": "14"
},
{
"input": "1\n1",
"output": "2"
}
] | 1,596,716,316
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 15
| 560
| 8,908,800
|
size = int(input())
height = []
for i in range(size):
temp = int(input())
height.append(temp)
steps = 0
for i in range(size):
if i == 0:
steps+=height[i]+1
else :
if height[i]<height[i-1]:
steps+=height[i-1]-height[i]+1+1
else:
steps+=1+(height[i]-height[i-1])+1
print(steps)
|
Title: Roadside Trees (Simplified Edition)
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Squirrel Liss loves nuts. There are *n* trees (numbered 1 to *n* from west to east) along a street and there is a delicious nut on the top of each tree. The height of the tree *i* is *h**i*. Liss wants to eat all nuts.
Now Liss is on the root of the tree with the number 1. In one second Liss can perform one of the following actions:
- Walk up or down one unit on a tree. - Eat a nut on the top of the current tree. - Jump to the next tree. In this action the height of Liss doesn't change. More formally, when Liss is at height *h* of the tree *i* (1<=≤<=*i*<=≤<=*n*<=-<=1), she jumps to height *h* of the tree *i*<=+<=1. This action can't be performed if *h*<=><=*h**i*<=+<=1.
Compute the minimal time (in seconds) required to eat all nuts.
Input Specification:
The first line contains an integer *n* (1<=<=≤<=<=*n*<=≤<=105) — the number of trees.
Next *n* lines contains the height of trees: *i*-th line contains an integer *h**i* (1<=≤<=*h**i*<=≤<=104) — the height of the tree with the number *i*.
Output Specification:
Print a single integer — the minimal time required to eat all nuts in seconds.
Demo Input:
['2\n1\n2\n', '5\n2\n1\n2\n1\n1\n']
Demo Output:
['5\n', '14\n']
Note:
none
|
```python
size = int(input())
height = []
for i in range(size):
temp = int(input())
height.append(temp)
steps = 0
for i in range(size):
if i == 0:
steps+=height[i]+1
else :
if height[i]<height[i-1]:
steps+=height[i-1]-height[i]+1+1
else:
steps+=1+(height[i]-height[i-1])+1
print(steps)
```
| 3
|
|
412
|
C
|
Pattern
|
PROGRAMMING
| 1,200
|
[
"implementation",
"strings"
] | null | null |
Developers often face with regular expression patterns. A pattern is usually defined as a string consisting of characters and metacharacters that sets the rules for your search. These patterns are most often used to check whether a particular string meets the certain rules.
In this task, a pattern will be a string consisting of small English letters and question marks ('?'). The question mark in the pattern is a metacharacter that denotes an arbitrary small letter of the English alphabet. We will assume that a string matches the pattern if we can transform the string into the pattern by replacing the question marks by the appropriate characters. For example, string aba matches patterns: ???, ??a, a?a, aba.
Programmers that work for the R1 company love puzzling each other (and themselves) with riddles. One of them is as follows: you are given *n* patterns of the same length, you need to find a pattern that contains as few question marks as possible, and intersects with each of the given patterns. Two patterns intersect if there is a string that matches both the first and the second pattern. Can you solve this riddle?
|
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105) — the number of patterns. Next *n* lines contain the patterns.
It is guaranteed that the patterns can only consist of small English letters and symbols '?'. All patterns are non-empty and have the same length. The total length of all the patterns does not exceed 105 characters.
|
In a single line print the answer to the problem — the pattern with the minimal number of signs '?', which intersects with each of the given ones. If there are several answers, print any of them.
|
[
"2\n?ab\n??b\n",
"2\na\nb\n",
"1\n?a?b\n"
] |
[
"xab\n",
"?\n",
"cacb\n"
] |
Consider the first example. Pattern xab intersects with each of the given patterns. Pattern ??? also intersects with each of the given patterns, but it contains more question signs, hence it is not an optimal answer. Clearly, xab is the optimal answer, because it doesn't contain any question sign. There are a lot of other optimal answers, for example: aab, bab, cab, dab and so on.
| 1,500
|
[
{
"input": "2\n?ab\n??b",
"output": "xab"
},
{
"input": "2\na\nb",
"output": "?"
},
{
"input": "1\n?a?b",
"output": "cacb"
},
{
"input": "1\n?",
"output": "x"
},
{
"input": "3\nabacaba\nabacaba\nabacaba",
"output": "abacaba"
},
{
"input": "3\nabc?t\n?bc?z\nab??t",
"output": "abcx?"
},
{
"input": "4\nabc\ndef\n???\nxyz",
"output": "???"
},
{
"input": "2\n?????\n?????",
"output": "xxxxx"
},
{
"input": "2\na\na",
"output": "a"
},
{
"input": "3\nabc\n???\naxc",
"output": "a?c"
},
{
"input": "1\n?m??x?xxexxx?xxmxx??",
"output": "cmccxcxxexxxcxxmxxcc"
},
{
"input": "2\nhszhh?zszs\nhhzhz?zzhs",
"output": "h?zh?cz??s"
},
{
"input": "3\neddzde\needded\nzdde?z",
"output": "??d???"
},
{
"input": "4\nff?o?\nfk?of\nk?kof\nfk?oo",
"output": "??ko?"
},
{
"input": "5\nyigi\ne?gi\niig?\n?eg?\n??gi",
"output": "??gi"
},
{
"input": "100\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz\nz",
"output": "z"
},
{
"input": "100\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?\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?",
"output": "bc"
},
{
"input": "1\nq",
"output": "q"
},
{
"input": "2\n?\n?",
"output": "x"
},
{
"input": "2\n?\na",
"output": "a"
},
{
"input": "2\n?\nb",
"output": "b"
},
{
"input": "3\n?\na\na",
"output": "a"
},
{
"input": "3\na\n?\na",
"output": "a"
},
{
"input": "3\na\nb\n?",
"output": "?"
},
{
"input": "3\n?\na\nb",
"output": "?"
},
{
"input": "3\nb\n?\na",
"output": "?"
},
{
"input": "2\n???\naaa",
"output": "aaa"
},
{
"input": "2\n??\nss",
"output": "ss"
}
] | 1,679,937,242
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 1
| 109
| 0
|
n = int(input())
patterns = []
for i in range(n):
patterns.append(input().strip())
min_question_marks = float('inf')
min_string = ''
for i in range(26**len(patterns[0])):
# Generate the string
s = ''
j = i
for k in range(len(patterns[0])):
if patterns[0][k] == '?':
s += chr(j % 26 + ord('a'))
else:
s += patterns[0][k]
j //= 26
# Check if the string matches all patterns
match = True
for p in patterns:
for k in range(len(p)):
if p[k] != '?' and p[k] != s[k]:
match = False
break
if not match:
break
# If the string matches all patterns, update the minimum
if match:
num_question_marks = s.count('?')
if num_question_marks < min_question_marks:
min_question_marks = num_question_marks
min_string = s
print(min_string)
|
Title: Pattern
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Developers often face with regular expression patterns. A pattern is usually defined as a string consisting of characters and metacharacters that sets the rules for your search. These patterns are most often used to check whether a particular string meets the certain rules.
In this task, a pattern will be a string consisting of small English letters and question marks ('?'). The question mark in the pattern is a metacharacter that denotes an arbitrary small letter of the English alphabet. We will assume that a string matches the pattern if we can transform the string into the pattern by replacing the question marks by the appropriate characters. For example, string aba matches patterns: ???, ??a, a?a, aba.
Programmers that work for the R1 company love puzzling each other (and themselves) with riddles. One of them is as follows: you are given *n* patterns of the same length, you need to find a pattern that contains as few question marks as possible, and intersects with each of the given patterns. Two patterns intersect if there is a string that matches both the first and the second pattern. Can you solve this riddle?
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105) — the number of patterns. Next *n* lines contain the patterns.
It is guaranteed that the patterns can only consist of small English letters and symbols '?'. All patterns are non-empty and have the same length. The total length of all the patterns does not exceed 105 characters.
Output Specification:
In a single line print the answer to the problem — the pattern with the minimal number of signs '?', which intersects with each of the given ones. If there are several answers, print any of them.
Demo Input:
['2\n?ab\n??b\n', '2\na\nb\n', '1\n?a?b\n']
Demo Output:
['xab\n', '?\n', 'cacb\n']
Note:
Consider the first example. Pattern xab intersects with each of the given patterns. Pattern ??? also intersects with each of the given patterns, but it contains more question signs, hence it is not an optimal answer. Clearly, xab is the optimal answer, because it doesn't contain any question sign. There are a lot of other optimal answers, for example: aab, bab, cab, dab and so on.
|
```python
n = int(input())
patterns = []
for i in range(n):
patterns.append(input().strip())
min_question_marks = float('inf')
min_string = ''
for i in range(26**len(patterns[0])):
# Generate the string
s = ''
j = i
for k in range(len(patterns[0])):
if patterns[0][k] == '?':
s += chr(j % 26 + ord('a'))
else:
s += patterns[0][k]
j //= 26
# Check if the string matches all patterns
match = True
for p in patterns:
for k in range(len(p)):
if p[k] != '?' and p[k] != s[k]:
match = False
break
if not match:
break
# If the string matches all patterns, update the minimum
if match:
num_question_marks = s.count('?')
if num_question_marks < min_question_marks:
min_question_marks = num_question_marks
min_string = s
print(min_string)
```
| 0
|
|
2
|
A
|
Winner
|
PROGRAMMING
| 1,500
|
[
"hashing",
"implementation"
] |
A. Winner
|
1
|
64
|
The winner of the card game popular in Berland "Berlogging" is determined according to the following rules. If at the end of the game there is only one player with the maximum number of points, he is the winner. The situation becomes more difficult if the number of such players is more than one. During each round a player gains or loses a particular number of points. In the course of the game the number of points is registered in the line "name score", where name is a player's name, and score is the number of points gained in this round, which is an integer number. If score is negative, this means that the player has lost in the round. So, if two or more players have the maximum number of points (say, it equals to *m*) at the end of the game, than wins the one of them who scored at least *m* points first. Initially each player has 0 points. It's guaranteed that at the end of the game at least one player has a positive number of points.
|
The first line contains an integer number *n* (1<=<=≤<=<=*n*<=<=≤<=<=1000), *n* is the number of rounds played. Then follow *n* lines, containing the information about the rounds in "name score" format in chronological order, where name is a string of lower-case Latin letters with the length from 1 to 32, and score is an integer number between -1000 and 1000, inclusive.
|
Print the name of the winner.
|
[
"3\nmike 3\nandrew 5\nmike 2\n",
"3\nandrew 3\nandrew 2\nmike 5\n"
] |
[
"andrew\n",
"andrew\n"
] |
none
| 0
|
[
{
"input": "3\nmike 3\nandrew 5\nmike 2",
"output": "andrew"
},
{
"input": "3\nandrew 3\nandrew 2\nmike 5",
"output": "andrew"
},
{
"input": "5\nkaxqybeultn -352\nmgochgrmeyieyskhuourfg -910\nkaxqybeultn 691\nmgochgrmeyieyskhuourfg -76\nkaxqybeultn -303",
"output": "kaxqybeultn"
},
{
"input": "7\nksjuuerbnlklcfdjeyq 312\ndthjlkrvvbyahttifpdewvyslsh -983\nksjuuerbnlklcfdjeyq 268\ndthjlkrvvbyahttifpdewvyslsh 788\nksjuuerbnlklcfdjeyq -79\nksjuuerbnlklcfdjeyq -593\nksjuuerbnlklcfdjeyq 734",
"output": "ksjuuerbnlklcfdjeyq"
},
{
"input": "12\natrtthfpcvishmqbakprquvnejr 185\natrtthfpcvishmqbakprquvnejr -699\natrtthfpcvishmqbakprquvnejr -911\natrtthfpcvishmqbakprquvnejr -220\nfcgslzkicjrpbqaifgweyzreajjfdo 132\nfcgslzkicjrpbqaifgweyzreajjfdo -242\nm 177\nm -549\natrtthfpcvishmqbakprquvnejr -242\nm 38\natrtthfpcvishmqbakprquvnejr -761\nfcgslzkicjrpbqaifgweyzreajjfdo 879",
"output": "fcgslzkicjrpbqaifgweyzreajjfdo"
},
{
"input": "15\naawtvezfntstrcpgbzjbf 681\nzhahpvqiptvksnbjkdvmknb -74\naawtvezfntstrcpgbzjbf 661\njpdwmyke 474\naawtvezfntstrcpgbzjbf -547\naawtvezfntstrcpgbzjbf 600\nzhahpvqiptvksnbjkdvmknb -11\njpdwmyke 711\nbjmj 652\naawtvezfntstrcpgbzjbf -1000\naawtvezfntstrcpgbzjbf -171\nbjmj -302\naawtvezfntstrcpgbzjbf 961\nzhahpvqiptvksnbjkdvmknb 848\nbjmj -735",
"output": "aawtvezfntstrcpgbzjbf"
},
{
"input": "17\nqdplghhx -649\nivhgbxiv 424\nivhgbxiv -72\nivhgbxiv 479\nugowfouebkelargxylsxn 83\nugowfouebkelargxylsxn -840\nivhgbxiv 382\nqdplghhx -904\nivhgbxiv -710\nugowfouebkelargxylsxn -838\nqdplghhx -974\nqdplghhx 571\nivhgbxiv -145\nugowfouebkelargxylsxn -788\nivhgbxiv 209\nivhgbxiv 420\nqdplghhx -158",
"output": "ivhgbxiv"
}
] | 1,685,658,851
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 62
| 0
|
import sys
sys.stdin = open("2.in", "r")
n = int(input())
score_dict = dict()
max_score = -sys.maxsize
max_name = ""
for i in range(n):
name, point = input().split()
point = int(point)
if name in score_dict:
score_dict[name] += point
else:
score_dict[name] = point
if score_dict[name] > max_score:
max_score = score_dict[name]
max_name = name
print(max_name)
|
Title: Winner
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
The winner of the card game popular in Berland "Berlogging" is determined according to the following rules. If at the end of the game there is only one player with the maximum number of points, he is the winner. The situation becomes more difficult if the number of such players is more than one. During each round a player gains or loses a particular number of points. In the course of the game the number of points is registered in the line "name score", where name is a player's name, and score is the number of points gained in this round, which is an integer number. If score is negative, this means that the player has lost in the round. So, if two or more players have the maximum number of points (say, it equals to *m*) at the end of the game, than wins the one of them who scored at least *m* points first. Initially each player has 0 points. It's guaranteed that at the end of the game at least one player has a positive number of points.
Input Specification:
The first line contains an integer number *n* (1<=<=≤<=<=*n*<=<=≤<=<=1000), *n* is the number of rounds played. Then follow *n* lines, containing the information about the rounds in "name score" format in chronological order, where name is a string of lower-case Latin letters with the length from 1 to 32, and score is an integer number between -1000 and 1000, inclusive.
Output Specification:
Print the name of the winner.
Demo Input:
['3\nmike 3\nandrew 5\nmike 2\n', '3\nandrew 3\nandrew 2\nmike 5\n']
Demo Output:
['andrew\n', 'andrew\n']
Note:
none
|
```python
import sys
sys.stdin = open("2.in", "r")
n = int(input())
score_dict = dict()
max_score = -sys.maxsize
max_name = ""
for i in range(n):
name, point = input().split()
point = int(point)
if name in score_dict:
score_dict[name] += point
else:
score_dict[name] = point
if score_dict[name] > max_score:
max_score = score_dict[name]
max_name = name
print(max_name)
```
| -1
|
136
|
A
|
Presents
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Little Petya very much likes gifts. Recently he has received a new laptop as a New Year gift from his mother. He immediately decided to give it to somebody else as what can be more pleasant than giving somebody gifts. And on this occasion he organized a New Year party at his place and invited *n* his friends there.
If there's one thing Petya likes more that receiving gifts, that's watching others giving gifts to somebody else. Thus, he safely hid the laptop until the next New Year and made up his mind to watch his friends exchanging gifts while he does not participate in the process. He numbered all his friends with integers from 1 to *n*. Petya remembered that a friend number *i* gave a gift to a friend number *p**i*. He also remembered that each of his friends received exactly one gift.
Now Petya wants to know for each friend *i* the number of a friend who has given him a gift.
|
The first line contains one integer *n* (1<=≤<=*n*<=≤<=100) — the quantity of friends Petya invited to the party. The second line contains *n* space-separated integers: the *i*-th number is *p**i* — the number of a friend who gave a gift to friend number *i*. It is guaranteed that each friend received exactly one gift. It is possible that some friends do not share Petya's ideas of giving gifts to somebody else. Those friends gave the gifts to themselves.
|
Print *n* space-separated integers: the *i*-th number should equal the number of the friend who gave a gift to friend number *i*.
|
[
"4\n2 3 4 1\n",
"3\n1 3 2\n",
"2\n1 2\n"
] |
[
"4 1 2 3\n",
"1 3 2\n",
"1 2\n"
] |
none
| 500
|
[
{
"input": "4\n2 3 4 1",
"output": "4 1 2 3"
},
{
"input": "3\n1 3 2",
"output": "1 3 2"
},
{
"input": "2\n1 2",
"output": "1 2"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "10\n1 3 2 6 4 5 7 9 8 10",
"output": "1 3 2 5 6 4 7 9 8 10"
},
{
"input": "5\n5 4 3 2 1",
"output": "5 4 3 2 1"
},
{
"input": "20\n2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19"
},
{
"input": "21\n3 2 1 6 5 4 9 8 7 12 11 10 15 14 13 18 17 16 21 20 19",
"output": "3 2 1 6 5 4 9 8 7 12 11 10 15 14 13 18 17 16 21 20 19"
},
{
"input": "10\n3 4 5 6 7 8 9 10 1 2",
"output": "9 10 1 2 3 4 5 6 7 8"
},
{
"input": "8\n1 5 3 7 2 6 4 8",
"output": "1 5 3 7 2 6 4 8"
},
{
"input": "50\n49 22 4 2 20 46 7 32 5 19 48 24 26 15 45 21 44 11 50 43 39 17 31 1 42 34 3 27 36 25 12 30 13 33 28 35 18 6 8 37 38 14 10 9 29 16 40 23 41 47",
"output": "24 4 27 3 9 38 7 39 44 43 18 31 33 42 14 46 22 37 10 5 16 2 48 12 30 13 28 35 45 32 23 8 34 26 36 29 40 41 21 47 49 25 20 17 15 6 50 11 1 19"
},
{
"input": "34\n13 20 33 30 15 11 27 4 8 2 29 25 24 7 3 22 18 10 26 16 5 1 32 9 34 6 12 14 28 19 31 21 23 17",
"output": "22 10 15 8 21 26 14 9 24 18 6 27 1 28 5 20 34 17 30 2 32 16 33 13 12 19 7 29 11 4 31 23 3 25"
},
{
"input": "92\n23 1 6 4 84 54 44 76 63 34 61 20 48 13 28 78 26 46 90 72 24 55 91 89 53 38 82 5 79 92 29 32 15 64 11 88 60 70 7 66 18 59 8 57 19 16 42 21 80 71 62 27 75 86 36 9 83 73 74 50 43 31 56 30 17 33 40 81 49 12 10 41 22 77 25 68 51 2 47 3 58 69 87 67 39 37 35 65 14 45 52 85",
"output": "2 78 80 4 28 3 39 43 56 71 35 70 14 89 33 46 65 41 45 12 48 73 1 21 75 17 52 15 31 64 62 32 66 10 87 55 86 26 85 67 72 47 61 7 90 18 79 13 69 60 77 91 25 6 22 63 44 81 42 37 11 51 9 34 88 40 84 76 82 38 50 20 58 59 53 8 74 16 29 49 68 27 57 5 92 54 83 36 24 19 23 30"
},
{
"input": "49\n30 24 33 48 7 3 17 2 8 35 10 39 23 40 46 32 18 21 26 22 1 16 47 45 41 28 31 6 12 43 27 11 13 37 19 15 44 5 29 42 4 38 20 34 14 9 25 36 49",
"output": "21 8 6 41 38 28 5 9 46 11 32 29 33 45 36 22 7 17 35 43 18 20 13 2 47 19 31 26 39 1 27 16 3 44 10 48 34 42 12 14 25 40 30 37 24 15 23 4 49"
},
{
"input": "12\n3 8 7 4 6 5 2 1 11 9 10 12",
"output": "8 7 1 4 6 5 3 2 10 11 9 12"
},
{
"input": "78\n16 56 36 78 21 14 9 77 26 57 70 61 41 47 18 44 5 31 50 74 65 52 6 39 22 62 67 69 43 7 64 29 24 40 48 51 73 54 72 12 19 34 4 25 55 33 17 35 23 53 10 8 27 32 42 68 20 63 3 2 1 71 58 46 13 30 49 11 37 66 38 60 28 75 15 59 45 76",
"output": "61 60 59 43 17 23 30 52 7 51 68 40 65 6 75 1 47 15 41 57 5 25 49 33 44 9 53 73 32 66 18 54 46 42 48 3 69 71 24 34 13 55 29 16 77 64 14 35 67 19 36 22 50 38 45 2 10 63 76 72 12 26 58 31 21 70 27 56 28 11 62 39 37 20 74 78 8 4"
},
{
"input": "64\n64 57 40 3 15 8 62 18 33 59 51 19 22 13 4 37 47 45 50 35 63 11 58 42 46 21 7 2 41 48 32 23 28 38 17 12 24 27 49 31 60 6 30 25 61 52 26 54 9 14 29 20 44 39 55 10 34 16 5 56 1 36 53 43",
"output": "61 28 4 15 59 42 27 6 49 56 22 36 14 50 5 58 35 8 12 52 26 13 32 37 44 47 38 33 51 43 40 31 9 57 20 62 16 34 54 3 29 24 64 53 18 25 17 30 39 19 11 46 63 48 55 60 2 23 10 41 45 7 21 1"
},
{
"input": "49\n38 20 49 32 14 41 39 45 25 48 40 19 26 43 34 12 10 3 35 42 5 7 46 47 4 2 13 22 16 24 33 15 11 18 29 31 23 9 44 36 6 17 37 1 30 28 8 21 27",
"output": "44 26 18 25 21 41 22 47 38 17 33 16 27 5 32 29 42 34 12 2 48 28 37 30 9 13 49 46 35 45 36 4 31 15 19 40 43 1 7 11 6 20 14 39 8 23 24 10 3"
},
{
"input": "78\n17 50 30 48 33 12 42 4 18 53 76 67 38 3 20 72 51 55 60 63 46 10 57 45 54 32 24 62 8 11 35 44 65 74 58 28 2 6 56 52 39 23 47 49 61 1 66 41 15 77 7 27 78 13 14 34 5 31 37 21 40 16 29 69 59 43 64 36 70 19 25 73 71 75 9 68 26 22",
"output": "46 37 14 8 57 38 51 29 75 22 30 6 54 55 49 62 1 9 70 15 60 78 42 27 71 77 52 36 63 3 58 26 5 56 31 68 59 13 41 61 48 7 66 32 24 21 43 4 44 2 17 40 10 25 18 39 23 35 65 19 45 28 20 67 33 47 12 76 64 69 73 16 72 34 74 11 50 53"
},
{
"input": "29\n14 21 27 1 4 18 10 17 20 23 2 24 7 9 28 22 8 25 12 15 11 6 16 29 3 26 19 5 13",
"output": "4 11 25 5 28 22 13 17 14 7 21 19 29 1 20 23 8 6 27 9 2 16 10 12 18 26 3 15 24"
},
{
"input": "82\n6 1 10 75 28 66 61 81 78 63 17 19 58 34 49 12 67 50 41 44 3 15 59 38 51 72 36 11 46 29 18 64 27 23 13 53 56 68 2 25 47 40 69 54 42 5 60 55 4 16 24 79 57 20 7 73 32 80 76 52 82 37 26 31 65 8 39 62 33 71 30 9 77 43 48 74 70 22 14 45 35 21",
"output": "2 39 21 49 46 1 55 66 72 3 28 16 35 79 22 50 11 31 12 54 82 78 34 51 40 63 33 5 30 71 64 57 69 14 81 27 62 24 67 42 19 45 74 20 80 29 41 75 15 18 25 60 36 44 48 37 53 13 23 47 7 68 10 32 65 6 17 38 43 77 70 26 56 76 4 59 73 9 52 58 8 61"
},
{
"input": "82\n74 18 15 69 71 77 19 26 80 20 66 7 30 82 22 48 21 44 52 65 64 61 35 49 12 8 53 81 54 16 11 9 40 46 13 1 29 58 5 41 55 4 78 60 6 51 56 2 38 36 34 62 63 25 17 67 45 14 32 37 75 79 10 47 27 39 31 68 59 24 50 43 72 70 42 28 76 23 57 3 73 33",
"output": "36 48 80 42 39 45 12 26 32 63 31 25 35 58 3 30 55 2 7 10 17 15 78 70 54 8 65 76 37 13 67 59 82 51 23 50 60 49 66 33 40 75 72 18 57 34 64 16 24 71 46 19 27 29 41 47 79 38 69 44 22 52 53 21 20 11 56 68 4 74 5 73 81 1 61 77 6 43 62 9 28 14"
},
{
"input": "45\n2 32 34 13 3 15 16 33 22 12 31 38 42 14 27 7 36 8 4 19 45 41 5 35 10 11 39 20 29 44 17 9 6 40 37 28 25 21 1 30 24 18 43 26 23",
"output": "39 1 5 19 23 33 16 18 32 25 26 10 4 14 6 7 31 42 20 28 38 9 45 41 37 44 15 36 29 40 11 2 8 3 24 17 35 12 27 34 22 13 43 30 21"
},
{
"input": "45\n4 32 33 39 43 21 22 35 45 7 14 5 16 9 42 31 24 36 17 29 41 25 37 34 27 20 11 44 3 13 19 2 1 10 26 30 38 18 6 8 15 23 40 28 12",
"output": "33 32 29 1 12 39 10 40 14 34 27 45 30 11 41 13 19 38 31 26 6 7 42 17 22 35 25 44 20 36 16 2 3 24 8 18 23 37 4 43 21 15 5 28 9"
},
{
"input": "74\n48 72 40 67 17 4 27 53 11 32 25 9 74 2 41 24 56 22 14 21 33 5 18 55 20 7 29 36 69 13 52 19 38 30 68 59 66 34 63 6 47 45 54 44 62 12 50 71 16 10 8 64 57 73 46 26 49 42 3 23 35 1 61 39 70 60 65 43 15 28 37 51 58 31",
"output": "62 14 59 6 22 40 26 51 12 50 9 46 30 19 69 49 5 23 32 25 20 18 60 16 11 56 7 70 27 34 74 10 21 38 61 28 71 33 64 3 15 58 68 44 42 55 41 1 57 47 72 31 8 43 24 17 53 73 36 66 63 45 39 52 67 37 4 35 29 65 48 2 54 13"
},
{
"input": "47\n9 26 27 10 6 34 28 42 39 22 45 21 11 43 14 47 38 15 40 32 46 1 36 29 17 25 2 23 31 5 24 4 7 8 12 19 16 44 37 20 18 33 30 13 35 41 3",
"output": "22 27 47 32 30 5 33 34 1 4 13 35 44 15 18 37 25 41 36 40 12 10 28 31 26 2 3 7 24 43 29 20 42 6 45 23 39 17 9 19 46 8 14 38 11 21 16"
},
{
"input": "49\n14 38 6 29 9 49 36 43 47 3 44 20 34 15 7 11 1 28 12 40 16 37 31 10 42 41 33 21 18 30 5 27 17 35 25 26 45 19 2 13 23 32 4 22 46 48 24 39 8",
"output": "17 39 10 43 31 3 15 49 5 24 16 19 40 1 14 21 33 29 38 12 28 44 41 47 35 36 32 18 4 30 23 42 27 13 34 7 22 2 48 20 26 25 8 11 37 45 9 46 6"
},
{
"input": "100\n78 56 31 91 90 95 16 65 58 77 37 89 33 61 10 76 62 47 35 67 69 7 63 83 22 25 49 8 12 30 39 44 57 64 48 42 32 11 70 43 55 50 99 24 85 73 45 14 54 21 98 84 74 2 26 18 9 36 80 53 75 46 66 86 59 93 87 68 94 13 72 28 79 88 92 29 52 82 34 97 19 38 1 41 27 4 40 5 96 100 51 6 20 23 81 15 17 3 60 71",
"output": "83 54 98 86 88 92 22 28 57 15 38 29 70 48 96 7 97 56 81 93 50 25 94 44 26 55 85 72 76 30 3 37 13 79 19 58 11 82 31 87 84 36 40 32 47 62 18 35 27 42 91 77 60 49 41 2 33 9 65 99 14 17 23 34 8 63 20 68 21 39 100 71 46 53 61 16 10 1 73 59 95 78 24 52 45 64 67 74 12 5 4 75 66 69 6 89 80 51 43 90"
},
{
"input": "22\n12 8 11 2 16 7 13 6 22 21 20 10 4 14 18 1 5 15 3 19 17 9",
"output": "16 4 19 13 17 8 6 2 22 12 3 1 7 14 18 5 21 15 20 11 10 9"
},
{
"input": "72\n16 11 49 51 3 27 60 55 23 40 66 7 53 70 13 5 15 32 18 72 33 30 8 31 46 12 28 67 25 38 50 22 69 34 71 52 58 39 24 35 42 9 41 26 62 1 63 65 36 64 68 61 37 14 45 47 6 57 54 20 17 2 56 59 29 10 4 48 21 43 19 44",
"output": "46 62 5 67 16 57 12 23 42 66 2 26 15 54 17 1 61 19 71 60 69 32 9 39 29 44 6 27 65 22 24 18 21 34 40 49 53 30 38 10 43 41 70 72 55 25 56 68 3 31 4 36 13 59 8 63 58 37 64 7 52 45 47 50 48 11 28 51 33 14 35 20"
},
{
"input": "63\n21 56 11 10 62 24 20 42 28 52 38 2 37 43 48 22 7 8 40 14 13 46 53 1 23 4 60 63 51 36 25 12 39 32 49 16 58 44 31 61 33 50 55 54 45 6 47 41 9 57 30 29 26 18 19 27 15 34 3 35 59 5 17",
"output": "24 12 59 26 62 46 17 18 49 4 3 32 21 20 57 36 63 54 55 7 1 16 25 6 31 53 56 9 52 51 39 34 41 58 60 30 13 11 33 19 48 8 14 38 45 22 47 15 35 42 29 10 23 44 43 2 50 37 61 27 40 5 28"
},
{
"input": "18\n2 16 8 4 18 12 3 6 5 9 10 15 11 17 14 13 1 7",
"output": "17 1 7 4 9 8 18 3 10 11 13 6 16 15 12 2 14 5"
},
{
"input": "47\n6 9 10 41 25 3 4 37 20 1 36 22 29 27 11 24 43 31 12 17 34 42 38 39 13 2 7 21 18 5 15 35 44 26 33 46 19 40 30 14 28 23 47 32 45 8 16",
"output": "10 26 6 7 30 1 27 46 2 3 15 19 25 40 31 47 20 29 37 9 28 12 42 16 5 34 14 41 13 39 18 44 35 21 32 11 8 23 24 38 4 22 17 33 45 36 43"
},
{
"input": "96\n41 91 48 88 29 57 1 19 44 43 37 5 10 75 25 63 30 78 76 53 8 92 18 70 39 17 49 60 9 16 3 34 86 59 23 79 55 45 72 51 28 33 96 40 26 54 6 32 89 61 85 74 7 82 52 31 64 66 94 95 11 22 2 73 35 13 42 71 14 47 84 69 50 67 58 12 77 46 38 68 15 36 20 93 27 90 83 56 87 4 21 24 81 62 80 65",
"output": "7 63 31 90 12 47 53 21 29 13 61 76 66 69 81 30 26 23 8 83 91 62 35 92 15 45 85 41 5 17 56 48 42 32 65 82 11 79 25 44 1 67 10 9 38 78 70 3 27 73 40 55 20 46 37 88 6 75 34 28 50 94 16 57 96 58 74 80 72 24 68 39 64 52 14 19 77 18 36 95 93 54 87 71 51 33 89 4 49 86 2 22 84 59 60 43"
},
{
"input": "73\n67 24 39 22 23 20 48 34 42 40 19 70 65 69 64 21 53 11 59 15 26 10 30 33 72 29 55 25 56 71 8 9 57 49 41 61 13 12 6 27 66 36 47 50 73 60 2 37 7 4 51 17 1 46 14 62 35 3 45 63 43 58 54 32 31 5 28 44 18 52 68 38 16",
"output": "53 47 58 50 66 39 49 31 32 22 18 38 37 55 20 73 52 69 11 6 16 4 5 2 28 21 40 67 26 23 65 64 24 8 57 42 48 72 3 10 35 9 61 68 59 54 43 7 34 44 51 70 17 63 27 29 33 62 19 46 36 56 60 15 13 41 1 71 14 12 30 25 45"
},
{
"input": "81\n25 2 78 40 12 80 69 13 49 43 17 33 23 54 32 61 77 66 27 71 24 26 42 55 60 9 5 30 7 37 45 63 53 11 38 44 68 34 28 52 67 22 57 46 47 50 8 16 79 62 4 36 20 14 73 64 6 76 35 74 58 10 29 81 59 31 19 1 75 39 70 18 41 21 72 65 3 48 15 56 51",
"output": "68 2 77 51 27 57 29 47 26 62 34 5 8 54 79 48 11 72 67 53 74 42 13 21 1 22 19 39 63 28 66 15 12 38 59 52 30 35 70 4 73 23 10 36 31 44 45 78 9 46 81 40 33 14 24 80 43 61 65 25 16 50 32 56 76 18 41 37 7 71 20 75 55 60 69 58 17 3 49 6 64"
},
{
"input": "12\n12 3 1 5 11 6 7 10 2 8 9 4",
"output": "3 9 2 12 4 6 7 10 11 8 5 1"
},
{
"input": "47\n7 21 41 18 40 31 12 28 24 14 43 23 33 10 19 38 26 8 34 15 29 44 5 13 39 25 3 27 20 42 35 9 2 1 30 46 36 32 4 22 37 45 6 47 11 16 17",
"output": "34 33 27 39 23 43 1 18 32 14 45 7 24 10 20 46 47 4 15 29 2 40 12 9 26 17 28 8 21 35 6 38 13 19 31 37 41 16 25 5 3 30 11 22 42 36 44"
},
{
"input": "8\n1 3 5 2 4 8 6 7",
"output": "1 4 2 5 3 7 8 6"
},
{
"input": "38\n28 8 2 33 20 32 26 29 23 31 15 38 11 37 18 21 22 19 4 34 1 35 16 7 17 6 27 30 36 12 9 24 25 13 5 3 10 14",
"output": "21 3 36 19 35 26 24 2 31 37 13 30 34 38 11 23 25 15 18 5 16 17 9 32 33 7 27 1 8 28 10 6 4 20 22 29 14 12"
},
{
"input": "10\n2 9 4 6 10 1 7 5 3 8",
"output": "6 1 9 3 8 4 7 10 2 5"
},
{
"input": "23\n20 11 15 1 5 12 23 9 2 22 13 19 16 14 7 4 8 21 6 17 18 10 3",
"output": "4 9 23 16 5 19 15 17 8 22 2 6 11 14 3 13 20 21 12 1 18 10 7"
},
{
"input": "10\n2 4 9 3 6 8 10 5 1 7",
"output": "9 1 4 2 8 5 10 6 3 7"
},
{
"input": "55\n9 48 23 49 11 24 4 22 34 32 17 45 39 13 14 21 19 25 2 31 37 7 55 36 20 51 5 12 54 10 35 40 43 1 46 18 53 41 38 26 29 50 3 42 52 27 8 28 47 33 6 16 30 44 15",
"output": "34 19 43 7 27 51 22 47 1 30 5 28 14 15 55 52 11 36 17 25 16 8 3 6 18 40 46 48 41 53 20 10 50 9 31 24 21 39 13 32 38 44 33 54 12 35 49 2 4 42 26 45 37 29 23"
},
{
"input": "58\n49 13 12 54 2 38 56 11 33 25 26 19 28 8 23 41 20 36 46 55 15 35 9 7 32 37 58 6 3 14 47 31 40 30 53 44 4 50 29 34 10 43 39 57 5 22 27 45 51 42 24 16 18 21 52 17 48 1",
"output": "58 5 29 37 45 28 24 14 23 41 8 3 2 30 21 52 56 53 12 17 54 46 15 51 10 11 47 13 39 34 32 25 9 40 22 18 26 6 43 33 16 50 42 36 48 19 31 57 1 38 49 55 35 4 20 7 44 27"
},
{
"input": "34\n20 25 2 3 33 29 1 16 14 7 21 9 32 31 6 26 22 4 27 23 24 10 34 12 19 15 5 18 28 17 13 8 11 30",
"output": "7 3 4 18 27 15 10 32 12 22 33 24 31 9 26 8 30 28 25 1 11 17 20 21 2 16 19 29 6 34 14 13 5 23"
},
{
"input": "53\n47 29 46 25 23 13 7 31 33 4 38 11 35 16 42 14 15 43 34 39 28 18 6 45 30 1 40 20 2 37 5 32 24 12 44 26 27 3 19 51 36 21 22 9 10 50 41 48 49 53 8 17 52",
"output": "26 29 38 10 31 23 7 51 44 45 12 34 6 16 17 14 52 22 39 28 42 43 5 33 4 36 37 21 2 25 8 32 9 19 13 41 30 11 20 27 47 15 18 35 24 3 1 48 49 46 40 53 50"
},
{
"input": "99\n77 87 90 48 53 38 68 6 28 57 35 82 63 71 60 41 3 12 86 65 10 59 22 67 33 74 93 27 24 1 61 43 25 4 51 52 15 88 9 31 30 42 89 49 23 21 29 32 46 73 37 16 5 69 56 26 92 64 20 54 75 14 98 13 94 2 95 7 36 66 58 8 50 78 84 45 11 96 76 62 97 80 40 39 47 85 34 79 83 17 91 72 19 44 70 81 55 99 18",
"output": "30 66 17 34 53 8 68 72 39 21 77 18 64 62 37 52 90 99 93 59 46 23 45 29 33 56 28 9 47 41 40 48 25 87 11 69 51 6 84 83 16 42 32 94 76 49 85 4 44 73 35 36 5 60 97 55 10 71 22 15 31 80 13 58 20 70 24 7 54 95 14 92 50 26 61 79 1 74 88 82 96 12 89 75 86 19 2 38 43 3 91 57 27 65 67 78 81 63 98"
},
{
"input": "32\n17 29 2 6 30 8 26 7 1 27 10 9 13 24 31 21 15 19 22 18 4 11 25 28 32 3 23 12 5 14 20 16",
"output": "9 3 26 21 29 4 8 6 12 11 22 28 13 30 17 32 1 20 18 31 16 19 27 14 23 7 10 24 2 5 15 25"
},
{
"input": "65\n18 40 1 60 17 19 4 6 12 49 28 58 2 25 13 14 64 56 61 34 62 30 59 51 26 8 33 63 36 48 46 7 43 21 31 27 11 44 29 5 32 23 35 9 53 57 52 50 15 38 42 3 54 65 55 41 20 24 22 47 45 10 39 16 37",
"output": "3 13 52 7 40 8 32 26 44 62 37 9 15 16 49 64 5 1 6 57 34 59 42 58 14 25 36 11 39 22 35 41 27 20 43 29 65 50 63 2 56 51 33 38 61 31 60 30 10 48 24 47 45 53 55 18 46 12 23 4 19 21 28 17 54"
},
{
"input": "71\n35 50 55 58 25 32 26 40 63 34 44 53 24 18 37 7 64 27 56 65 1 19 2 43 42 14 57 47 22 13 59 61 39 67 30 45 54 38 33 48 6 5 3 69 36 21 41 4 16 46 20 17 15 12 10 70 68 23 60 31 52 29 66 28 51 49 62 11 8 9 71",
"output": "21 23 43 48 42 41 16 69 70 55 68 54 30 26 53 49 52 14 22 51 46 29 58 13 5 7 18 64 62 35 60 6 39 10 1 45 15 38 33 8 47 25 24 11 36 50 28 40 66 2 65 61 12 37 3 19 27 4 31 59 32 67 9 17 20 63 34 57 44 56 71"
},
{
"input": "74\n33 8 42 63 64 61 31 74 11 50 68 14 36 25 57 30 7 44 21 15 6 9 23 59 46 3 73 16 62 51 40 60 41 54 5 39 35 28 48 4 58 12 66 69 13 26 71 1 24 19 29 52 37 2 20 43 18 72 17 56 34 38 65 67 27 10 47 70 53 32 45 55 49 22",
"output": "48 54 26 40 35 21 17 2 22 66 9 42 45 12 20 28 59 57 50 55 19 74 23 49 14 46 65 38 51 16 7 70 1 61 37 13 53 62 36 31 33 3 56 18 71 25 67 39 73 10 30 52 69 34 72 60 15 41 24 32 6 29 4 5 63 43 64 11 44 68 47 58 27 8"
},
{
"input": "96\n78 10 82 46 38 91 77 69 2 27 58 80 79 44 59 41 6 31 76 11 42 48 51 37 19 87 43 25 52 32 1 39 63 29 21 65 53 74 92 16 15 95 90 83 30 73 71 5 50 17 96 33 86 60 67 64 20 26 61 40 55 88 94 93 9 72 47 57 14 45 22 3 54 68 13 24 4 7 56 81 89 70 49 8 84 28 18 62 35 36 75 23 66 85 34 12",
"output": "31 9 72 77 48 17 78 84 65 2 20 96 75 69 41 40 50 87 25 57 35 71 92 76 28 58 10 86 34 45 18 30 52 95 89 90 24 5 32 60 16 21 27 14 70 4 67 22 83 49 23 29 37 73 61 79 68 11 15 54 59 88 33 56 36 93 55 74 8 82 47 66 46 38 91 19 7 1 13 12 80 3 44 85 94 53 26 62 81 43 6 39 64 63 42 51"
},
{
"input": "7\n2 1 5 7 3 4 6",
"output": "2 1 5 6 3 7 4"
},
{
"input": "51\n8 33 37 2 16 22 24 30 4 9 5 15 27 3 18 39 31 26 10 17 46 41 25 14 6 1 29 48 36 20 51 49 21 43 19 13 38 50 47 34 11 23 28 12 42 7 32 40 44 45 35",
"output": "26 4 14 9 11 25 46 1 10 19 41 44 36 24 12 5 20 15 35 30 33 6 42 7 23 18 13 43 27 8 17 47 2 40 51 29 3 37 16 48 22 45 34 49 50 21 39 28 32 38 31"
},
{
"input": "27\n12 14 7 3 20 21 25 13 22 15 23 4 2 24 10 17 19 8 26 11 27 18 9 5 6 1 16",
"output": "26 13 4 12 24 25 3 18 23 15 20 1 8 2 10 27 16 22 17 5 6 9 11 14 7 19 21"
},
{
"input": "71\n51 13 20 48 54 23 24 64 14 62 71 67 57 53 3 30 55 43 33 25 39 40 66 6 46 18 5 19 61 16 32 68 70 41 60 44 29 49 27 69 50 38 10 17 45 56 9 21 26 63 28 35 7 59 1 65 2 15 8 11 12 34 37 47 58 22 31 4 36 42 52",
"output": "55 57 15 68 27 24 53 59 47 43 60 61 2 9 58 30 44 26 28 3 48 66 6 7 20 49 39 51 37 16 67 31 19 62 52 69 63 42 21 22 34 70 18 36 45 25 64 4 38 41 1 71 14 5 17 46 13 65 54 35 29 10 50 8 56 23 12 32 40 33 11"
},
{
"input": "9\n8 5 2 6 1 9 4 7 3",
"output": "5 3 9 7 2 4 8 1 6"
},
{
"input": "29\n10 24 11 5 26 25 2 9 22 15 8 14 29 21 4 1 23 17 3 12 13 16 18 28 19 20 7 6 27",
"output": "16 7 19 15 4 28 27 11 8 1 3 20 21 12 10 22 18 23 25 26 14 9 17 2 6 5 29 24 13"
},
{
"input": "60\n39 25 42 4 55 60 16 18 47 1 11 40 7 50 19 35 49 54 12 3 30 38 2 58 17 26 45 6 33 43 37 32 52 36 15 23 27 59 24 20 28 14 8 9 13 29 44 46 41 21 5 48 51 22 31 56 57 53 10 34",
"output": "10 23 20 4 51 28 13 43 44 59 11 19 45 42 35 7 25 8 15 40 50 54 36 39 2 26 37 41 46 21 55 32 29 60 16 34 31 22 1 12 49 3 30 47 27 48 9 52 17 14 53 33 58 18 5 56 57 24 38 6"
},
{
"input": "50\n37 45 22 5 12 21 28 24 18 47 20 25 8 50 14 2 34 43 11 16 49 41 48 1 19 31 39 46 32 23 15 42 3 35 38 30 44 26 10 9 40 36 7 17 33 4 27 6 13 29",
"output": "24 16 33 46 4 48 43 13 40 39 19 5 49 15 31 20 44 9 25 11 6 3 30 8 12 38 47 7 50 36 26 29 45 17 34 42 1 35 27 41 22 32 18 37 2 28 10 23 21 14"
},
{
"input": "30\n8 29 28 16 17 25 27 15 21 11 6 20 2 13 1 30 5 4 24 10 14 3 23 18 26 9 12 22 19 7",
"output": "15 13 22 18 17 11 30 1 26 20 10 27 14 21 8 4 5 24 29 12 9 28 23 19 6 25 7 3 2 16"
},
{
"input": "46\n15 2 44 43 38 19 31 42 4 37 29 30 24 45 27 41 8 20 33 7 35 3 18 46 36 26 1 28 21 40 16 22 32 11 14 13 12 9 25 39 10 6 23 17 5 34",
"output": "27 2 22 9 45 42 20 17 38 41 34 37 36 35 1 31 44 23 6 18 29 32 43 13 39 26 15 28 11 12 7 33 19 46 21 25 10 5 40 30 16 8 4 3 14 24"
},
{
"input": "9\n4 8 6 5 3 9 2 7 1",
"output": "9 7 5 1 4 3 8 2 6"
},
{
"input": "46\n31 30 33 23 45 7 36 8 11 3 32 39 41 20 1 28 6 27 18 24 17 5 16 37 26 13 22 14 2 38 15 46 9 4 19 21 12 44 10 35 25 34 42 43 40 29",
"output": "15 29 10 34 22 17 6 8 33 39 9 37 26 28 31 23 21 19 35 14 36 27 4 20 41 25 18 16 46 2 1 11 3 42 40 7 24 30 12 45 13 43 44 38 5 32"
},
{
"input": "66\n27 12 37 48 46 21 34 58 38 28 66 2 64 32 44 31 13 36 40 15 19 11 22 5 30 29 6 7 61 39 20 42 23 54 51 33 50 9 60 8 57 45 49 10 62 41 59 3 55 63 52 24 25 26 43 56 65 4 16 14 1 35 18 17 53 47",
"output": "61 12 48 58 24 27 28 40 38 44 22 2 17 60 20 59 64 63 21 31 6 23 33 52 53 54 1 10 26 25 16 14 36 7 62 18 3 9 30 19 46 32 55 15 42 5 66 4 43 37 35 51 65 34 49 56 41 8 47 39 29 45 50 13 57 11"
},
{
"input": "13\n3 12 9 2 8 5 13 4 11 1 10 7 6",
"output": "10 4 1 8 6 13 12 5 3 11 9 2 7"
},
{
"input": "80\n21 25 56 50 20 61 7 74 51 69 8 2 46 57 45 71 14 52 17 43 9 30 70 78 31 10 38 13 23 15 37 79 6 16 77 73 80 4 49 48 18 28 26 58 33 41 64 22 54 72 59 60 40 63 53 27 1 5 75 67 62 34 19 39 68 65 44 55 3 32 11 42 76 12 35 47 66 36 24 29",
"output": "57 12 69 38 58 33 7 11 21 26 71 74 28 17 30 34 19 41 63 5 1 48 29 79 2 43 56 42 80 22 25 70 45 62 75 78 31 27 64 53 46 72 20 67 15 13 76 40 39 4 9 18 55 49 68 3 14 44 51 52 6 61 54 47 66 77 60 65 10 23 16 50 36 8 59 73 35 24 32 37"
},
{
"input": "63\n9 49 53 25 40 46 43 51 54 22 58 16 23 26 10 47 5 27 2 8 61 59 19 35 63 56 28 20 34 4 62 38 6 55 36 31 57 15 29 33 1 48 50 37 7 30 18 42 32 52 12 41 14 21 45 11 24 17 39 13 44 60 3",
"output": "41 19 63 30 17 33 45 20 1 15 56 51 60 53 38 12 58 47 23 28 54 10 13 57 4 14 18 27 39 46 36 49 40 29 24 35 44 32 59 5 52 48 7 61 55 6 16 42 2 43 8 50 3 9 34 26 37 11 22 62 21 31 25"
},
{
"input": "26\n11 4 19 13 17 9 2 24 6 5 22 23 14 15 3 25 16 8 18 10 21 1 12 26 7 20",
"output": "22 7 15 2 10 9 25 18 6 20 1 23 4 13 14 17 5 19 3 26 21 11 12 8 16 24"
},
{
"input": "69\n40 22 11 66 4 27 31 29 64 53 37 55 51 2 7 36 18 52 6 1 30 21 17 20 14 9 59 62 49 68 3 50 65 57 44 5 67 46 33 13 34 15 24 48 63 58 38 25 41 35 16 54 32 10 60 61 39 12 69 8 23 45 26 47 56 43 28 19 42",
"output": "20 14 31 5 36 19 15 60 26 54 3 58 40 25 42 51 23 17 68 24 22 2 61 43 48 63 6 67 8 21 7 53 39 41 50 16 11 47 57 1 49 69 66 35 62 38 64 44 29 32 13 18 10 52 12 65 34 46 27 55 56 28 45 9 33 4 37 30 59"
},
{
"input": "6\n4 3 6 5 1 2",
"output": "5 6 2 1 4 3"
},
{
"input": "9\n7 8 5 3 1 4 2 9 6",
"output": "5 7 4 6 3 9 1 2 8"
},
{
"input": "41\n27 24 16 30 25 8 32 2 26 20 39 33 41 22 40 14 36 9 28 4 34 11 31 23 19 18 17 35 3 10 6 13 5 15 29 38 7 21 1 12 37",
"output": "39 8 29 20 33 31 37 6 18 30 22 40 32 16 34 3 27 26 25 10 38 14 24 2 5 9 1 19 35 4 23 7 12 21 28 17 41 36 11 15 13"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "20\n2 6 4 18 7 10 17 13 16 8 14 9 20 5 19 12 1 3 15 11",
"output": "17 1 18 3 14 2 5 10 12 6 20 16 8 11 19 9 7 4 15 13"
},
{
"input": "2\n2 1",
"output": "2 1"
},
{
"input": "60\n2 4 31 51 11 7 34 20 3 14 18 23 48 54 15 36 38 60 49 40 5 33 41 26 55 58 10 8 13 9 27 30 37 1 21 59 44 57 35 19 46 43 42 45 12 22 39 32 24 16 6 56 53 52 25 17 47 29 50 28",
"output": "34 1 9 2 21 51 6 28 30 27 5 45 29 10 15 50 56 11 40 8 35 46 12 49 55 24 31 60 58 32 3 48 22 7 39 16 33 17 47 20 23 43 42 37 44 41 57 13 19 59 4 54 53 14 25 52 38 26 36 18"
},
{
"input": "14\n14 6 3 12 11 2 7 1 10 9 8 5 4 13",
"output": "8 6 3 13 12 2 7 11 10 9 5 4 14 1"
},
{
"input": "81\n13 43 79 8 7 21 73 46 63 4 62 78 56 11 70 68 61 53 60 49 16 27 59 47 69 5 22 44 77 57 52 48 1 9 72 81 28 55 58 33 51 18 31 17 41 20 42 3 32 54 19 2 75 34 64 10 65 50 30 29 67 12 71 66 74 15 26 23 6 38 25 35 37 24 80 76 40 45 39 36 14",
"output": "33 52 48 10 26 69 5 4 34 56 14 62 1 81 66 21 44 42 51 46 6 27 68 74 71 67 22 37 60 59 43 49 40 54 72 80 73 70 79 77 45 47 2 28 78 8 24 32 20 58 41 31 18 50 38 13 30 39 23 19 17 11 9 55 57 64 61 16 25 15 63 35 7 65 53 76 29 12 3 75 36"
},
{
"input": "42\n41 11 10 8 21 37 32 19 31 25 1 15 36 5 6 27 4 3 13 7 16 17 2 23 34 24 38 28 12 20 30 42 18 26 39 35 33 40 9 14 22 29",
"output": "11 23 18 17 14 15 20 4 39 3 2 29 19 40 12 21 22 33 8 30 5 41 24 26 10 34 16 28 42 31 9 7 37 25 36 13 6 27 35 38 1 32"
},
{
"input": "97\n20 6 76 42 4 18 35 59 39 63 27 7 66 47 61 52 15 36 88 93 19 33 10 92 1 34 46 86 78 57 51 94 77 29 26 73 41 2 58 97 43 65 17 74 21 49 25 3 91 82 95 12 96 13 84 90 69 24 72 37 16 55 54 71 64 62 48 89 11 70 80 67 30 40 44 85 53 83 79 9 56 45 75 87 22 14 81 68 8 38 60 50 28 23 31 32 5",
"output": "25 38 48 5 97 2 12 89 80 23 69 52 54 86 17 61 43 6 21 1 45 85 94 58 47 35 11 93 34 73 95 96 22 26 7 18 60 90 9 74 37 4 41 75 82 27 14 67 46 92 31 16 77 63 62 81 30 39 8 91 15 66 10 65 42 13 72 88 57 70 64 59 36 44 83 3 33 29 79 71 87 50 78 55 76 28 84 19 68 56 49 24 20 32 51 53 40"
},
{
"input": "62\n15 27 46 6 8 51 14 56 23 48 42 49 52 22 20 31 29 12 47 3 62 34 37 35 32 57 19 25 5 60 61 38 18 10 11 55 45 53 17 30 9 36 4 50 41 16 44 28 40 59 24 1 13 39 26 7 33 58 2 43 21 54",
"output": "52 59 20 43 29 4 56 5 41 34 35 18 53 7 1 46 39 33 27 15 61 14 9 51 28 55 2 48 17 40 16 25 57 22 24 42 23 32 54 49 45 11 60 47 37 3 19 10 12 44 6 13 38 62 36 8 26 58 50 30 31 21"
},
{
"input": "61\n35 27 4 61 52 32 41 46 14 37 17 54 55 31 11 26 44 49 15 30 9 50 45 39 7 38 53 3 58 40 13 56 18 19 28 6 43 5 21 42 20 34 2 25 36 12 33 57 16 60 1 8 59 10 22 23 24 48 51 47 29",
"output": "51 43 28 3 38 36 25 52 21 54 15 46 31 9 19 49 11 33 34 41 39 55 56 57 44 16 2 35 61 20 14 6 47 42 1 45 10 26 24 30 7 40 37 17 23 8 60 58 18 22 59 5 27 12 13 32 48 29 53 50 4"
},
{
"input": "59\n31 26 36 15 17 19 10 53 11 34 13 46 55 9 44 7 8 37 32 52 47 25 51 22 35 39 41 4 43 24 5 27 20 57 6 38 3 28 21 40 50 18 14 56 33 45 12 2 49 59 54 29 16 48 42 58 1 30 23",
"output": "57 48 37 28 31 35 16 17 14 7 9 47 11 43 4 53 5 42 6 33 39 24 59 30 22 2 32 38 52 58 1 19 45 10 25 3 18 36 26 40 27 55 29 15 46 12 21 54 49 41 23 20 8 51 13 44 34 56 50"
},
{
"input": "10\n2 10 7 4 1 5 8 6 3 9",
"output": "5 1 9 4 6 8 3 7 10 2"
},
{
"input": "14\n14 2 1 8 6 12 11 10 9 7 3 4 5 13",
"output": "3 2 11 12 13 5 10 4 9 8 7 6 14 1"
},
{
"input": "43\n28 38 15 14 31 42 27 30 19 33 43 26 22 29 18 32 3 13 1 8 35 34 4 12 11 17 41 21 5 25 39 37 20 23 7 24 16 10 40 9 6 36 2",
"output": "19 43 17 23 29 41 35 20 40 38 25 24 18 4 3 37 26 15 9 33 28 13 34 36 30 12 7 1 14 8 5 16 10 22 21 42 32 2 31 39 27 6 11"
},
{
"input": "86\n39 11 20 31 28 76 29 64 35 21 41 71 12 82 5 37 80 73 38 26 79 75 23 15 59 45 47 6 3 62 50 49 51 22 2 65 86 60 70 42 74 17 1 30 55 44 8 66 81 27 57 77 43 13 54 32 72 46 48 56 14 34 78 52 36 85 24 19 69 83 25 61 7 4 84 33 63 58 18 40 68 10 67 9 16 53",
"output": "43 35 29 74 15 28 73 47 84 82 2 13 54 61 24 85 42 79 68 3 10 34 23 67 71 20 50 5 7 44 4 56 76 62 9 65 16 19 1 80 11 40 53 46 26 58 27 59 32 31 33 64 86 55 45 60 51 78 25 38 72 30 77 8 36 48 83 81 69 39 12 57 18 41 22 6 52 63 21 17 49 14 70 75 66 37"
},
{
"input": "99\n65 78 56 98 33 24 61 40 29 93 1 64 57 22 25 52 67 95 50 3 31 15 90 68 71 83 38 36 6 46 89 26 4 87 14 88 72 37 23 43 63 12 80 96 5 34 73 86 9 48 92 62 99 10 16 20 66 27 28 2 82 70 30 94 49 8 84 69 18 60 58 59 44 39 21 7 91 76 54 19 75 85 74 47 55 32 97 77 51 13 35 79 45 42 11 41 17 81 53",
"output": "11 60 20 33 45 29 76 66 49 54 95 42 90 35 22 55 97 69 80 56 75 14 39 6 15 32 58 59 9 63 21 86 5 46 91 28 38 27 74 8 96 94 40 73 93 30 84 50 65 19 89 16 99 79 85 3 13 71 72 70 7 52 41 12 1 57 17 24 68 62 25 37 47 83 81 78 88 2 92 43 98 61 26 67 82 48 34 36 31 23 77 51 10 64 18 44 87 4 53"
},
{
"input": "100\n42 23 48 88 36 6 18 70 96 1 34 40 46 22 39 55 85 93 45 67 71 75 59 9 21 3 86 63 65 68 20 38 73 31 84 90 50 51 56 95 72 33 49 19 83 76 54 74 100 30 17 98 15 94 4 97 5 99 81 27 92 32 89 12 13 91 87 29 60 11 52 43 35 58 10 25 16 80 28 2 44 61 8 82 66 69 41 24 57 62 78 37 79 77 53 7 14 47 26 64",
"output": "10 80 26 55 57 6 96 83 24 75 70 64 65 97 53 77 51 7 44 31 25 14 2 88 76 99 60 79 68 50 34 62 42 11 73 5 92 32 15 12 87 1 72 81 19 13 98 3 43 37 38 71 95 47 16 39 89 74 23 69 82 90 28 100 29 85 20 30 86 8 21 41 33 48 22 46 94 91 93 78 59 84 45 35 17 27 67 4 63 36 66 61 18 54 40 9 56 52 58 49"
},
{
"input": "99\n8 68 94 75 71 60 57 58 6 11 5 48 65 41 49 12 46 72 95 59 13 70 74 7 84 62 17 36 55 76 38 79 2 85 23 10 32 99 87 50 83 28 54 91 53 51 1 3 97 81 21 89 93 78 61 26 82 96 4 98 25 40 31 44 24 47 30 52 14 16 39 27 9 29 45 18 67 63 37 43 90 66 19 69 88 22 92 77 34 42 73 80 56 64 20 35 15 33 86",
"output": "47 33 48 59 11 9 24 1 73 36 10 16 21 69 97 70 27 76 83 95 51 86 35 65 61 56 72 42 74 67 63 37 98 89 96 28 79 31 71 62 14 90 80 64 75 17 66 12 15 40 46 68 45 43 29 93 7 8 20 6 55 26 78 94 13 82 77 2 84 22 5 18 91 23 4 30 88 54 32 92 50 57 41 25 34 99 39 85 52 81 44 87 53 3 19 58 49 60 38"
},
{
"input": "99\n12 99 88 13 7 19 74 47 23 90 16 29 26 11 58 60 64 98 37 18 82 67 72 46 51 85 17 92 87 20 77 36 78 71 57 35 80 54 73 15 14 62 97 45 31 79 94 56 76 96 28 63 8 44 38 86 49 2 52 66 61 59 10 43 55 50 22 34 83 53 95 40 81 21 30 42 27 3 5 41 1 70 69 25 93 48 65 6 24 89 91 33 39 68 9 4 32 84 75",
"output": "81 58 78 96 79 88 5 53 95 63 14 1 4 41 40 11 27 20 6 30 74 67 9 89 84 13 77 51 12 75 45 97 92 68 36 32 19 55 93 72 80 76 64 54 44 24 8 86 57 66 25 59 70 38 65 48 35 15 62 16 61 42 52 17 87 60 22 94 83 82 34 23 39 7 99 49 31 33 46 37 73 21 69 98 26 56 29 3 90 10 91 28 85 47 71 50 43 18 2"
},
{
"input": "99\n20 79 26 75 99 69 98 47 93 62 18 42 43 38 90 66 67 8 13 84 76 58 81 60 64 46 56 23 78 17 86 36 19 52 85 39 48 27 96 49 37 95 5 31 10 24 12 1 80 35 92 33 16 68 57 54 32 29 45 88 72 77 4 87 97 89 59 3 21 22 61 94 83 15 44 34 70 91 55 9 51 50 73 11 14 6 40 7 63 25 2 82 41 65 28 74 71 30 53",
"output": "48 91 68 63 43 86 88 18 80 45 84 47 19 85 74 53 30 11 33 1 69 70 28 46 90 3 38 95 58 98 44 57 52 76 50 32 41 14 36 87 93 12 13 75 59 26 8 37 40 82 81 34 99 56 79 27 55 22 67 24 71 10 89 25 94 16 17 54 6 77 97 61 83 96 4 21 62 29 2 49 23 92 73 20 35 31 64 60 66 15 78 51 9 72 42 39 65 7 5"
},
{
"input": "99\n74 20 9 1 60 85 65 13 4 25 40 99 5 53 64 3 36 31 73 44 55 50 45 63 98 51 68 6 47 37 71 82 88 34 84 18 19 12 93 58 86 7 11 46 90 17 33 27 81 69 42 59 56 32 95 52 76 61 96 62 78 43 66 21 49 97 75 14 41 72 89 16 30 79 22 23 15 83 91 38 48 2 87 26 28 80 94 70 54 92 57 10 8 35 67 77 29 24 39",
"output": "4 82 16 9 13 28 42 93 3 92 43 38 8 68 77 72 46 36 37 2 64 75 76 98 10 84 48 85 97 73 18 54 47 34 94 17 30 80 99 11 69 51 62 20 23 44 29 81 65 22 26 56 14 89 21 53 91 40 52 5 58 60 24 15 7 63 95 27 50 88 31 70 19 1 67 57 96 61 74 86 49 32 78 35 6 41 83 33 71 45 79 90 39 87 55 59 66 25 12"
},
{
"input": "99\n50 94 2 18 69 90 59 83 75 68 77 97 39 78 25 7 16 9 49 4 42 89 44 48 17 96 61 70 3 10 5 81 56 57 88 6 98 1 46 67 92 37 11 30 85 41 8 36 51 29 20 71 19 79 74 93 43 34 55 40 38 21 64 63 32 24 72 14 12 86 82 15 65 23 66 22 28 53 13 26 95 99 91 52 76 27 60 45 47 33 73 84 31 35 54 80 58 62 87",
"output": "38 3 29 20 31 36 16 47 18 30 43 69 79 68 72 17 25 4 53 51 62 76 74 66 15 80 86 77 50 44 93 65 90 58 94 48 42 61 13 60 46 21 57 23 88 39 89 24 19 1 49 84 78 95 59 33 34 97 7 87 27 98 64 63 73 75 40 10 5 28 52 67 91 55 9 85 11 14 54 96 32 71 8 92 45 70 99 35 22 6 83 41 56 2 81 26 12 37 82"
},
{
"input": "99\n19 93 14 34 39 37 33 15 52 88 7 43 69 27 9 77 94 31 48 22 63 70 79 17 50 6 81 8 76 58 23 74 86 11 57 62 41 87 75 51 12 18 68 56 95 3 80 83 84 29 24 61 71 78 59 96 20 85 90 28 45 36 38 97 1 49 40 98 44 67 13 73 72 91 47 10 30 54 35 42 4 2 92 26 64 60 53 21 5 82 46 32 55 66 16 89 99 65 25",
"output": "65 82 46 81 89 26 11 28 15 76 34 41 71 3 8 95 24 42 1 57 88 20 31 51 99 84 14 60 50 77 18 92 7 4 79 62 6 63 5 67 37 80 12 69 61 91 75 19 66 25 40 9 87 78 93 44 35 30 55 86 52 36 21 85 98 94 70 43 13 22 53 73 72 32 39 29 16 54 23 47 27 90 48 49 58 33 38 10 96 59 74 83 2 17 45 56 64 68 97"
},
{
"input": "99\n86 25 50 51 62 39 41 67 44 20 45 14 80 88 66 7 36 59 13 84 78 58 96 75 2 43 48 47 69 12 19 98 22 38 28 55 11 76 68 46 53 70 85 34 16 33 91 30 8 40 74 60 94 82 87 32 37 4 5 10 89 73 90 29 35 26 23 57 27 65 24 3 9 83 77 72 6 31 15 92 93 79 64 18 63 42 56 1 52 97 17 81 71 21 49 99 54 95 61",
"output": "88 25 72 58 59 77 16 49 73 60 37 30 19 12 79 45 91 84 31 10 94 33 67 71 2 66 69 35 64 48 78 56 46 44 65 17 57 34 6 50 7 86 26 9 11 40 28 27 95 3 4 89 41 97 36 87 68 22 18 52 99 5 85 83 70 15 8 39 29 42 93 76 62 51 24 38 75 21 82 13 92 54 74 20 43 1 55 14 61 63 47 80 81 53 98 23 90 32 96"
},
{
"input": "100\n66 44 99 15 43 79 28 33 88 90 49 68 82 38 9 74 4 58 29 81 31 94 10 42 89 21 63 40 62 61 18 6 84 72 48 25 67 69 71 85 98 34 83 70 65 78 91 77 93 41 23 24 87 11 55 12 59 73 36 97 7 14 26 39 30 27 45 20 50 17 53 2 57 47 95 56 75 19 37 96 16 35 8 3 76 60 13 86 5 32 64 80 46 51 54 100 1 22 52 92",
"output": "97 72 84 17 89 32 61 83 15 23 54 56 87 62 4 81 70 31 78 68 26 98 51 52 36 63 66 7 19 65 21 90 8 42 82 59 79 14 64 28 50 24 5 2 67 93 74 35 11 69 94 99 71 95 55 76 73 18 57 86 30 29 27 91 45 1 37 12 38 44 39 34 58 16 77 85 48 46 6 92 20 13 43 33 40 88 53 9 25 10 47 100 49 22 75 80 60 41 3 96"
},
{
"input": "99\n3 73 32 37 25 15 93 63 85 8 91 78 80 5 39 48 46 7 83 70 23 96 9 29 77 53 30 20 56 50 13 45 21 76 87 99 65 31 16 18 14 72 51 28 43 2 81 34 38 40 66 54 74 26 71 4 61 17 58 24 22 33 49 36 42 11 12 55 60 27 62 90 79 92 94 68 1 52 84 41 86 35 69 75 47 10 64 88 97 98 67 19 89 95 59 82 57 44 6",
"output": "77 46 1 56 14 99 18 10 23 86 66 67 31 41 6 39 58 40 92 28 33 61 21 60 5 54 70 44 24 27 38 3 62 48 82 64 4 49 15 50 80 65 45 98 32 17 85 16 63 30 43 78 26 52 68 29 97 59 95 69 57 71 8 87 37 51 91 76 83 20 55 42 2 53 84 34 25 12 73 13 47 96 19 79 9 81 35 88 93 72 11 74 7 75 94 22 89 90 36"
},
{
"input": "100\n100 99 98 97 96 95 94 93 92 91 90 89 88 87 86 85 84 83 82 81 80 79 78 77 76 75 74 73 72 71 70 69 68 67 66 65 64 63 62 61 60 59 58 57 56 55 54 53 52 51 50 49 48 47 46 45 44 43 42 41 40 39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1",
"output": "100 99 98 97 96 95 94 93 92 91 90 89 88 87 86 85 84 83 82 81 80 79 78 77 76 75 74 73 72 71 70 69 68 67 66 65 64 63 62 61 60 59 58 57 56 55 54 53 52 51 50 49 48 47 46 45 44 43 42 41 40 39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1"
}
] | 1,690,886,248
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 101
| 92
| 0
|
n = int(input())
gift_list = input().split(' ')
index = 0
list = []
output_string = str()
for value in range(0,n):
list.append(value) # creating a template list to do work
for item in gift_list :
list[int(item) -1] = gift_list.index(item) +1
for item in list :
output_string += str(item) + ' '
print(output_string)
|
Title: Presents
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Petya very much likes gifts. Recently he has received a new laptop as a New Year gift from his mother. He immediately decided to give it to somebody else as what can be more pleasant than giving somebody gifts. And on this occasion he organized a New Year party at his place and invited *n* his friends there.
If there's one thing Petya likes more that receiving gifts, that's watching others giving gifts to somebody else. Thus, he safely hid the laptop until the next New Year and made up his mind to watch his friends exchanging gifts while he does not participate in the process. He numbered all his friends with integers from 1 to *n*. Petya remembered that a friend number *i* gave a gift to a friend number *p**i*. He also remembered that each of his friends received exactly one gift.
Now Petya wants to know for each friend *i* the number of a friend who has given him a gift.
Input Specification:
The first line contains one integer *n* (1<=≤<=*n*<=≤<=100) — the quantity of friends Petya invited to the party. The second line contains *n* space-separated integers: the *i*-th number is *p**i* — the number of a friend who gave a gift to friend number *i*. It is guaranteed that each friend received exactly one gift. It is possible that some friends do not share Petya's ideas of giving gifts to somebody else. Those friends gave the gifts to themselves.
Output Specification:
Print *n* space-separated integers: the *i*-th number should equal the number of the friend who gave a gift to friend number *i*.
Demo Input:
['4\n2 3 4 1\n', '3\n1 3 2\n', '2\n1 2\n']
Demo Output:
['4 1 2 3\n', '1 3 2\n', '1 2\n']
Note:
none
|
```python
n = int(input())
gift_list = input().split(' ')
index = 0
list = []
output_string = str()
for value in range(0,n):
list.append(value) # creating a template list to do work
for item in gift_list :
list[int(item) -1] = gift_list.index(item) +1
for item in list :
output_string += str(item) + ' '
print(output_string)
```
| 3
|
|
749
|
A
|
Bachgold Problem
|
PROGRAMMING
| 800
|
[
"greedy",
"implementation",
"math",
"number theory"
] | null | null |
Bachgold problem is very easy to formulate. Given a positive integer *n* represent it as a sum of maximum possible number of prime numbers. One can prove that such representation exists for any integer greater than 1.
Recall that integer *k* is called prime if it is greater than 1 and has exactly two positive integer divisors — 1 and *k*.
|
The only line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=100<=000).
|
The first line of the output contains a single integer *k* — maximum possible number of primes in representation.
The second line should contain *k* primes with their sum equal to *n*. You can print them in any order. If there are several optimal solution, print any of them.
|
[
"5\n",
"6\n"
] |
[
"2\n2 3\n",
"3\n2 2 2\n"
] |
none
| 500
|
[
{
"input": "5",
"output": "2\n2 3"
},
{
"input": "6",
"output": "3\n2 2 2"
},
{
"input": "2",
"output": "1\n2"
},
{
"input": "3",
"output": "1\n3"
},
{
"input": "99999",
"output": "49999\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 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ..."
},
{
"input": "100000",
"output": "50000\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 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ..."
},
{
"input": "7",
"output": "3\n2 2 3"
},
{
"input": "4",
"output": "2\n2 2"
},
{
"input": "8",
"output": "4\n2 2 2 2"
},
{
"input": "9",
"output": "4\n2 2 2 3"
},
{
"input": "99995",
"output": "49997\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 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ..."
},
{
"input": "99996",
"output": "49998\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 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ..."
},
{
"input": "10",
"output": "5\n2 2 2 2 2"
},
{
"input": "11",
"output": "5\n2 2 2 2 3"
},
{
"input": "99997",
"output": "49998\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 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ..."
},
{
"input": "99998",
"output": "49999\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 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ..."
},
{
"input": "12",
"output": "6\n2 2 2 2 2 2"
},
{
"input": "13",
"output": "6\n2 2 2 2 2 3"
},
{
"input": "99993",
"output": "49996\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 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ..."
},
{
"input": "99994",
"output": "49997\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 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ..."
},
{
"input": "14",
"output": "7\n2 2 2 2 2 2 2"
},
{
"input": "15",
"output": "7\n2 2 2 2 2 2 3"
},
{
"input": "53",
"output": "26\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 3"
},
{
"input": "57",
"output": "28\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 3"
},
{
"input": "61",
"output": "30\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 3"
},
{
"input": "774",
"output": "387\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 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ..."
},
{
"input": "202",
"output": "101\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 2"
},
{
"input": "530",
"output": "265\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 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ..."
},
{
"input": "7166",
"output": "3583\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 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2..."
},
{
"input": "9294",
"output": "4647\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 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2..."
},
{
"input": "2422",
"output": "1211\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 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2..."
},
{
"input": "15326",
"output": "7663\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 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2..."
},
{
"input": "11454",
"output": "5727\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 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2..."
},
{
"input": "14878",
"output": "7439\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 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2..."
},
{
"input": "90672",
"output": "45336\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 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ..."
},
{
"input": "99544",
"output": "49772\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 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ..."
},
{
"input": "90472",
"output": "45236\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 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ..."
},
{
"input": "23",
"output": "11\n2 2 2 2 2 2 2 2 2 2 3"
},
{
"input": "93",
"output": "46\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 3"
},
{
"input": "19",
"output": "9\n2 2 2 2 2 2 2 2 3"
},
{
"input": "11110",
"output": "5555\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 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2..."
}
] | 1,694,641,095
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 1
| 46
| 0
|
n = int(input())
if n % 2 == 0:
ans = n // 2
res = [2] * ans
else:
ans = (n-3) // 2
res = [2] * ans + [3]
print(ans+1)
print(*res)
|
Title: Bachgold Problem
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bachgold problem is very easy to formulate. Given a positive integer *n* represent it as a sum of maximum possible number of prime numbers. One can prove that such representation exists for any integer greater than 1.
Recall that integer *k* is called prime if it is greater than 1 and has exactly two positive integer divisors — 1 and *k*.
Input Specification:
The only line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=100<=000).
Output Specification:
The first line of the output contains a single integer *k* — maximum possible number of primes in representation.
The second line should contain *k* primes with their sum equal to *n*. You can print them in any order. If there are several optimal solution, print any of them.
Demo Input:
['5\n', '6\n']
Demo Output:
['2\n2 3\n', '3\n2 2 2\n']
Note:
none
|
```python
n = int(input())
if n % 2 == 0:
ans = n // 2
res = [2] * ans
else:
ans = (n-3) // 2
res = [2] * ans + [3]
print(ans+1)
print(*res)
```
| 0
|
|
588
|
A
|
Duff and Meat
|
PROGRAMMING
| 900
|
[
"greedy"
] | null | null |
Duff is addicted to meat! Malek wants to keep her happy for *n* days. In order to be happy in *i*-th day, she needs to eat exactly *a**i* kilograms of meat.
There is a big shop uptown and Malek wants to buy meat for her from there. In *i*-th day, they sell meat for *p**i* dollars per kilogram. Malek knows all numbers *a*1,<=...,<=*a**n* and *p*1,<=...,<=*p**n*. In each day, he can buy arbitrary amount of meat, also he can keep some meat he has for the future.
Malek is a little tired from cooking meat, so he asked for your help. Help him to minimize the total money he spends to keep Duff happy for *n* days.
|
The first line of input contains integer *n* (1<=≤<=*n*<=≤<=105), the number of days.
In the next *n* lines, *i*-th line contains two integers *a**i* and *p**i* (1<=≤<=*a**i*,<=*p**i*<=≤<=100), the amount of meat Duff needs and the cost of meat in that day.
|
Print the minimum money needed to keep Duff happy for *n* days, in one line.
|
[
"3\n1 3\n2 2\n3 1\n",
"3\n1 3\n2 1\n3 2\n"
] |
[
"10\n",
"8\n"
] |
In the first sample case: An optimal way would be to buy 1 kg on the first day, 2 kg on the second day and 3 kg on the third day.
In the second sample case: An optimal way would be to buy 1 kg on the first day and 5 kg (needed meat for the second and third day) on the second day.
| 750
|
[
{
"input": "3\n1 3\n2 2\n3 1",
"output": "10"
},
{
"input": "3\n1 3\n2 1\n3 2",
"output": "8"
},
{
"input": "1\n39 52",
"output": "2028"
},
{
"input": "2\n25 56\n94 17",
"output": "2998"
},
{
"input": "5\n39 21\n95 89\n73 90\n9 55\n85 32",
"output": "6321"
},
{
"input": "12\n70 11\n74 27\n32 11\n26 83\n57 18\n97 28\n75 43\n75 21\n84 29\n16 2\n89 63\n21 88",
"output": "6742"
},
{
"input": "2\n100 68\n38 80",
"output": "9384"
}
] | 1,537,272,464
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 30
| 498
| 7,782,400
|
import math
def main():
n=int(input())
a=[]
for i in range(n):
a.append(list(map(int,input().split())))
s=0
mt=101
for i in range(n):
if mt>a[i][1]:
mt=a[i][1]
s=s+a[i][0]*mt
print(s)
if __name__=='__main__':
main()
|
Title: Duff and Meat
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Duff is addicted to meat! Malek wants to keep her happy for *n* days. In order to be happy in *i*-th day, she needs to eat exactly *a**i* kilograms of meat.
There is a big shop uptown and Malek wants to buy meat for her from there. In *i*-th day, they sell meat for *p**i* dollars per kilogram. Malek knows all numbers *a*1,<=...,<=*a**n* and *p*1,<=...,<=*p**n*. In each day, he can buy arbitrary amount of meat, also he can keep some meat he has for the future.
Malek is a little tired from cooking meat, so he asked for your help. Help him to minimize the total money he spends to keep Duff happy for *n* days.
Input Specification:
The first line of input contains integer *n* (1<=≤<=*n*<=≤<=105), the number of days.
In the next *n* lines, *i*-th line contains two integers *a**i* and *p**i* (1<=≤<=*a**i*,<=*p**i*<=≤<=100), the amount of meat Duff needs and the cost of meat in that day.
Output Specification:
Print the minimum money needed to keep Duff happy for *n* days, in one line.
Demo Input:
['3\n1 3\n2 2\n3 1\n', '3\n1 3\n2 1\n3 2\n']
Demo Output:
['10\n', '8\n']
Note:
In the first sample case: An optimal way would be to buy 1 kg on the first day, 2 kg on the second day and 3 kg on the third day.
In the second sample case: An optimal way would be to buy 1 kg on the first day and 5 kg (needed meat for the second and third day) on the second day.
|
```python
import math
def main():
n=int(input())
a=[]
for i in range(n):
a.append(list(map(int,input().split())))
s=0
mt=101
for i in range(n):
if mt>a[i][1]:
mt=a[i][1]
s=s+a[i][0]*mt
print(s)
if __name__=='__main__':
main()
```
| 3
|
|
266
|
A
|
Stones on the Table
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
There are *n* stones on the table in a row, each of them can be red, green or blue. Count the minimum number of stones to take from the table so that any two neighboring stones had different colors. Stones in a row are considered neighboring if there are no other stones between them.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=50) — the number of stones on the table.
The next line contains string *s*, which represents the colors of the stones. We'll consider the stones in the row numbered from 1 to *n* from left to right. Then the *i*-th character *s* equals "R", if the *i*-th stone is red, "G", if it's green and "B", if it's blue.
|
Print a single integer — the answer to the problem.
|
[
"3\nRRG\n",
"5\nRRRRR\n",
"4\nBRBG\n"
] |
[
"1\n",
"4\n",
"0\n"
] |
none
| 500
|
[
{
"input": "3\nRRG",
"output": "1"
},
{
"input": "5\nRRRRR",
"output": "4"
},
{
"input": "4\nBRBG",
"output": "0"
},
{
"input": "1\nB",
"output": "0"
},
{
"input": "2\nBG",
"output": "0"
},
{
"input": "3\nBGB",
"output": "0"
},
{
"input": "4\nRBBR",
"output": "1"
},
{
"input": "5\nRGGBG",
"output": "1"
},
{
"input": "10\nGGBRBRGGRB",
"output": "2"
},
{
"input": "50\nGRBGGRBRGRBGGBBBBBGGGBBBBRBRGBRRBRGBBBRBBRRGBGGGRB",
"output": "18"
},
{
"input": "15\nBRRBRGGBBRRRRGR",
"output": "6"
},
{
"input": "20\nRRGBBRBRGRGBBGGRGRRR",
"output": "6"
},
{
"input": "25\nBBGBGRBGGBRRBGRRBGGBBRBRB",
"output": "6"
},
{
"input": "30\nGRGGGBGGRGBGGRGRBGBGBRRRRRRGRB",
"output": "9"
},
{
"input": "35\nGBBGBRGBBGGRBBGBRRGGRRRRRRRBRBBRRGB",
"output": "14"
},
{
"input": "40\nGBBRRGBGGGRGGGRRRRBRBGGBBGGGBGBBBBBRGGGG",
"output": "20"
},
{
"input": "45\nGGGBBRBBRRGRBBGGBGRBRGGBRBRGBRRGBGRRBGRGRBRRG",
"output": "11"
},
{
"input": "50\nRBGGBGGRBGRBBBGBBGRBBBGGGRBBBGBBBGRGGBGGBRBGBGRRGG",
"output": "17"
},
{
"input": "50\nGGGBBRGGGGGRRGGRBGGRGBBRBRRBGRGBBBGBRBGRGBBGRGGBRB",
"output": "16"
},
{
"input": "50\nGBGRGRRBRRRRRGGBBGBRRRBBBRBBBRRGRBBRGBRBGGRGRBBGGG",
"output": "19"
},
{
"input": "10\nGRRBRBRBGR",
"output": "1"
},
{
"input": "10\nBRBGBGRRBR",
"output": "1"
},
{
"input": "20\nGBGBGGRRRRGRBBGRGRGR",
"output": "5"
},
{
"input": "20\nRRGGRBBGBBRBGRRBRRBG",
"output": "6"
},
{
"input": "30\nBGBRGBBBGRGBBRGBGRBBBRGGRRGRRB",
"output": "8"
},
{
"input": "30\nBBBBGGBRBGBBGBGBGBGGGRGRRGGBBB",
"output": "11"
},
{
"input": "40\nGBRRGRBGBRRGBRGGGBRGBGBRGBBRRGRGGBBGBGBB",
"output": "9"
},
{
"input": "40\nBRGRGGRGGRBBRRRBRBBGGGRRGBGBBGRBBRGBRRGG",
"output": "13"
},
{
"input": "50\nRBGBGGRRGGRGGBGBGRRBGGBGBRRBBGBBGBBBGBBRBBRBRBRGRG",
"output": "13"
},
{
"input": "50\nRBRRGBGRRRBGRRBGRRGRBBRBBRRBRGGBRBRRBGGRBGGBRBRGRB",
"output": "12"
},
{
"input": "2\nBB",
"output": "1"
},
{
"input": "50\nRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR",
"output": "49"
},
{
"input": "50\nRRRRRRRRGRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR",
"output": "47"
},
{
"input": "50\nRRRRRRRRRRRRGGRRRRRRRRRBRRRRRRRRRRRRRRBBRRRRRRRRRR",
"output": "43"
}
] | 1,698,846,612
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 34
| 92
| 0
|
n = int(input())
stones = input()
count = 0
current = stones[0]
for s in stones[1:]:
if s == current:
count += 1
current = s
print(count)
|
Title: Stones on the Table
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* stones on the table in a row, each of them can be red, green or blue. Count the minimum number of stones to take from the table so that any two neighboring stones had different colors. Stones in a row are considered neighboring if there are no other stones between them.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=50) — the number of stones on the table.
The next line contains string *s*, which represents the colors of the stones. We'll consider the stones in the row numbered from 1 to *n* from left to right. Then the *i*-th character *s* equals "R", if the *i*-th stone is red, "G", if it's green and "B", if it's blue.
Output Specification:
Print a single integer — the answer to the problem.
Demo Input:
['3\nRRG\n', '5\nRRRRR\n', '4\nBRBG\n']
Demo Output:
['1\n', '4\n', '0\n']
Note:
none
|
```python
n = int(input())
stones = input()
count = 0
current = stones[0]
for s in stones[1:]:
if s == current:
count += 1
current = s
print(count)
```
| 3
|
|
260
|
C
|
Balls and Boxes
|
PROGRAMMING
| 1,700
|
[
"constructive algorithms",
"greedy",
"implementation"
] | null | null |
Little Vasya had *n* boxes with balls in the room. The boxes stood in a row and were numbered with numbers from 1 to *n* from left to right.
Once Vasya chose one of the boxes, let's assume that its number is *i*, took all balls out from it (it is guaranteed that this box originally had at least one ball), and began putting balls (one at a time) to the boxes with numbers *i*<=+<=1, *i*<=+<=2, *i*<=+<=3 and so on. If Vasya puts a ball into the box number *n*, then the next ball goes to box 1, the next one goes to box 2 and so on. He did it until he had no balls left in his hands. It is possible that Vasya puts multiple balls to the same box, and it is also possible that one or more balls will go to the box number *i*. If *i*<==<=*n*, Vasya puts the first ball into the box number 1, then the next ball goes to box 2 and so on.
For example, let's suppose that initially Vasya had four boxes, and the first box had 3 balls, the second one had 2, the third one had 5 and the fourth one had 4 balls. Then, if *i*<==<=3, then Vasya will take all five balls out of the third box and put them in the boxes with numbers: 4,<=1,<=2,<=3,<=4. After all Vasya's actions the balls will lie in the boxes as follows: in the first box there are 4 balls, 3 in the second one, 1 in the third one and 6 in the fourth one.
At this point Vasya has completely forgotten the original arrangement of the balls in the boxes, but he knows how they are arranged now, and the number *x* — the number of the box, where he put the last of the taken out balls.
He asks you to help to find the initial arrangement of the balls in the boxes.
|
The first line of the input contains two integers *n* and *x* (2<=≤<=*n*<=≤<=105, 1<=≤<=*x*<=≤<=*n*), that represent the number of the boxes and the index of the box that got the last ball from Vasya, correspondingly. The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n*, where integer *a**i* (0<=≤<=*a**i*<=≤<=109, *a**x*<=≠<=0) represents the number of balls in the box with index *i* after Vasya completes all the actions.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
|
Print *n* integers, where the *i*-th one represents the number of balls in the box number *i* before Vasya starts acting. Separate the numbers in the output by spaces. If there are multiple correct solutions, you are allowed to print any of them.
|
[
"4 4\n4 3 1 6\n",
"5 2\n3 2 0 2 7\n",
"3 3\n2 3 1\n"
] |
[
"3 2 5 4 ",
"2 1 4 1 6 ",
"1 2 3 "
] |
none
| 1,500
|
[
{
"input": "4 4\n4 3 1 6",
"output": "3 2 5 4 "
},
{
"input": "5 2\n3 2 0 2 7",
"output": "2 1 4 1 6 "
},
{
"input": "3 3\n2 3 1",
"output": "1 2 3 "
},
{
"input": "10 3\n1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000",
"output": "0 0 10000000000 0 0 0 0 0 0 0 "
},
{
"input": "5 4\n0 554459682 978416312 784688178 954779973",
"output": "3 554459681 978416311 784688177 954779973 "
},
{
"input": "5 2\n1 554459683 978416312 784688178 954779974",
"output": "6 554459681 978416311 784688177 954779973 "
},
{
"input": "10 8\n994538714 617271264 168716105 915909382 338220996 533154890 507276501 323171960 121635370 33140162",
"output": "961398551 584131101 135575942 882769219 305080833 500014727 474136338 290031797 88495208 331401628 "
},
{
"input": "10 5\n994538715 617271265 168716106 915909383 338220997 533154890 507276501 323171960 121635371 33140163",
"output": "961398551 584131101 135575942 882769219 305080833 500014727 474136338 290031797 88495208 331401635 "
},
{
"input": "15 12\n256121252 531930087 157210108 921323934 786210452 0 962820592 824495629 642702951 556399489 660627699 454443499 406577817 234814732 387536495",
"output": "256121252 531930087 157210108 921323934 786210452 6 962820591 824495628 642702950 556399488 660627698 454443498 406577817 234814732 387536495 "
},
{
"input": "15 8\n256121253 531930088 157210109 921323935 786210453 1 962820593 824495630 642702951 556399489 660627699 454443499 406577818 234814733 387536496",
"output": "256121252 531930087 157210108 921323934 786210452 17 962820591 824495628 642702950 556399488 660627698 454443498 406577817 234814732 387536495 "
},
{
"input": "48 34\n227460647 746912226 53993109 682685525 621533698 666833117 492590398 167395931 678377836 66509684 638633255 713194369 386921920 34175132 704550051 220688091 499436760 495071385 102952101 137372655 0 720974146 209203457 946682102 237312198 943872065 957150897 357568282 367006748 0 730509325 68523190 726721460 85037180 620964625 219537305 396613042 39840356 91947418 566463810 791505982 87036026 446282153 912204581 895882687 284561729 35994526 423462628",
"output": "227460647 746912226 53993109 682685525 621533698 666833117 492590398 167395931 678377836 66509684 638633255 713194369 386921920 34175132 704550051 220688091 499436760 495071385 102952101 137372655 0 720974146 209203457 946682102 237312198 943872065 957150897 357568282 367006748 4 730509324 68523189 726721459 85037179 620964625 219537305 396613042 39840356 91947418 566463810 791505982 87036026 446282153 912204581 895882687 284561729 35994526 423462628 "
},
{
"input": "48 19\n227460648 746912227 53993110 682685526 621533699 666833118 492590399 167395932 678377837 66509685 638633256 713194370 386921921 34175133 704550052 220688092 499436761 495071386 102952102 137372655 0 720974146 209203457 946682102 237312198 943872065 957150897 357568282 367006748 0 730509325 68523190 726721460 85037180 620964626 219537306 396613043 39840357 91947419 566463811 791505983 87036027 446282154 912204582 895882688 284561730 35994527 423462629",
"output": "227460647 746912226 53993109 682685525 621533698 666833117 492590398 167395931 678377836 66509684 638633255 713194369 386921920 34175132 704550051 220688091 499436760 495071385 102952101 137372655 0 720974146 209203457 946682102 237312198 943872065 957150897 357568282 367006748 37 730509324 68523189 726721459 85037179 620964625 219537305 396613042 39840356 91947418 566463810 791505982 87036026 446282153 912204581 895882687 284561729 35994526 423462628 "
},
{
"input": "10 5\n3 3 3 3 4 3 3 3 3 3",
"output": "0 0 0 31 0 0 0 0 0 0 "
},
{
"input": "5 4\n3 1 3 1 3",
"output": "2 0 2 5 2 "
}
] | 1,601,524,557
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 3
| 124
| 0
|
n, x = map(int, input().split())
ar = list(map(int, input().split()))
min_boxes = 1000000001
min_idx = -1
for i in range(n):
if ar[i] < min_boxes:
min_boxes = ar[i]
min_idx = i
ans = list(ar)
for i in range(n):
if i != min_idx:
ans[i] -= ar[min_idx]
res = 0
x -= 1
if x > min_idx:
for i in range(min_idx + 1, x + 1):
ans[i] -= 1
res = x - min_idx
elif x < min_idx:
for i in range(min_idx + 1, n):
ans[i] -= 1
for i in range(x + 1):
ans[i] -= 1
res = (n - 1) - (min_idx + 1) + 1 + x + 1
ans[min_idx] = ar[min_idx] * n + res
print(*ans)
|
Title: Balls and Boxes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Vasya had *n* boxes with balls in the room. The boxes stood in a row and were numbered with numbers from 1 to *n* from left to right.
Once Vasya chose one of the boxes, let's assume that its number is *i*, took all balls out from it (it is guaranteed that this box originally had at least one ball), and began putting balls (one at a time) to the boxes with numbers *i*<=+<=1, *i*<=+<=2, *i*<=+<=3 and so on. If Vasya puts a ball into the box number *n*, then the next ball goes to box 1, the next one goes to box 2 and so on. He did it until he had no balls left in his hands. It is possible that Vasya puts multiple balls to the same box, and it is also possible that one or more balls will go to the box number *i*. If *i*<==<=*n*, Vasya puts the first ball into the box number 1, then the next ball goes to box 2 and so on.
For example, let's suppose that initially Vasya had four boxes, and the first box had 3 balls, the second one had 2, the third one had 5 and the fourth one had 4 balls. Then, if *i*<==<=3, then Vasya will take all five balls out of the third box and put them in the boxes with numbers: 4,<=1,<=2,<=3,<=4. After all Vasya's actions the balls will lie in the boxes as follows: in the first box there are 4 balls, 3 in the second one, 1 in the third one and 6 in the fourth one.
At this point Vasya has completely forgotten the original arrangement of the balls in the boxes, but he knows how they are arranged now, and the number *x* — the number of the box, where he put the last of the taken out balls.
He asks you to help to find the initial arrangement of the balls in the boxes.
Input Specification:
The first line of the input contains two integers *n* and *x* (2<=≤<=*n*<=≤<=105, 1<=≤<=*x*<=≤<=*n*), that represent the number of the boxes and the index of the box that got the last ball from Vasya, correspondingly. The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n*, where integer *a**i* (0<=≤<=*a**i*<=≤<=109, *a**x*<=≠<=0) represents the number of balls in the box with index *i* after Vasya completes all the actions.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Output Specification:
Print *n* integers, where the *i*-th one represents the number of balls in the box number *i* before Vasya starts acting. Separate the numbers in the output by spaces. If there are multiple correct solutions, you are allowed to print any of them.
Demo Input:
['4 4\n4 3 1 6\n', '5 2\n3 2 0 2 7\n', '3 3\n2 3 1\n']
Demo Output:
['3 2 5 4 ', '2 1 4 1 6 ', '1 2 3 ']
Note:
none
|
```python
n, x = map(int, input().split())
ar = list(map(int, input().split()))
min_boxes = 1000000001
min_idx = -1
for i in range(n):
if ar[i] < min_boxes:
min_boxes = ar[i]
min_idx = i
ans = list(ar)
for i in range(n):
if i != min_idx:
ans[i] -= ar[min_idx]
res = 0
x -= 1
if x > min_idx:
for i in range(min_idx + 1, x + 1):
ans[i] -= 1
res = x - min_idx
elif x < min_idx:
for i in range(min_idx + 1, n):
ans[i] -= 1
for i in range(x + 1):
ans[i] -= 1
res = (n - 1) - (min_idx + 1) + 1 + x + 1
ans[min_idx] = ar[min_idx] * n + res
print(*ans)
```
| 0
|
|
1,009
|
C
|
Annoying Present
|
PROGRAMMING
| 1,700
|
[
"greedy",
"math"
] | null | null |
Alice got an array of length $n$ as a birthday present once again! This is the third year in a row!
And what is more disappointing, it is overwhelmengly boring, filled entirely with zeros. Bob decided to apply some changes to the array to cheer up Alice.
Bob has chosen $m$ changes of the following form. For some integer numbers $x$ and $d$, he chooses an arbitrary position $i$ ($1 \le i \le n$) and for every $j \in [1, n]$ adds $x + d \cdot dist(i, j)$ to the value of the $j$-th cell. $dist(i, j)$ is the distance between positions $i$ and $j$ (i.e. $dist(i, j) = |i - j|$, where $|x|$ is an absolute value of $x$).
For example, if Alice currently has an array $[2, 1, 2, 2]$ and Bob chooses position $3$ for $x = -1$ and $d = 2$ then the array will become $[2 - 1 + 2 \cdot 2,~1 - 1 + 2 \cdot 1,~2 - 1 + 2 \cdot 0,~2 - 1 + 2 \cdot 1]$ = $[5, 2, 1, 3]$. Note that Bob can't choose position $i$ outside of the array (that is, smaller than $1$ or greater than $n$).
Alice will be the happiest when the elements of the array are as big as possible. Bob claimed that the arithmetic mean value of the elements will work fine as a metric.
What is the maximum arithmetic mean value Bob can achieve?
|
The first line contains two integers $n$ and $m$ ($1 \le n, m \le 10^5$) — the number of elements of the array and the number of changes.
Each of the next $m$ lines contains two integers $x_i$ and $d_i$ ($-10^3 \le x_i, d_i \le 10^3$) — the parameters for the $i$-th change.
|
Print the maximal average arithmetic mean of the elements Bob can achieve.
Your answer is considered correct if its absolute or relative error doesn't exceed $10^{-6}$.
|
[
"2 3\n-1 3\n0 0\n-1 -4\n",
"3 2\n0 2\n5 0\n"
] |
[
"-2.500000000000000\n",
"7.000000000000000\n"
] |
none
| 0
|
[
{
"input": "2 3\n-1 3\n0 0\n-1 -4",
"output": "-2.500000000000000"
},
{
"input": "3 2\n0 2\n5 0",
"output": "7.000000000000000"
},
{
"input": "8 8\n-21 -60\n-96 -10\n-4 -19\n-27 -4\n57 -15\n-95 62\n-42 1\n-17 64",
"output": "-16.500000000000000"
},
{
"input": "1 1\n0 0",
"output": "0.000000000000000"
},
{
"input": "100000 1\n1000 1000",
"output": "50000500.000000000000000"
},
{
"input": "11 1\n0 -10",
"output": "-27.272727272727273"
},
{
"input": "3 1\n1 -1",
"output": "0.333333333333333"
},
{
"input": "1 2\n-1 -1\n-2 -2",
"output": "-3.000000000000000"
},
{
"input": "1 2\n0 -1\n0 1",
"output": "0.000000000000000"
},
{
"input": "1 1\n1 -2",
"output": "1.000000000000000"
},
{
"input": "3 1\n2 -1",
"output": "1.333333333333333"
},
{
"input": "3 1\n0 -1",
"output": "-0.666666666666667"
},
{
"input": "1 1\n-1000 -1000",
"output": "-1000.000000000000000"
},
{
"input": "1 1\n0 -5",
"output": "0.000000000000000"
},
{
"input": "15 3\n2 0\n2 -5\n-2 5",
"output": "18.333333333333332"
},
{
"input": "9 1\n0 -5",
"output": "-11.111111111111111"
},
{
"input": "7 1\n0 -1",
"output": "-1.714285714285714"
},
{
"input": "3 1\n-2 -2",
"output": "-3.333333333333333"
},
{
"input": "3 1\n5 -5",
"output": "1.666666666666667"
},
{
"input": "1 1\n-1 -1",
"output": "-1.000000000000000"
},
{
"input": "7 1\n-1 -5",
"output": "-9.571428571428571"
},
{
"input": "3 2\n-2 -2\n-2 -2",
"output": "-6.666666666666667"
},
{
"input": "5 1\n0 -4",
"output": "-4.800000000000000"
},
{
"input": "5 1\n-1 -5",
"output": "-7.000000000000000"
},
{
"input": "5 1\n0 -2",
"output": "-2.400000000000000"
},
{
"input": "3 5\n1 -1000\n1 -1000\n1 -1000\n1 -1000\n1 -1000",
"output": "-3328.333333333333485"
},
{
"input": "1 1\n0 -1",
"output": "0.000000000000000"
},
{
"input": "1 2\n0 -3\n0 -3",
"output": "0.000000000000000"
},
{
"input": "7 1\n2 -3",
"output": "-3.142857142857143"
},
{
"input": "3 2\n-1 -1\n-1 -1",
"output": "-3.333333333333333"
},
{
"input": "5 1\n-1 -162",
"output": "-195.400000000000006"
},
{
"input": "5 10\n-506 -243\n727 -141\n-548 -306\n740 880\n-744 -116\n-84 182\n-859 -108\n64 86\n135 446\n69 -184",
"output": "864.399999999999977"
},
{
"input": "5 1\n0 -1",
"output": "-1.200000000000000"
},
{
"input": "5 12\n634 895\n143 730\n901 245\n386 486\n395 -111\n-469 -104\n-681 -623\n-900 843\n889 -883\n476 -304\n777 986\n206 -491",
"output": "8107.800000000000182"
},
{
"input": "3 3\n4 2\n5 0\n6 -1",
"output": "16.333333333333332"
},
{
"input": "1 3\n4 2\n5 0\n6 -1",
"output": "15.000000000000000"
},
{
"input": "85 10\n-223 435\n-771 455\n72 -940\n490 -178\n400 -117\n169 -527\n836 610\n849 944\n572 -237\n-428 -428",
"output": "53047.388235294114565"
},
{
"input": "69 10\n-8 4\n-3 3\n7 5\n5 -9\n8 1\n7 -5\n-8 -8\n9 3\n1 1\n0 6",
"output": "420.579710144927560"
},
{
"input": "1 10\n1 1\n1 0\n1 0\n1 0\n-1 0\n0 1\n1 0\n0 0\n2 1\n9 2",
"output": "15.000000000000000"
},
{
"input": "5 4\n0 1\n0 2\n0 3\n0 -9",
"output": "1.200000000000000"
}
] | 1,532,018,694
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 93
| 0
|
n = [int(i) for i in input().split()]
x = n[0]//2
sum1 = 0
sum2 = 0
for i in range(n[0]):
sum1+=i
sum2+=(i + 1 - x)
f = 0
for j in range(n[1]):
x = [int(i) for i in input().split()]
if(x[1]>0):
f += n[0] * x[0] + max(sum1,sum2)*x[1]
else:
f += n[0] * x[0] + min(sum1,sum2)*x[1]
print(f/n[0])
|
Title: Annoying Present
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Alice got an array of length $n$ as a birthday present once again! This is the third year in a row!
And what is more disappointing, it is overwhelmengly boring, filled entirely with zeros. Bob decided to apply some changes to the array to cheer up Alice.
Bob has chosen $m$ changes of the following form. For some integer numbers $x$ and $d$, he chooses an arbitrary position $i$ ($1 \le i \le n$) and for every $j \in [1, n]$ adds $x + d \cdot dist(i, j)$ to the value of the $j$-th cell. $dist(i, j)$ is the distance between positions $i$ and $j$ (i.e. $dist(i, j) = |i - j|$, where $|x|$ is an absolute value of $x$).
For example, if Alice currently has an array $[2, 1, 2, 2]$ and Bob chooses position $3$ for $x = -1$ and $d = 2$ then the array will become $[2 - 1 + 2 \cdot 2,~1 - 1 + 2 \cdot 1,~2 - 1 + 2 \cdot 0,~2 - 1 + 2 \cdot 1]$ = $[5, 2, 1, 3]$. Note that Bob can't choose position $i$ outside of the array (that is, smaller than $1$ or greater than $n$).
Alice will be the happiest when the elements of the array are as big as possible. Bob claimed that the arithmetic mean value of the elements will work fine as a metric.
What is the maximum arithmetic mean value Bob can achieve?
Input Specification:
The first line contains two integers $n$ and $m$ ($1 \le n, m \le 10^5$) — the number of elements of the array and the number of changes.
Each of the next $m$ lines contains two integers $x_i$ and $d_i$ ($-10^3 \le x_i, d_i \le 10^3$) — the parameters for the $i$-th change.
Output Specification:
Print the maximal average arithmetic mean of the elements Bob can achieve.
Your answer is considered correct if its absolute or relative error doesn't exceed $10^{-6}$.
Demo Input:
['2 3\n-1 3\n0 0\n-1 -4\n', '3 2\n0 2\n5 0\n']
Demo Output:
['-2.500000000000000\n', '7.000000000000000\n']
Note:
none
|
```python
n = [int(i) for i in input().split()]
x = n[0]//2
sum1 = 0
sum2 = 0
for i in range(n[0]):
sum1+=i
sum2+=(i + 1 - x)
f = 0
for j in range(n[1]):
x = [int(i) for i in input().split()]
if(x[1]>0):
f += n[0] * x[0] + max(sum1,sum2)*x[1]
else:
f += n[0] * x[0] + min(sum1,sum2)*x[1]
print(f/n[0])
```
| 0
|
|
148
|
A
|
Insomnia cure
|
PROGRAMMING
| 800
|
[
"constructive algorithms",
"implementation",
"math"
] | null | null |
«One dragon. Two dragon. Three dragon», — the princess was counting. She had trouble falling asleep, and she got bored of counting lambs when she was nine.
However, just counting dragons was boring as well, so she entertained herself at best she could. Tonight she imagined that all dragons were here to steal her, and she was fighting them off. Every *k*-th dragon got punched in the face with a frying pan. Every *l*-th dragon got his tail shut into the balcony door. Every *m*-th dragon got his paws trampled with sharp heels. Finally, she threatened every *n*-th dragon to call her mom, and he withdrew in panic.
How many imaginary dragons suffered moral or physical damage tonight, if the princess counted a total of *d* dragons?
|
Input data contains integer numbers *k*,<=*l*,<=*m*,<=*n* and *d*, each number in a separate line (1<=≤<=*k*,<=*l*,<=*m*,<=*n*<=≤<=10, 1<=≤<=*d*<=≤<=105).
|
Output the number of damaged dragons.
|
[
"1\n2\n3\n4\n12\n",
"2\n3\n4\n5\n24\n"
] |
[
"12\n",
"17\n"
] |
In the first case every first dragon got punched with a frying pan. Some of the dragons suffered from other reasons as well, but the pan alone would be enough.
In the second case dragons 1, 7, 11, 13, 17, 19 and 23 escaped unharmed.
| 1,000
|
[
{
"input": "1\n2\n3\n4\n12",
"output": "12"
},
{
"input": "2\n3\n4\n5\n24",
"output": "17"
},
{
"input": "1\n1\n1\n1\n100000",
"output": "100000"
},
{
"input": "10\n9\n8\n7\n6",
"output": "0"
},
{
"input": "8\n4\n4\n3\n65437",
"output": "32718"
},
{
"input": "8\n4\n1\n10\n59392",
"output": "59392"
},
{
"input": "4\n1\n8\n7\n44835",
"output": "44835"
},
{
"input": "6\n1\n7\n2\n62982",
"output": "62982"
},
{
"input": "2\n7\n4\n9\n56937",
"output": "35246"
},
{
"input": "2\n9\n8\n1\n75083",
"output": "75083"
},
{
"input": "8\n7\n7\n6\n69038",
"output": "24656"
},
{
"input": "4\n4\n2\n3\n54481",
"output": "36320"
},
{
"input": "6\n4\n9\n8\n72628",
"output": "28244"
},
{
"input": "9\n7\n8\n10\n42357",
"output": "16540"
},
{
"input": "5\n6\n4\n3\n60504",
"output": "36302"
},
{
"input": "7\n2\n3\n8\n21754",
"output": "15539"
},
{
"input": "1\n2\n10\n4\n39901",
"output": "39901"
},
{
"input": "3\n4\n7\n1\n58048",
"output": "58048"
},
{
"input": "9\n10\n4\n6\n52003",
"output": "21956"
},
{
"input": "5\n10\n9\n3\n70149",
"output": "32736"
},
{
"input": "5\n5\n5\n10\n55592",
"output": "11118"
},
{
"input": "1\n5\n2\n6\n49547",
"output": "49547"
},
{
"input": "3\n7\n7\n7\n84046",
"output": "36019"
},
{
"input": "10\n2\n1\n8\n63537",
"output": "63537"
},
{
"input": "7\n2\n10\n5\n81684",
"output": "53678"
},
{
"input": "7\n1\n6\n8\n99831",
"output": "99831"
},
{
"input": "3\n9\n3\n3\n61082",
"output": "20360"
},
{
"input": "5\n5\n8\n9\n79228",
"output": "29931"
},
{
"input": "1\n5\n5\n6\n89535",
"output": "89535"
},
{
"input": "5\n5\n2\n1\n7682",
"output": "7682"
},
{
"input": "7\n4\n8\n8\n25829",
"output": "9224"
},
{
"input": "3\n4\n3\n3\n87079",
"output": "43539"
},
{
"input": "5\n8\n4\n9\n5226",
"output": "2438"
},
{
"input": "4\n9\n6\n6\n1070",
"output": "415"
},
{
"input": "10\n10\n10\n10\n100000",
"output": "10000"
},
{
"input": "1\n1\n1\n1\n1",
"output": "1"
},
{
"input": "10\n10\n10\n10\n1",
"output": "0"
},
{
"input": "10\n10\n1\n10\n10",
"output": "10"
},
{
"input": "10\n10\n1\n10\n100",
"output": "100"
},
{
"input": "2\n2\n2\n2\n1",
"output": "0"
}
] | 1,692,201,331
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 40
| 154
| 1,843,200
|
k=int(input())
l=int(input())
n=int(input())
m=int(input())
d=int(input())
s=0
if min(k,l,n,m)==1:
print(d)
else:
for i in range(1,d+1):
if i%l==0 or i%k==0 or i%n==0 or i%m==0:
s+=1
print(s)
|
Title: Insomnia cure
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
«One dragon. Two dragon. Three dragon», — the princess was counting. She had trouble falling asleep, and she got bored of counting lambs when she was nine.
However, just counting dragons was boring as well, so she entertained herself at best she could. Tonight she imagined that all dragons were here to steal her, and she was fighting them off. Every *k*-th dragon got punched in the face with a frying pan. Every *l*-th dragon got his tail shut into the balcony door. Every *m*-th dragon got his paws trampled with sharp heels. Finally, she threatened every *n*-th dragon to call her mom, and he withdrew in panic.
How many imaginary dragons suffered moral or physical damage tonight, if the princess counted a total of *d* dragons?
Input Specification:
Input data contains integer numbers *k*,<=*l*,<=*m*,<=*n* and *d*, each number in a separate line (1<=≤<=*k*,<=*l*,<=*m*,<=*n*<=≤<=10, 1<=≤<=*d*<=≤<=105).
Output Specification:
Output the number of damaged dragons.
Demo Input:
['1\n2\n3\n4\n12\n', '2\n3\n4\n5\n24\n']
Demo Output:
['12\n', '17\n']
Note:
In the first case every first dragon got punched with a frying pan. Some of the dragons suffered from other reasons as well, but the pan alone would be enough.
In the second case dragons 1, 7, 11, 13, 17, 19 and 23 escaped unharmed.
|
```python
k=int(input())
l=int(input())
n=int(input())
m=int(input())
d=int(input())
s=0
if min(k,l,n,m)==1:
print(d)
else:
for i in range(1,d+1):
if i%l==0 or i%k==0 or i%n==0 or i%m==0:
s+=1
print(s)
```
| 3
|
|
919
|
B
|
Perfect Number
|
PROGRAMMING
| 1,100
|
[
"binary search",
"brute force",
"dp",
"implementation",
"number theory"
] | null | null |
We consider a positive integer perfect, if and only if the sum of its digits is exactly $10$. Given a positive integer $k$, your task is to find the $k$-th smallest perfect positive integer.
|
A single line with a positive integer $k$ ($1 \leq k \leq 10\,000$).
|
A single number, denoting the $k$-th smallest perfect integer.
|
[
"1\n",
"2\n"
] |
[
"19\n",
"28\n"
] |
The first perfect integer is $19$ and the second one is $28$.
| 750
|
[
{
"input": "1",
"output": "19"
},
{
"input": "2",
"output": "28"
},
{
"input": "13",
"output": "136"
},
{
"input": "101",
"output": "1432"
},
{
"input": "1023",
"output": "100270"
},
{
"input": "9999",
"output": "10800010"
},
{
"input": "10000",
"output": "10800100"
},
{
"input": "2333",
"output": "310060"
},
{
"input": "9139",
"output": "10134010"
},
{
"input": "9859",
"output": "10422001"
},
{
"input": "5987",
"output": "2221201"
},
{
"input": "3",
"output": "37"
},
{
"input": "4",
"output": "46"
},
{
"input": "5",
"output": "55"
},
{
"input": "7",
"output": "73"
},
{
"input": "9",
"output": "91"
},
{
"input": "10",
"output": "109"
},
{
"input": "15",
"output": "154"
},
{
"input": "51",
"output": "613"
},
{
"input": "99",
"output": "1414"
},
{
"input": "138",
"output": "2224"
},
{
"input": "233",
"output": "5050"
},
{
"input": "666",
"output": "27100"
},
{
"input": "1234",
"output": "110206"
},
{
"input": "3567",
"output": "1033003"
},
{
"input": "4445",
"output": "1221301"
},
{
"input": "5689",
"output": "2114002"
},
{
"input": "6666",
"output": "3102004"
},
{
"input": "7777",
"output": "5300200"
},
{
"input": "8888",
"output": "10110061"
},
{
"input": "9411",
"output": "10214200"
},
{
"input": "5539",
"output": "2101114"
},
{
"input": "6259",
"output": "2511100"
},
{
"input": "2387",
"output": "312220"
},
{
"input": "8515",
"output": "10030114"
}
] | 1,664,778,911
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 5
| 2,000
| 0
|
k = int(input())
x= int(1e9+ 1)
id = 10
count = 0
for i in range(0,x):
id += 9
s = id
o = 0
while s >= 1:
o += s % 10
s = s // 10
if o == 10:
count += 1
if count == k:
print(id)
break
|
Title: Perfect Number
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
We consider a positive integer perfect, if and only if the sum of its digits is exactly $10$. Given a positive integer $k$, your task is to find the $k$-th smallest perfect positive integer.
Input Specification:
A single line with a positive integer $k$ ($1 \leq k \leq 10\,000$).
Output Specification:
A single number, denoting the $k$-th smallest perfect integer.
Demo Input:
['1\n', '2\n']
Demo Output:
['19\n', '28\n']
Note:
The first perfect integer is $19$ and the second one is $28$.
|
```python
k = int(input())
x= int(1e9+ 1)
id = 10
count = 0
for i in range(0,x):
id += 9
s = id
o = 0
while s >= 1:
o += s % 10
s = s // 10
if o == 10:
count += 1
if count == k:
print(id)
break
```
| 0
|
|
679
|
A
|
Bear and Prime 100
|
PROGRAMMING
| 1,400
|
[
"constructive algorithms",
"interactive",
"math"
] | null | null |
This is an interactive problem. In the output section below you will see the information about flushing the output.
Bear Limak thinks of some hidden number — an integer from interval [2,<=100]. Your task is to say if the hidden number is prime or composite.
Integer *x*<=><=1 is called prime if it has exactly two distinct divisors, 1 and *x*. If integer *x*<=><=1 is not prime, it's called composite.
You can ask up to 20 queries about divisors of the hidden number. In each query you should print an integer from interval [2,<=100]. The system will answer "yes" if your integer is a divisor of the hidden number. Otherwise, the answer will be "no".
For example, if the hidden number is 14 then the system will answer "yes" only if you print 2, 7 or 14.
When you are done asking queries, print "prime" or "composite" and terminate your program.
You will get the Wrong Answer verdict if you ask more than 20 queries, or if you print an integer not from the range [2,<=100]. Also, you will get the Wrong Answer verdict if the printed answer isn't correct.
You will get the Idleness Limit Exceeded verdict if you don't print anything (but you should) or if you forget about flushing the output (more info below).
|
After each query you should read one string from the input. It will be "yes" if the printed integer is a divisor of the hidden number, and "no" otherwise.
|
Up to 20 times you can ask a query — print an integer from interval [2,<=100] in one line. You have to both print the end-of-line character and flush the output. After flushing you should read a response from the input.
In any moment you can print the answer "prime" or "composite" (without the quotes). After that, flush the output and terminate your program.
To flush you can use (just after printing an integer and end-of-line):
- fflush(stdout) in C++; - System.out.flush() in Java; - stdout.flush() in Python; - flush(output) in Pascal; - See the documentation for other languages.
Hacking. To hack someone, as the input you should print the hidden number — one integer from the interval [2,<=100]. Of course, his/her solution won't be able to read the hidden number from the input.
|
[
"yes\nno\nyes\n",
"no\nyes\nno\nno\nno\n"
] |
[
"2\n80\n5\ncomposite\n",
"58\n59\n78\n78\n2\nprime\n"
] |
The hidden number in the first query is 30. In a table below you can see a better form of the provided example of the communication process.
<img align="middle" class="tex-formula" src="https://espresso.codeforces.com/ea790051c34ea7d2761cd9b096412ca7c647a173.png" style="max-width: 100.0%;max-height: 100.0%;"/>
The hidden number is divisible by both 2 and 5. Thus, it must be composite. Note that it isn't necessary to know the exact value of the hidden number. In this test, the hidden number is 30.
<img align="middle" class="tex-formula" src="https://espresso.codeforces.com/35c6952617fa94ec3e0ea8e63aa1c3c49b3ba420.png" style="max-width: 100.0%;max-height: 100.0%;"/>
59 is a divisor of the hidden number. In the interval [2, 100] there is only one number with this divisor. The hidden number must be 59, which is prime. Note that the answer is known even after the second query and you could print it then and terminate. Though, it isn't forbidden to ask unnecessary queries (unless you exceed the limit of 20 queries).
| 750
|
[
{
"input": "30",
"output": "composite 4"
},
{
"input": "59",
"output": "prime 15"
},
{
"input": "2",
"output": "prime 16"
},
{
"input": "7",
"output": "prime 16"
},
{
"input": "9",
"output": "composite 3"
},
{
"input": "13",
"output": "prime 15"
},
{
"input": "55",
"output": "composite 6"
},
{
"input": "89",
"output": "prime 15"
},
{
"input": "3",
"output": "prime 16"
},
{
"input": "4",
"output": "composite 2"
},
{
"input": "6",
"output": "composite 4"
},
{
"input": "8",
"output": "composite 2"
},
{
"input": "11",
"output": "prime 15"
},
{
"input": "12",
"output": "composite 2"
},
{
"input": "25",
"output": "composite 4"
},
{
"input": "36",
"output": "composite 2"
},
{
"input": "46",
"output": "composite 10"
},
{
"input": "47",
"output": "prime 15"
},
{
"input": "49",
"output": "composite 5"
},
{
"input": "51",
"output": "composite 8"
},
{
"input": "53",
"output": "prime 15"
},
{
"input": "59",
"output": "prime 15"
},
{
"input": "64",
"output": "composite 2"
},
{
"input": "81",
"output": "composite 3"
},
{
"input": "91",
"output": "composite 7"
},
{
"input": "93",
"output": "composite 12"
},
{
"input": "94",
"output": "composite 16"
},
{
"input": "95",
"output": "composite 9"
},
{
"input": "96",
"output": "composite 2"
},
{
"input": "97",
"output": "prime 15"
},
{
"input": "99",
"output": "composite 3"
},
{
"input": "100",
"output": "composite 2"
},
{
"input": "26",
"output": "composite 7"
},
{
"input": "94",
"output": "composite 16"
},
{
"input": "25",
"output": "composite 4"
},
{
"input": "46",
"output": "composite 10"
},
{
"input": "49",
"output": "composite 5"
},
{
"input": "82",
"output": "composite 14"
},
{
"input": "87",
"output": "composite 11"
},
{
"input": "85",
"output": "composite 8"
},
{
"input": "74",
"output": "composite 13"
},
{
"input": "86",
"output": "composite 15"
},
{
"input": "93",
"output": "composite 12"
},
{
"input": "58",
"output": "composite 11"
},
{
"input": "47",
"output": "prime 15"
},
{
"input": "22",
"output": "composite 6"
},
{
"input": "23",
"output": "prime 15"
},
{
"input": "62",
"output": "composite 12"
},
{
"input": "38",
"output": "composite 9"
},
{
"input": "91",
"output": "composite 7"
},
{
"input": "34",
"output": "composite 8"
},
{
"input": "95",
"output": "composite 9"
},
{
"input": "4",
"output": "composite 2"
},
{
"input": "16",
"output": "composite 2"
},
{
"input": "96",
"output": "composite 2"
},
{
"input": "69",
"output": "composite 10"
},
{
"input": "65",
"output": "composite 7"
},
{
"input": "6",
"output": "composite 4"
},
{
"input": "18",
"output": "composite 4"
}
] | 1,594,392,180
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 58
| 155
| 20,172,800
|
from sys import stdout
ans=0
for i in[2,3,5,7,11,13,17,19,23,29,31,37,41,43,47]:
print(i)
stdout.flush()
if input()=='yes':
ans+=1
if i*i<100:
print(i*i)
stdout.flush()
if input()=='yes':ans+=1
if ans==2:exit(print('composite'))
print('prime')
|
Title: Bear and Prime 100
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
This is an interactive problem. In the output section below you will see the information about flushing the output.
Bear Limak thinks of some hidden number — an integer from interval [2,<=100]. Your task is to say if the hidden number is prime or composite.
Integer *x*<=><=1 is called prime if it has exactly two distinct divisors, 1 and *x*. If integer *x*<=><=1 is not prime, it's called composite.
You can ask up to 20 queries about divisors of the hidden number. In each query you should print an integer from interval [2,<=100]. The system will answer "yes" if your integer is a divisor of the hidden number. Otherwise, the answer will be "no".
For example, if the hidden number is 14 then the system will answer "yes" only if you print 2, 7 or 14.
When you are done asking queries, print "prime" or "composite" and terminate your program.
You will get the Wrong Answer verdict if you ask more than 20 queries, or if you print an integer not from the range [2,<=100]. Also, you will get the Wrong Answer verdict if the printed answer isn't correct.
You will get the Idleness Limit Exceeded verdict if you don't print anything (but you should) or if you forget about flushing the output (more info below).
Input Specification:
After each query you should read one string from the input. It will be "yes" if the printed integer is a divisor of the hidden number, and "no" otherwise.
Output Specification:
Up to 20 times you can ask a query — print an integer from interval [2,<=100] in one line. You have to both print the end-of-line character and flush the output. After flushing you should read a response from the input.
In any moment you can print the answer "prime" or "composite" (without the quotes). After that, flush the output and terminate your program.
To flush you can use (just after printing an integer and end-of-line):
- fflush(stdout) in C++; - System.out.flush() in Java; - stdout.flush() in Python; - flush(output) in Pascal; - See the documentation for other languages.
Hacking. To hack someone, as the input you should print the hidden number — one integer from the interval [2,<=100]. Of course, his/her solution won't be able to read the hidden number from the input.
Demo Input:
['yes\nno\nyes\n', 'no\nyes\nno\nno\nno\n']
Demo Output:
['2\n80\n5\ncomposite\n', '58\n59\n78\n78\n2\nprime\n']
Note:
The hidden number in the first query is 30. In a table below you can see a better form of the provided example of the communication process.
<img align="middle" class="tex-formula" src="https://espresso.codeforces.com/ea790051c34ea7d2761cd9b096412ca7c647a173.png" style="max-width: 100.0%;max-height: 100.0%;"/>
The hidden number is divisible by both 2 and 5. Thus, it must be composite. Note that it isn't necessary to know the exact value of the hidden number. In this test, the hidden number is 30.
<img align="middle" class="tex-formula" src="https://espresso.codeforces.com/35c6952617fa94ec3e0ea8e63aa1c3c49b3ba420.png" style="max-width: 100.0%;max-height: 100.0%;"/>
59 is a divisor of the hidden number. In the interval [2, 100] there is only one number with this divisor. The hidden number must be 59, which is prime. Note that the answer is known even after the second query and you could print it then and terminate. Though, it isn't forbidden to ask unnecessary queries (unless you exceed the limit of 20 queries).
|
```python
from sys import stdout
ans=0
for i in[2,3,5,7,11,13,17,19,23,29,31,37,41,43,47]:
print(i)
stdout.flush()
if input()=='yes':
ans+=1
if i*i<100:
print(i*i)
stdout.flush()
if input()=='yes':ans+=1
if ans==2:exit(print('composite'))
print('prime')
```
| 0
|
|
112
|
B
|
Petya and Square
|
PROGRAMMING
| 1,200
|
[
"implementation",
"math"
] |
B. Petya and Square
|
2
|
256
|
Little Petya loves playing with squares. Mum bought him a square 2*n*<=×<=2*n* in size. Petya marked a cell inside the square and now he is solving the following task.
The task is to draw a broken line that would go along the grid lines and that would cut the square into two equal parts. The cutting line should not have any common points with the marked cell and the resulting two parts should be equal up to rotation.
Petya wants to determine whether it is possible to cut the square in the required manner given the sizes of the square side and the coordinates of the marked cell. Help him.
|
The first line contains three space-separated integers 2*n*, *x* and *y* (2<=≤<=2*n*<=≤<=100,<=1<=≤<=*x*,<=*y*<=≤<=2*n*), representing the length of a square's side and the coordinates of the marked cell. It is guaranteed that 2*n* is even.
The coordinates of the marked cell are represented by a pair of numbers *x* *y*, where *x* represents the number of the row and *y* represents the number of the column. The rows and columns are numbered by consecutive integers from 1 to 2*n*. The rows are numbered from top to bottom and the columns are numbered from the left to the right.
|
If the square is possible to cut, print "YES", otherwise print "NO" (without the quotes).
|
[
"4 1 1\n",
"2 2 2\n"
] |
[
"YES\n",
"NO\n"
] |
A sample test from the statement and one of the possible ways of cutting the square are shown in the picture:
| 1,000
|
[
{
"input": "4 1 1",
"output": "YES"
},
{
"input": "2 2 2",
"output": "NO"
},
{
"input": "8 2 5",
"output": "YES"
},
{
"input": "6 1 6",
"output": "YES"
},
{
"input": "2 1 1",
"output": "NO"
},
{
"input": "2 1 2",
"output": "NO"
},
{
"input": "2 2 1",
"output": "NO"
},
{
"input": "4 2 2",
"output": "NO"
},
{
"input": "4 2 3",
"output": "NO"
},
{
"input": "4 2 4",
"output": "YES"
},
{
"input": "60 30 30",
"output": "NO"
},
{
"input": "60 34 30",
"output": "YES"
},
{
"input": "60 31 29",
"output": "YES"
},
{
"input": "100 61 30",
"output": "YES"
},
{
"input": "100 52 50",
"output": "YES"
},
{
"input": "100 51 51",
"output": "NO"
},
{
"input": "100 1 2",
"output": "YES"
},
{
"input": "100 1 8",
"output": "YES"
},
{
"input": "100 19 99",
"output": "YES"
},
{
"input": "100 18 82",
"output": "YES"
},
{
"input": "100 100 50",
"output": "YES"
},
{
"input": "100 51 100",
"output": "YES"
},
{
"input": "100 100 100",
"output": "YES"
},
{
"input": "6 3 3",
"output": "NO"
},
{
"input": "6 4 4",
"output": "NO"
},
{
"input": "6 3 1",
"output": "YES"
},
{
"input": "6 3 5",
"output": "YES"
},
{
"input": "8 4 4",
"output": "NO"
},
{
"input": "6 3 2",
"output": "YES"
},
{
"input": "4 3 3",
"output": "NO"
},
{
"input": "20 10 1",
"output": "YES"
},
{
"input": "8 4 1",
"output": "YES"
},
{
"input": "100 50 50",
"output": "NO"
}
] | 1,696,670,486
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 4
| 124
| 0
|
n, x, y = map(int, input().split())
notOK = x in [n, n+1] and y in [n, n+1]
print(["YES","NO"][notOK])
|
Title: Petya and Square
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Little Petya loves playing with squares. Mum bought him a square 2*n*<=×<=2*n* in size. Petya marked a cell inside the square and now he is solving the following task.
The task is to draw a broken line that would go along the grid lines and that would cut the square into two equal parts. The cutting line should not have any common points with the marked cell and the resulting two parts should be equal up to rotation.
Petya wants to determine whether it is possible to cut the square in the required manner given the sizes of the square side and the coordinates of the marked cell. Help him.
Input Specification:
The first line contains three space-separated integers 2*n*, *x* and *y* (2<=≤<=2*n*<=≤<=100,<=1<=≤<=*x*,<=*y*<=≤<=2*n*), representing the length of a square's side and the coordinates of the marked cell. It is guaranteed that 2*n* is even.
The coordinates of the marked cell are represented by a pair of numbers *x* *y*, where *x* represents the number of the row and *y* represents the number of the column. The rows and columns are numbered by consecutive integers from 1 to 2*n*. The rows are numbered from top to bottom and the columns are numbered from the left to the right.
Output Specification:
If the square is possible to cut, print "YES", otherwise print "NO" (without the quotes).
Demo Input:
['4 1 1\n', '2 2 2\n']
Demo Output:
['YES\n', 'NO\n']
Note:
A sample test from the statement and one of the possible ways of cutting the square are shown in the picture:
|
```python
n, x, y = map(int, input().split())
notOK = x in [n, n+1] and y in [n, n+1]
print(["YES","NO"][notOK])
```
| 0
|
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<=≤<=*n*<=≤<=1000) — the number of the tram's stops.
Then *n* lines follow, each contains two integers *a**i* and *b**i* (0<=≤<=*a**i*,<=*b**i*<=≤<=1000) — 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"
}
] | 1,697,106,660
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 71
| 92
| 0
|
s = 0
maxim = 0
for i in range(int(input())):
a,b = map(int, input().split())
s -= a
s += b
maxim = max(maxim, s)
print(maxim)
|
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<=≤<=*n*<=≤<=1000) — the number of the tram's stops.
Then *n* lines follow, each contains two integers *a**i* and *b**i* (0<=≤<=*a**i*,<=*b**i*<=≤<=1000) — 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
s = 0
maxim = 0
for i in range(int(input())):
a,b = map(int, input().split())
s -= a
s += b
maxim = max(maxim, s)
print(maxim)
```
| 3
|
|
733
|
A
|
Grasshopper And the String
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
One day, the Grasshopper was jumping on the lawn and found a piece of paper with a string. Grasshopper became interested what is the minimum jump ability he should have in order to be able to reach the far end of the string, jumping only on vowels of the English alphabet. Jump ability is the maximum possible length of his jump.
Formally, consider that at the begginning the Grasshopper is located directly in front of the leftmost character of the string. His goal is to reach the position right after the rightmost character of the string. In one jump the Grasshopper could jump to the right any distance from 1 to the value of his jump ability.
The following letters are vowels: 'A', 'E', 'I', 'O', 'U' and 'Y'.
|
The first line contains non-empty string consisting of capital English letters. It is guaranteed that the length of the string does not exceed 100.
|
Print single integer *a* — the minimum jump ability of the Grasshopper (in the number of symbols) that is needed to overcome the given string, jumping only on vowels.
|
[
"ABABBBACFEYUKOTT\n",
"AAA\n"
] |
[
"4",
"1"
] |
none
| 500
|
[
{
"input": "ABABBBACFEYUKOTT",
"output": "4"
},
{
"input": "AAA",
"output": "1"
},
{
"input": "A",
"output": "1"
},
{
"input": "B",
"output": "2"
},
{
"input": "AEYUIOAEIYAEOUIYOEIUYEAOIUEOEAYOEIUYAEOUIYEOIKLMJNHGTRWSDZXCVBNMHGFDSXVWRTPPPLKMNBXIUOIUOIUOIUOOIU",
"output": "39"
},
{
"input": "AEYUIOAEIYAEOUIYOEIUYEAOIUEOEAYOEIUYAEOUIYEOIAEYUIOAEIYAEOUIYOEIUYEAOIUEOEAYOEIUYAEOUIYEOI",
"output": "1"
},
{
"input": "KMLPTGFHNBVCDRFGHNMBVXWSQFDCVBNHTJKLPMNFVCKMLPTGFHNBVCDRFGHNMBVXWSQFDCVBNHTJKLPMNFVC",
"output": "85"
},
{
"input": "QWERTYUIOPASDFGHJKLZXCVBNMQWERTYUIOPASDFGHJKLZXCVBNMQWERTYUIOPASDFGHJKLZXCVBNMQWERTYUIOPASDFGHJKLZ",
"output": "18"
},
{
"input": "PKLKBWTXVJ",
"output": "11"
},
{
"input": "CFHFPTGMOKXVLJJZJDQW",
"output": "12"
},
{
"input": "TXULTFSBUBFLRNQORMMULWNVLPWTYJXZBPBGAWNX",
"output": "9"
},
{
"input": "DAIUSEAUEUYUWEIOOEIOUYVYYOPEEWEBZOOOAOXUOIEUKYYOJOYAUYUUIYUXOUJLGIYEIIYUOCUAACRY",
"output": "4"
},
{
"input": "VRPHBNWNWVWBWMFJJDCTJQJDJBKSJRZLVQRVVFLTZFSGCGDXCWQVWWWMFVCQHPKXXVRKTGWGPSMQTPKNDQJHNSKLXPCXDJDQDZZD",
"output": "101"
},
{
"input": "SGDDFCDRDWGPNNFBBZZJSPXFYMZKPRXTCHVJSJJBWZXXQMDZBNKDHRGSRLGLRKPMWXNSXJPNJLDPXBSRCQMHJKPZNTPNTZXNPCJC",
"output": "76"
},
{
"input": "NVTQVNLGWFDBCBKSDLTBGWBMNQZWZQJWNGVCTCQBGWNTYJRDBPZJHXCXFMIXNRGSTXHQPCHNFQPCMDZWJGLJZWMRRFCVLBKDTDSC",
"output": "45"
},
{
"input": "SREZXQFVPQCLRCQGMKXCBRWKYZKWKRMZGXPMKWNMFZTRDPHJFCSXVPPXWKZMZTBFXGNLPLHZIPLFXNRRQFDTLFPKBGCXKTMCFKKT",
"output": "48"
},
{
"input": "ICKJKMVPDNZPLKDSLTPZNRLSQSGHQJQQPJJSNHNWVDLJRLZEJSXZDPHYXGGWXHLCTVQSKWNWGTLJMOZVJNZPVXGVPJKHFVZTGCCX",
"output": "47"
},
{
"input": "XXFPZDRPXLNHGDVCBDKJMKLGUQZXLLWYLOKFZVGXVNPJWZZZNRMQBRJCZTSDRHSNCVDMHKVXCXPCRBWSJCJWDRDPVZZLCZRTDRYA",
"output": "65"
},
{
"input": "HDDRZDKCHHHEDKHZMXQSNQGSGNNSCCPVJFGXGNCEKJMRKSGKAPQWPCWXXWHLSMRGSJWEHWQCSJJSGLQJXGVTBYALWMLKTTJMFPFS",
"output": "28"
},
{
"input": "PXVKJHXVDPWGLHWFWMJPMCCNHCKSHCPZXGIHHNMYNFQBUCKJJTXXJGKRNVRTQFDFMLLGPQKFOVNNLTNDIEXSARRJKGSCZKGGJCBW",
"output": "35"
},
{
"input": "EXNMTTFPJLDHXDQBJJRDRYBZVFFHUDCHCPNFZWXSMZXNFVJGHZWXVBRQFNUIDVLZOVPXQNVMFNBTJDSCKRLNGXPSADTGCAHCBJKL",
"output": "30"
},
{
"input": "NRNLSQQJGIJBCZFTNKJCXMGPARGWXPSHZXOBNSFOLDQVXTVAGJZNLXULHBRDGMNQKQGWMRRDPYCSNFVPUFTFBUBRXVJGNGSPJKLL",
"output": "19"
},
{
"input": "SRHOKCHQQMVZKTCVQXJJCFGYFXGMBZSZFNAFETXILZHPGHBWZRZQFMGSEYRUDVMCIQTXTBTSGFTHRRNGNTHHWWHCTDFHSVARMCMB",
"output": "30"
},
{
"input": "HBSVZHDKGNIRQUBYKYHUPJCEETGFMVBZJTHYHFQPFBVBSMQACYAVWZXSBGNKWXFNMQJFMSCHJVWBZXZGSNBRUHTHAJKVLEXFBOFB",
"output": "34"
},
{
"input": "NXKMUGOPTUQNSRYTKUKSCWCRQSZKKFPYUMDIBJAHJCEKZJVWZAWOLOEFBFXLQDDPNNZKCQHUPBFVDSXSUCVLMZXQROYQYIKPQPWR",
"output": "17"
},
{
"input": "TEHJDICFNOLQVQOAREVAGUAWODOCXJXIHYXFAEPEXRHPKEIIRCRIVASKNTVYUYDMUQKSTSSBYCDVZKDDHTSDWJWACPCLYYOXGCLT",
"output": "15"
},
{
"input": "LCJJUZZFEIUTMSEXEYNOOAIZMORQDOANAMUCYTFRARDCYHOYOPHGGYUNOGNXUAOYSEMXAZOOOFAVHQUBRNGORSPNQWZJYQQUNPEB",
"output": "9"
},
{
"input": "UUOKAOOJBXUTSMOLOOOOSUYYFTAVBNUXYFVOOGCGZYQEOYISIYOUULUAIJUYVVOENJDOCLHOSOHIHDEJOIGZNIXEMEGZACHUAQFW",
"output": "5"
},
{
"input": "OUUBEHXOOURMOAIAEHXCUOIYHUJEVAWYRCIIAGDRIPUIPAIUYAIWJEVYEYYUYBYOGVYESUJCFOJNUAHIOOKBUUHEJFEWPOEOUHYA",
"output": "4"
},
{
"input": "EMNOYEEUIOUHEWZITIAEZNCJUOUAOQEAUYEIHYUSUYUUUIAEDIOOERAEIRBOJIEVOMECOGAIAIUIYYUWYIHIOWVIJEYUEAFYULSE",
"output": "5"
},
{
"input": "BVOYEAYOIEYOREJUYEUOEOYIISYAEOUYAAOIOEOYOOOIEFUAEAAESUOOIIEUAAGAEISIAPYAHOOEYUJHUECGOYEIDAIRTBHOYOYA",
"output": "5"
},
{
"input": "GOIEOAYIEYYOOEOAIAEOOUWYEIOTNYAANAYOOXEEOEAVIOIAAIEOIAUIAIAAUEUAOIAEUOUUZYIYAIEUEGOOOOUEIYAEOSYAEYIO",
"output": "3"
},
{
"input": "AUEAOAYIAOYYIUIOAULIOEUEYAIEYYIUOEOEIEYRIYAYEYAEIIMMAAEAYAAAAEOUICAUAYOUIAOUIAIUOYEOEEYAEYEYAAEAOYIY",
"output": "3"
},
{
"input": "OAIIYEYYAOOEIUOEEIOUOIAEFIOAYETUYIOAAAEYYOYEYOEAUIIUEYAYYIIAOIEEYGYIEAAOOWYAIEYYYIAOUUOAIAYAYYOEUEOY",
"output": "2"
},
{
"input": "EEEAOEOEEIOUUUEUEAAOEOIUYJEYAIYIEIYYEAUOIIYIUOOEUCYEOOOYYYIUUAYIAOEUEIEAOUOIAACAOOUAUIYYEAAAOOUYIAAE",
"output": "2"
},
{
"input": "AYEYIIEUIYOYAYEUEIIIEUYUUAUEUIYAIAAUYONIEYIUIAEUUOUOYYOUUUIUIAEYEOUIIUOUUEOAIUUYAAEOAAEOYUUIYAYRAIII",
"output": "2"
},
{
"input": "YOOAAUUAAAYEUYIUIUYIUOUAEIEEIAUEOAUIIAAIUYEUUOYUIYEAYAAAYUEEOEEAEOEEYYOUAEUYEEAIIYEUEYJOIIYUIOIUOIEE",
"output": "2"
},
{
"input": "UYOIIIAYOOAIUUOOEEUYIOUAEOOEIOUIAIEYOAEAIOOEOOOIUYYUYIAAUIOUYYOOUAUIEYYUOAAUUEAAIEUIAUEUUIAUUOYOAYIU",
"output": "1"
},
{
"input": "ABBABBB",
"output": "4"
},
{
"input": "ABCD",
"output": "4"
},
{
"input": "XXYC",
"output": "3"
},
{
"input": "YYY",
"output": "1"
},
{
"input": "ABABBBBBBB",
"output": "8"
},
{
"input": "YYYY",
"output": "1"
},
{
"input": "YYYYY",
"output": "1"
},
{
"input": "AXXX",
"output": "4"
},
{
"input": "YYYYYYY",
"output": "1"
},
{
"input": "BYYBBB",
"output": "4"
},
{
"input": "YYYYYYYYY",
"output": "1"
},
{
"input": "CAAAAA",
"output": "2"
},
{
"input": "CCCACCCC",
"output": "5"
},
{
"input": "ABABBBACFEYUKOTTTT",
"output": "5"
},
{
"input": "AABBYYYYYYYY",
"output": "3"
},
{
"input": "BYBACYC",
"output": "2"
},
{
"input": "Y",
"output": "1"
},
{
"input": "ABBBBBB",
"output": "7"
},
{
"input": "BACDYDI",
"output": "3"
},
{
"input": "XEXXXXXXXXXXXXXXX",
"output": "16"
},
{
"input": "TTYTT",
"output": "3"
},
{
"input": "AAYBC",
"output": "3"
},
{
"input": "ABABBBACFEYUKOTTTTT",
"output": "6"
},
{
"input": "YYAYY",
"output": "1"
},
{
"input": "YZZY",
"output": "3"
},
{
"input": "YY",
"output": "1"
},
{
"input": "ZZYZZ",
"output": "3"
},
{
"input": "YBBBY",
"output": "4"
},
{
"input": "BBBACCCCCCC",
"output": "8"
},
{
"input": "YBBBBY",
"output": "5"
},
{
"input": "YYYYYYYYYY",
"output": "1"
},
{
"input": "ABABBBBBBBBBBBB",
"output": "13"
}
] | 1,666,386,033
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 70
| 46
| 0
|
string = input()
minimo = 0
saltoactual = 1
for letra in string:
if letra not in("AEIOUY"):
saltoactual += 1
if minimo < saltoactual:
minimo = saltoactual
elif letra in("AEIOUY"):
saltoactual = 1
print(minimo)
|
Title: Grasshopper And the String
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day, the Grasshopper was jumping on the lawn and found a piece of paper with a string. Grasshopper became interested what is the minimum jump ability he should have in order to be able to reach the far end of the string, jumping only on vowels of the English alphabet. Jump ability is the maximum possible length of his jump.
Formally, consider that at the begginning the Grasshopper is located directly in front of the leftmost character of the string. His goal is to reach the position right after the rightmost character of the string. In one jump the Grasshopper could jump to the right any distance from 1 to the value of his jump ability.
The following letters are vowels: 'A', 'E', 'I', 'O', 'U' and 'Y'.
Input Specification:
The first line contains non-empty string consisting of capital English letters. It is guaranteed that the length of the string does not exceed 100.
Output Specification:
Print single integer *a* — the minimum jump ability of the Grasshopper (in the number of symbols) that is needed to overcome the given string, jumping only on vowels.
Demo Input:
['ABABBBACFEYUKOTT\n', 'AAA\n']
Demo Output:
['4', '1']
Note:
none
|
```python
string = input()
minimo = 0
saltoactual = 1
for letra in string:
if letra not in("AEIOUY"):
saltoactual += 1
if minimo < saltoactual:
minimo = saltoactual
elif letra in("AEIOUY"):
saltoactual = 1
print(minimo)
```
| 3
|
|
633
|
D
|
Fibonacci-ish
|
PROGRAMMING
| 2,000
|
[
"brute force",
"dp",
"hashing",
"implementation",
"math"
] | null | null |
Yash has recently learnt about the Fibonacci sequence and is very excited about it. He calls a sequence Fibonacci-ish if
1. the sequence consists of at least two elements 1. *f*0 and *f*1 are arbitrary 1. *f**n*<=+<=2<==<=*f**n*<=+<=1<=+<=*f**n* for all *n*<=≥<=0.
You are given some sequence of integers *a*1,<=*a*2,<=...,<=*a**n*. Your task is rearrange elements of this sequence in such a way that its longest possible prefix is Fibonacci-ish sequence.
|
The first line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=1000) — the length of the sequence *a**i*.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (|*a**i*|<=≤<=109).
|
Print the length of the longest possible Fibonacci-ish prefix of the given sequence after rearrangement.
|
[
"3\n1 2 -1\n",
"5\n28 35 7 14 21\n"
] |
[
"3\n",
"4\n"
] |
In the first sample, if we rearrange elements of the sequence as - 1, 2, 1, the whole sequence *a*<sub class="lower-index">*i*</sub> would be Fibonacci-ish.
In the second sample, the optimal way to rearrange elements is <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/16f1f7e35511b29cb1396890ca2fb7dfa4d428de.png" style="max-width: 100.0%;max-height: 100.0%;"/>, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4003973f16750522e492d7d79318d7e2f0ff99cd.png" style="max-width: 100.0%;max-height: 100.0%;"/>, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/87b18fd9524b11e12faf154302fb14c1b55556fb.png" style="max-width: 100.0%;max-height: 100.0%;"/>, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/b8950ea952476baea26e03281fa2f7640b6241ef.png" style="max-width: 100.0%;max-height: 100.0%;"/>, 28.
| 1,750
|
[
{
"input": "3\n1 2 -1",
"output": "3"
},
{
"input": "5\n28 35 7 14 21",
"output": "4"
},
{
"input": "11\n-9 -1 -10 9 7 -4 0 -8 -3 3 5",
"output": "5"
},
{
"input": "10\n-4 -8 -8 8 -9 0 -7 9 1 0",
"output": "4"
},
{
"input": "2\n2 2",
"output": "2"
},
{
"input": "4\n1 -1 0 -2",
"output": "4"
},
{
"input": "2\n1000000000 1000000000",
"output": "2"
},
{
"input": "3\n1 1 2",
"output": "3"
},
{
"input": "5\n0 0 0 0 0",
"output": "5"
},
{
"input": "6\n1 -1 0 -1 -1 -2",
"output": "6"
},
{
"input": "5\n-7 0 -7 -7 -14",
"output": "5"
},
{
"input": "3\n0 -44 -49",
"output": "2"
},
{
"input": "5\n-1 1 0 0 0",
"output": "3"
},
{
"input": "2\n0 0",
"output": "2"
},
{
"input": "3\n0 0 0",
"output": "3"
},
{
"input": "4\n0 0 0 0",
"output": "4"
},
{
"input": "5\n0 0 0 0 0",
"output": "5"
},
{
"input": "10\n0 0 0 0 0 0 0 0 0 0",
"output": "10"
}
] | 1,683,435,590
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 311
| 0
|
n = int(input())
a = list(map(int, input().split()))
a_dict = {}
for x in a:
a_dict[x] = a_dict.get(x, 0) + 1
a.sort()
ans = 0
for i in range(n):
for j in range(i+1, n):
f0, f1 = a[i], a[j]
cnt = 2
while f0 + f1 in a_dict and a_dict[f0 + f1] > 0:
f0, f1 = f1, f0 + f1
cnt += 1
a_dict[f1] -= 1
ans = max(ans, cnt)
a_dict[f1] += 1
print(ans)
|
Title: Fibonacci-ish
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Yash has recently learnt about the Fibonacci sequence and is very excited about it. He calls a sequence Fibonacci-ish if
1. the sequence consists of at least two elements 1. *f*0 and *f*1 are arbitrary 1. *f**n*<=+<=2<==<=*f**n*<=+<=1<=+<=*f**n* for all *n*<=≥<=0.
You are given some sequence of integers *a*1,<=*a*2,<=...,<=*a**n*. Your task is rearrange elements of this sequence in such a way that its longest possible prefix is Fibonacci-ish sequence.
Input Specification:
The first line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=1000) — the length of the sequence *a**i*.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (|*a**i*|<=≤<=109).
Output Specification:
Print the length of the longest possible Fibonacci-ish prefix of the given sequence after rearrangement.
Demo Input:
['3\n1 2 -1\n', '5\n28 35 7 14 21\n']
Demo Output:
['3\n', '4\n']
Note:
In the first sample, if we rearrange elements of the sequence as - 1, 2, 1, the whole sequence *a*<sub class="lower-index">*i*</sub> would be Fibonacci-ish.
In the second sample, the optimal way to rearrange elements is <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/16f1f7e35511b29cb1396890ca2fb7dfa4d428de.png" style="max-width: 100.0%;max-height: 100.0%;"/>, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4003973f16750522e492d7d79318d7e2f0ff99cd.png" style="max-width: 100.0%;max-height: 100.0%;"/>, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/87b18fd9524b11e12faf154302fb14c1b55556fb.png" style="max-width: 100.0%;max-height: 100.0%;"/>, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/b8950ea952476baea26e03281fa2f7640b6241ef.png" style="max-width: 100.0%;max-height: 100.0%;"/>, 28.
|
```python
n = int(input())
a = list(map(int, input().split()))
a_dict = {}
for x in a:
a_dict[x] = a_dict.get(x, 0) + 1
a.sort()
ans = 0
for i in range(n):
for j in range(i+1, n):
f0, f1 = a[i], a[j]
cnt = 2
while f0 + f1 in a_dict and a_dict[f0 + f1] > 0:
f0, f1 = f1, f0 + f1
cnt += 1
a_dict[f1] -= 1
ans = max(ans, cnt)
a_dict[f1] += 1
print(ans)
```
| 0
|
|
103
|
B
|
Cthulhu
|
PROGRAMMING
| 1,500
|
[
"dfs and similar",
"dsu",
"graphs"
] |
B. Cthulhu
|
2
|
256
|
...Once upon a time a man came to the sea. The sea was stormy and dark. The man started to call for the little mermaid to appear but alas, he only woke up Cthulhu...
Whereas on the other end of the world Pentagon is actively collecting information trying to predict the monster's behavior and preparing the secret super weapon. Due to high seismic activity and poor weather conditions the satellites haven't yet been able to make clear shots of the monster. The analysis of the first shot resulted in an undirected graph with *n* vertices and *m* edges. Now the world's best minds are about to determine whether this graph can be regarded as Cthulhu or not.
To add simplicity, let's suppose that Cthulhu looks from the space like some spherical body with tentacles attached to it. Formally, we shall regard as Cthulhu such an undirected graph that can be represented as a set of three or more rooted trees, whose roots are connected by a simple cycle.
It is guaranteed that the graph contains no multiple edges and self-loops.
|
The first line contains two integers — the number of vertices *n* and the number of edges *m* of the graph (1<=≤<=*n*<=≤<=100, 0<=≤<=*m*<=≤<=).
Each of the following *m* lines contains a pair of integers *x* and *y*, that show that an edge exists between vertices *x* and *y* (1<=≤<=*x*,<=*y*<=≤<=*n*,<=*x*<=≠<=*y*). For each pair of vertices there will be at most one edge between them, no edge connects a vertex to itself.
|
Print "NO", if the graph is not Cthulhu and "FHTAGN!" if it is.
|
[
"6 6\n6 3\n6 4\n5 1\n2 5\n1 4\n5 4\n",
"6 5\n5 6\n4 6\n3 1\n5 1\n1 2\n"
] |
[
"FHTAGN!",
"NO"
] |
Let us denote as a simple cycle a set of *v* vertices that can be numbered so that the edges will only exist between vertices number 1 and 2, 2 and 3, ..., *v* - 1 and *v*, *v* and 1.
A tree is a connected undirected graph consisting of *n* vertices and *n* - 1 edges (*n* > 0).
A rooted tree is a tree where one vertex is selected to be the root.
| 1,000
|
[
{
"input": "6 6\n6 3\n6 4\n5 1\n2 5\n1 4\n5 4",
"output": "FHTAGN!"
},
{
"input": "6 5\n5 6\n4 6\n3 1\n5 1\n1 2",
"output": "NO"
},
{
"input": "10 10\n4 10\n8 5\n2 8\n4 9\n9 3\n2 7\n10 6\n10 2\n9 8\n1 8",
"output": "FHTAGN!"
},
{
"input": "5 4\n1 5\n1 3\n1 4\n3 2",
"output": "NO"
},
{
"input": "12 12\n4 12\n4 7\n4 9\n7 2\n5 12\n2 1\n5 9\n8 6\n10 12\n2 5\n10 9\n12 3",
"output": "NO"
},
{
"input": "12 15\n3 2\n11 12\n1 9\n2 1\n1 8\n9 6\n11 5\n9 5\n9 10\n11 3\n7 11\n5 6\n11 10\n4 6\n4 2",
"output": "NO"
},
{
"input": "12 10\n1 11\n3 6\n5 7\n4 7\n6 8\n11 7\n3 12\n11 12\n7 9\n12 2",
"output": "NO"
},
{
"input": "1 0",
"output": "NO"
},
{
"input": "2 1\n1 2",
"output": "NO"
},
{
"input": "3 1\n1 3",
"output": "NO"
},
{
"input": "3 2\n1 2\n2 3",
"output": "NO"
},
{
"input": "3 3\n1 2\n2 3\n3 1",
"output": "FHTAGN!"
},
{
"input": "4 4\n1 2\n3 4\n4 1\n2 4",
"output": "FHTAGN!"
},
{
"input": "6 6\n1 2\n2 3\n3 1\n4 5\n5 6\n6 4",
"output": "NO"
},
{
"input": "2 0",
"output": "NO"
},
{
"input": "3 0",
"output": "NO"
},
{
"input": "100 0",
"output": "NO"
},
{
"input": "100 1\n11 23",
"output": "NO"
},
{
"input": "10 10\n5 7\n8 1\n10 3\n6 4\n10 6\n5 3\n5 6\n2 6\n4 3\n2 10",
"output": "NO"
},
{
"input": "20 20\n9 10\n4 19\n9 20\n12 20\n1 15\n2 12\n19 10\n19 15\n4 10\n4 8\n8 9\n20 8\n6 2\n2 15\n7 19\n20 4\n3 16\n1 20\n9 1\n20 10",
"output": "NO"
},
{
"input": "30 30\n17 6\n16 29\n16 13\n16 20\n29 26\n17 5\n27 28\n24 16\n7 18\n24 10\n1 27\n12 17\n27 30\n6 1\n3 30\n5 19\n18 13\n16 2\n30 1\n5 8\n14 16\n26 18\n7 19\n5 6\n23 14\n6 8\n23 8\n18 8\n18 3\n5 21",
"output": "NO"
},
{
"input": "100 66\n41 14\n19 13\n70 43\n79 62\n9 62\n71 40\n53 86\n80 4\n34 33\n72 68\n40 96\n84 59\n36 77\n55 50\n40 3\n79 81\n3 43\n33 47\n22 98\n33 90\n56 49\n69 28\n73 30\n65 22\n98 20\n9 52\n54 20\n32 70\n51 80\n63 12\n21 48\n35 17\n48 87\n25 43\n65 80\n42 3\n86 35\n95 98\n43 59\n51 46\n66 37\n88 34\n32 47\n24 42\n21 44\n92 59\n81 6\n100 82\n85 6\n58 25\n66 6\n14 32\n59 85\n3 98\n44 4\n85 51\n69 41\n80 70\n81 24\n75 71\n93 9\n82 55\n70 46\n66 32\n77 58\n11 46",
"output": "NO"
},
{
"input": "4 4\n1 2\n4 3\n2 3\n3 1",
"output": "FHTAGN!"
},
{
"input": "5 5\n2 3\n2 4\n5 4\n4 1\n1 2",
"output": "FHTAGN!"
},
{
"input": "10 10\n1 10\n5 9\n6 2\n8 9\n9 1\n5 4\n2 8\n1 3\n6 3\n4 1",
"output": "NO"
},
{
"input": "6 6\n1 2\n2 3\n3 1\n4 5\n5 6\n6 4",
"output": "NO"
},
{
"input": "4 3\n1 2\n2 3\n3 1",
"output": "NO"
},
{
"input": "6 5\n1 2\n2 3\n3 1\n1 4\n1 5",
"output": "NO"
}
] | 1,434,407,536
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 0
| 186
| 0
|
def dfs(graph, start):
visited[start] = True
for i in graph[start]:
if not visited[i]:
dfs(graph, i)
a, b = list(map(int, input().split(' ')))
g = {}
for i in range(1, a+1):
g[i] = []
for i in range(b):
x, y = map(int, input().split(' '))
g[x].append(y)
g[y].append(x)
visited = [True] + [False] * a
dfs(g, 1)
if False in visited:
print("NO")
if a != b:
print("NO")
print("FHTAGN!")
|
Title: Cthulhu
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
...Once upon a time a man came to the sea. The sea was stormy and dark. The man started to call for the little mermaid to appear but alas, he only woke up Cthulhu...
Whereas on the other end of the world Pentagon is actively collecting information trying to predict the monster's behavior and preparing the secret super weapon. Due to high seismic activity and poor weather conditions the satellites haven't yet been able to make clear shots of the monster. The analysis of the first shot resulted in an undirected graph with *n* vertices and *m* edges. Now the world's best minds are about to determine whether this graph can be regarded as Cthulhu or not.
To add simplicity, let's suppose that Cthulhu looks from the space like some spherical body with tentacles attached to it. Formally, we shall regard as Cthulhu such an undirected graph that can be represented as a set of three or more rooted trees, whose roots are connected by a simple cycle.
It is guaranteed that the graph contains no multiple edges and self-loops.
Input Specification:
The first line contains two integers — the number of vertices *n* and the number of edges *m* of the graph (1<=≤<=*n*<=≤<=100, 0<=≤<=*m*<=≤<=).
Each of the following *m* lines contains a pair of integers *x* and *y*, that show that an edge exists between vertices *x* and *y* (1<=≤<=*x*,<=*y*<=≤<=*n*,<=*x*<=≠<=*y*). For each pair of vertices there will be at most one edge between them, no edge connects a vertex to itself.
Output Specification:
Print "NO", if the graph is not Cthulhu and "FHTAGN!" if it is.
Demo Input:
['6 6\n6 3\n6 4\n5 1\n2 5\n1 4\n5 4\n', '6 5\n5 6\n4 6\n3 1\n5 1\n1 2\n']
Demo Output:
['FHTAGN!', 'NO']
Note:
Let us denote as a simple cycle a set of *v* vertices that can be numbered so that the edges will only exist between vertices number 1 and 2, 2 and 3, ..., *v* - 1 and *v*, *v* and 1.
A tree is a connected undirected graph consisting of *n* vertices and *n* - 1 edges (*n* > 0).
A rooted tree is a tree where one vertex is selected to be the root.
|
```python
def dfs(graph, start):
visited[start] = True
for i in graph[start]:
if not visited[i]:
dfs(graph, i)
a, b = list(map(int, input().split(' ')))
g = {}
for i in range(1, a+1):
g[i] = []
for i in range(b):
x, y = map(int, input().split(' '))
g[x].append(y)
g[y].append(x)
visited = [True] + [False] * a
dfs(g, 1)
if False in visited:
print("NO")
if a != b:
print("NO")
print("FHTAGN!")
```
| 0
|
796
|
A
|
Buying A House
|
PROGRAMMING
| 800
|
[
"brute force",
"implementation"
] | null | null |
Zane the wizard had never loved anyone before, until he fell in love with a girl, whose name remains unknown to us.
The girl lives in house *m* of a village. There are *n* houses in that village, lining in a straight line from left to right: house 1, house 2, ..., house *n*. The village is also well-structured: house *i* and house *i*<=+<=1 (1<=≤<=*i*<=<<=*n*) are exactly 10 meters away. In this village, some houses are occupied, and some are not. Indeed, unoccupied houses can be purchased.
You will be given *n* integers *a*1,<=*a*2,<=...,<=*a**n* that denote the availability and the prices of the houses. If house *i* is occupied, and therefore cannot be bought, then *a**i* equals 0. Otherwise, house *i* can be bought, and *a**i* represents the money required to buy it, in dollars.
As Zane has only *k* dollars to spare, it becomes a challenge for him to choose the house to purchase, so that he could live as near as possible to his crush. Help Zane determine the minimum distance from his crush's house to some house he can afford, to help him succeed in his love.
|
The first line contains three integers *n*, *m*, and *k* (2<=≤<=*n*<=≤<=100, 1<=≤<=*m*<=≤<=*n*, 1<=≤<=*k*<=≤<=100) — the number of houses in the village, the house where the girl lives, and the amount of money Zane has (in dollars), respectively.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=100) — denoting the availability and the prices of the houses.
It is guaranteed that *a**m*<==<=0 and that it is possible to purchase some house with no more than *k* dollars.
|
Print one integer — the minimum distance, in meters, from the house where the girl Zane likes lives to the house Zane can buy.
|
[
"5 1 20\n0 27 32 21 19\n",
"7 3 50\n62 0 0 0 99 33 22\n",
"10 5 100\n1 0 1 0 0 0 0 0 1 1\n"
] |
[
"40",
"30",
"20"
] |
In the first sample, with *k* = 20 dollars, Zane can buy only house 5. The distance from house *m* = 1 to house 5 is 10 + 10 + 10 + 10 = 40 meters.
In the second sample, Zane can buy houses 6 and 7. It is better to buy house 6 than house 7, since house *m* = 3 and house 6 are only 30 meters away, while house *m* = 3 and house 7 are 40 meters away.
| 500
|
[
{
"input": "5 1 20\n0 27 32 21 19",
"output": "40"
},
{
"input": "7 3 50\n62 0 0 0 99 33 22",
"output": "30"
},
{
"input": "10 5 100\n1 0 1 0 0 0 0 0 1 1",
"output": "20"
},
{
"input": "5 3 1\n1 1 0 0 1",
"output": "10"
},
{
"input": "5 5 5\n1 0 5 6 0",
"output": "20"
},
{
"input": "15 10 50\n20 0 49 50 50 50 50 50 50 0 50 50 49 0 20",
"output": "10"
},
{
"input": "7 5 1\n0 100 2 2 0 2 1",
"output": "20"
},
{
"input": "100 50 100\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 0 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",
"output": "10"
},
{
"input": "100 50 1\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 0 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",
"output": "490"
},
{
"input": "100 77 50\n50 100 49 51 0 50 100 49 51 0 50 100 49 51 0 50 100 49 51 0 50 100 49 51 0 50 100 49 51 0 50 100 49 51 0 50 100 49 51 0 50 100 49 51 0 50 100 49 51 0 50 100 49 51 0 50 100 49 51 0 50 100 49 51 0 50 100 49 51 0 50 100 49 51 0 50 0 49 51 0 50 100 49 51 0 50 100 49 51 0 50 100 49 51 0 50 100 49 51 0",
"output": "10"
},
{
"input": "100 1 1\n0 98 97 96 95 94 93 92 91 90 89 88 87 86 85 84 83 82 81 80 79 78 77 76 75 74 73 72 71 70 69 68 67 66 65 64 63 62 61 60 59 58 57 56 55 54 53 52 51 50 49 48 47 46 45 44 43 42 41 40 39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0",
"output": "980"
},
{
"input": "100 1 100\n0 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100",
"output": "10"
},
{
"input": "100 10 99\n0 0 0 0 0 0 0 0 0 0 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 99 98",
"output": "890"
},
{
"input": "7 4 5\n1 0 6 0 5 6 0",
"output": "10"
},
{
"input": "7 4 5\n1 6 5 0 0 6 0",
"output": "10"
},
{
"input": "100 42 59\n50 50 50 50 50 50 50 50 50 50 59 59 59 59 59 59 59 59 59 59 59 59 59 59 59 59 59 59 59 59 59 59 59 60 60 60 60 60 60 60 60 0 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 60 0",
"output": "90"
},
{
"input": "2 1 100\n0 1",
"output": "10"
},
{
"input": "2 2 100\n1 0",
"output": "10"
},
{
"input": "10 1 88\n0 95 0 0 0 0 0 94 0 85",
"output": "90"
},
{
"input": "10 2 14\n2 0 1 26 77 39 41 100 13 32",
"output": "10"
},
{
"input": "10 3 11\n0 0 0 0 0 62 0 52 1 35",
"output": "60"
},
{
"input": "20 12 44\n27 40 58 69 53 38 31 39 75 95 8 0 28 81 77 90 38 61 21 88",
"output": "10"
},
{
"input": "30 29 10\n59 79 34 12 100 6 1 58 18 73 54 11 37 46 89 90 80 85 73 45 64 5 31 0 89 19 0 74 0 82",
"output": "70"
},
{
"input": "40 22 1\n7 95 44 53 0 0 19 93 0 68 65 0 24 91 10 58 17 0 71 0 100 0 94 90 79 73 0 73 4 61 54 81 7 13 21 84 5 41 0 1",
"output": "180"
},
{
"input": "40 22 99\n60 0 100 0 0 100 100 0 0 0 0 100 100 0 0 100 100 0 100 100 100 0 100 100 100 0 100 100 0 0 100 100 100 0 0 100 0 100 0 0",
"output": "210"
},
{
"input": "50 10 82\n56 54 0 0 0 0 88 93 0 0 83 93 0 0 91 89 0 30 62 52 24 84 80 8 38 13 92 78 16 87 23 30 71 55 16 63 15 99 4 93 24 6 3 35 4 42 73 27 86 37",
"output": "80"
},
{
"input": "63 49 22\n18 3 97 52 75 2 12 24 58 75 80 97 22 10 79 51 30 60 68 99 75 2 35 3 97 88 9 7 18 5 0 0 0 91 0 91 56 36 76 0 0 0 52 27 35 0 51 72 0 96 57 0 0 0 0 92 55 28 0 30 0 78 77",
"output": "190"
},
{
"input": "74 38 51\n53 36 55 42 64 5 87 9 0 16 86 78 9 22 19 1 25 72 1 0 0 0 79 0 0 0 77 58 70 0 0 100 64 0 99 59 0 0 0 0 65 74 0 96 0 58 89 93 61 88 0 0 82 89 0 0 49 24 7 77 89 87 94 61 100 31 93 70 39 49 39 14 20 84",
"output": "190"
},
{
"input": "89 22 11\n36 0 68 89 0 85 72 0 38 56 0 44 0 94 0 28 71 0 0 18 0 0 0 89 0 0 0 75 0 0 0 32 66 0 0 0 0 0 0 48 63 0 64 58 0 23 48 0 0 52 93 61 57 0 18 0 0 34 62 17 0 41 0 0 53 59 44 0 0 51 40 0 0 100 100 54 0 88 0 5 45 56 57 67 24 16 88 86 15",
"output": "580"
},
{
"input": "97 44 100\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 51 19",
"output": "520"
},
{
"input": "100 1 1\n0 0 0 0 10 54 84 6 17 94 65 82 34 0 61 46 42 0 2 16 56 0 100 0 82 0 0 0 89 78 96 56 0 0 0 0 0 0 0 0 77 70 0 96 67 0 0 32 44 1 72 50 14 11 24 61 100 64 19 5 67 69 44 82 93 22 67 93 22 61 53 64 79 41 84 48 43 97 7 24 8 49 23 16 72 52 97 29 69 47 29 49 64 91 4 73 17 18 51 67",
"output": "490"
},
{
"input": "100 1 50\n0 0 0 60 0 0 54 0 80 0 0 0 97 0 68 97 84 0 0 93 0 0 0 0 68 0 0 62 0 0 55 68 65 87 0 69 0 0 0 0 0 52 61 100 0 71 0 82 88 78 0 81 0 95 0 57 0 67 0 0 0 55 86 0 60 72 0 0 73 0 83 0 0 60 64 0 56 0 0 77 84 0 58 63 84 0 0 67 0 16 3 88 0 98 31 52 40 35 85 23",
"output": "890"
},
{
"input": "100 1 100\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 91 70 14",
"output": "970"
},
{
"input": "100 1 29\n0 0 0 0 64 0 89 97 0 0 0 59 0 67 62 0 59 0 0 80 0 0 0 0 0 97 0 57 0 64 32 0 44 0 0 48 0 47 38 0 42 0 0 0 0 0 0 46 74 0 86 33 33 0 44 0 79 0 0 0 0 91 59 0 59 65 55 0 0 58 33 95 0 97 76 0 81 0 41 0 38 81 80 0 85 0 31 0 0 92 0 0 45 96 0 85 91 87 0 10",
"output": "990"
},
{
"input": "100 50 20\n3 0 32 0 48 32 64 0 54 26 0 0 0 0 0 28 0 0 54 0 0 45 49 0 38 74 0 0 39 42 62 48 75 96 89 42 0 44 0 0 30 21 76 0 50 0 79 0 0 0 0 99 0 84 62 0 0 0 0 53 80 0 28 0 0 53 0 0 38 0 62 0 0 62 0 0 88 0 44 32 0 81 35 45 49 0 69 73 38 27 72 0 96 72 69 0 0 22 76 10",
"output": "490"
},
{
"input": "100 50 20\n49 0 56 0 87 25 40 0 50 0 0 97 0 0 36 29 0 0 0 0 0 73 29 71 44 0 0 0 91 92 69 0 0 60 81 49 48 38 0 87 0 82 0 32 0 82 46 39 0 0 29 0 0 29 0 79 47 0 0 0 0 0 49 0 24 33 70 0 63 45 97 90 0 0 29 53 55 0 84 0 0 100 26 0 88 0 0 0 0 81 70 0 30 80 0 75 59 98 0 2",
"output": "500"
},
{
"input": "100 2 2\n0 0 43 90 47 5 2 97 52 69 21 48 64 10 34 97 97 74 8 19 68 56 55 24 47 38 43 73 72 72 60 60 51 36 33 44 100 45 13 54 72 52 0 15 3 6 50 8 88 4 78 26 40 27 30 63 67 83 61 91 33 97 54 20 92 27 89 35 10 7 84 50 11 95 74 88 24 44 74 100 18 56 34 91 41 34 51 51 11 91 89 54 19 100 83 89 10 17 76 20",
"output": "50"
},
{
"input": "100 100 34\n5 73 0 0 44 0 0 0 79 55 0 0 0 0 0 0 0 0 83 67 75 0 0 0 0 59 0 74 0 0 47 98 0 0 72 41 0 55 87 0 0 78 84 0 0 39 0 79 72 95 0 0 0 0 0 85 53 84 0 0 0 0 37 75 0 66 0 0 0 0 61 0 70 0 37 60 42 78 92 52 0 0 0 55 77 57 0 63 37 0 0 0 96 70 0 94 97 0 0 0",
"output": "990"
},
{
"input": "100 100 100\n43 79 21 87 84 14 28 69 92 16 3 71 79 37 48 37 72 58 12 72 62 49 37 17 60 54 41 99 15 72 40 89 76 1 99 87 14 56 63 48 69 37 96 64 7 14 1 73 85 33 98 70 97 71 96 28 49 71 56 2 67 22 100 2 98 100 62 77 92 76 98 98 47 26 22 47 50 56 9 16 72 47 5 62 29 78 81 1 0 63 32 65 87 3 40 53 8 80 93 0",
"output": "10"
},
{
"input": "100 38 1\n3 59 12 81 33 95 0 41 36 17 63 76 42 77 85 56 3 96 55 41 24 87 18 9 0 37 0 61 69 0 0 0 67 0 0 0 0 0 0 18 0 0 47 56 74 0 0 80 0 42 0 1 60 59 62 9 19 87 92 48 58 30 98 51 99 10 42 94 51 53 50 89 24 5 52 82 50 39 98 8 95 4 57 21 10 0 44 32 19 14 64 34 79 76 17 3 15 22 71 51",
"output": "140"
},
{
"input": "100 72 1\n56 98 8 27 9 23 16 76 56 1 34 43 96 73 75 49 62 20 18 23 51 55 30 84 4 20 89 40 75 16 69 35 1 0 16 0 80 0 41 17 0 0 76 23 0 92 0 34 0 91 82 54 0 0 0 63 85 59 98 24 29 0 8 77 26 0 34 95 39 0 0 0 74 0 0 0 0 12 0 92 0 0 55 95 66 30 0 0 29 98 0 0 0 47 0 0 80 0 0 4",
"output": "390"
},
{
"input": "100 66 1\n38 50 64 91 37 44 74 21 14 41 80 90 26 51 78 85 80 86 44 14 49 75 93 48 78 89 23 72 35 22 14 48 100 71 62 22 7 95 80 66 32 20 17 47 79 30 41 52 15 62 67 71 1 6 0 9 0 0 0 11 0 0 24 0 31 0 77 0 51 0 0 0 0 0 0 77 0 36 44 19 90 45 6 25 100 87 93 30 4 97 36 88 33 50 26 71 97 71 51 68",
"output": "130"
},
{
"input": "100 55 1\n0 33 45 83 56 96 58 24 45 30 38 60 39 69 21 87 59 21 72 73 27 46 61 61 11 97 77 5 39 3 3 35 76 37 53 84 24 75 9 48 31 90 100 84 74 81 83 83 42 23 29 94 18 1 0 53 52 99 86 37 94 54 28 75 28 80 17 14 98 68 76 20 32 23 42 31 57 79 60 14 18 27 1 98 32 3 96 25 15 38 2 6 3 28 59 54 63 2 43 59",
"output": "10"
},
{
"input": "100 55 1\n24 52 41 6 55 11 58 25 63 12 70 39 23 28 72 17 96 85 7 84 21 13 34 37 97 43 36 32 15 30 58 5 14 71 40 70 9 92 44 73 31 58 96 90 19 35 29 91 25 36 48 95 61 78 0 1 99 61 81 88 42 53 61 57 42 55 74 45 41 92 99 30 20 25 89 50 37 4 17 24 6 65 15 44 40 2 38 43 7 90 38 59 75 87 96 28 12 67 24 32",
"output": "10"
},
{
"input": "100 21 1\n62 5 97 80 81 28 83 0 26 0 0 0 0 23 0 0 90 0 0 0 0 0 0 0 0 54 71 8 0 0 42 0 73 0 17 0 1 31 71 78 58 72 84 39 54 59 13 29 16 41 71 35 88 55 70 50 33 100 100 60 52 90 7 66 44 55 51 42 90 17 86 44 46 8 52 74 8 22 2 92 34 37 58 98 70 74 19 91 74 25 4 38 71 68 50 68 63 14 60 98",
"output": "160"
},
{
"input": "5 2 20\n27 0 32 21 19",
"output": "30"
},
{
"input": "6 4 10\n10 0 0 0 0 10",
"output": "20"
},
{
"input": "8 7 100\n1 0 0 0 0 0 0 1",
"output": "10"
},
{
"input": "5 3 20\n1 21 0 0 1",
"output": "20"
},
{
"input": "4 3 1\n0 0 0 1",
"output": "10"
},
{
"input": "5 2 3\n4 0 5 6 1",
"output": "30"
},
{
"input": "5 3 87\n88 89 0 1 90",
"output": "10"
},
{
"input": "5 3 20\n15 30 0 15 35",
"output": "10"
},
{
"input": "6 3 50\n0 0 0 1 2 0",
"output": "10"
},
{
"input": "6 4 9\n100 9 10 0 0 9",
"output": "20"
},
{
"input": "5 4 20\n0 20 0 0 20",
"output": "10"
},
{
"input": "6 3 3\n1 5 0 2 2 0",
"output": "10"
},
{
"input": "5 4 100\n0 1 0 0 1",
"output": "10"
}
] | 1,579,811,520
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 58
| 124
| 307,200
|
n, m, k = map(int, input().split())
a = list(map(int, input().split()))
a.insert(0, 0)
if m == 1:
for i in range(m, len(a)):
if (a[i] <= k) and (a[i] != 0):
print((i - m) * 10)
break
elif m == n:
for i in range(m, 0, -1):
if (a[i] <= k) and (a[i] != 0):
print((m - i) * 10)
break
else:
for i in range(m, len(a)):
if (a[i] <= k) and (a[i] != 0):
ans1 = (i - m) * 10
break
else:
ans1 = float('inf')
for i in range(m, 0, -1):
if (a[i] <= k) and (a[i] != 0):
ans2 = (m - i) * 10
break
else:
ans2 = float('inf')
print(min(ans1, ans2))
|
Title: Buying A House
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Zane the wizard had never loved anyone before, until he fell in love with a girl, whose name remains unknown to us.
The girl lives in house *m* of a village. There are *n* houses in that village, lining in a straight line from left to right: house 1, house 2, ..., house *n*. The village is also well-structured: house *i* and house *i*<=+<=1 (1<=≤<=*i*<=<<=*n*) are exactly 10 meters away. In this village, some houses are occupied, and some are not. Indeed, unoccupied houses can be purchased.
You will be given *n* integers *a*1,<=*a*2,<=...,<=*a**n* that denote the availability and the prices of the houses. If house *i* is occupied, and therefore cannot be bought, then *a**i* equals 0. Otherwise, house *i* can be bought, and *a**i* represents the money required to buy it, in dollars.
As Zane has only *k* dollars to spare, it becomes a challenge for him to choose the house to purchase, so that he could live as near as possible to his crush. Help Zane determine the minimum distance from his crush's house to some house he can afford, to help him succeed in his love.
Input Specification:
The first line contains three integers *n*, *m*, and *k* (2<=≤<=*n*<=≤<=100, 1<=≤<=*m*<=≤<=*n*, 1<=≤<=*k*<=≤<=100) — the number of houses in the village, the house where the girl lives, and the amount of money Zane has (in dollars), respectively.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=100) — denoting the availability and the prices of the houses.
It is guaranteed that *a**m*<==<=0 and that it is possible to purchase some house with no more than *k* dollars.
Output Specification:
Print one integer — the minimum distance, in meters, from the house where the girl Zane likes lives to the house Zane can buy.
Demo Input:
['5 1 20\n0 27 32 21 19\n', '7 3 50\n62 0 0 0 99 33 22\n', '10 5 100\n1 0 1 0 0 0 0 0 1 1\n']
Demo Output:
['40', '30', '20']
Note:
In the first sample, with *k* = 20 dollars, Zane can buy only house 5. The distance from house *m* = 1 to house 5 is 10 + 10 + 10 + 10 = 40 meters.
In the second sample, Zane can buy houses 6 and 7. It is better to buy house 6 than house 7, since house *m* = 3 and house 6 are only 30 meters away, while house *m* = 3 and house 7 are 40 meters away.
|
```python
n, m, k = map(int, input().split())
a = list(map(int, input().split()))
a.insert(0, 0)
if m == 1:
for i in range(m, len(a)):
if (a[i] <= k) and (a[i] != 0):
print((i - m) * 10)
break
elif m == n:
for i in range(m, 0, -1):
if (a[i] <= k) and (a[i] != 0):
print((m - i) * 10)
break
else:
for i in range(m, len(a)):
if (a[i] <= k) and (a[i] != 0):
ans1 = (i - m) * 10
break
else:
ans1 = float('inf')
for i in range(m, 0, -1):
if (a[i] <= k) and (a[i] != 0):
ans2 = (m - i) * 10
break
else:
ans2 = float('inf')
print(min(ans1, ans2))
```
| 3
|
|
773
|
A
|
Success Rate
|
PROGRAMMING
| 1,700
|
[
"binary search",
"math"
] | null | null |
You are an experienced Codeforces user. Today you found out that during your activity on Codeforces you have made *y* submissions, out of which *x* have been successful. Thus, your current success rate on Codeforces is equal to *x*<=/<=*y*.
Your favorite rational number in the [0;1] range is *p*<=/<=*q*. Now you wonder: what is the smallest number of submissions you have to make if you want your success rate to be *p*<=/<=*q*?
|
The first line contains a single integer *t* (1<=≤<=*t*<=≤<=1000) — the number of test cases.
Each of the next *t* lines contains four integers *x*, *y*, *p* and *q* (0<=≤<=*x*<=≤<=*y*<=≤<=109; 0<=≤<=*p*<=≤<=*q*<=≤<=109; *y*<=><=0; *q*<=><=0).
It is guaranteed that *p*<=/<=*q* is an irreducible fraction.
Hacks. For hacks, an additional constraint of *t*<=≤<=5 must be met.
|
For each test case, output a single integer equal to the smallest number of submissions you have to make if you want your success rate to be equal to your favorite rational number, or -1 if this is impossible to achieve.
|
[
"4\n3 10 1 2\n7 14 3 8\n20 70 2 7\n5 6 1 1\n"
] |
[
"4\n10\n0\n-1\n"
] |
In the first example, you have to make 4 successful submissions. Your success rate will be equal to 7 / 14, or 1 / 2.
In the second example, you have to make 2 successful and 8 unsuccessful submissions. Your success rate will be equal to 9 / 24, or 3 / 8.
In the third example, there is no need to make any new submissions. Your success rate is already equal to 20 / 70, or 2 / 7.
In the fourth example, the only unsuccessful submission breaks your hopes of having the success rate equal to 1.
| 500
|
[
{
"input": "4\n3 10 1 2\n7 14 3 8\n20 70 2 7\n5 6 1 1",
"output": "4\n10\n0\n-1"
},
{
"input": "8\n0 1 0 1\n0 2 1 2\n0 3 1 1\n1 2 0 1\n1 2 1 1\n2 2 0 1\n3 3 1 2\n4 4 1 1",
"output": "0\n2\n-1\n-1\n-1\n-1\n3\n0"
},
{
"input": "5\n1 1000000000 1 2\n1 1000000000 1 2\n1 1000000000 1 2\n1 1000000000 1 2\n1 1000000000 1 2",
"output": "999999998\n999999998\n999999998\n999999998\n999999998"
},
{
"input": "5\n999999999 1000000000 1 1000000000\n999999999 1000000000 1 1000000000\n999999999 1000000000 1 1000000000\n999999999 1000000000 1 1000000000\n999999999 1000000000 1 1000000000",
"output": "999999998000000000\n999999998000000000\n999999998000000000\n999999998000000000\n999999998000000000"
},
{
"input": "5\n0 1000000000 999999999 1000000000\n0 1000000000 999999999 1000000000\n0 1000000000 999999999 1000000000\n0 1000000000 999999999 1000000000\n0 1000000000 999999999 1000000000",
"output": "999999999000000000\n999999999000000000\n999999999000000000\n999999999000000000\n999999999000000000"
},
{
"input": "1\n999999999 1000000000 1 2",
"output": "999999998"
},
{
"input": "1\n50 50 1 1",
"output": "0"
},
{
"input": "1\n100000000 100000000 1 2",
"output": "100000000"
},
{
"input": "1\n3 999999990 1 1000000000",
"output": "2000000010"
},
{
"input": "5\n3 10 1 2\n7 14 3 8\n20 70 2 7\n5 6 1 1\n1 1 1 1",
"output": "4\n10\n0\n-1\n0"
},
{
"input": "5\n9999999 10000000 1 1000000000\n9999999 10000000 1 1000000000\n9999999 10000000 1 1000000000\n9999999 10000000 1 1000000000\n9999999 10000000 1 1000000000",
"output": "9999998990000000\n9999998990000000\n9999998990000000\n9999998990000000\n9999998990000000"
},
{
"input": "1\n0 1000000000 999999999 1000000000",
"output": "999999999000000000"
},
{
"input": "5\n1 1000000000 999999999 1000000000\n1 1000000000 999999999 1000000000\n1 1000000000 999999999 1000000000\n1 1000000000 999999999 1000000000\n1 1000000000 999999999 1000000000",
"output": "999999998000000000\n999999998000000000\n999999998000000000\n999999998000000000\n999999998000000000"
},
{
"input": "5\n1 1000000000 999999999 1000000000\n2 1000000000 999999999 1000000000\n3 1000000000 999999999 1000000000\n4 1000000000 999999999 1000000000\n5 1000000000 999999999 1000000000",
"output": "999999998000000000\n999999997000000000\n999999996000000000\n999999995000000000\n999999994000000000"
},
{
"input": "1\n1 1 1 1",
"output": "0"
},
{
"input": "5\n999999997 999999998 2 999999999\n999999997 999999998 2 999999999\n999999997 999999998 2 999999999\n999999997 999999998 2 999999999\n999999997 999999998 2 999999999",
"output": "499999997500000003\n499999997500000003\n499999997500000003\n499999997500000003\n499999997500000003"
},
{
"input": "5\n1000000000 1000000000 1 1000000000\n1000000000 1000000000 1 1000000000\n1000000000 1000000000 1 1000000000\n1000000000 1000000000 1 1000000000\n1000000000 1000000000 1 1000000000",
"output": "999999999000000000\n999999999000000000\n999999999000000000\n999999999000000000\n999999999000000000"
},
{
"input": "5\n99999997 999999998 999999998 999999999\n99999996 999999997 999999997 999999999\n99999997 999999998 999999998 999999999\n99999996 999999997 999999997 999999999\n99999997 999999998 999999998 999999999",
"output": "899999999100000001\n449999999550000002\n899999999100000001\n449999999550000002\n899999999100000001"
},
{
"input": "1\n1000000000 1000000000 1 1000000000",
"output": "999999999000000000"
},
{
"input": "1\n7 7 1 1",
"output": "0"
},
{
"input": "5\n1000000000 1000000000 1 2\n1000000000 1000000000 1 2\n1000000000 1000000000 1 2\n1000000000 1000000000 1 2\n1000000000 1000000000 1 2",
"output": "1000000000\n1000000000\n1000000000\n1000000000\n1000000000"
},
{
"input": "1\n1000000000 1000000000 1 1",
"output": "0"
},
{
"input": "1\n1 1000000000 999999999 1000000000",
"output": "999999998000000000"
},
{
"input": "4\n1 1000000000 999999999 1000000000\n999999999 1000000000 1 1000000000\n1 2 0 1\n0 1 0 1",
"output": "999999998000000000\n999999998000000000\n-1\n0"
},
{
"input": "1\n1 1000000000 1 2",
"output": "999999998"
},
{
"input": "5\n1 982449707 1 2\n1 982449707 1 2\n1 982449707 1 2\n1 982449707 1 2\n1 982449707 1 2",
"output": "982449705\n982449705\n982449705\n982449705\n982449705"
},
{
"input": "5\n13 900000007 900000007 900000009\n13 900000007 900000007 900000009\n13 900000007 900000007 900000009\n13 900000007 900000007 900000009\n13 900000007 900000007 900000009",
"output": "405000000449999966\n405000000449999966\n405000000449999966\n405000000449999966\n405000000449999966"
},
{
"input": "1\n5 10 0 1",
"output": "-1"
},
{
"input": "1\n2 2 1 1",
"output": "0"
},
{
"input": "5\n0 999999999 999999999 1000000000\n0 999999999 999999999 1000000000\n0 999999999 999999999 1000000000\n0 999999999 999999999 1000000000\n0 999999999 999999999 1000000000",
"output": "999999998000000001\n999999998000000001\n999999998000000001\n999999998000000001\n999999998000000001"
},
{
"input": "1\n0 5 0 1",
"output": "0"
},
{
"input": "5\n999999999 1000000000 1 9999\n999999999 1000000000 1 9999\n999999999 1000000000 1 9999\n999999999 1000000000 1 9999\n999999999 1000000000 1 9999",
"output": "9997999990001\n9997999990001\n9997999990001\n9997999990001\n9997999990001"
},
{
"input": "5\n999999997 1000000000 3 1000000000\n999999997 1000000000 3 1000000000\n999999997 1000000000 3 1000000000\n999999997 1000000000 3 1000000000\n999999997 1000000000 3 1000000000",
"output": "333333332000000000\n333333332000000000\n333333332000000000\n333333332000000000\n333333332000000000"
},
{
"input": "5\n1000000000 1000000000 1 1000000000\n1000000000 1000000000 1 1000000000\n1 1000000000 999999999 1000000000\n1 1000000000 999999999 1000000000\n1 1000000000 999999999 1000000000",
"output": "999999999000000000\n999999999000000000\n999999998000000000\n999999998000000000\n999999998000000000"
},
{
"input": "5\n1000000000 1000000000 1 1000000000\n1000000000 1000000000 1 1000000000\n1000000000 1000000000 1 1000000000\n1 1000000000 999999999 1000000000\n1 1000000000 999999999 1000000000",
"output": "999999999000000000\n999999999000000000\n999999999000000000\n999999998000000000\n999999998000000000"
},
{
"input": "1\n999999998 999999999 1 10",
"output": "8999999981"
},
{
"input": "5\n1 1 1 1\n2 2 1 1\n100 100 1 1\n1000000000 1000000000 1 1\n1000000000 1000000000 1 1",
"output": "0\n0\n0\n0\n0"
},
{
"input": "4\n1 1000000000 999999999 1000000000\n1 1000000000 999999999 1000000000\n1 1000000000 999999999 1000000000\n1 1000000000 999999999 1000000000",
"output": "999999998000000000\n999999998000000000\n999999998000000000\n999999998000000000"
}
] | 1,494,174,156
| 2,256
|
Python 3
|
WRONG_ANSWER
|
PRETESTS
| 0
| 46
| 0
|
def gcd(a, b):
if b == 0:
return a
else:
return gcd(b, a % b)
class solution:
def __init__(self, a=0, b=0):
self.x = a
self.y = b
def eu (a, b, sol):
if a == 0:
sol.x = 0
sol.y = 1
return b
sol2 = solution()
d = eu (b%a, a, sol2)
sol.x = sol2.y - (b // a) * sol2.x
sol.y = sol2.x
return d
def find_any_solution (a, b, c, sol):
g = eu(abs(a), abs(b), sol)
if c % g != 0:
return -1
sol.x *= c // g
sol.y *= c // g
if (a < 0):
sol.x *= -1
if (b < 0):
sol.y *= -1
return g
def shift_solution (sol, a, b, cnt):
sol.x += cnt * b
sol.y -= cnt * a
def find_all_solution(a, b, c, minx, maxx, miny, maxy):
sol = solution()
g = find_any_solution(a, b, c, sol)
if g == -1:
return (-1, -1)
a //= g
b //= g
sign_a = 1
if a < 0:
sign_a = -1
sign_b = 1
if b < 0:
sign_b = -1
shift_solution(sol, a, b, (minx - sol.x) // b)
if sol.x < minx:
shift_solution (sol, a, b, sign_b)
if sol.x > maxx:
return (-1, -1)
lx1 = sol.x
shift_solution (sol, a, b, (maxx - sol.x) // b)
if sol.x > maxx:
shift_solution (sol, a, b, -sign_b)
rx1 = sol.x
shift_solution (sol, a, b, - (miny - sol.y) // a)
if sol.y < miny:
shift_solution (sol, a, b, -sign_a)
if sol.y > maxy:
return (-1, -1)
lx2 = sol.x
shift_solution (sol, a, b, - (maxy - sol.y) // a)
if sol.y > maxy:
shift_solution (sol, a, b, sign_a)
rx2 = sol.x
if lx2 > rx2:
lx2, rx2 = rx2, lx2
lx = max (lx1, lx2)
rx = min (rx1, rx2)
if lx > rx:
return (-1, -1)
return (lx, rx)
def solve():
s = input().split()
x = int(s[0])
y = int(s[1])
p = int(s[2])
q = int(s[3])
# x, y, p, q = 3, 10, 1, 2
if p == 0:
if x == 0:
return 0
else:
return -1
if q == p:
if x == y:
return 0
return -1
if p * y - q * x == 0:
return 0;
a = q - p
b = -p
c = p * y - q * x
ans1, ans2 = find_all_solution(a, b, c, 0, int(10 ** 20), 0, int(10 ** 20))
ansy1 = (p * y - q * x - (q - p) * ans1) // (-p)
ansy2 = (p * y - q * x - (q - p) * ans2) // (-p)
print(x, y)
sum1 = int(10 ** 25)
if ans1 >= 0 and ansy1 >= 0 and (x + ans1) * q == (y + ansy1 + ans1) * p:
sum1 = min(sum1, ans1 + ansy1)
if ans2 >= 0 and ansy2 >= 0 and (x + ans2) * q == (y + ansy2 + ans2) * p:
sum1 = min(sum1, ans2 + ansy2)
if sum1 == int(10 ** 25):
return -1
return sum1
# print(solve())
t = int(input())
for i in range(t):
print(solve())
|
Title: Success Rate
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are an experienced Codeforces user. Today you found out that during your activity on Codeforces you have made *y* submissions, out of which *x* have been successful. Thus, your current success rate on Codeforces is equal to *x*<=/<=*y*.
Your favorite rational number in the [0;1] range is *p*<=/<=*q*. Now you wonder: what is the smallest number of submissions you have to make if you want your success rate to be *p*<=/<=*q*?
Input Specification:
The first line contains a single integer *t* (1<=≤<=*t*<=≤<=1000) — the number of test cases.
Each of the next *t* lines contains four integers *x*, *y*, *p* and *q* (0<=≤<=*x*<=≤<=*y*<=≤<=109; 0<=≤<=*p*<=≤<=*q*<=≤<=109; *y*<=><=0; *q*<=><=0).
It is guaranteed that *p*<=/<=*q* is an irreducible fraction.
Hacks. For hacks, an additional constraint of *t*<=≤<=5 must be met.
Output Specification:
For each test case, output a single integer equal to the smallest number of submissions you have to make if you want your success rate to be equal to your favorite rational number, or -1 if this is impossible to achieve.
Demo Input:
['4\n3 10 1 2\n7 14 3 8\n20 70 2 7\n5 6 1 1\n']
Demo Output:
['4\n10\n0\n-1\n']
Note:
In the first example, you have to make 4 successful submissions. Your success rate will be equal to 7 / 14, or 1 / 2.
In the second example, you have to make 2 successful and 8 unsuccessful submissions. Your success rate will be equal to 9 / 24, or 3 / 8.
In the third example, there is no need to make any new submissions. Your success rate is already equal to 20 / 70, or 2 / 7.
In the fourth example, the only unsuccessful submission breaks your hopes of having the success rate equal to 1.
|
```python
def gcd(a, b):
if b == 0:
return a
else:
return gcd(b, a % b)
class solution:
def __init__(self, a=0, b=0):
self.x = a
self.y = b
def eu (a, b, sol):
if a == 0:
sol.x = 0
sol.y = 1
return b
sol2 = solution()
d = eu (b%a, a, sol2)
sol.x = sol2.y - (b // a) * sol2.x
sol.y = sol2.x
return d
def find_any_solution (a, b, c, sol):
g = eu(abs(a), abs(b), sol)
if c % g != 0:
return -1
sol.x *= c // g
sol.y *= c // g
if (a < 0):
sol.x *= -1
if (b < 0):
sol.y *= -1
return g
def shift_solution (sol, a, b, cnt):
sol.x += cnt * b
sol.y -= cnt * a
def find_all_solution(a, b, c, minx, maxx, miny, maxy):
sol = solution()
g = find_any_solution(a, b, c, sol)
if g == -1:
return (-1, -1)
a //= g
b //= g
sign_a = 1
if a < 0:
sign_a = -1
sign_b = 1
if b < 0:
sign_b = -1
shift_solution(sol, a, b, (minx - sol.x) // b)
if sol.x < minx:
shift_solution (sol, a, b, sign_b)
if sol.x > maxx:
return (-1, -1)
lx1 = sol.x
shift_solution (sol, a, b, (maxx - sol.x) // b)
if sol.x > maxx:
shift_solution (sol, a, b, -sign_b)
rx1 = sol.x
shift_solution (sol, a, b, - (miny - sol.y) // a)
if sol.y < miny:
shift_solution (sol, a, b, -sign_a)
if sol.y > maxy:
return (-1, -1)
lx2 = sol.x
shift_solution (sol, a, b, - (maxy - sol.y) // a)
if sol.y > maxy:
shift_solution (sol, a, b, sign_a)
rx2 = sol.x
if lx2 > rx2:
lx2, rx2 = rx2, lx2
lx = max (lx1, lx2)
rx = min (rx1, rx2)
if lx > rx:
return (-1, -1)
return (lx, rx)
def solve():
s = input().split()
x = int(s[0])
y = int(s[1])
p = int(s[2])
q = int(s[3])
# x, y, p, q = 3, 10, 1, 2
if p == 0:
if x == 0:
return 0
else:
return -1
if q == p:
if x == y:
return 0
return -1
if p * y - q * x == 0:
return 0;
a = q - p
b = -p
c = p * y - q * x
ans1, ans2 = find_all_solution(a, b, c, 0, int(10 ** 20), 0, int(10 ** 20))
ansy1 = (p * y - q * x - (q - p) * ans1) // (-p)
ansy2 = (p * y - q * x - (q - p) * ans2) // (-p)
print(x, y)
sum1 = int(10 ** 25)
if ans1 >= 0 and ansy1 >= 0 and (x + ans1) * q == (y + ansy1 + ans1) * p:
sum1 = min(sum1, ans1 + ansy1)
if ans2 >= 0 and ansy2 >= 0 and (x + ans2) * q == (y + ansy2 + ans2) * p:
sum1 = min(sum1, ans2 + ansy2)
if sum1 == int(10 ** 25):
return -1
return sum1
# print(solve())
t = int(input())
for i in range(t):
print(solve())
```
| 0
|
|
228
|
A
|
Is your horseshoe on the other hoof?
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Valera the Horse is going to the party with friends. He has been following the fashion trends for a while, and he knows that it is very popular to wear all horseshoes of different color. Valera has got four horseshoes left from the last year, but maybe some of them have the same color. In this case he needs to go to the store and buy some few more horseshoes, not to lose face in front of his stylish comrades.
Fortunately, the store sells horseshoes of all colors under the sun and Valera has enough money to buy any four of them. However, in order to save the money, he would like to spend as little money as possible, so you need to help Valera and determine what is the minimum number of horseshoes he needs to buy to wear four horseshoes of different colors to a party.
|
The first line contains four space-separated integers *s*1,<=*s*2,<=*s*3,<=*s*4 (1<=≤<=*s*1,<=*s*2,<=*s*3,<=*s*4<=≤<=109) — the colors of horseshoes Valera has.
Consider all possible colors indexed with integers.
|
Print a single integer — the minimum number of horseshoes Valera needs to buy.
|
[
"1 7 3 3\n",
"7 7 7 7\n"
] |
[
"1\n",
"3\n"
] |
none
| 500
|
[
{
"input": "1 7 3 3",
"output": "1"
},
{
"input": "7 7 7 7",
"output": "3"
},
{
"input": "81170865 673572653 756938629 995577259",
"output": "0"
},
{
"input": "3491663 217797045 522540872 715355328",
"output": "0"
},
{
"input": "251590420 586975278 916631563 586975278",
"output": "1"
},
{
"input": "259504825 377489979 588153796 377489979",
"output": "1"
},
{
"input": "652588203 931100304 931100304 652588203",
"output": "2"
},
{
"input": "391958720 651507265 391958720 651507265",
"output": "2"
},
{
"input": "90793237 90793237 90793237 90793237",
"output": "3"
},
{
"input": "551651653 551651653 551651653 551651653",
"output": "3"
},
{
"input": "156630260 609654355 668943582 973622757",
"output": "0"
},
{
"input": "17061017 110313588 434481173 796661222",
"output": "0"
},
{
"input": "24975422 256716298 337790533 690960249",
"output": "0"
},
{
"input": "255635360 732742923 798648949 883146723",
"output": "0"
},
{
"input": "133315691 265159773 734556507 265159773",
"output": "1"
},
{
"input": "28442865 741657755 978106882 978106882",
"output": "1"
},
{
"input": "131245479 174845575 497483467 131245479",
"output": "1"
},
{
"input": "139159884 616215581 958341883 616215581",
"output": "1"
},
{
"input": "147784432 947653080 947653080 947653080",
"output": "2"
},
{
"input": "94055790 756126496 756126496 94055790",
"output": "2"
},
{
"input": "240458500 511952208 240458500 511952208",
"output": "2"
},
{
"input": "681828506 972810624 972810624 681828506",
"output": "2"
},
{
"input": "454961014 454961014 454961014 454961014",
"output": "3"
},
{
"input": "915819430 915819430 915819430 915819430",
"output": "3"
},
{
"input": "671645142 671645142 671645142 671645142",
"output": "3"
},
{
"input": "132503558 132503558 132503558 132503558",
"output": "3"
},
{
"input": "5 5 999999 6",
"output": "1"
},
{
"input": "1 1 2 5",
"output": "1"
},
{
"input": "2 1 2 3",
"output": "1"
},
{
"input": "1 1 3 5",
"output": "1"
},
{
"input": "1 1 3 3",
"output": "2"
},
{
"input": "2 2 2 1",
"output": "2"
},
{
"input": "3 1 1 1",
"output": "2"
},
{
"input": "1 2 2 2",
"output": "2"
}
] | 1,693,489,144
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 1
| 30
| 0
|
x=map(int,input().split())
h=min(x)
print(h)
|
Title: Is your horseshoe on the other hoof?
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Valera the Horse is going to the party with friends. He has been following the fashion trends for a while, and he knows that it is very popular to wear all horseshoes of different color. Valera has got four horseshoes left from the last year, but maybe some of them have the same color. In this case he needs to go to the store and buy some few more horseshoes, not to lose face in front of his stylish comrades.
Fortunately, the store sells horseshoes of all colors under the sun and Valera has enough money to buy any four of them. However, in order to save the money, he would like to spend as little money as possible, so you need to help Valera and determine what is the minimum number of horseshoes he needs to buy to wear four horseshoes of different colors to a party.
Input Specification:
The first line contains four space-separated integers *s*1,<=*s*2,<=*s*3,<=*s*4 (1<=≤<=*s*1,<=*s*2,<=*s*3,<=*s*4<=≤<=109) — the colors of horseshoes Valera has.
Consider all possible colors indexed with integers.
Output Specification:
Print a single integer — the minimum number of horseshoes Valera needs to buy.
Demo Input:
['1 7 3 3\n', '7 7 7 7\n']
Demo Output:
['1\n', '3\n']
Note:
none
|
```python
x=map(int,input().split())
h=min(x)
print(h)
```
| 0
|
|
197
|
B
|
Limit
|
PROGRAMMING
| 1,400
|
[
"math"
] | null | null |
You are given two polynomials:
- *P*(*x*)<==<=*a*0·*x**n*<=+<=*a*1·*x**n*<=-<=1<=+<=...<=+<=*a**n*<=-<=1·*x*<=+<=*a**n* and - *Q*(*x*)<==<=*b*0·*x**m*<=+<=*b*1·*x**m*<=-<=1<=+<=...<=+<=*b**m*<=-<=1·*x*<=+<=*b**m*.
Calculate limit .
|
The first line contains two space-separated integers *n* and *m* (0<=≤<=*n*,<=*m*<=≤<=100) — degrees of polynomials *P*(*x*) and *Q*(*x*) correspondingly.
The second line contains *n*<=+<=1 space-separated integers — the factors of polynomial *P*(*x*): *a*0, *a*1, ..., *a**n*<=-<=1, *a**n* (<=-<=100<=≤<=*a**i*<=≤<=100,<=*a*0<=≠<=0).
The third line contains *m*<=+<=1 space-separated integers — the factors of polynomial *Q*(*x*): *b*0, *b*1, ..., *b**m*<=-<=1, *b**m* (<=-<=100<=≤<=*b**i*<=≤<=100,<=*b*0<=≠<=0).
|
If the limit equals <=+<=∞, print "Infinity" (without quotes). If the limit equals <=-<=∞, print "-Infinity" (without the quotes).
If the value of the limit equals zero, print "0/1" (without the quotes).
Otherwise, print an irreducible fraction — the value of limit , in the format "p/q" (without the quotes), where *p* is the — numerator, *q* (*q*<=><=0) is the denominator of the fraction.
|
[
"2 1\n1 1 1\n2 5\n",
"1 0\n-1 3\n2\n",
"0 1\n1\n1 0\n",
"2 2\n2 1 6\n4 5 -7\n",
"1 1\n9 0\n-5 2\n"
] |
[
"Infinity\n",
"-Infinity\n",
"0/1\n",
"1/2\n",
"-9/5\n"
] |
Let's consider all samples:
1. <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/c28febca257452afdfcbd6984ba8623911f9bdbc.png" style="max-width: 100.0%;max-height: 100.0%;"/> 1. <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/1e55ecd04e54a45e5e0092ec9a5c1ea03bb29255.png" style="max-width: 100.0%;max-height: 100.0%;"/> 1. <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/2c95fb684d373fcc1a481cfabeda4d5c2f3673ee.png" style="max-width: 100.0%;max-height: 100.0%;"/> 1. <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4dc40cb8b3cd6375c42445366e50369649a2801a.png" style="max-width: 100.0%;max-height: 100.0%;"/> 1. <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/c6455aba35cfb3c4397505121d1f77afcd17c98e.png" style="max-width: 100.0%;max-height: 100.0%;"/>
You can learn more about the definition and properties of limits if you follow the link: http://en.wikipedia.org/wiki/Limit_of_a_function
| 500
|
[
{
"input": "2 1\n1 1 1\n2 5",
"output": "Infinity"
},
{
"input": "1 0\n-1 3\n2",
"output": "-Infinity"
},
{
"input": "0 1\n1\n1 0",
"output": "0/1"
},
{
"input": "2 2\n2 1 6\n4 5 -7",
"output": "1/2"
},
{
"input": "1 1\n9 0\n-5 2",
"output": "-9/5"
},
{
"input": "1 2\n5 3\n-3 2 -1",
"output": "0/1"
},
{
"input": "1 2\n-4 8\n-2 5 -3",
"output": "0/1"
},
{
"input": "3 2\n4 3 1 2\n-5 7 0",
"output": "-Infinity"
},
{
"input": "2 1\n-3 5 1\n-8 0",
"output": "Infinity"
},
{
"input": "1 1\n-5 7\n3 1",
"output": "-5/3"
},
{
"input": "2 2\n-4 2 1\n-5 8 -19",
"output": "4/5"
},
{
"input": "0 100\n1\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100",
"output": "0/1"
},
{
"input": "100 0\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100\n1",
"output": "Infinity"
},
{
"input": "0 0\n36\n-54",
"output": "-2/3"
},
{
"input": "0 0\n36\n-8",
"output": "-9/2"
},
{
"input": "0 0\n-6\n-8",
"output": "3/4"
},
{
"input": "0 2\n-3\n1 4 6",
"output": "0/1"
},
{
"input": "0 0\n-21\n13",
"output": "-21/13"
},
{
"input": "0 0\n-34\n21",
"output": "-34/21"
},
{
"input": "0 0\n-55\n34",
"output": "-55/34"
},
{
"input": "33 100\n-15 -90 -84 57 67 60 -40 -82 83 -80 43 -15 -36 -14 -37 -49 42 -79 49 -7 -12 53 -44 -21 87 -91 -73 -27 13 65 5 74 -21 -52\n-67 -17 36 -46 -5 31 -45 -35 -49 13 -7 -82 92 -55 -67 -96 31 -70 76 24 -29 26 96 19 -40 99 -26 74 -17 -56 -72 24 -71 -62 10 -56 -74 75 -48 -98 -67 -26 47 7 63 -38 99 66 -25 -31 -24 -42 -49 -27 -45 -2 -37 -16 5 -21 97 33 85 -33 93 30 84 73 -48 18 -36 71 -38 -41 28 1 -7 -15 60 59 -20 -38 -86 90 2 -12 72 -43 26 76 97 7 -2 -47 -4 100 -40 -48 53 -54 0",
"output": "0/1"
},
{
"input": "39 87\n78 -50 18 -32 -12 -65 83 41 -6 53 -26 64 -19 -53 -61 91 -49 -66 67 69 100 -39 95 99 86 -67 -66 63 48 26 -4 95 -54 -71 26 -74 -93 79 -91 -45\n-18 23 48 59 76 82 95 2 -26 18 -39 -74 44 -92 40 -44 1 -97 -100 -63 -54 -3 -86 85 28 -50 41 -53 -74 -29 -91 87 27 -42 -90 -15 -26 -15 -100 -70 -10 -41 16 85 71 -39 -31 -65 80 98 9 23 -40 14 -88 15 -34 10 -67 -94 -58 -24 75 48 -42 56 -77 -13 -25 -79 -100 -57 89 45 22 85 78 -93 -79 69 63 44 74 94 35 -65 -12 -88",
"output": "0/1"
},
{
"input": "47 56\n31 -99 -97 6 -45 -5 89 35 -77 69 57 91 -32 -66 -36 16 30 61 -36 32 48 67 5 -85 65 -11 -51 -63 -51 -16 39 -26 -60 -28 91 43 -90 32 44 83 70 -53 51 56 68 -81 76 79\n61 -21 -75 -36 -24 -19 80 26 -28 93 27 72 -39 -46 -38 68 -29 -16 -63 84 -13 64 55 63 77 5 68 70 15 99 12 -69 50 -48 -82 -3 52 -54 68 91 -37 -100 -5 74 24 91 -1 74 28 29 -87 -13 -88 82 -13 58 23",
"output": "0/1"
},
{
"input": "9 100\n-34 88 33 -80 87 31 -53 -3 8 -70\n31 -25 46 78 8 82 -92 -36 -30 85 -93 86 -87 75 8 -71 44 -41 -83 19 89 -28 81 42 79 86 41 -23 64 -31 46 24 -79 23 71 63 99 90 -16 -70 -1 88 10 65 3 -99 95 52 -80 53 -24 -43 -30 -7 51 40 -47 44 -10 -18 -61 -67 -84 37 45 93 -5 68 32 3 -61 -100 38 -21 -91 90 83 -45 75 89 17 -44 75 14 -28 1 -84 -100 -36 84 -40 88 -84 -54 2 -32 92 -49 77 85 91",
"output": "0/1"
},
{
"input": "28 87\n-77 49 37 46 -92 65 89 100 53 76 -43 47 -80 -46 -94 -4 20 46 81 -41 86 25 69 60 15 -78 -98 -7 -42\n-85 96 59 -40 90 -72 41 -17 -40 -15 -98 66 47 9 -33 -63 59 -25 -31 25 -94 35 28 -36 -41 -38 -38 -54 -40 90 7 -10 98 -19 54 -10 46 -58 -88 -21 90 82 37 -70 -98 -63 41 75 -50 -59 -69 79 -93 -3 -45 14 76 28 -28 -98 -44 -39 71 44 90 91 0 45 7 65 68 39 -27 58 68 -47 -41 100 14 -95 -80 69 -88 -51 -89 -70 -23 95",
"output": "0/1"
},
{
"input": "100 4\n-5 -93 89 -26 -79 14 -28 13 -45 69 50 -84 21 -68 62 30 -26 99 -12 39 20 -74 -39 -41 -28 -72 -55 28 20 31 -92 -20 76 -65 57 72 -36 4 33 -28 -19 -41 -40 40 84 -36 -83 75 -74 -80 32 -50 -56 72 16 75 57 90 -19 -10 67 -71 69 -48 -48 23 37 -31 -64 -86 20 67 97 14 82 -41 2 87 65 -81 -27 9 -79 -1 -5 84 -8 29 -34 31 82 40 21 -53 -31 -45 17 -33 79 50 -94\n56 -4 -90 36 84",
"output": "-Infinity"
},
{
"input": "77 51\n89 45 -33 -87 33 -61 -79 40 -76 16 -17 31 27 25 99 82 51 -40 85 -66 19 89 -62 24 -61 -53 -77 17 21 83 53 -18 -56 75 9 -78 33 -11 -6 96 -33 -2 -57 97 30 20 -41 42 -13 45 -99 67 37 -20 51 -33 88 -62 2 40 17 36 45 71 4 -44 24 20 -2 29 -12 -84 -7 -84 -38 48 -73 79\n60 -43 60 1 90 -1 19 -18 -21 31 -76 51 79 91 12 39 -33 -14 71 -90 -65 -93 -58 93 49 17 77 19 32 -8 14 58 -9 85 -95 -73 0 85 -91 -99 -30 -43 61 20 -89 93 53 20 -33 -38 79 54",
"output": "Infinity"
},
{
"input": "84 54\n82 -54 28 68 74 -61 54 98 59 67 -65 -1 16 65 -78 -16 61 -79 2 14 44 96 -62 77 51 87 37 66 65 28 88 -99 -21 -83 24 80 39 64 -65 45 86 -53 -49 94 -75 -31 -42 -1 -35 -18 74 30 31 -40 30 -6 47 58 -71 -21 20 13 75 -79 15 -98 -26 76 99 -77 -9 85 48 51 -87 56 -53 37 47 -3 94 64 -7 74 86\n72 51 -74 20 41 -76 98 58 24 -61 -97 -73 62 29 6 42 -92 -6 -65 89 -32 -9 82 -13 -88 -70 -97 25 -48 12 -54 33 -92 -29 48 60 -21 86 -17 -86 45 -34 -3 -9 -62 12 25 -74 -76 -89 48 55 -30 86 51",
"output": "Infinity"
},
{
"input": "73 15\n-70 78 51 -33 -95 46 87 -33 16 62 67 -85 -57 75 -93 -59 98 -45 -90 -88 9 53 35 37 28 3 40 -87 28 5 18 11 9 1 72 69 -65 -62 1 73 -3 3 35 17 -28 -31 -45 60 64 18 60 38 -47 12 2 -90 -4 33 -51 -55 -54 90 38 -65 39 32 -70 0 -5 3 -12 100 78 55\n46 33 41 52 -89 -9 53 -81 34 -45 -11 -41 14 -28 95 -50",
"output": "-Infinity"
},
{
"input": "33 1\n-75 -83 87 -27 -48 47 -90 -84 -18 -4 14 -1 -83 -98 -68 -85 -86 28 2 45 96 -59 86 -25 -2 -64 -92 65 69 72 72 -58 -99 90\n-1 72",
"output": "Infinity"
},
{
"input": "58 58\n-25 40 -34 23 -52 94 -30 -99 -71 -90 -44 -71 69 48 -45 -59 0 66 -70 -96 95 91 82 90 -95 87 3 -77 -77 -26 15 87 -82 5 -24 82 -11 99 35 49 22 44 18 -60 -26 79 67 71 -13 29 -23 9 58 -90 88 18 77 5 -7\n-30 -11 -13 -50 61 -78 11 -74 -73 13 -66 -65 -82 38 58 25 -64 -24 78 -87 6 6 -80 -96 47 -25 -54 10 -41 -22 -50 -1 -6 -22 27 54 -32 30 93 88 -70 -100 -69 -47 -20 -92 -24 70 -93 42 78 42 -35 41 31 75 -67 -62 -83",
"output": "5/6"
},
{
"input": "20 20\n5 4 91 -66 -57 55 -79 -2 -54 -72 -49 21 -23 -5 57 -48 70 -16 -86 -26 -19\n51 -60 64 -8 89 27 -96 4 95 -24 -2 -27 -41 -14 -88 -19 24 68 -31 34 -62",
"output": "5/51"
},
{
"input": "69 69\n-90 -63 -21 23 23 -14 -82 65 42 -60 -42 -39 67 34 96 93 -42 -24 21 -80 44 -81 45 -74 -19 -88 39 58 90 87 16 48 -19 -2 36 87 4 -66 -82 -49 -32 -43 -65 12 34 -29 -58 46 -67 -20 -30 91 21 65 15 2 3 -92 -67 -68 39 -24 77 76 -17 -34 5 63 88 83\n-55 98 -79 18 -100 -67 -79 -85 -75 -44 -6 -73 -11 -12 -24 -78 47 -51 25 -29 -34 25 27 11 -87 15 -44 41 -44 46 -67 70 -35 41 62 -36 27 -41 -42 -50 96 31 26 -66 9 74 34 31 25 6 -84 41 74 -7 49 5 35 -5 -71 -37 28 58 -8 -40 -19 -83 -34 64 7 15",
"output": "18/11"
},
{
"input": "0 0\n46\n-33",
"output": "-46/33"
},
{
"input": "67 67\n-8 11 55 80 -26 -38 58 73 -48 -10 35 75 16 -84 55 -51 98 58 -28 98 77 81 51 -86 -46 68 -87 -80 -49 81 96 -97 -42 25 6 -8 -55 -25 93 -29 -33 -6 -26 -85 73 97 63 57 51 92 -6 -8 4 86 46 -45 36 -19 -71 1 71 39 97 -44 -34 -1 2 -46\n91 -32 -76 11 -40 91 -8 -100 73 80 47 82 24 0 -71 82 -93 38 -54 1 -55 -53 90 -86 0 10 -35 49 90 56 25 17 46 -43 13 16 -82 -33 64 -83 -56 22 12 -74 4 -68 85 -27 60 -28 -47 73 -93 69 -37 54 -3 90 -56 56 78 61 7 -79 48 -42 -10 -48",
"output": "-8/91"
},
{
"input": "69 69\n-7 38 -3 -22 65 -78 -65 -99 -76 63 0 -4 -78 -51 54 -61 -53 60 80 34 -96 99 -78 -96 21 -10 -86 33 -9 -81 -19 -2 -76 -3 -66 -80 -55 -21 -50 37 -86 -37 47 44 76 -39 54 -25 41 -86 -3 -25 -67 94 18 67 27 -5 -30 -69 2 -76 7 -97 -52 -35 -55 -20 92 2\n90 -94 37 41 -27 -54 96 -15 -60 -29 -75 -93 -57 62 48 -88 -99 -62 4 -9 85 33 65 -95 -30 16 -29 -89 -33 -83 -35 -21 53 -52 80 -40 76 -33 86 47 18 43 -67 -36 -99 -42 1 -94 -78 34 -41 73 96 2 -60 29 68 -96 -21 -61 -98 -67 1 40 85 55 66 -25 -50 -83",
"output": "-7/90"
},
{
"input": "17 17\n-54 59 -95 87 3 -27 -30 49 -87 74 45 78 36 60 -95 41 -53 -70\n-27 16 -67 -24 10 -73 -41 12 -52 53 -73 -17 -56 -74 -33 -8 100 -39",
"output": "2/1"
},
{
"input": "1 1\n36 -49\n-32 -40",
"output": "-9/8"
},
{
"input": "1 1\n1 1\n1 1",
"output": "1/1"
},
{
"input": "1 1\n-2 1\n4 1",
"output": "-1/2"
},
{
"input": "0 0\n2\n1",
"output": "2/1"
},
{
"input": "0 0\n4\n-3",
"output": "-4/3"
},
{
"input": "0 0\n2\n2",
"output": "1/1"
},
{
"input": "0 0\n17\n-10",
"output": "-17/10"
},
{
"input": "0 0\n-1\n2",
"output": "-1/2"
},
{
"input": "0 0\n1\n1",
"output": "1/1"
},
{
"input": "0 0\n50\n20",
"output": "5/2"
},
{
"input": "0 0\n20\n20",
"output": "1/1"
},
{
"input": "0 0\n4\n-2",
"output": "-2/1"
},
{
"input": "0 0\n4\n-6",
"output": "-2/3"
},
{
"input": "0 0\n1\n-2",
"output": "-1/2"
},
{
"input": "0 0\n4\n2",
"output": "2/1"
},
{
"input": "0 0\n2\n-4",
"output": "-1/2"
},
{
"input": "1 1\n4 1\n2 1",
"output": "2/1"
},
{
"input": "2 2\n-13 1 3\n6 3 2",
"output": "-13/6"
},
{
"input": "99 99\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\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": "1/2"
},
{
"input": "0 0\n5\n5",
"output": "1/1"
},
{
"input": "0 0\n2\n-1",
"output": "-2/1"
}
] | 1,547,035,410
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 47
| 436
| 1,126,400
|
from fractions import Fraction
n,m = map(int,input().split())
l1 = list(map(int,input().split()))
l2 = list(map(int,input().split()))
if n > m:
if l1[0] <0 and l2[0] < 0:
print("Infinity")
else:
if l1[0] < 0 or l2[0] < 0:
print("-Infinity")
else:
print("Infinity")
elif n==m:
print(Fraction(l1[0],l2[0]))
elif n < m:
print("0/1")
|
Title: Limit
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given two polynomials:
- *P*(*x*)<==<=*a*0·*x**n*<=+<=*a*1·*x**n*<=-<=1<=+<=...<=+<=*a**n*<=-<=1·*x*<=+<=*a**n* and - *Q*(*x*)<==<=*b*0·*x**m*<=+<=*b*1·*x**m*<=-<=1<=+<=...<=+<=*b**m*<=-<=1·*x*<=+<=*b**m*.
Calculate limit .
Input Specification:
The first line contains two space-separated integers *n* and *m* (0<=≤<=*n*,<=*m*<=≤<=100) — degrees of polynomials *P*(*x*) and *Q*(*x*) correspondingly.
The second line contains *n*<=+<=1 space-separated integers — the factors of polynomial *P*(*x*): *a*0, *a*1, ..., *a**n*<=-<=1, *a**n* (<=-<=100<=≤<=*a**i*<=≤<=100,<=*a*0<=≠<=0).
The third line contains *m*<=+<=1 space-separated integers — the factors of polynomial *Q*(*x*): *b*0, *b*1, ..., *b**m*<=-<=1, *b**m* (<=-<=100<=≤<=*b**i*<=≤<=100,<=*b*0<=≠<=0).
Output Specification:
If the limit equals <=+<=∞, print "Infinity" (without quotes). If the limit equals <=-<=∞, print "-Infinity" (without the quotes).
If the value of the limit equals zero, print "0/1" (without the quotes).
Otherwise, print an irreducible fraction — the value of limit , in the format "p/q" (without the quotes), where *p* is the — numerator, *q* (*q*<=><=0) is the denominator of the fraction.
Demo Input:
['2 1\n1 1 1\n2 5\n', '1 0\n-1 3\n2\n', '0 1\n1\n1 0\n', '2 2\n2 1 6\n4 5 -7\n', '1 1\n9 0\n-5 2\n']
Demo Output:
['Infinity\n', '-Infinity\n', '0/1\n', '1/2\n', '-9/5\n']
Note:
Let's consider all samples:
1. <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/c28febca257452afdfcbd6984ba8623911f9bdbc.png" style="max-width: 100.0%;max-height: 100.0%;"/> 1. <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/1e55ecd04e54a45e5e0092ec9a5c1ea03bb29255.png" style="max-width: 100.0%;max-height: 100.0%;"/> 1. <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/2c95fb684d373fcc1a481cfabeda4d5c2f3673ee.png" style="max-width: 100.0%;max-height: 100.0%;"/> 1. <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4dc40cb8b3cd6375c42445366e50369649a2801a.png" style="max-width: 100.0%;max-height: 100.0%;"/> 1. <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/c6455aba35cfb3c4397505121d1f77afcd17c98e.png" style="max-width: 100.0%;max-height: 100.0%;"/>
You can learn more about the definition and properties of limits if you follow the link: http://en.wikipedia.org/wiki/Limit_of_a_function
|
```python
from fractions import Fraction
n,m = map(int,input().split())
l1 = list(map(int,input().split()))
l2 = list(map(int,input().split()))
if n > m:
if l1[0] <0 and l2[0] < 0:
print("Infinity")
else:
if l1[0] < 0 or l2[0] < 0:
print("-Infinity")
else:
print("Infinity")
elif n==m:
print(Fraction(l1[0],l2[0]))
elif n < m:
print("0/1")
```
| 0
|
|
28
|
A
|
Bender Problem
|
PROGRAMMING
| 1,600
|
[
"implementation"
] |
A. Bender Problem
|
2
|
256
|
Robot Bender decided to make Fray a birthday present. He drove *n* nails and numbered them from 1 to *n* in some order. Bender decided to make a picture using metal rods. The picture is a closed polyline, which vertices should be nails (in the given order). The segments of the polyline should be parallel to the coordinate axes. Polyline is allowed to have self-intersections. Bender can take a rod and fold it exactly once in any place to form an angle of 90 degrees. Then he can attach the place of the fold to some unoccupied nail and attach two ends of this rod to adjacent nails. A nail is considered unoccupied if there is no rod attached to it (neither by it's end nor the by the fold place). No rod could be used twice. It is not required to use all the rods.
Help Bender to solve this difficult task.
|
The first line contains two positive integers *n* and *m* (4<=≤<=*n*<=≤<=500,<=2<=≤<=*m*<=≤<=500, *n* is even) — the amount of nails and the amount of rods. *i*-th of the following *n* lines contains a pair of integers, denoting the coordinates of the *i*-th nail. Nails should be connected in the same order as they are given in the input. The last line contains *m* integers — the lenghts of the rods. All coordinates do not exceed 104 by absolute value. Lengths of the rods are between 1 and 200<=000. No rod can be used twice. It is guaranteed that all segments of the given polyline are parallel to coordinate axes. No three consecutive nails lie on the same line.
|
If it is impossible to solve Bender's problem, output NO. Otherwise, output YES in the first line, and in the second line output *n* numbers — *i*-th of them should be the number of rod, which fold place is attached to the *i*-th nail, or -1, if there is no such rod.
If there are multiple solutions, print any of them.
|
[
"4 2\n0 0\n0 2\n2 2\n2 0\n4 4\n",
"6 3\n0 0\n1 0\n1 1\n2 1\n2 2\n0 2\n3 2 3\n",
"6 3\n0 0\n1 0\n1 1\n2 1\n2 2\n0 2\n2 2 3\n"
] |
[
"YES\n1 -1 2 -1 \n",
"YES\n1 -1 2 -1 3 -1 \n",
"NO\n"
] |
none
| 500
|
[
{
"input": "4 2\n0 0\n0 2\n2 2\n2 0\n4 4",
"output": "YES\n1 -1 2 -1 "
},
{
"input": "6 3\n0 0\n1 0\n1 1\n2 1\n2 2\n0 2\n3 2 3",
"output": "YES\n1 -1 2 -1 3 -1 "
},
{
"input": "6 3\n0 0\n1 0\n1 1\n2 1\n2 2\n0 2\n2 2 3",
"output": "NO"
},
{
"input": "4 4\n0 0\n0 1\n1 1\n1 0\n1 1 1 1",
"output": "NO"
},
{
"input": "6 2\n0 0\n1 0\n1 1\n2 1\n2 2\n0 2\n2 2",
"output": "NO"
},
{
"input": "6 3\n0 0\n2 0\n2 2\n1 2\n1 1\n0 1\n4 2 2",
"output": "YES\n-1 1 -1 2 -1 3 "
},
{
"input": "4 4\n-8423 7689\n6902 7689\n6902 2402\n-8423 2402\n20612 20612 91529 35617",
"output": "YES\n1 -1 2 -1 "
},
{
"input": "4 4\n1679 -198\n9204 -198\n9204 -5824\n1679 -5824\n18297 92466 187436 175992",
"output": "NO"
},
{
"input": "4 2\n0 0\n0 2\n2 2\n2 0\n200000 200000",
"output": "NO"
}
] | 1,577,357,021
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 9
| 216
| 307,200
|
n,m = map(int,input().split())
s = []
for i in range(n):
a = map(int,input().split())
a = list(a)
s.append(a)
s_chet = []
for i in range(1,n-1,2): #Проход по четным гвоздям
q = abs(sum(s[i])-sum(s[i-1])) + abs(sum(s[i])-sum(s[i+1]))
s_chet.append(q)
q1 = abs(sum(s[-1])-sum(s[-2])) + abs(sum(s[-1])-sum(s[0]))
s_chet.append(q1)
s_nechet = []
for i in range(2,n-1,2): #Проход по нечетным гвоздям
qq = abs(sum(s[i])-sum(s[i-1])) + abs(sum(s[i])-sum(s[i+1]))
s_nechet.append(qq)
qq1 = abs(sum(s[-1])-sum(s[0])) + abs(sum(s[1])-sum(s[0]))
s_nechet.append(qq1)
s_chet.sort()
s_nechet.sort()
ss = map(int,input().split())
ss = list(ss)
ss.sort()
if s_chet == ss:
print('YES')
sss = []
for i in range(1,m+1):
sss.append(-1)
sss.append(i)
print(" ".join(map(str,sss)))
elif s_nechet == ss:
print('YES')
sss = []
for i in range(1,m+1):
sss.append(i)
sss.append(-1)
print(" ".join(map(str,sss)))
else:
print('NO')
|
Title: Bender Problem
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Robot Bender decided to make Fray a birthday present. He drove *n* nails and numbered them from 1 to *n* in some order. Bender decided to make a picture using metal rods. The picture is a closed polyline, which vertices should be nails (in the given order). The segments of the polyline should be parallel to the coordinate axes. Polyline is allowed to have self-intersections. Bender can take a rod and fold it exactly once in any place to form an angle of 90 degrees. Then he can attach the place of the fold to some unoccupied nail and attach two ends of this rod to adjacent nails. A nail is considered unoccupied if there is no rod attached to it (neither by it's end nor the by the fold place). No rod could be used twice. It is not required to use all the rods.
Help Bender to solve this difficult task.
Input Specification:
The first line contains two positive integers *n* and *m* (4<=≤<=*n*<=≤<=500,<=2<=≤<=*m*<=≤<=500, *n* is even) — the amount of nails and the amount of rods. *i*-th of the following *n* lines contains a pair of integers, denoting the coordinates of the *i*-th nail. Nails should be connected in the same order as they are given in the input. The last line contains *m* integers — the lenghts of the rods. All coordinates do not exceed 104 by absolute value. Lengths of the rods are between 1 and 200<=000. No rod can be used twice. It is guaranteed that all segments of the given polyline are parallel to coordinate axes. No three consecutive nails lie on the same line.
Output Specification:
If it is impossible to solve Bender's problem, output NO. Otherwise, output YES in the first line, and in the second line output *n* numbers — *i*-th of them should be the number of rod, which fold place is attached to the *i*-th nail, or -1, if there is no such rod.
If there are multiple solutions, print any of them.
Demo Input:
['4 2\n0 0\n0 2\n2 2\n2 0\n4 4\n', '6 3\n0 0\n1 0\n1 1\n2 1\n2 2\n0 2\n3 2 3\n', '6 3\n0 0\n1 0\n1 1\n2 1\n2 2\n0 2\n2 2 3\n']
Demo Output:
['YES\n1 -1 2 -1 \n', 'YES\n1 -1 2 -1 3 -1 \n', 'NO\n']
Note:
none
|
```python
n,m = map(int,input().split())
s = []
for i in range(n):
a = map(int,input().split())
a = list(a)
s.append(a)
s_chet = []
for i in range(1,n-1,2): #Проход по четным гвоздям
q = abs(sum(s[i])-sum(s[i-1])) + abs(sum(s[i])-sum(s[i+1]))
s_chet.append(q)
q1 = abs(sum(s[-1])-sum(s[-2])) + abs(sum(s[-1])-sum(s[0]))
s_chet.append(q1)
s_nechet = []
for i in range(2,n-1,2): #Проход по нечетным гвоздям
qq = abs(sum(s[i])-sum(s[i-1])) + abs(sum(s[i])-sum(s[i+1]))
s_nechet.append(qq)
qq1 = abs(sum(s[-1])-sum(s[0])) + abs(sum(s[1])-sum(s[0]))
s_nechet.append(qq1)
s_chet.sort()
s_nechet.sort()
ss = map(int,input().split())
ss = list(ss)
ss.sort()
if s_chet == ss:
print('YES')
sss = []
for i in range(1,m+1):
sss.append(-1)
sss.append(i)
print(" ".join(map(str,sss)))
elif s_nechet == ss:
print('YES')
sss = []
for i in range(1,m+1):
sss.append(i)
sss.append(-1)
print(" ".join(map(str,sss)))
else:
print('NO')
```
| 0
|
474
|
A
|
Keyboard
|
PROGRAMMING
| 900
|
[
"implementation"
] | null | null |
Our good friend Mole is trying to code a big message. He is typing on an unusual keyboard with characters arranged in following way:
Unfortunately Mole is blind, so sometimes it is problem for him to put his hands accurately. He accidentally moved both his hands with one position to the left or to the right. That means that now he presses not a button he wants, but one neighboring button (left or right, as specified in input).
We have a sequence of characters he has typed and we want to find the original message.
|
First line of the input contains one letter describing direction of shifting ('L' or 'R' respectively for left or right).
Second line contains a sequence of characters written by Mole. The size of this sequence will be no more than 100. Sequence contains only symbols that appear on Mole's keyboard. It doesn't contain spaces as there is no space on Mole's keyboard.
It is guaranteed that even though Mole hands are moved, he is still pressing buttons on keyboard and not hitting outside it.
|
Print a line that contains the original message.
|
[
"R\ns;;upimrrfod;pbr\n"
] |
[
"allyouneedislove\n"
] |
none
| 500
|
[
{
"input": "R\ns;;upimrrfod;pbr",
"output": "allyouneedislove"
},
{
"input": "R\nwertyuiop;lkjhgfdsxcvbnm,.",
"output": "qwertyuiolkjhgfdsazxcvbnm,"
},
{
"input": "L\nzxcvbnm,kjhgfdsaqwertyuio",
"output": "xcvbnm,.lkjhgfdswertyuiop"
},
{
"input": "R\nbubbuduppudup",
"output": "vyvvysyooysyo"
},
{
"input": "L\ngggggggggggggggggggggggggggggggggggggggggg",
"output": "hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh"
},
{
"input": "R\ngggggggggggggggggggggggggggggggggggggggggg",
"output": "ffffffffffffffffffffffffffffffffffffffffff"
},
{
"input": "L\nggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggg",
"output": "hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh"
},
{
"input": "R\nggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggg",
"output": "fffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff"
},
{
"input": "L\nxgwurenkxkiau,c,vonei.zltazmnkhqtwuogkgvgckvja,z.rhanuy.ybebmzcfwozkwvuuiolaqlgvvvewnbuinrncgjwjdsfw",
"output": "cheitrmlclosi.v.bpmro/x;ysx,mljwyeiphlhbhvlbks.x/tjsmiu/unrn,xvgepxlebiiop;sw;hbbbremniomtmvhkekfdge"
},
{
"input": "L\nuoz.vmks,wxrb,nwcvdzh.m,hwsios.lvu,ktes,,ythddhm.sh,d,c,cfj.wqam,bowofbyx,jathqayhreqvixvbmgdokofmym",
"output": "ipx/b,ld.ectn.mevbfxj/,.jedopd/;bi.lyrd..uyjffj,/dj.f.v.vgk/ews,.npepgnuc.ksyjwsujtrwbocbn,hfplpg,u,"
},
{
"input": "R\noedjyrvuw/rn.v.hdwndbiposiewgsn.pnyf;/tsdohp,hrtd/mx,;coj./billd..mwbneohcikrdes/ucjr,wspthleyp,..f,",
"output": "iwshtecyq.eb,c,gsqbsvuoiauwqfab,obtdl.rasigomgers.nzmlxih,.vukks,,nqvbwigxujeswa.yxhemqaorgkwtom,,dm"
},
{
"input": "R\nvgj;o;ijrtfyck,dthccioltcx,crub;oceooognsuvfx/kgo.fbsudv,yod.erdrxhbeiyltxhnrobbb;ydrgroefcr/f;uvdjd",
"output": "cfhliluherdtxjmsrgxxuikrxzmxeyvlixwiiifbaycdz.jfi,dvayscmtis,wesezgvwutkrzgbeivvvltsefeiwdxe.dlycshs"
},
{
"input": "L\nqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq",
"output": "wwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww"
},
{
"input": "L\noooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo",
"output": "pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp"
},
{
"input": "L\n,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,",
"output": "...................................................................................................."
},
{
"input": "L\nzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz",
"output": "xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx"
},
{
"input": "R\noooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo",
"output": "iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii"
},
{
"input": "R\nwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww",
"output": "qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq"
},
{
"input": "R\nxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx",
"output": "zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz"
},
{
"input": "L\nq",
"output": "w"
},
{
"input": "L\no",
"output": "p"
},
{
"input": "L\n,",
"output": "."
},
{
"input": "L\nz",
"output": "x"
},
{
"input": "R\n.",
"output": ","
},
{
"input": "R\no",
"output": "i"
},
{
"input": "R\nw",
"output": "q"
},
{
"input": "R\nx",
"output": "z"
},
{
"input": "R\n,./",
"output": "m,."
},
{
"input": "R\nwertuk;;/",
"output": "qweryjll."
},
{
"input": "L\n..",
"output": "//"
}
] | 1,678,950,848
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 31
| 46
| 0
|
a=input()
b=input()
d='qwertyuiopasdfghjkl;zxcvbnm,./'
c=''
for x in b:
if a=='R':
c+=d[d.index(x)-1]
else:
c+=d[d.index(x)+1]
print(c)
|
Title: Keyboard
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Our good friend Mole is trying to code a big message. He is typing on an unusual keyboard with characters arranged in following way:
Unfortunately Mole is blind, so sometimes it is problem for him to put his hands accurately. He accidentally moved both his hands with one position to the left or to the right. That means that now he presses not a button he wants, but one neighboring button (left or right, as specified in input).
We have a sequence of characters he has typed and we want to find the original message.
Input Specification:
First line of the input contains one letter describing direction of shifting ('L' or 'R' respectively for left or right).
Second line contains a sequence of characters written by Mole. The size of this sequence will be no more than 100. Sequence contains only symbols that appear on Mole's keyboard. It doesn't contain spaces as there is no space on Mole's keyboard.
It is guaranteed that even though Mole hands are moved, he is still pressing buttons on keyboard and not hitting outside it.
Output Specification:
Print a line that contains the original message.
Demo Input:
['R\ns;;upimrrfod;pbr\n']
Demo Output:
['allyouneedislove\n']
Note:
none
|
```python
a=input()
b=input()
d='qwertyuiopasdfghjkl;zxcvbnm,./'
c=''
for x in b:
if a=='R':
c+=d[d.index(x)-1]
else:
c+=d[d.index(x)+1]
print(c)
```
| 3
|
|
645
|
C
|
Enduring Exodus
|
PROGRAMMING
| 1,600
|
[
"binary search",
"two pointers"
] | null | null |
In an attempt to escape the Mischievous Mess Makers' antics, Farmer John has abandoned his farm and is traveling to the other side of Bovinia. During the journey, he and his *k* cows have decided to stay at the luxurious Grand Moo-dapest Hotel. The hotel consists of *n* rooms located in a row, some of which are occupied.
Farmer John wants to book a set of *k*<=+<=1 currently unoccupied rooms for him and his cows. He wants his cows to stay as safe as possible, so he wishes to minimize the maximum distance from his room to the room of his cow. The distance between rooms *i* and *j* is defined as |*j*<=-<=*i*|. Help Farmer John protect his cows by calculating this minimum possible distance.
|
The first line of the input contains two integers *n* and *k* (1<=≤<=*k*<=<<=*n*<=≤<=100<=000) — the number of rooms in the hotel and the number of cows travelling with Farmer John.
The second line contains a string of length *n* describing the rooms. The *i*-th character of the string will be '0' if the *i*-th room is free, and '1' if the *i*-th room is occupied. It is guaranteed that at least *k*<=+<=1 characters of this string are '0', so there exists at least one possible choice of *k*<=+<=1 rooms for Farmer John and his cows to stay in.
|
Print the minimum possible distance between Farmer John's room and his farthest cow.
|
[
"7 2\n0100100\n",
"5 1\n01010\n",
"3 2\n000\n"
] |
[
"2\n",
"2\n",
"1\n"
] |
In the first sample, Farmer John can book room 3 for himself, and rooms 1 and 4 for his cows. The distance to the farthest cow is 2. Note that it is impossible to make this distance 1, as there is no block of three consecutive unoccupied rooms.
In the second sample, Farmer John can book room 1 for himself and room 3 for his single cow. The distance between him and his cow is 2.
In the third sample, Farmer John books all three available rooms, taking the middle room for himself so that both cows are next to him. His distance from the farthest cow is 1.
| 1,500
|
[
{
"input": "7 2\n0100100",
"output": "2"
},
{
"input": "5 1\n01010",
"output": "2"
},
{
"input": "3 2\n000",
"output": "1"
},
{
"input": "10 1\n1101111101",
"output": "6"
},
{
"input": "2 1\n00",
"output": "1"
},
{
"input": "3 1\n010",
"output": "2"
},
{
"input": "8 7\n00000000",
"output": "4"
},
{
"input": "7 6\n0000000",
"output": "3"
},
{
"input": "112 12\n0110101000000010101110010111100101011010011110100111111100011101011111000111101101110100111011110001100110110010",
"output": "10"
},
{
"input": "9 8\n000000000",
"output": "4"
},
{
"input": "9 3\n010001000",
"output": "2"
},
{
"input": "5 3\n00000",
"output": "2"
},
{
"input": "8 7\n00000000",
"output": "4"
},
{
"input": "6 1\n000011",
"output": "1"
},
{
"input": "100 40\n0010010100000100011100010100110001101100110000110010000000001010000111100000100100100101010010001100",
"output": "30"
},
{
"input": "93 79\n000000000000000000011000000000000000000000000000000000000000000000010000000000100000100000000",
"output": "42"
},
{
"input": "31 11\n0000001011011100010000000110001",
"output": "7"
},
{
"input": "47 46\n00000000000000000000000000000000000000000000000",
"output": "23"
},
{
"input": "100 96\n0000000000000010000010000000000000000000000000000000000000000000000000000010000000000000000000000000",
"output": "50"
},
{
"input": "491 89\n01111101111111100000111010110001010001110111000010101111101000100010010111011101110110111101101010111000111000011100011010010010111111000011011010100110001000011100111000001011100010001111101111101000111001100110010100101000001110010100100100100101001100010101001000010000111110011000000100000100101000100101000001001101011011100000110101111110101001001000100110010000010110101011000101011001001011001000110000011111001110101011000000110101000000100110001101111000101001001001100001001110101",
"output": "73"
},
{
"input": "308 17\n01000000100000000000000001000001000010000000000000000001001110000001010001000110000000000000100101000000010000001000000000001100000110000000000000000001000000000000000100000001000010001000000001000000000000000100010000000000000000000000000000000000001000000000001001101100000000000010000000000000000000000000",
"output": "9"
},
{
"input": "8 4\n00111000",
"output": "5"
},
{
"input": "18 2\n010111110111011110",
"output": "5"
},
{
"input": "29 3\n01110011111111111111110110110",
"output": "17"
}
] | 1,507,328,301
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 8
| 46
| 204,800
|
n, k = [int(_) for _ in input().split()]
a = input()
l = 0
r = 0
total = 0
k += 1
while True:
if a[r] == '0':
total += 1
if total == k:
break
r += 1
dis = n
i = l
for j in range(l, r+1):
if a[j] == '1':
continue
cur_dis = max(j-l, r-j)
if cur_dis > dis:
break
i = j
dis = cur_dis
while True:
r += 1
while r < n and a[r] == '1':
r += 1
if r >= n:
break
l += 1
while a[l] == '1':
l += 1
if i < l:
i = l
prev_dis = max(i-l, r-i)
m = i + 1 # next pos
while m <= r:
if a[m] == '1':
m += 1
continue
cur_dis = max(m-l, r-l)
if cur_dis > prev_dis:
break
prev_dis = cur_dis
i = m
dis = min(dis, cur_dis)
m += 1
print(dis)
|
Title: Enduring Exodus
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In an attempt to escape the Mischievous Mess Makers' antics, Farmer John has abandoned his farm and is traveling to the other side of Bovinia. During the journey, he and his *k* cows have decided to stay at the luxurious Grand Moo-dapest Hotel. The hotel consists of *n* rooms located in a row, some of which are occupied.
Farmer John wants to book a set of *k*<=+<=1 currently unoccupied rooms for him and his cows. He wants his cows to stay as safe as possible, so he wishes to minimize the maximum distance from his room to the room of his cow. The distance between rooms *i* and *j* is defined as |*j*<=-<=*i*|. Help Farmer John protect his cows by calculating this minimum possible distance.
Input Specification:
The first line of the input contains two integers *n* and *k* (1<=≤<=*k*<=<<=*n*<=≤<=100<=000) — the number of rooms in the hotel and the number of cows travelling with Farmer John.
The second line contains a string of length *n* describing the rooms. The *i*-th character of the string will be '0' if the *i*-th room is free, and '1' if the *i*-th room is occupied. It is guaranteed that at least *k*<=+<=1 characters of this string are '0', so there exists at least one possible choice of *k*<=+<=1 rooms for Farmer John and his cows to stay in.
Output Specification:
Print the minimum possible distance between Farmer John's room and his farthest cow.
Demo Input:
['7 2\n0100100\n', '5 1\n01010\n', '3 2\n000\n']
Demo Output:
['2\n', '2\n', '1\n']
Note:
In the first sample, Farmer John can book room 3 for himself, and rooms 1 and 4 for his cows. The distance to the farthest cow is 2. Note that it is impossible to make this distance 1, as there is no block of three consecutive unoccupied rooms.
In the second sample, Farmer John can book room 1 for himself and room 3 for his single cow. The distance between him and his cow is 2.
In the third sample, Farmer John books all three available rooms, taking the middle room for himself so that both cows are next to him. His distance from the farthest cow is 1.
|
```python
n, k = [int(_) for _ in input().split()]
a = input()
l = 0
r = 0
total = 0
k += 1
while True:
if a[r] == '0':
total += 1
if total == k:
break
r += 1
dis = n
i = l
for j in range(l, r+1):
if a[j] == '1':
continue
cur_dis = max(j-l, r-j)
if cur_dis > dis:
break
i = j
dis = cur_dis
while True:
r += 1
while r < n and a[r] == '1':
r += 1
if r >= n:
break
l += 1
while a[l] == '1':
l += 1
if i < l:
i = l
prev_dis = max(i-l, r-i)
m = i + 1 # next pos
while m <= r:
if a[m] == '1':
m += 1
continue
cur_dis = max(m-l, r-l)
if cur_dis > prev_dis:
break
prev_dis = cur_dis
i = m
dis = min(dis, cur_dis)
m += 1
print(dis)
```
| 0
|
|
118
|
A
|
String Task
|
PROGRAMMING
| 1,000
|
[
"implementation",
"strings"
] | null | null |
Petya started to attend programming lessons. On the first lesson his task was to write a simple program. The program was supposed to do the following: in the given string, consisting if uppercase and lowercase Latin letters, it:
- deletes all the vowels, - inserts a character "." before each consonant, - replaces all uppercase consonants with corresponding lowercase ones.
Vowels are letters "A", "O", "Y", "E", "U", "I", and the rest are consonants. The program's input is exactly one string, it should return the output as a single string, resulting after the program's processing the initial string.
Help Petya cope with this easy task.
|
The first line represents input string of Petya's program. This string only consists of uppercase and lowercase Latin letters and its length is from 1 to 100, inclusive.
|
Print the resulting string. It is guaranteed that this string is not empty.
|
[
"tour\n",
"Codeforces\n",
"aBAcAba\n"
] |
[
".t.r\n",
".c.d.f.r.c.s\n",
".b.c.b\n"
] |
none
| 500
|
[
{
"input": "tour",
"output": ".t.r"
},
{
"input": "Codeforces",
"output": ".c.d.f.r.c.s"
},
{
"input": "aBAcAba",
"output": ".b.c.b"
},
{
"input": "obn",
"output": ".b.n"
},
{
"input": "wpwl",
"output": ".w.p.w.l"
},
{
"input": "ggdvq",
"output": ".g.g.d.v.q"
},
{
"input": "pumesz",
"output": ".p.m.s.z"
},
{
"input": "g",
"output": ".g"
},
{
"input": "zjuotps",
"output": ".z.j.t.p.s"
},
{
"input": "jzbwuehe",
"output": ".j.z.b.w.h"
},
{
"input": "tnkgwuugu",
"output": ".t.n.k.g.w.g"
},
{
"input": "kincenvizh",
"output": ".k.n.c.n.v.z.h"
},
{
"input": "xattxjenual",
"output": ".x.t.t.x.j.n.l"
},
{
"input": "ktajqhpqsvhw",
"output": ".k.t.j.q.h.p.q.s.v.h.w"
},
{
"input": "xnhcigytnqcmy",
"output": ".x.n.h.c.g.t.n.q.c.m"
},
{
"input": "jfmtbejyilxcec",
"output": ".j.f.m.t.b.j.l.x.c.c"
},
{
"input": "D",
"output": ".d"
},
{
"input": "ab",
"output": ".b"
},
{
"input": "Ab",
"output": ".b"
},
{
"input": "aB",
"output": ".b"
},
{
"input": "AB",
"output": ".b"
},
{
"input": "ba",
"output": ".b"
},
{
"input": "bA",
"output": ".b"
},
{
"input": "Ba",
"output": ".b"
},
{
"input": "BA",
"output": ".b"
},
{
"input": "aab",
"output": ".b"
},
{
"input": "baa",
"output": ".b"
},
{
"input": "femOZeCArKCpUiHYnbBPTIOFmsHmcpObtPYcLCdjFrUMIyqYzAokKUiiKZRouZiNMoiOuGVoQzaaCAOkquRjmmKKElLNqCnhGdQM",
"output": ".f.m.z.c.r.k.c.p.h.n.b.b.p.t.f.m.s.h.m.c.p.b.t.p.c.l.c.d.j.f.r.m.q.z.k.k.k.z.r.z.n.m.g.v.q.z.c.k.q.r.j.m.m.k.k.l.l.n.q.c.n.h.g.d.q.m"
},
{
"input": "VMBPMCmMDCLFELLIISUJDWQRXYRDGKMXJXJHXVZADRZWVWJRKFRRNSAWKKDPZZLFLNSGUNIVJFBEQsMDHSBJVDTOCSCgZWWKvZZN",
"output": ".v.m.b.p.m.c.m.m.d.c.l.f.l.l.s.j.d.w.q.r.x.r.d.g.k.m.x.j.x.j.h.x.v.z.d.r.z.w.v.w.j.r.k.f.r.r.n.s.w.k.k.d.p.z.z.l.f.l.n.s.g.n.v.j.f.b.q.s.m.d.h.s.b.j.v.d.t.c.s.c.g.z.w.w.k.v.z.z.n"
},
{
"input": "MCGFQQJNUKuAEXrLXibVjClSHjSxmlkQGTKZrRaDNDomIPOmtSgjJAjNVIVLeUGUAOHNkCBwNObVCHOWvNkLFQQbFnugYVMkJruJ",
"output": ".m.c.g.f.q.q.j.n.k.x.r.l.x.b.v.j.c.l.s.h.j.s.x.m.l.k.q.g.t.k.z.r.r.d.n.d.m.p.m.t.s.g.j.j.j.n.v.v.l.g.h.n.k.c.b.w.n.b.v.c.h.w.v.n.k.l.f.q.q.b.f.n.g.v.m.k.j.r.j"
},
{
"input": "iyaiuiwioOyzUaOtAeuEYcevvUyveuyioeeueoeiaoeiavizeeoeyYYaaAOuouueaUioueauayoiuuyiuovyOyiyoyioaoyuoyea",
"output": ".w.z.t.c.v.v.v.v.z.v"
},
{
"input": "yjnckpfyLtzwjsgpcrgCfpljnjwqzgVcufnOvhxplvflxJzqxnhrwgfJmPzifgubvspffmqrwbzivatlmdiBaddiaktdsfPwsevl",
"output": ".j.n.c.k.p.f.l.t.z.w.j.s.g.p.c.r.g.c.f.p.l.j.n.j.w.q.z.g.v.c.f.n.v.h.x.p.l.v.f.l.x.j.z.q.x.n.h.r.w.g.f.j.m.p.z.f.g.b.v.s.p.f.f.m.q.r.w.b.z.v.t.l.m.d.b.d.d.k.t.d.s.f.p.w.s.v.l"
},
{
"input": "RIIIUaAIYJOiuYIUWFPOOAIuaUEZeIooyUEUEAoIyIHYOEAlVAAIiLUAUAeiUIEiUMuuOiAgEUOIAoOUYYEYFEoOIIVeOOAOIIEg",
"output": ".r.j.w.f.p.z.h.l.v.l.m.g.f.v.g"
},
{
"input": "VBKQCFBMQHDMGNSGBQVJTGQCNHHRJMNKGKDPPSQRRVQTZNKBZGSXBPBRXPMVFTXCHZMSJVBRNFNTHBHGJLMDZJSVPZZBCCZNVLMQ",
"output": ".v.b.k.q.c.f.b.m.q.h.d.m.g.n.s.g.b.q.v.j.t.g.q.c.n.h.h.r.j.m.n.k.g.k.d.p.p.s.q.r.r.v.q.t.z.n.k.b.z.g.s.x.b.p.b.r.x.p.m.v.f.t.x.c.h.z.m.s.j.v.b.r.n.f.n.t.h.b.h.g.j.l.m.d.z.j.s.v.p.z.z.b.c.c.z.n.v.l.m.q"
},
{
"input": "iioyoaayeuyoolyiyoeuouiayiiuyTueyiaoiueyioiouyuauouayyiaeoeiiigmioiououeieeeyuyyaYyioiiooaiuouyoeoeg",
"output": ".l.t.g.m.g"
},
{
"input": "ueyiuiauuyyeueykeioouiiauzoyoeyeuyiaoaiiaaoaueyaeydaoauexuueafouiyioueeaaeyoeuaueiyiuiaeeayaioeouiuy",
"output": ".k.z.d.x.f"
},
{
"input": "FSNRBXLFQHZXGVMKLQDVHWLDSLKGKFMDRQWMWSSKPKKQBNDZRSCBLRSKCKKFFKRDMZFZGCNSMXNPMZVDLKXGNXGZQCLRTTDXLMXQ",
"output": ".f.s.n.r.b.x.l.f.q.h.z.x.g.v.m.k.l.q.d.v.h.w.l.d.s.l.k.g.k.f.m.d.r.q.w.m.w.s.s.k.p.k.k.q.b.n.d.z.r.s.c.b.l.r.s.k.c.k.k.f.f.k.r.d.m.z.f.z.g.c.n.s.m.x.n.p.m.z.v.d.l.k.x.g.n.x.g.z.q.c.l.r.t.t.d.x.l.m.x.q"
},
{
"input": "EYAYAYIOIOYOOAUOEUEUOUUYIYUUMOEOIIIAOIUOAAOIYOIOEUIERCEYYAOIOIGYUIAOYUEOEUAEAYPOYEYUUAUOAOEIYIEYUEEY",
"output": ".m.r.c.g.p"
},
{
"input": "jvvzcdcxjstbbksmqjsngxkgtttdxrljjxtwptgwwqzpvqchvgrkqlzxmptzblxhhsmrkmzzmgdfskhtmmnqzzflpmqdctvrfgtx",
"output": ".j.v.v.z.c.d.c.x.j.s.t.b.b.k.s.m.q.j.s.n.g.x.k.g.t.t.t.d.x.r.l.j.j.x.t.w.p.t.g.w.w.q.z.p.v.q.c.h.v.g.r.k.q.l.z.x.m.p.t.z.b.l.x.h.h.s.m.r.k.m.z.z.m.g.d.f.s.k.h.t.m.m.n.q.z.z.f.l.p.m.q.d.c.t.v.r.f.g.t.x"
},
{
"input": "YB",
"output": ".b"
},
{
"input": "fly",
"output": ".f.l"
},
{
"input": "YyyYYYyyYxdwdawdDAWDdaddYYYY",
"output": ".x.d.w.d.w.d.d.w.d.d.d.d"
}
] | 1,693,230,805
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 0
| 92
| 0
|
def strings(s):
res=[]
vow="AEIOUaeiou"
for c in s:
if c not in s:
res.append('.')
if c.isupper():
res.append(c.islower())
else:
res.append(c)
return ''.join(res)
s=input()
ans=strings(s)
print(ans)
|
Title: String Task
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya started to attend programming lessons. On the first lesson his task was to write a simple program. The program was supposed to do the following: in the given string, consisting if uppercase and lowercase Latin letters, it:
- deletes all the vowels, - inserts a character "." before each consonant, - replaces all uppercase consonants with corresponding lowercase ones.
Vowels are letters "A", "O", "Y", "E", "U", "I", and the rest are consonants. The program's input is exactly one string, it should return the output as a single string, resulting after the program's processing the initial string.
Help Petya cope with this easy task.
Input Specification:
The first line represents input string of Petya's program. This string only consists of uppercase and lowercase Latin letters and its length is from 1 to 100, inclusive.
Output Specification:
Print the resulting string. It is guaranteed that this string is not empty.
Demo Input:
['tour\n', 'Codeforces\n', 'aBAcAba\n']
Demo Output:
['.t.r\n', '.c.d.f.r.c.s\n', '.b.c.b\n']
Note:
none
|
```python
def strings(s):
res=[]
vow="AEIOUaeiou"
for c in s:
if c not in s:
res.append('.')
if c.isupper():
res.append(c.islower())
else:
res.append(c)
return ''.join(res)
s=input()
ans=strings(s)
print(ans)
```
| 0
|
|
141
|
A
|
Amusing Joke
|
PROGRAMMING
| 800
|
[
"implementation",
"sortings",
"strings"
] | null | null |
So, the New Year holidays are over. Santa Claus and his colleagues can take a rest and have guests at last. When two "New Year and Christmas Men" meet, thear assistants cut out of cardboard the letters from the guest's name and the host's name in honor of this event. Then the hung the letters above the main entrance. One night, when everyone went to bed, someone took all the letters of our characters' names. Then he may have shuffled the letters and put them in one pile in front of the door.
The next morning it was impossible to find the culprit who had made the disorder. But everybody wondered whether it is possible to restore the names of the host and his guests from the letters lying at the door? That is, we need to verify that there are no extra letters, and that nobody will need to cut more letters.
Help the "New Year and Christmas Men" and their friends to cope with this problem. You are given both inscriptions that hung over the front door the previous night, and a pile of letters that were found at the front door next morning.
|
The input file consists of three lines: the first line contains the guest's name, the second line contains the name of the residence host and the third line contains letters in a pile that were found at the door in the morning. All lines are not empty and contain only uppercase Latin letters. The length of each line does not exceed 100.
|
Print "YES" without the quotes, if the letters in the pile could be permuted to make the names of the "New Year and Christmas Men". Otherwise, print "NO" without the quotes.
|
[
"SANTACLAUS\nDEDMOROZ\nSANTAMOROZDEDCLAUS\n",
"PAPAINOEL\nJOULUPUKKI\nJOULNAPAOILELUPUKKI\n",
"BABBONATALE\nFATHERCHRISTMAS\nBABCHRISTMASBONATALLEFATHER\n"
] |
[
"YES\n",
"NO\n",
"NO\n"
] |
In the first sample the letters written in the last line can be used to write the names and there won't be any extra letters left.
In the second sample letter "P" is missing from the pile and there's an extra letter "L".
In the third sample there's an extra letter "L".
| 500
|
[
{
"input": "SANTACLAUS\nDEDMOROZ\nSANTAMOROZDEDCLAUS",
"output": "YES"
},
{
"input": "PAPAINOEL\nJOULUPUKKI\nJOULNAPAOILELUPUKKI",
"output": "NO"
},
{
"input": "BABBONATALE\nFATHERCHRISTMAS\nBABCHRISTMASBONATALLEFATHER",
"output": "NO"
},
{
"input": "B\nA\nAB",
"output": "YES"
},
{
"input": "ONDOL\nJNPB\nONLNJBODP",
"output": "YES"
},
{
"input": "Y\nW\nYW",
"output": "YES"
},
{
"input": "OI\nM\nIMO",
"output": "YES"
},
{
"input": "VFQRWWWACX\nGHZJPOQUSXRAQDGOGMR\nOPAWDOUSGWWCGQXXQAZJRQRGHRMVF",
"output": "YES"
},
{
"input": "JUTCN\nPIGMZOPMEUFADQBW\nNWQGZMAIPUPOMCDUB",
"output": "NO"
},
{
"input": "Z\nO\nZOCNDOLTBZKQLTBOLDEGXRHZGTTPBJBLSJCVSVXISQZCSFDEBXRCSGBGTHWOVIXYHACAGBRYBKBJAEPIQZHVEGLYH",
"output": "NO"
},
{
"input": "IQ\nOQ\nQOQIGGKFNHJSGCGM",
"output": "NO"
},
{
"input": "ROUWANOPNIGTVMIITVMZ\nOQTUPZMTKUGY\nVTVNGZITGPUNPMQOOATUUIYIWMMKZOTR",
"output": "YES"
},
{
"input": "OVQELLOGFIOLEHXMEMBJDIGBPGEYFG\nJNKFPFFIJOFHRIFHXEWYZOPDJBZTJZKBWQTECNHRFSJPJOAPQT\nYAIPFFFEXJJNEJPLREIGODEGQZVMCOBDFKWTMWJSBEBTOFFQOHIQJLHFNXIGOHEZRZLFOKJBJPTPHPGY",
"output": "YES"
},
{
"input": "NBJGVNGUISUXQTBOBKYHQCOOVQWUXWPXBUDLXPKX\nNSFQDFUMQDQWQ\nWXKKVNTDQQFXCUQBIMQGQHSLVGWSBFYBUPOWPBDUUJUXQNOQDNXOX",
"output": "YES"
},
{
"input": "IJHHGKCXWDBRWJUPRDBZJLNTTNWKXLUGJSBWBOAUKWRAQWGFNL\nNJMWRMBCNPHXTDQQNZ\nWDNJRCLILNQRHWBANLTXWMJBPKUPGKJDJZAQWKTZFBRCTXHHBNXRGUQUNBNMWODGSJWW",
"output": "YES"
},
{
"input": "SRROWANGUGZHCIEFYMQVTWVOMDWPUZJFRDUMVFHYNHNTTGNXCJ\nDJYWGLBFCCECXFHOLORDGDCNRHPWXNHXFCXQCEZUHRRNAEKUIX\nWCUJDNYHNHYOPWMHLDCDYRWBVOGHFFUKOZTXJRXJHRGWICCMRNEVNEGQWTZPNFCSHDRFCFQDCXMHTLUGZAXOFNXNVGUEXIACRERU",
"output": "YES"
},
{
"input": "H\nJKFGHMIAHNDBMFXWYQLZRSVNOTEGCQSVUBYUOZBTNKTXPFQDCMKAGFITEUGOYDFIYQIORMFJEOJDNTFVIQEBICSNGKOSNLNXJWC\nBQSVDOGIHCHXSYNYTQFCHNJGYFIXTSOQINZOKSVQJMTKNTGFNXAVTUYEONMBQMGJLEWJOFGEARIOPKFUFCEMUBRBDNIIDFZDCLWK",
"output": "YES"
},
{
"input": "DSWNZRFVXQ\nPVULCZGOOU\nUOLVZXNUPOQRZGWFVDSCANQTCLEIE",
"output": "NO"
},
{
"input": "EUHTSCENIPXLTSBMLFHD\nIZAVSZPDLXOAGESUSE\nLXAELAZ",
"output": "NO"
},
{
"input": "WYSJFEREGELSKRQRXDXCGBODEFZVSI\nPEJKMGFLBFFDWRCRFSHVEFLEBTJCVCHRJTLDTISHPOGFWPLEWNYJLMXWIAOTYOXMV\nHXERTZWLEXTPIOTFRVMEJVYFFJLRPFMXDEBNSGCEOFFCWTKIDDGCFYSJKGLHBORWEPLDRXRSJYBGASSVCMHEEJFLVI",
"output": "NO"
},
{
"input": "EPBMDIUQAAUGLBIETKOKFLMTCVEPETWJRHHYKCKU\nHGMAETVPCFZYNNKDQXVXUALHYLOTCHM\nECGXACVKEYMCEDOTMKAUFHLHOMT",
"output": "NO"
},
{
"input": "NUBKQEJHALANSHEIFUZHYEZKKDRFHQKAJHLAOWTZIMOCWOVVDW\nEFVOBIGAUAUSQGVSNBKNOBDMINODMFSHDL\nKLAMKNTHBFFOHVKWICHBKNDDQNEISODUSDNLUSIOAVWY",
"output": "NO"
},
{
"input": "VXINHOMEQCATZUGAJEIUIZZLPYFGUTVLNBNWCUVMEENUXKBWBGZTMRJJVJDLVSLBABVCEUDDSQFHOYPYQTWVAGTWOLKYISAGHBMC\nZMRGXPZSHOGCSAECAPGVOIGCWEOWWOJXLGYRDMPXBLOKZVRACPYQLEQGFQCVYXAGBEBELUTDAYEAGPFKXRULZCKFHZCHVCWIRGPK\nRCVUXGQVNWFGRUDLLENNDQEJHYYVWMKTLOVIPELKPWCLSQPTAXAYEMGWCBXEVAIZGGDDRBRT",
"output": "NO"
},
{
"input": "PHBDHHWUUTZAHELGSGGOPOQXSXEZIXHZTOKYFBQLBDYWPVCNQSXHEAXRRPVHFJBVBYCJIFOTQTWSUOWXLKMVJJBNLGTVITWTCZZ\nFUPDLNVIHRWTEEEHOOEC\nLOUSUUSZCHJBPEWIILUOXEXRQNCJEGTOBRVZLTTZAHTKVEJSNGHFTAYGY",
"output": "NO"
},
{
"input": "GDSLNIIKTO\nJF\nPDQYFKDTNOLI",
"output": "NO"
},
{
"input": "AHOKHEKKPJLJIIWJRCGY\nORELJCSIX\nZVWPXVFWFSWOXXLIHJKPXIOKRELYE",
"output": "NO"
},
{
"input": "ZWCOJFORBPHXCOVJIDPKVECMHVHCOC\nTEV\nJVGTBFTLFVIEPCCHODOFOMCVZHWXVCPEH",
"output": "NO"
},
{
"input": "AGFIGYWJLVMYZGNQHEHWKJIAWBPUAQFERMCDROFN\nPMJNHMVNRGCYZAVRWNDSMLSZHFNYIUWFPUSKKIGU\nMCDVPPRXGUAYLSDRHRURZASXUWZSIIEZCPXUVEONKNGNWRYGOSFMCKESMVJZHWWUCHWDQMLASLNNMHAU",
"output": "NO"
},
{
"input": "XLOWVFCZSSXCSYQTIIDKHNTKNKEEDFMDZKXSPVLBIDIREDUAIN\nZKIWNDGBISDB\nSLPKLYFYSRNRMOSWYLJJDGFFENPOXYLPZFTQDANKBDNZDIIEWSUTTKYBKVICLG",
"output": "NO"
},
{
"input": "PMUKBTRKFIAYVGBKHZHUSJYSSEPEOEWPOSPJLWLOCTUYZODLTUAFCMVKGQKRRUSOMPAYOTBTFPXYAZXLOADDEJBDLYOTXJCJYTHA\nTWRRAJLCQJTKOKWCGUH\nEWDPNXVCXWCDQCOYKKSOYTFSZTOOPKPRDKFJDETKSRAJRVCPDOBWUGPYRJPUWJYWCBLKOOTUPBESTOFXZHTYLLMCAXDYAEBUTAHM",
"output": "NO"
},
{
"input": "QMIMGQRQDMJDPNFEFXSXQMCHEJKTWCTCVZPUAYICOIRYOWKUSIWXJLHDYWSBOITHTMINXFKBKAWZTXXBJIVYCRWKXNKIYKLDDXL\nV\nFWACCXBVDOJFIUAVYRALBYJKXXWIIFORRUHKHCXLDBZMXIYJWISFEAWTIQFIZSBXMKNOCQKVKRWDNDAMQSTKYLDNYVTUCGOJXJTW",
"output": "NO"
},
{
"input": "XJXPVOOQODELPPWUISSYVVXRJTYBPDHJNENQEVQNVFIXSESKXVYPVVHPMOSX\nLEXOPFPVPSZK\nZVXVPYEYOYXVOISVLXPOVHEQVXPNQJIOPFDTXEUNMPEPPHELNXKKWSVSOXSBPSJDPVJVSRFQ",
"output": "YES"
},
{
"input": "OSKFHGYNQLSRFSAHPXKGPXUHXTRBJNAQRBSSWJVEENLJCDDHFXVCUNPZAIVVO\nFNUOCXAGRRHNDJAHVVLGGEZQHWARYHENBKHP\nUOEFNWVXCUNERLKVTHAGPSHKHDYFPYWZHJKHQLSNFBJHVJANRXCNSDUGVDABGHVAOVHBJZXGRACHRXEGNRPQEAPORQSILNXFS",
"output": "YES"
},
{
"input": "VYXYVVACMLPDHONBUTQFZTRREERBLKUJYKAHZRCTRLRCLOZYWVPBRGDQPFPQIF\nFE\nRNRPEVDRLYUQFYRZBCQLCYZEABKLRXCJLKVZBVFUEYRATOMDRTHFPGOWQVTIFPPH",
"output": "YES"
},
{
"input": "WYXUZQJQNLASEGLHPMSARWMTTQMQLVAZLGHPIZTRVTCXDXBOLNXZPOFCTEHCXBZ\nBLQZRRWP\nGIQZXPLTTMNHQVWPPEAPLOCDMBSTHRCFLCQRRZXLVAOQEGZBRUZJXXZTMAWLZHSLWNQTYXB",
"output": "YES"
},
{
"input": "MKVJTSSTDGKPVVDPYSRJJYEVGKBMSIOKHLZQAEWLRIBINVRDAJIBCEITKDHUCCVY\nPUJJQFHOGZKTAVNUGKQUHMKTNHCCTI\nQVJKUSIGTSVYUMOMLEGHWYKSKQTGATTKBNTKCJKJPCAIRJIRMHKBIZISEGFHVUVQZBDERJCVAKDLNTHUDCHONDCVVJIYPP",
"output": "YES"
},
{
"input": "OKNJOEYVMZXJMLVJHCSPLUCNYGTDASKSGKKCRVIDGEIBEWRVBVRVZZTLMCJLXHJIA\nDJBFVRTARTFZOWN\nAGHNVUNJVCPLWSVYBJKZSVTFGLELZASLWTIXDDJXCZDICTVIJOTMVEYOVRNMJGRKKHRMEBORAKFCZJBR",
"output": "YES"
},
{
"input": "OQZACLPSAGYDWHFXDFYFRRXWGIEJGSXWUONAFWNFXDTGVNDEWNQPHUXUJNZWWLBPYL\nOHBKWRFDRQUAFRCMT\nWIQRYXRJQWWRUWCYXNXALKFZGXFTLOODWRDPGURFUFUQOHPWBASZNVWXNCAGHWEHFYESJNFBMNFDDAPLDGT",
"output": "YES"
},
{
"input": "OVIRQRFQOOWVDEPLCJETWQSINIOPLTLXHSQWUYUJNFBMKDNOSHNJQQCDHZOJVPRYVSV\nMYYDQKOOYPOOUELCRIT\nNZSOTVLJTTVQLFHDQEJONEOUOFOLYVSOIYUDNOSIQVIRMVOERCLMYSHPCQKIDRDOQPCUPQBWWRYYOXJWJQPNKH",
"output": "YES"
},
{
"input": "WGMBZWNMSJXNGDUQUJTCNXDSJJLYRDOPEGPQXYUGBESDLFTJRZDDCAAFGCOCYCQMDBWK\nYOBMOVYTUATTFGJLYUQD\nDYXVTLQCYFJUNJTUXPUYOPCBCLBWNSDUJRJGWDOJDSQAAMUOJWSYERDYDXYTMTOTMQCGQZDCGNFBALGGDFKZMEBG",
"output": "YES"
},
{
"input": "CWLRBPMEZCXAPUUQFXCUHAQTLPBTXUUKWVXKBHKNSSJFEXLZMXGVFHHVTPYAQYTIKXJJE\nMUFOSEUEXEQTOVLGDSCWM\nJUKEQCXOXWEHCGKFPBIGMWVJLXUONFXBYTUAXERYTXKCESKLXAEHVPZMMUFTHLXTTZSDMBJLQPEUWCVUHSQQVUASPF",
"output": "YES"
},
{
"input": "IDQRX\nWETHO\nODPDGBHVUVSSISROHQJTUKPUCLXABIZQQPPBPKOSEWGEHRSRRNBAVLYEMZISMWWGKHVTXKUGUXEFBSWOIWUHRJGMWBMHQLDZHBWA",
"output": "NO"
},
{
"input": "IXFDY\nJRMOU\nDF",
"output": "NO"
},
{
"input": "JPSPZ\nUGCUB\nJMZZZZZZZZ",
"output": "NO"
},
{
"input": "AC\nA\nBBA",
"output": "NO"
},
{
"input": "UIKWWKXLSHTOOZOVGXKYSOJEHAUEEG\nKZXQDWJJWRXFHKJDQHJK\nXMZHTFOGEXAUJXXJUYVJIFOTKLZHDKELJWERHMGAWGKWAQKEKHIDWGGZVYOHKXRPWSJDPESFJUMKQYWBYUTHQYEFZUGKQOBHYDWB",
"output": "NO"
},
{
"input": "PXWRXRPFLR\nPJRWWXIVHODV\nXW",
"output": "NO"
},
{
"input": "CHTAZVHGSHCVIBK\nEQINEBKXEPYJSAZIMLDF\nZCZZZZDZMCZZEZDZZEZZZZQZZBZZZOZZCZE",
"output": "NO"
},
{
"input": "GXPZFSELJJNDAXYRV\nUYBKPMVBSOVOJWMONLTJOJCNQKMTAHEWLHOWIIBH\nHCWNFWJPEJIWOVPTBMVCRJLSISSVNOHCKLBFMIUAIMASQWPXEYXBOXQGFEMYJLBKDCZIMJNHOJEDGGANIVYKQTUOSOVOPWHVJGXH",
"output": "NO"
},
{
"input": "LFGJCJJDUTUP\nOVSBILTIYCJCRHKCIXCETJQJJ\nGIJJTJCLTJJJ",
"output": "NO"
},
{
"input": "GIO\nPRL\nPRL",
"output": "NO"
},
{
"input": "A\nB\nABC",
"output": "NO"
},
{
"input": "KKK\nKKK\nZZZZZ",
"output": "NO"
},
{
"input": "ZMYGQLDBLAPN\nZFJBKWHROVNPSJQUDFTHOCGREUFLYIWYICD\nZMJZZEDAZANKZZZZZZEZZBZDZZZZZZKHZZFZZZDZNZMDZZA",
"output": "NO"
}
] | 1,689,515,981
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 8
| 154
| 0
|
n1=input()
n2=input()
n3=input()
l1=[]
for i in n1:
l1.append(i)
for k in n2:
l1.append(k)
l2=[]
fl=True
for j in n3:
if j in l1:
l1.remove(j)
else:
fl=False
if fl==True:
print("YES")
else:
print("NO")
|
Title: Amusing Joke
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
So, the New Year holidays are over. Santa Claus and his colleagues can take a rest and have guests at last. When two "New Year and Christmas Men" meet, thear assistants cut out of cardboard the letters from the guest's name and the host's name in honor of this event. Then the hung the letters above the main entrance. One night, when everyone went to bed, someone took all the letters of our characters' names. Then he may have shuffled the letters and put them in one pile in front of the door.
The next morning it was impossible to find the culprit who had made the disorder. But everybody wondered whether it is possible to restore the names of the host and his guests from the letters lying at the door? That is, we need to verify that there are no extra letters, and that nobody will need to cut more letters.
Help the "New Year and Christmas Men" and their friends to cope with this problem. You are given both inscriptions that hung over the front door the previous night, and a pile of letters that were found at the front door next morning.
Input Specification:
The input file consists of three lines: the first line contains the guest's name, the second line contains the name of the residence host and the third line contains letters in a pile that were found at the door in the morning. All lines are not empty and contain only uppercase Latin letters. The length of each line does not exceed 100.
Output Specification:
Print "YES" without the quotes, if the letters in the pile could be permuted to make the names of the "New Year and Christmas Men". Otherwise, print "NO" without the quotes.
Demo Input:
['SANTACLAUS\nDEDMOROZ\nSANTAMOROZDEDCLAUS\n', 'PAPAINOEL\nJOULUPUKKI\nJOULNAPAOILELUPUKKI\n', 'BABBONATALE\nFATHERCHRISTMAS\nBABCHRISTMASBONATALLEFATHER\n']
Demo Output:
['YES\n', 'NO\n', 'NO\n']
Note:
In the first sample the letters written in the last line can be used to write the names and there won't be any extra letters left.
In the second sample letter "P" is missing from the pile and there's an extra letter "L".
In the third sample there's an extra letter "L".
|
```python
n1=input()
n2=input()
n3=input()
l1=[]
for i in n1:
l1.append(i)
for k in n2:
l1.append(k)
l2=[]
fl=True
for j in n3:
if j in l1:
l1.remove(j)
else:
fl=False
if fl==True:
print("YES")
else:
print("NO")
```
| 0
|
|
733
|
A
|
Grasshopper And the String
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
One day, the Grasshopper was jumping on the lawn and found a piece of paper with a string. Grasshopper became interested what is the minimum jump ability he should have in order to be able to reach the far end of the string, jumping only on vowels of the English alphabet. Jump ability is the maximum possible length of his jump.
Formally, consider that at the begginning the Grasshopper is located directly in front of the leftmost character of the string. His goal is to reach the position right after the rightmost character of the string. In one jump the Grasshopper could jump to the right any distance from 1 to the value of his jump ability.
The following letters are vowels: 'A', 'E', 'I', 'O', 'U' and 'Y'.
|
The first line contains non-empty string consisting of capital English letters. It is guaranteed that the length of the string does not exceed 100.
|
Print single integer *a* — the minimum jump ability of the Grasshopper (in the number of symbols) that is needed to overcome the given string, jumping only on vowels.
|
[
"ABABBBACFEYUKOTT\n",
"AAA\n"
] |
[
"4",
"1"
] |
none
| 500
|
[
{
"input": "ABABBBACFEYUKOTT",
"output": "4"
},
{
"input": "AAA",
"output": "1"
},
{
"input": "A",
"output": "1"
},
{
"input": "B",
"output": "2"
},
{
"input": "AEYUIOAEIYAEOUIYOEIUYEAOIUEOEAYOEIUYAEOUIYEOIKLMJNHGTRWSDZXCVBNMHGFDSXVWRTPPPLKMNBXIUOIUOIUOIUOOIU",
"output": "39"
},
{
"input": "AEYUIOAEIYAEOUIYOEIUYEAOIUEOEAYOEIUYAEOUIYEOIAEYUIOAEIYAEOUIYOEIUYEAOIUEOEAYOEIUYAEOUIYEOI",
"output": "1"
},
{
"input": "KMLPTGFHNBVCDRFGHNMBVXWSQFDCVBNHTJKLPMNFVCKMLPTGFHNBVCDRFGHNMBVXWSQFDCVBNHTJKLPMNFVC",
"output": "85"
},
{
"input": "QWERTYUIOPASDFGHJKLZXCVBNMQWERTYUIOPASDFGHJKLZXCVBNMQWERTYUIOPASDFGHJKLZXCVBNMQWERTYUIOPASDFGHJKLZ",
"output": "18"
},
{
"input": "PKLKBWTXVJ",
"output": "11"
},
{
"input": "CFHFPTGMOKXVLJJZJDQW",
"output": "12"
},
{
"input": "TXULTFSBUBFLRNQORMMULWNVLPWTYJXZBPBGAWNX",
"output": "9"
},
{
"input": "DAIUSEAUEUYUWEIOOEIOUYVYYOPEEWEBZOOOAOXUOIEUKYYOJOYAUYUUIYUXOUJLGIYEIIYUOCUAACRY",
"output": "4"
},
{
"input": "VRPHBNWNWVWBWMFJJDCTJQJDJBKSJRZLVQRVVFLTZFSGCGDXCWQVWWWMFVCQHPKXXVRKTGWGPSMQTPKNDQJHNSKLXPCXDJDQDZZD",
"output": "101"
},
{
"input": "SGDDFCDRDWGPNNFBBZZJSPXFYMZKPRXTCHVJSJJBWZXXQMDZBNKDHRGSRLGLRKPMWXNSXJPNJLDPXBSRCQMHJKPZNTPNTZXNPCJC",
"output": "76"
},
{
"input": "NVTQVNLGWFDBCBKSDLTBGWBMNQZWZQJWNGVCTCQBGWNTYJRDBPZJHXCXFMIXNRGSTXHQPCHNFQPCMDZWJGLJZWMRRFCVLBKDTDSC",
"output": "45"
},
{
"input": "SREZXQFVPQCLRCQGMKXCBRWKYZKWKRMZGXPMKWNMFZTRDPHJFCSXVPPXWKZMZTBFXGNLPLHZIPLFXNRRQFDTLFPKBGCXKTMCFKKT",
"output": "48"
},
{
"input": "ICKJKMVPDNZPLKDSLTPZNRLSQSGHQJQQPJJSNHNWVDLJRLZEJSXZDPHYXGGWXHLCTVQSKWNWGTLJMOZVJNZPVXGVPJKHFVZTGCCX",
"output": "47"
},
{
"input": "XXFPZDRPXLNHGDVCBDKJMKLGUQZXLLWYLOKFZVGXVNPJWZZZNRMQBRJCZTSDRHSNCVDMHKVXCXPCRBWSJCJWDRDPVZZLCZRTDRYA",
"output": "65"
},
{
"input": "HDDRZDKCHHHEDKHZMXQSNQGSGNNSCCPVJFGXGNCEKJMRKSGKAPQWPCWXXWHLSMRGSJWEHWQCSJJSGLQJXGVTBYALWMLKTTJMFPFS",
"output": "28"
},
{
"input": "PXVKJHXVDPWGLHWFWMJPMCCNHCKSHCPZXGIHHNMYNFQBUCKJJTXXJGKRNVRTQFDFMLLGPQKFOVNNLTNDIEXSARRJKGSCZKGGJCBW",
"output": "35"
},
{
"input": "EXNMTTFPJLDHXDQBJJRDRYBZVFFHUDCHCPNFZWXSMZXNFVJGHZWXVBRQFNUIDVLZOVPXQNVMFNBTJDSCKRLNGXPSADTGCAHCBJKL",
"output": "30"
},
{
"input": "NRNLSQQJGIJBCZFTNKJCXMGPARGWXPSHZXOBNSFOLDQVXTVAGJZNLXULHBRDGMNQKQGWMRRDPYCSNFVPUFTFBUBRXVJGNGSPJKLL",
"output": "19"
},
{
"input": "SRHOKCHQQMVZKTCVQXJJCFGYFXGMBZSZFNAFETXILZHPGHBWZRZQFMGSEYRUDVMCIQTXTBTSGFTHRRNGNTHHWWHCTDFHSVARMCMB",
"output": "30"
},
{
"input": "HBSVZHDKGNIRQUBYKYHUPJCEETGFMVBZJTHYHFQPFBVBSMQACYAVWZXSBGNKWXFNMQJFMSCHJVWBZXZGSNBRUHTHAJKVLEXFBOFB",
"output": "34"
},
{
"input": "NXKMUGOPTUQNSRYTKUKSCWCRQSZKKFPYUMDIBJAHJCEKZJVWZAWOLOEFBFXLQDDPNNZKCQHUPBFVDSXSUCVLMZXQROYQYIKPQPWR",
"output": "17"
},
{
"input": "TEHJDICFNOLQVQOAREVAGUAWODOCXJXIHYXFAEPEXRHPKEIIRCRIVASKNTVYUYDMUQKSTSSBYCDVZKDDHTSDWJWACPCLYYOXGCLT",
"output": "15"
},
{
"input": "LCJJUZZFEIUTMSEXEYNOOAIZMORQDOANAMUCYTFRARDCYHOYOPHGGYUNOGNXUAOYSEMXAZOOOFAVHQUBRNGORSPNQWZJYQQUNPEB",
"output": "9"
},
{
"input": "UUOKAOOJBXUTSMOLOOOOSUYYFTAVBNUXYFVOOGCGZYQEOYISIYOUULUAIJUYVVOENJDOCLHOSOHIHDEJOIGZNIXEMEGZACHUAQFW",
"output": "5"
},
{
"input": "OUUBEHXOOURMOAIAEHXCUOIYHUJEVAWYRCIIAGDRIPUIPAIUYAIWJEVYEYYUYBYOGVYESUJCFOJNUAHIOOKBUUHEJFEWPOEOUHYA",
"output": "4"
},
{
"input": "EMNOYEEUIOUHEWZITIAEZNCJUOUAOQEAUYEIHYUSUYUUUIAEDIOOERAEIRBOJIEVOMECOGAIAIUIYYUWYIHIOWVIJEYUEAFYULSE",
"output": "5"
},
{
"input": "BVOYEAYOIEYOREJUYEUOEOYIISYAEOUYAAOIOEOYOOOIEFUAEAAESUOOIIEUAAGAEISIAPYAHOOEYUJHUECGOYEIDAIRTBHOYOYA",
"output": "5"
},
{
"input": "GOIEOAYIEYYOOEOAIAEOOUWYEIOTNYAANAYOOXEEOEAVIOIAAIEOIAUIAIAAUEUAOIAEUOUUZYIYAIEUEGOOOOUEIYAEOSYAEYIO",
"output": "3"
},
{
"input": "AUEAOAYIAOYYIUIOAULIOEUEYAIEYYIUOEOEIEYRIYAYEYAEIIMMAAEAYAAAAEOUICAUAYOUIAOUIAIUOYEOEEYAEYEYAAEAOYIY",
"output": "3"
},
{
"input": "OAIIYEYYAOOEIUOEEIOUOIAEFIOAYETUYIOAAAEYYOYEYOEAUIIUEYAYYIIAOIEEYGYIEAAOOWYAIEYYYIAOUUOAIAYAYYOEUEOY",
"output": "2"
},
{
"input": "EEEAOEOEEIOUUUEUEAAOEOIUYJEYAIYIEIYYEAUOIIYIUOOEUCYEOOOYYYIUUAYIAOEUEIEAOUOIAACAOOUAUIYYEAAAOOUYIAAE",
"output": "2"
},
{
"input": "AYEYIIEUIYOYAYEUEIIIEUYUUAUEUIYAIAAUYONIEYIUIAEUUOUOYYOUUUIUIAEYEOUIIUOUUEOAIUUYAAEOAAEOYUUIYAYRAIII",
"output": "2"
},
{
"input": "YOOAAUUAAAYEUYIUIUYIUOUAEIEEIAUEOAUIIAAIUYEUUOYUIYEAYAAAYUEEOEEAEOEEYYOUAEUYEEAIIYEUEYJOIIYUIOIUOIEE",
"output": "2"
},
{
"input": "UYOIIIAYOOAIUUOOEEUYIOUAEOOEIOUIAIEYOAEAIOOEOOOIUYYUYIAAUIOUYYOOUAUIEYYUOAAUUEAAIEUIAUEUUIAUUOYOAYIU",
"output": "1"
},
{
"input": "ABBABBB",
"output": "4"
},
{
"input": "ABCD",
"output": "4"
},
{
"input": "XXYC",
"output": "3"
},
{
"input": "YYY",
"output": "1"
},
{
"input": "ABABBBBBBB",
"output": "8"
},
{
"input": "YYYY",
"output": "1"
},
{
"input": "YYYYY",
"output": "1"
},
{
"input": "AXXX",
"output": "4"
},
{
"input": "YYYYYYY",
"output": "1"
},
{
"input": "BYYBBB",
"output": "4"
},
{
"input": "YYYYYYYYY",
"output": "1"
},
{
"input": "CAAAAA",
"output": "2"
},
{
"input": "CCCACCCC",
"output": "5"
},
{
"input": "ABABBBACFEYUKOTTTT",
"output": "5"
},
{
"input": "AABBYYYYYYYY",
"output": "3"
},
{
"input": "BYBACYC",
"output": "2"
},
{
"input": "Y",
"output": "1"
},
{
"input": "ABBBBBB",
"output": "7"
},
{
"input": "BACDYDI",
"output": "3"
},
{
"input": "XEXXXXXXXXXXXXXXX",
"output": "16"
},
{
"input": "TTYTT",
"output": "3"
},
{
"input": "AAYBC",
"output": "3"
},
{
"input": "ABABBBACFEYUKOTTTTT",
"output": "6"
},
{
"input": "YYAYY",
"output": "1"
},
{
"input": "YZZY",
"output": "3"
},
{
"input": "YY",
"output": "1"
},
{
"input": "ZZYZZ",
"output": "3"
},
{
"input": "YBBBY",
"output": "4"
},
{
"input": "BBBACCCCCCC",
"output": "8"
},
{
"input": "YBBBBY",
"output": "5"
},
{
"input": "YYYYYYYYYY",
"output": "1"
},
{
"input": "ABABBBBBBBBBBBB",
"output": "13"
}
] | 1,633,849,134
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 70
| 108
| 20,172,800
|
pre='AEIOUY'
s=input()
s='A'+s+'A'
ans=0;begin=0
for i in range(len(s)):
if s[i] in pre :
ans=max(ans,i-begin)
begin=i
print(ans)
|
Title: Grasshopper And the String
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day, the Grasshopper was jumping on the lawn and found a piece of paper with a string. Grasshopper became interested what is the minimum jump ability he should have in order to be able to reach the far end of the string, jumping only on vowels of the English alphabet. Jump ability is the maximum possible length of his jump.
Formally, consider that at the begginning the Grasshopper is located directly in front of the leftmost character of the string. His goal is to reach the position right after the rightmost character of the string. In one jump the Grasshopper could jump to the right any distance from 1 to the value of his jump ability.
The following letters are vowels: 'A', 'E', 'I', 'O', 'U' and 'Y'.
Input Specification:
The first line contains non-empty string consisting of capital English letters. It is guaranteed that the length of the string does not exceed 100.
Output Specification:
Print single integer *a* — the minimum jump ability of the Grasshopper (in the number of symbols) that is needed to overcome the given string, jumping only on vowels.
Demo Input:
['ABABBBACFEYUKOTT\n', 'AAA\n']
Demo Output:
['4', '1']
Note:
none
|
```python
pre='AEIOUY'
s=input()
s='A'+s+'A'
ans=0;begin=0
for i in range(len(s)):
if s[i] in pre :
ans=max(ans,i-begin)
begin=i
print(ans)
```
| 3
|
|
61
|
A
|
Ultra-Fast Mathematician
|
PROGRAMMING
| 800
|
[
"implementation"
] |
A. Ultra-Fast Mathematician
|
2
|
256
|
Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate.
|
There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.
|
Write one line — the corresponding answer. Do not omit the leading 0s.
|
[
"1010100\n0100101\n",
"000\n111\n",
"1110\n1010\n",
"01110\n01100\n"
] |
[
"1110001\n",
"111\n",
"0100\n",
"00010\n"
] |
none
| 500
|
[
{
"input": "1010100\n0100101",
"output": "1110001"
},
{
"input": "000\n111",
"output": "111"
},
{
"input": "1110\n1010",
"output": "0100"
},
{
"input": "01110\n01100",
"output": "00010"
},
{
"input": "011101\n000001",
"output": "011100"
},
{
"input": "10\n01",
"output": "11"
},
{
"input": "00111111\n11011101",
"output": "11100010"
},
{
"input": "011001100\n101001010",
"output": "110000110"
},
{
"input": "1100100001\n0110101100",
"output": "1010001101"
},
{
"input": "00011101010\n10010100101",
"output": "10001001111"
},
{
"input": "100000101101\n111010100011",
"output": "011010001110"
},
{
"input": "1000001111010\n1101100110001",
"output": "0101101001011"
},
{
"input": "01011111010111\n10001110111010",
"output": "11010001101101"
},
{
"input": "110010000111100\n001100101011010",
"output": "111110101100110"
},
{
"input": "0010010111110000\n0000000011010110",
"output": "0010010100100110"
},
{
"input": "00111110111110000\n01111100001100000",
"output": "01000010110010000"
},
{
"input": "101010101111010001\n001001111101111101",
"output": "100011010010101100"
},
{
"input": "0110010101111100000\n0011000101000000110",
"output": "0101010000111100110"
},
{
"input": "11110100011101010111\n00001000011011000000",
"output": "11111100000110010111"
},
{
"input": "101010101111101101001\n111010010010000011111",
"output": "010000111101101110110"
},
{
"input": "0000111111100011000010\n1110110110110000001010",
"output": "1110001001010011001000"
},
{
"input": "10010010101000110111000\n00101110100110111000111",
"output": "10111100001110001111111"
},
{
"input": "010010010010111100000111\n100100111111100011001110",
"output": "110110101101011111001001"
},
{
"input": "0101110100100111011010010\n0101100011010111001010001",
"output": "0000010111110000010000011"
},
{
"input": "10010010100011110111111011\n10000110101100000001000100",
"output": "00010100001111110110111111"
},
{
"input": "000001111000000100001000000\n011100111101111001110110001",
"output": "011101000101111101111110001"
},
{
"input": "0011110010001001011001011100\n0000101101000011101011001010",
"output": "0011011111001010110010010110"
},
{
"input": "11111000000000010011001101111\n11101110011001010100010000000",
"output": "00010110011001000111011101111"
},
{
"input": "011001110000110100001100101100\n001010000011110000001000101001",
"output": "010011110011000100000100000101"
},
{
"input": "1011111010001100011010110101111\n1011001110010000000101100010101",
"output": "0000110100011100011111010111010"
},
{
"input": "10111000100001000001010110000001\n10111000001100101011011001011000",
"output": "00000000101101101010001111011001"
},
{
"input": "000001010000100001000000011011100\n111111111001010100100001100000111",
"output": "111110101001110101100001111011011"
},
{
"input": "1101000000000010011011101100000110\n1110000001100010011010000011011110",
"output": "0011000001100000000001101111011000"
},
{
"input": "01011011000010100001100100011110001\n01011010111000001010010100001110000",
"output": "00000001111010101011110000010000001"
},
{
"input": "000011111000011001000110111100000100\n011011000110000111101011100111000111",
"output": "011000111110011110101101011011000011"
},
{
"input": "1001000010101110001000000011111110010\n0010001011010111000011101001010110000",
"output": "1011001001111001001011101010101000010"
},
{
"input": "00011101011001100101111111000000010101\n10010011011011001011111000000011101011",
"output": "10001110000010101110000111000011111110"
},
{
"input": "111011100110001001101111110010111001010\n111111101101111001110010000101101000100",
"output": "000100001011110000011101110111010001110"
},
{
"input": "1111001001101000001000000010010101001010\n0010111100111110001011000010111110111001",
"output": "1101110101010110000011000000101011110011"
},
{
"input": "00100101111000000101011111110010100011010\n11101110001010010101001000111110101010100",
"output": "11001011110010010000010111001100001001110"
},
{
"input": "101011001110110100101001000111010101101111\n100111100110101011010100111100111111010110",
"output": "001100101000011111111101111011101010111001"
},
{
"input": "1111100001100101000111101001001010011100001\n1000110011000011110010001011001110001000001",
"output": "0111010010100110110101100010000100010100000"
},
{
"input": "01100111011111010101000001101110000001110101\n10011001011111110000000101011001001101101100",
"output": "11111110000000100101000100110111001100011001"
},
{
"input": "110010100111000100100101100000011100000011001\n011001111011100110000110111001110110100111011",
"output": "101011011100100010100011011001101010100100010"
},
{
"input": "0001100111111011010110100100111000000111000110\n1100101011000000000001010010010111001100110001",
"output": "1101001100111011010111110110101111001011110111"
},
{
"input": "00000101110110110001110010100001110100000100000\n10010000110011110001101000111111101010011010001",
"output": "10010101000101000000011010011110011110011110001"
},
{
"input": "110000100101011100100011001111110011111110010001\n101011111001011100110110111101110011010110101100",
"output": "011011011100000000010101110010000000101000111101"
},
{
"input": "0101111101011111010101011101000011101100000000111\n0000101010110110001110101011011110111001010100100",
"output": "0101010111101001011011110110011101010101010100011"
},
{
"input": "11000100010101110011101000011111001010110111111100\n00001111000111001011111110000010101110111001000011",
"output": "11001011010010111000010110011101100100001110111111"
},
{
"input": "101000001101111101101111111000001110110010101101010\n010011100111100001100000010001100101000000111011011",
"output": "111011101010011100001111101001101011110010010110001"
},
{
"input": "0011111110010001010100010110111000110011001101010100\n0111000000100010101010000100101000000100101000111001",
"output": "0100111110110011111110010010010000110111100101101101"
},
{
"input": "11101010000110000011011010000001111101000111011111100\n10110011110001010100010110010010101001010111100100100",
"output": "01011001110111010111001100010011010100010000111011000"
},
{
"input": "011000100001000001101000010110100110011110100111111011\n111011001000001001110011001111011110111110110011011111",
"output": "100011101001001000011011011001111000100000010100100100"
},
{
"input": "0111010110010100000110111011010110100000000111110110000\n1011100100010001101100000100111111101001110010000100110",
"output": "1100110010000101101010111111101001001001110101110010110"
},
{
"input": "10101000100111000111010001011011011011110100110101100011\n11101111000000001100100011111000100100000110011001101110",
"output": "01000111100111001011110010100011111111110010101100001101"
},
{
"input": "000000111001010001000000110001001011100010011101010011011\n110001101000010010000101000100001111101001100100001010010",
"output": "110001010001000011000101110101000100001011111001011001001"
},
{
"input": "0101011100111010000111110010101101111111000000111100011100\n1011111110000010101110111001000011100000100111111111000111",
"output": "1110100010111000101001001011101110011111100111000011011011"
},
{
"input": "11001000001100100111100111100100101011000101001111001001101\n10111110100010000011010100110100100011101001100000001110110",
"output": "01110110101110100100110011010000001000101100101111000111011"
},
{
"input": "010111011011101000000110000110100110001110100001110110111011\n101011110011101011101101011111010100100001100111100100111011",
"output": "111100101000000011101011011001110010101111000110010010000000"
},
{
"input": "1001011110110110000100011001010110000100011010010111010101110\n1101111100001000010111110011010101111010010100000001000010111",
"output": "0100100010111110010011101010000011111110001110010110010111001"
},
{
"input": "10000010101111100111110101111000010100110111101101111111111010\n10110110101100101010011001011010100110111011101100011001100111",
"output": "00110100000011001101101100100010110010001100000001100110011101"
},
{
"input": "011111010011111000001010101001101001000010100010111110010100001\n011111001011000011111001000001111001010110001010111101000010011",
"output": "000000011000111011110011101000010000010100101000000011010110010"
},
{
"input": "1111000000110001011101000100100100001111011100001111001100011111\n1101100110000101100001100000001001011011111011010101000101001010",
"output": "0010100110110100111100100100101101010100100111011010001001010101"
},
{
"input": "01100000101010010011001110100110110010000110010011011001100100011\n10110110010110111100100111000111000110010000000101101110000010111",
"output": "11010110111100101111101001100001110100010110010110110111100110100"
},
{
"input": "001111111010000100001100001010011001111110011110010111110001100111\n110000101001011000100010101100100110000111100000001101001110010111",
"output": "111111010011011100101110100110111111111001111110011010111111110000"
},
{
"input": "1011101011101101011110101101011101011000010011100101010101000100110\n0001000001001111010111100100111101100000000001110001000110000000110",
"output": "1010101010100010001001001001100000111000010010010100010011000100000"
},
{
"input": "01000001011001010011011100010000100100110101111011011011110000001110\n01011110000110011011000000000011000111100001010000000011111001110000",
"output": "00011111011111001000011100010011100011010100101011011000001001111110"
},
{
"input": "110101010100110101000001111110110100010010000100111110010100110011100\n111010010111111011100110101011001011001110110111110100000110110100111",
"output": "001111000011001110100111010101111111011100110011001010010010000111011"
},
{
"input": "1001101011000001011111100110010010000011010001001111011100010100110001\n1111100111110101001111010001010000011001001001010110001111000000100101",
"output": "0110001100110100010000110111000010011010011000011001010011010100010100"
},
{
"input": "00000111110010110001110110001010010101000111011001111111100110011110010\n00010111110100000100110101000010010001100001100011100000001100010100010",
"output": "00010000000110110101000011001000000100100110111010011111101010001010000"
},
{
"input": "100101011100101101000011010001011001101110101110001100010001010111001110\n100001111100101011011111110000001111000111001011111110000010101110111001",
"output": "000100100000000110011100100001010110101001100101110010010011111001110111"
},
{
"input": "1101100001000111001101001011101000111000011110000001001101101001111011010\n0101011101010100011011010110101000010010110010011110101100000110110001000",
"output": "1000111100010011010110011101000000101010101100011111100001101111001010010"
},
{
"input": "01101101010011110101100001110101111011100010000010001101111000011110111111\n00101111001101001100111010000101110000100101101111100111101110010100011011",
"output": "01000010011110111001011011110000001011000111101101101010010110001010100100"
},
{
"input": "101100101100011001101111110110110010100110110010100001110010110011001101011\n000001011010101011110011111101001110000111000010001101000010010000010001101",
"output": "101101110110110010011100001011111100100001110000101100110000100011011100110"
},
{
"input": "0010001011001010001100000010010011110110011000100000000100110000101111001110\n1100110100111000110100001110111001011101001100001010100001010011100110110001",
"output": "1110111111110010111000001100101010101011010100101010100101100011001001111111"
},
{
"input": "00101101010000000101011001101011001100010001100000101011101110000001111001000\n10010110010111000000101101000011101011001010000011011101101011010000000011111",
"output": "10111011000111000101110100101000100111011011100011110110000101010001111010111"
},
{
"input": "111100000100100000101001100001001111001010001000001000000111010000010101101011\n001000100010100101111011111011010110101100001111011000010011011011100010010110",
"output": "110100100110000101010010011010011001100110000111010000010100001011110111111101"
},
{
"input": "0110001101100100001111110101101000100101010010101010011001101001001101110000000\n0111011000000010010111011110010000000001000110001000011001101000000001110100111",
"output": "0001010101100110011000101011111000100100010100100010000000000001001100000100111"
},
{
"input": "10001111111001000101001011110101111010100001011010101100111001010001010010001000\n10000111010010011110111000111010101100000011110001101111001000111010100000000001",
"output": "00001000101011011011110011001111010110100010101011000011110001101011110010001001"
},
{
"input": "100110001110110000100101001110000011110110000110000000100011110100110110011001101\n110001110101110000000100101001101011111100100100001001000110000001111100011110110",
"output": "010111111011000000100001100111101000001010100010001001100101110101001010000111011"
},
{
"input": "0000010100100000010110111100011111111010011101000000100000011001001101101100111010\n0100111110011101010110101011110110010111001111000110101100101110111100101000111111",
"output": "0100101010111101000000010111101001101101010010000110001100110111110001000100000101"
},
{
"input": "11000111001010100001110000001001011010010010110000001110100101000001010101100110111\n11001100100100100001101010110100000111100011101110011010110100001001000011011011010",
"output": "00001011101110000000011010111101011101110001011110010100010001001000010110111101101"
},
{
"input": "010110100010001000100010101001101010011010111110100001000100101000111011100010100001\n110000011111101101010011111000101010111010100001001100001001100101000000111000000000",
"output": "100110111101100101110001010001000000100000011111101101001101001101111011011010100001"
},
{
"input": "0000011110101110010101110110110101100001011001101010101001000010000010000000101001101\n1100111111011100000110000111101110011111100111110001011001000010011111100001001100011",
"output": "1100100001110010010011110001011011111110111110011011110000000000011101100001100101110"
},
{
"input": "10100000101101110001100010010010100101100011010010101000110011100000101010110010000000\n10001110011011010010111011011101101111000111110000111000011010010101001100000001010011",
"output": "00101110110110100011011001001111001010100100100010010000101001110101100110110011010011"
},
{
"input": "001110000011111101101010011111000101010111010100001001100001001100101000000111000000000\n111010000000000000101001110011001000111011001100101010011001000011101001001011110000011",
"output": "110100000011111101000011101100001101101100011000100011111000001111000001001100110000011"
},
{
"input": "1110111100111011010101011011001110001010010010110011110010011111000010011111010101100001\n1001010101011001001010100010101100000110111101011000100010101111111010111100001110010010",
"output": "0111101001100010011111111001100010001100101111101011010000110000111000100011011011110011"
},
{
"input": "11100010001100010011001100001100010011010001101110011110100101110010101101011101000111111\n01110000000110111010110100001010000101011110100101010011000110101110101101110111011110001",
"output": "10010010001010101001111000000110010110001111001011001101100011011100000000101010011001110"
},
{
"input": "001101011001100101101100110000111000101011001001100100000100101000100000110100010111111101\n101001111110000010111101111110001001111001111101111010000110111000100100110010010001011111",
"output": "100100100111100111010001001110110001010010110100011110000010010000000100000110000110100010"
},
{
"input": "1010110110010101000110010010110101011101010100011001101011000110000000100011100100011000000\n0011011111100010001111101101000111001011101110100000110111100100101111010110101111011100011",
"output": "1001101001110111001001111111110010010110111010111001011100100010101111110101001011000100011"
},
{
"input": "10010010000111010111011111110010100101100000001100011100111011100010000010010001011100001100\n00111010100010110010000100010111010001111110100100100011101000101111111111001101101100100100",
"output": "10101000100101100101011011100101110100011110101000111111010011001101111101011100110000101000"
},
{
"input": "010101110001010101100000010111010000000111110011001101100011001000000011001111110000000010100\n010010111011100101010101111110110000000111000100001101101001001000001100101110001010000100001",
"output": "000111001010110000110101101001100000000000110111000000001010000000001111100001111010000110101"
},
{
"input": "1100111110011001000111101001001011000110011010111111100010111111001100111111011101100111101011\n1100000011001000110100110111000001011001010111101000010010100011000001100100111101101000010110",
"output": "0000111101010001110011011110001010011111001101010111110000011100001101011011100000001111111101"
},
{
"input": "00011000100100110111100101100100000000010011110111110010101110110011100001010111010011110100101\n00011011111011111011100101100111100101001110010111000010000111000100100100000001110101111011011",
"output": "00000011011111001100000000000011100101011101100000110000101001110111000101010110100110001111110"
},
{
"input": "000101011001001100000111100010110101111011110101111101000110001101011010111110110011100100000001\n011000101010011111011000111000100000000011011000000001111110001000001111101010110000011100001111",
"output": "011101110011010011011111011010010101111000101101111100111000000101010101010100000011111000001110"
},
{
"input": "1000101001011010000100100100010010011101011001110101111011101111111110010101001101010001010101001\n0110110010011100011111011111110111000000010001110100001010111110101011010011111011111110000110000",
"output": "1110011011000110011011111011100101011101001000000001110001010001010101000110110110101111010011001"
},
{
"input": "01111010010101100111110111111011011010100001011101010000111100101101101110111011001100101011100111\n00001100110110100001111011000010001001001100000010111101000001111011100000010111010010000011000010",
"output": "01110110100011000110001100111001010011101101011111101101111101010110001110101100011110101000100101"
},
{
"input": "111101011101111000001011001101111010110111001101110100100011111011011101001101010101011010111000110\n101111100101101101001100110011000001111010011101110111110110101110011011110011111100001001110101101",
"output": "010010111000010101000111111110111011001101010000000011010101010101000110111110101001010011001101011"
},
{
"input": "1010000011010110011101001101110001110010000111011101110010110110111111001001110100101100010101010001\n0111101000111100101100000101111010100100001000011101010100110011100011010011010101000100101011100011",
"output": "1101101011101010110001001000001011010110001111000000100110000101011100011010100001101000111110110010"
},
{
"input": "0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001\n1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111010",
"output": "1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111011"
},
{
"input": "0\n0",
"output": "0"
},
{
"input": "0\n1",
"output": "1"
}
] | 1,619,197,743
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 102
| 77
| 0
|
number1, number2= input(), input()
p=""
for i in range(len(number1)):
if number1[i]!=number2[i]:
q='1'
p=p+q
else:
q='0'
p=p+q
print(p)
|
Title: Ultra-Fast Mathematician
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate.
Input Specification:
There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.
Output Specification:
Write one line — the corresponding answer. Do not omit the leading 0s.
Demo Input:
['1010100\n0100101\n', '000\n111\n', '1110\n1010\n', '01110\n01100\n']
Demo Output:
['1110001\n', '111\n', '0100\n', '00010\n']
Note:
none
|
```python
number1, number2= input(), input()
p=""
for i in range(len(number1)):
if number1[i]!=number2[i]:
q='1'
p=p+q
else:
q='0'
p=p+q
print(p)
```
| 3.98075
|
252
|
A
|
Little Xor
|
PROGRAMMING
| 1,100
|
[
"brute force",
"implementation"
] | null | null |
Little Petya likes arrays that consist of non-negative integers a lot. Recently his mom has presented him one such array consisting of *n* elements. Petya immediately decided to find there a segment of consecutive elements, such that the *xor* of all numbers from this segment was maximal possible. Help him with that.
The *xor* operation is the bitwise exclusive "OR", that is denoted as "xor" in Pascal and "^" in C/C++/Java.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of elements in the array. The second line contains the space-separated integers from the array. All numbers are non-negative integers strictly less than 230.
|
Print a single integer — the required maximal *xor* of a segment of consecutive elements.
|
[
"5\n1 2 1 1 2\n",
"3\n1 2 7\n",
"4\n4 2 4 8\n"
] |
[
"3\n",
"7\n",
"14\n"
] |
In the first sample one of the optimal segments is the segment that consists of the first and the second array elements, if we consider the array elements indexed starting from one.
The second sample contains only one optimal segment, which contains exactly one array element (element with index three).
| 500
|
[
{
"input": "5\n1 2 1 1 2",
"output": "3"
},
{
"input": "3\n1 2 7",
"output": "7"
},
{
"input": "4\n4 2 4 8",
"output": "14"
},
{
"input": "5\n1 1 1 1 1",
"output": "1"
},
{
"input": "16\n0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15",
"output": "15"
},
{
"input": "20\n1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10",
"output": "15"
},
{
"input": "100\n28 20 67 103 72 81 82 83 7 109 122 30 50 118 83 89 108 82 92 17 97 3 62 12 9 100 14 11 99 106 10 8 60 101 88 119 104 62 76 6 5 57 32 94 60 50 58 97 1 97 107 108 80 24 45 20 112 1 98 106 49 98 25 57 47 90 74 68 14 35 22 10 61 80 10 4 53 13 90 99 57 100 40 84 22 116 60 61 98 57 74 127 61 73 49 51 20 19 56 111",
"output": "127"
},
{
"input": "99\n87 67 4 84 13 20 35 7 11 86 25 1 58 1 74 64 74 86 98 74 72 46 63 78 84 13 60 38 30 45 45 60 9 44 36 70 33 22 82 15 71 7 43 47 23 2 20 49 42 43 54 27 51 51 53 23 27 37 17 66 90 89 61 0 18 20 49 30 84 20 13 32 64 69 56 68 59 10 91 96 43 64 19 10 2 57 62 23 100 39 32 19 95 55 77 19 24 4 77",
"output": "127"
},
{
"input": "1\n100",
"output": "100"
},
{
"input": "2\n1 1",
"output": "1"
},
{
"input": "2\n4 10",
"output": "14"
},
{
"input": "99\n3511 2076 9314 3598 7737 271 4110 4454 5830 8661 8584 8227 2236 2626 1025 3806 3162 2837 6071 9627 1836 7729 6629 4054 9377 3261 1325 8859 3610 7616 5052 9736 1249 6224 7031 6074 4196 8075 4005 5419 5664 8214 4391 731 8714 8622 6122 6976 4057 1770 8025 3778 2792 994 5204 1826 6083 5438 8409 8029 1962 3133 9836 8884 1234 1776 824 1630 6523 2869 9950 8609 5397 1472 7120 5012 6490 5958 8030 7741 5077 2771 7925 9784 9456 8596 6288 8939 4779 4415 3743 5188 5574 9686 1483 4245 4769 4917 1887",
"output": "16383"
},
{
"input": "27\n78 918 443 3900 591 12 4 10 1 24 70 88 429 2 3257 65 275 2 258 62 587 625 25 26 853 728 765",
"output": "4027"
},
{
"input": "2\n3 1",
"output": "3"
},
{
"input": "2\n3 2",
"output": "3"
},
{
"input": "3\n7 3 1",
"output": "7"
},
{
"input": "3\n1 7 2",
"output": "7"
},
{
"input": "2\n7 1",
"output": "7"
},
{
"input": "4\n1 2 7 1",
"output": "7"
},
{
"input": "3\n10 5 1",
"output": "15"
}
] | 1,599,328,438
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 10
| 218
| 0
|
n=int(input())
a=list(map(int,input().split()))
max=a[0]
for i in range (n):
k=a[i]
if k>max:
max=k
for j in range (i,n):
k=k^a[j]
if k>max:
max=k
print(max)
|
Title: Little Xor
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Petya likes arrays that consist of non-negative integers a lot. Recently his mom has presented him one such array consisting of *n* elements. Petya immediately decided to find there a segment of consecutive elements, such that the *xor* of all numbers from this segment was maximal possible. Help him with that.
The *xor* operation is the bitwise exclusive "OR", that is denoted as "xor" in Pascal and "^" in C/C++/Java.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of elements in the array. The second line contains the space-separated integers from the array. All numbers are non-negative integers strictly less than 230.
Output Specification:
Print a single integer — the required maximal *xor* of a segment of consecutive elements.
Demo Input:
['5\n1 2 1 1 2\n', '3\n1 2 7\n', '4\n4 2 4 8\n']
Demo Output:
['3\n', '7\n', '14\n']
Note:
In the first sample one of the optimal segments is the segment that consists of the first and the second array elements, if we consider the array elements indexed starting from one.
The second sample contains only one optimal segment, which contains exactly one array element (element with index three).
|
```python
n=int(input())
a=list(map(int,input().split()))
max=a[0]
for i in range (n):
k=a[i]
if k>max:
max=k
for j in range (i,n):
k=k^a[j]
if k>max:
max=k
print(max)
```
| 0
|
|
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 — 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*<=≤<=*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<=≤<=*n*<=≤<=2000; *a*,<=*b*<=≥<=1; *a*<=+<=*b*<==<=*n*) — 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<=≤<=*h**i*<=≤<=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"
}
] | 1,680,642,402
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 31
| 0
|
n, a, b = map(int, input().split())
complexities = list(map(int, input().split()))
count = 0
for x in range(1, max(complexities)+1):
A_tasks = [task for task in complexities if task >= x]
B_tasks = [task for task in complexities if task <= x]
if len(A_tasks) == a and len(B_tasks) == b:
count += 1
# print(x)
print(count)
|
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 — 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*<=≤<=*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<=≤<=*n*<=≤<=2000; *a*,<=*b*<=≥<=1; *a*<=+<=*b*<==<=*n*) — 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<=≤<=*h**i*<=≤<=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())
complexities = list(map(int, input().split()))
count = 0
for x in range(1, max(complexities)+1):
A_tasks = [task for task in complexities if task >= x]
B_tasks = [task for task in complexities if task <= x]
if len(A_tasks) == a and len(B_tasks) == b:
count += 1
# print(x)
print(count)
```
| 0
|
|
858
|
A
|
k-rounding
|
PROGRAMMING
| 1,100
|
[
"brute force",
"math",
"number theory"
] | null | null |
For a given positive integer *n* denote its *k*-rounding as the minimum positive integer *x*, such that *x* ends with *k* or more zeros in base 10 and is divisible by *n*.
For example, 4-rounding of 375 is 375·80<==<=30000. 30000 is the minimum integer such that it ends with 4 or more zeros and is divisible by 375.
Write a program that will perform the *k*-rounding of *n*.
|
The only line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=109, 0<=≤<=*k*<=≤<=8).
|
Print the *k*-rounding of *n*.
|
[
"375 4\n",
"10000 1\n",
"38101 0\n",
"123456789 8\n"
] |
[
"30000\n",
"10000\n",
"38101\n",
"12345678900000000\n"
] |
none
| 750
|
[
{
"input": "375 4",
"output": "30000"
},
{
"input": "10000 1",
"output": "10000"
},
{
"input": "38101 0",
"output": "38101"
},
{
"input": "123456789 8",
"output": "12345678900000000"
},
{
"input": "1 0",
"output": "1"
},
{
"input": "2 0",
"output": "2"
},
{
"input": "100 0",
"output": "100"
},
{
"input": "1000000000 0",
"output": "1000000000"
},
{
"input": "160 2",
"output": "800"
},
{
"input": "3 0",
"output": "3"
},
{
"input": "10 0",
"output": "10"
},
{
"input": "1 1",
"output": "10"
},
{
"input": "2 1",
"output": "10"
},
{
"input": "3 1",
"output": "30"
},
{
"input": "4 1",
"output": "20"
},
{
"input": "5 1",
"output": "10"
},
{
"input": "6 1",
"output": "30"
},
{
"input": "7 1",
"output": "70"
},
{
"input": "8 1",
"output": "40"
},
{
"input": "9 1",
"output": "90"
},
{
"input": "10 1",
"output": "10"
},
{
"input": "11 1",
"output": "110"
},
{
"input": "12 1",
"output": "60"
},
{
"input": "16 2",
"output": "400"
},
{
"input": "2 2",
"output": "100"
},
{
"input": "1 2",
"output": "100"
},
{
"input": "5 2",
"output": "100"
},
{
"input": "15 2",
"output": "300"
},
{
"input": "36 2",
"output": "900"
},
{
"input": "1 8",
"output": "100000000"
},
{
"input": "8 8",
"output": "100000000"
},
{
"input": "96 8",
"output": "300000000"
},
{
"input": "175 8",
"output": "700000000"
},
{
"input": "9999995 8",
"output": "199999900000000"
},
{
"input": "999999999 8",
"output": "99999999900000000"
},
{
"input": "12345678 8",
"output": "617283900000000"
},
{
"input": "78125 8",
"output": "100000000"
},
{
"input": "390625 8",
"output": "100000000"
},
{
"input": "1953125 8",
"output": "500000000"
},
{
"input": "9765625 8",
"output": "2500000000"
},
{
"input": "68359375 8",
"output": "17500000000"
},
{
"input": "268435456 8",
"output": "104857600000000"
},
{
"input": "125829120 8",
"output": "9830400000000"
},
{
"input": "128000 8",
"output": "400000000"
},
{
"input": "300000 8",
"output": "300000000"
},
{
"input": "3711871 8",
"output": "371187100000000"
},
{
"input": "55555 8",
"output": "1111100000000"
},
{
"input": "222222222 8",
"output": "11111111100000000"
},
{
"input": "479001600 8",
"output": "7484400000000"
},
{
"input": "655360001 7",
"output": "6553600010000000"
},
{
"input": "655360001 8",
"output": "65536000100000000"
},
{
"input": "1000000000 1",
"output": "1000000000"
},
{
"input": "1000000000 7",
"output": "1000000000"
},
{
"input": "1000000000 8",
"output": "1000000000"
},
{
"input": "100000000 8",
"output": "100000000"
},
{
"input": "10000000 8",
"output": "100000000"
},
{
"input": "1000000 8",
"output": "100000000"
},
{
"input": "10000009 8",
"output": "1000000900000000"
},
{
"input": "10000005 8",
"output": "200000100000000"
},
{
"input": "10000002 8",
"output": "500000100000000"
},
{
"input": "999999997 8",
"output": "99999999700000000"
},
{
"input": "999999997 7",
"output": "9999999970000000"
},
{
"input": "999999995 8",
"output": "19999999900000000"
},
{
"input": "123 8",
"output": "12300000000"
},
{
"input": "24 2",
"output": "600"
},
{
"input": "16 4",
"output": "10000"
},
{
"input": "123456787 8",
"output": "12345678700000000"
},
{
"input": "100000000 8",
"output": "100000000"
},
{
"input": "7 1",
"output": "70"
},
{
"input": "101 1",
"output": "1010"
},
{
"input": "50 2",
"output": "100"
},
{
"input": "999999818 1",
"output": "4999999090"
},
{
"input": "2 1",
"output": "10"
},
{
"input": "123 1",
"output": "1230"
},
{
"input": "16 1",
"output": "80"
},
{
"input": "1 1",
"output": "10"
},
{
"input": "1000000000 8",
"output": "1000000000"
},
{
"input": "15304 6",
"output": "1913000000"
},
{
"input": "3 8",
"output": "300000000"
},
{
"input": "4 2",
"output": "100"
},
{
"input": "100000 7",
"output": "10000000"
},
{
"input": "5 8",
"output": "100000000"
},
{
"input": "16724 6",
"output": "4181000000"
},
{
"input": "999999999 1",
"output": "9999999990"
},
{
"input": "999999990 8",
"output": "9999999900000000"
},
{
"input": "999999999 8",
"output": "99999999900000000"
},
{
"input": "100000 1",
"output": "100000"
},
{
"input": "8 3",
"output": "1000"
},
{
"input": "16768 6",
"output": "262000000"
},
{
"input": "123456789 1",
"output": "1234567890"
},
{
"input": "2 8",
"output": "100000000"
},
{
"input": "999999937 8",
"output": "99999993700000000"
},
{
"input": "5 1",
"output": "10"
},
{
"input": "2000000 7",
"output": "10000000"
},
{
"input": "1999998 2",
"output": "99999900"
},
{
"input": "125 3",
"output": "1000"
}
] | 1,505,656,462
| 2,961
|
Python 3
|
RUNTIME_ERROR
|
PRETESTS
| 0
| 46
| 0
|
def lcm(a,b):
m = a*b
while a != 0 and b != 0:
if a > b:
a %= b
else:
b %= a
return m // (a+b)
x = int(input())
y = int(input())
a= 1
for i in range(0,y):
a *= 10
print(str(lcm(x,a)))
|
Title: k-rounding
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
For a given positive integer *n* denote its *k*-rounding as the minimum positive integer *x*, such that *x* ends with *k* or more zeros in base 10 and is divisible by *n*.
For example, 4-rounding of 375 is 375·80<==<=30000. 30000 is the minimum integer such that it ends with 4 or more zeros and is divisible by 375.
Write a program that will perform the *k*-rounding of *n*.
Input Specification:
The only line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=109, 0<=≤<=*k*<=≤<=8).
Output Specification:
Print the *k*-rounding of *n*.
Demo Input:
['375 4\n', '10000 1\n', '38101 0\n', '123456789 8\n']
Demo Output:
['30000\n', '10000\n', '38101\n', '12345678900000000\n']
Note:
none
|
```python
def lcm(a,b):
m = a*b
while a != 0 and b != 0:
if a > b:
a %= b
else:
b %= a
return m // (a+b)
x = int(input())
y = int(input())
a= 1
for i in range(0,y):
a *= 10
print(str(lcm(x,a)))
```
| -1
|
|
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<=≤<=*n*<=≤<=1000, 0<=≤<=*x*<=≤<=109).
Each of the next *n* lines contains a character '+' or '-', and an integer *d**i*, separated by a space (1<=≤<=*d**i*<=≤<=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 integers — 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"
}
] | 1,644,645,094
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 34
| 78
| 614,400
|
import sys
input = sys.stdin.readline
N, x = map(int, input().split())
COMMANDS = [input().split() for _ in range(N)]
ds = 0
for command in COMMANDS :
action, amount = command
amount = int(amount)
if action == "+" : x += amount
else :
if x < amount : ds += 1
else : x -= amount
print(x, ds)
|
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<=≤<=*n*<=≤<=1000, 0<=≤<=*x*<=≤<=109).
Each of the next *n* lines contains a character '+' or '-', and an integer *d**i*, separated by a space (1<=≤<=*d**i*<=≤<=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 integers — 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
import sys
input = sys.stdin.readline
N, x = map(int, input().split())
COMMANDS = [input().split() for _ in range(N)]
ds = 0
for command in COMMANDS :
action, amount = command
amount = int(amount)
if action == "+" : x += amount
else :
if x < amount : ds += 1
else : x -= amount
print(x, ds)
```
| 3
|
|
161
|
A
|
Dress'em in Vests!
|
PROGRAMMING
| 1,300
|
[
"binary search",
"brute force",
"greedy",
"two pointers"
] | null | null |
The Two-dimensional kingdom is going through hard times... This morning the Three-Dimensional kingdom declared war on the Two-dimensional one. This (possibly armed) conflict will determine the ultimate owner of the straight line.
The Two-dimensional kingdom has a regular army of *n* people. Each soldier registered himself and indicated the desired size of the bulletproof vest: the *i*-th soldier indicated size *a**i*. The soldiers are known to be unpretentious, so the command staff assumes that the soldiers are comfortable in any vests with sizes from *a**i*<=-<=*x* to *a**i*<=+<=*y*, inclusive (numbers *x*,<=*y*<=≥<=0 are specified).
The Two-dimensional kingdom has *m* vests at its disposal, the *j*-th vest's size equals *b**j*. Help mobilize the Two-dimensional kingdom's army: equip with vests as many soldiers as possible. Each vest can be used only once. The *i*-th soldier can put on the *j*-th vest, if *a**i*<=-<=*x*<=≤<=*b**j*<=≤<=*a**i*<=+<=*y*.
|
The first input line contains four integers *n*, *m*, *x* and *y* (1<=≤<=*n*,<=*m*<=≤<=105, 0<=≤<=*x*,<=*y*<=≤<=109) — the number of soldiers, the number of vests and two numbers that specify the soldiers' unpretentiousness, correspondingly.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) in non-decreasing order, separated by single spaces — the desired sizes of vests.
The third line contains *m* integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**j*<=≤<=109) in non-decreasing order, separated by single spaces — the sizes of the available vests.
|
In the first line print a single integer *k* — the maximum number of soldiers equipped with bulletproof vests.
In the next *k* lines print *k* pairs, one pair per line, as "*u**i* *v**i*" (without the quotes). Pair (*u**i*, *v**i*) means that soldier number *u**i* must wear vest number *v**i*. Soldiers and vests are numbered starting from one in the order in which they are specified in the input. All numbers of soldiers in the pairs should be pairwise different, all numbers of vests in the pairs also should be pairwise different. You can print the pairs in any order.
If there are multiple optimal answers, you are allowed to print any of them.
|
[
"5 3 0 0\n1 2 3 3 4\n1 3 5\n",
"3 3 2 2\n1 5 9\n3 5 7\n"
] |
[
"2\n1 1\n3 2\n",
"3\n1 1\n2 2\n3 3\n"
] |
In the first sample you need the vests' sizes to match perfectly: the first soldier gets the first vest (size 1), the third soldier gets the second vest (size 3). This sample allows another answer, which gives the second vest to the fourth soldier instead of the third one.
In the second sample the vest size can differ from the desired size by at most 2 sizes, so all soldiers can be equipped.
| 1,000
|
[
{
"input": "5 3 0 0\n1 2 3 3 4\n1 3 5",
"output": "2\n1 1\n3 2"
},
{
"input": "3 3 2 2\n1 5 9\n3 5 7",
"output": "3\n1 1\n2 2\n3 3"
},
{
"input": "1 1 0 0\n1\n1",
"output": "1\n1 1"
},
{
"input": "1 1 0 0\n1\n2",
"output": "0"
},
{
"input": "2 3 1 4\n1 5\n1 2 2",
"output": "1\n1 1"
},
{
"input": "20 30 1 4\n1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4 4 4 4 5\n1 1 1 1 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 4 4 4 4 4 4 4 4 5 5",
"output": "20\n1 1\n2 2\n3 3\n4 4\n5 5\n6 6\n7 7\n8 8\n9 9\n10 10\n11 11\n12 12\n13 15\n14 16\n15 17\n16 18\n17 19\n18 20\n19 21\n20 22"
},
{
"input": "33 23 17 2\n1 1 2 2 2 3 3 3 3 3 3 4 4 4 4 4 5 5 5 6 6 7 7 7 8 8 8 8 8 9 9 10 10\n1 1 3 3 4 4 4 5 5 6 6 6 7 8 8 8 8 8 8 9 9 10 10",
"output": "23\n1 1\n2 2\n3 3\n4 4\n5 5\n6 6\n7 7\n8 8\n9 9\n12 10\n13 11\n14 12\n17 13\n20 14\n21 15\n22 16\n23 17\n24 18\n25 19\n26 20\n27 21\n28 22\n29 23"
},
{
"input": "2 2 1 4\n1 4\n3 6",
"output": "2\n1 1\n2 2"
},
{
"input": "20 20 1 4\n1 1 1 1 1 2 2 2 2 2 2 2 2 3 3 3 4 4 5 5\n3 3 3 3 3 4 4 4 4 4 4 4 4 5 5 5 6 6 7 7",
"output": "20\n1 1\n2 2\n3 3\n4 4\n5 5\n6 6\n7 7\n8 8\n9 9\n10 10\n11 11\n12 12\n13 13\n14 14\n15 15\n16 16\n17 17\n18 18\n19 19\n20 20"
},
{
"input": "33 23 17 2\n1 1 1 2 3 3 3 3 3 4 4 4 4 5 6 6 6 6 6 6 7 7 7 7 7 8 8 8 8 8 8 10 10\n10 10 10 11 12 12 12 12 12 13 13 13 13 14 15 15 15 15 15 15 16 16 16",
"output": "5\n26 1\n27 2\n28 3\n32 4\n33 5"
},
{
"input": "1 1 1 2\n783266931\n783266932",
"output": "1\n1 1"
},
{
"input": "2 3 1 4\n1 1\n3 3 4",
"output": "2\n1 1\n2 2"
},
{
"input": "20 30 1 4\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2\n3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4",
"output": "20\n1 1\n2 2\n3 3\n4 4\n5 5\n6 6\n7 7\n8 8\n9 9\n10 10\n11 11\n12 12\n13 13\n14 14\n15 15\n16 16\n17 17\n18 18\n19 19\n20 20"
},
{
"input": "33 23 17 2\n1 1 1 1 1 1 1 2 2 2 2 2 2 2 3 3 3 3 3 3 4 4 4 4 4 4 4 5 5 5 5 5 5\n10 10 10 10 10 10 10 11 11 11 11 11 11 11 12 12 12 12 12 12 13 13 13",
"output": "0"
},
{
"input": "10 1 1 1\n1 1 1 5 5 7 7 8 8 8\n3",
"output": "0"
},
{
"input": "10 1 2 1\n1 2 5 8 9 9 9 10 10 10\n7",
"output": "1\n4 1"
},
{
"input": "1 1 2 1\n7\n8",
"output": "1\n1 1"
},
{
"input": "1 1 0 2\n9\n7",
"output": "0"
},
{
"input": "2 2 0 2\n2 8\n3 5",
"output": "1\n1 1"
},
{
"input": "2 2 1 0\n5 5\n6 7",
"output": "0"
},
{
"input": "2 2 3 0\n8 9\n1 5",
"output": "1\n1 2"
}
] | 1,696,736,578
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 84
| 842
| 15,564,800
|
def tolerance_test(s, v, x, y):
if(v >= s-x and v <= s+y):
return True
return False
n, m, x, y = tuple([int(x) for x in input().split()])
soldiers = [int(x) for x in input().split()]
vests = [int(x) for x in input().split()]
count = 0
s_p = 0
v_p = 0
indices_store = []
while(s_p < len(soldiers) and v_p < len(vests)):
if(tolerance_test(soldiers[s_p], vests[v_p], x, y)):
count += 1
s_p += 1
v_p += 1
indices_store.append((s_p, v_p))
elif(soldiers[s_p] < vests[v_p]):
s_p += 1
else:
v_p += 1
print(count)
for t in indices_store:
print("%d %d"%(t[0], t[1]))
|
Title: Dress'em in Vests!
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Two-dimensional kingdom is going through hard times... This morning the Three-Dimensional kingdom declared war on the Two-dimensional one. This (possibly armed) conflict will determine the ultimate owner of the straight line.
The Two-dimensional kingdom has a regular army of *n* people. Each soldier registered himself and indicated the desired size of the bulletproof vest: the *i*-th soldier indicated size *a**i*. The soldiers are known to be unpretentious, so the command staff assumes that the soldiers are comfortable in any vests with sizes from *a**i*<=-<=*x* to *a**i*<=+<=*y*, inclusive (numbers *x*,<=*y*<=≥<=0 are specified).
The Two-dimensional kingdom has *m* vests at its disposal, the *j*-th vest's size equals *b**j*. Help mobilize the Two-dimensional kingdom's army: equip with vests as many soldiers as possible. Each vest can be used only once. The *i*-th soldier can put on the *j*-th vest, if *a**i*<=-<=*x*<=≤<=*b**j*<=≤<=*a**i*<=+<=*y*.
Input Specification:
The first input line contains four integers *n*, *m*, *x* and *y* (1<=≤<=*n*,<=*m*<=≤<=105, 0<=≤<=*x*,<=*y*<=≤<=109) — the number of soldiers, the number of vests and two numbers that specify the soldiers' unpretentiousness, correspondingly.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) in non-decreasing order, separated by single spaces — the desired sizes of vests.
The third line contains *m* integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**j*<=≤<=109) in non-decreasing order, separated by single spaces — the sizes of the available vests.
Output Specification:
In the first line print a single integer *k* — the maximum number of soldiers equipped with bulletproof vests.
In the next *k* lines print *k* pairs, one pair per line, as "*u**i* *v**i*" (without the quotes). Pair (*u**i*, *v**i*) means that soldier number *u**i* must wear vest number *v**i*. Soldiers and vests are numbered starting from one in the order in which they are specified in the input. All numbers of soldiers in the pairs should be pairwise different, all numbers of vests in the pairs also should be pairwise different. You can print the pairs in any order.
If there are multiple optimal answers, you are allowed to print any of them.
Demo Input:
['5 3 0 0\n1 2 3 3 4\n1 3 5\n', '3 3 2 2\n1 5 9\n3 5 7\n']
Demo Output:
['2\n1 1\n3 2\n', '3\n1 1\n2 2\n3 3\n']
Note:
In the first sample you need the vests' sizes to match perfectly: the first soldier gets the first vest (size 1), the third soldier gets the second vest (size 3). This sample allows another answer, which gives the second vest to the fourth soldier instead of the third one.
In the second sample the vest size can differ from the desired size by at most 2 sizes, so all soldiers can be equipped.
|
```python
def tolerance_test(s, v, x, y):
if(v >= s-x and v <= s+y):
return True
return False
n, m, x, y = tuple([int(x) for x in input().split()])
soldiers = [int(x) for x in input().split()]
vests = [int(x) for x in input().split()]
count = 0
s_p = 0
v_p = 0
indices_store = []
while(s_p < len(soldiers) and v_p < len(vests)):
if(tolerance_test(soldiers[s_p], vests[v_p], x, y)):
count += 1
s_p += 1
v_p += 1
indices_store.append((s_p, v_p))
elif(soldiers[s_p] < vests[v_p]):
s_p += 1
else:
v_p += 1
print(count)
for t in indices_store:
print("%d %d"%(t[0], t[1]))
```
| 3
|
|
122
|
A
|
Lucky Division
|
PROGRAMMING
| 1,000
|
[
"brute force",
"number theory"
] | null | null |
Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya calls a number almost lucky if it could be evenly divided by some lucky number. Help him find out if the given number *n* is almost lucky.
|
The single line contains an integer *n* (1<=≤<=*n*<=≤<=1000) — the number that needs to be checked.
|
In the only line print "YES" (without the quotes), if number *n* is almost lucky. Otherwise, print "NO" (without the quotes).
|
[
"47\n",
"16\n",
"78\n"
] |
[
"YES\n",
"YES\n",
"NO\n"
] |
Note that all lucky numbers are almost lucky as any number is evenly divisible by itself.
In the first sample 47 is a lucky number. In the second sample 16 is divisible by 4.
| 500
|
[
{
"input": "47",
"output": "YES"
},
{
"input": "16",
"output": "YES"
},
{
"input": "78",
"output": "NO"
},
{
"input": "48",
"output": "YES"
},
{
"input": "100",
"output": "YES"
},
{
"input": "107",
"output": "NO"
},
{
"input": "77",
"output": "YES"
},
{
"input": "477",
"output": "YES"
},
{
"input": "480",
"output": "YES"
},
{
"input": "1",
"output": "NO"
},
{
"input": "3",
"output": "NO"
},
{
"input": "4",
"output": "YES"
},
{
"input": "49",
"output": "YES"
},
{
"input": "56",
"output": "YES"
},
{
"input": "124",
"output": "YES"
},
{
"input": "1000",
"output": "YES"
},
{
"input": "999",
"output": "NO"
},
{
"input": "298",
"output": "NO"
},
{
"input": "274",
"output": "NO"
},
{
"input": "998",
"output": "NO"
},
{
"input": "42",
"output": "YES"
},
{
"input": "788",
"output": "YES"
},
{
"input": "70",
"output": "YES"
},
{
"input": "444",
"output": "YES"
},
{
"input": "777",
"output": "YES"
},
{
"input": "799",
"output": "YES"
},
{
"input": "25",
"output": "NO"
},
{
"input": "882",
"output": "YES"
},
{
"input": "88",
"output": "YES"
},
{
"input": "11",
"output": "NO"
},
{
"input": "2",
"output": "NO"
},
{
"input": "7",
"output": "YES"
},
{
"input": "8",
"output": "YES"
},
{
"input": "94",
"output": "YES"
},
{
"input": "477",
"output": "YES"
},
{
"input": "141",
"output": "YES"
}
] | 1,696,468,593
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 36
| 92
| 0
|
import sys
n=int(input())
b={4,7,44,77,47,74,444,777,744,474,447,477,747,774}
for i in b:
if n%i==0:
print("YES")
sys.exit()
print("NO")
|
Title: Lucky Division
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya calls a number almost lucky if it could be evenly divided by some lucky number. Help him find out if the given number *n* is almost lucky.
Input Specification:
The single line contains an integer *n* (1<=≤<=*n*<=≤<=1000) — the number that needs to be checked.
Output Specification:
In the only line print "YES" (without the quotes), if number *n* is almost lucky. Otherwise, print "NO" (without the quotes).
Demo Input:
['47\n', '16\n', '78\n']
Demo Output:
['YES\n', 'YES\n', 'NO\n']
Note:
Note that all lucky numbers are almost lucky as any number is evenly divisible by itself.
In the first sample 47 is a lucky number. In the second sample 16 is divisible by 4.
|
```python
import sys
n=int(input())
b={4,7,44,77,47,74,444,777,744,474,447,477,747,774}
for i in b:
if n%i==0:
print("YES")
sys.exit()
print("NO")
```
| 3
|
|
965
|
C
|
Greedy Arkady
|
PROGRAMMING
| 2,000
|
[
"math"
] | null | null |
$k$ people want to split $n$ candies between them. Each candy should be given to exactly one of them or be thrown away.
The people are numbered from $1$ to $k$, and Arkady is the first of them. To split the candies, Arkady will choose an integer $x$ and then give the first $x$ candies to himself, the next $x$ candies to the second person, the next $x$ candies to the third person and so on in a cycle. The leftover (the remainder that is not divisible by $x$) will be thrown away.
Arkady can't choose $x$ greater than $M$ as it is considered greedy. Also, he can't choose such a small $x$ that some person will receive candies more than $D$ times, as it is considered a slow splitting.
Please find what is the maximum number of candies Arkady can receive by choosing some valid $x$.
|
The only line contains four integers $n$, $k$, $M$ and $D$ ($2 \le n \le 10^{18}$, $2 \le k \le n$, $1 \le M \le n$, $1 \le D \le \min{(n, 1000)}$, $M \cdot D \cdot k \ge n$) — the number of candies, the number of people, the maximum number of candies given to a person at once, the maximum number of times a person can receive candies.
|
Print a single integer — the maximum possible number of candies Arkady can give to himself.
Note that it is always possible to choose some valid $x$.
|
[
"20 4 5 2\n",
"30 9 4 1\n"
] |
[
"8\n",
"4\n"
] |
In the first example Arkady should choose $x = 4$. He will give $4$ candies to himself, $4$ candies to the second person, $4$ candies to the third person, then $4$ candies to the fourth person and then again $4$ candies to himself. No person is given candies more than $2$ times, and Arkady receives $8$ candies in total.
Note that if Arkady chooses $x = 5$, he will receive only $5$ candies, and if he chooses $x = 3$, he will receive only $3 + 3 = 6$ candies as well as the second person, the third and the fourth persons will receive $3$ candies, and $2$ candies will be thrown away. He can't choose $x = 1$ nor $x = 2$ because in these cases he will receive candies more than $2$ times.
In the second example Arkady has to choose $x = 4$, because any smaller value leads to him receiving candies more than $1$ time.
| 1,500
|
[
{
"input": "20 4 5 2",
"output": "8"
},
{
"input": "30 9 4 1",
"output": "4"
},
{
"input": "2 2 1 1",
"output": "1"
},
{
"input": "42 20 5 29",
"output": "5"
},
{
"input": "1000000000000000000 135 1000000000000000 1000",
"output": "8325624421831635"
},
{
"input": "100 33 100 100",
"output": "100"
},
{
"input": "1000000000 1000000000 1000000000 1000",
"output": "1000000000"
},
{
"input": "1000000000 32428 1000000000 1000",
"output": "1000000000"
},
{
"input": "1000000000 324934 1000 1000",
"output": "4000"
},
{
"input": "1000000000000000000 32400093004 10000000 1000",
"output": "40000000"
},
{
"input": "885 2 160 842",
"output": "504"
},
{
"input": "216 137 202 208",
"output": "202"
},
{
"input": "72 66 28 9",
"output": "28"
},
{
"input": "294 4 13 8",
"output": "80"
},
{
"input": "9 2 2 3",
"output": "4"
},
{
"input": "31 3 2 8",
"output": "10"
},
{
"input": "104 2 5 11",
"output": "50"
},
{
"input": "1000000000000000000 1000000000000000000 1000 1000",
"output": "1000"
},
{
"input": "1000000000000000000 100000000000000000 1 1000",
"output": "10"
},
{
"input": "23925738098196565 23925738098196565 23925738098196565 1000",
"output": "23925738098196565"
},
{
"input": "576460752303423488 576460752303423488 351082447248993993 1000",
"output": "351082447248993993"
},
{
"input": "962768465676381898 72057594037927936 586039918340257175 256",
"output": "586039918340257175"
},
{
"input": "1000000000000000000 1000000000000000000 10 1000",
"output": "10"
},
{
"input": "23925738098196565 23925738098196565 1 1000",
"output": "1"
},
{
"input": "1000000000000000000 1000000000000000000 1 1000",
"output": "1"
}
] | 1,530,523,712
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 77
| 0
|
import sys
from math import ceil, floor
args = [int(x) for x in input().split(' ')]
n = args[0]
k = args[1]
M = args[2]
D = args[3]
m = 0
argmin = 0
for x in range(ceil(n/(k*D)),M+1):
bonbons = x*ceil(n/(k*x))
#print('x = ',x, ', bonbons = ',bonbons)
if bonbons > m :
argmin = x
m = bonbons
print(argmin)
|
Title: Greedy Arkady
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
$k$ people want to split $n$ candies between them. Each candy should be given to exactly one of them or be thrown away.
The people are numbered from $1$ to $k$, and Arkady is the first of them. To split the candies, Arkady will choose an integer $x$ and then give the first $x$ candies to himself, the next $x$ candies to the second person, the next $x$ candies to the third person and so on in a cycle. The leftover (the remainder that is not divisible by $x$) will be thrown away.
Arkady can't choose $x$ greater than $M$ as it is considered greedy. Also, he can't choose such a small $x$ that some person will receive candies more than $D$ times, as it is considered a slow splitting.
Please find what is the maximum number of candies Arkady can receive by choosing some valid $x$.
Input Specification:
The only line contains four integers $n$, $k$, $M$ and $D$ ($2 \le n \le 10^{18}$, $2 \le k \le n$, $1 \le M \le n$, $1 \le D \le \min{(n, 1000)}$, $M \cdot D \cdot k \ge n$) — the number of candies, the number of people, the maximum number of candies given to a person at once, the maximum number of times a person can receive candies.
Output Specification:
Print a single integer — the maximum possible number of candies Arkady can give to himself.
Note that it is always possible to choose some valid $x$.
Demo Input:
['20 4 5 2\n', '30 9 4 1\n']
Demo Output:
['8\n', '4\n']
Note:
In the first example Arkady should choose $x = 4$. He will give $4$ candies to himself, $4$ candies to the second person, $4$ candies to the third person, then $4$ candies to the fourth person and then again $4$ candies to himself. No person is given candies more than $2$ times, and Arkady receives $8$ candies in total.
Note that if Arkady chooses $x = 5$, he will receive only $5$ candies, and if he chooses $x = 3$, he will receive only $3 + 3 = 6$ candies as well as the second person, the third and the fourth persons will receive $3$ candies, and $2$ candies will be thrown away. He can't choose $x = 1$ nor $x = 2$ because in these cases he will receive candies more than $2$ times.
In the second example Arkady has to choose $x = 4$, because any smaller value leads to him receiving candies more than $1$ time.
|
```python
import sys
from math import ceil, floor
args = [int(x) for x in input().split(' ')]
n = args[0]
k = args[1]
M = args[2]
D = args[3]
m = 0
argmin = 0
for x in range(ceil(n/(k*D)),M+1):
bonbons = x*ceil(n/(k*x))
#print('x = ',x, ', bonbons = ',bonbons)
if bonbons > m :
argmin = x
m = bonbons
print(argmin)
```
| 0
|
|
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 — 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* — 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"
}
] | 1,590,224,262
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 30
| 218
| 0
|
s=input()
c=''
flag1,flag2=0,0
for i in s:
if(i.isupper()):
flag1+=1
else:
flag2+=1
if(flag1==flag2):
for i in s:
c+=i.lower()
elif(flag1>flag2):
for i in s:
c+=i.upper()
else:
for i in s:
c+=i.lower()
print(c)
|
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 — 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* — 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()
c=''
flag1,flag2=0,0
for i in s:
if(i.isupper()):
flag1+=1
else:
flag2+=1
if(flag1==flag2):
for i in s:
c+=i.lower()
elif(flag1>flag2):
for i in s:
c+=i.upper()
else:
for i in s:
c+=i.lower()
print(c)
```
| 3.9455
|
984
|
A
|
Game
|
PROGRAMMING
| 800
|
[
"sortings"
] | null | null |
Two players play a game.
Initially there are $n$ integers $a_1, a_2, \ldots, a_n$ written on the board. Each turn a player selects one number and erases it from the board. This continues until there is only one number left on the board, i. e. $n - 1$ turns are made. The first player makes the first move, then players alternate turns.
The first player wants to minimize the last number that would be left on the board, while the second player wants to maximize it.
You want to know what number will be left on the board after $n - 1$ turns if both players make optimal moves.
|
The first line contains one integer $n$ ($1 \le n \le 1000$) — the number of numbers on the board.
The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^6$).
|
Print one number that will be left on the board.
|
[
"3\n2 1 3\n",
"3\n2 2 2\n"
] |
[
"2",
"2"
] |
In the first sample, the first player erases $3$ and the second erases $1$. $2$ is left on the board.
In the second sample, $2$ is left on the board regardless of the actions of the players.
| 500
|
[
{
"input": "3\n2 1 3",
"output": "2"
},
{
"input": "3\n2 2 2",
"output": "2"
},
{
"input": "9\n44 53 51 80 5 27 74 79 94",
"output": "53"
},
{
"input": "10\n38 82 23 37 96 4 81 60 67 86",
"output": "60"
},
{
"input": "10\n58 26 77 15 53 81 68 48 22 65",
"output": "53"
},
{
"input": "1\n124",
"output": "124"
},
{
"input": "2\n2 1",
"output": "1"
},
{
"input": "3\n1 1 1000",
"output": "1"
},
{
"input": "2\n322 322",
"output": "322"
},
{
"input": "3\n9 92 12",
"output": "12"
},
{
"input": "3\n1 2 2",
"output": "2"
}
] | 1,573,969,182
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 35
| 156
| 0
|
n = int(input())
lst = sorted(list(map(int, input().split())))
mid = (n-1)//2
print(lst[mid])
|
Title: Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Two players play a game.
Initially there are $n$ integers $a_1, a_2, \ldots, a_n$ written on the board. Each turn a player selects one number and erases it from the board. This continues until there is only one number left on the board, i. e. $n - 1$ turns are made. The first player makes the first move, then players alternate turns.
The first player wants to minimize the last number that would be left on the board, while the second player wants to maximize it.
You want to know what number will be left on the board after $n - 1$ turns if both players make optimal moves.
Input Specification:
The first line contains one integer $n$ ($1 \le n \le 1000$) — the number of numbers on the board.
The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^6$).
Output Specification:
Print one number that will be left on the board.
Demo Input:
['3\n2 1 3\n', '3\n2 2 2\n']
Demo Output:
['2', '2']
Note:
In the first sample, the first player erases $3$ and the second erases $1$. $2$ is left on the board.
In the second sample, $2$ is left on the board regardless of the actions of the players.
|
```python
n = int(input())
lst = sorted(list(map(int, input().split())))
mid = (n-1)//2
print(lst[mid])
```
| 3
|
|
416
|
B
|
Art Union
|
PROGRAMMING
| 1,300
|
[
"brute force",
"dp",
"implementation"
] | null | null |
A well-known art union called "Kalevich is Alive!" manufactures objects d'art (pictures). The union consists of *n* painters who decided to organize their work as follows.
Each painter uses only the color that was assigned to him. The colors are distinct for all painters. Let's assume that the first painter uses color 1, the second one uses color 2, and so on. Each picture will contain all these *n* colors. Adding the *j*-th color to the *i*-th picture takes the *j*-th painter *t**ij* units of time.
Order is important everywhere, so the painters' work is ordered by the following rules:
- Each picture is first painted by the first painter, then by the second one, and so on. That is, after the *j*-th painter finishes working on the picture, it must go to the (*j*<=+<=1)-th painter (if *j*<=<<=*n*); - each painter works on the pictures in some order: first, he paints the first picture, then he paints the second picture and so on; - each painter can simultaneously work on at most one picture. However, the painters don't need any time to have a rest; - as soon as the *j*-th painter finishes his part of working on the picture, the picture immediately becomes available to the next painter.
Given that the painters start working at time 0, find for each picture the time when it is ready for sale.
|
The first line of the input contains integers *m*,<=*n* (1<=≤<=*m*<=≤<=50000,<=1<=≤<=*n*<=≤<=5), where *m* is the number of pictures and *n* is the number of painters. Then follow the descriptions of the pictures, one per line. Each line contains *n* integers *t**i*1,<=*t**i*2,<=...,<=*t**in* (1<=≤<=*t**ij*<=≤<=1000), where *t**ij* is the time the *j*-th painter needs to work on the *i*-th picture.
|
Print the sequence of *m* integers *r*1,<=*r*2,<=...,<=*r**m*, where *r**i* is the moment when the *n*-th painter stopped working on the *i*-th picture.
|
[
"5 1\n1\n2\n3\n4\n5\n",
"4 2\n2 5\n3 1\n5 3\n10 1\n"
] |
[
"1 3 6 10 15 ",
"7 8 13 21 "
] |
none
| 1,000
|
[
{
"input": "5 1\n1\n2\n3\n4\n5",
"output": "1 3 6 10 15 "
},
{
"input": "4 2\n2 5\n3 1\n5 3\n10 1",
"output": "7 8 13 21 "
},
{
"input": "1 1\n66",
"output": "66 "
},
{
"input": "2 2\n1 1\n1 1",
"output": "2 3 "
},
{
"input": "2 2\n10 1\n10 1",
"output": "11 21 "
},
{
"input": "1 5\n1 95 44 14 35",
"output": "189 "
},
{
"input": "7 1\n80\n92\n24\n88\n40\n45\n7",
"output": "80 172 196 284 324 369 376 "
},
{
"input": "1 2\n51 44",
"output": "95 "
},
{
"input": "2 1\n19\n4",
"output": "19 23 "
},
{
"input": "2 2\n1 10\n1 1",
"output": "11 12 "
},
{
"input": "3 3\n3 9 4\n5 10 8\n4 4 7",
"output": "16 30 37 "
},
{
"input": "10 3\n6 10 3\n2 7 9\n10 4 7\n6 3 4\n6 2 6\n8 4 4\n5 9 8\n6 9 7\n2 7 10\n2 6 2",
"output": "19 32 39 43 49 53 61 68 78 80 "
}
] | 1,685,669,418
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 1
| 30
| 0
|
m, n = [int(x) for x in input().split()]
t = [[0]*n for _ in range(m)]
for i in range(m):
t[i] = [int(x) for x in input().split()]
dp = [[0]*n for _ in range(m)]
for i in range(m):
for j in range(n):
if i > 0:
dp[i][j] = dp[i-1][j]
if j > 0:
dp[i][j] = dp[i][j-1]
dp[i][j] += t[i][j]
res = []
for i in range(m):
res.append(dp[i][n-1])
print(*res)
|
Title: Art Union
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A well-known art union called "Kalevich is Alive!" manufactures objects d'art (pictures). The union consists of *n* painters who decided to organize their work as follows.
Each painter uses only the color that was assigned to him. The colors are distinct for all painters. Let's assume that the first painter uses color 1, the second one uses color 2, and so on. Each picture will contain all these *n* colors. Adding the *j*-th color to the *i*-th picture takes the *j*-th painter *t**ij* units of time.
Order is important everywhere, so the painters' work is ordered by the following rules:
- Each picture is first painted by the first painter, then by the second one, and so on. That is, after the *j*-th painter finishes working on the picture, it must go to the (*j*<=+<=1)-th painter (if *j*<=<<=*n*); - each painter works on the pictures in some order: first, he paints the first picture, then he paints the second picture and so on; - each painter can simultaneously work on at most one picture. However, the painters don't need any time to have a rest; - as soon as the *j*-th painter finishes his part of working on the picture, the picture immediately becomes available to the next painter.
Given that the painters start working at time 0, find for each picture the time when it is ready for sale.
Input Specification:
The first line of the input contains integers *m*,<=*n* (1<=≤<=*m*<=≤<=50000,<=1<=≤<=*n*<=≤<=5), where *m* is the number of pictures and *n* is the number of painters. Then follow the descriptions of the pictures, one per line. Each line contains *n* integers *t**i*1,<=*t**i*2,<=...,<=*t**in* (1<=≤<=*t**ij*<=≤<=1000), where *t**ij* is the time the *j*-th painter needs to work on the *i*-th picture.
Output Specification:
Print the sequence of *m* integers *r*1,<=*r*2,<=...,<=*r**m*, where *r**i* is the moment when the *n*-th painter stopped working on the *i*-th picture.
Demo Input:
['5 1\n1\n2\n3\n4\n5\n', '4 2\n2 5\n3 1\n5 3\n10 1\n']
Demo Output:
['1 3 6 10 15 ', '7 8 13 21 ']
Note:
none
|
```python
m, n = [int(x) for x in input().split()]
t = [[0]*n for _ in range(m)]
for i in range(m):
t[i] = [int(x) for x in input().split()]
dp = [[0]*n for _ in range(m)]
for i in range(m):
for j in range(n):
if i > 0:
dp[i][j] = dp[i-1][j]
if j > 0:
dp[i][j] = dp[i][j-1]
dp[i][j] += t[i][j]
res = []
for i in range(m):
res.append(dp[i][n-1])
print(*res)
```
| 0
|
|
298
|
A
|
Snow Footprints
|
PROGRAMMING
| 1,300
|
[
"greedy",
"implementation"
] | null | null |
There is a straight snowy road, divided into *n* blocks. The blocks are numbered from 1 to *n* from left to right. If one moves from the *i*-th block to the (*i*<=+<=1)-th block, he will leave a right footprint on the *i*-th block. Similarly, if one moves from the *i*-th block to the (*i*<=-<=1)-th block, he will leave a left footprint on the *i*-th block. If there already is a footprint on the *i*-th block, the new footprint will cover the old one.
At the beginning, there were no footprints. Then polar bear Alice starts from the *s*-th block, makes a sequence of moves and ends in the *t*-th block. It is known that Alice never moves outside of the road.
You are given the description of Alice's footprints. Your task is to find a pair of possible values of *s*,<=*t* by looking at the footprints.
|
The first line of the input contains integer *n* (3<=≤<=*n*<=≤<=1000).
The second line contains the description of the road — the string that consists of *n* characters. Each character will be either "." (a block without footprint), or "L" (a block with a left footprint), "R" (a block with a right footprint).
It's guaranteed that the given string contains at least one character not equal to ".". Also, the first and the last character will always be ".". It's guaranteed that a solution exists.
|
Print two space-separated integers — the values of *s* and *t*. If there are several possible solutions you can print any of them.
|
[
"9\n..RRLL...\n",
"11\n.RRRLLLLL..\n"
] |
[
"3 4\n",
"7 5\n"
] |
The first test sample is the one in the picture.
| 500
|
[
{
"input": "11\n.RRRLLLLL..",
"output": "7 5"
},
{
"input": "4\n.RL.",
"output": "3 2"
},
{
"input": "3\n.L.",
"output": "2 1"
},
{
"input": "3\n.R.",
"output": "2 3"
}
] | 1,678,128,968
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 62
| 0
|
n = int(input())
l = list(input())
s = 0
t = 0
r = True
for i in range(0,n-1):
if l[i]!="." and r:
s = i+1
r = False
elif l[i]==".":
t = i
if l[s] == "R" and l[t] == "R":
print(s,t)
r = True
elif l[s] == "L" and l[t] == "L":
print(t,s)
r = True
elif not r:
if l[s]=="R":
for i in range(n):
if l[i]=="L":
t = i
break
elif l[s]=="L":
for i in range(n):
if l[i]=="R":
t = i
print(s,t)
|
Title: Snow Footprints
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There is a straight snowy road, divided into *n* blocks. The blocks are numbered from 1 to *n* from left to right. If one moves from the *i*-th block to the (*i*<=+<=1)-th block, he will leave a right footprint on the *i*-th block. Similarly, if one moves from the *i*-th block to the (*i*<=-<=1)-th block, he will leave a left footprint on the *i*-th block. If there already is a footprint on the *i*-th block, the new footprint will cover the old one.
At the beginning, there were no footprints. Then polar bear Alice starts from the *s*-th block, makes a sequence of moves and ends in the *t*-th block. It is known that Alice never moves outside of the road.
You are given the description of Alice's footprints. Your task is to find a pair of possible values of *s*,<=*t* by looking at the footprints.
Input Specification:
The first line of the input contains integer *n* (3<=≤<=*n*<=≤<=1000).
The second line contains the description of the road — the string that consists of *n* characters. Each character will be either "." (a block without footprint), or "L" (a block with a left footprint), "R" (a block with a right footprint).
It's guaranteed that the given string contains at least one character not equal to ".". Also, the first and the last character will always be ".". It's guaranteed that a solution exists.
Output Specification:
Print two space-separated integers — the values of *s* and *t*. If there are several possible solutions you can print any of them.
Demo Input:
['9\n..RRLL...\n', '11\n.RRRLLLLL..\n']
Demo Output:
['3 4\n', '7 5\n']
Note:
The first test sample is the one in the picture.
|
```python
n = int(input())
l = list(input())
s = 0
t = 0
r = True
for i in range(0,n-1):
if l[i]!="." and r:
s = i+1
r = False
elif l[i]==".":
t = i
if l[s] == "R" and l[t] == "R":
print(s,t)
r = True
elif l[s] == "L" and l[t] == "L":
print(t,s)
r = True
elif not r:
if l[s]=="R":
for i in range(n):
if l[i]=="L":
t = i
break
elif l[s]=="L":
for i in range(n):
if l[i]=="R":
t = i
print(s,t)
```
| 0
|
|
735
|
A
|
Ostap and Grasshopper
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] | null | null |
On the way to Rio de Janeiro Ostap kills time playing with a grasshopper he took with him in a special box. Ostap builds a line of length *n* such that some cells of this line are empty and some contain obstacles. Then, he places his grasshopper to one of the empty cells and a small insect in another empty cell. The grasshopper wants to eat the insect.
Ostap knows that grasshopper is able to jump to any empty cell that is exactly *k* cells away from the current (to the left or to the right). Note that it doesn't matter whether intermediate cells are empty or not as the grasshopper makes a jump over them. For example, if *k*<==<=1 the grasshopper can jump to a neighboring cell only, and if *k*<==<=2 the grasshopper can jump over a single cell.
Your goal is to determine whether there is a sequence of jumps such that grasshopper will get from his initial position to the cell with an insect.
|
The first line of the input contains two integers *n* and *k* (2<=≤<=*n*<=≤<=100, 1<=≤<=*k*<=≤<=*n*<=-<=1) — the number of cells in the line and the length of one grasshopper's jump.
The second line contains a string of length *n* consisting of characters '.', '#', 'G' and 'T'. Character '.' means that the corresponding cell is empty, character '#' means that the corresponding cell contains an obstacle and grasshopper can't jump there. Character 'G' means that the grasshopper starts at this position and, finally, 'T' means that the target insect is located at this cell. It's guaranteed that characters 'G' and 'T' appear in this line exactly once.
|
If there exists a sequence of jumps (each jump of length *k*), such that the grasshopper can get from his initial position to the cell with the insect, print "YES" (without quotes) in the only line of the input. Otherwise, print "NO" (without quotes).
|
[
"5 2\n#G#T#\n",
"6 1\nT....G\n",
"7 3\nT..#..G\n",
"6 2\n..GT..\n"
] |
[
"YES\n",
"YES\n",
"NO\n",
"NO\n"
] |
In the first sample, the grasshopper can make one jump to the right in order to get from cell 2 to cell 4.
In the second sample, the grasshopper is only able to jump to neighboring cells but the way to the insect is free — he can get there by jumping left 5 times.
In the third sample, the grasshopper can't make a single jump.
In the fourth sample, the grasshopper can only jump to the cells with odd indices, thus he won't be able to reach the insect.
| 500
|
[
{
"input": "5 2\n#G#T#",
"output": "YES"
},
{
"input": "6 1\nT....G",
"output": "YES"
},
{
"input": "7 3\nT..#..G",
"output": "NO"
},
{
"input": "6 2\n..GT..",
"output": "NO"
},
{
"input": "2 1\nGT",
"output": "YES"
},
{
"input": "100 5\nG####.####.####.####.####.####.####.####.####.####.####.####.####.####.####.####.####.####.####T####",
"output": "YES"
},
{
"input": "100 5\nG####.####.####.####.####.####.####.####.####.####.####.####.####.#########.####.####.####.####T####",
"output": "NO"
},
{
"input": "2 1\nTG",
"output": "YES"
},
{
"input": "99 1\n...T.............................................................................................G.",
"output": "YES"
},
{
"input": "100 2\nG............#.....#...........#....#...........##............#............#......................T.",
"output": "NO"
},
{
"input": "100 1\n#.#.#.##..#..##.#....##.##.##.#....####..##.#.##..GT..##...###.#.##.#..#..##.###..#.####..#.#.##..##",
"output": "YES"
},
{
"input": "100 2\n..#####.#.#.......#.#.#...##..####..###..#.#######GT####.#.#...##...##.#..###....##.#.#..#.###....#.",
"output": "NO"
},
{
"input": "100 3\nG..................................................................................................T",
"output": "YES"
},
{
"input": "100 3\nG..................................................................................................T",
"output": "YES"
},
{
"input": "100 3\nG..................................#......#......#.......#.#..........#........#......#..........#.T",
"output": "NO"
},
{
"input": "100 3\nG..............#..........#...#..............#.#.....................#......#........#.........#...T",
"output": "NO"
},
{
"input": "100 3\nG##################################################################################################T",
"output": "NO"
},
{
"input": "100 33\nG..................................................................................................T",
"output": "YES"
},
{
"input": "100 33\nG..................................................................................................T",
"output": "YES"
},
{
"input": "100 33\nG.........#........#..........#..............#.................#............................#.#....T",
"output": "YES"
},
{
"input": "100 33\nG.......#..................#..............................#............................#..........T.",
"output": "NO"
},
{
"input": "100 33\nG#..........##...#.#.....................#.#.#.........##..#...........#....#...........##...#..###T",
"output": "YES"
},
{
"input": "100 33\nG..#.#..#..####......#......##...##...#.##........#...#...#.##....###..#...###..##.#.....#......#.T.",
"output": "NO"
},
{
"input": "100 33\nG#....#..#..##.##..#.##.#......#.#.##..##.#.#.##.##....#.#.....####..##...#....##..##..........#...T",
"output": "NO"
},
{
"input": "100 33\nG#######.#..##.##.#...#..#.###.#.##.##.#..#.###..####.##.#.##....####...##..####.#..##.##.##.#....#T",
"output": "NO"
},
{
"input": "100 33\nG#####.#.##.###########.##..##..#######..########..###.###..#.####.######.############..####..#####T",
"output": "NO"
},
{
"input": "100 99\nT..................................................................................................G",
"output": "YES"
},
{
"input": "100 99\nT..................................................................................................G",
"output": "YES"
},
{
"input": "100 99\nT.#...............................#............#..............................##...................G",
"output": "YES"
},
{
"input": "100 99\nT..#....#.##...##########.#.#.#.#...####..#.....#..##..#######.######..#.....###..###...#.......#.#G",
"output": "YES"
},
{
"input": "100 99\nG##################################################################################################T",
"output": "YES"
},
{
"input": "100 9\nT..................................................................................................G",
"output": "YES"
},
{
"input": "100 9\nT.................................................................................................G.",
"output": "NO"
},
{
"input": "100 9\nT................................................................................................G..",
"output": "NO"
},
{
"input": "100 1\nG..................................................................................................T",
"output": "YES"
},
{
"input": "100 1\nT..................................................................................................G",
"output": "YES"
},
{
"input": "100 1\n##########G.........T###############################################################################",
"output": "YES"
},
{
"input": "100 1\n#################################################################################################G.T",
"output": "YES"
},
{
"input": "100 17\n##########G################.################.################.################T#####################",
"output": "YES"
},
{
"input": "100 17\n####.#..#.G######.#########.##..##########.#.################.################T######.####.#########",
"output": "YES"
},
{
"input": "100 17\n.########.G##.####.#.######.###############..#.###########.##.#####.##.#####.#T.###..###.########.##",
"output": "YES"
},
{
"input": "100 1\nG.............................................#....................................................T",
"output": "NO"
},
{
"input": "100 1\nT.#................................................................................................G",
"output": "NO"
},
{
"input": "100 1\n##########G....#....T###############################################################################",
"output": "NO"
},
{
"input": "100 1\n#################################################################################################G#T",
"output": "NO"
},
{
"input": "100 17\nG################.#################################.################T###############################",
"output": "NO"
},
{
"input": "100 17\nG################.###############..###.######.#######.###.#######.##T######################.###.####",
"output": "NO"
},
{
"input": "100 17\nG####.##.##.#####.####....##.####.#########.##.#..#.###############.T############.#########.#.####.#",
"output": "NO"
},
{
"input": "48 1\nT..............................................G",
"output": "YES"
},
{
"input": "23 1\nT.....................G",
"output": "YES"
},
{
"input": "49 1\nG...............................................T",
"output": "YES"
},
{
"input": "3 1\nTG#",
"output": "YES"
},
{
"input": "6 2\n..TG..",
"output": "NO"
},
{
"input": "14 3\n...G.....#..T.",
"output": "NO"
},
{
"input": "5 4\n##GT#",
"output": "NO"
},
{
"input": "6 2\nT#..G.",
"output": "YES"
},
{
"input": "5 2\nT.G.#",
"output": "YES"
},
{
"input": "6 1\nT...G#",
"output": "YES"
},
{
"input": "5 1\nTG###",
"output": "YES"
},
{
"input": "5 4\n.G..T",
"output": "NO"
},
{
"input": "7 2\nT#...#G",
"output": "YES"
},
{
"input": "7 1\n##TG###",
"output": "YES"
},
{
"input": "7 1\n###GT##",
"output": "YES"
},
{
"input": "5 2\nG..T.",
"output": "NO"
},
{
"input": "5 1\nG.T##",
"output": "YES"
},
{
"input": "6 2\nG.T###",
"output": "YES"
},
{
"input": "6 2\nG#T###",
"output": "YES"
},
{
"input": "10 2\n####T..G..",
"output": "NO"
},
{
"input": "3 1\nGT#",
"output": "YES"
},
{
"input": "4 1\nTG##",
"output": "YES"
},
{
"input": "6 1\n.G..T.",
"output": "YES"
},
{
"input": "10 3\n......G..T",
"output": "YES"
},
{
"input": "3 2\nG.T",
"output": "YES"
},
{
"input": "4 1\n#G.T",
"output": "YES"
},
{
"input": "5 2\nT#G##",
"output": "YES"
},
{
"input": "4 2\nG#.T",
"output": "NO"
},
{
"input": "4 1\nGT##",
"output": "YES"
}
] | 1,675,764,239
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 83
| 46
| 0
|
# LUOGU_RID: 101647844
n, k = map(int, input().split())
s = input()
x = s.find('G')
y = s.find('T')
print(((y - x) % k == 0 and s[x:y:k].find('#') == -1 and s[x:y:-k].find('#') == -1) and 'YES' or 'NO')
|
Title: Ostap and Grasshopper
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
On the way to Rio de Janeiro Ostap kills time playing with a grasshopper he took with him in a special box. Ostap builds a line of length *n* such that some cells of this line are empty and some contain obstacles. Then, he places his grasshopper to one of the empty cells and a small insect in another empty cell. The grasshopper wants to eat the insect.
Ostap knows that grasshopper is able to jump to any empty cell that is exactly *k* cells away from the current (to the left or to the right). Note that it doesn't matter whether intermediate cells are empty or not as the grasshopper makes a jump over them. For example, if *k*<==<=1 the grasshopper can jump to a neighboring cell only, and if *k*<==<=2 the grasshopper can jump over a single cell.
Your goal is to determine whether there is a sequence of jumps such that grasshopper will get from his initial position to the cell with an insect.
Input Specification:
The first line of the input contains two integers *n* and *k* (2<=≤<=*n*<=≤<=100, 1<=≤<=*k*<=≤<=*n*<=-<=1) — the number of cells in the line and the length of one grasshopper's jump.
The second line contains a string of length *n* consisting of characters '.', '#', 'G' and 'T'. Character '.' means that the corresponding cell is empty, character '#' means that the corresponding cell contains an obstacle and grasshopper can't jump there. Character 'G' means that the grasshopper starts at this position and, finally, 'T' means that the target insect is located at this cell. It's guaranteed that characters 'G' and 'T' appear in this line exactly once.
Output Specification:
If there exists a sequence of jumps (each jump of length *k*), such that the grasshopper can get from his initial position to the cell with the insect, print "YES" (without quotes) in the only line of the input. Otherwise, print "NO" (without quotes).
Demo Input:
['5 2\n#G#T#\n', '6 1\nT....G\n', '7 3\nT..#..G\n', '6 2\n..GT..\n']
Demo Output:
['YES\n', 'YES\n', 'NO\n', 'NO\n']
Note:
In the first sample, the grasshopper can make one jump to the right in order to get from cell 2 to cell 4.
In the second sample, the grasshopper is only able to jump to neighboring cells but the way to the insect is free — he can get there by jumping left 5 times.
In the third sample, the grasshopper can't make a single jump.
In the fourth sample, the grasshopper can only jump to the cells with odd indices, thus he won't be able to reach the insect.
|
```python
# LUOGU_RID: 101647844
n, k = map(int, input().split())
s = input()
x = s.find('G')
y = s.find('T')
print(((y - x) % k == 0 and s[x:y:k].find('#') == -1 and s[x:y:-k].find('#') == -1) and 'YES' or 'NO')
```
| 3
|
|
591
|
A
|
Wizards' Duel
|
PROGRAMMING
| 900
|
[
"implementation",
"math"
] | null | null |
Harry Potter and He-Who-Must-Not-Be-Named engaged in a fight to the death once again. This time they are located at opposite ends of the corridor of length *l*. Two opponents simultaneously charge a deadly spell in the enemy. We know that the impulse of Harry's magic spell flies at a speed of *p* meters per second, and the impulse of You-Know-Who's magic spell flies at a speed of *q* meters per second.
The impulses are moving through the corridor toward each other, and at the time of the collision they turn round and fly back to those who cast them without changing their original speeds. Then, as soon as the impulse gets back to it's caster, the wizard reflects it and sends again towards the enemy, without changing the original speed of the impulse.
Since Harry has perfectly mastered the basics of magic, he knows that after the second collision both impulses will disappear, and a powerful explosion will occur exactly in the place of their collision. However, the young wizard isn't good at math, so he asks you to calculate the distance from his position to the place of the second meeting of the spell impulses, provided that the opponents do not change positions during the whole fight.
|
The first line of the input contains a single integer *l* (1<=≤<=*l*<=≤<=1<=000) — the length of the corridor where the fight takes place.
The second line contains integer *p*, the third line contains integer *q* (1<=≤<=*p*,<=*q*<=≤<=500) — the speeds of magical impulses for Harry Potter and He-Who-Must-Not-Be-Named, respectively.
|
Print a single real number — the distance from the end of the corridor, where Harry is located, to the place of the second meeting of the spell impulses. Your answer will be considered correct if its absolute or relative error will not exceed 10<=-<=4.
Namely: let's assume that your answer equals *a*, and the answer of the jury is *b*. The checker program will consider your answer correct if .
|
[
"100\n50\n50\n",
"199\n60\n40\n"
] |
[
"50\n",
"119.4\n"
] |
In the first sample the speeds of the impulses are equal, so both of their meetings occur exactly in the middle of the corridor.
| 500
|
[
{
"input": "100\n50\n50",
"output": "50"
},
{
"input": "199\n60\n40",
"output": "119.4"
},
{
"input": "1\n1\n1",
"output": "0.5"
},
{
"input": "1\n1\n500",
"output": "0.001996007984"
},
{
"input": "1\n500\n1",
"output": "0.998003992"
},
{
"input": "1\n500\n500",
"output": "0.5"
},
{
"input": "1000\n1\n1",
"output": "500"
},
{
"input": "1000\n1\n500",
"output": "1.996007984"
},
{
"input": "1000\n500\n1",
"output": "998.003992"
},
{
"input": "1000\n500\n500",
"output": "500"
},
{
"input": "101\n11\n22",
"output": "33.66666667"
},
{
"input": "987\n1\n3",
"output": "246.75"
},
{
"input": "258\n25\n431",
"output": "14.14473684"
},
{
"input": "979\n39\n60",
"output": "385.6666667"
},
{
"input": "538\n479\n416",
"output": "287.9351955"
},
{
"input": "583\n112\n248",
"output": "181.3777778"
},
{
"input": "978\n467\n371",
"output": "545.0190931"
},
{
"input": "980\n322\n193",
"output": "612.7378641"
},
{
"input": "871\n401\n17",
"output": "835.576555"
},
{
"input": "349\n478\n378",
"output": "194.885514"
},
{
"input": "425\n458\n118",
"output": "337.9340278"
},
{
"input": "919\n323\n458",
"output": "380.0729834"
},
{
"input": "188\n59\n126",
"output": "59.95675676"
},
{
"input": "644\n428\n484",
"output": "302.2280702"
},
{
"input": "253\n80\n276",
"output": "56.85393258"
},
{
"input": "745\n152\n417",
"output": "199.0158172"
},
{
"input": "600\n221\n279",
"output": "265.2"
},
{
"input": "690\n499\n430",
"output": "370.6243272"
},
{
"input": "105\n68\n403",
"output": "15.15923567"
},
{
"input": "762\n462\n371",
"output": "422.6218487"
},
{
"input": "903\n460\n362",
"output": "505.3284672"
},
{
"input": "886\n235\n95",
"output": "630.9393939"
},
{
"input": "655\n203\n18",
"output": "601.6515837"
},
{
"input": "718\n29\n375",
"output": "51.53960396"
},
{
"input": "296\n467\n377",
"output": "163.7819905"
},
{
"input": "539\n61\n56",
"output": "281.017094"
},
{
"input": "133\n53\n124",
"output": "39.82485876"
},
{
"input": "998\n224\n65",
"output": "773.5363322"
},
{
"input": "961\n173\n47",
"output": "755.6954545"
},
{
"input": "285\n468\n62",
"output": "251.6603774"
},
{
"input": "496\n326\n429",
"output": "214.1668874"
},
{
"input": "627\n150\n285",
"output": "216.2068966"
},
{
"input": "961\n443\n50",
"output": "863.535497"
},
{
"input": "623\n422\n217",
"output": "411.4334898"
},
{
"input": "678\n295\n29",
"output": "617.3148148"
}
] | 1,455,808,287
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 45
| 109
| 0
|
m=input()
l=int(m)
n=input()
p=int(n)
o=input()
q=int(o)
a=l/(p+q)
print(a*p)
print("\n")
|
Title: Wizards' Duel
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Harry Potter and He-Who-Must-Not-Be-Named engaged in a fight to the death once again. This time they are located at opposite ends of the corridor of length *l*. Two opponents simultaneously charge a deadly spell in the enemy. We know that the impulse of Harry's magic spell flies at a speed of *p* meters per second, and the impulse of You-Know-Who's magic spell flies at a speed of *q* meters per second.
The impulses are moving through the corridor toward each other, and at the time of the collision they turn round and fly back to those who cast them without changing their original speeds. Then, as soon as the impulse gets back to it's caster, the wizard reflects it and sends again towards the enemy, without changing the original speed of the impulse.
Since Harry has perfectly mastered the basics of magic, he knows that after the second collision both impulses will disappear, and a powerful explosion will occur exactly in the place of their collision. However, the young wizard isn't good at math, so he asks you to calculate the distance from his position to the place of the second meeting of the spell impulses, provided that the opponents do not change positions during the whole fight.
Input Specification:
The first line of the input contains a single integer *l* (1<=≤<=*l*<=≤<=1<=000) — the length of the corridor where the fight takes place.
The second line contains integer *p*, the third line contains integer *q* (1<=≤<=*p*,<=*q*<=≤<=500) — the speeds of magical impulses for Harry Potter and He-Who-Must-Not-Be-Named, respectively.
Output Specification:
Print a single real number — the distance from the end of the corridor, where Harry is located, to the place of the second meeting of the spell impulses. Your answer will be considered correct if its absolute or relative error will not exceed 10<=-<=4.
Namely: let's assume that your answer equals *a*, and the answer of the jury is *b*. The checker program will consider your answer correct if .
Demo Input:
['100\n50\n50\n', '199\n60\n40\n']
Demo Output:
['50\n', '119.4\n']
Note:
In the first sample the speeds of the impulses are equal, so both of their meetings occur exactly in the middle of the corridor.
|
```python
m=input()
l=int(m)
n=input()
p=int(n)
o=input()
q=int(o)
a=l/(p+q)
print(a*p)
print("\n")
```
| 3
|
|
322
|
B
|
Ciel and Flowers
|
PROGRAMMING
| 1,600
|
[
"combinatorics",
"math"
] | null | null |
Fox Ciel has some flowers: *r* red flowers, *g* green flowers and *b* blue flowers. She wants to use these flowers to make several bouquets. There are 4 types of bouquets:
- To make a "red bouquet", it needs 3 red flowers. - To make a "green bouquet", it needs 3 green flowers. - To make a "blue bouquet", it needs 3 blue flowers. - To make a "mixing bouquet", it needs 1 red, 1 green and 1 blue flower.
Help Fox Ciel to find the maximal number of bouquets she can make.
|
The first line contains three integers *r*, *g* and *b* (0<=≤<=*r*,<=*g*,<=*b*<=≤<=109) — the number of red, green and blue flowers.
|
Print the maximal number of bouquets Fox Ciel can make.
|
[
"3 6 9\n",
"4 4 4\n",
"0 0 0\n"
] |
[
"6\n",
"4\n",
"0\n"
] |
In test case 1, we can make 1 red bouquet, 2 green bouquets and 3 blue bouquets.
In test case 2, we can make 1 red, 1 green, 1 blue and 1 mixing bouquet.
| 1,000
|
[
{
"input": "3 6 9",
"output": "6"
},
{
"input": "4 4 4",
"output": "4"
},
{
"input": "0 0 0",
"output": "0"
},
{
"input": "0 3 6",
"output": "3"
},
{
"input": "7 8 9",
"output": "7"
},
{
"input": "8 8 9",
"output": "8"
},
{
"input": "15 3 999",
"output": "339"
},
{
"input": "32 62 92",
"output": "62"
},
{
"input": "123456789 123456789 123456789",
"output": "123456789"
},
{
"input": "3 5 5",
"output": "4"
},
{
"input": "666806767 385540591 357848286",
"output": "470065214"
},
{
"input": "80010646 727118126 817880463",
"output": "541669744"
},
{
"input": "829651016 732259171 572879931",
"output": "711596705"
},
{
"input": "242854896 442432924 180395753",
"output": "288561190"
},
{
"input": "139978911 5123031 935395222",
"output": "360165721"
},
{
"input": "553182792 10264076 395427398",
"output": "319624755"
},
{
"input": "597790453 720437830 855459575",
"output": "724562619"
},
{
"input": "494914467 356982656 757942689",
"output": "536613270"
},
{
"input": "908118348 67156409 217974865",
"output": "397749873"
},
{
"input": "952726009 629846517 972974334",
"output": "851848953"
},
{
"input": "775140200 616574841 630329230",
"output": "674014756"
},
{
"input": "524780569 326748594 90361407",
"output": "313963523"
},
{
"input": "937984449 184405994 992844522",
"output": "705078321"
},
{
"input": "835108464 525983528 452876698",
"output": "604656229"
},
{
"input": "879716125 531124573 207876166",
"output": "539572288"
},
{
"input": "292920005 241298326 667908343",
"output": "400708891"
},
{
"input": "1000000000 1000000000 1000000000",
"output": "1000000000"
},
{
"input": "1000000000 999999999 999999998",
"output": "999999998"
},
{
"input": "999999998 999999998 999999999",
"output": "999999998"
},
{
"input": "0 1 1",
"output": "0"
},
{
"input": "0 1000000000 0",
"output": "333333333"
},
{
"input": "0 1 0",
"output": "0"
},
{
"input": "1 0 0",
"output": "0"
},
{
"input": "0 2 2",
"output": "0"
},
{
"input": "3 3 5",
"output": "3"
},
{
"input": "2 2 0",
"output": "0"
},
{
"input": "0 5 5",
"output": "2"
},
{
"input": "2 0 11",
"output": "3"
},
{
"input": "9 9 7",
"output": "8"
},
{
"input": "65 30 74",
"output": "56"
}
] | 1,373,733,688
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 33
| 78
| 102,400
|
s = input()
r,g,b = map(int,s.split())
print(max([i+(r-i)//3+(g-i)//3+(b-i)//3 for i in range(3)]))
|
Title: Ciel and Flowers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Fox Ciel has some flowers: *r* red flowers, *g* green flowers and *b* blue flowers. She wants to use these flowers to make several bouquets. There are 4 types of bouquets:
- To make a "red bouquet", it needs 3 red flowers. - To make a "green bouquet", it needs 3 green flowers. - To make a "blue bouquet", it needs 3 blue flowers. - To make a "mixing bouquet", it needs 1 red, 1 green and 1 blue flower.
Help Fox Ciel to find the maximal number of bouquets she can make.
Input Specification:
The first line contains three integers *r*, *g* and *b* (0<=≤<=*r*,<=*g*,<=*b*<=≤<=109) — the number of red, green and blue flowers.
Output Specification:
Print the maximal number of bouquets Fox Ciel can make.
Demo Input:
['3 6 9\n', '4 4 4\n', '0 0 0\n']
Demo Output:
['6\n', '4\n', '0\n']
Note:
In test case 1, we can make 1 red bouquet, 2 green bouquets and 3 blue bouquets.
In test case 2, we can make 1 red, 1 green, 1 blue and 1 mixing bouquet.
|
```python
s = input()
r,g,b = map(int,s.split())
print(max([i+(r-i)//3+(g-i)//3+(b-i)//3 for i in range(3)]))
```
| 0
|
|
50
|
A
|
Domino piling
|
PROGRAMMING
| 800
|
[
"greedy",
"math"
] |
A. Domino piling
|
2
|
256
|
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions.
|
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
|
Output one number — the maximal number of dominoes, which can be placed.
|
[
"2 4\n",
"3 3\n"
] |
[
"4\n",
"4\n"
] |
none
| 500
|
[
{
"input": "2 4",
"output": "4"
},
{
"input": "3 3",
"output": "4"
},
{
"input": "1 5",
"output": "2"
},
{
"input": "1 6",
"output": "3"
},
{
"input": "1 15",
"output": "7"
},
{
"input": "1 16",
"output": "8"
},
{
"input": "2 5",
"output": "5"
},
{
"input": "2 6",
"output": "6"
},
{
"input": "2 7",
"output": "7"
},
{
"input": "2 14",
"output": "14"
},
{
"input": "2 15",
"output": "15"
},
{
"input": "1 4",
"output": "2"
},
{
"input": "2 16",
"output": "16"
},
{
"input": "3 5",
"output": "7"
},
{
"input": "3 6",
"output": "9"
},
{
"input": "3 10",
"output": "15"
},
{
"input": "3 14",
"output": "21"
},
{
"input": "3 15",
"output": "22"
},
{
"input": "3 16",
"output": "24"
},
{
"input": "5 7",
"output": "17"
},
{
"input": "16 16",
"output": "128"
},
{
"input": "15 16",
"output": "120"
},
{
"input": "2 3",
"output": "3"
},
{
"input": "15 15",
"output": "112"
},
{
"input": "14 16",
"output": "112"
},
{
"input": "11 13",
"output": "71"
},
{
"input": "5 16",
"output": "40"
},
{
"input": "8 15",
"output": "60"
},
{
"input": "2 2",
"output": "2"
},
{
"input": "3 4",
"output": "6"
},
{
"input": "4 4",
"output": "8"
},
{
"input": "1 1",
"output": "0"
},
{
"input": "1 2",
"output": "1"
},
{
"input": "1 3",
"output": "1"
},
{
"input": "14 15",
"output": "105"
}
] | 1,617,276,314
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 92
| 0
|
first_num=int(input(""))
second_num=int(input(""))
if first_num == 1 and second_num ==2:
print(first_num)
elif first_num == 1 :
print(second_num//2)
elif second_num ==2 :
print(first_num)
else:
print(first_num//2 + second_num)
|
Title: Domino piling
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions.
Input Specification:
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
Output Specification:
Output one number — the maximal number of dominoes, which can be placed.
Demo Input:
['2 4\n', '3 3\n']
Demo Output:
['4\n', '4\n']
Note:
none
|
```python
first_num=int(input(""))
second_num=int(input(""))
if first_num == 1 and second_num ==2:
print(first_num)
elif first_num == 1 :
print(second_num//2)
elif second_num ==2 :
print(first_num)
else:
print(first_num//2 + second_num)
```
| -1
|
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 — 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"
}
] | 1,591,515,239
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 88
| 2,000
| 1,638,400
|
a = input()
b = input()
def judge(left, right):
if left == right:
return 1
n = len(left)
if n % 2:
return 0
mid = n//2
l1 = left[:mid]
l2 = left[mid:]
r1 = right[:mid]
r2 = right[mid:]
if (judge(l1, r1) and judge(l2, r2)) or (judge(l1, r2) and judge(l2, r1)):
return 1
return 0
if judge(a, b):
print('YES')
else:
print('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 — 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
a = input()
b = input()
def judge(left, right):
if left == right:
return 1
n = len(left)
if n % 2:
return 0
mid = n//2
l1 = left[:mid]
l2 = left[mid:]
r1 = right[:mid]
r2 = right[mid:]
if (judge(l1, r1) and judge(l2, r2)) or (judge(l1, r2) and judge(l2, r1)):
return 1
return 0
if judge(a, b):
print('YES')
else:
print('NO')
```
| 0
|
|
331
|
C1
|
The Great Julya Calendar
|
PROGRAMMING
| 1,100
|
[
"dp"
] | null | null |
Yet another Armageddon is coming! This time the culprit is the Julya tribe calendar.
The beavers in this tribe knew math very well. Smart Beaver, an archaeologist, got a sacred plate with a magic integer on it. The translation from Old Beaverish is as follows:
"May the Great Beaver bless you! May your chacres open and may your third eye never turn blind from beholding the Truth! Take the magic number, subtract a digit from it (the digit must occur in the number) and get a new magic number. Repeat this operation until a magic number equals zero. The Earth will stand on Three Beavers for the time, equal to the number of subtractions you perform!"
Distinct subtraction sequences can obviously get you different number of operations. But the Smart Beaver is ready to face the worst and is asking you to count the minimum number of operations he needs to reduce the magic number to zero.
|
The single line contains the magic integer *n*, 0<=≤<=*n*.
- to get 20 points, you need to solve the problem with constraints: *n*<=≤<=106 (subproblem C1); - to get 40 points, you need to solve the problem with constraints: *n*<=≤<=1012 (subproblems C1+C2); - to get 100 points, you need to solve the problem with constraints: *n*<=≤<=1018 (subproblems C1+C2+C3).
|
Print a single integer — the minimum number of subtractions that turns the magic number to a zero.
|
[
"24\n"
] |
[
"5"
] |
In the first test sample the minimum number of operations can be reached by the following sequence of subtractions:
| 20
|
[
{
"input": "24",
"output": "5"
},
{
"input": "0",
"output": "0"
},
{
"input": "3",
"output": "1"
},
{
"input": "8",
"output": "1"
},
{
"input": "9",
"output": "1"
},
{
"input": "10",
"output": "2"
},
{
"input": "31",
"output": "6"
},
{
"input": "701",
"output": "116"
},
{
"input": "222",
"output": "39"
},
{
"input": "156",
"output": "28"
},
{
"input": "12343",
"output": "1778"
},
{
"input": "7984",
"output": "1178"
},
{
"input": "1000000",
"output": "128207"
},
{
"input": "1",
"output": "1"
},
{
"input": "3",
"output": "1"
},
{
"input": "5",
"output": "1"
},
{
"input": "11",
"output": "3"
},
{
"input": "77",
"output": "14"
},
{
"input": "216",
"output": "37"
},
{
"input": "1468",
"output": "233"
},
{
"input": "1995",
"output": "305"
},
{
"input": "11010",
"output": "1568"
},
{
"input": "47320",
"output": "6746"
},
{
"input": "258634",
"output": "35024"
}
] | 1,575,239,222
| 2,147,483,647
|
PyPy 3
|
RUNTIME_ERROR
|
TESTS1
| 12
| 496
| 6,553,600
|
# https://codeforces.com/problemset/problem/331/C1
import math
def to_digits(n):
return [int(d) for d in str(n)]
def to_integer(digits):
accum = 0
factor = 1
for d in reversed(digits):
accum += factor * d
factor *= 10
return accum
def split(digits):
d = 0
index = 0
for i in range(len(digits)):
if digits[i] >= d:
d = digits[i]
index = i
# print(str(d) + " " + str(digits[index:]))
return (d, digits[i:])
memoize = {}
def num(x):
if x > 0:
digits = to_digits(x)
d, r = split(digits)
rint = to_integer(r)
n = rint // d
mod = rint % d
if mod == 0 and n != 0:
n = n
elif rint < d or d < rint:
n = n + 1
return n + num(x - n * d)
return 0
x = int(input())
print(num(x))
|
Title: The Great Julya Calendar
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Yet another Armageddon is coming! This time the culprit is the Julya tribe calendar.
The beavers in this tribe knew math very well. Smart Beaver, an archaeologist, got a sacred plate with a magic integer on it. The translation from Old Beaverish is as follows:
"May the Great Beaver bless you! May your chacres open and may your third eye never turn blind from beholding the Truth! Take the magic number, subtract a digit from it (the digit must occur in the number) and get a new magic number. Repeat this operation until a magic number equals zero. The Earth will stand on Three Beavers for the time, equal to the number of subtractions you perform!"
Distinct subtraction sequences can obviously get you different number of operations. But the Smart Beaver is ready to face the worst and is asking you to count the minimum number of operations he needs to reduce the magic number to zero.
Input Specification:
The single line contains the magic integer *n*, 0<=≤<=*n*.
- to get 20 points, you need to solve the problem with constraints: *n*<=≤<=106 (subproblem C1); - to get 40 points, you need to solve the problem with constraints: *n*<=≤<=1012 (subproblems C1+C2); - to get 100 points, you need to solve the problem with constraints: *n*<=≤<=1018 (subproblems C1+C2+C3).
Output Specification:
Print a single integer — the minimum number of subtractions that turns the magic number to a zero.
Demo Input:
['24\n']
Demo Output:
['5']
Note:
In the first test sample the minimum number of operations can be reached by the following sequence of subtractions:
|
```python
# https://codeforces.com/problemset/problem/331/C1
import math
def to_digits(n):
return [int(d) for d in str(n)]
def to_integer(digits):
accum = 0
factor = 1
for d in reversed(digits):
accum += factor * d
factor *= 10
return accum
def split(digits):
d = 0
index = 0
for i in range(len(digits)):
if digits[i] >= d:
d = digits[i]
index = i
# print(str(d) + " " + str(digits[index:]))
return (d, digits[i:])
memoize = {}
def num(x):
if x > 0:
digits = to_digits(x)
d, r = split(digits)
rint = to_integer(r)
n = rint // d
mod = rint % d
if mod == 0 and n != 0:
n = n
elif rint < d or d < rint:
n = n + 1
return n + num(x - n * d)
return 0
x = int(input())
print(num(x))
```
| -1
|
|
758
|
B
|
Blown Garland
|
PROGRAMMING
| 1,100
|
[
"brute force",
"implementation",
"number theory"
] | null | null |
Nothing is eternal in the world, Kostya understood it on the 7-th of January when he saw partially dead four-color garland.
Now he has a goal to replace dead light bulbs, however he doesn't know how many light bulbs for each color are required. It is guaranteed that for each of four colors at least one light is working.
It is known that the garland contains light bulbs of four colors: red, blue, yellow and green. The garland is made as follows: if you take any four consecutive light bulbs then there will not be light bulbs with the same color among them. For example, the garland can look like "RYBGRYBGRY", "YBGRYBGRYBG", "BGRYB", but can not look like "BGRYG", "YBGRYBYGR" or "BGYBGY". Letters denote colors: 'R' — red, 'B' — blue, 'Y' — yellow, 'G' — green.
Using the information that for each color at least one light bulb still works count the number of dead light bulbs of each four colors.
|
The first and the only line contains the string *s* (4<=≤<=|*s*|<=≤<=100), which describes the garland, the *i*-th symbol of which describes the color of the *i*-th light bulb in the order from the beginning of garland:
- 'R' — the light bulb is red, - 'B' — the light bulb is blue, - 'Y' — the light bulb is yellow, - 'G' — the light bulb is green, - '!' — the light bulb is dead.
The string *s* can not contain other symbols except those five which were described.
It is guaranteed that in the given string at least once there is each of four letters 'R', 'B', 'Y' and 'G'.
It is guaranteed that the string *s* is correct garland with some blown light bulbs, it means that for example the line "GRBY!!!B" can not be in the input data.
|
In the only line print four integers *k**r*,<=*k**b*,<=*k**y*,<=*k**g* — the number of dead light bulbs of red, blue, yellow and green colors accordingly.
|
[
"RYBGRYBGR\n",
"!RGYB\n",
"!!!!YGRB\n",
"!GB!RG!Y!\n"
] |
[
"0 0 0 0",
"0 1 0 0",
"1 1 1 1",
"2 1 1 0"
] |
In the first example there are no dead light bulbs.
In the second example it is obvious that one blue bulb is blown, because it could not be light bulbs of other colors on its place according to the statements.
| 1,000
|
[
{
"input": "RYBGRYBGR",
"output": "0 0 0 0"
},
{
"input": "!RGYB",
"output": "0 1 0 0"
},
{
"input": "!!!!YGRB",
"output": "1 1 1 1"
},
{
"input": "!GB!RG!Y!",
"output": "2 1 1 0"
},
{
"input": "RYBG",
"output": "0 0 0 0"
},
{
"input": "!Y!!!Y!!G!!!G!!B!!R!!!!B!!!!!Y!!G!R!!!!!!!!!!!!B!!!!GY!B!!!!!YR!G!!!!!!B!Y!B!!!!!!R!G!!!!!!!G!R!!!!B",
"output": "20 18 19 18"
},
{
"input": "!R!GBRYG!RYGB!!G!!YG!!Y!!",
"output": "3 5 2 1"
},
{
"input": "RBGYRBGYRBGYRBGYRBGYRBGYRBGYRBGYRBGYRBGYRBGYRBGYRBGYRBGYRBGYRBGYRBGYRBGYRBGYRBGYRBGYRBGYRBGYRBGYRBGY",
"output": "0 0 0 0"
},
{
"input": "GYRB!",
"output": "0 0 0 1"
},
{
"input": "RBYGR",
"output": "0 0 0 0"
},
{
"input": "BRYGB",
"output": "0 0 0 0"
},
{
"input": "YRGBY",
"output": "0 0 0 0"
},
{
"input": "GBYRG",
"output": "0 0 0 0"
},
{
"input": "GBYR!!!!",
"output": "1 1 1 1"
},
{
"input": "!!!BRYG!!",
"output": "2 1 1 1"
},
{
"input": "!!!YBGR!!!",
"output": "1 2 1 2"
},
{
"input": "R!!Y!!B!!G!",
"output": "2 2 1 2"
},
{
"input": "!!!!BR!!!!GY",
"output": "2 2 2 2"
},
{
"input": "!!!!!!!!!!!!!!!!!!Y!!!!!!!!!!!!!!!!!!B!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!G!!R!!!!!!!!!!!!",
"output": "24 24 24 24"
},
{
"input": "!!G!!!G!!!G!!!G!!!GB!!G!!!G!!YG!!!G!!!G!!!G!!!G!!!G!!!G!!!G!!!G!!!G!!!G!!!G!!!G!!!G!!!G!!!G!R!G!!!G!",
"output": "24 24 24 0"
},
{
"input": "!!Y!!!Y!!!Y!!!Y!!!Y!!!Y!!!YR!!Y!!!Y!B!Y!!!Y!!!Y!!!Y!!!Y!!GY!!!Y!!!Y!!!Y!!!Y!!!Y!!!Y!!!Y!!!Y!!!Y!!!Y!",
"output": "24 24 0 24"
},
{
"input": "!B!!!B!!!B!!!B!!!B!!!B!G!B!!!B!!!B!!!B!!!B!!!B!!!BR!!B!!!B!!!B!!!B!!!B!!YB!!!B!!!B!!!B!!!B!!!B!!!B!!",
"output": "24 0 24 24"
},
{
"input": "YR!!!R!!!RB!!R!!!R!!!R!!!R!!!R!!!R!!!R!!!R!!!R!!!R!!!R!!!R!!!R!!!R!!!R!!!R!!!R!!!R!G!R!!!R!!!R!!!R!!",
"output": "0 24 24 24"
},
{
"input": "R!YBRGY!R!",
"output": "0 1 0 2"
},
{
"input": "B!RGB!!GBYR!B!R",
"output": "1 0 3 1"
},
{
"input": "Y!!GYB!G!!!!YB!G!!RG",
"output": "4 3 2 1"
},
{
"input": "R!!BRYG!!YG!R!!!R!!!!!G!R!!!!!",
"output": "3 6 6 4"
},
{
"input": "R!!!R!!!R!!!R!B!RGB!!G!!R!B!R!B!RG!YR!B!",
"output": "1 5 9 7"
},
{
"input": "!Y!R!Y!RB!G!BY!!!!!R!YG!!YGRB!!!!!!!BYGR!!!RBYGRBY",
"output": "5 7 5 7"
},
{
"input": "!!G!!!!!Y!!RYBGRY!!R!!!R!!!!!!!R!B!!!!!R!!!R!!!R!!!R!!!R!!!!",
"output": "5 13 12 13"
},
{
"input": "!!BG!!B!!RBG!!B!YRB!!!B!YRBG!!BG!!B!!!BG!!BG!RB!Y!!!!!B!Y!B!Y!!!!!B!!!",
"output": "14 2 13 11"
},
{
"input": "R!GBRYGBRYGBRYG!RY!BRYGBRYGBRYGBRYGBRYGBRYGBRYGBRYGBR!GBRY!BRY!BRYGBRYGBRYGBRYGB",
"output": "0 1 2 3"
},
{
"input": "!!!!B!!!!G!!B!R!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!YB!R!!!!!!G!!!!!!!!",
"output": "20 20 21 21"
},
{
"input": "G!!!GY!!GYBRGYB!GY!RG!B!GYBRGY!!GY!!GYBRGYBRGY!RGY!!GYBRGY!!G!BRGYB!GYBRGYB!GY!!G!!RGYB!GYB!G!B!GYB!",
"output": "15 10 5 0"
},
{
"input": "R!!!!!!Y!B!!!!!!!!!!!!!!R!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!Y!!G!!!!!!!!!!!!!!!!!!!!!!!!!!!!!",
"output": "23 24 23 24"
},
{
"input": "!!YR!!YR!!YR!!YR!!YR!BYR!!YR!!YR!!YRG!YR!!YR!!YR!!YR!!YR!!YR!!YR!!YR!!YR!!YR!!YR!!YR!!YR!!YR!!YR!!YR",
"output": "0 24 0 24"
},
{
"input": "!!!YR!B!!!B!R!!!R!!YR!BY!G!YR!B!R!BYRG!!!!BY!!!!!!B!!!B!R!!Y!!B!!GB!R!B!!!!!!G!!RG!!R!BYR!!!!!B!!!!!",
"output": "13 12 17 20"
},
{
"input": "B!RG!!R!B!R!B!R!B!R!!!R!B!RG!!RGB!R!!!RGB!!!!YR!B!!!!!RGB!R!B!R!B!!!!!RGBY!!B!RG!Y!GB!!!B!!GB!RGB!R!",
"output": "7 8 22 15"
},
{
"input": "!B!YR!!YR!!YRB!Y!B!Y!B!Y!!!YR!GYR!!YRB!YR!!Y!!!YR!!YRB!YR!!Y!B!Y!!!Y!!!YR!!Y!B!YRB!YR!!YR!!Y!B!Y!B!Y",
"output": "11 14 0 24"
},
{
"input": "!RBYGRBYGRBY!!!!GRBYGRB!GRBY!R!YGRBYG!BYGRBYG!!Y!!BYGRB!G!B!G!!!G!BY!RBYGRB!!R!!GR!YG!BY!!B!GR!Y!!!!",
"output": "10 8 9 8"
},
{
"input": "BRG!!RGYBRGYBRG!B!GY!!GYB!GY!!G!BRGY!RGYB!G!!RGYBRGYB!GY!!GYB!GYBRGY!!GYB!GY!!GYB!GY!!GYBRGY!!GYB!G",
"output": "15 10 4 0"
},
{
"input": "!Y!!!!!!!!!!!!!!!!!GB!!!!!!!!!!!!Y!!!!!!!!!!!!!!!!!!!!!!!!!!!!R!!!!!!!!!!!!!!!!!!!!!!!!!!!R!!!!!!!",
"output": "22 24 23 23"
},
{
"input": "!R!!Y!G!!!!BYR!!!!G!!!!!!R!!!!!!!!!B!!!B!R!BY!!B!!GB!!G!!!G!!!G!!!!!!R!!!!G!!!!!Y!!BY!!!!!!!Y!!!",
"output": "19 17 18 17"
},
{
"input": "!!GYRBGY!BGY!BGY!BGYR!G!RBGYRBGYR!G!RBGY!BGY!!GY!BGY!BGYRBGYR!GYRBGYR!G!!BGY!!GY!!GY!BGY!!GY!BG",
"output": "14 9 3 0"
},
{
"input": "!!!!!!!!Y!!!!!!!!!GR!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!B!!!!!!Y!!!!!!R!!!!!!!!!!!!!!!!!!!!!!!!!!",
"output": "21 23 22 22"
},
{
"input": "!B!!Y!!GY!RGY!!!!!!!!!!!Y!!!!!!!!!!!!!!!!!RG!BR!!!!!!!!G!!!!!B!!!!R!!B!G!B!!YB!!Y!!!!BRG!!!G",
"output": "18 16 17 16"
},
{
"input": "YB!!Y!GR!B!!YB!RYBG!!!!RY!GR!!!R!B!R!B!R!!!R!B!R!B!!!B!!YB!R!!G!YB!!Y!!R!BG!!!!!!B!!!!!R!!!",
"output": "10 10 15 18"
},
{
"input": "R!G!R!GBR!!BR!GB!!!B!!!BR!GBRYG!R!!!R!GBRYGBR!GBR!!BR!GBR!GBRY!B!!!!R!!BR!!BR!!!!!!B!!!BR!",
"output": "5 5 20 12"
},
{
"input": "YRB!Y!B!YRB!Y!!!Y!B!YR!!YR!!Y!!!YRB!YR!!YRB!Y!B!YRB!YR!!Y!!!YR!!YRB!YR!!Y!B!YRB!Y!!GYR!!Y",
"output": "8 11 0 21"
},
{
"input": "!!GBRY!!!YG!R!GBR!G!RY!B!YGB!!G!RYGBRYGB!Y!BR!G!RYGBRYGBRYGBRYGBRYGBRYGB!Y!B!YGBR!!BRYGB",
"output": "7 5 6 5"
},
{
"input": "G!!!!Y!!!!R!!!!B",
"output": "3 3 3 3"
},
{
"input": "!Y!!!!!!G!!!!!!!!!B!!!!!!!!!!!!R",
"output": "7 7 7 7"
},
{
"input": "RGBYRGBYRGBY",
"output": "0 0 0 0"
},
{
"input": "!!!!!!!!!GBYRGBY",
"output": "3 2 2 2"
},
{
"input": "RBYGRBYGRBYGRB!",
"output": "0 0 1 0"
},
{
"input": "R!!!!!!!!!!!!B!!!!!!!!!!!!Y!!!!!!!!!!!!G",
"output": "9 9 9 9"
},
{
"input": "GY!!!!R!!Y!B",
"output": "2 2 1 2"
},
{
"input": "R!!!!!!!!!!!!!!!!!!!!!!!!Y!!!!!!!!!!!!!!!!!!!!!!!!G!!!!!!!!!!!!!!!!!!!!!!!!B!!!!!!!!!!!!!!!!!!!!!!!!",
"output": "24 24 24 24"
},
{
"input": "R!!!!G!!!!B!!!!Y",
"output": "3 3 3 3"
},
{
"input": "R!!!!B!!!!Y!!!!G!!!!",
"output": "4 4 4 4"
},
{
"input": "!R!B!!!!G!Y",
"output": "2 1 2 2"
},
{
"input": "!!!!!R!!!!G!!!!B!!!!Y!!!!!!!!!",
"output": "7 6 7 6"
},
{
"input": "R!!!!B!!!!Y!!!!G",
"output": "3 3 3 3"
},
{
"input": "!!!!!R!!!!G!!!!B!!!!!!!!Y!!!!!!!!!",
"output": "8 7 8 7"
},
{
"input": "!!!!!!!!R!!!!!!!!B!!!!!!!!G!!!!!!!!Y!!!!!!!!",
"output": "10 10 10 10"
}
] | 1,694,521,374
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 65
| 62
| 0
|
s = input()
ans = ""
for j in range(4):
for i in s[j::4]:
if i!='!':
ans += i
break
arr = {}
arr['R'] = 0
arr['B'] = 0
arr['Y'] = 0
arr['G'] = 0
for j in range(4):
for i in s[j::4]:
if i=='!':
arr[ans[j]] += 1
for i in "RBYG":
print(arr[i] , end = " ")
|
Title: Blown Garland
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Nothing is eternal in the world, Kostya understood it on the 7-th of January when he saw partially dead four-color garland.
Now he has a goal to replace dead light bulbs, however he doesn't know how many light bulbs for each color are required. It is guaranteed that for each of four colors at least one light is working.
It is known that the garland contains light bulbs of four colors: red, blue, yellow and green. The garland is made as follows: if you take any four consecutive light bulbs then there will not be light bulbs with the same color among them. For example, the garland can look like "RYBGRYBGRY", "YBGRYBGRYBG", "BGRYB", but can not look like "BGRYG", "YBGRYBYGR" or "BGYBGY". Letters denote colors: 'R' — red, 'B' — blue, 'Y' — yellow, 'G' — green.
Using the information that for each color at least one light bulb still works count the number of dead light bulbs of each four colors.
Input Specification:
The first and the only line contains the string *s* (4<=≤<=|*s*|<=≤<=100), which describes the garland, the *i*-th symbol of which describes the color of the *i*-th light bulb in the order from the beginning of garland:
- 'R' — the light bulb is red, - 'B' — the light bulb is blue, - 'Y' — the light bulb is yellow, - 'G' — the light bulb is green, - '!' — the light bulb is dead.
The string *s* can not contain other symbols except those five which were described.
It is guaranteed that in the given string at least once there is each of four letters 'R', 'B', 'Y' and 'G'.
It is guaranteed that the string *s* is correct garland with some blown light bulbs, it means that for example the line "GRBY!!!B" can not be in the input data.
Output Specification:
In the only line print four integers *k**r*,<=*k**b*,<=*k**y*,<=*k**g* — the number of dead light bulbs of red, blue, yellow and green colors accordingly.
Demo Input:
['RYBGRYBGR\n', '!RGYB\n', '!!!!YGRB\n', '!GB!RG!Y!\n']
Demo Output:
['0 0 0 0', '0 1 0 0', '1 1 1 1', '2 1 1 0']
Note:
In the first example there are no dead light bulbs.
In the second example it is obvious that one blue bulb is blown, because it could not be light bulbs of other colors on its place according to the statements.
|
```python
s = input()
ans = ""
for j in range(4):
for i in s[j::4]:
if i!='!':
ans += i
break
arr = {}
arr['R'] = 0
arr['B'] = 0
arr['Y'] = 0
arr['G'] = 0
for j in range(4):
for i in s[j::4]:
if i=='!':
arr[ans[j]] += 1
for i in "RBYG":
print(arr[i] , end = " ")
```
| 3
|
|
339
|
D
|
Xenia and Bit Operations
|
PROGRAMMING
| 1,700
|
[
"data structures",
"trees"
] | null | null |
Xenia the beginner programmer has a sequence *a*, consisting of 2*n* non-negative integers: *a*1,<=*a*2,<=...,<=*a*2*n*. Xenia is currently studying bit operations. To better understand how they work, Xenia decided to calculate some value *v* for *a*.
Namely, it takes several iterations to calculate value *v*. At the first iteration, Xenia writes a new sequence *a*1 *or* *a*2,<=*a*3 *or* *a*4,<=...,<=*a*2*n*<=-<=1 *or* *a*2*n*, consisting of 2*n*<=-<=1 elements. In other words, she writes down the bit-wise OR of adjacent elements of sequence *a*. At the second iteration, Xenia writes the bitwise exclusive OR of adjacent elements of the sequence obtained after the first iteration. At the third iteration Xenia writes the bitwise OR of the adjacent elements of the sequence obtained after the second iteration. And so on; the operations of bitwise exclusive OR and bitwise OR alternate. In the end, she obtains a sequence consisting of one element, and that element is *v*.
Let's consider an example. Suppose that sequence *a*<==<=(1,<=2,<=3,<=4). Then let's write down all the transformations (1,<=2,<=3,<=4) <=→<= (1 *or* 2<==<=3,<=3 *or* 4<==<=7) <=→<= (3 *xor* 7<==<=4). The result is *v*<==<=4.
You are given Xenia's initial sequence. But to calculate value *v* for a given sequence would be too easy, so you are given additional *m* queries. Each query is a pair of integers *p*,<=*b*. Query *p*,<=*b* means that you need to perform the assignment *a**p*<==<=*b*. After each query, you need to print the new value *v* for the new sequence *a*.
|
The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=17,<=1<=≤<=*m*<=≤<=105). The next line contains 2*n* integers *a*1,<=*a*2,<=...,<=*a*2*n* (0<=≤<=*a**i*<=<<=230). Each of the next *m* lines contains queries. The *i*-th line contains integers *p**i*,<=*b**i* (1<=≤<=*p**i*<=≤<=2*n*,<=0<=≤<=*b**i*<=<<=230) — the *i*-th query.
|
Print *m* integers — the *i*-th integer denotes value *v* for sequence *a* after the *i*-th query.
|
[
"2 4\n1 6 3 5\n1 4\n3 4\n1 2\n1 2\n"
] |
[
"1\n3\n3\n3\n"
] |
For more information on the bit operations, you can follow this link: http://en.wikipedia.org/wiki/Bitwise_operation
| 2,000
|
[
{
"input": "2 4\n1 6 3 5\n1 4\n3 4\n1 2\n1 2",
"output": "1\n3\n3\n3"
},
{
"input": "1 1\n1 1\n1 1",
"output": "1"
},
{
"input": "1 10\n6 26\n1 11\n1 9\n1 31\n1 10\n2 12\n1 8\n2 10\n2 4\n2 18\n1 31",
"output": "27\n27\n31\n26\n14\n12\n10\n12\n26\n31"
},
{
"input": "1 10\n22 17\n2 15\n2 12\n1 6\n1 16\n2 24\n1 21\n2 19\n2 25\n2 19\n2 1",
"output": "31\n30\n14\n28\n24\n29\n23\n29\n23\n21"
},
{
"input": "2 10\n15 27 17 18\n2 12\n4 19\n2 3\n4 1\n3 15\n3 11\n2 5\n1 26\n1 17\n3 18",
"output": "28\n28\n28\n30\n0\n4\n4\n20\n30\n6"
},
{
"input": "2 10\n31 17 22 5\n2 15\n2 26\n1 9\n4 13\n3 28\n1 20\n1 26\n1 27\n2 20\n2 12",
"output": "8\n8\n12\n4\n6\n3\n7\n6\n2\n2"
}
] | 1,695,141,771
| 2,147,483,647
|
PyPy 3-64
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 19
| 2,000
| 19,251,200
|
import sys;input=sys.stdin.readline
n,m=map(int,input().split())
a=[*map(int,input().split())]
n=2**n
tree=[0]*4*n
def build(v,tl,tr):
if tl==tr:tree[v]=a[tl];return 0
tm=tl+tr>>1
t=build(v*2,tl,tm)^1
build(v*2+1,tm+1,tr)
if t:tree[v]=tree[v*2]|tree[v*2+1]
else:tree[v]=tree[v*2]^tree[v*2+1]
return t
build(1,0,n-1)
def upd(v,tl,tr,pos,val):
if tl==tr and tr==pos:tree[v]=val;return 0
else:
tm=tl+tr>>1
if pos<=tm:t=upd(v*2,tl,tm,pos,val)^1
else:t=upd(v*2+1,tm+1,tr,pos,val)^1
if t:tree[v]=tree[v*2]|tree[v*2+1]
else:tree[v]=tree[v*2]^tree[v*2+1]
return t
for i in range(m):
p,b=map(int,input().split())
upd(1,0,n-1,p-1,b)
print(tree[1])
|
Title: Xenia and Bit Operations
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Xenia the beginner programmer has a sequence *a*, consisting of 2*n* non-negative integers: *a*1,<=*a*2,<=...,<=*a*2*n*. Xenia is currently studying bit operations. To better understand how they work, Xenia decided to calculate some value *v* for *a*.
Namely, it takes several iterations to calculate value *v*. At the first iteration, Xenia writes a new sequence *a*1 *or* *a*2,<=*a*3 *or* *a*4,<=...,<=*a*2*n*<=-<=1 *or* *a*2*n*, consisting of 2*n*<=-<=1 elements. In other words, she writes down the bit-wise OR of adjacent elements of sequence *a*. At the second iteration, Xenia writes the bitwise exclusive OR of adjacent elements of the sequence obtained after the first iteration. At the third iteration Xenia writes the bitwise OR of the adjacent elements of the sequence obtained after the second iteration. And so on; the operations of bitwise exclusive OR and bitwise OR alternate. In the end, she obtains a sequence consisting of one element, and that element is *v*.
Let's consider an example. Suppose that sequence *a*<==<=(1,<=2,<=3,<=4). Then let's write down all the transformations (1,<=2,<=3,<=4) <=→<= (1 *or* 2<==<=3,<=3 *or* 4<==<=7) <=→<= (3 *xor* 7<==<=4). The result is *v*<==<=4.
You are given Xenia's initial sequence. But to calculate value *v* for a given sequence would be too easy, so you are given additional *m* queries. Each query is a pair of integers *p*,<=*b*. Query *p*,<=*b* means that you need to perform the assignment *a**p*<==<=*b*. After each query, you need to print the new value *v* for the new sequence *a*.
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=17,<=1<=≤<=*m*<=≤<=105). The next line contains 2*n* integers *a*1,<=*a*2,<=...,<=*a*2*n* (0<=≤<=*a**i*<=<<=230). Each of the next *m* lines contains queries. The *i*-th line contains integers *p**i*,<=*b**i* (1<=≤<=*p**i*<=≤<=2*n*,<=0<=≤<=*b**i*<=<<=230) — the *i*-th query.
Output Specification:
Print *m* integers — the *i*-th integer denotes value *v* for sequence *a* after the *i*-th query.
Demo Input:
['2 4\n1 6 3 5\n1 4\n3 4\n1 2\n1 2\n']
Demo Output:
['1\n3\n3\n3\n']
Note:
For more information on the bit operations, you can follow this link: http://en.wikipedia.org/wiki/Bitwise_operation
|
```python
import sys;input=sys.stdin.readline
n,m=map(int,input().split())
a=[*map(int,input().split())]
n=2**n
tree=[0]*4*n
def build(v,tl,tr):
if tl==tr:tree[v]=a[tl];return 0
tm=tl+tr>>1
t=build(v*2,tl,tm)^1
build(v*2+1,tm+1,tr)
if t:tree[v]=tree[v*2]|tree[v*2+1]
else:tree[v]=tree[v*2]^tree[v*2+1]
return t
build(1,0,n-1)
def upd(v,tl,tr,pos,val):
if tl==tr and tr==pos:tree[v]=val;return 0
else:
tm=tl+tr>>1
if pos<=tm:t=upd(v*2,tl,tm,pos,val)^1
else:t=upd(v*2+1,tm+1,tr,pos,val)^1
if t:tree[v]=tree[v*2]|tree[v*2+1]
else:tree[v]=tree[v*2]^tree[v*2+1]
return t
for i in range(m):
p,b=map(int,input().split())
upd(1,0,n-1,p-1,b)
print(tree[1])
```
| 0
|
|
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<=≤<=*x*1,<=*x*2,<=*x*3<=≤<=100) — the coordinates of the houses of the first, the second and the third friends respectively.
|
Print one integer — 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"
}
] | 1,675,218,985
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 48
| 62
| 0
|
x = list(map(int, input().split()))
maxi = max(x)
x.remove(maxi)
mini = min(x)
x.remove(mini)
result = maxi - x[0]
result += x[0] - mini
print(result)
|
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<=≤<=*x*1,<=*x*2,<=*x*3<=≤<=100) — the coordinates of the houses of the first, the second and the third friends respectively.
Output Specification:
Print one integer — 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
x = list(map(int, input().split()))
maxi = max(x)
x.remove(maxi)
mini = min(x)
x.remove(mini)
result = maxi - x[0]
result += x[0] - mini
print(result)
```
| 3
|
|
518
|
A
|
Vitaly and Strings
|
PROGRAMMING
| 1,600
|
[
"constructive algorithms",
"strings"
] | null | null |
Vitaly is a diligent student who never missed a lesson in his five years of studying in the university. He always does his homework on time and passes his exams in time.
During the last lesson the teacher has provided two strings *s* and *t* to Vitaly. The strings have the same length, they consist of lowercase English letters, string *s* is lexicographically smaller than string *t*. Vitaly wondered if there is such string that is lexicographically larger than string *s* and at the same is lexicographically smaller than string *t*. This string should also consist of lowercase English letters and have the length equal to the lengths of strings *s* and *t*.
Let's help Vitaly solve this easy problem!
|
The first line contains string *s* (1<=≤<=|*s*|<=≤<=100), consisting of lowercase English letters. Here, |*s*| denotes the length of the string.
The second line contains string *t* (|*t*|<==<=|*s*|), consisting of lowercase English letters.
It is guaranteed that the lengths of strings *s* and *t* are the same and string *s* is lexicographically less than string *t*.
|
If the string that meets the given requirements doesn't exist, print a single string "No such string" (without the quotes).
If such string exists, print it. If there are multiple valid strings, you may print any of them.
|
[
"a\nc\n",
"aaa\nzzz\n",
"abcdefg\nabcdefh\n"
] |
[
"b\n",
"kkk\n",
"No such string\n"
] |
String *s* = *s*<sub class="lower-index">1</sub>*s*<sub class="lower-index">2</sub>... *s*<sub class="lower-index">*n*</sub> is said to be lexicographically smaller than *t* = *t*<sub class="lower-index">1</sub>*t*<sub class="lower-index">2</sub>... *t*<sub class="lower-index">*n*</sub>, if there exists such *i*, that *s*<sub class="lower-index">1</sub> = *t*<sub class="lower-index">1</sub>, *s*<sub class="lower-index">2</sub> = *t*<sub class="lower-index">2</sub>, ... *s*<sub class="lower-index">*i* - 1</sub> = *t*<sub class="lower-index">*i* - 1</sub>, *s*<sub class="lower-index">*i*</sub> < *t*<sub class="lower-index">*i*</sub>.
| 500
|
[
{
"input": "a\nc",
"output": "b"
},
{
"input": "aaa\nzzz",
"output": "kkk"
},
{
"input": "abcdefg\nabcdefh",
"output": "No such string"
},
{
"input": "abcdefg\nabcfefg",
"output": "abcdefh"
},
{
"input": "frt\nfru",
"output": "No such string"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa\nzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz",
"output": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaab"
},
{
"input": "zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzx\nzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz",
"output": "zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzy"
},
{
"input": "q\nz",
"output": "r"
},
{
"input": "pnzcl\npnzdf",
"output": "pnzcm"
},
{
"input": "vklldrxnfgyorgfpfezvhbouyzzzzz\nvklldrxnfgyorgfpfezvhbouzaaadv",
"output": "vklldrxnfgyorgfpfezvhbouzaaaaa"
},
{
"input": "pkjlxzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz\npkjlyaaaaaaaaaaaaaaaaaaaaaaaaaaaahr",
"output": "pkjlyaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "exoudpymnspkocwszzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz\nexoudpymnspkocwtaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaabml",
"output": "exoudpymnspkocwtaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "anarzvsklmwvovozwnmhklkpcseeogdgauoppmzrukynbjjoxytuvsiecuzfquxnowewebhtuoxepocyeamqfrblpwqiokbcubil\nanarzvsklmwvovozwnmhklkpcseeogdgauoppmzrukynbjjoxytuvsiecuzfquxnowewebhtuoxepocyeamqfrblpwqiokbcubim",
"output": "No such string"
},
{
"input": "uqyugulumzwlxsjnxxkutzqayskrbjoaaekbhckjryhjjllzzz\nuqyugulumzwlxsjnxxkutzqayskrbjoaaekbhckjryhjjlmaaa",
"output": "No such string"
},
{
"input": "esfaeyxpblcrriizhnhfrxnbopqvhwtetgjqavlqdlxexaifgvkqfwzneibhxxdacbzzzzzzzzzzzzzz\nesfaeyxpblcrriizhnhfrxnbopqvhwtetgjqavlqdlxexaifgvkqfwzneibhxxdaccaaaaaaaaaaaatf",
"output": "esfaeyxpblcrriizhnhfrxnbopqvhwtetgjqavlqdlxexaifgvkqfwzneibhxxdaccaaaaaaaaaaaaaa"
},
{
"input": "oisjtilteipnzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz\noisjtilteipoaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaao",
"output": "oisjtilteipoaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "svpoxbsudndfnnpugbouawegyxgtmvqzbewxpcwhopdbwscimgzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz\nsvpoxbsudndfnnpugbouawegyxgtmvqzbewxpcwhopdbwscimhaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "No such string"
},
{
"input": "ddzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz\ndeaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaao",
"output": "deaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "xqzbhslocdbifnyzyjenlpctocieaccsycmwlcebkqqkeibatfvylbqlutvjijgjhdetqsjqnoipqbmjhhzxggdobyvpczdavdzz\nxqzbhslocdbifnyzyjenlpctocieaccsycmwlcebkqqkeibatfvylbqlutvjijgjhdetqsjqnoipqbmjhhzxggdobyvpczdavilj",
"output": "xqzbhslocdbifnyzyjenlpctocieaccsycmwlcebkqqkeibatfvylbqlutvjijgjhdetqsjqnoipqbmjhhzxggdobyvpczdaveaa"
},
{
"input": "poflpxucohdobeisxfsnkbdzwizjjhgngufssqhmfgmydmmrnuminrvxxamoebhczlwsfefdtnchaisfxkfcovxmvppxnrfawfoq\npoflpxucohdobeisxfsnkbdzwizjjhgngufssqhmfgmydmmrnuminrvxxamoebhczlwsfefdtnchaisfxkfcovxmvppxnrfawujg",
"output": "poflpxucohdobeisxfsnkbdzwizjjhgngufssqhmfgmydmmrnuminrvxxamoebhczlwsfefdtnchaisfxkfcovxmvppxnrfawfor"
},
{
"input": "vonggnmokmvmguwtobkxoqgxkuxtyjmxrygyliohlhwxuxjmlkqcfuxboxjnzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz\nvonggnmokmvmguwtobkxoqgxkuxtyjmxrygyliohlhwxuxjmlkqcfuxboxjoaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaac",
"output": "vonggnmokmvmguwtobkxoqgxkuxtyjmxrygyliohlhwxuxjmlkqcfuxboxjoaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "bqycw\nquhod",
"output": "bqycx"
},
{
"input": "hceslswecf\nnmxshuymaa",
"output": "hceslswecg"
},
{
"input": "awqtzslxowuaefe\nvujscakjpvxviki",
"output": "awqtzslxowuaeff"
},
{
"input": "lerlcnaogdravnogfogcyoxgi\nojrbithvjdqtempegvqxmgmmw",
"output": "lerlcnaogdravnogfogcyoxgj"
},
{
"input": "jbrhvicytqaivheqeourrlosvnsujsxdinryyawgalidsaufxv\noevvkhujmhagaholrmsatdjjyfmyblvgetpnxgjcilugjsncjs",
"output": "jbrhvicytqaivheqeourrlosvnsujsxdinryyawgalidsaufxw"
},
{
"input": "jrpogrcuhqdpmyzpuabuhaptlxaeiqjxhqkmuzsjbhqxvdtoocrkusaeasqdwlunomwzww\nspvgaswympzlscnumemgiznngnxqgccbubmxgqmaakbnyngkxlxjjsafricchhpecdjgxw",
"output": "jrpogrcuhqdpmyzpuabuhaptlxaeiqjxhqkmuzsjbhqxvdtoocrkusaeasqdwlunomwzwx"
},
{
"input": "mzmhjmfxaxaplzjmjkbyadeweltagyyuzpvrmnyvirjpdmebxyzjvdoezhnayfrvtnccryhkvhcvakcf\nohhhhkujfpjbgouebtmmbzizuhuumvrsqfniwpmxdtzhyiaivdyxhywnqzagicydixjtvbqbevhbqttu",
"output": "mzmhjmfxaxaplzjmjkbyadeweltagyyuzpvrmnyvirjpdmebxyzjvdoezhnayfrvtnccryhkvhcvakcg"
},
{
"input": "cdmwmzutsicpzhcokbbhwktqbomozxvvjlhwdgtiledgurxsfreisgczdwgupzxmjnfyjxcpdwzkggludkcmgppndl\nuvuqvyrnhtyubpevizhjxdvmpueittksrnosmfuuzbimnqussasdjufrthrgjbyzomauaxbvwferfvtmydmwmjaoxg",
"output": "cdmwmzutsicpzhcokbbhwktqbomozxvvjlhwdgtiledgurxsfreisgczdwgupzxmjnfyjxcpdwzkggludkcmgppndm"
},
{
"input": "dpnmrwpbgzvcmrcodwgvvfwpyagdwlngmhrazyvalszhruprxzmwltftxmujfyrrnwzvphgqlcphreumqkytswxziugburwrlyay\nqibcfxdfovoejutaeetbbwrgexdrvqywwmhipxgfrvhzovxkfawpfnpjvlhkyahessodqcclangxefcaixysqijnitevwmpalkzd",
"output": "dpnmrwpbgzvcmrcodwgvvfwpyagdwlngmhrazyvalszhruprxzmwltftxmujfyrrnwzvphgqlcphreumqkytswxziugburwrlyaz"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaab",
"output": "No such string"
},
{
"input": "phdvmuwqmvzyurtnshitcypuzbhpceovkibzbhhjwxkdtvqmbpoumeoiztxtvkvsjrlnhowsdmgftuiulzebdigmun\nphdvmuwqmvzyurtnshitcypuzbhpceovkibzbhhjwxkdtvqmbpoumeoiztxtvkvsjrlnhowsdmgftuiulzebdigmuo",
"output": "No such string"
},
{
"input": "hrsantdquixzjyjtqytcmnflnyehzbibkbgkqffgqpkgeuqmbmxzhbjwsnfkizvbcyoghyvnxxjavoahlqjxomtsouzoog\nhrsantdquixzjyjtqytcmnflnyehzbibkbgkqffgqpkgeuqmbmxzhbjwsnfkizvbcyoghyvnxxjavoahlqjxomtsouzooh",
"output": "No such string"
},
{
"input": "kexdbtpkjbwwyibjndbtmwqzolopqitgkomqggojevoankiepxirrcidxldlzsppehmoazdywltmjbxgsxgihwnwpmczjrcwpywl\nkexdbtpkjbwwyibjndbtmwqzolopqitgkomqggojevoankiepxirrcidxldlzsppehmoazdywltmjbxgsxgihwnwpmczjrcwpywm",
"output": "No such string"
},
{
"input": "yijdysvzfcnaedvnecswgoylhzgguxecmucepgstjbdkbjyfdlxxxejkrrxfiuwjpdmdhhqhlqeqzjwudtdryrfkpwfxdjlkowmk\nyijdysvzfcnaedvnecswgoylhzgguxecmucepgstjbdkbjyfdlxxxejkrrxfiuwjpdmdhhqhlqeqzjwudtdryrfkpwfxdjlkowml",
"output": "No such string"
},
{
"input": "a\nb",
"output": "No such string"
},
{
"input": "x\nz",
"output": "y"
},
{
"input": "a\nz",
"output": "b"
},
{
"input": "y\nz",
"output": "No such string"
},
{
"input": "zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzy\nzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz",
"output": "No such string"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaac",
"output": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaab"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa\nzaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaab"
},
{
"input": "yzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz\nzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz",
"output": "zaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "yyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyy\nzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz",
"output": "yyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyz"
},
{
"input": "aaa\naac",
"output": "aab"
},
{
"input": "aaa\naab",
"output": "No such string"
},
{
"input": "zzx\nzzz",
"output": "zzy"
},
{
"input": "zzy\nzzz",
"output": "No such string"
},
{
"input": "aba\naca",
"output": "abb"
},
{
"input": "aaa\nbbb",
"output": "aab"
},
{
"input": "azzz\ncaaa",
"output": "baaa"
}
] | 1,608,109,682
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 8
| 140
| 0
|
s=list(input())
s1=list(input())
ans = "No such string"
r=0
for i in range(len(s)):
if s[i]!=s1[i]:
r+=1
for i in range(len(s)):
if ord(s[i])+1<ord(s1[i]):
ans = ''.join(s[:i]) + chr(ord(s[i])+1) + ''.join(s[i+1:])
break
elif ord(s[i])+1==ord(s1[i]) and r>1:
ans = ''.join(s[:i]) + chr(ord(s[i])+1) + ''.join(s[i+1:])
break
print(ans)
|
Title: Vitaly and Strings
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vitaly is a diligent student who never missed a lesson in his five years of studying in the university. He always does his homework on time and passes his exams in time.
During the last lesson the teacher has provided two strings *s* and *t* to Vitaly. The strings have the same length, they consist of lowercase English letters, string *s* is lexicographically smaller than string *t*. Vitaly wondered if there is such string that is lexicographically larger than string *s* and at the same is lexicographically smaller than string *t*. This string should also consist of lowercase English letters and have the length equal to the lengths of strings *s* and *t*.
Let's help Vitaly solve this easy problem!
Input Specification:
The first line contains string *s* (1<=≤<=|*s*|<=≤<=100), consisting of lowercase English letters. Here, |*s*| denotes the length of the string.
The second line contains string *t* (|*t*|<==<=|*s*|), consisting of lowercase English letters.
It is guaranteed that the lengths of strings *s* and *t* are the same and string *s* is lexicographically less than string *t*.
Output Specification:
If the string that meets the given requirements doesn't exist, print a single string "No such string" (without the quotes).
If such string exists, print it. If there are multiple valid strings, you may print any of them.
Demo Input:
['a\nc\n', 'aaa\nzzz\n', 'abcdefg\nabcdefh\n']
Demo Output:
['b\n', 'kkk\n', 'No such string\n']
Note:
String *s* = *s*<sub class="lower-index">1</sub>*s*<sub class="lower-index">2</sub>... *s*<sub class="lower-index">*n*</sub> is said to be lexicographically smaller than *t* = *t*<sub class="lower-index">1</sub>*t*<sub class="lower-index">2</sub>... *t*<sub class="lower-index">*n*</sub>, if there exists such *i*, that *s*<sub class="lower-index">1</sub> = *t*<sub class="lower-index">1</sub>, *s*<sub class="lower-index">2</sub> = *t*<sub class="lower-index">2</sub>, ... *s*<sub class="lower-index">*i* - 1</sub> = *t*<sub class="lower-index">*i* - 1</sub>, *s*<sub class="lower-index">*i*</sub> < *t*<sub class="lower-index">*i*</sub>.
|
```python
s=list(input())
s1=list(input())
ans = "No such string"
r=0
for i in range(len(s)):
if s[i]!=s1[i]:
r+=1
for i in range(len(s)):
if ord(s[i])+1<ord(s1[i]):
ans = ''.join(s[:i]) + chr(ord(s[i])+1) + ''.join(s[i+1:])
break
elif ord(s[i])+1==ord(s1[i]) and r>1:
ans = ''.join(s[:i]) + chr(ord(s[i])+1) + ''.join(s[i+1:])
break
print(ans)
```
| 0
|
|
617
|
A
|
Elephant
|
PROGRAMMING
| 800
|
[
"math"
] | null | null |
An elephant decided to visit his friend. It turned out that the elephant's house is located at point 0 and his friend's house is located at point *x*(*x*<=><=0) of the coordinate line. In one step the elephant can move 1, 2, 3, 4 or 5 positions forward. Determine, what is the minimum number of steps he need to make in order to get to his friend's house.
|
The first line of the input contains an integer *x* (1<=≤<=*x*<=≤<=1<=000<=000) — The coordinate of the friend's house.
|
Print the minimum number of steps that elephant needs to make to get from point 0 to point *x*.
|
[
"5\n",
"12\n"
] |
[
"1\n",
"3\n"
] |
In the first sample the elephant needs to make one step of length 5 to reach the point *x*.
In the second sample the elephant can get to point *x* if he moves by 3, 5 and 4. There are other ways to get the optimal answer but the elephant cannot reach *x* in less than three moves.
| 500
|
[
{
"input": "5",
"output": "1"
},
{
"input": "12",
"output": "3"
},
{
"input": "999999",
"output": "200000"
},
{
"input": "41",
"output": "9"
},
{
"input": "1000000",
"output": "200000"
},
{
"input": "1",
"output": "1"
},
{
"input": "2",
"output": "1"
},
{
"input": "3",
"output": "1"
},
{
"input": "4",
"output": "1"
},
{
"input": "534204",
"output": "106841"
},
{
"input": "469569",
"output": "93914"
},
{
"input": "502877",
"output": "100576"
},
{
"input": "942212",
"output": "188443"
},
{
"input": "97",
"output": "20"
},
{
"input": "53",
"output": "11"
},
{
"input": "89",
"output": "18"
},
{
"input": "574",
"output": "115"
},
{
"input": "716",
"output": "144"
},
{
"input": "729",
"output": "146"
},
{
"input": "8901",
"output": "1781"
},
{
"input": "3645",
"output": "729"
},
{
"input": "4426",
"output": "886"
},
{
"input": "46573",
"output": "9315"
},
{
"input": "86380",
"output": "17276"
},
{
"input": "94190",
"output": "18838"
},
{
"input": "999990",
"output": "199998"
},
{
"input": "999991",
"output": "199999"
},
{
"input": "999992",
"output": "199999"
},
{
"input": "999993",
"output": "199999"
},
{
"input": "999994",
"output": "199999"
},
{
"input": "999995",
"output": "199999"
},
{
"input": "999996",
"output": "200000"
},
{
"input": "999997",
"output": "200000"
},
{
"input": "999998",
"output": "200000"
}
] | 1,698,007,154
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 34
| 46
| 0
|
x = int(input())
p = x//5 + (x % 5)//4 + (x % 5 % 4)//3 + (x %
5 % 4 % 3)//2 + (x % 5 % 4 % 3 % 2)
print(p)
|
Title: Elephant
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
An elephant decided to visit his friend. It turned out that the elephant's house is located at point 0 and his friend's house is located at point *x*(*x*<=><=0) of the coordinate line. In one step the elephant can move 1, 2, 3, 4 or 5 positions forward. Determine, what is the minimum number of steps he need to make in order to get to his friend's house.
Input Specification:
The first line of the input contains an integer *x* (1<=≤<=*x*<=≤<=1<=000<=000) — The coordinate of the friend's house.
Output Specification:
Print the minimum number of steps that elephant needs to make to get from point 0 to point *x*.
Demo Input:
['5\n', '12\n']
Demo Output:
['1\n', '3\n']
Note:
In the first sample the elephant needs to make one step of length 5 to reach the point *x*.
In the second sample the elephant can get to point *x* if he moves by 3, 5 and 4. There are other ways to get the optimal answer but the elephant cannot reach *x* in less than three moves.
|
```python
x = int(input())
p = x//5 + (x % 5)//4 + (x % 5 % 4)//3 + (x %
5 % 4 % 3)//2 + (x % 5 % 4 % 3 % 2)
print(p)
```
| 3
|
|
131
|
A
|
cAPS lOCK
|
PROGRAMMING
| 1,000
|
[
"implementation",
"strings"
] | null | null |
wHAT DO WE NEED cAPS LOCK FOR?
Caps lock is a computer keyboard key. Pressing it sets an input mode in which typed letters are capital by default. If it is pressed by accident, it leads to accidents like the one we had in the first passage.
Let's consider that a word has been typed with the Caps lock key accidentally switched on, if:
- either it only contains uppercase letters; - or all letters except for the first one are uppercase.
In this case we should automatically change the case of all letters. For example, the case of the letters that form words "hELLO", "HTTP", "z" should be changed.
Write a program that applies the rule mentioned above. If the rule cannot be applied, the program should leave the word unchanged.
|
The first line of the input data contains a word consisting of uppercase and lowercase Latin letters. The word's length is from 1 to 100 characters, inclusive.
|
Print the result of the given word's processing.
|
[
"cAPS\n",
"Lock\n"
] |
[
"Caps",
"Lock\n"
] |
none
| 500
|
[
{
"input": "cAPS",
"output": "Caps"
},
{
"input": "Lock",
"output": "Lock"
},
{
"input": "cAPSlOCK",
"output": "cAPSlOCK"
},
{
"input": "CAPs",
"output": "CAPs"
},
{
"input": "LoCK",
"output": "LoCK"
},
{
"input": "OOPS",
"output": "oops"
},
{
"input": "oops",
"output": "oops"
},
{
"input": "a",
"output": "A"
},
{
"input": "A",
"output": "a"
},
{
"input": "aA",
"output": "Aa"
},
{
"input": "Zz",
"output": "Zz"
},
{
"input": "Az",
"output": "Az"
},
{
"input": "zA",
"output": "Za"
},
{
"input": "AAA",
"output": "aaa"
},
{
"input": "AAa",
"output": "AAa"
},
{
"input": "AaR",
"output": "AaR"
},
{
"input": "Tdr",
"output": "Tdr"
},
{
"input": "aTF",
"output": "Atf"
},
{
"input": "fYd",
"output": "fYd"
},
{
"input": "dsA",
"output": "dsA"
},
{
"input": "fru",
"output": "fru"
},
{
"input": "hYBKF",
"output": "Hybkf"
},
{
"input": "XweAR",
"output": "XweAR"
},
{
"input": "mogqx",
"output": "mogqx"
},
{
"input": "eOhEi",
"output": "eOhEi"
},
{
"input": "nkdku",
"output": "nkdku"
},
{
"input": "zcnko",
"output": "zcnko"
},
{
"input": "lcccd",
"output": "lcccd"
},
{
"input": "vwmvg",
"output": "vwmvg"
},
{
"input": "lvchf",
"output": "lvchf"
},
{
"input": "IUNVZCCHEWENCHQQXQYPUJCRDZLUXCLJHXPHBXEUUGNXOOOPBMOBRIBHHMIRILYJGYYGFMTMFSVURGYHUWDRLQVIBRLPEVAMJQYO",
"output": "iunvzcchewenchqqxqypujcrdzluxcljhxphbxeuugnxooopbmobribhhmirilyjgyygfmtmfsvurgyhuwdrlqvibrlpevamjqyo"
},
{
"input": "OBHSZCAMDXEJWOZLKXQKIVXUUQJKJLMMFNBPXAEFXGVNSKQLJGXHUXHGCOTESIVKSFMVVXFVMTEKACRIWALAGGMCGFEXQKNYMRTG",
"output": "obhszcamdxejwozlkxqkivxuuqjkjlmmfnbpxaefxgvnskqljgxhuxhgcotesivksfmvvxfvmtekacriwalaggmcgfexqknymrtg"
},
{
"input": "IKJYZIKROIYUUCTHSVSKZTETNNOCMAUBLFJCEVANCADASMZRCNLBZPQRXESHEEMOMEPCHROSRTNBIDXYMEPJSIXSZQEBTEKKUHFS",
"output": "ikjyzikroiyuucthsvskztetnnocmaublfjcevancadasmzrcnlbzpqrxesheemomepchrosrtnbidxymepjsixszqebtekkuhfs"
},
{
"input": "cTKDZNWVYRTFPQLDAUUNSPKTDJTUPPFPRXRSINTVFVNNQNKXWUZUDHZBUSOKTABUEDQKUIVRTTVUREEOBJTSDKJKVEGFXVHXEYPE",
"output": "Ctkdznwvyrtfpqldauunspktdjtuppfprxrsintvfvnnqnkxwuzudhzbusoktabuedqkuivrttvureeobjtsdkjkvegfxvhxeype"
},
{
"input": "uCKJZRGZJCPPLEEYJTUNKOQSWGBMTBQEVPYFPIPEKRVYQNTDPANOIXKMPINNFUSZWCURGBDPYTEKBEKCPMVZPMWAOSHJYMGKOMBQ",
"output": "Uckjzrgzjcppleeyjtunkoqswgbmtbqevpyfpipekrvyqntdpanoixkmpinnfuszwcurgbdpytekbekcpmvzpmwaoshjymgkombq"
},
{
"input": "KETAXTSWAAOBKUOKUQREHIOMVMMRSAEWKGXZKRASwTVNSSFSNIWYNPSTMRADOADEEBURRHPOOBIEUIBGYDJCEKPNLEUCANZYJKMR",
"output": "KETAXTSWAAOBKUOKUQREHIOMVMMRSAEWKGXZKRASwTVNSSFSNIWYNPSTMRADOADEEBURRHPOOBIEUIBGYDJCEKPNLEUCANZYJKMR"
},
{
"input": "ZEKGDMWJPVUWFlNXRLUmWKLMMYSLRQQIBRWDPKWITUIMZYYKOEYGREKHHZRZZUFPVTNIHKGTCCTLOKSZITXXZDMPITHNZUIGDZLE",
"output": "ZEKGDMWJPVUWFlNXRLUmWKLMMYSLRQQIBRWDPKWITUIMZYYKOEYGREKHHZRZZUFPVTNIHKGTCCTLOKSZITXXZDMPITHNZUIGDZLE"
},
{
"input": "TcMbVPCFvnNkCEUUCIFLgBJeCOKuJhIGwXFrhAZjuAhBraMSchBfWwIuHAEbgJOFzGtxDLDXzDSaPCFujGGxgxdlHUIQYRrMFCgJ",
"output": "TcMbVPCFvnNkCEUUCIFLgBJeCOKuJhIGwXFrhAZjuAhBraMSchBfWwIuHAEbgJOFzGtxDLDXzDSaPCFujGGxgxdlHUIQYRrMFCgJ"
},
{
"input": "xFGqoLILNvxARKuIntPfeukFtMbvzDezKpPRAKkIoIvwqNXnehRVwkkXYvuRCeoieBaBfTjwsYhDeCLvBwktntyluoxCYVioXGdm",
"output": "xFGqoLILNvxARKuIntPfeukFtMbvzDezKpPRAKkIoIvwqNXnehRVwkkXYvuRCeoieBaBfTjwsYhDeCLvBwktntyluoxCYVioXGdm"
},
{
"input": "udvqolbxdwbkijwvhlyaelhynmnfgszbhgshlcwdkaibceqomzujndixuzivlsjyjqxzxodzbukxxhwwultvekdfntwpzlhhrIjm",
"output": "udvqolbxdwbkijwvhlyaelhynmnfgszbhgshlcwdkaibceqomzujndixuzivlsjyjqxzxodzbukxxhwwultvekdfntwpzlhhrIjm"
},
{
"input": "jgpwhetqqoncighgzbbaLwwwxkxivuwtokehrgprfgewzcwxkavwoflcgsgbhoeamzbefzoonwsyzisetoydrpufktzgbaycgaeg",
"output": "jgpwhetqqoncighgzbbaLwwwxkxivuwtokehrgprfgewzcwxkavwoflcgsgbhoeamzbefzoonwsyzisetoydrpufktzgbaycgaeg"
},
{
"input": "vyujsazdstbnkxeunedfbolicojzjpufgfemhtmdrswvmuhoivjvonacefqenbqudelmdegxqtbwezsbydmanzutvdgkgrjxzlnc",
"output": "vyujsazdstbnkxeunedfbolicojzjpufgfemhtmdrswvmuhoivjvonacefqenbqudelmdegxqtbwezsbydmanzutvdgkgrjxzlnc"
},
{
"input": "pivqnuqkaofcduvbttztjbuavrqwiqrwkfncmvatoxruelyoecnkpqraiahumiaiqeyjapbqyrsxcdgjbihivtqezvasfmzntdfv",
"output": "pivqnuqkaofcduvbttztjbuavrqwiqrwkfncmvatoxruelyoecnkpqraiahumiaiqeyjapbqyrsxcdgjbihivtqezvasfmzntdfv"
},
{
"input": "upvtbsxswbohxshdrbjxcungzquhuomgxwlryvshshsfvqbrxvcikbglietlpqkiwbhiqpmdwuqosbtdvyxekkaqiwrbsibpoect",
"output": "upvtbsxswbohxshdrbjxcungzquhuomgxwlryvshshsfvqbrxvcikbglietlpqkiwbhiqpmdwuqosbtdvyxekkaqiwrbsibpoect"
},
{
"input": "cTKDZNWVYRTFPQLDAUUNSPKTDJTUPPFPRXRSINTVFVNNQNKXWUZUDHZBUSOKTABUEDQKUIVRTTVUREEOBJTSDKJKVEGFXVHXEYPE",
"output": "Ctkdznwvyrtfpqldauunspktdjtuppfprxrsintvfvnnqnkxwuzudhzbusoktabuedqkuivrttvureeobjtsdkjkvegfxvhxeype"
},
{
"input": "JEGDOFJOFGJHFROGHOFGMHOFGHORGFHMOu",
"output": "JEGDOFJOFGJHFROGHOFGMHOFGHORGFHMOu"
},
{
"input": "aBACABa",
"output": "aBACABa"
},
{
"input": "AAAAAAAAAAAAAAAAAAAAAAAAaa",
"output": "AAAAAAAAAAAAAAAAAAAAAAAAaa"
},
{
"input": "aaAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA",
"output": "aaAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA"
},
{
"input": "dDDDDDDDDDDDDD",
"output": "Dddddddddddddd"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "z",
"output": "Z"
},
{
"input": "AZ",
"output": "az"
},
{
"input": "Z",
"output": "z"
},
{
"input": "aAAAA",
"output": "Aaaaa"
},
{
"input": "F",
"output": "f"
}
] | 1,697,527,281
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 8
| 30
| 0
|
s=input()
a=s[0]+s[1:].upper()
if len(s)==1:
print(s.upper())
elif s==s.lower() or s==s.capitalize() and len(s)>1:
print(s)
elif s==s.upper():
print(s.lower())
elif s==a:
print(s.capitalize())
else:
print(s)
|
Title: cAPS lOCK
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
wHAT DO WE NEED cAPS LOCK FOR?
Caps lock is a computer keyboard key. Pressing it sets an input mode in which typed letters are capital by default. If it is pressed by accident, it leads to accidents like the one we had in the first passage.
Let's consider that a word has been typed with the Caps lock key accidentally switched on, if:
- either it only contains uppercase letters; - or all letters except for the first one are uppercase.
In this case we should automatically change the case of all letters. For example, the case of the letters that form words "hELLO", "HTTP", "z" should be changed.
Write a program that applies the rule mentioned above. If the rule cannot be applied, the program should leave the word unchanged.
Input Specification:
The first line of the input data contains a word consisting of uppercase and lowercase Latin letters. The word's length is from 1 to 100 characters, inclusive.
Output Specification:
Print the result of the given word's processing.
Demo Input:
['cAPS\n', 'Lock\n']
Demo Output:
['Caps', 'Lock\n']
Note:
none
|
```python
s=input()
a=s[0]+s[1:].upper()
if len(s)==1:
print(s.upper())
elif s==s.lower() or s==s.capitalize() and len(s)>1:
print(s)
elif s==s.upper():
print(s.lower())
elif s==a:
print(s.capitalize())
else:
print(s)
```
| 0
|
|
833
|
A
|
The Meaningless Game
|
PROGRAMMING
| 1,700
|
[
"math",
"number theory"
] | null | null |
Slastyona and her loyal dog Pushok are playing a meaningless game that is indeed very interesting.
The game consists of multiple rounds. Its rules are very simple: in each round, a natural number *k* is chosen. Then, the one who says (or barks) it faster than the other wins the round. After that, the winner's score is multiplied by *k*2, and the loser's score is multiplied by *k*. In the beginning of the game, both Slastyona and Pushok have scores equal to one.
Unfortunately, Slastyona had lost her notepad where the history of all *n* games was recorded. She managed to recall the final results for each games, though, but all of her memories of them are vague. Help Slastyona verify their correctness, or, to put it another way, for each given pair of scores determine whether it was possible for a game to finish with such result or not.
|
In the first string, the number of games *n* (1<=≤<=*n*<=≤<=350000) is given.
Each game is represented by a pair of scores *a*, *b* (1<=≤<=*a*,<=*b*<=≤<=109) – the results of Slastyona and Pushok, correspondingly.
|
For each pair of scores, answer "Yes" if it's possible for a game to finish with given score, and "No" otherwise.
You can output each letter in arbitrary case (upper or lower).
|
[
"6\n2 4\n75 45\n8 8\n16 16\n247 994\n1000000000 1000000\n"
] |
[
"Yes\nYes\nYes\nNo\nNo\nYes\n"
] |
First game might have been consisted of one round, in which the number 2 would have been chosen and Pushok would have won.
The second game needs exactly two rounds to finish with such result: in the first one, Slastyona would have said the number 5, and in the second one, Pushok would have barked the number 3.
| 500
|
[
{
"input": "6\n2 4\n75 45\n8 8\n16 16\n247 994\n1000000000 1000000",
"output": "Yes\nYes\nYes\nNo\nNo\nYes"
},
{
"input": "3\n1 1\n8 27\n1000 1331",
"output": "Yes\nNo\nNo"
},
{
"input": "1\n12004 18012002",
"output": "Yes"
},
{
"input": "1\n3331 11095561",
"output": "Yes"
},
{
"input": "1\n2 3",
"output": "No"
},
{
"input": "1\n1062961 1031",
"output": "Yes"
},
{
"input": "1\n6 12",
"output": "No"
},
{
"input": "1\n3 1",
"output": "No"
},
{
"input": "1\n3 10",
"output": "No"
},
{
"input": "1\n31159 970883281",
"output": "Yes"
},
{
"input": "1\n9907 98148649",
"output": "Yes"
},
{
"input": "1\n16 8",
"output": "No"
},
{
"input": "1\n90 72",
"output": "No"
}
] | 1,626,396,955
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 1
| 108
| 21,606,400
|
import math
t = int(input())
def primefac(n):
pf = []
while not n % 2:
n //= 2
pf.append(2)
for i in range(3, int(math.sqrt(n))+1, 2):
while not n % i:
n //= i
pf.append(i)
if n > 2:
pf.append(n)
occ = {}
for i in pf:
occ[i] = occ.get(i, 0) + 1
return occ
for tc in range(t):
a, b = map(int, input().split())
occ1 = primefac(a)
occ2 = primefac(b)
flag = False
for i in occ1:
if not occ2.get(i):
flag = True
continue
if not (occ2[i] == occ1[i] / 2 or occ2[i] == occ1[i] * 2 or (occ1[i] == occ2[i] and (0 == occ1[i] % 3)) or occ1[i] == (2/3) * occ2[i] or occ2[i] == (2/3) * occ1[i]):
flag = True
if flag:
print("No")
else:
print("Yes")
#print(occ1, occ2)
|
Title: The Meaningless Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Slastyona and her loyal dog Pushok are playing a meaningless game that is indeed very interesting.
The game consists of multiple rounds. Its rules are very simple: in each round, a natural number *k* is chosen. Then, the one who says (or barks) it faster than the other wins the round. After that, the winner's score is multiplied by *k*2, and the loser's score is multiplied by *k*. In the beginning of the game, both Slastyona and Pushok have scores equal to one.
Unfortunately, Slastyona had lost her notepad where the history of all *n* games was recorded. She managed to recall the final results for each games, though, but all of her memories of them are vague. Help Slastyona verify their correctness, or, to put it another way, for each given pair of scores determine whether it was possible for a game to finish with such result or not.
Input Specification:
In the first string, the number of games *n* (1<=≤<=*n*<=≤<=350000) is given.
Each game is represented by a pair of scores *a*, *b* (1<=≤<=*a*,<=*b*<=≤<=109) – the results of Slastyona and Pushok, correspondingly.
Output Specification:
For each pair of scores, answer "Yes" if it's possible for a game to finish with given score, and "No" otherwise.
You can output each letter in arbitrary case (upper or lower).
Demo Input:
['6\n2 4\n75 45\n8 8\n16 16\n247 994\n1000000000 1000000\n']
Demo Output:
['Yes\nYes\nYes\nNo\nNo\nYes\n']
Note:
First game might have been consisted of one round, in which the number 2 would have been chosen and Pushok would have won.
The second game needs exactly two rounds to finish with such result: in the first one, Slastyona would have said the number 5, and in the second one, Pushok would have barked the number 3.
|
```python
import math
t = int(input())
def primefac(n):
pf = []
while not n % 2:
n //= 2
pf.append(2)
for i in range(3, int(math.sqrt(n))+1, 2):
while not n % i:
n //= i
pf.append(i)
if n > 2:
pf.append(n)
occ = {}
for i in pf:
occ[i] = occ.get(i, 0) + 1
return occ
for tc in range(t):
a, b = map(int, input().split())
occ1 = primefac(a)
occ2 = primefac(b)
flag = False
for i in occ1:
if not occ2.get(i):
flag = True
continue
if not (occ2[i] == occ1[i] / 2 or occ2[i] == occ1[i] * 2 or (occ1[i] == occ2[i] and (0 == occ1[i] % 3)) or occ1[i] == (2/3) * occ2[i] or occ2[i] == (2/3) * occ1[i]):
flag = True
if flag:
print("No")
else:
print("Yes")
#print(occ1, occ2)
```
| 0
|
|
609
|
A
|
USB Flash Drives
|
PROGRAMMING
| 800
|
[
"greedy",
"implementation",
"sortings"
] | null | null |
Sean is trying to save a large file to a USB flash drive. He has *n* USB flash drives with capacities equal to *a*1,<=*a*2,<=...,<=*a**n* megabytes. The file size is equal to *m* megabytes.
Find the minimum number of USB flash drives needed to write Sean's file, if he can split the file between drives.
|
The first line contains positive integer *n* (1<=≤<=*n*<=≤<=100) — the number of USB flash drives.
The second line contains positive integer *m* (1<=≤<=*m*<=≤<=105) — the size of Sean's file.
Each of the next *n* lines contains positive integer *a**i* (1<=≤<=*a**i*<=≤<=1000) — the sizes of USB flash drives in megabytes.
It is guaranteed that the answer exists, i. e. the sum of all *a**i* is not less than *m*.
|
Print the minimum number of USB flash drives to write Sean's file, if he can split the file between drives.
|
[
"3\n5\n2\n1\n3\n",
"3\n6\n2\n3\n2\n",
"2\n5\n5\n10\n"
] |
[
"2\n",
"3\n",
"1\n"
] |
In the first example Sean needs only two USB flash drives — the first and the third.
In the second example Sean needs all three USB flash drives.
In the third example Sean needs only one USB flash drive and he can use any available USB flash drive — the first or the second.
| 0
|
[
{
"input": "3\n5\n2\n1\n3",
"output": "2"
},
{
"input": "3\n6\n2\n3\n2",
"output": "3"
},
{
"input": "2\n5\n5\n10",
"output": "1"
},
{
"input": "5\n16\n8\n1\n3\n4\n9",
"output": "2"
},
{
"input": "10\n121\n10\n37\n74\n56\n42\n39\n6\n68\n8\n100",
"output": "2"
},
{
"input": "12\n4773\n325\n377\n192\n780\n881\n816\n839\n223\n215\n125\n952\n8",
"output": "7"
},
{
"input": "15\n7758\n182\n272\n763\n910\n24\n359\n583\n890\n735\n819\n66\n992\n440\n496\n227",
"output": "15"
},
{
"input": "30\n70\n6\n2\n10\n4\n7\n10\n5\n1\n8\n10\n4\n3\n5\n9\n3\n6\n6\n4\n2\n6\n5\n10\n1\n9\n7\n2\n1\n10\n7\n5",
"output": "8"
},
{
"input": "40\n15705\n702\n722\n105\n873\n417\n477\n794\n300\n869\n496\n572\n232\n456\n298\n473\n584\n486\n713\n934\n121\n303\n956\n934\n840\n358\n201\n861\n497\n131\n312\n957\n96\n914\n509\n60\n300\n722\n658\n820\n103",
"output": "21"
},
{
"input": "50\n18239\n300\n151\n770\n9\n200\n52\n247\n753\n523\n263\n744\n463\n540\n244\n608\n569\n771\n32\n425\n777\n624\n761\n628\n124\n405\n396\n726\n626\n679\n237\n229\n49\n512\n18\n671\n290\n768\n632\n739\n18\n136\n413\n117\n83\n413\n452\n767\n664\n203\n404",
"output": "31"
},
{
"input": "70\n149\n5\n3\n3\n4\n6\n1\n2\n9\n8\n3\n1\n8\n4\n4\n3\n6\n10\n7\n1\n10\n8\n4\n9\n3\n8\n3\n2\n5\n1\n8\n6\n9\n10\n4\n8\n6\n9\n9\n9\n3\n4\n2\n2\n5\n8\n9\n1\n10\n3\n4\n3\n1\n9\n3\n5\n1\n3\n7\n6\n9\n8\n9\n1\n7\n4\n4\n2\n3\n5\n7",
"output": "17"
},
{
"input": "70\n2731\n26\n75\n86\n94\n37\n25\n32\n35\n92\n1\n51\n73\n53\n66\n16\n80\n15\n81\n100\n87\n55\n48\n30\n71\n39\n87\n77\n25\n70\n22\n75\n23\n97\n16\n75\n95\n61\n61\n28\n10\n78\n54\n80\n51\n25\n24\n90\n58\n4\n77\n40\n54\n53\n47\n62\n30\n38\n71\n97\n71\n60\n58\n1\n21\n15\n55\n99\n34\n88\n99",
"output": "35"
},
{
"input": "70\n28625\n34\n132\n181\n232\n593\n413\n862\n887\n808\n18\n35\n89\n356\n640\n339\n280\n975\n82\n345\n398\n948\n372\n91\n755\n75\n153\n948\n603\n35\n694\n722\n293\n363\n884\n264\n813\n175\n169\n646\n138\n449\n488\n828\n417\n134\n84\n763\n288\n845\n801\n556\n972\n332\n564\n934\n699\n842\n942\n644\n203\n406\n140\n37\n9\n423\n546\n675\n491\n113\n587",
"output": "45"
},
{
"input": "80\n248\n3\n9\n4\n5\n10\n7\n2\n6\n2\n2\n8\n2\n1\n3\n7\n9\n2\n8\n4\n4\n8\n5\n4\n4\n10\n2\n1\n4\n8\n4\n10\n1\n2\n10\n2\n3\n3\n1\n1\n8\n9\n5\n10\n2\n8\n10\n5\n3\n6\n1\n7\n8\n9\n10\n5\n10\n10\n2\n10\n1\n2\n4\n1\n9\n4\n7\n10\n8\n5\n8\n1\n4\n2\n2\n3\n9\n9\n9\n10\n6",
"output": "27"
},
{
"input": "80\n2993\n18\n14\n73\n38\n14\n73\n77\n18\n81\n6\n96\n65\n77\n86\n76\n8\n16\n81\n83\n83\n34\n69\n58\n15\n19\n1\n16\n57\n95\n35\n5\n49\n8\n15\n47\n84\n99\n94\n93\n55\n43\n47\n51\n61\n57\n13\n7\n92\n14\n4\n83\n100\n60\n75\n41\n95\n74\n40\n1\n4\n95\n68\n59\n65\n15\n15\n75\n85\n46\n77\n26\n30\n51\n64\n75\n40\n22\n88\n68\n24",
"output": "38"
},
{
"input": "80\n37947\n117\n569\n702\n272\n573\n629\n90\n337\n673\n589\n576\n205\n11\n284\n645\n719\n777\n271\n567\n466\n251\n402\n3\n97\n288\n699\n208\n173\n530\n782\n266\n395\n957\n159\n463\n43\n316\n603\n197\n386\n132\n799\n778\n905\n784\n71\n851\n963\n883\n705\n454\n275\n425\n727\n223\n4\n870\n833\n431\n463\n85\n505\n800\n41\n954\n981\n242\n578\n336\n48\n858\n702\n349\n929\n646\n528\n993\n506\n274\n227",
"output": "70"
},
{
"input": "90\n413\n5\n8\n10\n7\n5\n7\n5\n7\n1\n7\n8\n4\n3\n9\n4\n1\n10\n3\n1\n10\n9\n3\n1\n8\n4\n7\n5\n2\n9\n3\n10\n10\n3\n6\n3\n3\n10\n7\n5\n1\n1\n2\n4\n8\n2\n5\n5\n3\n9\n5\n5\n3\n10\n2\n3\n8\n5\n9\n1\n3\n6\n5\n9\n2\n3\n7\n10\n3\n4\n4\n1\n5\n9\n2\n6\n9\n1\n1\n9\n9\n7\n7\n7\n8\n4\n5\n3\n4\n6\n9",
"output": "59"
},
{
"input": "90\n4226\n33\n43\n83\n46\n75\n14\n88\n36\n8\n25\n47\n4\n96\n19\n33\n49\n65\n17\n59\n72\n1\n55\n94\n92\n27\n33\n39\n14\n62\n79\n12\n89\n22\n86\n13\n19\n77\n53\n96\n74\n24\n25\n17\n64\n71\n81\n87\n52\n72\n55\n49\n74\n36\n65\n86\n91\n33\n61\n97\n38\n87\n61\n14\n73\n95\n43\n67\n42\n67\n22\n12\n62\n32\n96\n24\n49\n82\n46\n89\n36\n75\n91\n11\n10\n9\n33\n86\n28\n75\n39",
"output": "64"
},
{
"input": "90\n40579\n448\n977\n607\n745\n268\n826\n479\n59\n330\n609\n43\n301\n970\n726\n172\n632\n600\n181\n712\n195\n491\n312\n849\n722\n679\n682\n780\n131\n404\n293\n387\n567\n660\n54\n339\n111\n833\n612\n911\n869\n356\n884\n635\n126\n639\n712\n473\n663\n773\n435\n32\n973\n484\n662\n464\n699\n274\n919\n95\n904\n253\n589\n543\n454\n250\n349\n237\n829\n511\n536\n36\n45\n152\n626\n384\n199\n877\n941\n84\n781\n115\n20\n52\n726\n751\n920\n291\n571\n6\n199",
"output": "64"
},
{
"input": "100\n66\n7\n9\n10\n5\n2\n8\n6\n5\n4\n10\n10\n6\n5\n2\n2\n1\n1\n5\n8\n7\n8\n10\n5\n6\n6\n5\n9\n9\n6\n3\n8\n7\n10\n5\n9\n6\n7\n3\n5\n8\n6\n8\n9\n1\n1\n1\n2\n4\n5\n5\n1\n1\n2\n6\n7\n1\n5\n8\n7\n2\n1\n7\n10\n9\n10\n2\n4\n10\n4\n10\n10\n5\n3\n9\n1\n2\n1\n10\n5\n1\n7\n4\n4\n5\n7\n6\n10\n4\n7\n3\n4\n3\n6\n2\n5\n2\n4\n9\n5\n3",
"output": "7"
},
{
"input": "100\n4862\n20\n47\n85\n47\n76\n38\n48\n93\n91\n81\n31\n51\n23\n60\n59\n3\n73\n72\n57\n67\n54\n9\n42\n5\n32\n46\n72\n79\n95\n61\n79\n88\n33\n52\n97\n10\n3\n20\n79\n82\n93\n90\n38\n80\n18\n21\n43\n60\n73\n34\n75\n65\n10\n84\n100\n29\n94\n56\n22\n59\n95\n46\n22\n57\n69\n67\n90\n11\n10\n61\n27\n2\n48\n69\n86\n91\n69\n76\n36\n71\n18\n54\n90\n74\n69\n50\n46\n8\n5\n41\n96\n5\n14\n55\n85\n39\n6\n79\n75\n87",
"output": "70"
},
{
"input": "100\n45570\n14\n881\n678\n687\n993\n413\n760\n451\n426\n787\n503\n343\n234\n530\n294\n725\n941\n524\n574\n441\n798\n399\n360\n609\n376\n525\n229\n995\n478\n347\n47\n23\n468\n525\n749\n601\n235\n89\n995\n489\n1\n239\n415\n122\n671\n128\n357\n886\n401\n964\n212\n968\n210\n130\n871\n360\n661\n844\n414\n187\n21\n824\n266\n713\n126\n496\n916\n37\n193\n755\n894\n641\n300\n170\n176\n383\n488\n627\n61\n897\n33\n242\n419\n881\n698\n107\n391\n418\n774\n905\n87\n5\n896\n835\n318\n373\n916\n393\n91\n460",
"output": "78"
},
{
"input": "100\n522\n1\n5\n2\n4\n2\n6\n3\n4\n2\n10\n10\n6\n7\n9\n7\n1\n7\n2\n5\n3\n1\n5\n2\n3\n5\n1\n7\n10\n10\n4\n4\n10\n9\n10\n6\n2\n8\n2\n6\n10\n9\n2\n7\n5\n9\n4\n6\n10\n7\n3\n1\n1\n9\n5\n10\n9\n2\n8\n3\n7\n5\n4\n7\n5\n9\n10\n6\n2\n9\n2\n5\n10\n1\n7\n7\n10\n5\n6\n2\n9\n4\n7\n10\n10\n8\n3\n4\n9\n3\n6\n9\n10\n2\n9\n9\n3\n4\n1\n10\n2",
"output": "74"
},
{
"input": "100\n32294\n414\n116\n131\n649\n130\n476\n630\n605\n213\n117\n757\n42\n109\n85\n127\n635\n629\n994\n410\n764\n204\n161\n231\n577\n116\n936\n537\n565\n571\n317\n722\n819\n229\n284\n487\n649\n304\n628\n727\n816\n854\n91\n111\n549\n87\n374\n417\n3\n868\n882\n168\n743\n77\n534\n781\n75\n956\n910\n734\n507\n568\n802\n946\n891\n659\n116\n678\n375\n380\n430\n627\n873\n350\n930\n285\n6\n183\n96\n517\n81\n794\n235\n360\n551\n6\n28\n799\n226\n996\n894\n981\n551\n60\n40\n460\n479\n161\n318\n952\n433",
"output": "42"
},
{
"input": "100\n178\n71\n23\n84\n98\n8\n14\n4\n42\n56\n83\n87\n28\n22\n32\n50\n5\n96\n90\n1\n59\n74\n56\n96\n77\n88\n71\n38\n62\n36\n85\n1\n97\n98\n98\n32\n99\n42\n6\n81\n20\n49\n57\n71\n66\n9\n45\n41\n29\n28\n32\n68\n38\n29\n35\n29\n19\n27\n76\n85\n68\n68\n41\n32\n78\n72\n38\n19\n55\n83\n83\n25\n46\n62\n48\n26\n53\n14\n39\n31\n94\n84\n22\n39\n34\n96\n63\n37\n42\n6\n78\n76\n64\n16\n26\n6\n79\n53\n24\n29\n63",
"output": "2"
},
{
"input": "100\n885\n226\n266\n321\n72\n719\n29\n121\n533\n85\n672\n225\n830\n783\n822\n30\n791\n618\n166\n487\n922\n434\n814\n473\n5\n741\n947\n910\n305\n998\n49\n945\n588\n868\n809\n803\n168\n280\n614\n434\n634\n538\n591\n437\n540\n445\n313\n177\n171\n799\n778\n55\n617\n554\n583\n611\n12\n94\n599\n182\n765\n556\n965\n542\n35\n460\n177\n313\n485\n744\n384\n21\n52\n879\n792\n411\n614\n811\n565\n695\n428\n587\n631\n794\n461\n258\n193\n696\n936\n646\n756\n267\n55\n690\n730\n742\n734\n988\n235\n762\n440",
"output": "1"
},
{
"input": "100\n29\n9\n2\n10\n8\n6\n7\n7\n3\n3\n10\n4\n5\n2\n5\n1\n6\n3\n2\n5\n10\n10\n9\n1\n4\n5\n2\n2\n3\n1\n2\n2\n9\n6\n9\n7\n8\n8\n1\n5\n5\n3\n1\n5\n6\n1\n9\n2\n3\n8\n10\n8\n3\n2\n7\n1\n2\n1\n2\n8\n10\n5\n2\n3\n1\n10\n7\n1\n7\n4\n9\n6\n6\n4\n7\n1\n2\n7\n7\n9\n9\n7\n10\n4\n10\n8\n2\n1\n5\n5\n10\n5\n8\n1\n5\n6\n5\n1\n5\n6\n8",
"output": "3"
},
{
"input": "100\n644\n94\n69\n43\n36\n54\n93\n30\n74\n56\n95\n70\n49\n11\n36\n57\n30\n59\n3\n52\n59\n90\n82\n39\n67\n32\n8\n80\n64\n8\n65\n51\n48\n89\n90\n35\n4\n54\n66\n96\n68\n90\n30\n4\n13\n97\n41\n90\n85\n17\n45\n94\n31\n58\n4\n39\n76\n95\n92\n59\n67\n46\n96\n55\n82\n64\n20\n20\n83\n46\n37\n15\n60\n37\n79\n45\n47\n63\n73\n76\n31\n52\n36\n32\n49\n26\n61\n91\n31\n25\n62\n90\n65\n65\n5\n94\n7\n15\n97\n88\n68",
"output": "7"
},
{
"input": "100\n1756\n98\n229\n158\n281\n16\n169\n149\n239\n235\n182\n147\n215\n49\n270\n194\n242\n295\n289\n249\n19\n12\n144\n157\n92\n270\n122\n212\n97\n152\n14\n42\n12\n198\n98\n295\n154\n229\n191\n294\n5\n156\n43\n185\n184\n20\n125\n23\n10\n257\n244\n264\n79\n46\n277\n13\n22\n97\n212\n77\n293\n20\n51\n17\n109\n37\n68\n117\n51\n248\n10\n149\n179\n192\n239\n161\n13\n173\n297\n73\n43\n109\n288\n198\n81\n70\n254\n187\n277\n1\n295\n113\n95\n291\n293\n119\n205\n191\n37\n34\n116",
"output": "6"
},
{
"input": "100\n20562\n721\n452\n11\n703\n376\n183\n197\n203\n406\n642\n346\n446\n256\n760\n201\n360\n702\n707\n388\n779\n653\n610\n497\n768\n670\n134\n780\n306\n661\n180\n259\n256\n362\n6\n121\n415\n747\n170\n67\n439\n728\n193\n622\n481\n38\n225\n343\n303\n253\n436\n305\n68\n794\n247\n291\n600\n750\n188\n199\n757\n28\n776\n749\n253\n351\n53\n629\n129\n578\n209\n89\n651\n262\n638\n353\n469\n31\n144\n460\n176\n535\n562\n366\n639\n234\n577\n364\n761\n617\n303\n450\n778\n311\n289\n221\n274\n239\n626\n194\n36",
"output": "30"
},
{
"input": "100\n32630\n548\n21\n756\n138\n56\n719\n97\n86\n599\n531\n464\n137\n737\n239\n626\n179\n170\n271\n335\n178\n298\n597\n616\n104\n190\n216\n333\n606\n621\n310\n70\n286\n593\n655\n472\n556\n77\n625\n744\n395\n487\n762\n316\n469\n258\n475\n596\n651\n180\n737\n101\n511\n708\n163\n713\n475\n557\n649\n705\n583\n97\n447\n119\n91\n372\n194\n394\n667\n384\n635\n611\n746\n758\n613\n54\n508\n405\n188\n570\n254\n28\n599\n767\n115\n737\n194\n148\n686\n705\n236\n762\n613\n770\n633\n401\n773\n448\n274\n141\n718",
"output": "52"
},
{
"input": "1\n1\n1",
"output": "1"
}
] | 1,621,317,603
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 34
| 62
| 0
|
n=int(input())
m=int(input())
a=[]
for i in range(n):
a.append(int(input()))
a=list(sorted(a))
a=a[::-1]
su=0
for i in range(n):
su+=a[i]
if su>=m:
print(i+1)
break
|
Title: USB Flash Drives
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Sean is trying to save a large file to a USB flash drive. He has *n* USB flash drives with capacities equal to *a*1,<=*a*2,<=...,<=*a**n* megabytes. The file size is equal to *m* megabytes.
Find the minimum number of USB flash drives needed to write Sean's file, if he can split the file between drives.
Input Specification:
The first line contains positive integer *n* (1<=≤<=*n*<=≤<=100) — the number of USB flash drives.
The second line contains positive integer *m* (1<=≤<=*m*<=≤<=105) — the size of Sean's file.
Each of the next *n* lines contains positive integer *a**i* (1<=≤<=*a**i*<=≤<=1000) — the sizes of USB flash drives in megabytes.
It is guaranteed that the answer exists, i. e. the sum of all *a**i* is not less than *m*.
Output Specification:
Print the minimum number of USB flash drives to write Sean's file, if he can split the file between drives.
Demo Input:
['3\n5\n2\n1\n3\n', '3\n6\n2\n3\n2\n', '2\n5\n5\n10\n']
Demo Output:
['2\n', '3\n', '1\n']
Note:
In the first example Sean needs only two USB flash drives — the first and the third.
In the second example Sean needs all three USB flash drives.
In the third example Sean needs only one USB flash drive and he can use any available USB flash drive — the first or the second.
|
```python
n=int(input())
m=int(input())
a=[]
for i in range(n):
a.append(int(input()))
a=list(sorted(a))
a=a[::-1]
su=0
for i in range(n):
su+=a[i]
if su>=m:
print(i+1)
break
```
| 3
|
|
678
|
B
|
The Same Calendar
|
PROGRAMMING
| 1,600
|
[
"implementation"
] | null | null |
The girl Taylor has a beautiful calendar for the year *y*. In the calendar all days are given with their days of week: Monday, Tuesday, Wednesday, Thursday, Friday, Saturday and Sunday.
The calendar is so beautiful that she wants to know what is the next year after *y* when the calendar will be exactly the same. Help Taylor to find that year.
Note that leap years has 366 days. The year is leap if it is divisible by 400 or it is divisible by 4, but not by 100 ([https://en.wikipedia.org/wiki/Leap_year](https://en.wikipedia.org/wiki/Leap_year)).
|
The only line contains integer *y* (1000<=≤<=*y*<=<<=100'000) — the year of the calendar.
|
Print the only integer *y*' — the next year after *y* when the calendar will be the same. Note that you should find the first year after *y* with the same calendar.
|
[
"2016\n",
"2000\n",
"50501\n"
] |
[
"2044\n",
"2028\n",
"50507\n"
] |
Today is Monday, the 13th of June, 2016.
| 0
|
[
{
"input": "2016",
"output": "2044"
},
{
"input": "2000",
"output": "2028"
},
{
"input": "50501",
"output": "50507"
},
{
"input": "1000",
"output": "1006"
},
{
"input": "1900",
"output": "1906"
},
{
"input": "1899",
"output": "1905"
},
{
"input": "99999",
"output": "100010"
},
{
"input": "50000",
"output": "50028"
},
{
"input": "99900",
"output": "99906"
},
{
"input": "12345",
"output": "12351"
},
{
"input": "1004",
"output": "1032"
},
{
"input": "2100",
"output": "2106"
},
{
"input": "1313",
"output": "1319"
},
{
"input": "1872",
"output": "1912"
},
{
"input": "2098",
"output": "2110"
},
{
"input": "2072",
"output": "2112"
},
{
"input": "2002",
"output": "2013"
},
{
"input": "1179",
"output": "1190"
},
{
"input": "2096",
"output": "2108"
},
{
"input": "1096",
"output": "1108"
},
{
"input": "1796",
"output": "1808"
},
{
"input": "2014",
"output": "2025"
},
{
"input": "2006",
"output": "2017"
},
{
"input": "1874",
"output": "1885"
},
{
"input": "1884",
"output": "1924"
},
{
"input": "2342",
"output": "2353"
},
{
"input": "2010",
"output": "2021"
},
{
"input": "2097",
"output": "2109"
},
{
"input": "1072",
"output": "1112"
},
{
"input": "1191",
"output": "1202"
},
{
"input": "2896",
"output": "2908"
},
{
"input": "1797",
"output": "1809"
},
{
"input": "1002",
"output": "1013"
},
{
"input": "99988",
"output": "100016"
},
{
"input": "1788",
"output": "1828"
},
{
"input": "1994",
"output": "2005"
},
{
"input": "5094",
"output": "5100"
},
{
"input": "99996",
"output": "100024"
},
{
"input": "3998",
"output": "4009"
},
{
"input": "49376",
"output": "49416"
}
] | 1,507,027,805
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 40
| 77
| 512,000
|
import datetime
y=int(input())
y_temp=y%400
if y_temp==0:
print(y+28)
else:
y_2_28=datetime.datetime(y_temp,2,28).weekday()
y_3_1=datetime.datetime(y_temp,3,1).weekday()
count=1
while datetime.datetime(y_temp+count,2,28).weekday()!=y_2_28 or datetime.datetime(y_temp+count,3,1).weekday()!=y_3_1:
count+=1
print(y+count)
|
Title: The Same Calendar
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The girl Taylor has a beautiful calendar for the year *y*. In the calendar all days are given with their days of week: Monday, Tuesday, Wednesday, Thursday, Friday, Saturday and Sunday.
The calendar is so beautiful that she wants to know what is the next year after *y* when the calendar will be exactly the same. Help Taylor to find that year.
Note that leap years has 366 days. The year is leap if it is divisible by 400 or it is divisible by 4, but not by 100 ([https://en.wikipedia.org/wiki/Leap_year](https://en.wikipedia.org/wiki/Leap_year)).
Input Specification:
The only line contains integer *y* (1000<=≤<=*y*<=<<=100'000) — the year of the calendar.
Output Specification:
Print the only integer *y*' — the next year after *y* when the calendar will be the same. Note that you should find the first year after *y* with the same calendar.
Demo Input:
['2016\n', '2000\n', '50501\n']
Demo Output:
['2044\n', '2028\n', '50507\n']
Note:
Today is Monday, the 13th of June, 2016.
|
```python
import datetime
y=int(input())
y_temp=y%400
if y_temp==0:
print(y+28)
else:
y_2_28=datetime.datetime(y_temp,2,28).weekday()
y_3_1=datetime.datetime(y_temp,3,1).weekday()
count=1
while datetime.datetime(y_temp+count,2,28).weekday()!=y_2_28 or datetime.datetime(y_temp+count,3,1).weekday()!=y_3_1:
count+=1
print(y+count)
```
| 3
|
|
545
|
C
|
Woodcutters
|
PROGRAMMING
| 1,500
|
[
"dp",
"greedy"
] | null | null |
Little Susie listens to fairy tales before bed every day. Today's fairy tale was about wood cutters and the little girl immediately started imagining the choppers cutting wood. She imagined the situation that is described below.
There are *n* trees located along the road at points with coordinates *x*1,<=*x*2,<=...,<=*x**n*. Each tree has its height *h**i*. Woodcutters can cut down a tree and fell it to the left or to the right. After that it occupies one of the segments [*x**i*<=-<=*h**i*,<=*x**i*] or [*x**i*;*x**i*<=+<=*h**i*]. The tree that is not cut down occupies a single point with coordinate *x**i*. Woodcutters can fell a tree if the segment to be occupied by the fallen tree doesn't contain any occupied point. The woodcutters want to process as many trees as possible, so Susie wonders, what is the maximum number of trees to fell.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of trees.
Next *n* lines contain pairs of integers *x**i*,<=*h**i* (1<=≤<=*x**i*,<=*h**i*<=≤<=109) — the coordinate and the height of the *і*-th tree.
The pairs are given in the order of ascending *x**i*. No two trees are located at the point with the same coordinate.
|
Print a single number — the maximum number of trees that you can cut down by the given rules.
|
[
"5\n1 2\n2 1\n5 10\n10 9\n19 1\n",
"5\n1 2\n2 1\n5 10\n10 9\n20 1\n"
] |
[
"3\n",
"4\n"
] |
In the first sample you can fell the trees like that:
- fell the 1-st tree to the left — now it occupies segment [ - 1;1] - fell the 2-nd tree to the right — now it occupies segment [2;3] - leave the 3-rd tree — it occupies point 5 - leave the 4-th tree — it occupies point 10 - fell the 5-th tree to the right — now it occupies segment [19;20]
In the second sample you can also fell 4-th tree to the right, after that it will occupy segment [10;19].
| 1,750
|
[
{
"input": "5\n1 2\n2 1\n5 10\n10 9\n19 1",
"output": "3"
},
{
"input": "5\n1 2\n2 1\n5 10\n10 9\n20 1",
"output": "4"
},
{
"input": "4\n10 4\n15 1\n19 3\n20 1",
"output": "4"
},
{
"input": "35\n1 7\n3 11\n6 12\n7 6\n8 5\n9 11\n15 3\n16 10\n22 2\n23 3\n25 7\n27 3\n34 5\n35 10\n37 3\n39 4\n40 5\n41 1\n44 1\n47 7\n48 11\n50 6\n52 5\n57 2\n58 7\n60 4\n62 1\n67 3\n68 12\n69 8\n70 1\n71 5\n72 5\n73 6\n74 4",
"output": "10"
},
{
"input": "40\n1 1\n2 1\n3 1\n4 1\n5 1\n6 1\n7 1\n8 1\n9 1\n10 1\n11 1\n12 1\n13 1\n14 1\n15 1\n16 1\n17 1\n18 1\n19 1\n20 1\n21 1\n22 1\n23 1\n24 1\n25 1\n26 1\n27 1\n28 1\n29 1\n30 1\n31 1\n32 1\n33 1\n34 1\n35 1\n36 1\n37 1\n38 1\n39 1\n40 1",
"output": "2"
},
{
"input": "67\n1 1\n3 8\n4 10\n7 8\n9 2\n10 1\n11 5\n12 8\n13 4\n16 6\n18 3\n19 3\n22 5\n24 6\n27 5\n28 3\n29 3\n30 5\n32 5\n33 10\n34 7\n35 8\n36 5\n41 3\n42 2\n43 5\n46 4\n48 4\n49 9\n52 4\n53 9\n55 1\n56 4\n59 7\n68 7\n69 4\n71 9\n72 10\n74 5\n76 4\n77 9\n80 7\n81 9\n82 5\n83 5\n84 9\n85 7\n86 9\n87 4\n88 7\n89 10\n90 3\n91 5\n92 10\n93 5\n94 8\n95 4\n96 2\n97 10\n98 1\n99 3\n100 1\n101 5\n102 4\n103 8\n104 8\n105 8",
"output": "5"
},
{
"input": "1\n1000000000 1000000000",
"output": "1"
},
{
"input": "10\n7 12\n10 2\n12 2\n15 1\n19 2\n20 1\n53 25\n63 10\n75 12\n87 1",
"output": "9"
},
{
"input": "3\n1 1\n1000 1000\n1000000000 1000000000",
"output": "3"
},
{
"input": "2\n1 999999999\n1000000000 1000000000",
"output": "2"
},
{
"input": "10\n999999900 1000000000\n999999901 1000000000\n999999902 1000000000\n999999903 1000000000\n999999904 1000000000\n999999905 1000000000\n999999906 1000000000\n999999907 1000000000\n999999908 1000000000\n999999909 1000000000",
"output": "2"
},
{
"input": "2\n100000000 1000000000\n1000000000 1000000000",
"output": "2"
}
] | 1,697,527,821
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 67
| 280
| 18,227,200
|
n=int(input())
li=[[int(j) for j in input().split()] for _ in range(n)]
ans=2
for i in range(1,n-1):
x,h=li[i][0],li[i][1]
if x-li[i-1][0]>h:
ans+=1
elif x+h<li[i+1][0]:
ans+=1
li[i][0]+=h
print(ans if n>1 else 1)
|
Title: Woodcutters
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Susie listens to fairy tales before bed every day. Today's fairy tale was about wood cutters and the little girl immediately started imagining the choppers cutting wood. She imagined the situation that is described below.
There are *n* trees located along the road at points with coordinates *x*1,<=*x*2,<=...,<=*x**n*. Each tree has its height *h**i*. Woodcutters can cut down a tree and fell it to the left or to the right. After that it occupies one of the segments [*x**i*<=-<=*h**i*,<=*x**i*] or [*x**i*;*x**i*<=+<=*h**i*]. The tree that is not cut down occupies a single point with coordinate *x**i*. Woodcutters can fell a tree if the segment to be occupied by the fallen tree doesn't contain any occupied point. The woodcutters want to process as many trees as possible, so Susie wonders, what is the maximum number of trees to fell.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of trees.
Next *n* lines contain pairs of integers *x**i*,<=*h**i* (1<=≤<=*x**i*,<=*h**i*<=≤<=109) — the coordinate and the height of the *і*-th tree.
The pairs are given in the order of ascending *x**i*. No two trees are located at the point with the same coordinate.
Output Specification:
Print a single number — the maximum number of trees that you can cut down by the given rules.
Demo Input:
['5\n1 2\n2 1\n5 10\n10 9\n19 1\n', '5\n1 2\n2 1\n5 10\n10 9\n20 1\n']
Demo Output:
['3\n', '4\n']
Note:
In the first sample you can fell the trees like that:
- fell the 1-st tree to the left — now it occupies segment [ - 1;1] - fell the 2-nd tree to the right — now it occupies segment [2;3] - leave the 3-rd tree — it occupies point 5 - leave the 4-th tree — it occupies point 10 - fell the 5-th tree to the right — now it occupies segment [19;20]
In the second sample you can also fell 4-th tree to the right, after that it will occupy segment [10;19].
|
```python
n=int(input())
li=[[int(j) for j in input().split()] for _ in range(n)]
ans=2
for i in range(1,n-1):
x,h=li[i][0],li[i][1]
if x-li[i-1][0]>h:
ans+=1
elif x+h<li[i+1][0]:
ans+=1
li[i][0]+=h
print(ans if n>1 else 1)
```
| 3
|
|
305
|
A
|
Strange Addition
|
PROGRAMMING
| 1,600
|
[
"brute force",
"constructive algorithms",
"implementation"
] | null | null |
Unfortunately, Vasya can only sum pairs of integers (*a*, *b*), such that for any decimal place at least one number has digit 0 in this place. For example, Vasya can sum numbers 505 and 50, but he cannot sum 1 and 4.
Vasya has a set of *k* distinct non-negative integers *d*1,<=*d*2,<=...,<=*d**k*.
Vasya wants to choose some integers from this set so that he could sum any two chosen numbers. What maximal number of integers can he choose in the required manner?
|
The first input line contains integer *k* (1<=≤<=*k*<=≤<=100) — the number of integers.
The second line contains *k* distinct space-separated integers *d*1,<=*d*2,<=...,<=*d**k* (0<=≤<=*d**i*<=≤<=100).
|
In the first line print a single integer *n* the maximum number of the chosen integers. In the second line print *n* distinct non-negative integers — the required integers.
If there are multiple solutions, print any of them. You can print the numbers in any order.
|
[
"4\n100 10 1 0\n",
"3\n2 70 3\n"
] |
[
"4\n0 1 10 100 ",
"2\n2 70 "
] |
none
| 500
|
[
{
"input": "4\n100 10 1 0",
"output": "4\n0 1 10 100 "
},
{
"input": "3\n2 70 3",
"output": "2\n2 70 "
},
{
"input": "39\n16 72 42 70 17 36 32 40 47 94 27 30 100 55 23 77 67 28 49 50 53 83 38 33 60 65 62 64 6 66 69 86 96 75 85 0 89 73 29",
"output": "4\n0 6 30 100 "
},
{
"input": "50\n20 67 96 6 75 12 37 46 38 86 83 22 10 8 21 2 93 9 81 49 69 52 63 62 70 92 97 40 47 99 16 85 48 77 39 100 28 5 11 44 89 1 19 42 35 27 7 14 88 33",
"output": "3\n1 10 100 "
},
{
"input": "2\n1 2",
"output": "1\n1 "
},
{
"input": "73\n39 66 3 59 40 93 72 34 95 79 83 65 99 57 48 44 82 76 31 21 64 19 53 75 37 16 43 5 47 24 15 22 20 55 45 74 42 10 61 49 23 80 35 62 2 9 67 97 51 81 1 70 88 63 33 25 68 13 69 71 73 6 18 52 41 38 96 46 92 85 14 36 100",
"output": "3\n1 10 100 "
},
{
"input": "15\n74 90 73 47 36 44 81 21 66 92 2 38 62 72 49",
"output": "2\n2 90 "
},
{
"input": "96\n17 10 0 85 57 78 15 99 55 6 7 88 12 95 58 19 47 18 96 82 21 80 97 77 46 31 54 70 23 60 59 100 66 92 51 14 91 25 16 27 44 4 35 98 8 52 24 5 81 29 73 13 61 56 45 75 49 71 94 48 3 76 32 65 72 1 84 36 86 40 83 50 22 33 41 11 26 93 90 43 39 79 89 9 64 68 42 74 87 2 62 34 20 63 67 37",
"output": "4\n0 1 10 100 "
},
{
"input": "5\n23 75 38 47 70",
"output": "1\n23 "
},
{
"input": "12\n89 61 45 92 22 3 94 66 48 21 54 14",
"output": "1\n3 "
},
{
"input": "1\n99",
"output": "1\n99 "
},
{
"input": "1\n0",
"output": "1\n0 "
},
{
"input": "2\n100 1",
"output": "2\n1 100 "
},
{
"input": "3\n1 100 99",
"output": "2\n1 100 "
},
{
"input": "100\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",
"output": "3\n1 10 100 "
},
{
"input": "100\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 98 100 0",
"output": "4\n0 1 10 100 "
},
{
"input": "99\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": "2\n1 10 "
},
{
"input": "2\n5 6",
"output": "1\n5 "
},
{
"input": "81\n11 12 13 14 15 16 17 18 19 21 22 23 24 25 26 27 28 29 31 32 33 34 35 36 37 38 39 41 42 43 44 45 46 47 48 49 51 52 53 54 55 56 57 58 59 61 62 63 64 65 66 67 68 69 71 72 73 74 75 76 77 78 79 81 82 83 84 85 86 87 88 89 91 92 93 94 95 96 97 98 99",
"output": "1\n11 "
},
{
"input": "3\n99 10 6",
"output": "2\n6 10 "
},
{
"input": "4\n11 10 100 3",
"output": "3\n3 10 100 "
},
{
"input": "2\n99 6",
"output": "1\n6 "
},
{
"input": "3\n23 0 100",
"output": "3\n0 23 100 "
},
{
"input": "2\n43 0",
"output": "2\n0 43 "
},
{
"input": "4\n99 0 100 6",
"output": "3\n0 6 100 "
},
{
"input": "1\n100",
"output": "1\n100 "
},
{
"input": "2\n0 100",
"output": "2\n0 100 "
},
{
"input": "3\n0 100 10",
"output": "3\n0 10 100 "
},
{
"input": "3\n0 100 12",
"output": "3\n0 12 100 "
},
{
"input": "3\n0 100 1",
"output": "3\n0 1 100 "
},
{
"input": "4\n0 100 10 1",
"output": "4\n0 1 10 100 "
},
{
"input": "4\n0 100 10 99",
"output": "3\n0 10 100 "
},
{
"input": "1\n1",
"output": "1\n1 "
},
{
"input": "2\n10 12",
"output": "1\n10 "
},
{
"input": "2\n90 9",
"output": "2\n9 90 "
}
] | 1,530,469,596
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 35
| 218
| 0
|
n = int(input().strip())
lst = list(map(int, input().strip().split()))
l1 = ['0']*lst.count(0) + ['100']*lst.count(100)
l2 = []
for i in range(n):
if lst[i] > 0 and lst[i] < 10:
l2.append(str(lst[i]))
break
l3 = []
for i in range(n):
if lst[i] % 10 == 0 and lst[i] not in [0, 100]:
l3.append(str(lst[i]))
break
l4 = []
for i in range(n):
if lst[i] > 10 and lst[i] < 100 and lst[i] % 10 != 0:
l4.append(str(lst[i]))
break
if len(l2) == 0 and len(l3) == 0:
l = l1 + l4
else:
l = l1 + l2 + l3
print(len(l))
print(' '.join(l))
|
Title: Strange Addition
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Unfortunately, Vasya can only sum pairs of integers (*a*, *b*), such that for any decimal place at least one number has digit 0 in this place. For example, Vasya can sum numbers 505 and 50, but he cannot sum 1 and 4.
Vasya has a set of *k* distinct non-negative integers *d*1,<=*d*2,<=...,<=*d**k*.
Vasya wants to choose some integers from this set so that he could sum any two chosen numbers. What maximal number of integers can he choose in the required manner?
Input Specification:
The first input line contains integer *k* (1<=≤<=*k*<=≤<=100) — the number of integers.
The second line contains *k* distinct space-separated integers *d*1,<=*d*2,<=...,<=*d**k* (0<=≤<=*d**i*<=≤<=100).
Output Specification:
In the first line print a single integer *n* the maximum number of the chosen integers. In the second line print *n* distinct non-negative integers — the required integers.
If there are multiple solutions, print any of them. You can print the numbers in any order.
Demo Input:
['4\n100 10 1 0\n', '3\n2 70 3\n']
Demo Output:
['4\n0 1 10 100 ', '2\n2 70 ']
Note:
none
|
```python
n = int(input().strip())
lst = list(map(int, input().strip().split()))
l1 = ['0']*lst.count(0) + ['100']*lst.count(100)
l2 = []
for i in range(n):
if lst[i] > 0 and lst[i] < 10:
l2.append(str(lst[i]))
break
l3 = []
for i in range(n):
if lst[i] % 10 == 0 and lst[i] not in [0, 100]:
l3.append(str(lst[i]))
break
l4 = []
for i in range(n):
if lst[i] > 10 and lst[i] < 100 and lst[i] % 10 != 0:
l4.append(str(lst[i]))
break
if len(l2) == 0 and len(l3) == 0:
l = l1 + l4
else:
l = l1 + l2 + l3
print(len(l))
print(' '.join(l))
```
| 3
|
|
39
|
D
|
Cubical Planet
|
PROGRAMMING
| 1,100
|
[
"math"
] |
D. Cubical Planet
|
2
|
64
|
You can find anything whatsoever in our Galaxy! A cubical planet goes round an icosahedral star. Let us introduce a system of axes so that the edges of the cubical planet are parallel to the coordinate axes and two opposite vertices lay in the points (0,<=0,<=0) and (1,<=1,<=1). Two flies live on the planet. At the moment they are sitting on two different vertices of the cubical planet. Your task is to determine whether they see each other or not. The flies see each other when the vertices they occupy lie on the same face of the cube.
|
The first line contains three space-separated integers (0 or 1) — the coordinates of the first fly, the second line analogously contains the coordinates of the second fly.
|
Output "YES" (without quotes) if the flies see each other. Otherwise, output "NO".
|
[
"0 0 0\n0 1 0\n",
"1 1 0\n0 1 0\n",
"0 0 0\n1 1 1\n"
] |
[
"YES\n",
"YES\n",
"NO\n"
] |
none
| 0
|
[
{
"input": "0 0 0\n0 1 0",
"output": "YES"
},
{
"input": "1 1 0\n0 1 0",
"output": "YES"
},
{
"input": "0 0 0\n1 1 1",
"output": "NO"
},
{
"input": "0 0 0\n1 0 0",
"output": "YES"
},
{
"input": "0 0 0\n0 1 0",
"output": "YES"
},
{
"input": "0 0 0\n1 1 0",
"output": "YES"
},
{
"input": "0 0 0\n0 0 1",
"output": "YES"
},
{
"input": "0 0 0\n1 0 1",
"output": "YES"
},
{
"input": "0 0 0\n0 1 1",
"output": "YES"
},
{
"input": "0 0 0\n1 1 1",
"output": "NO"
},
{
"input": "1 0 0\n0 0 0",
"output": "YES"
},
{
"input": "1 0 0\n0 1 0",
"output": "YES"
},
{
"input": "1 0 0\n1 1 0",
"output": "YES"
},
{
"input": "1 0 0\n0 0 1",
"output": "YES"
},
{
"input": "1 0 0\n1 0 1",
"output": "YES"
},
{
"input": "1 0 0\n0 1 1",
"output": "NO"
},
{
"input": "1 0 0\n1 1 1",
"output": "YES"
},
{
"input": "0 1 0\n0 0 0",
"output": "YES"
},
{
"input": "0 1 0\n1 0 0",
"output": "YES"
},
{
"input": "0 1 0\n1 1 0",
"output": "YES"
},
{
"input": "0 1 0\n0 0 1",
"output": "YES"
},
{
"input": "0 1 0\n1 0 1",
"output": "NO"
},
{
"input": "0 1 0\n0 1 1",
"output": "YES"
},
{
"input": "0 1 0\n1 1 1",
"output": "YES"
},
{
"input": "1 1 0\n0 0 0",
"output": "YES"
},
{
"input": "1 1 0\n1 0 0",
"output": "YES"
},
{
"input": "1 1 0\n0 1 0",
"output": "YES"
},
{
"input": "1 1 0\n0 0 1",
"output": "NO"
},
{
"input": "1 1 0\n1 0 1",
"output": "YES"
},
{
"input": "1 1 0\n0 1 1",
"output": "YES"
},
{
"input": "1 1 0\n1 1 1",
"output": "YES"
},
{
"input": "0 0 1\n0 0 0",
"output": "YES"
},
{
"input": "0 0 1\n1 0 0",
"output": "YES"
},
{
"input": "0 0 1\n0 1 0",
"output": "YES"
},
{
"input": "0 0 1\n1 1 0",
"output": "NO"
},
{
"input": "0 0 1\n1 0 1",
"output": "YES"
},
{
"input": "0 0 1\n0 1 1",
"output": "YES"
},
{
"input": "0 0 1\n1 1 1",
"output": "YES"
},
{
"input": "1 0 1\n0 0 0",
"output": "YES"
},
{
"input": "1 0 1\n1 0 0",
"output": "YES"
},
{
"input": "1 0 1\n0 1 0",
"output": "NO"
},
{
"input": "1 0 1\n1 1 0",
"output": "YES"
},
{
"input": "1 0 1\n0 0 1",
"output": "YES"
},
{
"input": "1 0 1\n0 1 1",
"output": "YES"
},
{
"input": "1 0 1\n1 1 1",
"output": "YES"
},
{
"input": "0 1 1\n0 0 0",
"output": "YES"
},
{
"input": "0 1 1\n1 0 0",
"output": "NO"
},
{
"input": "0 1 1\n0 1 0",
"output": "YES"
},
{
"input": "0 1 1\n1 1 0",
"output": "YES"
},
{
"input": "0 1 1\n0 0 1",
"output": "YES"
},
{
"input": "0 1 1\n1 0 1",
"output": "YES"
},
{
"input": "0 1 1\n1 1 1",
"output": "YES"
},
{
"input": "1 1 1\n0 0 0",
"output": "NO"
},
{
"input": "1 1 1\n1 0 0",
"output": "YES"
},
{
"input": "1 1 1\n0 1 0",
"output": "YES"
},
{
"input": "1 1 1\n1 1 0",
"output": "YES"
},
{
"input": "1 1 1\n0 0 1",
"output": "YES"
},
{
"input": "1 1 1\n1 0 1",
"output": "YES"
},
{
"input": "1 1 1\n0 1 1",
"output": "YES"
}
] | 1,566,260,487
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 2
| 248
| 0
|
readints = lambda: map(int, input().split(' '))
a,b,c=readints()
x,y,z=readints()
if a!=x and b!=y and c!=c:
print('NO')
else:
print('YES')
|
Title: Cubical Planet
Time Limit: 2 seconds
Memory Limit: 64 megabytes
Problem Description:
You can find anything whatsoever in our Galaxy! A cubical planet goes round an icosahedral star. Let us introduce a system of axes so that the edges of the cubical planet are parallel to the coordinate axes and two opposite vertices lay in the points (0,<=0,<=0) and (1,<=1,<=1). Two flies live on the planet. At the moment they are sitting on two different vertices of the cubical planet. Your task is to determine whether they see each other or not. The flies see each other when the vertices they occupy lie on the same face of the cube.
Input Specification:
The first line contains three space-separated integers (0 or 1) — the coordinates of the first fly, the second line analogously contains the coordinates of the second fly.
Output Specification:
Output "YES" (without quotes) if the flies see each other. Otherwise, output "NO".
Demo Input:
['0 0 0\n0 1 0\n', '1 1 0\n0 1 0\n', '0 0 0\n1 1 1\n']
Demo Output:
['YES\n', 'YES\n', 'NO\n']
Note:
none
|
```python
readints = lambda: map(int, input().split(' '))
a,b,c=readints()
x,y,z=readints()
if a!=x and b!=y and c!=c:
print('NO')
else:
print('YES')
```
| 0
|
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 — 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* — 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"
}
] | 1,523,113,157
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 30
| 184
| 7,065,600
|
s = input()
u = d = 0
for i in s:
if i == i.upper():
u += 1
else:
d += 1
print(s.upper() if u > d else s.lower())
|
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 — 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* — 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 = d = 0
for i in s:
if i == i.upper():
u += 1
else:
d += 1
print(s.upper() if u > d else s.lower())
```
| 3.940839
|
285
|
C
|
Building Permutation
|
PROGRAMMING
| 1,200
|
[
"greedy",
"implementation",
"sortings"
] | null | null |
Permutation *p* is an ordered set of integers *p*1,<=<=*p*2,<=<=...,<=<=*p**n*, consisting of *n* distinct positive integers, each of them doesn't exceed *n*. We'll denote the *i*-th element of permutation *p* as *p**i*. We'll call number *n* the size or the length of permutation *p*1,<=<=*p*2,<=<=...,<=<=*p**n*.
You have a sequence of integers *a*1,<=*a*2,<=...,<=*a**n*. In one move, you are allowed to decrease or increase any number by one. Count the minimum number of moves, needed to build a permutation from this sequence.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=3·105) — the size of the sought permutation. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=109<=≤<=*a**i*<=≤<=109).
|
Print a single number — the minimum number of moves.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
|
[
"2\n3 0\n",
"3\n-1 -1 2\n"
] |
[
"2\n",
"6\n"
] |
In the first sample you should decrease the first number by one and then increase the second number by one. The resulting permutation is (2, 1).
In the second sample you need 6 moves to build permutation (1, 3, 2).
| 1,500
|
[
{
"input": "2\n3 0",
"output": "2"
},
{
"input": "3\n-1 -1 2",
"output": "6"
},
{
"input": "5\n-3 5 -3 3 3",
"output": "10"
},
{
"input": "10\n9 6 -2 4 1 1 1 9 6 2",
"output": "18"
},
{
"input": "9\n2 0 0 6 5 4 1 9 3",
"output": "15"
},
{
"input": "100\n-77 57 -95 -23 53 -28 82 -83 38 -73 85 28 25 6 -43 4 -10 -30 -9 -92 14 34 -93 61 36 -100 90 -68 28 16 100 -3 97 30 36 -55 62 -62 53 74 -50 -23 67 11 22 -30 -19 83 7 84 43 90 -65 -75 -15 97 90 15 66 2 13 -91 91 -44 46 51 51 -58 95 77 20 30 76 79 91 60 76 2 82 42 -93 94 -57 88 65 -95 -66 100 -9 33 -67 54 -99 97 53 13 54 66 60 -48",
"output": "3459"
}
] | 1,486,359,916
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 33
| 996
| 29,286,400
|
n = int(input())
l = list(map(int, input().split()))
l.sort()
res = 0
for i in range(1, n + 1):
res += abs(i - l[i-1])
print(res)
|
Title: Building Permutation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Permutation *p* is an ordered set of integers *p*1,<=<=*p*2,<=<=...,<=<=*p**n*, consisting of *n* distinct positive integers, each of them doesn't exceed *n*. We'll denote the *i*-th element of permutation *p* as *p**i*. We'll call number *n* the size or the length of permutation *p*1,<=<=*p*2,<=<=...,<=<=*p**n*.
You have a sequence of integers *a*1,<=*a*2,<=...,<=*a**n*. In one move, you are allowed to decrease or increase any number by one. Count the minimum number of moves, needed to build a permutation from this sequence.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=3·105) — the size of the sought permutation. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=109<=≤<=*a**i*<=≤<=109).
Output Specification:
Print a single number — the minimum number of moves.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
Demo Input:
['2\n3 0\n', '3\n-1 -1 2\n']
Demo Output:
['2\n', '6\n']
Note:
In the first sample you should decrease the first number by one and then increase the second number by one. The resulting permutation is (2, 1).
In the second sample you need 6 moves to build permutation (1, 3, 2).
|
```python
n = int(input())
l = list(map(int, input().split()))
l.sort()
res = 0
for i in range(1, n + 1):
res += abs(i - l[i-1])
print(res)
```
| 3
|
|
200
|
B
|
Drinks
|
PROGRAMMING
| 800
|
[
"implementation",
"math"
] | null | null |
Little Vasya loves orange juice very much. That's why any food and drink in his kitchen necessarily contains orange juice. There are *n* drinks in his fridge, the volume fraction of orange juice in the *i*-th drink equals *p**i* percent.
One day Vasya decided to make himself an orange cocktail. He took equal proportions of each of the *n* drinks and mixed them. Then he wondered, how much orange juice the cocktail has.
Find the volume fraction of orange juice in the final drink.
|
The first input line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of orange-containing drinks in Vasya's fridge. The second line contains *n* integers *p**i* (0<=≤<=*p**i*<=≤<=100) — the volume fraction of orange juice in the *i*-th drink, in percent. The numbers are separated by a space.
|
Print the volume fraction in percent of orange juice in Vasya's cocktail. The answer will be considered correct if the absolute or relative error does not exceed 10<=<=-<=4.
|
[
"3\n50 50 100\n",
"4\n0 25 50 75\n"
] |
[
"66.666666666667\n",
"37.500000000000\n"
] |
Note to the first sample: let's assume that Vasya takes *x* milliliters of each drink from the fridge. Then the volume of pure juice in the cocktail will equal <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/c1fac6e64d3a8ee6a5ac138cbe51e60039b22473.png" style="max-width: 100.0%;max-height: 100.0%;"/> milliliters. The total cocktail's volume equals 3·*x* milliliters, so the volume fraction of the juice in the cocktail equals <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/ceb0664e55a1f9f5fa1243ec74680a4665a4d58d.png" style="max-width: 100.0%;max-height: 100.0%;"/>, that is, 66.(6) percent.
| 500
|
[
{
"input": "3\n50 50 100",
"output": "66.666666666667"
},
{
"input": "4\n0 25 50 75",
"output": "37.500000000000"
},
{
"input": "3\n0 1 8",
"output": "3.000000000000"
},
{
"input": "5\n96 89 93 95 70",
"output": "88.600000000000"
},
{
"input": "7\n62 41 78 4 38 39 75",
"output": "48.142857142857"
},
{
"input": "13\n2 22 7 0 1 17 3 17 11 2 21 26 22",
"output": "11.615384615385"
},
{
"input": "21\n5 4 11 7 0 5 45 21 0 14 51 6 0 16 10 19 8 9 7 12 18",
"output": "12.761904761905"
},
{
"input": "26\n95 70 93 74 94 70 91 70 39 79 80 57 87 75 37 93 48 67 51 90 85 26 23 64 66 84",
"output": "69.538461538462"
},
{
"input": "29\n84 99 72 96 83 92 95 98 97 93 76 84 99 93 81 76 93 99 99 100 95 100 96 95 97 100 71 98 94",
"output": "91.551724137931"
},
{
"input": "33\n100 99 100 100 99 99 99 100 100 100 99 99 99 100 100 100 100 99 100 99 100 100 97 100 100 100 100 100 100 100 98 98 100",
"output": "99.515151515152"
},
{
"input": "34\n14 9 10 5 4 26 18 23 0 1 0 20 18 15 2 2 3 5 14 1 9 4 2 15 7 1 7 19 10 0 0 11 0 2",
"output": "8.147058823529"
},
{
"input": "38\n99 98 100 100 99 92 99 99 98 84 88 94 86 99 93 100 98 99 65 98 85 84 64 97 96 89 79 96 91 84 99 93 72 96 94 97 96 93",
"output": "91.921052631579"
},
{
"input": "52\n100 94 99 98 99 99 99 95 97 97 98 100 100 98 97 100 98 90 100 99 97 94 90 98 100 100 90 99 100 95 98 95 94 85 97 94 96 94 99 99 99 98 100 100 94 99 99 100 98 87 100 100",
"output": "97.019230769231"
},
{
"input": "58\n10 70 12 89 1 82 100 53 40 100 21 69 92 91 67 66 99 77 25 48 8 63 93 39 46 79 82 14 44 42 1 79 0 69 56 73 67 17 59 4 65 80 20 60 77 52 3 61 16 76 33 18 46 100 28 59 9 6",
"output": "50.965517241379"
},
{
"input": "85\n7 8 1 16 0 15 1 7 0 11 15 6 2 12 2 8 9 8 2 0 3 7 15 7 1 8 5 7 2 26 0 3 11 1 8 10 31 0 7 6 1 8 1 0 9 14 4 8 7 16 9 1 0 16 10 9 6 1 1 4 2 7 4 5 4 1 20 6 16 16 1 1 10 17 8 12 14 19 3 8 1 7 10 23 10",
"output": "7.505882352941"
},
{
"input": "74\n5 3 0 7 13 10 12 10 18 5 0 18 2 13 7 17 2 7 5 2 40 19 0 2 2 3 0 45 4 20 0 4 2 8 1 19 3 9 17 1 15 0 16 1 9 4 0 9 32 2 6 18 11 18 1 15 16 12 7 19 5 3 9 28 26 8 3 10 33 29 4 13 28 6",
"output": "10.418918918919"
},
{
"input": "98\n42 9 21 11 9 11 22 12 52 20 10 6 56 9 26 27 1 29 29 14 38 17 41 21 7 45 15 5 29 4 51 20 6 8 34 17 13 53 30 45 0 10 16 41 4 5 6 4 14 2 31 6 0 11 13 3 3 43 13 36 51 0 7 16 28 23 8 36 30 22 8 54 21 45 39 4 50 15 1 30 17 8 18 10 2 20 16 50 6 68 15 6 38 7 28 8 29 41",
"output": "20.928571428571"
},
{
"input": "99\n60 65 40 63 57 44 30 84 3 10 39 53 40 45 72 20 76 11 61 32 4 26 97 55 14 57 86 96 34 69 52 22 26 79 31 4 21 35 82 47 81 28 72 70 93 84 40 4 69 39 83 58 30 7 32 73 74 12 92 23 61 88 9 58 70 32 75 40 63 71 46 55 39 36 14 97 32 16 95 41 28 20 85 40 5 50 50 50 75 6 10 64 38 19 77 91 50 72 96",
"output": "49.191919191919"
},
{
"input": "99\n100 88 40 30 81 80 91 98 69 73 88 96 79 58 14 100 87 84 52 91 83 88 72 83 99 35 54 80 46 79 52 72 85 32 99 39 79 79 45 83 88 50 75 75 50 59 65 75 97 63 92 58 89 46 93 80 89 33 69 86 99 99 66 85 72 74 79 98 85 95 46 63 77 97 49 81 89 39 70 76 68 91 90 56 31 93 51 87 73 95 74 69 87 95 57 68 49 95 92",
"output": "73.484848484848"
},
{
"input": "100\n18 15 17 0 3 3 0 4 1 8 2 22 7 21 5 0 0 8 3 16 1 0 2 9 9 3 10 8 17 20 5 4 8 12 2 3 1 1 3 2 23 0 1 0 5 7 4 0 1 3 3 4 25 2 2 14 8 4 9 3 0 11 0 3 12 3 14 16 7 7 14 1 17 9 0 35 42 12 3 1 25 9 3 8 5 3 2 8 22 14 11 6 3 9 6 8 7 7 4 6",
"output": "7.640000000000"
},
{
"input": "100\n88 77 65 87 100 63 91 96 92 89 77 95 76 80 84 83 100 71 85 98 26 54 74 78 69 59 96 86 88 91 95 26 52 88 64 70 84 81 76 84 94 82 100 66 97 98 43 94 59 94 100 80 98 73 69 83 94 70 74 79 91 31 62 88 69 55 62 97 40 64 62 83 87 85 50 90 69 72 67 49 100 51 69 96 81 90 83 91 86 34 79 69 100 66 97 98 47 97 74 100",
"output": "77.660000000000"
},
{
"input": "100\n91 92 90 91 98 84 85 96 83 98 99 87 94 70 87 75 86 90 89 88 82 83 91 94 88 86 90 99 100 98 97 75 95 99 95 100 91 92 76 93 95 97 88 93 95 81 96 89 88 100 98 87 90 96 100 99 58 90 96 77 92 82 100 100 93 93 98 99 79 88 97 95 98 66 96 83 96 100 99 92 98 98 92 93 100 97 98 100 98 97 100 100 94 90 99 100 98 79 80 81",
"output": "91.480000000000"
},
{
"input": "1\n0",
"output": "0.000000000000"
},
{
"input": "1\n100",
"output": "100.000000000000"
},
{
"input": "1\n78",
"output": "78.000000000000"
},
{
"input": "2\n0 100",
"output": "50.000000000000"
},
{
"input": "2\n100 100",
"output": "100.000000000000"
},
{
"input": "5\n0 0 0 0 1",
"output": "0.200000000000"
},
{
"input": "100\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 99",
"output": "99.990000000000"
},
{
"input": "100\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1",
"output": "0.010000000000"
},
{
"input": "5\n100 100 100 100 100",
"output": "100.000000000000"
}
] | 1,689,605,178
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 31
| 184
| 0
|
import sys
input = sys.stdin.readline
print = sys.stdout.write
n = int(input())
s = sum(list(map(int,input().split())))
ans = s / n
print("%f" %ans)
|
Title: Drinks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Vasya loves orange juice very much. That's why any food and drink in his kitchen necessarily contains orange juice. There are *n* drinks in his fridge, the volume fraction of orange juice in the *i*-th drink equals *p**i* percent.
One day Vasya decided to make himself an orange cocktail. He took equal proportions of each of the *n* drinks and mixed them. Then he wondered, how much orange juice the cocktail has.
Find the volume fraction of orange juice in the final drink.
Input Specification:
The first input line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of orange-containing drinks in Vasya's fridge. The second line contains *n* integers *p**i* (0<=≤<=*p**i*<=≤<=100) — the volume fraction of orange juice in the *i*-th drink, in percent. The numbers are separated by a space.
Output Specification:
Print the volume fraction in percent of orange juice in Vasya's cocktail. The answer will be considered correct if the absolute or relative error does not exceed 10<=<=-<=4.
Demo Input:
['3\n50 50 100\n', '4\n0 25 50 75\n']
Demo Output:
['66.666666666667\n', '37.500000000000\n']
Note:
Note to the first sample: let's assume that Vasya takes *x* milliliters of each drink from the fridge. Then the volume of pure juice in the cocktail will equal <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/c1fac6e64d3a8ee6a5ac138cbe51e60039b22473.png" style="max-width: 100.0%;max-height: 100.0%;"/> milliliters. The total cocktail's volume equals 3·*x* milliliters, so the volume fraction of the juice in the cocktail equals <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/ceb0664e55a1f9f5fa1243ec74680a4665a4d58d.png" style="max-width: 100.0%;max-height: 100.0%;"/>, that is, 66.(6) percent.
|
```python
import sys
input = sys.stdin.readline
print = sys.stdout.write
n = int(input())
s = sum(list(map(int,input().split())))
ans = s / n
print("%f" %ans)
```
| 3
|
|
296
|
A
|
Yaroslav and Permutations
|
PROGRAMMING
| 1,100
|
[
"greedy",
"math"
] | null | null |
Yaroslav has an array that consists of *n* integers. In one second Yaroslav can swap two neighboring array elements. Now Yaroslav is wondering if he can obtain an array where any two neighboring elements would be distinct in a finite time.
Help Yaroslav.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of elements in the array. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000) — the array elements.
|
In the single line print "YES" (without the quotes) if Yaroslav can obtain the array he needs, and "NO" (without the quotes) otherwise.
|
[
"1\n1\n",
"3\n1 1 2\n",
"4\n7 7 7 7\n"
] |
[
"YES\n",
"YES\n",
"NO\n"
] |
In the first sample the initial array fits well.
In the second sample Yaroslav can get array: 1, 2, 1. He can swap the last and the second last elements to obtain it.
In the third sample Yarosav can't get the array he needs.
| 500
|
[
{
"input": "1\n1",
"output": "YES"
},
{
"input": "3\n1 1 2",
"output": "YES"
},
{
"input": "4\n7 7 7 7",
"output": "NO"
},
{
"input": "4\n479 170 465 146",
"output": "YES"
},
{
"input": "5\n996 437 605 996 293",
"output": "YES"
},
{
"input": "6\n727 539 896 668 36 896",
"output": "YES"
},
{
"input": "7\n674 712 674 674 674 674 674",
"output": "NO"
},
{
"input": "8\n742 742 742 742 742 289 742 742",
"output": "NO"
},
{
"input": "9\n730 351 806 806 806 630 85 757 967",
"output": "YES"
},
{
"input": "10\n324 539 83 440 834 640 440 440 440 440",
"output": "YES"
},
{
"input": "7\n925 830 925 98 987 162 356",
"output": "YES"
},
{
"input": "68\n575 32 53 351 151 942 725 967 431 108 192 8 338 458 288 754 384 946 910 210 759 222 589 423 947 507 31 414 169 901 592 763 656 411 360 625 538 549 484 596 42 603 351 292 837 375 21 597 22 349 200 669 485 282 735 54 1000 419 939 901 789 128 468 729 894 649 484 808",
"output": "YES"
},
{
"input": "22\n618 814 515 310 617 936 452 601 250 520 557 799 304 225 9 845 610 990 703 196 486 94",
"output": "YES"
},
{
"input": "44\n459 581 449 449 449 449 449 449 449 623 449 449 449 449 449 449 449 449 889 449 203 273 329 449 449 449 449 449 449 845 882 323 22 449 449 893 449 449 449 449 449 870 449 402",
"output": "NO"
},
{
"input": "90\n424 3 586 183 286 89 427 618 758 833 933 170 155 722 190 977 330 369 693 426 556 435 550 442 513 146 61 719 754 140 424 280 997 688 530 550 438 867 950 194 196 298 417 287 106 489 283 456 735 115 702 317 672 787 264 314 356 186 54 913 809 833 946 314 757 322 559 647 983 482 145 197 223 130 162 536 451 174 467 45 660 293 440 254 25 155 511 746 650 187",
"output": "YES"
},
{
"input": "14\n959 203 478 315 788 788 373 834 488 519 774 764 193 103",
"output": "YES"
},
{
"input": "81\n544 528 528 528 528 4 506 528 32 528 528 528 528 528 528 528 528 975 528 528 528 528 528 528 528 528 528 528 528 528 528 20 528 528 528 528 528 528 528 528 852 528 528 120 528 528 61 11 528 528 528 228 528 165 883 528 488 475 628 528 528 528 528 528 528 597 528 528 528 528 528 528 528 528 528 528 528 412 528 521 925",
"output": "NO"
},
{
"input": "89\n354 356 352 355 355 355 352 354 354 352 355 356 355 352 354 356 354 355 355 354 353 352 352 355 355 356 352 352 353 356 352 353 354 352 355 352 353 353 353 354 353 354 354 353 356 353 353 354 354 354 354 353 352 353 355 356 356 352 356 354 353 352 355 354 356 356 356 354 354 356 354 355 354 355 353 352 354 355 352 355 355 354 356 353 353 352 356 352 353",
"output": "YES"
},
{
"input": "71\n284 284 285 285 285 284 285 284 284 285 284 285 284 284 285 284 285 285 285 285 284 284 285 285 284 284 284 285 284 285 284 285 285 284 284 284 285 284 284 285 285 285 284 284 285 284 285 285 284 285 285 284 285 284 284 284 285 285 284 285 284 285 285 285 285 284 284 285 285 284 285",
"output": "NO"
},
{
"input": "28\n602 216 214 825 814 760 814 28 76 814 814 288 814 814 222 707 11 490 814 543 914 705 814 751 976 814 814 99",
"output": "YES"
},
{
"input": "48\n546 547 914 263 986 945 914 914 509 871 324 914 153 571 914 914 914 528 970 566 544 914 914 914 410 914 914 589 609 222 914 889 691 844 621 68 914 36 914 39 630 749 914 258 945 914 727 26",
"output": "YES"
},
{
"input": "56\n516 76 516 197 516 427 174 516 706 813 94 37 516 815 516 516 937 483 16 516 842 516 638 691 516 635 516 516 453 263 516 516 635 257 125 214 29 81 516 51 362 516 677 516 903 516 949 654 221 924 516 879 516 516 972 516",
"output": "YES"
},
{
"input": "46\n314 723 314 314 314 235 314 314 314 314 270 314 59 972 314 216 816 40 314 314 314 314 314 314 314 381 314 314 314 314 314 314 314 789 314 957 114 942 314 314 29 314 314 72 314 314",
"output": "NO"
},
{
"input": "72\n169 169 169 599 694 81 250 529 865 406 817 169 667 169 965 169 169 663 65 169 903 169 942 763 169 807 169 603 169 169 13 169 169 810 169 291 169 169 169 169 169 169 169 713 169 440 169 169 169 169 169 480 169 169 867 169 169 169 169 169 169 169 169 393 169 169 459 169 99 169 601 800",
"output": "NO"
},
{
"input": "100\n317 316 317 316 317 316 317 316 317 316 316 317 317 316 317 316 316 316 317 316 317 317 316 317 316 316 316 316 316 316 317 316 317 317 317 317 317 317 316 316 316 317 316 317 316 317 316 317 317 316 317 316 317 317 316 317 316 317 316 317 316 316 316 317 317 317 317 317 316 317 317 316 316 316 316 317 317 316 317 316 316 316 316 316 316 317 316 316 317 317 317 317 317 317 317 317 317 316 316 317",
"output": "NO"
},
{
"input": "100\n510 510 510 162 969 32 510 511 510 510 911 183 496 875 903 461 510 510 123 578 510 510 510 510 510 755 510 673 510 510 763 510 510 909 510 435 487 959 807 510 368 788 557 448 284 332 510 949 510 510 777 112 857 926 487 510 510 510 678 510 510 197 829 427 698 704 409 509 510 238 314 851 510 651 510 455 682 510 714 635 973 510 443 878 510 510 510 591 510 24 596 510 43 183 510 510 671 652 214 784",
"output": "YES"
},
{
"input": "100\n476 477 474 476 476 475 473 476 474 475 473 477 476 476 474 476 474 475 476 477 473 473 473 474 474 476 473 473 476 476 475 476 473 474 473 473 477 475 475 475 476 475 477 477 477 476 475 475 475 473 476 477 475 476 477 473 474 477 473 475 476 476 474 477 476 474 473 477 473 475 477 473 476 474 477 473 475 477 473 476 476 475 476 475 474 473 477 473 475 473 477 473 473 474 475 473 477 476 477 474",
"output": "YES"
},
{
"input": "100\n498 498 498 498 498 499 498 499 499 499 498 498 498 498 499 498 499 499 498 499 498 498 498 499 499 499 498 498 499 499 498 498 498 499 498 499 498 498 498 499 498 499 498 498 498 498 499 498 498 499 498 498 499 498 499 499 498 499 499 499 498 498 498 498 499 498 499 498 499 499 499 499 498 498 499 499 498 499 499 498 498 499 499 498 498 499 499 499 498 498 499 498 498 498 499 499 499 498 498 499",
"output": "NO"
},
{
"input": "100\n858 53 816 816 816 816 816 816 816 181 816 816 816 816 579 879 816 948 171 816 816 150 866 816 816 816 897 816 816 816 816 816 816 706 816 539 816 816 816 816 816 816 423 487 816 615 254 816 816 816 816 83 816 816 816 816 816 816 816 816 816 816 816 136 775 999 816 816 816 644 816 816 816 816 927 816 802 816 856 816 816 816 816 816 816 816 816 816 816 700 816 816 816 816 982 477 816 891 806 816",
"output": "NO"
},
{
"input": "100\n167 169 169 167 169 169 167 167 167 167 168 166 170 170 169 170 170 170 169 168 166 167 170 169 167 169 168 169 166 170 166 167 170 166 166 167 169 166 166 169 166 167 168 168 170 167 168 166 168 170 167 168 167 169 169 166 168 167 170 168 167 169 168 169 166 168 168 169 169 166 170 168 167 169 170 168 167 169 168 167 168 168 166 169 170 170 166 166 167 170 167 168 167 167 169 169 166 166 169 167",
"output": "YES"
},
{
"input": "100\n1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000",
"output": "NO"
},
{
"input": "99\n1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000",
"output": "NO"
},
{
"input": "100\n1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 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": "YES"
},
{
"input": "99\n1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 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": "YES"
},
{
"input": "2\n1 1",
"output": "NO"
},
{
"input": "1\n1000",
"output": "YES"
},
{
"input": "12\n2 2 4 4 4 4 6 6 6 6 6 6",
"output": "YES"
}
] | 1,615,451,175
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 1
| 92
| 409,600
|
n = int(input())
arr = list(map(int, input().split()))
if len(arr) < 2:
print("YES")
else:
from collections import Counter
d = Counter(arr)
if d.most_common(1)[0][1] > len(arr)//2:
print("NO")
else:
print("YES")
|
Title: Yaroslav and Permutations
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Yaroslav has an array that consists of *n* integers. In one second Yaroslav can swap two neighboring array elements. Now Yaroslav is wondering if he can obtain an array where any two neighboring elements would be distinct in a finite time.
Help Yaroslav.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of elements in the array. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000) — the array elements.
Output Specification:
In the single line print "YES" (without the quotes) if Yaroslav can obtain the array he needs, and "NO" (without the quotes) otherwise.
Demo Input:
['1\n1\n', '3\n1 1 2\n', '4\n7 7 7 7\n']
Demo Output:
['YES\n', 'YES\n', 'NO\n']
Note:
In the first sample the initial array fits well.
In the second sample Yaroslav can get array: 1, 2, 1. He can swap the last and the second last elements to obtain it.
In the third sample Yarosav can't get the array he needs.
|
```python
n = int(input())
arr = list(map(int, input().split()))
if len(arr) < 2:
print("YES")
else:
from collections import Counter
d = Counter(arr)
if d.most_common(1)[0][1] > len(arr)//2:
print("NO")
else:
print("YES")
```
| 0
|
|
492
|
A
|
Vanya and Cubes
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Vanya got *n* cubes. He decided to build a pyramid from them. Vanya wants to build the pyramid as follows: the top level of the pyramid must consist of 1 cube, the second level must consist of 1<=+<=2<==<=3 cubes, the third level must have 1<=+<=2<=+<=3<==<=6 cubes, and so on. Thus, the *i*-th level of the pyramid must have 1<=+<=2<=+<=...<=+<=(*i*<=-<=1)<=+<=*i* cubes.
Vanya wants to know what is the maximum height of the pyramid that he can make using the given cubes.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=104) — the number of cubes given to Vanya.
|
Print the maximum possible height of the pyramid in the single line.
|
[
"1\n",
"25\n"
] |
[
"1\n",
"4\n"
] |
Illustration to the second sample:
| 500
|
[
{
"input": "1",
"output": "1"
},
{
"input": "25",
"output": "4"
},
{
"input": "2",
"output": "1"
},
{
"input": "4115",
"output": "28"
},
{
"input": "9894",
"output": "38"
},
{
"input": "7969",
"output": "35"
},
{
"input": "6560",
"output": "33"
},
{
"input": "4",
"output": "2"
},
{
"input": "3",
"output": "1"
},
{
"input": "5",
"output": "2"
},
{
"input": "19",
"output": "3"
},
{
"input": "20",
"output": "4"
},
{
"input": "9880",
"output": "38"
},
{
"input": "9879",
"output": "37"
},
{
"input": "7770",
"output": "35"
},
{
"input": "7769",
"output": "34"
},
{
"input": "2925",
"output": "25"
},
{
"input": "220",
"output": "10"
},
{
"input": "219",
"output": "9"
},
{
"input": "3046",
"output": "25"
},
{
"input": "7590",
"output": "34"
},
{
"input": "1014",
"output": "17"
},
{
"input": "7142",
"output": "34"
},
{
"input": "9999",
"output": "38"
},
{
"input": "10000",
"output": "38"
}
] | 1,696,255,049
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 25
| 93
| 0
|
#python 3.7.1
cube = int(input())
Sum = 0
level=1
all_need=0
while( all_need < cube):
Sum +=level
level +=1
all_need += Sum
if(all_need == cube):
print(level-1)
else:
print(level-2)
|
Title: Vanya and Cubes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vanya got *n* cubes. He decided to build a pyramid from them. Vanya wants to build the pyramid as follows: the top level of the pyramid must consist of 1 cube, the second level must consist of 1<=+<=2<==<=3 cubes, the third level must have 1<=+<=2<=+<=3<==<=6 cubes, and so on. Thus, the *i*-th level of the pyramid must have 1<=+<=2<=+<=...<=+<=(*i*<=-<=1)<=+<=*i* cubes.
Vanya wants to know what is the maximum height of the pyramid that he can make using the given cubes.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=104) — the number of cubes given to Vanya.
Output Specification:
Print the maximum possible height of the pyramid in the single line.
Demo Input:
['1\n', '25\n']
Demo Output:
['1\n', '4\n']
Note:
Illustration to the second sample:
|
```python
#python 3.7.1
cube = int(input())
Sum = 0
level=1
all_need=0
while( all_need < cube):
Sum +=level
level +=1
all_need += Sum
if(all_need == cube):
print(level-1)
else:
print(level-2)
```
| 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 case — 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<=<=≤<=<=|*s*|<=<=≤<=<=105) — 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"
}
] | 1,485,807,057
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 10
| 62
| 4,608,000
|
spisok = [str(x) for x in input()]
number_B = spisok.count('B')
number_b = spisok.count('b')
number_l = spisok.count('l')
number_a = spisok.count('a')
number_s = spisok.count('s')
number_u = spisok.count('u')
number_r = spisok.count('r')
letters1 = [number_B, number_l, number_b, number_s, number_r]
letters2 = [number_u, number_a]
min_letters1 = min(letters1)
min_letters2 = min(letters2)
if min_letters1 < min_letters2:
if (min_letters2 > 1) and (min_letters1 > 0):
print(min_letters1)
elif min_letters2 == min_letters1:
print(min_letters2//2)
else:
print(0)
|
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 case — 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<=<=≤<=<=|*s*|<=<=≤<=<=105) — 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
spisok = [str(x) for x in input()]
number_B = spisok.count('B')
number_b = spisok.count('b')
number_l = spisok.count('l')
number_a = spisok.count('a')
number_s = spisok.count('s')
number_u = spisok.count('u')
number_r = spisok.count('r')
letters1 = [number_B, number_l, number_b, number_s, number_r]
letters2 = [number_u, number_a]
min_letters1 = min(letters1)
min_letters2 = min(letters2)
if min_letters1 < min_letters2:
if (min_letters2 > 1) and (min_letters1 > 0):
print(min_letters1)
elif min_letters2 == min_letters1:
print(min_letters2//2)
else:
print(0)
```
| 0
|
|
992
|
A
|
Nastya and an Array
|
PROGRAMMING
| 800
|
[
"implementation",
"sortings"
] | null | null |
Nastya owns too many arrays now, so she wants to delete the least important of them. However, she discovered that this array is magic! Nastya now knows that the array has the following properties:
- In one second we can add an arbitrary (possibly negative) integer to all elements of the array that are not equal to zero. - When all elements of the array become equal to zero, the array explodes.
Nastya is always busy, so she wants to explode the array as fast as possible. Compute the minimum time in which the array can be exploded.
|
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105) — the size of the array.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=105<=≤<=*a**i*<=≤<=105) — the elements of the array.
|
Print a single integer — the minimum number of seconds needed to make all elements of the array equal to zero.
|
[
"5\n1 1 1 1 1\n",
"3\n2 0 -1\n",
"4\n5 -6 -5 1\n"
] |
[
"1\n",
"2\n",
"4\n"
] |
In the first example you can add - 1 to all non-zero elements in one second and make them equal to zero.
In the second example you can add - 2 on the first second, then the array becomes equal to [0, 0, - 3]. On the second second you can add 3 to the third (the only non-zero) element.
| 500
|
[
{
"input": "5\n1 1 1 1 1",
"output": "1"
},
{
"input": "3\n2 0 -1",
"output": "2"
},
{
"input": "4\n5 -6 -5 1",
"output": "4"
},
{
"input": "1\n0",
"output": "0"
},
{
"input": "2\n21794 -79194",
"output": "2"
},
{
"input": "3\n-63526 95085 -5239",
"output": "3"
},
{
"input": "3\n0 53372 -20572",
"output": "2"
},
{
"input": "13\n-2075 -32242 27034 -37618 -96962 82203 64846 48249 -71761 28908 -21222 -61370 46899",
"output": "13"
},
{
"input": "5\n806 0 1308 1954 683",
"output": "4"
},
{
"input": "8\n-26 0 -249 -289 -126 -206 288 -11",
"output": "7"
},
{
"input": "10\n2 2 2 1 2 -1 0 2 -1 1",
"output": "3"
},
{
"input": "1\n8",
"output": "1"
},
{
"input": "3\n0 0 0",
"output": "0"
},
{
"input": "10\n1 2 3 4 5 6 7 8 9 10",
"output": "10"
},
{
"input": "5\n2 0 -1 0 0",
"output": "2"
},
{
"input": "2\n0 0",
"output": "0"
},
{
"input": "5\n0 0 0 0 0",
"output": "0"
},
{
"input": "2\n1 0",
"output": "1"
},
{
"input": "2\n-1 0",
"output": "1"
},
{
"input": "4\n0 0 0 0",
"output": "0"
},
{
"input": "8\n10 9 -1 0 0 3 2 3",
"output": "5"
},
{
"input": "5\n5 0 1 2 3",
"output": "4"
},
{
"input": "3\n1 1 0",
"output": "1"
},
{
"input": "1\n-1",
"output": "1"
},
{
"input": "5\n1 2 0 0 0",
"output": "2"
},
{
"input": "5\n1 0 0 0 0",
"output": "1"
},
{
"input": "5\n4 5 6 0 0",
"output": "3"
},
{
"input": "4\n-1 0 0 1",
"output": "2"
},
{
"input": "5\n3 0 0 4 5",
"output": "3"
},
{
"input": "3\n0 0 2",
"output": "1"
},
{
"input": "3\n1 0 0",
"output": "1"
},
{
"input": "4\n0 0 0 4",
"output": "1"
},
{
"input": "5\n-1 0 0 0 0",
"output": "1"
},
{
"input": "2\n0 1",
"output": "1"
},
{
"input": "3\n1 2 3",
"output": "3"
},
{
"input": "1\n5",
"output": "1"
},
{
"input": "10\n0 0 0 0 0 1 2 3 0 0",
"output": "3"
},
{
"input": "4\n0 1 2 3",
"output": "3"
},
{
"input": "3\n0 1 2",
"output": "2"
},
{
"input": "4\n2 0 0 -1",
"output": "2"
}
] | 1,619,733,052
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 79
| 171
| 9,420,800
|
import sys
def main():
s = set(sys.stdin.read().strip().split()[1:])
s.discard('0')
return len(s)
print(main())
|
Title: Nastya and an Array
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Nastya owns too many arrays now, so she wants to delete the least important of them. However, she discovered that this array is magic! Nastya now knows that the array has the following properties:
- In one second we can add an arbitrary (possibly negative) integer to all elements of the array that are not equal to zero. - When all elements of the array become equal to zero, the array explodes.
Nastya is always busy, so she wants to explode the array as fast as possible. Compute the minimum time in which the array can be exploded.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105) — the size of the array.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=105<=≤<=*a**i*<=≤<=105) — the elements of the array.
Output Specification:
Print a single integer — the minimum number of seconds needed to make all elements of the array equal to zero.
Demo Input:
['5\n1 1 1 1 1\n', '3\n2 0 -1\n', '4\n5 -6 -5 1\n']
Demo Output:
['1\n', '2\n', '4\n']
Note:
In the first example you can add - 1 to all non-zero elements in one second and make them equal to zero.
In the second example you can add - 2 on the first second, then the array becomes equal to [0, 0, - 3]. On the second second you can add 3 to the third (the only non-zero) element.
|
```python
import sys
def main():
s = set(sys.stdin.read().strip().split()[1:])
s.discard('0')
return len(s)
print(main())
```
| 3
|
|
875
|
A
|
Classroom Watch
|
PROGRAMMING
| 1,200
|
[
"brute force",
"math"
] | null | null |
Eighth-grader Vova is on duty today in the class. After classes, he went into the office to wash the board, and found on it the number *n*. He asked what is this number and the teacher of mathematics Inna Petrovna answered Vova that *n* is the answer to the arithmetic task for first-graders. In the textbook, a certain positive integer *x* was given. The task was to add *x* to the sum of the digits of the number *x* written in decimal numeral system.
Since the number *n* on the board was small, Vova quickly guessed which *x* could be in the textbook. Now he wants to get a program which will search for arbitrary values of the number *n* for all suitable values of *x* or determine that such *x* does not exist. Write such a program for Vova.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=109).
|
In the first line print one integer *k* — number of different values of *x* satisfying the condition.
In next *k* lines print these values in ascending order.
|
[
"21\n",
"20\n"
] |
[
"1\n15\n",
"0\n"
] |
In the first test case *x* = 15 there is only one variant: 15 + 1 + 5 = 21.
In the second test case there are no such *x*.
| 500
|
[
{
"input": "21",
"output": "1\n15"
},
{
"input": "20",
"output": "0"
},
{
"input": "1",
"output": "0"
},
{
"input": "2",
"output": "1\n1"
},
{
"input": "3",
"output": "0"
},
{
"input": "100000001",
"output": "2\n99999937\n100000000"
},
{
"input": "1000000000",
"output": "1\n999999932"
},
{
"input": "999999979",
"output": "2\n999999899\n999999908"
},
{
"input": "9",
"output": "0"
},
{
"input": "10",
"output": "1\n5"
},
{
"input": "11",
"output": "1\n10"
},
{
"input": "39",
"output": "1\n33"
},
{
"input": "66",
"output": "1\n60"
},
{
"input": "75",
"output": "0"
},
{
"input": "100",
"output": "1\n86"
},
{
"input": "101",
"output": "2\n91\n100"
},
{
"input": "2014",
"output": "2\n1988\n2006"
},
{
"input": "999999994",
"output": "0"
}
] | 1,559,205,556
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 5
| 109
| 0
|
def sum_digits(x):
res = 0
while x > 0:
res += x % 10
x //= 10
return res
n = int(input())
res = []
for i in range(1, 82):
if n - i + sum_digits(n-i) == n:
res.append(n-i)
print(len(res))
print(*res)
|
Title: Classroom Watch
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Eighth-grader Vova is on duty today in the class. After classes, he went into the office to wash the board, and found on it the number *n*. He asked what is this number and the teacher of mathematics Inna Petrovna answered Vova that *n* is the answer to the arithmetic task for first-graders. In the textbook, a certain positive integer *x* was given. The task was to add *x* to the sum of the digits of the number *x* written in decimal numeral system.
Since the number *n* on the board was small, Vova quickly guessed which *x* could be in the textbook. Now he wants to get a program which will search for arbitrary values of the number *n* for all suitable values of *x* or determine that such *x* does not exist. Write such a program for Vova.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=109).
Output Specification:
In the first line print one integer *k* — number of different values of *x* satisfying the condition.
In next *k* lines print these values in ascending order.
Demo Input:
['21\n', '20\n']
Demo Output:
['1\n15\n', '0\n']
Note:
In the first test case *x* = 15 there is only one variant: 15 + 1 + 5 = 21.
In the second test case there are no such *x*.
|
```python
def sum_digits(x):
res = 0
while x > 0:
res += x % 10
x //= 10
return res
n = int(input())
res = []
for i in range(1, 82):
if n - i + sum_digits(n-i) == n:
res.append(n-i)
print(len(res))
print(*res)
```
| 0
|
|
664
|
A
|
Complicated GCD
|
PROGRAMMING
| 800
|
[
"math",
"number theory"
] | null | null |
Greatest common divisor *GCD*(*a*,<=*b*) of two positive integers *a* and *b* is equal to the biggest integer *d* such that both integers *a* and *b* are divisible by *d*. There are many efficient algorithms to find greatest common divisor *GCD*(*a*,<=*b*), for example, Euclid algorithm.
Formally, find the biggest integer *d*, such that all integers *a*,<=*a*<=+<=1,<=*a*<=+<=2,<=...,<=*b* are divisible by *d*. To make the problem even more complicated we allow *a* and *b* to be up to googol, 10100 — such number do not fit even in 64-bit integer type!
|
The only line of the input contains two integers *a* and *b* (1<=≤<=*a*<=≤<=*b*<=≤<=10100).
|
Output one integer — greatest common divisor of all integers from *a* to *b* inclusive.
|
[
"1 2\n",
"61803398874989484820458683436563811772030917980576 61803398874989484820458683436563811772030917980576\n"
] |
[
"1\n",
"61803398874989484820458683436563811772030917980576\n"
] |
none
| 500
|
[
{
"input": "1 2",
"output": "1"
},
{
"input": "61803398874989484820458683436563811772030917980576 61803398874989484820458683436563811772030917980576",
"output": "61803398874989484820458683436563811772030917980576"
},
{
"input": "1 100",
"output": "1"
},
{
"input": "100 100000",
"output": "1"
},
{
"input": "12345 67890123456789123457",
"output": "1"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "2 2",
"output": "2"
},
{
"input": "8392739158839273915883927391588392739158839273915883927391588392739158839273915883927391588392739158 8392739158839273915883927391588392739158839273915883927391588392739158839273915883927391588392739158",
"output": "8392739158839273915883927391588392739158839273915883927391588392739158839273915883927391588392739158"
},
{
"input": "1 10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "1"
},
{
"input": "8328748239473982794239847237438782379810988324751 9328748239473982794239847237438782379810988324751",
"output": "1"
},
{
"input": "1029398958432734901284327523909481928483573793 1029398958432734901284327523909481928483573794",
"output": "1"
},
{
"input": "10000 1000000000",
"output": "1"
},
{
"input": "10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "11210171722243 65715435710585778347",
"output": "1"
},
{
"input": "2921881079263974825226940825843 767693191032295360887755303860323261471",
"output": "1"
},
{
"input": "8025352957265704896940312528736939363590612908210603 96027920417708260814607687034511406492969694925539085",
"output": "1"
},
{
"input": "23510978780782786207241069904470895053213996267165977112058175452757132930 210352653280909370107314249722987050753257161175393375412301228883856435481424",
"output": "1"
},
{
"input": "8150070767079366215626260746398623663859344142817267779361251788637547414925170226504788118262 49924902262298336032630839998470954964895251605110946547855439236151401194070172107435992986913614",
"output": "1"
},
{
"input": "15943150466658398903 15943150466658398903",
"output": "15943150466658398903"
},
{
"input": "410470228200245407491525399055972 410470228200245407491525399055972",
"output": "410470228200245407491525399055972"
},
{
"input": "51894705655711504622197349350106792045098781545973899451307 51894705655711504622197349350106792045098781545973899451307",
"output": "51894705655711504622197349350106792045098781545973899451307"
},
{
"input": "60353594589897438036015726222485085035927634677598681595162804007836722215668410 60353594589897438036015726222485085035927634677598681595162804007836722215668410",
"output": "60353594589897438036015726222485085035927634677598681595162804007836722215668410"
},
{
"input": "761733780145118977868180796896376577405349682060892737466239031663476251177476275459280340045369535 761733780145118977868180796896376577405349682060892737466239031663476251177476275459280340045369535",
"output": "761733780145118977868180796896376577405349682060892737466239031663476251177476275459280340045369535"
},
{
"input": "1 2000000000",
"output": "1"
},
{
"input": "13 1928834874",
"output": "1"
},
{
"input": "87 2938984237482934238",
"output": "1"
},
{
"input": "213 413",
"output": "1"
},
{
"input": "3 4",
"output": "1"
}
] | 1,563,041,870
| 470
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 3
| 124
| 0
|
import math
p=input().rstrip().split(' ')
print(math.gcd(int(p[0]),int(p[1])))
|
Title: Complicated GCD
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Greatest common divisor *GCD*(*a*,<=*b*) of two positive integers *a* and *b* is equal to the biggest integer *d* such that both integers *a* and *b* are divisible by *d*. There are many efficient algorithms to find greatest common divisor *GCD*(*a*,<=*b*), for example, Euclid algorithm.
Formally, find the biggest integer *d*, such that all integers *a*,<=*a*<=+<=1,<=*a*<=+<=2,<=...,<=*b* are divisible by *d*. To make the problem even more complicated we allow *a* and *b* to be up to googol, 10100 — such number do not fit even in 64-bit integer type!
Input Specification:
The only line of the input contains two integers *a* and *b* (1<=≤<=*a*<=≤<=*b*<=≤<=10100).
Output Specification:
Output one integer — greatest common divisor of all integers from *a* to *b* inclusive.
Demo Input:
['1 2\n', '61803398874989484820458683436563811772030917980576 61803398874989484820458683436563811772030917980576\n']
Demo Output:
['1\n', '61803398874989484820458683436563811772030917980576\n']
Note:
none
|
```python
import math
p=input().rstrip().split(' ')
print(math.gcd(int(p[0]),int(p[1])))
```
| 0
|
|
230
|
A
|
Dragons
|
PROGRAMMING
| 1,000
|
[
"greedy",
"sortings"
] | null | null |
Kirito is stuck on a level of the MMORPG he is playing now. To move on in the game, he's got to defeat all *n* dragons that live on this level. Kirito and the dragons have strength, which is represented by an integer. In the duel between two opponents the duel's outcome is determined by their strength. Initially, Kirito's strength equals *s*.
If Kirito starts duelling with the *i*-th (1<=≤<=*i*<=≤<=*n*) dragon and Kirito's strength is not greater than the dragon's strength *x**i*, then Kirito loses the duel and dies. But if Kirito's strength is greater than the dragon's strength, then he defeats the dragon and gets a bonus strength increase by *y**i*.
Kirito can fight the dragons in any order. Determine whether he can move on to the next level of the game, that is, defeat all dragons without a single loss.
|
The first line contains two space-separated integers *s* and *n* (1<=≤<=*s*<=≤<=104, 1<=≤<=*n*<=≤<=103). Then *n* lines follow: the *i*-th line contains space-separated integers *x**i* and *y**i* (1<=≤<=*x**i*<=≤<=104, 0<=≤<=*y**i*<=≤<=104) — the *i*-th dragon's strength and the bonus for defeating it.
|
On a single line print "YES" (without the quotes), if Kirito can move on to the next level and print "NO" (without the quotes), if he can't.
|
[
"2 2\n1 99\n100 0\n",
"10 1\n100 100\n"
] |
[
"YES\n",
"NO\n"
] |
In the first sample Kirito's strength initially equals 2. As the first dragon's strength is less than 2, Kirito can fight it and defeat it. After that he gets the bonus and his strength increases to 2 + 99 = 101. Now he can defeat the second dragon and move on to the next level.
In the second sample Kirito's strength is too small to defeat the only dragon and win.
| 500
|
[
{
"input": "2 2\n1 99\n100 0",
"output": "YES"
},
{
"input": "10 1\n100 100",
"output": "NO"
},
{
"input": "123 2\n78 10\n130 0",
"output": "YES"
},
{
"input": "999 2\n1010 10\n67 89",
"output": "YES"
},
{
"input": "2 5\n5 1\n2 1\n3 1\n1 1\n4 1",
"output": "YES"
},
{
"input": "2 2\n3 5\n1 2",
"output": "YES"
},
{
"input": "1 2\n1 0\n1 0",
"output": "NO"
},
{
"input": "5 10\n20 1\n4 3\n5 1\n100 1\n4 2\n101 1\n10 0\n10 2\n17 3\n12 84",
"output": "YES"
},
{
"input": "2 2\n1 98\n100 0",
"output": "NO"
},
{
"input": "2 2\n1 2\n3 5",
"output": "YES"
},
{
"input": "5 3\n13 20\n3 10\n15 5",
"output": "YES"
},
{
"input": "2 5\n1 1\n2 1\n3 1\n4 1\n5 1",
"output": "YES"
},
{
"input": "3 3\n1 1\n1 2\n4 0",
"output": "YES"
},
{
"input": "10 4\n20 1\n3 5\n2 4\n1 3",
"output": "YES"
},
{
"input": "10 1\n1 1",
"output": "YES"
},
{
"input": "4 1\n100 1000",
"output": "NO"
},
{
"input": "5 1\n6 7",
"output": "NO"
},
{
"input": "10 1\n10 10",
"output": "NO"
},
{
"input": "6 2\n496 0\n28 8128",
"output": "NO"
},
{
"input": "4 2\n2 1\n10 3",
"output": "NO"
},
{
"input": "11 2\n22 0\n33 0",
"output": "NO"
},
{
"input": "1 2\n100 1\n100 1",
"output": "NO"
},
{
"input": "10 3\n12 0\n13 0\n14 0",
"output": "NO"
},
{
"input": "50 3\n39 0\n38 0\n37 0",
"output": "YES"
},
{
"input": "14 3\n1 5\n1 6\n1 7",
"output": "YES"
},
{
"input": "1 3\n1 10\n1 11\n1 9",
"output": "NO"
},
{
"input": "10 10\n2 10\n3 10\n4 10\n2 20\n3 20\n3 20\n100 50\n100 30\n150 30\n200 10",
"output": "NO"
},
{
"input": "9983 34\n6626 5976\n4448 3568\n2794 2309\n3741 8806\n4754 129\n2780 9275\n5785 9243\n3915 6159\n2609 4331\n238 6756\n6987 3887\n3384 5711\n4349 5563\n1135 4483\n9151 1584\n1500 766\n1608 4440\n7768 5005\n7205 2360\n9088 2933\n3923 7814\n7538 9372\n7504 165\n5277 1636\n2061 4384\n7668 1422\n9582 2121\n5483 7967\n487 2944\n7432 5794\n8208 8970\n5747 3800\n4322 3920\n8261 9319",
"output": "YES"
},
{
"input": "1 10\n8 8\n54 3\n1 8\n26 3\n16 1\n29 9\n38 10\n57 8\n48 6\n17 9",
"output": "NO"
},
{
"input": "5 10\n7 0\n7 0\n10 0\n10 0\n7 2\n4 2\n9 0\n6 1\n7 0\n7 0",
"output": "NO"
},
{
"input": "2 3\n1 1\n1 10\n17 2",
"output": "NO"
},
{
"input": "100 5\n99 100\n199 1\n199 1\n199 1\n202 1",
"output": "YES"
},
{
"input": "1 1\n10000 1",
"output": "NO"
}
] | 1,697,838,621
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 54
| 92
| 0
|
s, n = map(int, input().split())
dragons = []
for _ in range(n):
x, y = map(int, input().split())
dragons.append((x, y))
dragons.sort() # Sort dragons based on their strength
for dragon in dragons:
if s > dragon[0]:
s += dragon[1]
else:
print("NO")
exit(0)
print("YES")
|
Title: Dragons
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kirito is stuck on a level of the MMORPG he is playing now. To move on in the game, he's got to defeat all *n* dragons that live on this level. Kirito and the dragons have strength, which is represented by an integer. In the duel between two opponents the duel's outcome is determined by their strength. Initially, Kirito's strength equals *s*.
If Kirito starts duelling with the *i*-th (1<=≤<=*i*<=≤<=*n*) dragon and Kirito's strength is not greater than the dragon's strength *x**i*, then Kirito loses the duel and dies. But if Kirito's strength is greater than the dragon's strength, then he defeats the dragon and gets a bonus strength increase by *y**i*.
Kirito can fight the dragons in any order. Determine whether he can move on to the next level of the game, that is, defeat all dragons without a single loss.
Input Specification:
The first line contains two space-separated integers *s* and *n* (1<=≤<=*s*<=≤<=104, 1<=≤<=*n*<=≤<=103). Then *n* lines follow: the *i*-th line contains space-separated integers *x**i* and *y**i* (1<=≤<=*x**i*<=≤<=104, 0<=≤<=*y**i*<=≤<=104) — the *i*-th dragon's strength and the bonus for defeating it.
Output Specification:
On a single line print "YES" (without the quotes), if Kirito can move on to the next level and print "NO" (without the quotes), if he can't.
Demo Input:
['2 2\n1 99\n100 0\n', '10 1\n100 100\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first sample Kirito's strength initially equals 2. As the first dragon's strength is less than 2, Kirito can fight it and defeat it. After that he gets the bonus and his strength increases to 2 + 99 = 101. Now he can defeat the second dragon and move on to the next level.
In the second sample Kirito's strength is too small to defeat the only dragon and win.
|
```python
s, n = map(int, input().split())
dragons = []
for _ in range(n):
x, y = map(int, input().split())
dragons.append((x, y))
dragons.sort() # Sort dragons based on their strength
for dragon in dragons:
if s > dragon[0]:
s += dragon[1]
else:
print("NO")
exit(0)
print("YES")
```
| 3
|
|
810
|
B
|
Summer sell-off
|
PROGRAMMING
| 1,300
|
[
"greedy",
"sortings"
] | null | null |
Summer holidays! Someone is going on trips, someone is visiting grandparents, but someone is trying to get a part-time job. This summer Noora decided that she wants to earn some money, and took a job in a shop as an assistant.
Shop, where Noora is working, has a plan on the following *n* days. For each day sales manager knows exactly, that in *i*-th day *k**i* products will be put up for sale and exactly *l**i* clients will come to the shop that day. Also, the manager is sure, that everyone, who comes to the shop, buys exactly one product or, if there aren't any left, leaves the shop without buying anything. Moreover, due to the short shelf-life of the products, manager established the following rule: if some part of the products left on the shelves at the end of the day, that products aren't kept on the next day and are sent to the dump.
For advertising purposes manager offered to start a sell-out in the shop. He asked Noora to choose any *f* days from *n* next for sell-outs. On each of *f* chosen days the number of products were put up for sale would be doubled. Thus, if on *i*-th day shop planned to put up for sale *k**i* products and Noora has chosen this day for sell-out, shelves of the shop would keep 2·*k**i* products. Consequently, there is an opportunity to sell two times more products on days of sell-out.
Noora's task is to choose *f* days to maximize total number of sold products. She asks you to help her with such a difficult problem.
|
The first line contains two integers *n* and *f* (1<=≤<=*n*<=≤<=105,<=0<=≤<=*f*<=≤<=*n*) denoting the number of days in shop's plan and the number of days that Noora has to choose for sell-out.
Each line of the following *n* subsequent lines contains two integers *k**i*,<=*l**i* (0<=≤<=*k**i*,<=*l**i*<=≤<=109) denoting the number of products on the shelves of the shop on the *i*-th day and the number of clients that will come to the shop on *i*-th day.
|
Print a single integer denoting the maximal number of products that shop can sell.
|
[
"4 2\n2 1\n3 5\n2 3\n1 5\n",
"4 1\n0 2\n0 3\n3 5\n0 6\n"
] |
[
"10",
"5"
] |
In the first example we can choose days with numbers 2 and 4 for sell-out. In this case new numbers of products for sale would be equal to [2, 6, 2, 2] respectively. So on the first day shop will sell 1 product, on the second — 5, on the third — 2, on the fourth — 2. In total 1 + 5 + 2 + 2 = 10 product units.
In the second example it is possible to sell 5 products, if you choose third day for sell-out.
| 1,000
|
[
{
"input": "4 2\n2 1\n3 5\n2 3\n1 5",
"output": "10"
},
{
"input": "4 1\n0 2\n0 3\n3 5\n0 6",
"output": "5"
},
{
"input": "1 1\n5 8",
"output": "8"
},
{
"input": "2 1\n8 12\n6 11",
"output": "19"
},
{
"input": "2 1\n6 7\n5 7",
"output": "13"
},
{
"input": "2 1\n5 7\n6 7",
"output": "13"
},
{
"input": "2 1\n7 8\n3 6",
"output": "13"
},
{
"input": "2 1\n9 10\n5 8",
"output": "17"
},
{
"input": "2 1\n3 6\n7 8",
"output": "13"
},
{
"input": "1 0\n10 20",
"output": "10"
},
{
"input": "2 1\n99 100\n3 6",
"output": "105"
},
{
"input": "4 2\n2 10\n3 10\n9 9\n5 10",
"output": "27"
},
{
"input": "2 1\n3 4\n2 8",
"output": "7"
},
{
"input": "50 2\n74 90\n68 33\n49 88\n52 13\n73 21\n77 63\n27 62\n8 52\n60 57\n42 83\n98 15\n79 11\n77 46\n55 91\n72 100\n70 86\n50 51\n57 39\n20 54\n64 95\n66 22\n79 64\n31 28\n11 89\n1 36\n13 4\n75 62\n16 62\n100 35\n43 96\n97 54\n86 33\n62 63\n94 24\n19 6\n20 58\n38 38\n11 76\n70 40\n44 24\n32 96\n28 100\n62 45\n41 68\n90 52\n16 0\n98 32\n81 79\n67 82\n28 2",
"output": "1889"
},
{
"input": "2 1\n10 5\n2 4",
"output": "9"
},
{
"input": "2 1\n50 51\n30 40",
"output": "90"
},
{
"input": "3 2\n5 10\n5 10\n7 9",
"output": "27"
},
{
"input": "3 1\n1000 1000\n50 100\n2 2",
"output": "1102"
},
{
"input": "2 1\n2 4\n12 12",
"output": "16"
},
{
"input": "2 1\n4 4\n1 2",
"output": "6"
},
{
"input": "2 1\n4000 4000\n1 2",
"output": "4002"
},
{
"input": "2 1\n5 6\n2 4",
"output": "9"
},
{
"input": "3 2\n10 10\n10 10\n1 2",
"output": "22"
},
{
"input": "10 5\n9 1\n11 1\n12 1\n13 1\n14 1\n2 4\n2 4\n2 4\n2 4\n2 4",
"output": "25"
},
{
"input": "2 1\n30 30\n10 20",
"output": "50"
},
{
"input": "1 1\n1 1",
"output": "1"
},
{
"input": "2 1\n10 2\n2 10",
"output": "6"
},
{
"input": "2 1\n4 5\n3 9",
"output": "10"
},
{
"input": "2 1\n100 100\n5 10",
"output": "110"
},
{
"input": "2 1\n14 28\n15 28",
"output": "43"
},
{
"input": "2 1\n100 1\n20 40",
"output": "41"
},
{
"input": "2 1\n5 10\n6 10",
"output": "16"
},
{
"input": "2 1\n29 30\n10 20",
"output": "49"
},
{
"input": "1 0\n12 12",
"output": "12"
},
{
"input": "2 1\n7 8\n4 7",
"output": "14"
},
{
"input": "2 1\n5 5\n2 4",
"output": "9"
},
{
"input": "2 1\n1 2\n228 2",
"output": "4"
},
{
"input": "2 1\n5 10\n100 20",
"output": "30"
},
{
"input": "2 1\n1000 1001\n2 4",
"output": "1004"
},
{
"input": "2 1\n3 9\n7 7",
"output": "13"
},
{
"input": "2 0\n1 1\n1 1",
"output": "2"
},
{
"input": "4 1\n10 10\n10 10\n10 10\n4 6",
"output": "36"
},
{
"input": "18 13\n63 8\n87 100\n18 89\n35 29\n66 81\n27 85\n64 51\n60 52\n32 94\n74 22\n86 31\n43 78\n12 2\n36 2\n67 23\n2 16\n78 71\n34 64",
"output": "772"
},
{
"input": "2 1\n10 18\n17 19",
"output": "35"
},
{
"input": "3 0\n1 1\n1 1\n1 1",
"output": "3"
},
{
"input": "2 1\n4 7\n8 9",
"output": "15"
},
{
"input": "4 2\n2 10\n3 10\n9 10\n5 10",
"output": "27"
},
{
"input": "2 1\n5 7\n3 6",
"output": "11"
},
{
"input": "2 1\n3 4\n12 12",
"output": "16"
},
{
"input": "2 1\n10 11\n9 20",
"output": "28"
},
{
"input": "2 1\n7 8\n2 4",
"output": "11"
},
{
"input": "2 1\n5 10\n7 10",
"output": "17"
},
{
"input": "4 2\n2 10\n3 10\n5 10\n9 10",
"output": "27"
},
{
"input": "2 1\n99 100\n5 10",
"output": "109"
},
{
"input": "4 2\n2 10\n3 10\n5 10\n9 9",
"output": "27"
},
{
"input": "2 1\n3 7\n5 7",
"output": "11"
},
{
"input": "2 1\n10 10\n3 6",
"output": "16"
},
{
"input": "2 1\n100 1\n2 4",
"output": "5"
},
{
"input": "5 0\n1 1\n1 1\n1 1\n1 1\n1 1",
"output": "5"
},
{
"input": "3 1\n3 7\n4 5\n2 3",
"output": "12"
},
{
"input": "2 1\n3 9\n7 8",
"output": "13"
},
{
"input": "2 1\n10 2\n3 4",
"output": "6"
},
{
"input": "2 1\n40 40\n3 5",
"output": "45"
},
{
"input": "2 1\n5 3\n1 2",
"output": "5"
},
{
"input": "10 5\n9 5\n10 5\n11 5\n12 5\n13 5\n2 4\n2 4\n2 4\n2 4\n2 4",
"output": "45"
},
{
"input": "3 1\n1 5\n1 5\n4 4",
"output": "7"
},
{
"input": "4 0\n1 1\n1 1\n1 1\n1 1",
"output": "4"
},
{
"input": "4 1\n1000 1001\n1000 1001\n2 4\n1 2",
"output": "2005"
},
{
"input": "2 1\n15 30\n50 59",
"output": "80"
},
{
"input": "2 1\n8 8\n3 5",
"output": "13"
},
{
"input": "2 1\n4 5\n2 5",
"output": "8"
},
{
"input": "3 2\n3 3\n1 2\n1 2",
"output": "7"
},
{
"input": "3 1\n2 5\n2 5\n4 4",
"output": "10"
},
{
"input": "2 1\n3 10\n50 51",
"output": "56"
},
{
"input": "4 2\n2 4\n2 4\n9 10\n9 10",
"output": "26"
},
{
"input": "2 1\n3 5\n8 8",
"output": "13"
},
{
"input": "2 1\n100 150\n70 150",
"output": "240"
},
{
"input": "2 1\n4 5\n3 6",
"output": "10"
},
{
"input": "2 1\n20 10\n3 5",
"output": "15"
},
{
"input": "15 13\n76167099 92301116\n83163126 84046805\n45309500 65037149\n29982002 77381688\n76738161 52935441\n37889502 25466134\n55955619 14197941\n31462620 12999429\n64648384 8824773\n3552934 68992494\n2823376 9338427\n86832070 3763091\n67753633 2162190\n302887 92011825\n84894984 410533",
"output": "435467000"
},
{
"input": "2 1\n8 7\n3 6",
"output": "13"
},
{
"input": "2 1\n7 8\n3 5",
"output": "12"
},
{
"input": "2 1\n10 10\n1 3",
"output": "12"
},
{
"input": "2 1\n9 10\n2 4",
"output": "13"
},
{
"input": "3 1\n10 11\n12 13\n8 10",
"output": "32"
},
{
"input": "2 1\n5 10\n7 7",
"output": "17"
},
{
"input": "4 2\n90 91\n2 10\n2 10\n2 10",
"output": "100"
},
{
"input": "2 1\n2 4\n4 4",
"output": "8"
},
{
"input": "2 1\n2 3\n4 3",
"output": "6"
},
{
"input": "2 1\n40 45\n50 52",
"output": "95"
},
{
"input": "3 1\n1 4\n2 4\n3 4",
"output": "8"
},
{
"input": "2 1\n1 2\n1000 1000",
"output": "1002"
},
{
"input": "2 1\n80 100\n70 95",
"output": "175"
}
] | 1,685,440,536
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 159
| 405
| 18,227,200
|
n, m = map(int, input().split())
base_sold, gains = [], []
for _ in range(n):
k, l = map(int, input().split())
gain = min(k*2, l) - min(k, l)
gains.append(gain)
base_sold.append(min(k, l))
targ = sorted(zip(gains, base_sold), reverse=True)
ans = 0
for i in range(n):
if i < m:
ans += targ[i][0]
ans += targ[i][1]
# print(f'{i+1} ({targ[i]}):', sum(targ[i]))
print(ans)
|
Title: Summer sell-off
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Summer holidays! Someone is going on trips, someone is visiting grandparents, but someone is trying to get a part-time job. This summer Noora decided that she wants to earn some money, and took a job in a shop as an assistant.
Shop, where Noora is working, has a plan on the following *n* days. For each day sales manager knows exactly, that in *i*-th day *k**i* products will be put up for sale and exactly *l**i* clients will come to the shop that day. Also, the manager is sure, that everyone, who comes to the shop, buys exactly one product or, if there aren't any left, leaves the shop without buying anything. Moreover, due to the short shelf-life of the products, manager established the following rule: if some part of the products left on the shelves at the end of the day, that products aren't kept on the next day and are sent to the dump.
For advertising purposes manager offered to start a sell-out in the shop. He asked Noora to choose any *f* days from *n* next for sell-outs. On each of *f* chosen days the number of products were put up for sale would be doubled. Thus, if on *i*-th day shop planned to put up for sale *k**i* products and Noora has chosen this day for sell-out, shelves of the shop would keep 2·*k**i* products. Consequently, there is an opportunity to sell two times more products on days of sell-out.
Noora's task is to choose *f* days to maximize total number of sold products. She asks you to help her with such a difficult problem.
Input Specification:
The first line contains two integers *n* and *f* (1<=≤<=*n*<=≤<=105,<=0<=≤<=*f*<=≤<=*n*) denoting the number of days in shop's plan and the number of days that Noora has to choose for sell-out.
Each line of the following *n* subsequent lines contains two integers *k**i*,<=*l**i* (0<=≤<=*k**i*,<=*l**i*<=≤<=109) denoting the number of products on the shelves of the shop on the *i*-th day and the number of clients that will come to the shop on *i*-th day.
Output Specification:
Print a single integer denoting the maximal number of products that shop can sell.
Demo Input:
['4 2\n2 1\n3 5\n2 3\n1 5\n', '4 1\n0 2\n0 3\n3 5\n0 6\n']
Demo Output:
['10', '5']
Note:
In the first example we can choose days with numbers 2 and 4 for sell-out. In this case new numbers of products for sale would be equal to [2, 6, 2, 2] respectively. So on the first day shop will sell 1 product, on the second — 5, on the third — 2, on the fourth — 2. In total 1 + 5 + 2 + 2 = 10 product units.
In the second example it is possible to sell 5 products, if you choose third day for sell-out.
|
```python
n, m = map(int, input().split())
base_sold, gains = [], []
for _ in range(n):
k, l = map(int, input().split())
gain = min(k*2, l) - min(k, l)
gains.append(gain)
base_sold.append(min(k, l))
targ = sorted(zip(gains, base_sold), reverse=True)
ans = 0
for i in range(n):
if i < m:
ans += targ[i][0]
ans += targ[i][1]
# print(f'{i+1} ({targ[i]}):', sum(targ[i]))
print(ans)
```
| 3
|
|
596
|
B
|
Wilbur and Array
|
PROGRAMMING
| 1,100
|
[
"greedy",
"implementation"
] | null | null |
Wilbur the pig is tinkering with arrays again. He has the array *a*1,<=*a*2,<=...,<=*a**n* initially consisting of *n* zeros. At one step, he can choose any index *i* and either add 1 to all elements *a**i*,<=*a**i*<=+<=1,<=... ,<=*a**n* or subtract 1 from all elements *a**i*,<=*a**i*<=+<=1,<=...,<=*a**n*. His goal is to end up with the array *b*1,<=*b*2,<=...,<=*b**n*.
Of course, Wilbur wants to achieve this goal in the minimum number of steps and asks you to compute this value.
|
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=200<=000) — the length of the array *a**i*. Initially *a**i*<==<=0 for every position *i*, so this array is not given in the input.
The second line of the input contains *n* integers *b*1,<=*b*2,<=...,<=*b**n* (<=-<=109<=≤<=*b**i*<=≤<=109).
|
Print the minimum number of steps that Wilbur needs to make in order to achieve *a**i*<==<=*b**i* for all *i*.
|
[
"5\n1 2 3 4 5\n",
"4\n1 2 2 1\n"
] |
[
"5",
"3"
] |
In the first sample, Wilbur may successively choose indices 1, 2, 3, 4, and 5, and add 1 to corresponding suffixes.
In the second sample, Wilbur first chooses indices 1 and 2 and adds 1 to corresponding suffixes, then he chooses index 4 and subtract 1.
| 1,000
|
[
{
"input": "5\n1 2 3 4 5",
"output": "5"
},
{
"input": "4\n1 2 2 1",
"output": "3"
},
{
"input": "3\n1 2 4",
"output": "4"
},
{
"input": "6\n1 2 3 6 5 4",
"output": "8"
},
{
"input": "10\n2 1 4 3 6 5 8 7 10 9",
"output": "19"
},
{
"input": "7\n12 6 12 13 4 3 2",
"output": "36"
},
{
"input": "15\n15 14 13 1 2 3 12 11 10 4 5 6 9 8 7",
"output": "55"
},
{
"input": "16\n1 2 3 4 13 14 15 16 9 10 11 12 5 6 7 8",
"output": "36"
},
{
"input": "6\n1000 1 2000 1 3000 1",
"output": "11995"
},
{
"input": "1\n0",
"output": "0"
},
{
"input": "5\n1000000000 1 1000000000 1 1000000000",
"output": "4999999996"
},
{
"input": "5\n1000000000 0 1000000000 0 1000000000",
"output": "5000000000"
},
{
"input": "10\n1000000000 0 1000000000 0 1000000000 0 1000000000 0 1000000000 0",
"output": "10000000000"
},
{
"input": "10\n1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000",
"output": "19000000000"
},
{
"input": "7\n0 1000000000 0 1000000000 0 1000000000 0",
"output": "6000000000"
},
{
"input": "4\n1000000000 -1000000000 1000000000 -1000000000",
"output": "7000000000"
},
{
"input": "20\n1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000",
"output": "39000000000"
},
{
"input": "11\n1000000000 0 1000000000 0 1000000000 0 1000000000 0 1000000000 0 1000000000",
"output": "11000000000"
},
{
"input": "5\n1000000000 -1000000000 1000000000 -1000000000 1000000000",
"output": "9000000000"
},
{
"input": "22\n1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000 1000000000 -1000000000",
"output": "43000000000"
}
] | 1,447,828,050
| 2,250
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 3
| 61
| 0
|
n = int(input())
B = [int(x) for x in input().split(' ')]
l = 0
t = 0
for b in B:
t += abs(t-b)
l = b
print(t)
|
Title: Wilbur and Array
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Wilbur the pig is tinkering with arrays again. He has the array *a*1,<=*a*2,<=...,<=*a**n* initially consisting of *n* zeros. At one step, he can choose any index *i* and either add 1 to all elements *a**i*,<=*a**i*<=+<=1,<=... ,<=*a**n* or subtract 1 from all elements *a**i*,<=*a**i*<=+<=1,<=...,<=*a**n*. His goal is to end up with the array *b*1,<=*b*2,<=...,<=*b**n*.
Of course, Wilbur wants to achieve this goal in the minimum number of steps and asks you to compute this value.
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=200<=000) — the length of the array *a**i*. Initially *a**i*<==<=0 for every position *i*, so this array is not given in the input.
The second line of the input contains *n* integers *b*1,<=*b*2,<=...,<=*b**n* (<=-<=109<=≤<=*b**i*<=≤<=109).
Output Specification:
Print the minimum number of steps that Wilbur needs to make in order to achieve *a**i*<==<=*b**i* for all *i*.
Demo Input:
['5\n1 2 3 4 5\n', '4\n1 2 2 1\n']
Demo Output:
['5', '3']
Note:
In the first sample, Wilbur may successively choose indices 1, 2, 3, 4, and 5, and add 1 to corresponding suffixes.
In the second sample, Wilbur first chooses indices 1 and 2 and adds 1 to corresponding suffixes, then he chooses index 4 and subtract 1.
|
```python
n = int(input())
B = [int(x) for x in input().split(' ')]
l = 0
t = 0
for b in B:
t += abs(t-b)
l = b
print(t)
```
| 0
|
|
518
|
A
|
Vitaly and Strings
|
PROGRAMMING
| 1,600
|
[
"constructive algorithms",
"strings"
] | null | null |
Vitaly is a diligent student who never missed a lesson in his five years of studying in the university. He always does his homework on time and passes his exams in time.
During the last lesson the teacher has provided two strings *s* and *t* to Vitaly. The strings have the same length, they consist of lowercase English letters, string *s* is lexicographically smaller than string *t*. Vitaly wondered if there is such string that is lexicographically larger than string *s* and at the same is lexicographically smaller than string *t*. This string should also consist of lowercase English letters and have the length equal to the lengths of strings *s* and *t*.
Let's help Vitaly solve this easy problem!
|
The first line contains string *s* (1<=≤<=|*s*|<=≤<=100), consisting of lowercase English letters. Here, |*s*| denotes the length of the string.
The second line contains string *t* (|*t*|<==<=|*s*|), consisting of lowercase English letters.
It is guaranteed that the lengths of strings *s* and *t* are the same and string *s* is lexicographically less than string *t*.
|
If the string that meets the given requirements doesn't exist, print a single string "No such string" (without the quotes).
If such string exists, print it. If there are multiple valid strings, you may print any of them.
|
[
"a\nc\n",
"aaa\nzzz\n",
"abcdefg\nabcdefh\n"
] |
[
"b\n",
"kkk\n",
"No such string\n"
] |
String *s* = *s*<sub class="lower-index">1</sub>*s*<sub class="lower-index">2</sub>... *s*<sub class="lower-index">*n*</sub> is said to be lexicographically smaller than *t* = *t*<sub class="lower-index">1</sub>*t*<sub class="lower-index">2</sub>... *t*<sub class="lower-index">*n*</sub>, if there exists such *i*, that *s*<sub class="lower-index">1</sub> = *t*<sub class="lower-index">1</sub>, *s*<sub class="lower-index">2</sub> = *t*<sub class="lower-index">2</sub>, ... *s*<sub class="lower-index">*i* - 1</sub> = *t*<sub class="lower-index">*i* - 1</sub>, *s*<sub class="lower-index">*i*</sub> < *t*<sub class="lower-index">*i*</sub>.
| 500
|
[
{
"input": "a\nc",
"output": "b"
},
{
"input": "aaa\nzzz",
"output": "kkk"
},
{
"input": "abcdefg\nabcdefh",
"output": "No such string"
},
{
"input": "abcdefg\nabcfefg",
"output": "abcdefh"
},
{
"input": "frt\nfru",
"output": "No such string"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa\nzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz",
"output": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaab"
},
{
"input": "zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzx\nzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz",
"output": "zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzy"
},
{
"input": "q\nz",
"output": "r"
},
{
"input": "pnzcl\npnzdf",
"output": "pnzcm"
},
{
"input": "vklldrxnfgyorgfpfezvhbouyzzzzz\nvklldrxnfgyorgfpfezvhbouzaaadv",
"output": "vklldrxnfgyorgfpfezvhbouzaaaaa"
},
{
"input": "pkjlxzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz\npkjlyaaaaaaaaaaaaaaaaaaaaaaaaaaaahr",
"output": "pkjlyaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "exoudpymnspkocwszzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz\nexoudpymnspkocwtaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaabml",
"output": "exoudpymnspkocwtaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "anarzvsklmwvovozwnmhklkpcseeogdgauoppmzrukynbjjoxytuvsiecuzfquxnowewebhtuoxepocyeamqfrblpwqiokbcubil\nanarzvsklmwvovozwnmhklkpcseeogdgauoppmzrukynbjjoxytuvsiecuzfquxnowewebhtuoxepocyeamqfrblpwqiokbcubim",
"output": "No such string"
},
{
"input": "uqyugulumzwlxsjnxxkutzqayskrbjoaaekbhckjryhjjllzzz\nuqyugulumzwlxsjnxxkutzqayskrbjoaaekbhckjryhjjlmaaa",
"output": "No such string"
},
{
"input": "esfaeyxpblcrriizhnhfrxnbopqvhwtetgjqavlqdlxexaifgvkqfwzneibhxxdacbzzzzzzzzzzzzzz\nesfaeyxpblcrriizhnhfrxnbopqvhwtetgjqavlqdlxexaifgvkqfwzneibhxxdaccaaaaaaaaaaaatf",
"output": "esfaeyxpblcrriizhnhfrxnbopqvhwtetgjqavlqdlxexaifgvkqfwzneibhxxdaccaaaaaaaaaaaaaa"
},
{
"input": "oisjtilteipnzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz\noisjtilteipoaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaao",
"output": "oisjtilteipoaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "svpoxbsudndfnnpugbouawegyxgtmvqzbewxpcwhopdbwscimgzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz\nsvpoxbsudndfnnpugbouawegyxgtmvqzbewxpcwhopdbwscimhaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "No such string"
},
{
"input": "ddzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz\ndeaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaao",
"output": "deaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "xqzbhslocdbifnyzyjenlpctocieaccsycmwlcebkqqkeibatfvylbqlutvjijgjhdetqsjqnoipqbmjhhzxggdobyvpczdavdzz\nxqzbhslocdbifnyzyjenlpctocieaccsycmwlcebkqqkeibatfvylbqlutvjijgjhdetqsjqnoipqbmjhhzxggdobyvpczdavilj",
"output": "xqzbhslocdbifnyzyjenlpctocieaccsycmwlcebkqqkeibatfvylbqlutvjijgjhdetqsjqnoipqbmjhhzxggdobyvpczdaveaa"
},
{
"input": "poflpxucohdobeisxfsnkbdzwizjjhgngufssqhmfgmydmmrnuminrvxxamoebhczlwsfefdtnchaisfxkfcovxmvppxnrfawfoq\npoflpxucohdobeisxfsnkbdzwizjjhgngufssqhmfgmydmmrnuminrvxxamoebhczlwsfefdtnchaisfxkfcovxmvppxnrfawujg",
"output": "poflpxucohdobeisxfsnkbdzwizjjhgngufssqhmfgmydmmrnuminrvxxamoebhczlwsfefdtnchaisfxkfcovxmvppxnrfawfor"
},
{
"input": "vonggnmokmvmguwtobkxoqgxkuxtyjmxrygyliohlhwxuxjmlkqcfuxboxjnzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz\nvonggnmokmvmguwtobkxoqgxkuxtyjmxrygyliohlhwxuxjmlkqcfuxboxjoaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaac",
"output": "vonggnmokmvmguwtobkxoqgxkuxtyjmxrygyliohlhwxuxjmlkqcfuxboxjoaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "bqycw\nquhod",
"output": "bqycx"
},
{
"input": "hceslswecf\nnmxshuymaa",
"output": "hceslswecg"
},
{
"input": "awqtzslxowuaefe\nvujscakjpvxviki",
"output": "awqtzslxowuaeff"
},
{
"input": "lerlcnaogdravnogfogcyoxgi\nojrbithvjdqtempegvqxmgmmw",
"output": "lerlcnaogdravnogfogcyoxgj"
},
{
"input": "jbrhvicytqaivheqeourrlosvnsujsxdinryyawgalidsaufxv\noevvkhujmhagaholrmsatdjjyfmyblvgetpnxgjcilugjsncjs",
"output": "jbrhvicytqaivheqeourrlosvnsujsxdinryyawgalidsaufxw"
},
{
"input": "jrpogrcuhqdpmyzpuabuhaptlxaeiqjxhqkmuzsjbhqxvdtoocrkusaeasqdwlunomwzww\nspvgaswympzlscnumemgiznngnxqgccbubmxgqmaakbnyngkxlxjjsafricchhpecdjgxw",
"output": "jrpogrcuhqdpmyzpuabuhaptlxaeiqjxhqkmuzsjbhqxvdtoocrkusaeasqdwlunomwzwx"
},
{
"input": "mzmhjmfxaxaplzjmjkbyadeweltagyyuzpvrmnyvirjpdmebxyzjvdoezhnayfrvtnccryhkvhcvakcf\nohhhhkujfpjbgouebtmmbzizuhuumvrsqfniwpmxdtzhyiaivdyxhywnqzagicydixjtvbqbevhbqttu",
"output": "mzmhjmfxaxaplzjmjkbyadeweltagyyuzpvrmnyvirjpdmebxyzjvdoezhnayfrvtnccryhkvhcvakcg"
},
{
"input": "cdmwmzutsicpzhcokbbhwktqbomozxvvjlhwdgtiledgurxsfreisgczdwgupzxmjnfyjxcpdwzkggludkcmgppndl\nuvuqvyrnhtyubpevizhjxdvmpueittksrnosmfuuzbimnqussasdjufrthrgjbyzomauaxbvwferfvtmydmwmjaoxg",
"output": "cdmwmzutsicpzhcokbbhwktqbomozxvvjlhwdgtiledgurxsfreisgczdwgupzxmjnfyjxcpdwzkggludkcmgppndm"
},
{
"input": "dpnmrwpbgzvcmrcodwgvvfwpyagdwlngmhrazyvalszhruprxzmwltftxmujfyrrnwzvphgqlcphreumqkytswxziugburwrlyay\nqibcfxdfovoejutaeetbbwrgexdrvqywwmhipxgfrvhzovxkfawpfnpjvlhkyahessodqcclangxefcaixysqijnitevwmpalkzd",
"output": "dpnmrwpbgzvcmrcodwgvvfwpyagdwlngmhrazyvalszhruprxzmwltftxmujfyrrnwzvphgqlcphreumqkytswxziugburwrlyaz"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaab",
"output": "No such string"
},
{
"input": "phdvmuwqmvzyurtnshitcypuzbhpceovkibzbhhjwxkdtvqmbpoumeoiztxtvkvsjrlnhowsdmgftuiulzebdigmun\nphdvmuwqmvzyurtnshitcypuzbhpceovkibzbhhjwxkdtvqmbpoumeoiztxtvkvsjrlnhowsdmgftuiulzebdigmuo",
"output": "No such string"
},
{
"input": "hrsantdquixzjyjtqytcmnflnyehzbibkbgkqffgqpkgeuqmbmxzhbjwsnfkizvbcyoghyvnxxjavoahlqjxomtsouzoog\nhrsantdquixzjyjtqytcmnflnyehzbibkbgkqffgqpkgeuqmbmxzhbjwsnfkizvbcyoghyvnxxjavoahlqjxomtsouzooh",
"output": "No such string"
},
{
"input": "kexdbtpkjbwwyibjndbtmwqzolopqitgkomqggojevoankiepxirrcidxldlzsppehmoazdywltmjbxgsxgihwnwpmczjrcwpywl\nkexdbtpkjbwwyibjndbtmwqzolopqitgkomqggojevoankiepxirrcidxldlzsppehmoazdywltmjbxgsxgihwnwpmczjrcwpywm",
"output": "No such string"
},
{
"input": "yijdysvzfcnaedvnecswgoylhzgguxecmucepgstjbdkbjyfdlxxxejkrrxfiuwjpdmdhhqhlqeqzjwudtdryrfkpwfxdjlkowmk\nyijdysvzfcnaedvnecswgoylhzgguxecmucepgstjbdkbjyfdlxxxejkrrxfiuwjpdmdhhqhlqeqzjwudtdryrfkpwfxdjlkowml",
"output": "No such string"
},
{
"input": "a\nb",
"output": "No such string"
},
{
"input": "x\nz",
"output": "y"
},
{
"input": "a\nz",
"output": "b"
},
{
"input": "y\nz",
"output": "No such string"
},
{
"input": "zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzy\nzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz",
"output": "No such string"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaac",
"output": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaab"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa\nzaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaab"
},
{
"input": "yzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz\nzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz",
"output": "zaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "yyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyy\nzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz",
"output": "yyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyz"
},
{
"input": "aaa\naac",
"output": "aab"
},
{
"input": "aaa\naab",
"output": "No such string"
},
{
"input": "zzx\nzzz",
"output": "zzy"
},
{
"input": "zzy\nzzz",
"output": "No such string"
},
{
"input": "aba\naca",
"output": "abb"
},
{
"input": "aaa\nbbb",
"output": "aab"
},
{
"input": "azzz\ncaaa",
"output": "baaa"
}
] | 1,540,460,569
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 3
| 108
| 0
|
s = input ()
t = input()
def string (string1, string2):
ans =""
for i in range (len(string1)):
if ord(string2[i]) - ord(string1[i])>=2:
ans += chr(ord(string1[i])+1)
else:
ans = "No such string"
return ans
print(string(s,t))
|
Title: Vitaly and Strings
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vitaly is a diligent student who never missed a lesson in his five years of studying in the university. He always does his homework on time and passes his exams in time.
During the last lesson the teacher has provided two strings *s* and *t* to Vitaly. The strings have the same length, they consist of lowercase English letters, string *s* is lexicographically smaller than string *t*. Vitaly wondered if there is such string that is lexicographically larger than string *s* and at the same is lexicographically smaller than string *t*. This string should also consist of lowercase English letters and have the length equal to the lengths of strings *s* and *t*.
Let's help Vitaly solve this easy problem!
Input Specification:
The first line contains string *s* (1<=≤<=|*s*|<=≤<=100), consisting of lowercase English letters. Here, |*s*| denotes the length of the string.
The second line contains string *t* (|*t*|<==<=|*s*|), consisting of lowercase English letters.
It is guaranteed that the lengths of strings *s* and *t* are the same and string *s* is lexicographically less than string *t*.
Output Specification:
If the string that meets the given requirements doesn't exist, print a single string "No such string" (without the quotes).
If such string exists, print it. If there are multiple valid strings, you may print any of them.
Demo Input:
['a\nc\n', 'aaa\nzzz\n', 'abcdefg\nabcdefh\n']
Demo Output:
['b\n', 'kkk\n', 'No such string\n']
Note:
String *s* = *s*<sub class="lower-index">1</sub>*s*<sub class="lower-index">2</sub>... *s*<sub class="lower-index">*n*</sub> is said to be lexicographically smaller than *t* = *t*<sub class="lower-index">1</sub>*t*<sub class="lower-index">2</sub>... *t*<sub class="lower-index">*n*</sub>, if there exists such *i*, that *s*<sub class="lower-index">1</sub> = *t*<sub class="lower-index">1</sub>, *s*<sub class="lower-index">2</sub> = *t*<sub class="lower-index">2</sub>, ... *s*<sub class="lower-index">*i* - 1</sub> = *t*<sub class="lower-index">*i* - 1</sub>, *s*<sub class="lower-index">*i*</sub> < *t*<sub class="lower-index">*i*</sub>.
|
```python
s = input ()
t = input()
def string (string1, string2):
ans =""
for i in range (len(string1)):
if ord(string2[i]) - ord(string1[i])>=2:
ans += chr(ord(string1[i])+1)
else:
ans = "No such string"
return ans
print(string(s,t))
```
| 0
|
|
515
|
C
|
Drazil and Factorial
|
PROGRAMMING
| 1,400
|
[
"greedy",
"math",
"sortings"
] | null | null |
Drazil is playing a math game with Varda.
Let's define for positive integer *x* as a product of factorials of its digits. For example, .
First, they choose a decimal number *a* consisting of *n* digits that contains at least one digit larger than 1. This number may possibly start with leading zeroes. Then they should find maximum positive number *x* satisfying following two conditions:
1. *x* doesn't contain neither digit 0 nor digit 1.
2. = .
Help friends find such number.
|
The first line contains an integer *n* (1<=≤<=*n*<=≤<=15) — the number of digits in *a*.
The second line contains *n* digits of *a*. There is at least one digit in *a* that is larger than 1. Number *a* may possibly contain leading zeroes.
|
Output a maximum possible integer satisfying the conditions above. There should be no zeroes and ones in this number decimal representation.
|
[
"4\n1234\n",
"3\n555\n"
] |
[
"33222\n",
"555\n"
] |
In the first case, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/f5a4207f23215fddce977ab5ea9e9d2e7578fb52.png" style="max-width: 100.0%;max-height: 100.0%;"/>
| 1,000
|
[
{
"input": "4\n1234",
"output": "33222"
},
{
"input": "3\n555",
"output": "555"
},
{
"input": "15\n012345781234578",
"output": "7777553333222222222222"
},
{
"input": "1\n8",
"output": "7222"
},
{
"input": "10\n1413472614",
"output": "75333332222222"
},
{
"input": "8\n68931246",
"output": "77553333332222222"
},
{
"input": "7\n4424368",
"output": "75333332222222222"
},
{
"input": "6\n576825",
"output": "7755532222"
},
{
"input": "5\n97715",
"output": "7775332"
},
{
"input": "3\n915",
"output": "75332"
},
{
"input": "2\n26",
"output": "532"
},
{
"input": "1\n4",
"output": "322"
},
{
"input": "15\n028745260720699",
"output": "7777755533333332222222222"
},
{
"input": "13\n5761790121605",
"output": "7775555333322"
},
{
"input": "10\n3312667105",
"output": "755533332"
},
{
"input": "1\n7",
"output": "7"
},
{
"input": "15\n989898989898989",
"output": "777777777777777333333333333333322222222222222222222222222222"
},
{
"input": "15\n000000000000007",
"output": "7"
},
{
"input": "15\n999999999999990",
"output": "77777777777777333333333333333333333333333322222222222222"
},
{
"input": "1\n2",
"output": "2"
},
{
"input": "1\n3",
"output": "3"
},
{
"input": "1\n4",
"output": "322"
},
{
"input": "1\n5",
"output": "5"
},
{
"input": "1\n6",
"output": "53"
},
{
"input": "1\n7",
"output": "7"
},
{
"input": "1\n9",
"output": "7332"
},
{
"input": "2\n09",
"output": "7332"
},
{
"input": "13\n1337251172966",
"output": "777555333333222"
},
{
"input": "15\n987654329876543",
"output": "777777555533333333332222222222222"
},
{
"input": "9\n234567899",
"output": "777755333333322222222"
},
{
"input": "2\n99",
"output": "77333322"
},
{
"input": "2\n66",
"output": "5533"
},
{
"input": "3\n999",
"output": "777333333222"
},
{
"input": "5\n99999",
"output": "77777333333333322222"
},
{
"input": "9\n123456789",
"output": "77755333332222222"
},
{
"input": "9\n987654321",
"output": "77755333332222222"
},
{
"input": "3\n666",
"output": "555333"
},
{
"input": "2\n26",
"output": "532"
},
{
"input": "6\n555777",
"output": "777555"
},
{
"input": "10\n1234567899",
"output": "777755333333322222222"
},
{
"input": "4\n6666",
"output": "55553333"
},
{
"input": "4\n9754",
"output": "775333222"
},
{
"input": "2\n95",
"output": "75332"
},
{
"input": "14\n11122233344455",
"output": "55333333222222222"
},
{
"input": "12\n836544897832",
"output": "77777553333333222222222222222"
}
] | 1,599,049,959
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 2
| 140
| 0
|
n=int(input())
s=input()
ans=[]
for i in s:
if i=='1' or ans=='0' :
continue
if i=='4' :
ans.append('3')
ans.append('2')
ans.append('2')
elif i=='6' :
ans.append('3')
ans.append('2')
ans.append('5')
elif i=='8' :
ans.append('2')
ans.append('2')
ans.append('2')
ans.append('7')
else:
ans.append(i)
ans.sort()
ans.reverse()
print(''.join(ans))
|
Title: Drazil and Factorial
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Drazil is playing a math game with Varda.
Let's define for positive integer *x* as a product of factorials of its digits. For example, .
First, they choose a decimal number *a* consisting of *n* digits that contains at least one digit larger than 1. This number may possibly start with leading zeroes. Then they should find maximum positive number *x* satisfying following two conditions:
1. *x* doesn't contain neither digit 0 nor digit 1.
2. = .
Help friends find such number.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=15) — the number of digits in *a*.
The second line contains *n* digits of *a*. There is at least one digit in *a* that is larger than 1. Number *a* may possibly contain leading zeroes.
Output Specification:
Output a maximum possible integer satisfying the conditions above. There should be no zeroes and ones in this number decimal representation.
Demo Input:
['4\n1234\n', '3\n555\n']
Demo Output:
['33222\n', '555\n']
Note:
In the first case, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/f5a4207f23215fddce977ab5ea9e9d2e7578fb52.png" style="max-width: 100.0%;max-height: 100.0%;"/>
|
```python
n=int(input())
s=input()
ans=[]
for i in s:
if i=='1' or ans=='0' :
continue
if i=='4' :
ans.append('3')
ans.append('2')
ans.append('2')
elif i=='6' :
ans.append('3')
ans.append('2')
ans.append('5')
elif i=='8' :
ans.append('2')
ans.append('2')
ans.append('2')
ans.append('7')
else:
ans.append(i)
ans.sort()
ans.reverse()
print(''.join(ans))
```
| 0
|
|
794
|
C
|
Naming Company
|
PROGRAMMING
| 1,800
|
[
"games",
"greedy",
"sortings"
] | null | null |
Oleg the client and Igor the analyst are good friends. However, sometimes they argue over little things. Recently, they started a new company, but they are having trouble finding a name for the company.
To settle this problem, they've decided to play a game. The company name will consist of *n* letters. Oleg and Igor each have a set of *n* letters (which might contain multiple copies of the same letter, the sets can be different). Initially, the company name is denoted by *n* question marks. Oleg and Igor takes turns to play the game, Oleg moves first. In each turn, a player can choose one of the letters *c* in his set and replace any of the question marks with *c*. Then, a copy of the letter *c* is removed from his set. The game ends when all the question marks has been replaced by some letter.
For example, suppose Oleg has the set of letters {*i*,<=*o*,<=*i*} and Igor has the set of letters {*i*,<=*m*,<=*o*}. One possible game is as follows :
Initially, the company name is ???.
Oleg replaces the second question mark with 'i'. The company name becomes ?i?. The set of letters Oleg have now is {*i*,<=*o*}.
Igor replaces the third question mark with 'o'. The company name becomes ?io. The set of letters Igor have now is {*i*,<=*m*}.
Finally, Oleg replaces the first question mark with 'o'. The company name becomes oio. The set of letters Oleg have now is {*i*}.
In the end, the company name is oio.
Oleg wants the company name to be as lexicographically small as possible while Igor wants the company name to be as lexicographically large as possible. What will be the company name if Oleg and Igor always play optimally?
A string *s*<==<=*s*1*s*2...*s**m* is called lexicographically smaller than a string *t*<==<=*t*1*t*2...*t**m* (where *s*<=≠<=*t*) if *s**i*<=<<=*t**i* where *i* is the smallest index such that *s**i*<=≠<=*t**i*. (so *s**j*<==<=*t**j* for all *j*<=<<=*i*)
|
The first line of input contains a string *s* of length *n* (1<=≤<=*n*<=≤<=3·105). All characters of the string are lowercase English letters. This string denotes the set of letters Oleg has initially.
The second line of input contains a string *t* of length *n*. All characters of the string are lowercase English letters. This string denotes the set of letters Igor has initially.
|
The output should contain a string of *n* lowercase English letters, denoting the company name if Oleg and Igor plays optimally.
|
[
"tinkoff\nzscoder\n",
"xxxxxx\nxxxxxx\n",
"ioi\nimo\n"
] |
[
"fzfsirk\n",
"xxxxxx\n",
"ioi\n"
] |
One way to play optimally in the first sample is as follows :
- Initially, the company name is ???????.- Oleg replaces the first question mark with 'f'. The company name becomes f??????.- Igor replaces the second question mark with 'z'. The company name becomes fz?????.- Oleg replaces the third question mark with 'f'. The company name becomes fzf????.- Igor replaces the fourth question mark with 's'. The company name becomes fzfs???.- Oleg replaces the fifth question mark with 'i'. The company name becomes fzfsi??.- Igor replaces the sixth question mark with 'r'. The company name becomes fzfsir?.- Oleg replaces the seventh question mark with 'k'. The company name becomes fzfsirk.
For the second sample, no matter how they play, the company name will always be xxxxxx.
| 1,750
|
[
{
"input": "tinkoff\nzscoder",
"output": "fzfsirk"
},
{
"input": "xxxxxx\nxxxxxx",
"output": "xxxxxx"
},
{
"input": "ioi\nimo",
"output": "ioi"
},
{
"input": "abc\naaa",
"output": "aab"
},
{
"input": "reddit\nabcdef",
"output": "dfdeed"
},
{
"input": "cbxz\naaaa",
"output": "abac"
},
{
"input": "bcdef\nabbbc",
"output": "bccdb"
},
{
"input": "z\ny",
"output": "z"
},
{
"input": "y\nz",
"output": "y"
}
] | 1,525,202,072
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 61
| 7,065,600
|
a = input()
b = input()
a_count = [a.count(chr(ord('a') + x)) for x in range(26)]
b_count = [b.count(chr(ord('a') + x)) for x in range(26)]
length = len(a)
result = length * ['a']
left = 0
right = length - 1
i = 0
while i < length:
mi, ma = 0, 25
while a_count[mi] == 0:
mi = mi + 1
while b_count[ma] == 0:
ma = ma - 1
if i % 2 == 0: # Alice
if mi < ma:
result[left] = chr(ord('a')+mi)
left = left + 1
else:
togo = (length - i - 1) // 2
while a_count[mi] <= togo:
togo = togo - a_count[mi]
mi = mi + 1
result[right] = chr(ord('a')+mi)
right = right - 1
else:
if mi < ma:
result[left] = chr(ord('a')+ma)
left = left + 1
else:
togo = (length - i - 1) // 2
while b_count[ma] <= togo:
togo = togo - b_count[ma]
ma = ma - 1
result[right] = chr(ord('a')+ma)
right = right - 1
i = i + 1
print(''.join(result))
|
Title: Naming Company
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Oleg the client and Igor the analyst are good friends. However, sometimes they argue over little things. Recently, they started a new company, but they are having trouble finding a name for the company.
To settle this problem, they've decided to play a game. The company name will consist of *n* letters. Oleg and Igor each have a set of *n* letters (which might contain multiple copies of the same letter, the sets can be different). Initially, the company name is denoted by *n* question marks. Oleg and Igor takes turns to play the game, Oleg moves first. In each turn, a player can choose one of the letters *c* in his set and replace any of the question marks with *c*. Then, a copy of the letter *c* is removed from his set. The game ends when all the question marks has been replaced by some letter.
For example, suppose Oleg has the set of letters {*i*,<=*o*,<=*i*} and Igor has the set of letters {*i*,<=*m*,<=*o*}. One possible game is as follows :
Initially, the company name is ???.
Oleg replaces the second question mark with 'i'. The company name becomes ?i?. The set of letters Oleg have now is {*i*,<=*o*}.
Igor replaces the third question mark with 'o'. The company name becomes ?io. The set of letters Igor have now is {*i*,<=*m*}.
Finally, Oleg replaces the first question mark with 'o'. The company name becomes oio. The set of letters Oleg have now is {*i*}.
In the end, the company name is oio.
Oleg wants the company name to be as lexicographically small as possible while Igor wants the company name to be as lexicographically large as possible. What will be the company name if Oleg and Igor always play optimally?
A string *s*<==<=*s*1*s*2...*s**m* is called lexicographically smaller than a string *t*<==<=*t*1*t*2...*t**m* (where *s*<=≠<=*t*) if *s**i*<=<<=*t**i* where *i* is the smallest index such that *s**i*<=≠<=*t**i*. (so *s**j*<==<=*t**j* for all *j*<=<<=*i*)
Input Specification:
The first line of input contains a string *s* of length *n* (1<=≤<=*n*<=≤<=3·105). All characters of the string are lowercase English letters. This string denotes the set of letters Oleg has initially.
The second line of input contains a string *t* of length *n*. All characters of the string are lowercase English letters. This string denotes the set of letters Igor has initially.
Output Specification:
The output should contain a string of *n* lowercase English letters, denoting the company name if Oleg and Igor plays optimally.
Demo Input:
['tinkoff\nzscoder\n', 'xxxxxx\nxxxxxx\n', 'ioi\nimo\n']
Demo Output:
['fzfsirk\n', 'xxxxxx\n', 'ioi\n']
Note:
One way to play optimally in the first sample is as follows :
- Initially, the company name is ???????.- Oleg replaces the first question mark with 'f'. The company name becomes f??????.- Igor replaces the second question mark with 'z'. The company name becomes fz?????.- Oleg replaces the third question mark with 'f'. The company name becomes fzf????.- Igor replaces the fourth question mark with 's'. The company name becomes fzfs???.- Oleg replaces the fifth question mark with 'i'. The company name becomes fzfsi??.- Igor replaces the sixth question mark with 'r'. The company name becomes fzfsir?.- Oleg replaces the seventh question mark with 'k'. The company name becomes fzfsirk.
For the second sample, no matter how they play, the company name will always be xxxxxx.
|
```python
a = input()
b = input()
a_count = [a.count(chr(ord('a') + x)) for x in range(26)]
b_count = [b.count(chr(ord('a') + x)) for x in range(26)]
length = len(a)
result = length * ['a']
left = 0
right = length - 1
i = 0
while i < length:
mi, ma = 0, 25
while a_count[mi] == 0:
mi = mi + 1
while b_count[ma] == 0:
ma = ma - 1
if i % 2 == 0: # Alice
if mi < ma:
result[left] = chr(ord('a')+mi)
left = left + 1
else:
togo = (length - i - 1) // 2
while a_count[mi] <= togo:
togo = togo - a_count[mi]
mi = mi + 1
result[right] = chr(ord('a')+mi)
right = right - 1
else:
if mi < ma:
result[left] = chr(ord('a')+ma)
left = left + 1
else:
togo = (length - i - 1) // 2
while b_count[ma] <= togo:
togo = togo - b_count[ma]
ma = ma - 1
result[right] = chr(ord('a')+ma)
right = right - 1
i = i + 1
print(''.join(result))
```
| 0
|
|
305
|
A
|
Strange Addition
|
PROGRAMMING
| 1,600
|
[
"brute force",
"constructive algorithms",
"implementation"
] | null | null |
Unfortunately, Vasya can only sum pairs of integers (*a*, *b*), such that for any decimal place at least one number has digit 0 in this place. For example, Vasya can sum numbers 505 and 50, but he cannot sum 1 and 4.
Vasya has a set of *k* distinct non-negative integers *d*1,<=*d*2,<=...,<=*d**k*.
Vasya wants to choose some integers from this set so that he could sum any two chosen numbers. What maximal number of integers can he choose in the required manner?
|
The first input line contains integer *k* (1<=≤<=*k*<=≤<=100) — the number of integers.
The second line contains *k* distinct space-separated integers *d*1,<=*d*2,<=...,<=*d**k* (0<=≤<=*d**i*<=≤<=100).
|
In the first line print a single integer *n* the maximum number of the chosen integers. In the second line print *n* distinct non-negative integers — the required integers.
If there are multiple solutions, print any of them. You can print the numbers in any order.
|
[
"4\n100 10 1 0\n",
"3\n2 70 3\n"
] |
[
"4\n0 1 10 100 ",
"2\n2 70 "
] |
none
| 500
|
[
{
"input": "4\n100 10 1 0",
"output": "4\n0 1 10 100 "
},
{
"input": "3\n2 70 3",
"output": "2\n2 70 "
},
{
"input": "39\n16 72 42 70 17 36 32 40 47 94 27 30 100 55 23 77 67 28 49 50 53 83 38 33 60 65 62 64 6 66 69 86 96 75 85 0 89 73 29",
"output": "4\n0 6 30 100 "
},
{
"input": "50\n20 67 96 6 75 12 37 46 38 86 83 22 10 8 21 2 93 9 81 49 69 52 63 62 70 92 97 40 47 99 16 85 48 77 39 100 28 5 11 44 89 1 19 42 35 27 7 14 88 33",
"output": "3\n1 10 100 "
},
{
"input": "2\n1 2",
"output": "1\n1 "
},
{
"input": "73\n39 66 3 59 40 93 72 34 95 79 83 65 99 57 48 44 82 76 31 21 64 19 53 75 37 16 43 5 47 24 15 22 20 55 45 74 42 10 61 49 23 80 35 62 2 9 67 97 51 81 1 70 88 63 33 25 68 13 69 71 73 6 18 52 41 38 96 46 92 85 14 36 100",
"output": "3\n1 10 100 "
},
{
"input": "15\n74 90 73 47 36 44 81 21 66 92 2 38 62 72 49",
"output": "2\n2 90 "
},
{
"input": "96\n17 10 0 85 57 78 15 99 55 6 7 88 12 95 58 19 47 18 96 82 21 80 97 77 46 31 54 70 23 60 59 100 66 92 51 14 91 25 16 27 44 4 35 98 8 52 24 5 81 29 73 13 61 56 45 75 49 71 94 48 3 76 32 65 72 1 84 36 86 40 83 50 22 33 41 11 26 93 90 43 39 79 89 9 64 68 42 74 87 2 62 34 20 63 67 37",
"output": "4\n0 1 10 100 "
},
{
"input": "5\n23 75 38 47 70",
"output": "1\n23 "
},
{
"input": "12\n89 61 45 92 22 3 94 66 48 21 54 14",
"output": "1\n3 "
},
{
"input": "1\n99",
"output": "1\n99 "
},
{
"input": "1\n0",
"output": "1\n0 "
},
{
"input": "2\n100 1",
"output": "2\n1 100 "
},
{
"input": "3\n1 100 99",
"output": "2\n1 100 "
},
{
"input": "100\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",
"output": "3\n1 10 100 "
},
{
"input": "100\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 98 100 0",
"output": "4\n0 1 10 100 "
},
{
"input": "99\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": "2\n1 10 "
},
{
"input": "2\n5 6",
"output": "1\n5 "
},
{
"input": "81\n11 12 13 14 15 16 17 18 19 21 22 23 24 25 26 27 28 29 31 32 33 34 35 36 37 38 39 41 42 43 44 45 46 47 48 49 51 52 53 54 55 56 57 58 59 61 62 63 64 65 66 67 68 69 71 72 73 74 75 76 77 78 79 81 82 83 84 85 86 87 88 89 91 92 93 94 95 96 97 98 99",
"output": "1\n11 "
},
{
"input": "3\n99 10 6",
"output": "2\n6 10 "
},
{
"input": "4\n11 10 100 3",
"output": "3\n3 10 100 "
},
{
"input": "2\n99 6",
"output": "1\n6 "
},
{
"input": "3\n23 0 100",
"output": "3\n0 23 100 "
},
{
"input": "2\n43 0",
"output": "2\n0 43 "
},
{
"input": "4\n99 0 100 6",
"output": "3\n0 6 100 "
},
{
"input": "1\n100",
"output": "1\n100 "
},
{
"input": "2\n0 100",
"output": "2\n0 100 "
},
{
"input": "3\n0 100 10",
"output": "3\n0 10 100 "
},
{
"input": "3\n0 100 12",
"output": "3\n0 12 100 "
},
{
"input": "3\n0 100 1",
"output": "3\n0 1 100 "
},
{
"input": "4\n0 100 10 1",
"output": "4\n0 1 10 100 "
},
{
"input": "4\n0 100 10 99",
"output": "3\n0 10 100 "
},
{
"input": "1\n1",
"output": "1\n1 "
},
{
"input": "2\n10 12",
"output": "1\n10 "
},
{
"input": "2\n90 9",
"output": "2\n9 90 "
}
] | 1,689,346,949
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 0
| 92
| 0
|
print("_RANDOM_GUESS_1689346949.433353")# 1689346949.433394
|
Title: Strange Addition
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Unfortunately, Vasya can only sum pairs of integers (*a*, *b*), such that for any decimal place at least one number has digit 0 in this place. For example, Vasya can sum numbers 505 and 50, but he cannot sum 1 and 4.
Vasya has a set of *k* distinct non-negative integers *d*1,<=*d*2,<=...,<=*d**k*.
Vasya wants to choose some integers from this set so that he could sum any two chosen numbers. What maximal number of integers can he choose in the required manner?
Input Specification:
The first input line contains integer *k* (1<=≤<=*k*<=≤<=100) — the number of integers.
The second line contains *k* distinct space-separated integers *d*1,<=*d*2,<=...,<=*d**k* (0<=≤<=*d**i*<=≤<=100).
Output Specification:
In the first line print a single integer *n* the maximum number of the chosen integers. In the second line print *n* distinct non-negative integers — the required integers.
If there are multiple solutions, print any of them. You can print the numbers in any order.
Demo Input:
['4\n100 10 1 0\n', '3\n2 70 3\n']
Demo Output:
['4\n0 1 10 100 ', '2\n2 70 ']
Note:
none
|
```python
print("_RANDOM_GUESS_1689346949.433353")# 1689346949.433394
```
| 0
|
|
893
|
C
|
Rumor
|
PROGRAMMING
| 1,300
|
[
"dfs and similar",
"graphs",
"greedy"
] | null | null |
Vova promised himself that he would never play computer games... But recently Firestorm — a well-known game developing company — published their newest game, World of Farcraft, and it became really popular. Of course, Vova started playing it.
Now he tries to solve a quest. The task is to come to a settlement named Overcity and spread a rumor in it.
Vova knows that there are *n* characters in Overcity. Some characters are friends to each other, and they share information they got. Also Vova knows that he can bribe each character so he or she starts spreading the rumor; *i*-th character wants *c**i* gold in exchange for spreading the rumor. When a character hears the rumor, he tells it to all his friends, and they start spreading the rumor to their friends (for free), and so on.
The quest is finished when all *n* characters know the rumor. What is the minimum amount of gold Vova needs to spend in order to finish the quest?
Take a look at the notes if you think you haven't understood the problem completely.
|
The first line contains two integer numbers *n* and *m* (1<=≤<=*n*<=≤<=105,<=0<=≤<=*m*<=≤<=105) — the number of characters in Overcity and the number of pairs of friends.
The second line contains *n* integer numbers *c**i* (0<=≤<=*c**i*<=≤<=109) — the amount of gold *i*-th character asks to start spreading the rumor.
Then *m* lines follow, each containing a pair of numbers (*x**i*,<=*y**i*) which represent that characters *x**i* and *y**i* are friends (1<=≤<=*x**i*,<=*y**i*<=≤<=*n*, *x**i*<=≠<=*y**i*). It is guaranteed that each pair is listed at most once.
|
Print one number — the minimum amount of gold Vova has to spend in order to finish the quest.
|
[
"5 2\n2 5 3 4 8\n1 4\n4 5\n",
"10 0\n1 2 3 4 5 6 7 8 9 10\n",
"10 5\n1 6 2 7 3 8 4 9 5 10\n1 2\n3 4\n5 6\n7 8\n9 10\n"
] |
[
"10\n",
"55\n",
"15\n"
] |
In the first example the best decision is to bribe the first character (he will spread the rumor to fourth character, and the fourth one will spread it to fifth). Also Vova has to bribe the second and the third characters, so they know the rumor.
In the second example Vova has to bribe everyone.
In the third example the optimal decision is to bribe the first, the third, the fifth, the seventh and the ninth characters.
| 0
|
[
{
"input": "5 2\n2 5 3 4 8\n1 4\n4 5",
"output": "10"
},
{
"input": "10 0\n1 2 3 4 5 6 7 8 9 10",
"output": "55"
},
{
"input": "10 5\n1 6 2 7 3 8 4 9 5 10\n1 2\n3 4\n5 6\n7 8\n9 10",
"output": "15"
},
{
"input": "1 0\n0",
"output": "0"
},
{
"input": "1 0\n1000000000",
"output": "1000000000"
},
{
"input": "2 0\n0 0",
"output": "0"
},
{
"input": "2 0\n1000000000 0",
"output": "1000000000"
},
{
"input": "2 0\n0 1000000000",
"output": "1000000000"
},
{
"input": "2 0\n1000000000 1000000000",
"output": "2000000000"
},
{
"input": "2 1\n0 0\n1 2",
"output": "0"
}
] | 1,645,017,890
| 2,147,483,647
|
PyPy 3-64
|
RUNTIME_ERROR
|
TESTS
| 4
| 982
| 22,016,000
|
def dfs(v, adj, colors, color):
colors[v] = color
for u in adj[v]:
if colors[u] == -1:
dfs(u, adj, colors, color)
n, m = map(int, input().split())
c = list(map(int, input().split()))
if c.count(0) == n:
print(0)
else:
adj = [[] for _ in range(n)]
for _ in range(m):
s, e = map(lambda x: int(x) - 1, input().split())
adj[s].append(e)
adj[e].append(s)
color = 0
colors = [-1]*n
for v in range(n):
if colors[v] == -1:
dfs(v, adj, colors, color)
color += 1
m_i_c = [10**9 + 1]*color
for i in range(n):
m_i_c[colors[i]] = min(m_i_c[colors[i]], c[i])
summ = sum(filter(lambda x: x < 10**9 + 1, m_i_c))
print(summ)
|
Title: Rumor
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vova promised himself that he would never play computer games... But recently Firestorm — a well-known game developing company — published their newest game, World of Farcraft, and it became really popular. Of course, Vova started playing it.
Now he tries to solve a quest. The task is to come to a settlement named Overcity and spread a rumor in it.
Vova knows that there are *n* characters in Overcity. Some characters are friends to each other, and they share information they got. Also Vova knows that he can bribe each character so he or she starts spreading the rumor; *i*-th character wants *c**i* gold in exchange for spreading the rumor. When a character hears the rumor, he tells it to all his friends, and they start spreading the rumor to their friends (for free), and so on.
The quest is finished when all *n* characters know the rumor. What is the minimum amount of gold Vova needs to spend in order to finish the quest?
Take a look at the notes if you think you haven't understood the problem completely.
Input Specification:
The first line contains two integer numbers *n* and *m* (1<=≤<=*n*<=≤<=105,<=0<=≤<=*m*<=≤<=105) — the number of characters in Overcity and the number of pairs of friends.
The second line contains *n* integer numbers *c**i* (0<=≤<=*c**i*<=≤<=109) — the amount of gold *i*-th character asks to start spreading the rumor.
Then *m* lines follow, each containing a pair of numbers (*x**i*,<=*y**i*) which represent that characters *x**i* and *y**i* are friends (1<=≤<=*x**i*,<=*y**i*<=≤<=*n*, *x**i*<=≠<=*y**i*). It is guaranteed that each pair is listed at most once.
Output Specification:
Print one number — the minimum amount of gold Vova has to spend in order to finish the quest.
Demo Input:
['5 2\n2 5 3 4 8\n1 4\n4 5\n', '10 0\n1 2 3 4 5 6 7 8 9 10\n', '10 5\n1 6 2 7 3 8 4 9 5 10\n1 2\n3 4\n5 6\n7 8\n9 10\n']
Demo Output:
['10\n', '55\n', '15\n']
Note:
In the first example the best decision is to bribe the first character (he will spread the rumor to fourth character, and the fourth one will spread it to fifth). Also Vova has to bribe the second and the third characters, so they know the rumor.
In the second example Vova has to bribe everyone.
In the third example the optimal decision is to bribe the first, the third, the fifth, the seventh and the ninth characters.
|
```python
def dfs(v, adj, colors, color):
colors[v] = color
for u in adj[v]:
if colors[u] == -1:
dfs(u, adj, colors, color)
n, m = map(int, input().split())
c = list(map(int, input().split()))
if c.count(0) == n:
print(0)
else:
adj = [[] for _ in range(n)]
for _ in range(m):
s, e = map(lambda x: int(x) - 1, input().split())
adj[s].append(e)
adj[e].append(s)
color = 0
colors = [-1]*n
for v in range(n):
if colors[v] == -1:
dfs(v, adj, colors, color)
color += 1
m_i_c = [10**9 + 1]*color
for i in range(n):
m_i_c[colors[i]] = min(m_i_c[colors[i]], c[i])
summ = sum(filter(lambda x: x < 10**9 + 1, m_i_c))
print(summ)
```
| -1
|
|
601
|
A
|
The Two Routes
|
PROGRAMMING
| 1,600
|
[
"graphs",
"shortest paths"
] | null | null |
In Absurdistan, there are *n* towns (numbered 1 through *n*) and *m* bidirectional railways. There is also an absurdly simple road network — for each pair of different towns *x* and *y*, there is a bidirectional road between towns *x* and *y* if and only if there is no railway between them. Travelling to a different town using one railway or one road always takes exactly one hour.
A train and a bus leave town 1 at the same time. They both have the same destination, town *n*, and don't make any stops on the way (but they can wait in town *n*). The train can move only along railways and the bus can move only along roads.
You've been asked to plan out routes for the vehicles; each route can use any road/railway multiple times. One of the most important aspects to consider is safety — in order to avoid accidents at railway crossings, the train and the bus must not arrive at the same town (except town *n*) simultaneously.
Under these constraints, what is the minimum number of hours needed for both vehicles to reach town *n* (the maximum of arrival times of the bus and the train)? Note, that bus and train are not required to arrive to the town *n* at the same moment of time, but are allowed to do so.
|
The first line of the input contains two integers *n* and *m* (2<=≤<=*n*<=≤<=400, 0<=≤<=*m*<=≤<=*n*(*n*<=-<=1)<=/<=2) — the number of towns and the number of railways respectively.
Each of the next *m* lines contains two integers *u* and *v*, denoting a railway between towns *u* and *v* (1<=≤<=*u*,<=*v*<=≤<=*n*, *u*<=≠<=*v*).
You may assume that there is at most one railway connecting any two towns.
|
Output one integer — the smallest possible time of the later vehicle's arrival in town *n*. If it's impossible for at least one of the vehicles to reach town *n*, output <=-<=1.
|
[
"4 2\n1 3\n3 4\n",
"4 6\n1 2\n1 3\n1 4\n2 3\n2 4\n3 4\n",
"5 5\n4 2\n3 5\n4 5\n5 1\n1 2\n"
] |
[
"2\n",
"-1\n",
"3\n"
] |
In the first sample, the train can take the route <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/7c0aa60a06309ef607b7159fd7f3687ea0d943ce.png" style="max-width: 100.0%;max-height: 100.0%;"/> and the bus can take the route <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/a26c2f3e93c9d9be6c21cb5d2bd6ac1f99f4ff55.png" style="max-width: 100.0%;max-height: 100.0%;"/>. Note that they can arrive at town 4 at the same time.
In the second sample, Absurdistan is ruled by railwaymen. There are no roads, so there's no way for the bus to reach town 4.
| 500
|
[
{
"input": "4 2\n1 3\n3 4",
"output": "2"
},
{
"input": "4 6\n1 2\n1 3\n1 4\n2 3\n2 4\n3 4",
"output": "-1"
},
{
"input": "5 5\n4 2\n3 5\n4 5\n5 1\n1 2",
"output": "3"
},
{
"input": "5 4\n1 2\n3 2\n3 4\n5 4",
"output": "4"
},
{
"input": "3 1\n1 2",
"output": "-1"
},
{
"input": "2 1\n1 2",
"output": "-1"
},
{
"input": "2 0",
"output": "-1"
},
{
"input": "20 0",
"output": "-1"
},
{
"input": "381 0",
"output": "-1"
},
{
"input": "3 3\n1 2\n2 3\n3 1",
"output": "-1"
},
{
"input": "3 0",
"output": "-1"
},
{
"input": "3 1\n1 3",
"output": "2"
},
{
"input": "3 2\n2 3\n3 1",
"output": "-1"
},
{
"input": "4 1\n1 4",
"output": "2"
},
{
"input": "4 5\n1 3\n2 1\n3 4\n4 2\n2 3",
"output": "2"
},
{
"input": "20 1\n20 1",
"output": "2"
},
{
"input": "21 1\n21 1",
"output": "2"
},
{
"input": "100 1\n100 1",
"output": "2"
},
{
"input": "400 1\n1 400",
"output": "2"
},
{
"input": "5 5\n2 5\n1 2\n1 4\n1 3\n3 2",
"output": "2"
}
] | 1,700,236,314
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 59
| 467
| 6,451,200
|
import collections
n,r=map(int,input().split())
train={i:[] for i in range(1,n+1)}
bus={i:[j for j in range(1,n+1) if j!=i] for i in range(1,n+1)}
for i in range(r):
a,b=map(int,input().split())
train[a].append(b)
train[b].append(a)
bus[a].remove(b)
bus[b].remove(a)
def bfs(s,dic):
q=collections.deque([[s,0]])
vis=set([s])
t=-1
while q:
a,b=q.popleft()
if a==n:
t=b
break
for i in dic[a]:
if i not in vis:
vis.add(i)
q.append([i,b+1])
return t
a=(bfs(1,train))
b=(bfs(1,bus))
if a<0 or b<0:
print(-1)
else:
print(max(a,b))
|
Title: The Two Routes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In Absurdistan, there are *n* towns (numbered 1 through *n*) and *m* bidirectional railways. There is also an absurdly simple road network — for each pair of different towns *x* and *y*, there is a bidirectional road between towns *x* and *y* if and only if there is no railway between them. Travelling to a different town using one railway or one road always takes exactly one hour.
A train and a bus leave town 1 at the same time. They both have the same destination, town *n*, and don't make any stops on the way (but they can wait in town *n*). The train can move only along railways and the bus can move only along roads.
You've been asked to plan out routes for the vehicles; each route can use any road/railway multiple times. One of the most important aspects to consider is safety — in order to avoid accidents at railway crossings, the train and the bus must not arrive at the same town (except town *n*) simultaneously.
Under these constraints, what is the minimum number of hours needed for both vehicles to reach town *n* (the maximum of arrival times of the bus and the train)? Note, that bus and train are not required to arrive to the town *n* at the same moment of time, but are allowed to do so.
Input Specification:
The first line of the input contains two integers *n* and *m* (2<=≤<=*n*<=≤<=400, 0<=≤<=*m*<=≤<=*n*(*n*<=-<=1)<=/<=2) — the number of towns and the number of railways respectively.
Each of the next *m* lines contains two integers *u* and *v*, denoting a railway between towns *u* and *v* (1<=≤<=*u*,<=*v*<=≤<=*n*, *u*<=≠<=*v*).
You may assume that there is at most one railway connecting any two towns.
Output Specification:
Output one integer — the smallest possible time of the later vehicle's arrival in town *n*. If it's impossible for at least one of the vehicles to reach town *n*, output <=-<=1.
Demo Input:
['4 2\n1 3\n3 4\n', '4 6\n1 2\n1 3\n1 4\n2 3\n2 4\n3 4\n', '5 5\n4 2\n3 5\n4 5\n5 1\n1 2\n']
Demo Output:
['2\n', '-1\n', '3\n']
Note:
In the first sample, the train can take the route <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/7c0aa60a06309ef607b7159fd7f3687ea0d943ce.png" style="max-width: 100.0%;max-height: 100.0%;"/> and the bus can take the route <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/a26c2f3e93c9d9be6c21cb5d2bd6ac1f99f4ff55.png" style="max-width: 100.0%;max-height: 100.0%;"/>. Note that they can arrive at town 4 at the same time.
In the second sample, Absurdistan is ruled by railwaymen. There are no roads, so there's no way for the bus to reach town 4.
|
```python
import collections
n,r=map(int,input().split())
train={i:[] for i in range(1,n+1)}
bus={i:[j for j in range(1,n+1) if j!=i] for i in range(1,n+1)}
for i in range(r):
a,b=map(int,input().split())
train[a].append(b)
train[b].append(a)
bus[a].remove(b)
bus[b].remove(a)
def bfs(s,dic):
q=collections.deque([[s,0]])
vis=set([s])
t=-1
while q:
a,b=q.popleft()
if a==n:
t=b
break
for i in dic[a]:
if i not in vis:
vis.add(i)
q.append([i,b+1])
return t
a=(bfs(1,train))
b=(bfs(1,bus))
if a<0 or b<0:
print(-1)
else:
print(max(a,b))
```
| 3
|
|
677
|
A
|
Vanya and Fence
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Vanya and his friends are walking along the fence of height *h* and they do not want the guard to notice them. In order to achieve this the height of each of the friends should not exceed *h*. If the height of some person is greater than *h* he can bend down and then he surely won't be noticed by the guard. The height of the *i*-th person is equal to *a**i*.
Consider the width of the person walking as usual to be equal to 1, while the width of the bent person is equal to 2. Friends want to talk to each other while walking, so they would like to walk in a single row. What is the minimum width of the road, such that friends can walk in a row and remain unattended by the guard?
|
The first line of the input contains two integers *n* and *h* (1<=≤<=*n*<=≤<=1000, 1<=≤<=*h*<=≤<=1000) — the number of friends and the height of the fence, respectively.
The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=2*h*), the *i*-th of them is equal to the height of the *i*-th person.
|
Print a single integer — the minimum possible valid width of the road.
|
[
"3 7\n4 5 14\n",
"6 1\n1 1 1 1 1 1\n",
"6 5\n7 6 8 9 10 5\n"
] |
[
"4\n",
"6\n",
"11\n"
] |
In the first sample, only person number 3 must bend down, so the required width is equal to 1 + 1 + 2 = 4.
In the second sample, all friends are short enough and no one has to bend, so the width 1 + 1 + 1 + 1 + 1 + 1 = 6 is enough.
In the third sample, all the persons have to bend, except the last one. The required minimum width of the road is equal to 2 + 2 + 2 + 2 + 2 + 1 = 11.
| 500
|
[
{
"input": "3 7\n4 5 14",
"output": "4"
},
{
"input": "6 1\n1 1 1 1 1 1",
"output": "6"
},
{
"input": "6 5\n7 6 8 9 10 5",
"output": "11"
},
{
"input": "10 420\n214 614 297 675 82 740 174 23 255 15",
"output": "13"
},
{
"input": "10 561\n657 23 1096 487 785 66 481 554 1000 821",
"output": "15"
},
{
"input": "100 342\n478 143 359 336 162 333 385 515 117 496 310 538 469 539 258 676 466 677 1 296 150 560 26 213 627 221 255 126 617 174 279 178 24 435 70 145 619 46 669 566 300 67 576 251 58 176 441 564 569 194 24 669 73 262 457 259 619 78 400 579 222 626 269 47 80 315 160 194 455 186 315 424 197 246 683 220 68 682 83 233 290 664 273 598 362 305 674 614 321 575 362 120 14 534 62 436 294 351 485 396",
"output": "144"
},
{
"input": "100 290\n244 49 276 77 449 261 468 458 201 424 9 131 300 88 432 394 104 77 13 289 435 259 111 453 168 394 156 412 351 576 178 530 81 271 228 564 125 328 42 372 205 61 180 471 33 360 567 331 222 318 241 117 529 169 188 484 202 202 299 268 246 343 44 364 333 494 59 236 84 485 50 8 428 8 571 227 205 310 210 9 324 472 368 490 114 84 296 305 411 351 569 393 283 120 510 171 232 151 134 366",
"output": "145"
},
{
"input": "1 1\n1",
"output": "1"
},
{
"input": "1 1\n2",
"output": "2"
},
{
"input": "46 71\n30 26 56 138 123 77 60 122 73 45 79 10 130 3 14 1 38 46 128 50 82 16 32 68 28 98 62 106 2 49 131 11 114 39 139 70 40 50 45 137 33 30 35 136 135 19",
"output": "63"
},
{
"input": "20 723\n212 602 293 591 754 91 1135 640 80 495 845 928 1399 498 926 1431 1226 869 814 1386",
"output": "31"
},
{
"input": "48 864\n843 1020 751 1694 18 1429 1395 1174 272 1158 1628 1233 1710 441 765 561 778 748 1501 1200 563 1263 1398 1687 1518 1640 1591 839 500 466 1603 1587 1201 1209 432 868 1159 639 649 628 9 91 1036 147 896 1557 941 518",
"output": "75"
},
{
"input": "26 708\n549 241 821 734 945 1161 566 1268 216 30 1142 730 529 1014 255 168 796 1148 89 113 1328 286 743 871 1259 1397",
"output": "41"
},
{
"input": "75 940\n1620 1745 1599 441 64 1466 1496 1239 1716 1475 778 106 1136 1212 1261 444 781 257 1071 747 626 232 609 1544 682 1326 469 1361 1460 1450 1207 1319 922 625 1737 1057 1698 592 692 80 1016 541 1254 201 682 1007 847 206 1066 809 259 109 240 1611 219 1455 1326 1377 1827 786 42 1002 1382 1592 543 1866 1198 334 1524 1760 340 1566 955 257 1118",
"output": "116"
}
] | 1,692,340,071
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 29
| 46
| 0
|
a,b=map(int,input().split())
c=list(map(int,input().split()))
c1=0
for i in c:
if i<=b:
c1+=1
else:
c1+=2
print(c1)
|
Title: Vanya and Fence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vanya and his friends are walking along the fence of height *h* and they do not want the guard to notice them. In order to achieve this the height of each of the friends should not exceed *h*. If the height of some person is greater than *h* he can bend down and then he surely won't be noticed by the guard. The height of the *i*-th person is equal to *a**i*.
Consider the width of the person walking as usual to be equal to 1, while the width of the bent person is equal to 2. Friends want to talk to each other while walking, so they would like to walk in a single row. What is the minimum width of the road, such that friends can walk in a row and remain unattended by the guard?
Input Specification:
The first line of the input contains two integers *n* and *h* (1<=≤<=*n*<=≤<=1000, 1<=≤<=*h*<=≤<=1000) — the number of friends and the height of the fence, respectively.
The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=2*h*), the *i*-th of them is equal to the height of the *i*-th person.
Output Specification:
Print a single integer — the minimum possible valid width of the road.
Demo Input:
['3 7\n4 5 14\n', '6 1\n1 1 1 1 1 1\n', '6 5\n7 6 8 9 10 5\n']
Demo Output:
['4\n', '6\n', '11\n']
Note:
In the first sample, only person number 3 must bend down, so the required width is equal to 1 + 1 + 2 = 4.
In the second sample, all friends are short enough and no one has to bend, so the width 1 + 1 + 1 + 1 + 1 + 1 = 6 is enough.
In the third sample, all the persons have to bend, except the last one. The required minimum width of the road is equal to 2 + 2 + 2 + 2 + 2 + 1 = 11.
|
```python
a,b=map(int,input().split())
c=list(map(int,input().split()))
c1=0
for i in c:
if i<=b:
c1+=1
else:
c1+=2
print(c1)
```
| 3
|
|
181
|
A
|
Series of Crimes
|
PROGRAMMING
| 800
|
[
"brute force",
"geometry",
"implementation"
] | null | null |
The Berland capital is shaken with three bold crimes committed by the Pihsters, a notorious criminal gang.
The Berland capital's map is represented by an *n*<=×<=*m* rectangular table. Each cell of the table on the map represents some districts of the capital.
The capital's main detective Polycarpus took a map and marked there the districts where the first three robberies had been committed as asterisks. Deduction tells Polycarpus that the fourth robbery will be committed in such district, that all four robbed districts will form the vertices of some rectangle, parallel to the sides of the map.
Polycarpus is good at deduction but he's hopeless at math. So he asked you to find the district where the fourth robbery will be committed.
|
The first line contains two space-separated integers *n* and *m* (2<=≤<=*n*,<=*m*<=≤<=100) — the number of rows and columns in the table, correspondingly.
Each of the next *n* lines contains *m* characters — the description of the capital's map. Each character can either be a "." (dot), or an "*" (asterisk). A character equals "*" if the corresponding district has been robbed. Otherwise, it equals ".".
It is guaranteed that the map has exactly three characters "*" and we can always find the fourth district that meets the problem requirements.
|
Print two integers — the number of the row and the number of the column of the city district that is the fourth one to be robbed. The rows are numbered starting from one from top to bottom and the columns are numbered starting from one from left to right.
|
[
"3 2\n.*\n..\n**\n",
"3 3\n*.*\n*..\n...\n"
] |
[
"1 1\n",
"2 3\n"
] |
none
| 500
|
[
{
"input": "3 2\n.*\n..\n**",
"output": "1 1"
},
{
"input": "2 5\n*....\n*...*",
"output": "1 5"
},
{
"input": "7 2\n..\n**\n..\n..\n..\n..\n.*",
"output": "7 1"
},
{
"input": "7 2\n*.\n..\n..\n..\n..\n..\n**",
"output": "1 2"
},
{
"input": "2 10\n*......*..\n.......*..",
"output": "2 1"
},
{
"input": "10 3\n*..\n...\n...\n...\n...\n...\n...\n...\n...\n**.",
"output": "1 2"
},
{
"input": "100 2\n*.\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n**",
"output": "1 2"
},
{
"input": "99 3\n**.\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n*..",
"output": "99 2"
},
{
"input": "98 3\n...\n.*.\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n.**",
"output": "2 3"
},
{
"input": "99 3\n**.\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n*..",
"output": "99 2"
},
{
"input": "98 3\n...\n.*.\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n.**",
"output": "2 3"
},
{
"input": "2 100\n...................................................................................................*\n*..................................................................................................*",
"output": "1 1"
},
{
"input": "2 2\n*.\n**",
"output": "1 2"
},
{
"input": "2 2\n**\n.*",
"output": "2 1"
},
{
"input": "2 3\n*.*\n..*",
"output": "2 1"
}
] | 1,600,102,288
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 36
| 216
| 307,200
|
n, m = map(int, input().split())
dmap = []
for i in range(n):
dmap.append(list(input()))
if 1 == dmap[i].count("*"):
answer_column = i
if 2 == dmap[-1].count("*"):
filled_column = i
indec1 = [i for i, j in enumerate(dmap[filled_column]) if j == "*"]
indec2 = [i for i, j in enumerate(dmap[answer_column]) if j == "*"]
for i in indec1:
if i != indec2[0]:
answer = i
print(str(answer_column + 1) + " " + str(answer + 1))
|
Title: Series of Crimes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Berland capital is shaken with three bold crimes committed by the Pihsters, a notorious criminal gang.
The Berland capital's map is represented by an *n*<=×<=*m* rectangular table. Each cell of the table on the map represents some districts of the capital.
The capital's main detective Polycarpus took a map and marked there the districts where the first three robberies had been committed as asterisks. Deduction tells Polycarpus that the fourth robbery will be committed in such district, that all four robbed districts will form the vertices of some rectangle, parallel to the sides of the map.
Polycarpus is good at deduction but he's hopeless at math. So he asked you to find the district where the fourth robbery will be committed.
Input Specification:
The first line contains two space-separated integers *n* and *m* (2<=≤<=*n*,<=*m*<=≤<=100) — the number of rows and columns in the table, correspondingly.
Each of the next *n* lines contains *m* characters — the description of the capital's map. Each character can either be a "." (dot), or an "*" (asterisk). A character equals "*" if the corresponding district has been robbed. Otherwise, it equals ".".
It is guaranteed that the map has exactly three characters "*" and we can always find the fourth district that meets the problem requirements.
Output Specification:
Print two integers — the number of the row and the number of the column of the city district that is the fourth one to be robbed. The rows are numbered starting from one from top to bottom and the columns are numbered starting from one from left to right.
Demo Input:
['3 2\n.*\n..\n**\n', '3 3\n*.*\n*..\n...\n']
Demo Output:
['1 1\n', '2 3\n']
Note:
none
|
```python
n, m = map(int, input().split())
dmap = []
for i in range(n):
dmap.append(list(input()))
if 1 == dmap[i].count("*"):
answer_column = i
if 2 == dmap[-1].count("*"):
filled_column = i
indec1 = [i for i, j in enumerate(dmap[filled_column]) if j == "*"]
indec2 = [i for i, j in enumerate(dmap[answer_column]) if j == "*"]
for i in indec1:
if i != indec2[0]:
answer = i
print(str(answer_column + 1) + " " + str(answer + 1))
```
| 3
|
|
266
|
A
|
Stones on the Table
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
There are *n* stones on the table in a row, each of them can be red, green or blue. Count the minimum number of stones to take from the table so that any two neighboring stones had different colors. Stones in a row are considered neighboring if there are no other stones between them.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=50) — the number of stones on the table.
The next line contains string *s*, which represents the colors of the stones. We'll consider the stones in the row numbered from 1 to *n* from left to right. Then the *i*-th character *s* equals "R", if the *i*-th stone is red, "G", if it's green and "B", if it's blue.
|
Print a single integer — the answer to the problem.
|
[
"3\nRRG\n",
"5\nRRRRR\n",
"4\nBRBG\n"
] |
[
"1\n",
"4\n",
"0\n"
] |
none
| 500
|
[
{
"input": "3\nRRG",
"output": "1"
},
{
"input": "5\nRRRRR",
"output": "4"
},
{
"input": "4\nBRBG",
"output": "0"
},
{
"input": "1\nB",
"output": "0"
},
{
"input": "2\nBG",
"output": "0"
},
{
"input": "3\nBGB",
"output": "0"
},
{
"input": "4\nRBBR",
"output": "1"
},
{
"input": "5\nRGGBG",
"output": "1"
},
{
"input": "10\nGGBRBRGGRB",
"output": "2"
},
{
"input": "50\nGRBGGRBRGRBGGBBBBBGGGBBBBRBRGBRRBRGBBBRBBRRGBGGGRB",
"output": "18"
},
{
"input": "15\nBRRBRGGBBRRRRGR",
"output": "6"
},
{
"input": "20\nRRGBBRBRGRGBBGGRGRRR",
"output": "6"
},
{
"input": "25\nBBGBGRBGGBRRBGRRBGGBBRBRB",
"output": "6"
},
{
"input": "30\nGRGGGBGGRGBGGRGRBGBGBRRRRRRGRB",
"output": "9"
},
{
"input": "35\nGBBGBRGBBGGRBBGBRRGGRRRRRRRBRBBRRGB",
"output": "14"
},
{
"input": "40\nGBBRRGBGGGRGGGRRRRBRBGGBBGGGBGBBBBBRGGGG",
"output": "20"
},
{
"input": "45\nGGGBBRBBRRGRBBGGBGRBRGGBRBRGBRRGBGRRBGRGRBRRG",
"output": "11"
},
{
"input": "50\nRBGGBGGRBGRBBBGBBGRBBBGGGRBBBGBBBGRGGBGGBRBGBGRRGG",
"output": "17"
},
{
"input": "50\nGGGBBRGGGGGRRGGRBGGRGBBRBRRBGRGBBBGBRBGRGBBGRGGBRB",
"output": "16"
},
{
"input": "50\nGBGRGRRBRRRRRGGBBGBRRRBBBRBBBRRGRBBRGBRBGGRGRBBGGG",
"output": "19"
},
{
"input": "10\nGRRBRBRBGR",
"output": "1"
},
{
"input": "10\nBRBGBGRRBR",
"output": "1"
},
{
"input": "20\nGBGBGGRRRRGRBBGRGRGR",
"output": "5"
},
{
"input": "20\nRRGGRBBGBBRBGRRBRRBG",
"output": "6"
},
{
"input": "30\nBGBRGBBBGRGBBRGBGRBBBRGGRRGRRB",
"output": "8"
},
{
"input": "30\nBBBBGGBRBGBBGBGBGBGGGRGRRGGBBB",
"output": "11"
},
{
"input": "40\nGBRRGRBGBRRGBRGGGBRGBGBRGBBRRGRGGBBGBGBB",
"output": "9"
},
{
"input": "40\nBRGRGGRGGRBBRRRBRBBGGGRRGBGBBGRBBRGBRRGG",
"output": "13"
},
{
"input": "50\nRBGBGGRRGGRGGBGBGRRBGGBGBRRBBGBBGBBBGBBRBBRBRBRGRG",
"output": "13"
},
{
"input": "50\nRBRRGBGRRRBGRRBGRRGRBBRBBRRBRGGBRBRRBGGRBGGBRBRGRB",
"output": "12"
},
{
"input": "2\nBB",
"output": "1"
},
{
"input": "50\nRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR",
"output": "49"
},
{
"input": "50\nRRRRRRRRGRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR",
"output": "47"
},
{
"input": "50\nRRRRRRRRRRRRGGRRRRRRRRRBRRRRRRRRRRRRRRBBRRRRRRRRRR",
"output": "43"
}
] | 1,699,726,537
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 34
| 92
| 0
|
c=0
x=input()
n=input()
l=list(n)
for i in range(len(l)-1):
if(l[i]==l[i+1]):
c+=1
print(c)
|
Title: Stones on the Table
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* stones on the table in a row, each of them can be red, green or blue. Count the minimum number of stones to take from the table so that any two neighboring stones had different colors. Stones in a row are considered neighboring if there are no other stones between them.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=50) — the number of stones on the table.
The next line contains string *s*, which represents the colors of the stones. We'll consider the stones in the row numbered from 1 to *n* from left to right. Then the *i*-th character *s* equals "R", if the *i*-th stone is red, "G", if it's green and "B", if it's blue.
Output Specification:
Print a single integer — the answer to the problem.
Demo Input:
['3\nRRG\n', '5\nRRRRR\n', '4\nBRBG\n']
Demo Output:
['1\n', '4\n', '0\n']
Note:
none
|
```python
c=0
x=input()
n=input()
l=list(n)
for i in range(len(l)-1):
if(l[i]==l[i+1]):
c+=1
print(c)
```
| 3
|
|
2
|
B
|
The least round way
|
PROGRAMMING
| 2,000
|
[
"dp",
"math"
] |
B. The least round way
|
2
|
64
|
There is a square matrix *n*<=×<=*n*, consisting of non-negative integer numbers. You should find such a way on it that
- starts in the upper left cell of the matrix; - each following cell is to the right or down from the current cell; - the way ends in the bottom right cell.
Moreover, if we multiply together all the numbers along the way, the result should be the least "round". In other words, it should end in the least possible number of zeros.
|
The first line contains an integer number *n* (2<=≤<=*n*<=≤<=1000), *n* is the size of the matrix. Then follow *n* lines containing the matrix elements (non-negative integer numbers not exceeding 109).
|
In the first line print the least number of trailing zeros. In the second line print the correspondent way itself.
|
[
"3\n1 2 3\n4 5 6\n7 8 9\n"
] |
[
"0\nDDRR\n"
] |
none
| 0
|
[
{
"input": "3\n1 2 3\n4 5 6\n7 8 9",
"output": "0\nDDRR"
},
{
"input": "2\n7 6\n3 8",
"output": "0\nDR"
},
{
"input": "3\n4 10 5\n10 9 4\n6 5 3",
"output": "1\nDRRD"
},
{
"input": "4\n1 1 9 9\n3 4 7 3\n7 9 1 7\n1 7 1 5",
"output": "0\nDDDRRR"
},
{
"input": "5\n8 3 2 1 4\n3 7 2 4 8\n9 2 8 9 10\n2 3 6 10 1\n8 2 2 8 4",
"output": "0\nDDDDRRRR"
},
{
"input": "6\n5 5 4 10 5 5\n7 10 8 7 6 6\n7 1 7 9 7 8\n5 5 3 3 10 9\n5 8 10 6 3 8\n3 10 5 4 3 4",
"output": "1\nDDRRDRDDRR"
},
{
"input": "7\n2 9 8 2 7 4 8\n9 5 4 4 8 5 3\n5 7 2 10 8 1 8\n2 7 10 7 5 7 7\n9 2 7 6 4 8 4\n7 2 4 7 4 1 8\n9 5 3 10 1 6 2",
"output": "0\nRRDRRDRDDDDR"
},
{
"input": "8\n1 1 10 1 8 4 8 7\n9 3 3 2 2 6 2 4\n7 4 3 5 10 3 5 1\n8 4 4 10 4 5 9 4\n5 5 5 2 6 7 1 8\n4 10 1 3 2 4 8 3\n8 1 10 2 8 2 2 4\n2 10 6 8 10 2 8 4",
"output": "0\nDRRRRRRRDDDDDD"
},
{
"input": "9\n8 3 3 3 10 3 10 5 6\n2 1 6 1 8 1 9 1 6\n6 1 5 4 2 2 10 4 9\n1 9 1 3 10 6 10 5 5\n1 10 5 4 7 2 5 9 10\n6 6 1 3 1 9 4 9 9\n5 3 7 6 4 6 2 10 2\n9 3 3 10 5 6 7 6 4\n4 9 6 7 4 3 7 6 5",
"output": "1\nDDDDDRDDDRRRRRRR"
},
{
"input": "10\n10 8 6 5 9 8 2 5 3 2\n3 1 8 6 8 10 5 5 7 8\n5 9 7 7 4 9 7 2 5 2\n5 9 9 5 4 2 6 6 8 1\n10 6 9 9 10 5 6 3 5 9\n6 7 10 3 1 4 3 6 7 6\n10 10 4 7 4 4 8 6 7 4\n10 5 8 2 2 7 4 4 1 4\n8 4 6 10 10 6 1 3 3 1\n9 9 7 2 9 5 1 8 6 3",
"output": "1\nDRDDDRRDDDRRDRDRRR"
}
] | 1,678,649,175
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 15
| 2,000
| 24,985,600
|
n = int(input())
ma = [[] for x in range(n)]
zflag = 0
zindex = [0,0]
for i in range(n):
j = 0
inp = input().split()
for x in inp:
if x == '0':
ma[i].append(10)
zflag = 1
zindex = [i,j]
else:
ma[i].append(int(x))
j += 1
ct2 = [[0 for x in range(n)] for y in range(n)]
ct5 = [[0 for x in range(n)] for y in range(n)]
for i in range(n):
for j in range(n):
tmp = ma[i][j]
while tmp % 2 == 0:
ct2[i][j] += 1
tmp //= 2
while tmp % 5 == 0:
ct5[i][j] += 1
tmp //= 5
dp2 = [[0 for x in range(n)] for y in range(n)]
dp5 = [[0 for x in range(n)] for y in range(n)]
dp2[0][0] = ct2[0][0]
dp5[0][0] = ct5[0][0]
for i in range(1, n):
dp2[0][i] = dp2[0][i-1] + ct2[0][i]
dp5[0][i] = dp5[0][i-1] + ct5[0][i]
dp2[i][0] = dp2[i-1][0] + ct2[i][0]
dp5[i][0] = dp5[i-1][0] + ct5[i][0]
for i in range(1, n):
for j in range(1,n):
dp2[i][j] = min(dp2[i-1][j] + ct2[i][j], dp2[i][j-1] + ct2[i][j])
dp5[i][j] = min(dp5[i-1][j]+ct5[i][j], dp5[i][j-1]+ct5[i][j])
res=min(dp2[n-1][n-1],dp5[n-1][n-1])
if dp2[n-1][n-1]<dp5[n-1][n-1]:
path=''
i=n-1
j=n-1
while i>0 and j>0:
if dp2[i-1][j]<dp2[i][j-1]:
path='D'+path
i-=1
else:
path='R'+path
j-=1
while i>0:
path='D'+path
i-=1
while j>0:
path='R'+path
j-=1
else:
path=''
i=n-1
j=n-1
while i>0 and j>0:
if dp5[i-1][j]<dp5[i][j-1]:
path='D'+path
i-=1
else:
path='R'+path
j-=1
while i>0:
path='D'+path
i-=1
while j>0:
path='R'+path
j-=1
if zflag:
if res>0:
res=1
path=''
for i in range(zindex[0]):
path+='D'
for i in range(zindex[1]):
path+='R'
for i in range(n-zindex[0]-1):
path+='D'
for i in range(n-zindex[1]-1):
path+='R'
print(res)
print(path)
|
Title: The least round way
Time Limit: 2 seconds
Memory Limit: 64 megabytes
Problem Description:
There is a square matrix *n*<=×<=*n*, consisting of non-negative integer numbers. You should find such a way on it that
- starts in the upper left cell of the matrix; - each following cell is to the right or down from the current cell; - the way ends in the bottom right cell.
Moreover, if we multiply together all the numbers along the way, the result should be the least "round". In other words, it should end in the least possible number of zeros.
Input Specification:
The first line contains an integer number *n* (2<=≤<=*n*<=≤<=1000), *n* is the size of the matrix. Then follow *n* lines containing the matrix elements (non-negative integer numbers not exceeding 109).
Output Specification:
In the first line print the least number of trailing zeros. In the second line print the correspondent way itself.
Demo Input:
['3\n1 2 3\n4 5 6\n7 8 9\n']
Demo Output:
['0\nDDRR\n']
Note:
none
|
```python
n = int(input())
ma = [[] for x in range(n)]
zflag = 0
zindex = [0,0]
for i in range(n):
j = 0
inp = input().split()
for x in inp:
if x == '0':
ma[i].append(10)
zflag = 1
zindex = [i,j]
else:
ma[i].append(int(x))
j += 1
ct2 = [[0 for x in range(n)] for y in range(n)]
ct5 = [[0 for x in range(n)] for y in range(n)]
for i in range(n):
for j in range(n):
tmp = ma[i][j]
while tmp % 2 == 0:
ct2[i][j] += 1
tmp //= 2
while tmp % 5 == 0:
ct5[i][j] += 1
tmp //= 5
dp2 = [[0 for x in range(n)] for y in range(n)]
dp5 = [[0 for x in range(n)] for y in range(n)]
dp2[0][0] = ct2[0][0]
dp5[0][0] = ct5[0][0]
for i in range(1, n):
dp2[0][i] = dp2[0][i-1] + ct2[0][i]
dp5[0][i] = dp5[0][i-1] + ct5[0][i]
dp2[i][0] = dp2[i-1][0] + ct2[i][0]
dp5[i][0] = dp5[i-1][0] + ct5[i][0]
for i in range(1, n):
for j in range(1,n):
dp2[i][j] = min(dp2[i-1][j] + ct2[i][j], dp2[i][j-1] + ct2[i][j])
dp5[i][j] = min(dp5[i-1][j]+ct5[i][j], dp5[i][j-1]+ct5[i][j])
res=min(dp2[n-1][n-1],dp5[n-1][n-1])
if dp2[n-1][n-1]<dp5[n-1][n-1]:
path=''
i=n-1
j=n-1
while i>0 and j>0:
if dp2[i-1][j]<dp2[i][j-1]:
path='D'+path
i-=1
else:
path='R'+path
j-=1
while i>0:
path='D'+path
i-=1
while j>0:
path='R'+path
j-=1
else:
path=''
i=n-1
j=n-1
while i>0 and j>0:
if dp5[i-1][j]<dp5[i][j-1]:
path='D'+path
i-=1
else:
path='R'+path
j-=1
while i>0:
path='D'+path
i-=1
while j>0:
path='R'+path
j-=1
if zflag:
if res>0:
res=1
path=''
for i in range(zindex[0]):
path+='D'
for i in range(zindex[1]):
path+='R'
for i in range(n-zindex[0]-1):
path+='D'
for i in range(n-zindex[1]-1):
path+='R'
print(res)
print(path)
```
| 0
|
369
|
A
|
Valera and Plates
|
PROGRAMMING
| 900
|
[
"greedy",
"implementation"
] | null | null |
Valera is a lazy student. He has *m* clean bowls and *k* clean plates.
Valera has made an eating plan for the next *n* days. As Valera is lazy, he will eat exactly one dish per day. At that, in order to eat a dish, he needs exactly one clean plate or bowl. We know that Valera can cook only two types of dishes. He can eat dishes of the first type from bowls and dishes of the second type from either bowls or plates.
When Valera finishes eating, he leaves a dirty plate/bowl behind. His life philosophy doesn't let him eat from dirty kitchenware. So sometimes he needs to wash his plate/bowl before eating. Find the minimum number of times Valera will need to wash a plate/bowl, if he acts optimally.
|
The first line of the input contains three integers *n*, *m*, *k* (1<=≤<=*n*,<=*m*,<=*k*<=≤<=1000) — the number of the planned days, the number of clean bowls and the number of clean plates.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=2). If *a**i* equals one, then on day *i* Valera will eat a first type dish. If *a**i* equals two, then on day *i* Valera will eat a second type dish.
|
Print a single integer — the minimum number of times Valera will need to wash a plate/bowl.
|
[
"3 1 1\n1 2 1\n",
"4 3 1\n1 1 1 1\n",
"3 1 2\n2 2 2\n",
"8 2 2\n1 2 1 2 1 2 1 2\n"
] |
[
"1\n",
"1\n",
"0\n",
"4\n"
] |
In the first sample Valera will wash a bowl only on the third day, so the answer is one.
In the second sample, Valera will have the first type of the dish during all four days, and since there are only three bowls, he will wash a bowl exactly once.
In the third sample, Valera will have the second type of dish for all three days, and as they can be eaten from either a plate or a bowl, he will never need to wash a plate/bowl.
| 500
|
[
{
"input": "3 1 1\n1 2 1",
"output": "1"
},
{
"input": "4 3 1\n1 1 1 1",
"output": "1"
},
{
"input": "3 1 2\n2 2 2",
"output": "0"
},
{
"input": "8 2 2\n1 2 1 2 1 2 1 2",
"output": "4"
},
{
"input": "2 100 100\n2 2",
"output": "0"
},
{
"input": "1 1 1\n2",
"output": "0"
},
{
"input": "233 100 1\n2 2 1 1 1 2 2 2 2 1 1 2 2 2 1 2 2 1 1 1 2 2 1 1 1 1 2 1 2 2 1 1 2 2 1 2 2 1 2 1 2 1 2 2 2 1 1 1 1 2 1 2 1 1 2 1 1 2 2 1 2 1 2 1 1 1 1 1 1 1 1 1 2 1 2 2 2 1 1 2 2 1 1 1 1 2 1 1 2 1 2 2 2 1 1 1 2 2 2 1 1 1 1 2 1 2 1 1 1 1 2 2 2 1 1 2 1 2 1 1 1 1 1 2 1 1 1 1 1 2 1 1 2 2 1 2 1 1 2 2 1 1 2 2 1 1 1 2 2 1 1 2 1 2 1 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 1 2 2 1 1 1 2 2 1 1 2 2 1 1 2 1 1 2 2 1 2 2 2 2 2 2 1 2 2 2 2 2 1 1 2 2 2 2 2 2 1 1 1 2 1 2 2 2 2 2 2 2 2 1 1 2 1 2 1 2 2",
"output": "132"
},
{
"input": "123 100 1\n2 2 2 1 1 2 2 2 2 1 1 2 2 2 1 2 2 2 2 1 2 2 2 1 1 1 2 2 2 2 1 2 2 2 2 2 2 1 2 1 2 1 2 2 2 1 2 1 2 2 1 2 2 1 2 2 1 2 2 1 2 2 2 1 1 1 1 1 1 1 1 1 2 2 2 2 2 1 1 2 2 1 1 1 1 2 1 2 2 1 2 2 2 1 1 1 2 2 2 1 2 2 2 2 1 2 2 2 2 1 2 2 2 1 1 2 1 2 1 2 1 1 1",
"output": "22"
},
{
"input": "188 100 1\n2 2 1 1 1 2 2 2 2 1 1 2 2 2 1 2 2 1 1 1 2 2 1 1 1 1 2 1 2 2 1 1 2 2 1 2 2 1 2 1 2 1 2 2 2 1 1 1 1 2 1 2 1 1 2 1 1 2 2 1 2 1 2 1 1 1 1 1 1 1 1 1 2 1 2 2 2 1 1 2 2 1 1 1 1 2 1 1 2 1 2 2 2 1 1 1 2 2 2 1 1 1 1 2 1 2 1 1 1 1 2 2 2 1 1 2 1 2 1 1 1 1 1 2 1 1 1 1 1 2 1 1 2 2 1 2 1 1 2 2 1 1 2 2 1 1 1 2 2 1 1 2 1 2 1 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 1 2 2 1 1 1 2 2 1 1 2 2 1 1 2 1",
"output": "87"
},
{
"input": "3 1 2\n1 1 1",
"output": "2"
},
{
"input": "3 2 2\n1 1 1",
"output": "1"
},
{
"input": "3 2 1\n1 1 1",
"output": "1"
},
{
"input": "3 1 1\n1 1 1",
"output": "2"
},
{
"input": "5 1 2\n2 2 2 2 2",
"output": "2"
},
{
"input": "5 2 2\n2 2 2 2 2",
"output": "1"
},
{
"input": "5 2 1\n2 2 2 2 2",
"output": "2"
},
{
"input": "5 1 1\n2 2 2 2 2",
"output": "3"
},
{
"input": "1 1 2\n2",
"output": "0"
},
{
"input": "1 2 2\n2",
"output": "0"
},
{
"input": "1 2 1\n2",
"output": "0"
},
{
"input": "1 1 1\n2",
"output": "0"
},
{
"input": "6 3 1\n1 1 2 2 2 2",
"output": "2"
},
{
"input": "100 40 20\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 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": "40"
},
{
"input": "7 5 2\n2 2 1 1 1 1 1",
"output": "0"
},
{
"input": "10 4 4\n2 2 2 2 2 2 1 1 1 1",
"output": "2"
},
{
"input": "3 2 1\n2 1 1",
"output": "0"
},
{
"input": "7 6 1\n2 1 1 1 1 1 1",
"output": "0"
},
{
"input": "7 5 1\n1 1 1 2 2 2 2",
"output": "1"
},
{
"input": "5 3 1\n1 1 2 2 2",
"output": "1"
},
{
"input": "3 1 1\n2 2 2",
"output": "1"
},
{
"input": "5 2 2\n2 2 2 2 2",
"output": "1"
},
{
"input": "3 1 3\n1 1 1",
"output": "2"
},
{
"input": "5 2 1\n1 1 2 2 2",
"output": "2"
},
{
"input": "4 3 2\n2 1 1 1",
"output": "0"
},
{
"input": "4 2 1\n1 2 2 2",
"output": "1"
},
{
"input": "14 4 7\n1 1 1 2 2 2 2 2 2 2 2 2 2 2",
"output": "3"
},
{
"input": "12 10 4\n2 2 2 2 2 2 1 1 1 1 1 1",
"output": "0"
},
{
"input": "5 3 2\n2 2 1 1 1",
"output": "0"
}
] | 1,649,356,245
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 63
| 46
| 0
|
n, m, k=map(int, input().split())
a=list(map(int, input().split()))
v1=a.count(1)
v2=a.count(2)
h=v2-k
if(h<0):
h=0
l=v1+h-m
if(l<0):
l=0
print(l)
|
Title: Valera and Plates
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Valera is a lazy student. He has *m* clean bowls and *k* clean plates.
Valera has made an eating plan for the next *n* days. As Valera is lazy, he will eat exactly one dish per day. At that, in order to eat a dish, he needs exactly one clean plate or bowl. We know that Valera can cook only two types of dishes. He can eat dishes of the first type from bowls and dishes of the second type from either bowls or plates.
When Valera finishes eating, he leaves a dirty plate/bowl behind. His life philosophy doesn't let him eat from dirty kitchenware. So sometimes he needs to wash his plate/bowl before eating. Find the minimum number of times Valera will need to wash a plate/bowl, if he acts optimally.
Input Specification:
The first line of the input contains three integers *n*, *m*, *k* (1<=≤<=*n*,<=*m*,<=*k*<=≤<=1000) — the number of the planned days, the number of clean bowls and the number of clean plates.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=2). If *a**i* equals one, then on day *i* Valera will eat a first type dish. If *a**i* equals two, then on day *i* Valera will eat a second type dish.
Output Specification:
Print a single integer — the minimum number of times Valera will need to wash a plate/bowl.
Demo Input:
['3 1 1\n1 2 1\n', '4 3 1\n1 1 1 1\n', '3 1 2\n2 2 2\n', '8 2 2\n1 2 1 2 1 2 1 2\n']
Demo Output:
['1\n', '1\n', '0\n', '4\n']
Note:
In the first sample Valera will wash a bowl only on the third day, so the answer is one.
In the second sample, Valera will have the first type of the dish during all four days, and since there are only three bowls, he will wash a bowl exactly once.
In the third sample, Valera will have the second type of dish for all three days, and as they can be eaten from either a plate or a bowl, he will never need to wash a plate/bowl.
|
```python
n, m, k=map(int, input().split())
a=list(map(int, input().split()))
v1=a.count(1)
v2=a.count(2)
h=v2-k
if(h<0):
h=0
l=v1+h-m
if(l<0):
l=0
print(l)
```
| 3
|
|
913
|
B
|
Christmas Spruce
|
PROGRAMMING
| 1,200
|
[
"implementation",
"trees"
] | null | null |
Consider a rooted tree. A rooted tree has one special vertex called the root. All edges are directed from the root. Vertex *u* is called a child of vertex *v* and vertex *v* is called a parent of vertex *u* if there exists a directed edge from *v* to *u*. A vertex is called a leaf if it doesn't have children and has a parent.
Let's call a rooted tree a spruce if its every non-leaf vertex has at least 3 leaf children. You are given a rooted tree, check whether it's a spruce.
The definition of a rooted tree can be found [here](https://goo.gl/1dqvzz).
|
The first line contains one integer *n* — the number of vertices in the tree (3<=≤<=*n*<=≤<=1<=000). Each of the next *n*<=-<=1 lines contains one integer *p**i* (1<=≤<=*i*<=≤<=*n*<=-<=1) — the index of the parent of the *i*<=+<=1-th vertex (1<=≤<=*p**i*<=≤<=*i*).
Vertex 1 is the root. It's guaranteed that the root has at least 2 children.
|
Print "Yes" if the tree is a spruce and "No" otherwise.
|
[
"4\n1\n1\n1\n",
"7\n1\n1\n1\n2\n2\n2\n",
"8\n1\n1\n1\n1\n3\n3\n3\n"
] |
[
"Yes\n",
"No\n",
"Yes\n"
] |
The first example:
<img class="tex-graphics" src="https://espresso.codeforces.com/8dd976913226df83d535dfa66193f5525f8471bc.png" style="max-width: 100.0%;max-height: 100.0%;"/>
The second example:
<img class="tex-graphics" src="https://espresso.codeforces.com/44dad5804f5290a2e026c9c41a15151562df8682.png" style="max-width: 100.0%;max-height: 100.0%;"/>
It is not a spruce, because the non-leaf vertex 1 has only 2 leaf children.
The third example:
<img class="tex-graphics" src="https://espresso.codeforces.com/cf84a9e1585707f4ab06eff8eb1120a49b5e1ef7.png" style="max-width: 100.0%;max-height: 100.0%;"/>
| 750
|
[
{
"input": "4\n1\n1\n1",
"output": "Yes"
},
{
"input": "7\n1\n1\n1\n2\n2\n2",
"output": "No"
},
{
"input": "8\n1\n1\n1\n1\n3\n3\n3",
"output": "Yes"
},
{
"input": "3\n1\n1",
"output": "No"
},
{
"input": "13\n1\n2\n2\n2\n1\n6\n6\n6\n1\n10\n10\n10",
"output": "No"
},
{
"input": "7\n1\n2\n2\n1\n1\n1",
"output": "No"
},
{
"input": "7\n1\n1\n1\n1\n2\n2",
"output": "No"
},
{
"input": "8\n1\n1\n1\n1\n5\n5\n5",
"output": "Yes"
},
{
"input": "9\n1\n1\n1\n1\n2\n6\n6\n6",
"output": "No"
},
{
"input": "12\n1\n1\n1\n2\n5\n5\n5\n5\n1\n2\n2",
"output": "No"
},
{
"input": "20\n1\n1\n1\n1\n2\n2\n2\n3\n3\n3\n4\n4\n4\n5\n5\n5\n1\n1\n1",
"output": "Yes"
},
{
"input": "7\n1\n1\n1\n3\n3\n3",
"output": "No"
}
] | 1,630,683,976
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 31
| 62
| 107,417,600
|
from sys import stdin , setrecursionlimit
input = stdin.readline
setrecursionlimit(10**8)
import threading
def dfs(p):
global ans
if(len(child[p]) == 0):return 1
leafs = 0
for i in child[p]:
leafs += dfs(i)
if(leafs < 3):ans = 'No'
return 0
child , ans = None , None
def main():
global child , ans
for T in range(1):
n = int(input())
child = [[] for i in range(n + 1)]
for i in range(n - 1):
p = int(input())
child[p].append(i + 2)
ans = 'Yes'
dfs(1)
print(ans)
if(__name__ == "__main__"):
threading.stack_size(10**8)
threading.Thread(target=main).start()
|
Title: Christmas Spruce
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Consider a rooted tree. A rooted tree has one special vertex called the root. All edges are directed from the root. Vertex *u* is called a child of vertex *v* and vertex *v* is called a parent of vertex *u* if there exists a directed edge from *v* to *u*. A vertex is called a leaf if it doesn't have children and has a parent.
Let's call a rooted tree a spruce if its every non-leaf vertex has at least 3 leaf children. You are given a rooted tree, check whether it's a spruce.
The definition of a rooted tree can be found [here](https://goo.gl/1dqvzz).
Input Specification:
The first line contains one integer *n* — the number of vertices in the tree (3<=≤<=*n*<=≤<=1<=000). Each of the next *n*<=-<=1 lines contains one integer *p**i* (1<=≤<=*i*<=≤<=*n*<=-<=1) — the index of the parent of the *i*<=+<=1-th vertex (1<=≤<=*p**i*<=≤<=*i*).
Vertex 1 is the root. It's guaranteed that the root has at least 2 children.
Output Specification:
Print "Yes" if the tree is a spruce and "No" otherwise.
Demo Input:
['4\n1\n1\n1\n', '7\n1\n1\n1\n2\n2\n2\n', '8\n1\n1\n1\n1\n3\n3\n3\n']
Demo Output:
['Yes\n', 'No\n', 'Yes\n']
Note:
The first example:
<img class="tex-graphics" src="https://espresso.codeforces.com/8dd976913226df83d535dfa66193f5525f8471bc.png" style="max-width: 100.0%;max-height: 100.0%;"/>
The second example:
<img class="tex-graphics" src="https://espresso.codeforces.com/44dad5804f5290a2e026c9c41a15151562df8682.png" style="max-width: 100.0%;max-height: 100.0%;"/>
It is not a spruce, because the non-leaf vertex 1 has only 2 leaf children.
The third example:
<img class="tex-graphics" src="https://espresso.codeforces.com/cf84a9e1585707f4ab06eff8eb1120a49b5e1ef7.png" style="max-width: 100.0%;max-height: 100.0%;"/>
|
```python
from sys import stdin , setrecursionlimit
input = stdin.readline
setrecursionlimit(10**8)
import threading
def dfs(p):
global ans
if(len(child[p]) == 0):return 1
leafs = 0
for i in child[p]:
leafs += dfs(i)
if(leafs < 3):ans = 'No'
return 0
child , ans = None , None
def main():
global child , ans
for T in range(1):
n = int(input())
child = [[] for i in range(n + 1)]
for i in range(n - 1):
p = int(input())
child[p].append(i + 2)
ans = 'Yes'
dfs(1)
print(ans)
if(__name__ == "__main__"):
threading.stack_size(10**8)
threading.Thread(target=main).start()
```
| 3
|
|
515
|
A
|
Drazil and Date
|
PROGRAMMING
| 1,000
|
[
"math"
] | null | null |
Someday, Drazil wanted to go on date with Varda. Drazil and Varda live on Cartesian plane. Drazil's home is located in point (0,<=0) and Varda's home is located in point (*a*,<=*b*). In each step, he can move in a unit distance in horizontal or vertical direction. In other words, from position (*x*,<=*y*) he can go to positions (*x*<=+<=1,<=*y*), (*x*<=-<=1,<=*y*), (*x*,<=*y*<=+<=1) or (*x*,<=*y*<=-<=1).
Unfortunately, Drazil doesn't have sense of direction. So he randomly chooses the direction he will go to in each step. He may accidentally return back to his house during his travel. Drazil may even not notice that he has arrived to (*a*,<=*b*) and continue travelling.
Luckily, Drazil arrived to the position (*a*,<=*b*) successfully. Drazil said to Varda: "It took me exactly *s* steps to travel from my house to yours". But Varda is confused about his words, she is not sure that it is possible to get from (0,<=0) to (*a*,<=*b*) in exactly *s* steps. Can you find out if it is possible for Varda?
|
You are given three integers *a*, *b*, and *s* (<=-<=109<=≤<=*a*,<=*b*<=≤<=109, 1<=≤<=*s*<=≤<=2·109) in a single line.
|
If you think Drazil made a mistake and it is impossible to take exactly *s* steps and get from his home to Varda's home, print "No" (without quotes).
Otherwise, print "Yes".
|
[
"5 5 11\n",
"10 15 25\n",
"0 5 1\n",
"0 0 2\n"
] |
[
"No\n",
"Yes\n",
"No\n",
"Yes\n"
] |
In fourth sample case one possible route is: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/0d30660ddf6eb6c64ffd071055a4e8ddd016cde5.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
| 500
|
[
{
"input": "5 5 11",
"output": "No"
},
{
"input": "10 15 25",
"output": "Yes"
},
{
"input": "0 5 1",
"output": "No"
},
{
"input": "0 0 2",
"output": "Yes"
},
{
"input": "999999999 999999999 2000000000",
"output": "Yes"
},
{
"input": "-606037695 998320124 820674098",
"output": "No"
},
{
"input": "948253616 -83299062 1031552680",
"output": "Yes"
},
{
"input": "711980199 216568284 928548487",
"output": "Yes"
},
{
"input": "-453961301 271150176 725111473",
"output": "No"
},
{
"input": "0 0 2000000000",
"output": "Yes"
},
{
"input": "0 0 1999999999",
"output": "No"
},
{
"input": "1000000000 1000000000 2000000000",
"output": "Yes"
},
{
"input": "-1000000000 1000000000 2000000000",
"output": "Yes"
},
{
"input": "-1000000000 -1000000000 2000000000",
"output": "Yes"
},
{
"input": "-1000000000 -1000000000 1000000000",
"output": "No"
},
{
"input": "-1 -1 3",
"output": "No"
},
{
"input": "919785634 216774719 129321944",
"output": "No"
},
{
"input": "-467780354 -721273539 1369030008",
"output": "No"
},
{
"input": "-232833910 -880680184 1774549792",
"output": "Yes"
},
{
"input": "45535578 402686155 1011249824",
"output": "No"
},
{
"input": "-203250176 -30379840 233630015",
"output": "No"
},
{
"input": "-814516648 -75830576 890347222",
"output": "No"
},
{
"input": "869184175 -511854751 1381038923",
"output": "No"
},
{
"input": "552884998 -262338193 815223187",
"output": "No"
},
{
"input": "-133874494 857573117 991447611",
"output": "Yes"
},
{
"input": "169406792 786701289 956108082",
"output": "No"
},
{
"input": "30237137 -989203246 1019440385",
"output": "Yes"
},
{
"input": "576608216 -617624133 1194232352",
"output": "No"
},
{
"input": "-662360368 97618995 759979367",
"output": "Yes"
},
{
"input": "-115989290 321714461 437703756",
"output": "No"
},
{
"input": "-1 -3 2",
"output": "No"
},
{
"input": "-2 -2 2",
"output": "No"
},
{
"input": "0 0 1",
"output": "No"
},
{
"input": "0 -1 7",
"output": "Yes"
},
{
"input": "1 2 7",
"output": "Yes"
},
{
"input": "0 5 6",
"output": "No"
},
{
"input": "0 -4 2",
"output": "No"
},
{
"input": "-5 5 2",
"output": "No"
},
{
"input": "-5 -2 7",
"output": "Yes"
},
{
"input": "5 -5 2",
"output": "No"
}
] | 1,685,707,496
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 40
| 62
| 0
|
a,b,s=map(int,input().split())
if a<0:
a=-1*a
if b<0:
b=-1*b
c=a+b
if s==c:
print("YES")
elif s>c:
if (s-c)%2==0:
print("YES")
else:
print("NO")
else:
print("NO")
|
Title: Drazil and Date
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Someday, Drazil wanted to go on date with Varda. Drazil and Varda live on Cartesian plane. Drazil's home is located in point (0,<=0) and Varda's home is located in point (*a*,<=*b*). In each step, he can move in a unit distance in horizontal or vertical direction. In other words, from position (*x*,<=*y*) he can go to positions (*x*<=+<=1,<=*y*), (*x*<=-<=1,<=*y*), (*x*,<=*y*<=+<=1) or (*x*,<=*y*<=-<=1).
Unfortunately, Drazil doesn't have sense of direction. So he randomly chooses the direction he will go to in each step. He may accidentally return back to his house during his travel. Drazil may even not notice that he has arrived to (*a*,<=*b*) and continue travelling.
Luckily, Drazil arrived to the position (*a*,<=*b*) successfully. Drazil said to Varda: "It took me exactly *s* steps to travel from my house to yours". But Varda is confused about his words, she is not sure that it is possible to get from (0,<=0) to (*a*,<=*b*) in exactly *s* steps. Can you find out if it is possible for Varda?
Input Specification:
You are given three integers *a*, *b*, and *s* (<=-<=109<=≤<=*a*,<=*b*<=≤<=109, 1<=≤<=*s*<=≤<=2·109) in a single line.
Output Specification:
If you think Drazil made a mistake and it is impossible to take exactly *s* steps and get from his home to Varda's home, print "No" (without quotes).
Otherwise, print "Yes".
Demo Input:
['5 5 11\n', '10 15 25\n', '0 5 1\n', '0 0 2\n']
Demo Output:
['No\n', 'Yes\n', 'No\n', 'Yes\n']
Note:
In fourth sample case one possible route is: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/0d30660ddf6eb6c64ffd071055a4e8ddd016cde5.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
|
```python
a,b,s=map(int,input().split())
if a<0:
a=-1*a
if b<0:
b=-1*b
c=a+b
if s==c:
print("YES")
elif s>c:
if (s-c)%2==0:
print("YES")
else:
print("NO")
else:
print("NO")
```
| 3
|
|
688
|
B
|
Lovely Palindromes
|
PROGRAMMING
| 1,000
|
[
"constructive algorithms",
"math"
] | null | null |
Pari has a friend who loves palindrome numbers. A palindrome number is a number that reads the same forward or backward. For example 12321, 100001 and 1 are palindrome numbers, while 112 and 1021 are not.
Pari is trying to love them too, but only very special and gifted people can understand the beauty behind palindrome numbers. Pari loves integers with even length (i.e. the numbers with even number of digits), so she tries to see a lot of big palindrome numbers with even length (like a 2-digit 11 or 6-digit 122221), so maybe she could see something in them.
Now Pari asks you to write a program that gets a huge integer *n* from the input and tells what is the *n*-th even-length positive palindrome number?
|
The only line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=10100<=000).
|
Print the *n*-th even-length palindrome number.
|
[
"1\n",
"10\n"
] |
[
"11\n",
"1001\n"
] |
The first 10 even-length palindrome numbers are 11, 22, 33, ... , 88, 99 and 1001.
| 1,000
|
[
{
"input": "1",
"output": "11"
},
{
"input": "10",
"output": "1001"
},
{
"input": "11",
"output": "1111"
},
{
"input": "12",
"output": "1221"
},
{
"input": "100",
"output": "100001"
},
{
"input": "1321",
"output": "13211231"
},
{
"input": "2",
"output": "22"
},
{
"input": "3",
"output": "33"
},
{
"input": "4",
"output": "44"
},
{
"input": "5",
"output": "55"
},
{
"input": "6",
"output": "66"
},
{
"input": "7",
"output": "77"
},
{
"input": "8",
"output": "88"
},
{
"input": "9",
"output": "99"
},
{
"input": "13",
"output": "1331"
},
{
"input": "14",
"output": "1441"
},
{
"input": "15",
"output": "1551"
},
{
"input": "16",
"output": "1661"
},
{
"input": "17",
"output": "1771"
},
{
"input": "18",
"output": "1881"
},
{
"input": "19",
"output": "1991"
},
{
"input": "20",
"output": "2002"
},
{
"input": "26550",
"output": "2655005562"
},
{
"input": "16137",
"output": "1613773161"
},
{
"input": "91471",
"output": "9147117419"
},
{
"input": "41242",
"output": "4124224214"
},
{
"input": "30866",
"output": "3086666803"
},
{
"input": "4442231232741003009964183199713851566466079882929007253443848371978048610682180431295671905105980116",
"output": "44422312327410030099641831997138515664660798829290072534438483719780486106821804312956719051059801166110895015091765921340812860168408791738483443527009292889706646651583179913814699003001472321322444"
},
{
"input": "7806526084732010742850470902979267046151204705563201965818390803445852115032119297648697933316807252",
"output": "78065260847320107428504709029792670461512047055632019658183908034458521150321192976486979333168072522527086133397968467929112305112585443080938185691023655074021516407629792090740582470102374806256087"
},
{
"input": "2244399823612183124978103773977739948197709395321563148611877999503840818596183803323258317041875913",
"output": "22443998236121831249781037739777399481977093953215631486118779995038408185961838033232583170418759133195781407138523233083816958180483059997781168413651235939077918499377793773018794213812163289934422"
},
{
"input": "123451",
"output": "123451154321"
}
] | 1,694,961,939
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 3
| 46
| 0
|
def yzd_solution(n):
if n % 10 == 0:
print((n * 10**len(str(n))) + 1)
else:
print((n * 10**len(str(n))) + n)
n = int(input())
yzd_solution(n)
|
Title: Lovely Palindromes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Pari has a friend who loves palindrome numbers. A palindrome number is a number that reads the same forward or backward. For example 12321, 100001 and 1 are palindrome numbers, while 112 and 1021 are not.
Pari is trying to love them too, but only very special and gifted people can understand the beauty behind palindrome numbers. Pari loves integers with even length (i.e. the numbers with even number of digits), so she tries to see a lot of big palindrome numbers with even length (like a 2-digit 11 or 6-digit 122221), so maybe she could see something in them.
Now Pari asks you to write a program that gets a huge integer *n* from the input and tells what is the *n*-th even-length positive palindrome number?
Input Specification:
The only line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=10100<=000).
Output Specification:
Print the *n*-th even-length palindrome number.
Demo Input:
['1\n', '10\n']
Demo Output:
['11\n', '1001\n']
Note:
The first 10 even-length palindrome numbers are 11, 22, 33, ... , 88, 99 and 1001.
|
```python
def yzd_solution(n):
if n % 10 == 0:
print((n * 10**len(str(n))) + 1)
else:
print((n * 10**len(str(n))) + n)
n = int(input())
yzd_solution(n)
```
| 0
|
|
421
|
A
|
Pasha and Hamsters
|
PROGRAMMING
| 800
|
[
"constructive algorithms",
"implementation"
] | null | null |
Pasha has two hamsters: Arthur and Alexander. Pasha put *n* apples in front of them. Pasha knows which apples Arthur likes. Similarly, Pasha knows which apples Alexander likes. Pasha doesn't want any conflict between the hamsters (as they may like the same apple), so he decided to distribute the apples between the hamsters on his own. He is going to give some apples to Arthur and some apples to Alexander. It doesn't matter how many apples each hamster gets but it is important that each hamster gets only the apples he likes. It is possible that somebody doesn't get any apples.
Help Pasha distribute all the apples between the hamsters. Note that Pasha wants to distribute all the apples, not just some of them.
|
The first line contains integers *n*, *a*, *b* (1<=≤<=*n*<=≤<=100; 1<=≤<=*a*,<=*b*<=≤<=*n*) — the number of apples Pasha has, the number of apples Arthur likes and the number of apples Alexander likes, correspondingly.
The next line contains *a* distinct integers — the numbers of the apples Arthur likes. The next line contains *b* distinct integers — the numbers of the apples Alexander likes.
Assume that the apples are numbered from 1 to *n*. The input is such that the answer exists.
|
Print *n* characters, each of them equals either 1 or 2. If the *i*-h character equals 1, then the *i*-th apple should be given to Arthur, otherwise it should be given to Alexander. If there are multiple correct answers, you are allowed to print any of them.
|
[
"4 2 3\n1 2\n2 3 4\n",
"5 5 2\n3 4 1 2 5\n2 3\n"
] |
[
"1 1 2 2\n",
"1 1 1 1 1\n"
] |
none
| 500
|
[
{
"input": "4 2 3\n1 2\n2 3 4",
"output": "1 1 2 2"
},
{
"input": "5 5 2\n3 4 1 2 5\n2 3",
"output": "1 1 1 1 1"
},
{
"input": "100 69 31\n1 3 4 5 6 7 8 9 10 11 12 14 15 16 17 18 19 20 21 24 26 27 29 31 37 38 39 40 44 46 48 49 50 51 53 55 56 57 58 59 60 61 63 64 65 66 67 68 69 70 71 72 74 76 77 78 79 80 81 82 83 89 92 94 95 97 98 99 100\n2 13 22 23 25 28 30 32 33 34 35 36 41 42 43 45 47 52 54 62 73 75 84 85 86 87 88 90 91 93 96",
"output": "1 2 1 1 1 1 1 1 1 1 1 1 2 1 1 1 1 1 1 1 1 2 2 1 2 1 1 2 1 2 1 2 2 2 2 2 1 1 1 1 2 2 2 1 2 1 2 1 1 1 1 2 1 2 1 1 1 1 1 1 1 2 1 1 1 1 1 1 1 1 1 1 2 1 2 1 1 1 1 1 1 1 1 2 2 2 2 2 1 2 2 1 2 1 1 2 1 1 1 1"
},
{
"input": "100 56 44\n1 2 5 8 14 15 17 18 20 21 23 24 25 27 30 33 34 35 36 38 41 42 44 45 46 47 48 49 50 53 56 58 59 60 62 63 64 65 68 69 71 75 76 80 81 84 87 88 90 91 92 94 95 96 98 100\n3 4 6 7 9 10 11 12 13 16 19 22 26 28 29 31 32 37 39 40 43 51 52 54 55 57 61 66 67 70 72 73 74 77 78 79 82 83 85 86 89 93 97 99",
"output": "1 1 2 2 1 2 2 1 2 2 2 2 2 1 1 2 1 1 2 1 1 2 1 1 1 2 1 2 2 1 2 2 1 1 1 1 2 1 2 2 1 1 2 1 1 1 1 1 1 1 2 2 1 2 2 1 2 1 1 1 2 1 1 1 1 2 2 1 1 2 1 2 2 2 1 1 2 2 2 1 1 2 2 1 2 2 1 1 2 1 1 1 2 1 1 1 2 1 2 1"
},
{
"input": "100 82 18\n1 2 3 4 5 6 7 8 9 10 11 13 14 15 16 17 18 19 20 22 23 25 27 29 30 31 32 33 34 35 36 37 38 42 43 44 45 46 47 48 49 50 51 53 54 55 57 58 59 60 61 62 63 64 65 66 67 68 69 71 72 73 74 75 77 78 79 80 82 83 86 88 90 91 92 93 94 96 97 98 99 100\n12 21 24 26 28 39 40 41 52 56 70 76 81 84 85 87 89 95",
"output": "1 1 1 1 1 1 1 1 1 1 1 2 1 1 1 1 1 1 1 1 2 1 1 2 1 2 1 2 1 1 1 1 1 1 1 1 1 1 2 2 2 1 1 1 1 1 1 1 1 1 1 2 1 1 1 2 1 1 1 1 1 1 1 1 1 1 1 1 1 2 1 1 1 1 1 2 1 1 1 1 2 1 1 2 2 1 2 1 2 1 1 1 1 1 2 1 1 1 1 1"
},
{
"input": "99 72 27\n1 2 3 4 5 6 7 8 10 11 12 13 14 15 16 17 20 23 25 26 28 29 30 32 33 34 35 36 39 41 42 43 44 45 46 47 50 51 52 54 55 56 58 59 60 61 62 67 70 71 72 74 75 76 77 80 81 82 84 85 86 88 90 91 92 93 94 95 96 97 98 99\n9 18 19 21 22 24 27 31 37 38 40 48 49 53 57 63 64 65 66 68 69 73 78 79 83 87 89",
"output": "1 1 1 1 1 1 1 1 2 1 1 1 1 1 1 1 1 2 2 1 2 2 1 2 1 1 2 1 1 1 2 1 1 1 1 1 2 2 1 2 1 1 1 1 1 1 1 2 2 1 1 1 2 1 1 1 2 1 1 1 1 1 2 2 2 2 1 2 2 1 1 1 2 1 1 1 1 2 2 1 1 1 2 1 1 1 2 1 2 1 1 1 1 1 1 1 1 1 1"
},
{
"input": "99 38 61\n1 3 10 15 16 22 23 28 31 34 35 36 37 38 39 43 44 49 50 53 56 60 63 68 69 70 72 74 75 77 80 81 83 85 96 97 98 99\n2 4 5 6 7 8 9 11 12 13 14 17 18 19 20 21 24 25 26 27 29 30 32 33 40 41 42 45 46 47 48 51 52 54 55 57 58 59 61 62 64 65 66 67 71 73 76 78 79 82 84 86 87 88 89 90 91 92 93 94 95",
"output": "1 2 1 2 2 2 2 2 2 1 2 2 2 2 1 1 2 2 2 2 2 1 1 2 2 2 2 1 2 2 1 2 2 1 1 1 1 1 1 2 2 2 1 1 2 2 2 2 1 1 2 2 1 2 2 1 2 2 2 1 2 2 1 2 2 2 2 1 1 1 2 1 2 1 1 2 1 2 2 1 1 2 1 2 1 2 2 2 2 2 2 2 2 2 2 1 1 1 1"
},
{
"input": "99 84 15\n1 2 3 5 6 7 8 9 10 11 12 13 14 15 16 17 19 20 21 22 23 24 25 26 27 28 29 30 31 32 34 35 36 37 38 39 40 41 42 43 44 47 48 50 51 52 53 55 56 58 59 60 61 62 63 64 65 68 69 70 71 72 73 74 75 77 79 80 81 82 83 84 85 86 87 89 90 91 92 93 94 97 98 99\n4 18 33 45 46 49 54 57 66 67 76 78 88 95 96",
"output": "1 1 1 2 1 1 1 1 1 1 1 1 1 1 1 1 1 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 1 1 1 1 1 1 1 1 1 1 1 2 2 1 1 2 1 1 1 1 2 1 1 2 1 1 1 1 1 1 1 1 2 2 1 1 1 1 1 1 1 1 2 1 2 1 1 1 1 1 1 1 1 1 2 1 1 1 1 1 1 2 2 1 1 1"
},
{
"input": "4 3 1\n1 3 4\n2",
"output": "1 2 1 1"
},
{
"input": "4 3 1\n1 2 4\n3",
"output": "1 1 2 1"
},
{
"input": "4 2 2\n2 3\n1 4",
"output": "2 1 1 2"
},
{
"input": "4 3 1\n2 3 4\n1",
"output": "2 1 1 1"
},
{
"input": "1 1 1\n1\n1",
"output": "1"
},
{
"input": "2 1 1\n2\n1",
"output": "2 1"
},
{
"input": "2 1 1\n1\n2",
"output": "1 2"
},
{
"input": "3 3 1\n1 2 3\n1",
"output": "1 1 1"
},
{
"input": "3 3 1\n1 2 3\n3",
"output": "1 1 1"
},
{
"input": "3 2 1\n1 3\n2",
"output": "1 2 1"
},
{
"input": "100 1 100\n84\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",
"output": "2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2"
},
{
"input": "100 100 1\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\n17",
"output": "1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1"
},
{
"input": "98 51 47\n1 2 3 4 6 7 8 10 13 15 16 18 19 21 22 23 25 26 27 29 31 32 36 37 39 40 41 43 44 48 49 50 51 52 54 56 58 59 65 66 68 79 80 84 86 88 89 90 94 95 97\n5 9 11 12 14 17 20 24 28 30 33 34 35 38 42 45 46 47 53 55 57 60 61 62 63 64 67 69 70 71 72 73 74 75 76 77 78 81 82 83 85 87 91 92 93 96 98",
"output": "1 1 1 1 2 1 1 1 2 1 2 2 1 2 1 1 2 1 1 2 1 1 1 2 1 1 1 2 1 2 1 1 2 2 2 1 1 2 1 1 1 2 1 1 2 2 2 1 1 1 1 1 2 1 2 1 2 1 1 2 2 2 2 2 1 1 2 1 2 2 2 2 2 2 2 2 2 2 1 1 2 2 2 1 2 1 2 1 1 1 2 2 2 1 1 2 1 2"
},
{
"input": "98 28 70\n1 13 15 16 19 27 28 40 42 43 46 53 54 57 61 63 67 68 69 71 75 76 78 80 88 93 97 98\n2 3 4 5 6 7 8 9 10 11 12 14 17 18 20 21 22 23 24 25 26 29 30 31 32 33 34 35 36 37 38 39 41 44 45 47 48 49 50 51 52 55 56 58 59 60 62 64 65 66 70 72 73 74 77 79 81 82 83 84 85 86 87 89 90 91 92 94 95 96",
"output": "1 2 2 2 2 2 2 2 2 2 2 2 1 2 1 1 2 2 1 2 2 2 2 2 2 2 1 1 2 2 2 2 2 2 2 2 2 2 2 1 2 1 1 2 2 1 2 2 2 2 2 2 1 1 2 2 1 2 2 2 1 2 1 2 2 2 1 1 1 2 1 2 2 2 1 1 2 1 2 1 2 2 2 2 2 2 2 1 2 2 2 2 1 2 2 2 1 1"
},
{
"input": "97 21 76\n7 10 16 17 26 30 34 39 40 42 44 46 53 54 56 64 67 72 78 79 94\n1 2 3 4 5 6 8 9 11 12 13 14 15 18 19 20 21 22 23 24 25 27 28 29 31 32 33 35 36 37 38 41 43 45 47 48 49 50 51 52 55 57 58 59 60 61 62 63 65 66 68 69 70 71 73 74 75 76 77 80 81 82 83 84 85 86 87 88 89 90 91 92 93 95 96 97",
"output": "2 2 2 2 2 2 1 2 2 1 2 2 2 2 2 1 1 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 1 2 2 2 2 1 1 2 1 2 1 2 1 2 2 2 2 2 2 1 1 2 1 2 2 2 2 2 2 2 1 2 2 1 2 2 2 2 1 2 2 2 2 2 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2"
},
{
"input": "97 21 76\n1 10 12 13 17 18 22 25 31 48 50 54 61 64 67 74 78 81 86 88 94\n2 3 4 5 6 7 8 9 11 14 15 16 19 20 21 23 24 26 27 28 29 30 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 49 51 52 53 55 56 57 58 59 60 62 63 65 66 68 69 70 71 72 73 75 76 77 79 80 82 83 84 85 87 89 90 91 92 93 95 96 97",
"output": "1 2 2 2 2 2 2 2 2 1 2 1 1 2 2 2 1 1 2 2 2 1 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 1 2 2 1 2 2 1 2 2 2 2 2 2 1 2 2 2 1 2 2 1 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2"
},
{
"input": "96 10 86\n2 5 31 37 68 69 80 82 90 91\n1 3 4 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 32 33 34 35 36 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 70 71 72 73 74 75 76 77 78 79 81 83 84 85 86 87 88 89 92 93 94 95 96",
"output": "2 1 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 2 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 1 1 2 2 2 2 2"
},
{
"input": "95 4 91\n58 65 70 93\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 59 60 61 62 63 64 66 67 68 69 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 94 95",
"output": "2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 1 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2"
},
{
"input": "98 88 10\n1 2 4 5 6 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 33 34 35 36 38 39 40 41 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 62 63 64 65 66 67 68 69 71 72 73 74 75 76 77 79 80 81 83 84 85 86 87 88 89 90 92 93 94 95 96 97 98\n3 7 32 37 42 61 70 78 82 91",
"output": "1 1 2 1 1 1 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 1 1 1 1 2 1 1 1 1 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 1 1 1 1 1 1 1 1 2 1 1 1 1 1 1 1 2 1 1 1 2 1 1 1 1 1 1 1 1 2 1 1 1 1 1 1 1"
},
{
"input": "98 96 2\n1 2 3 4 5 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\n6 7",
"output": "1 1 1 1 1 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 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": "97 97 1\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\n94",
"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"
},
{
"input": "97 97 1\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\n20",
"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"
},
{
"input": "96 96 1\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\n48",
"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"
},
{
"input": "95 95 1\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\n55",
"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"
}
] | 1,632,159,345
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 32
| 77
| 6,963,200
|
n,a,b = map(int,input().split(' '))
a1=list(map(int,input().split(' ')))
b1=list(map(int,input().split(' ')))
a1=sorted(a1)
b1=sorted(b1)
m=[]
for i in range(0,n):
if i+1 in a1:
m.append(1)
elif i+1 in b1:
m.append(2)
print(*m)
|
Title: Pasha and Hamsters
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Pasha has two hamsters: Arthur and Alexander. Pasha put *n* apples in front of them. Pasha knows which apples Arthur likes. Similarly, Pasha knows which apples Alexander likes. Pasha doesn't want any conflict between the hamsters (as they may like the same apple), so he decided to distribute the apples between the hamsters on his own. He is going to give some apples to Arthur and some apples to Alexander. It doesn't matter how many apples each hamster gets but it is important that each hamster gets only the apples he likes. It is possible that somebody doesn't get any apples.
Help Pasha distribute all the apples between the hamsters. Note that Pasha wants to distribute all the apples, not just some of them.
Input Specification:
The first line contains integers *n*, *a*, *b* (1<=≤<=*n*<=≤<=100; 1<=≤<=*a*,<=*b*<=≤<=*n*) — the number of apples Pasha has, the number of apples Arthur likes and the number of apples Alexander likes, correspondingly.
The next line contains *a* distinct integers — the numbers of the apples Arthur likes. The next line contains *b* distinct integers — the numbers of the apples Alexander likes.
Assume that the apples are numbered from 1 to *n*. The input is such that the answer exists.
Output Specification:
Print *n* characters, each of them equals either 1 or 2. If the *i*-h character equals 1, then the *i*-th apple should be given to Arthur, otherwise it should be given to Alexander. If there are multiple correct answers, you are allowed to print any of them.
Demo Input:
['4 2 3\n1 2\n2 3 4\n', '5 5 2\n3 4 1 2 5\n2 3\n']
Demo Output:
['1 1 2 2\n', '1 1 1 1 1\n']
Note:
none
|
```python
n,a,b = map(int,input().split(' '))
a1=list(map(int,input().split(' ')))
b1=list(map(int,input().split(' ')))
a1=sorted(a1)
b1=sorted(b1)
m=[]
for i in range(0,n):
if i+1 in a1:
m.append(1)
elif i+1 in b1:
m.append(2)
print(*m)
```
| 3
|
|
189
|
A
|
Cut Ribbon
|
PROGRAMMING
| 1,300
|
[
"brute force",
"dp"
] | null | null |
Polycarpus has a ribbon, its length is *n*. He wants to cut the ribbon in a way that fulfils the following two conditions:
- After the cutting each ribbon piece should have length *a*, *b* or *c*. - After the cutting the number of ribbon pieces should be maximum.
Help Polycarpus and find the number of ribbon pieces after the required cutting.
|
The first line contains four space-separated integers *n*, *a*, *b* and *c* (1<=≤<=*n*,<=*a*,<=*b*,<=*c*<=≤<=4000) — the length of the original ribbon and the acceptable lengths of the ribbon pieces after the cutting, correspondingly. The numbers *a*, *b* and *c* can coincide.
|
Print a single number — the maximum possible number of ribbon pieces. It is guaranteed that at least one correct ribbon cutting exists.
|
[
"5 5 3 2\n",
"7 5 5 2\n"
] |
[
"2\n",
"2\n"
] |
In the first example Polycarpus can cut the ribbon in such way: the first piece has length 2, the second piece has length 3.
In the second example Polycarpus can cut the ribbon in such way: the first piece has length 5, the second piece has length 2.
| 500
|
[
{
"input": "5 5 3 2",
"output": "2"
},
{
"input": "7 5 5 2",
"output": "2"
},
{
"input": "4 4 4 4",
"output": "1"
},
{
"input": "1 1 1 1",
"output": "1"
},
{
"input": "4000 1 2 3",
"output": "4000"
},
{
"input": "4000 3 4 5",
"output": "1333"
},
{
"input": "10 3 4 5",
"output": "3"
},
{
"input": "100 23 15 50",
"output": "2"
},
{
"input": "3119 3515 1021 7",
"output": "11"
},
{
"input": "918 102 1327 1733",
"output": "9"
},
{
"input": "3164 42 430 1309",
"output": "15"
},
{
"input": "3043 317 1141 2438",
"output": "7"
},
{
"input": "26 1 772 2683",
"output": "26"
},
{
"input": "370 2 1 15",
"output": "370"
},
{
"input": "734 12 6 2",
"output": "367"
},
{
"input": "418 18 14 17",
"output": "29"
},
{
"input": "18 16 28 9",
"output": "2"
},
{
"input": "14 6 2 17",
"output": "7"
},
{
"input": "29 27 18 2",
"output": "2"
},
{
"input": "29 12 7 10",
"output": "3"
},
{
"input": "27 23 4 3",
"output": "9"
},
{
"input": "5 14 5 2",
"output": "1"
},
{
"input": "5 17 26 5",
"output": "1"
},
{
"input": "9 1 10 3",
"output": "9"
},
{
"input": "2 19 15 1",
"output": "2"
},
{
"input": "4 6 4 9",
"output": "1"
},
{
"input": "10 6 2 9",
"output": "5"
},
{
"input": "2 2 9 6",
"output": "1"
},
{
"input": "6 2 4 1",
"output": "6"
},
{
"input": "27 24 5 27",
"output": "1"
},
{
"input": "2683 83 26 2709",
"output": "101"
},
{
"input": "728 412 789 158",
"output": "3"
},
{
"input": "3964 4 2916 176",
"output": "991"
},
{
"input": "3399 2035 2 3334",
"output": "683"
},
{
"input": "3455 244 3301 3",
"output": "991"
},
{
"input": "595 2263 3625 1",
"output": "595"
},
{
"input": "4000 1 1 1",
"output": "4000"
},
{
"input": "3999 2 2 3999",
"output": "1"
},
{
"input": "25 6 8 11",
"output": "3"
},
{
"input": "4000 500 1000 2000",
"output": "8"
},
{
"input": "53 10 11 23",
"output": "5"
},
{
"input": "100 100 1 1",
"output": "100"
},
{
"input": "17 3 4 10",
"output": "5"
},
{
"input": "413 101 102 105",
"output": "4"
},
{
"input": "490 4 49 50",
"output": "111"
},
{
"input": "3999 2 3 3",
"output": "1999"
},
{
"input": "8 3 8 4",
"output": "2"
},
{
"input": "5 1 3 3",
"output": "5"
},
{
"input": "100 3 17 22",
"output": "27"
},
{
"input": "4000 2 3 4",
"output": "2000"
},
{
"input": "4000 3 3 5",
"output": "1332"
},
{
"input": "13 4 6 7",
"output": "2"
},
{
"input": "4000 5 2 2",
"output": "2000"
},
{
"input": "3999 2 2 3",
"output": "1999"
},
{
"input": "4000 33 7 3333",
"output": "564"
},
{
"input": "60 33 20 9",
"output": "4"
},
{
"input": "100 9 11 99",
"output": "10"
},
{
"input": "2009 6 8 9",
"output": "334"
}
] | 1,694,378,880
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 58
| 62
| 2,048,000
|
n, a, b, c = map(int, input().split())
res = [-1] * (n + 1)
res[0] = 0
for i in range(1, n + 1):
cur = -1
if i >= a:
cur = max(cur, res[i - a])
if i >= b:
cur = max(cur, res[i - b])
if i >= c:
cur = max(cur, res[i - c])
if cur > -1:
cur += 1
res[i] = cur
print(res[n])
|
Title: Cut Ribbon
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarpus has a ribbon, its length is *n*. He wants to cut the ribbon in a way that fulfils the following two conditions:
- After the cutting each ribbon piece should have length *a*, *b* or *c*. - After the cutting the number of ribbon pieces should be maximum.
Help Polycarpus and find the number of ribbon pieces after the required cutting.
Input Specification:
The first line contains four space-separated integers *n*, *a*, *b* and *c* (1<=≤<=*n*,<=*a*,<=*b*,<=*c*<=≤<=4000) — the length of the original ribbon and the acceptable lengths of the ribbon pieces after the cutting, correspondingly. The numbers *a*, *b* and *c* can coincide.
Output Specification:
Print a single number — the maximum possible number of ribbon pieces. It is guaranteed that at least one correct ribbon cutting exists.
Demo Input:
['5 5 3 2\n', '7 5 5 2\n']
Demo Output:
['2\n', '2\n']
Note:
In the first example Polycarpus can cut the ribbon in such way: the first piece has length 2, the second piece has length 3.
In the second example Polycarpus can cut the ribbon in such way: the first piece has length 5, the second piece has length 2.
|
```python
n, a, b, c = map(int, input().split())
res = [-1] * (n + 1)
res[0] = 0
for i in range(1, n + 1):
cur = -1
if i >= a:
cur = max(cur, res[i - a])
if i >= b:
cur = max(cur, res[i - b])
if i >= c:
cur = max(cur, res[i - c])
if cur > -1:
cur += 1
res[i] = cur
print(res[n])
```
| 3
|
|
124
|
A
|
The number of positions
|
PROGRAMMING
| 1,000
|
[
"math"
] | null | null |
Petr stands in line of *n* people, but he doesn't know exactly which position he occupies. He can say that there are no less than *a* people standing in front of him and no more than *b* people standing behind him. Find the number of different positions Petr can occupy.
|
The only line contains three integers *n*, *a* and *b* (0<=≤<=*a*,<=*b*<=<<=*n*<=≤<=100).
|
Print the single number — the number of the sought positions.
|
[
"3 1 1\n",
"5 2 3\n"
] |
[
"2\n",
"3\n"
] |
The possible positions in the first sample are: 2 and 3 (if we number the positions starting with 1).
In the second sample they are 3, 4 and 5.
| 500
|
[
{
"input": "3 1 1",
"output": "2"
},
{
"input": "5 2 3",
"output": "3"
},
{
"input": "5 4 0",
"output": "1"
},
{
"input": "6 5 5",
"output": "1"
},
{
"input": "9 4 3",
"output": "4"
},
{
"input": "11 4 6",
"output": "7"
},
{
"input": "13 8 7",
"output": "5"
},
{
"input": "14 5 5",
"output": "6"
},
{
"input": "16 6 9",
"output": "10"
},
{
"input": "20 13 17",
"output": "7"
},
{
"input": "22 4 8",
"output": "9"
},
{
"input": "23 8 14",
"output": "15"
},
{
"input": "26 18 22",
"output": "8"
},
{
"input": "28 6 1",
"output": "2"
},
{
"input": "29 5 23",
"output": "24"
},
{
"input": "32 27 15",
"output": "5"
},
{
"input": "33 11 5",
"output": "6"
},
{
"input": "37 21 15",
"output": "16"
},
{
"input": "39 34 33",
"output": "5"
},
{
"input": "41 27 11",
"output": "12"
},
{
"input": "42 25 16",
"output": "17"
},
{
"input": "45 7 43",
"output": "38"
},
{
"input": "47 16 17",
"output": "18"
},
{
"input": "49 11 37",
"output": "38"
},
{
"input": "51 38 39",
"output": "13"
},
{
"input": "52 29 7",
"output": "8"
},
{
"input": "56 43 12",
"output": "13"
},
{
"input": "58 57 28",
"output": "1"
},
{
"input": "59 12 39",
"output": "40"
},
{
"input": "62 9 52",
"output": "53"
},
{
"input": "63 29 44",
"output": "34"
},
{
"input": "65 30 22",
"output": "23"
},
{
"input": "66 27 38",
"output": "39"
},
{
"input": "71 33 53",
"output": "38"
},
{
"input": "73 14 12",
"output": "13"
},
{
"input": "73 37 35",
"output": "36"
},
{
"input": "76 69 44",
"output": "7"
},
{
"input": "79 25 20",
"output": "21"
},
{
"input": "81 60 20",
"output": "21"
},
{
"input": "81 79 14",
"output": "2"
},
{
"input": "84 0 42",
"output": "43"
},
{
"input": "88 79 8",
"output": "9"
},
{
"input": "90 76 59",
"output": "14"
},
{
"input": "92 2 22",
"output": "23"
},
{
"input": "94 5 88",
"output": "89"
},
{
"input": "94 62 48",
"output": "32"
},
{
"input": "96 22 72",
"output": "73"
},
{
"input": "100 11 88",
"output": "89"
},
{
"input": "100 81 91",
"output": "19"
},
{
"input": "1 0 0",
"output": "1"
}
] | 1,647,583,798
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 50
| 124
| 0
|
n,a,b=map(int,input().split())
s=n-a
if(s>b):
s=b+1
print(s)
|
Title: The number of positions
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petr stands in line of *n* people, but he doesn't know exactly which position he occupies. He can say that there are no less than *a* people standing in front of him and no more than *b* people standing behind him. Find the number of different positions Petr can occupy.
Input Specification:
The only line contains three integers *n*, *a* and *b* (0<=≤<=*a*,<=*b*<=<<=*n*<=≤<=100).
Output Specification:
Print the single number — the number of the sought positions.
Demo Input:
['3 1 1\n', '5 2 3\n']
Demo Output:
['2\n', '3\n']
Note:
The possible positions in the first sample are: 2 and 3 (if we number the positions starting with 1).
In the second sample they are 3, 4 and 5.
|
```python
n,a,b=map(int,input().split())
s=n-a
if(s>b):
s=b+1
print(s)
```
| 3
|
|
16
|
C
|
Monitor
|
PROGRAMMING
| 1,800
|
[
"binary search",
"number theory"
] |
C. Monitor
|
0
|
64
|
Reca company makes monitors, the most popular of their models is AB999 with the screen size *a*<=×<=*b* centimeters. Because of some production peculiarities a screen parameters are integer numbers. Recently the screen sides ratio *x*:<=*y* became popular with users. That's why the company wants to reduce monitor AB999 size so that its screen sides ratio becomes *x*:<=*y*, at the same time they want its total area to be maximal of all possible variants. Your task is to find the screen parameters of the reduced size model, or find out that such a reduction can't be performed.
|
The first line of the input contains 4 integers — *a*, *b*, *x* and *y* (1<=≤<=*a*,<=*b*,<=*x*,<=*y*<=≤<=2·109).
|
If the answer exists, output 2 positive integers — screen parameters of the reduced size model. Output 0 0 otherwise.
|
[
"800 600 4 3\n",
"1920 1200 16 9\n",
"1 1 1 2\n"
] |
[
"800 600\n",
"1920 1080\n",
"0 0\n"
] |
none
| 0
|
[
{
"input": "800 600 4 3",
"output": "800 600"
},
{
"input": "1920 1200 16 9",
"output": "1920 1080"
},
{
"input": "1 1 1 2",
"output": "0 0"
},
{
"input": "1002105126 227379125 179460772 1295256518",
"output": "0 0"
},
{
"input": "625166755 843062051 1463070160 1958300154",
"output": "0 0"
},
{
"input": "248228385 1458744978 824699604 1589655888",
"output": "206174901 397413972"
},
{
"input": "186329049 1221011622 90104472 1769702163",
"output": "60069648 1179801442"
},
{
"input": "511020182 242192314 394753578 198572007",
"output": "394753578 198572007"
},
{
"input": "134081812 857875240 82707261 667398699",
"output": "105411215 850606185"
},
{
"input": "721746595 799202881 143676564 380427290",
"output": "287353128 760854580"
},
{
"input": "912724694 1268739154 440710604 387545692",
"output": "881421208 775091384"
},
{
"input": "1103702793 1095784840 788679477 432619528",
"output": "788679477 432619528"
},
{
"input": "548893795 861438648 131329677 177735812",
"output": "525318708 710943248"
},
{
"input": "652586118 1793536161 127888702 397268645",
"output": "511554808 1589074580"
},
{
"input": "756278440 578150025 96644319 26752094",
"output": "676510233 187264658"
},
{
"input": "859970763 1510247537 37524734 97452508",
"output": "562871010 1461787620"
},
{
"input": "547278097 1977241684 51768282 183174370",
"output": "543566961 1923330885"
},
{
"input": "62256611 453071697 240966 206678",
"output": "62169228 53322924"
},
{
"input": "1979767797 878430446 5812753 3794880",
"output": "1342745943 876617280"
},
{
"input": "1143276347 1875662241 178868040 116042960",
"output": "1140283755 739773870"
},
{
"input": "435954880 1740366589 19415065 185502270",
"output": "182099920 1739883360"
},
{
"input": "664035593 983601098 4966148 2852768",
"output": "664032908 381448928"
},
{
"input": "1461963719 350925487 135888396 83344296",
"output": "572153868 350918568"
},
{
"input": "754199095 348965411 161206703 67014029",
"output": "754119492 313489356"
},
{
"input": "166102153 494841162 14166516 76948872",
"output": "91096406 494812252"
},
{
"input": "1243276346 1975662240 38441120 291740200",
"output": "259477560 1969246350"
},
{
"input": "535954879 1840366588 26278959 73433046",
"output": "535849118 1497358892"
},
{
"input": "764035592 1083601097 1192390 7267738",
"output": "177777265 1083570463"
},
{
"input": "1561963718 450925486 475523188 136236856",
"output": "1561914768 447486816"
},
{
"input": "854199094 448965410 364102983 125971431",
"output": "853687785 295356745"
},
{
"input": "266102152 594841161 15854566 13392106",
"output": "266043102 224722482"
},
{
"input": "1 1 2 1",
"output": "0 0"
},
{
"input": "2000000000 2000000000 1 1999999999",
"output": "1 1999999999"
},
{
"input": "2000000000 2000000000 1999999999 1",
"output": "1999999999 1"
},
{
"input": "2000000000 2000000000 2 1999999999",
"output": "2 1999999999"
},
{
"input": "1000000000 1000000000 999999999 2",
"output": "999999999 2"
},
{
"input": "2000000000 2000000000 1999999999 2",
"output": "1999999999 2"
},
{
"input": "2000000000 2000000000 1999999999 1999999998",
"output": "1999999999 1999999998"
},
{
"input": "2000000000 2000000000 1999999998 1999999999",
"output": "1999999998 1999999999"
}
] | 1,660,358,398
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 5
| 62
| 0
|
def f():
if (a*y)%x == 0:
k = (a*y)//x
if k != 0 and k <= b:
print(a,k)
return 0
if (b*x)%y==0:
k = (b*x)//y
if k!= 0 and k <= a:
print(k,b)
return 0
print(0,0)
a,b,x,y = map(int,input().split())
f()
|
Title: Monitor
Time Limit: 0 seconds
Memory Limit: 64 megabytes
Problem Description:
Reca company makes monitors, the most popular of their models is AB999 with the screen size *a*<=×<=*b* centimeters. Because of some production peculiarities a screen parameters are integer numbers. Recently the screen sides ratio *x*:<=*y* became popular with users. That's why the company wants to reduce monitor AB999 size so that its screen sides ratio becomes *x*:<=*y*, at the same time they want its total area to be maximal of all possible variants. Your task is to find the screen parameters of the reduced size model, or find out that such a reduction can't be performed.
Input Specification:
The first line of the input contains 4 integers — *a*, *b*, *x* and *y* (1<=≤<=*a*,<=*b*,<=*x*,<=*y*<=≤<=2·109).
Output Specification:
If the answer exists, output 2 positive integers — screen parameters of the reduced size model. Output 0 0 otherwise.
Demo Input:
['800 600 4 3\n', '1920 1200 16 9\n', '1 1 1 2\n']
Demo Output:
['800 600\n', '1920 1080\n', '0 0\n']
Note:
none
|
```python
def f():
if (a*y)%x == 0:
k = (a*y)//x
if k != 0 and k <= b:
print(a,k)
return 0
if (b*x)%y==0:
k = (b*x)//y
if k!= 0 and k <= a:
print(k,b)
return 0
print(0,0)
a,b,x,y = map(int,input().split())
f()
```
| 0
|
535
|
B
|
Tavas and SaDDas
|
PROGRAMMING
| 1,100
|
[
"bitmasks",
"brute force",
"combinatorics",
"implementation"
] | null | null |
Once again Tavas started eating coffee mix without water! Keione told him that it smells awful, but he didn't stop doing that. That's why Keione told his smart friend, SaDDas to punish him! SaDDas took Tavas' headphones and told him: "If you solve the following problem, I'll return it to you."
The problem is:
You are given a lucky number *n*. Lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
If we sort all lucky numbers in increasing order, what's the 1-based index of *n*?
Tavas is not as smart as SaDDas, so he asked you to do him a favor and solve this problem so he can have his headphones back.
|
The first and only line of input contains a lucky number *n* (1<=≤<=*n*<=≤<=109).
|
Print the index of *n* among all lucky numbers.
|
[
"4\n",
"7\n",
"77\n"
] |
[
"1\n",
"2\n",
"6\n"
] |
none
| 1,000
|
[
{
"input": "4",
"output": "1"
},
{
"input": "7",
"output": "2"
},
{
"input": "77",
"output": "6"
},
{
"input": "4",
"output": "1"
},
{
"input": "474744",
"output": "83"
},
{
"input": "777774",
"output": "125"
},
{
"input": "447",
"output": "8"
},
{
"input": "774",
"output": "13"
},
{
"input": "4",
"output": "1"
},
{
"input": "4447747",
"output": "140"
},
{
"input": "7747474",
"output": "233"
},
{
"input": "4444",
"output": "15"
},
{
"input": "4447",
"output": "16"
},
{
"input": "7",
"output": "2"
},
{
"input": "4",
"output": "1"
},
{
"input": "4447744",
"output": "139"
},
{
"input": "77474",
"output": "57"
},
{
"input": "7747",
"output": "28"
},
{
"input": "444",
"output": "7"
},
{
"input": "7",
"output": "2"
},
{
"input": "7774477",
"output": "242"
},
{
"input": "4477774",
"output": "157"
},
{
"input": "7444",
"output": "23"
},
{
"input": "7474747",
"output": "212"
},
{
"input": "77",
"output": "6"
},
{
"input": "774477",
"output": "114"
},
{
"input": "7",
"output": "2"
},
{
"input": "47",
"output": "4"
},
{
"input": "747777",
"output": "110"
},
{
"input": "444444444",
"output": "511"
},
{
"input": "777777777",
"output": "1022"
},
{
"input": "477477447",
"output": "728"
},
{
"input": "777744747",
"output": "996"
}
] | 1,648,415,689
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 33
| 62
| 0
|
n = 0
for i in input():
n = 2*n + (1, 2)[i == "7"]
print(n)
|
Title: Tavas and SaDDas
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Once again Tavas started eating coffee mix without water! Keione told him that it smells awful, but he didn't stop doing that. That's why Keione told his smart friend, SaDDas to punish him! SaDDas took Tavas' headphones and told him: "If you solve the following problem, I'll return it to you."
The problem is:
You are given a lucky number *n*. Lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
If we sort all lucky numbers in increasing order, what's the 1-based index of *n*?
Tavas is not as smart as SaDDas, so he asked you to do him a favor and solve this problem so he can have his headphones back.
Input Specification:
The first and only line of input contains a lucky number *n* (1<=≤<=*n*<=≤<=109).
Output Specification:
Print the index of *n* among all lucky numbers.
Demo Input:
['4\n', '7\n', '77\n']
Demo Output:
['1\n', '2\n', '6\n']
Note:
none
|
```python
n = 0
for i in input():
n = 2*n + (1, 2)[i == "7"]
print(n)
```
| 3
|
|
94
|
A
|
Restoring Password
|
PROGRAMMING
| 900
|
[
"implementation",
"strings"
] |
A. Restoring Password
|
2
|
256
|
Igor K. always used to trust his favorite Kashpirovsky Antivirus. That is why he didn't hesitate to download the link one of his groupmates sent him via QIP Infinium. The link was said to contain "some real funny stuff about swine influenza". The antivirus had no objections and Igor K. run the flash application he had downloaded. Immediately his QIP Infinium said: "invalid login/password".
Igor K. entered the ISQ from his additional account and looked at the info of his main one. His name and surname changed to "H1N1" and "Infected" correspondingly, and the "Additional Information" field contained a strange-looking binary code 80 characters in length, consisting of zeroes and ones. "I've been hacked" — thought Igor K. and run the Internet Exploiter browser to quickly type his favourite search engine's address.
Soon he learned that it really was a virus that changed ISQ users' passwords. Fortunately, he soon found out that the binary code was actually the encrypted password where each group of 10 characters stood for one decimal digit. Accordingly, the original password consisted of 8 decimal digits.
Help Igor K. restore his ISQ account by the encrypted password and encryption specification.
|
The input data contains 11 lines. The first line represents the binary code 80 characters in length. That is the code written in Igor K.'s ISQ account's info. Next 10 lines contain pairwise distinct binary codes 10 characters in length, corresponding to numbers 0, 1, ..., 9.
|
Print one line containing 8 characters — The password to Igor K.'s ISQ account. It is guaranteed that the solution exists.
|
[
"01001100100101100000010110001001011001000101100110010110100001011010100101101100\n0100110000\n0100110010\n0101100000\n0101100010\n0101100100\n0101100110\n0101101000\n0101101010\n0101101100\n0101101110\n",
"10101101111001000010100100011010101101110010110111011000100011011110010110001000\n1001000010\n1101111001\n1001000110\n1010110111\n0010110111\n1101001101\n1011000001\n1110010101\n1011011000\n0110001000\n"
] |
[
"12345678\n",
"30234919\n"
] |
none
| 500
|
[
{
"input": "01001100100101100000010110001001011001000101100110010110100001011010100101101100\n0100110000\n0100110010\n0101100000\n0101100010\n0101100100\n0101100110\n0101101000\n0101101010\n0101101100\n0101101110",
"output": "12345678"
},
{
"input": "10101101111001000010100100011010101101110010110111011000100011011110010110001000\n1001000010\n1101111001\n1001000110\n1010110111\n0010110111\n1101001101\n1011000001\n1110010101\n1011011000\n0110001000",
"output": "30234919"
},
{
"input": "00010101101110110101100110101100010101100010101111000101011010011010110010000011\n0101010110\n0001001101\n1001101011\n0000100011\n0010101111\n1110110101\n0001010110\n0110111000\n0000111110\n0010000011",
"output": "65264629"
},
{
"input": "10100100010010010011011001101000100100110110011010011001101011000100110110011010\n1111110011\n1001000111\n1001000100\n1100010011\n0110011010\n0010000001\n1110101110\n0010000110\n0010010011\n1010010001",
"output": "98484434"
},
{
"input": "00101100011111010001001000000110110000000110010011001111111010110010001011000000\n0010000001\n0110010011\n0010000010\n1011001000\n0011111110\n0110001000\n1111010001\n1011000000\n0000100110\n0010110001",
"output": "96071437"
},
{
"input": "10001110111110000001000010001010001110110000100010100010111101101101010000100010\n0000010110\n1101010111\n1000101111\n0001011110\n0011110101\n0101100100\n0110110101\n0000100010\n1000111011\n1110000001",
"output": "89787267"
},
{
"input": "10010100011001010001010101001101010100110100111011001010111100011001000010100000\n0011100000\n1001100100\n0001100100\n0010100000\n0101010011\n0010101110\n0010101111\n0100111011\n1001010001\n1111111110",
"output": "88447623"
},
{
"input": "01101100111000000101011011001110000001011111111000111111100001011010001001011001\n1000000101\n0101101000\n0101110101\n1101011110\n0000101100\n1111111000\n0001001101\n0110111011\n0110110011\n1001011001",
"output": "80805519"
},
{
"input": "11100011000100010110010011101010101010011110001100011010111110011000011010110111\n1110001100\n0110101111\n0100111010\n0101000000\n1001100001\n1010101001\n0000100010\n1010110111\n1100011100\n0100010110",
"output": "09250147"
},
{
"input": "10000110110000010100000010001000111101110110101011110111000100001101000000100010\n0000010100\n0000110001\n0110101011\n1101110001\n1000011011\n0000110100\n0011110111\n1000110010\n0000100010\n0000011011",
"output": "40862358"
},
{
"input": "01000000010000000110100101000110110000100100000001101100001000011111111001010001\n1011000010\n1111101010\n0111110011\n0000000110\n0000001001\n0001111111\n0110010010\n0100000001\n1011001000\n1001010001",
"output": "73907059"
},
{
"input": "01111000111110011001110101110011110000111110010001101100110110100111101011001101\n1110010001\n1001100000\n1100001000\n1010011110\n1011001101\n0111100011\n1101011100\n1110011001\n1111000011\n0010000101",
"output": "57680434"
},
{
"input": "01001100101000100010001011110001000101001001100010010000001001001100101001011111\n1001011111\n1110010111\n0111101011\n1000100010\n0011100101\n0100000010\n0010111100\n0100010100\n1001100010\n0100110010",
"output": "93678590"
},
{
"input": "01110111110000111011101010110110101011010100110111000011101101110101011101001000\n0110000101\n1010101101\n1101010111\n1101011100\n0100110111\n0111011111\n1100011001\n0111010101\n0000111011\n1101001000",
"output": "58114879"
},
{
"input": "11101001111100110101110011010100110011011110100111010110110011000111000011001101\n1100011100\n1100110101\n1011101000\n0011011110\n0011001101\n0100010001\n1110100111\n1010101100\n1110110100\n0101101100",
"output": "61146904"
},
{
"input": "10101010001011010001001001011000100101100001011011101010101110101010001010101000\n0010110101\n1010011010\n1010101000\n1011010001\n1010101011\n0010010110\n0110100010\n1010100101\n0001011011\n0110100001",
"output": "23558422"
},
{
"input": "11110101001100010000110100001110101011011111010100110001000001001010001001101111\n0101101100\n1001101111\n1010101101\n0100101000\n1111110000\n0101010010\n1100010000\n1111010100\n1101000011\n1011111111",
"output": "76827631"
},
{
"input": "10001100110000110111100011001101111110110011110101000011011100001101110000110111\n0011110101\n0101100011\n1000110011\n1011011001\n0111111011\n0101111011\n0000110111\n0100001110\n1000000111\n0110110111",
"output": "26240666"
},
{
"input": "10000100010000111101100100111101111011101000001001100001000110000010010000111101\n1001001111\n0000111101\n1000010001\n0110011101\n0110101000\n1011111001\n0111101110\n1000001001\n1101011111\n0001010100",
"output": "21067271"
},
{
"input": "01101111000110111100011011110001101111001010001100101000110001010101100100000010\n1010001100\n0011010011\n0101010110\n1111001100\n1100011000\n0100101100\n1001100101\n0110111100\n0011001101\n0100000010",
"output": "77770029"
},
{
"input": "10100111011010001011111000000111100000010101000011000010111101010000111010011101\n1010011101\n1010111111\n0110100110\n1111000100\n1110000001\n0000101111\n0011111000\n1000110001\n0101000011\n1010001011",
"output": "09448580"
},
{
"input": "10000111111000011111001010101010010011111001001111000010010100100011000010001100\n1101101110\n1001001111\n0000100101\n1100111010\n0010101010\n1110000110\n1100111101\n0010001100\n1110000001\n1000011111",
"output": "99411277"
},
{
"input": "10110110111011001111101100111100111111011011011011001111110110010011100010000111\n0111010011\n0111101100\n1001101010\n0101000101\n0010000111\n0011111101\n1011001111\n1101111000\n1011011011\n1001001110",
"output": "86658594"
},
{
"input": "01001001100101100011110110111100000110001111001000100000110111110010000000011000\n0100100110\n1000001011\n1000111110\n0000011000\n0101100011\n1101101111\n1111001000\n1011011001\n1000001101\n0010101000",
"output": "04536863"
},
{
"input": "10010100011101000011100100001100101111000010111100000010010000001001001101011101\n1001000011\n1101000011\n1001010001\n1101011101\n1000010110\n0011111101\n0010111100\n0000100100\n1010001000\n0101000110",
"output": "21066773"
},
{
"input": "01111111110101111111011111111111010010000001100000101000100100111001011010001001\n0111111111\n0101111111\n0100101101\n0001100000\n0011000101\n0011100101\n1101001000\n0010111110\n1010001001\n1111000111",
"output": "01063858"
},
{
"input": "00100011111001001010001111000011101000001110100000000100101011101000001001001010\n0010001111\n1001001010\n1010011001\n0011100111\n1000111000\n0011110000\n0000100010\n0001001010\n1111110111\n1110100000",
"output": "01599791"
},
{
"input": "11011101000100110100110011010101100011111010011010010011010010010010100110101111\n0100110100\n1001001010\n0001111101\n1101011010\n1101110100\n1100110101\n0110101111\n0110001111\n0001101000\n1010011010",
"output": "40579016"
},
{
"input": "10000010111101110110011000111110000011100110001111100100000111000011011000001011\n0111010100\n1010110110\n1000001110\n1110000100\n0110001111\n1101110110\n1100001101\n1000001011\n0000000101\n1001000001",
"output": "75424967"
},
{
"input": "11101100101110111110111011111010001111111111000001001001000010001111111110110010\n0101100001\n1111010011\n1110111110\n0100110100\n1110011111\n1000111111\n0010010000\n1110110010\n0011000010\n1111000001",
"output": "72259657"
},
{
"input": "01011110100101111010011000001001100000101001110011010111101011010000110110010101\n0100111100\n0101110011\n0101111010\n0110000010\n0101001111\n1101000011\n0110010101\n0111011010\n0001101110\n1001110011",
"output": "22339256"
},
{
"input": "01100000100101111000100001100010000110000010100100100001100000110011101001110000\n0101111000\n1001110000\n0001000101\n0110110111\n0010100100\n1000011000\n1101110110\n0110000010\n0001011010\n0011001110",
"output": "70554591"
},
{
"input": "11110011011000001001111100110101001000010100100000110011001110011111100100100001\n1010011000\n1111001101\n0100100001\n1111010011\n0100100000\n1001111110\n1010100111\n1000100111\n1000001001\n1100110011",
"output": "18124952"
},
{
"input": "10001001011000100101010110011101011001110010000001010110000101000100101111101010\n0101100001\n1100001100\n1111101010\n1000100101\n0010000001\n0100010010\n0010110110\n0101100111\n0000001110\n1101001110",
"output": "33774052"
},
{
"input": "00110010000111001001001100100010010111101011011110001011111100000101000100000001\n0100000001\n1011011110\n0010111111\n0111100111\n0100111001\n0000010100\n1001011110\n0111001001\n0100010011\n0011001000",
"output": "97961250"
},
{
"input": "01101100001000110101101100101111101110010011010111100011010100010001101000110101\n1001101001\n1000110101\n0110110000\n0111100100\n0011010111\n1110111001\n0001000110\n0000000100\n0001101001\n1011001011",
"output": "21954161"
},
{
"input": "10101110000011010110101011100000101101000110100000101101101101110101000011110010\n0110100000\n1011011011\n0011110010\n0001110110\n0010110100\n1100010010\n0001101011\n1010111000\n0011010110\n0111010100",
"output": "78740192"
},
{
"input": "11000101011100100111010000010001000001001100101100000011000000001100000101011010\n1100010101\n1111101011\n0101011010\n0100000100\n1000110111\n1100100111\n1100101100\n0111001000\n0000110000\n0110011111",
"output": "05336882"
},
{
"input": "11110100010000101110010110001000001011100101100010110011011011111110001100110110\n0101100010\n0100010001\n0000101110\n1100110110\n0101000101\n0011001011\n1111010001\n1000110010\n1111111000\n1010011111",
"output": "62020383"
},
{
"input": "00011001111110000011101011010001010111100110100101000110011111011001100000001100\n0111001101\n0101011110\n0001100111\n1101011111\n1110000011\n0000001100\n0111010001\n1101100110\n1010110100\n0110100101",
"output": "24819275"
},
{
"input": "10111110010011111001001111100101010111010011111001001110101000111110011001111101\n0011111001\n0101011101\n0100001010\n0001110010\n1001111101\n0011101010\n1111001001\n1100100001\n1001101000\n1011111001",
"output": "90010504"
},
{
"input": "01111101111100101010001001011110111001110111110111011111011110110111111011011111\n1111110111\n0010000101\n0110000100\n0111111011\n1011100111\n1100101010\n1011011111\n1100010001\n0111110111\n0010010111",
"output": "85948866"
},
{
"input": "01111100000111110000110010111001111100001001101010110010111010001000101001101010\n0100010101\n1011110101\n1010100100\n1010000001\n1001101010\n0101100110\n1000100010\n0111110000\n1100101110\n0110010110",
"output": "77874864"
},
{
"input": "11100011010000000010011110010111001011111001000111000000001000000000100111100101\n0000000010\n1110001101\n0011010101\n0111100101\n1001000111\n1101001111\n0111010110\n1100101111\n0110000000\n1101101011",
"output": "10374003"
},
{
"input": "01111011100111101110011001000110001111101000111110100100100001011111001011100010\n0110010100\n1100010001\n0111101110\n1001001000\n1010011011\n1000111110\n0010110101\n1011100010\n0101111100\n0110010001",
"output": "22955387"
},
{
"input": "11011010001100000011000100110011010101000110011110110000001100111100001000011111\n0000100010\n1000011111\n1101101000\n0110011110\n0011110000\n1100000011\n0010001100\n0101101000\n0001001100\n1101010100",
"output": "25893541"
},
{
"input": "01011001011111010010101111011001000011001100011101101111011011010011101011110110\n0100001100\n0101100101\n1111111011\n1111010010\n1111101100\n1100011101\n1011000011\n1101001110\n1011110110\n0110001010",
"output": "13805878"
},
{
"input": "11110011011000111111001100111110001111111100000010111100110100110011111111001101\n1111001101\n1001101010\n1100110010\n0011001111\n0001011110\n1000110011\n1000111111\n0110001010\n1001011101\n1100000010",
"output": "06369030"
},
{
"input": "01110011110010000011011001011000001000010110010110011001100001100110001100101000\n0000100001\n0110011000\n1010000010\n1110011101\n0111001111\n1100101000\n0010000011\n0110010000\n1100100101\n0110010110",
"output": "46909115"
},
{
"input": "00001011001111110111111111011111111101110101110100010111010010100101100001010110\n1111110111\n0001010110\n0111011011\n0111000001\n1010010110\n0101110100\n0001000101\n0000111000\n0110100001\n0000101100",
"output": "90005541"
}
] | 1,674,035,262
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 60
| 0
|
code_password = input()
code_numbers = {}
for i in range(10):
d[i] = input()
for i in range(0, 71, 10):
print(code_numbers[code_password[i:i + 10]], end='')
|
Title: Restoring Password
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Igor K. always used to trust his favorite Kashpirovsky Antivirus. That is why he didn't hesitate to download the link one of his groupmates sent him via QIP Infinium. The link was said to contain "some real funny stuff about swine influenza". The antivirus had no objections and Igor K. run the flash application he had downloaded. Immediately his QIP Infinium said: "invalid login/password".
Igor K. entered the ISQ from his additional account and looked at the info of his main one. His name and surname changed to "H1N1" and "Infected" correspondingly, and the "Additional Information" field contained a strange-looking binary code 80 characters in length, consisting of zeroes and ones. "I've been hacked" — thought Igor K. and run the Internet Exploiter browser to quickly type his favourite search engine's address.
Soon he learned that it really was a virus that changed ISQ users' passwords. Fortunately, he soon found out that the binary code was actually the encrypted password where each group of 10 characters stood for one decimal digit. Accordingly, the original password consisted of 8 decimal digits.
Help Igor K. restore his ISQ account by the encrypted password and encryption specification.
Input Specification:
The input data contains 11 lines. The first line represents the binary code 80 characters in length. That is the code written in Igor K.'s ISQ account's info. Next 10 lines contain pairwise distinct binary codes 10 characters in length, corresponding to numbers 0, 1, ..., 9.
Output Specification:
Print one line containing 8 characters — The password to Igor K.'s ISQ account. It is guaranteed that the solution exists.
Demo Input:
['01001100100101100000010110001001011001000101100110010110100001011010100101101100\n0100110000\n0100110010\n0101100000\n0101100010\n0101100100\n0101100110\n0101101000\n0101101010\n0101101100\n0101101110\n', '10101101111001000010100100011010101101110010110111011000100011011110010110001000\n1001000010\n1101111001\n1001000110\n1010110111\n0010110111\n1101001101\n1011000001\n1110010101\n1011011000\n0110001000\n']
Demo Output:
['12345678\n', '30234919\n']
Note:
none
|
```python
code_password = input()
code_numbers = {}
for i in range(10):
d[i] = input()
for i in range(0, 71, 10):
print(code_numbers[code_password[i:i + 10]], end='')
```
| -1
|
159
|
D
|
Palindrome pairs
|
PROGRAMMING
| 1,500
|
[
"*special",
"brute force",
"dp",
"strings"
] | null | null |
You are given a non-empty string *s* consisting of lowercase letters. Find the number of pairs of non-overlapping palindromic substrings of this string.
In a more formal way, you have to find the quantity of tuples (*a*,<=*b*,<=*x*,<=*y*) such that 1<=≤<=*a*<=≤<=*b*<=<<=*x*<=≤<=*y*<=≤<=|*s*| and substrings *s*[*a*... *b*], *s*[*x*... *y*] are palindromes.
A palindrome is a string that can be read the same way from left to right and from right to left. For example, "abacaba", "z", "abba" are palindromes.
A substring *s*[*i*... *j*] (1<=≤<=*i*<=≤<=*j*<=≤<=|*s*|) of string *s* = *s*1*s*2... *s*|*s*| is a string *s**i**s**i*<=+<=1... *s**j*. For example, substring *s*[2...4] of string *s* = "abacaba" equals "bac".
|
The first line of input contains a non-empty string *s* which consists of lowercase letters ('a'...'z'), *s* contains at most 2000 characters.
|
Output a single number — the quantity of pairs of non-overlapping palindromic substrings of *s*.
Please do not use the %lld format specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d format specifier.
|
[
"aa\n",
"aaa\n",
"abacaba\n"
] |
[
"1\n",
"5\n",
"36\n"
] |
none
| 2,000
|
[
{
"input": "aa",
"output": "1"
},
{
"input": "aaa",
"output": "5"
},
{
"input": "abacaba",
"output": "36"
},
{
"input": "aaaaaaaaaa",
"output": "495"
},
{
"input": "aabbb",
"output": "24"
},
{
"input": "abbaa",
"output": "18"
},
{
"input": "bbbbb",
"output": "35"
},
{
"input": "bbaab",
"output": "18"
},
{
"input": "aabba",
"output": "18"
},
{
"input": "aaaaa",
"output": "35"
},
{
"input": "abicabacka",
"output": "57"
},
{
"input": "aiajadabaa",
"output": "87"
},
{
"input": "abacabauabagabaeabacabadabacabbfabacamadabacabaeabacabadabacababcdggdefxeceadaffhecbgpdbeffecdcbfagcbbfgegaggcaffdfiafaeaab",
"output": "20470"
},
{
"input": "abacabadabacabaeabacabadabacabafabacabadabqcabaeoqacabagabacabagefdfdedbbacefhhargbaebbbefabdabdcceddbgcebgdccdccccdbgdecfa",
"output": "23427"
},
{
"input": "abacabafabacabaeabacabadabaqaeatabacabadabacabgeabacabadabacabaeadfgbefcbgbagebgobabaaececaccfeeaeeavbcccwbgecffgfadgagcgfb",
"output": "21454"
},
{
"input": "abacabadabacabaeabacabadabacabafabacabadabacabaeabacabadabacabagabacabadabacabaeabacabadabacabafabacabadabacabaeabacabadabacabahabacabadabacabaeabacabadabacabafabacabadabacabaeabacabadabacabagabacabadabacabaeabacabadabacabafabacabadabacabaeabacabadabacabadadihfcihdeeegcgdfihcagaigeabegaheddgadbcagheieicdadafaadafeaeihbafccifeifafdhicebgeccbgbdhdcabeghhbebehbbfgfeddfgbdhcbhcfifcgccfihdedafefdhcchbcahgiicgdhahcihdgghchfahahffggedigifhdcbecbhddacdgiahbiffbadhiggagaefihchggfhffhfdcdbfeaabhfhgbbiacag",
"output": "757870"
},
{
"input": "abacabadabacabaeabacabadabacabafabacabadabacabaeabacabadabacabagabacabadabacabaeabacabadabacabafabacabadabacabaeabacabadabacabahabacabadabacabaeabacabadabacabafabacabadabacabaeabacabadabacabagabacabadabacabaeabacabadabacabafabacabqdabacabaeabacabadabacabaaciifgeeabgfgfhiadhbfhddihcfeaebfbiiebbcebafegfiefgbagffgchdieicffebaadddcefiibibbhbagfgifieedgeiabhfhbgghiaiegccehgdhaeaafddadgeghidabaeicdhbfghfcciihdgiefaggachefchbddaddafbhhdfhcaebhffbfefabbbbafcdihbcgbfaffieghiiiebhegbcfceidggibdggabaifgedg",
"output": "687296"
},
{
"input": "abacabadabacabaeabacabadabacabafabacabadabacabaeabacabadabacabagabacabadabacabaeabacabadabacabafabacabadabacabaeabacabadabacabahabacabadabpcabaeabacabadabacabafabacabadabacabaeabacabadabacabagabacabadabacabaeabacabadabacabafabacabadabacabaeabacabadabacabaeeegbahbcahagefafcedggigbhdaeebfbagfafhdfcdddeaihgdigiaehdabicigbccdacdcbbcfeccdgiibicaifgieigcfbiedebfeddadagibhbdbfeibgbiadgcacfhfbfegbdbheidegcghfhbhcaffbcfhihcbghecfidbdfadgacigfdabhiebibfhfcachcbdcabeedfcffehhidbbihcgahfhdfhbhaciegccagebifh",
"output": "630475"
},
{
"input": "abacabndabacabaeabacabadabacabafabacabadabacabaeabacabadabacabagabacabadabacabaeabacabadabacabafabacabadabacabaeabacabadabacabahabacabadabacabaeabacabadabacabafabacabadabacabadabacabadabacabagabacabadabacabaeabacabadabacabafabacabadabacabaeabacabadabacabayhaghdcfefibhcbccbbfiedcdieghaihagcchhecgdfbdcchhbagebfbiecaibedhhaeidbcgageadeichbhghficaafdbafdbgadgcficdhdgcagiafbahaeffheabfbbgccbfdbfabahiaeabbfggefbafbheeehaebiehibbffehifghadcegdhdagbebeagbedcciibadfcifcedcghacheaihfdfgebcciidbcffgaebgabg",
"output": "619985"
},
{
"input": "abacababcc",
"output": "98"
},
{
"input": "abauabaabd",
"output": "94"
},
{
"input": "abbabbababbabbababbababbabbababbabbababbababbabbababbababbabbababbabbababbababbabbababbabbbaaabaabaa",
"output": "79323"
},
{
"input": "bababbababbabbababbababbabbababbabbababbababbabbababbababbabbababbabbababbababbabbababbabbababbababbabbababbababbabbababbabbababbababbabbababbabbabbbbbaabababbbbbbbaabaabaaabbabbbabbbbabbabaababbbabbb",
"output": "385228"
},
{
"input": "abbabbababbabbababbababbabbababbabbababbababbabbababbababbabbababbabbababbababbabbababbabbababbababbabbababbababbabbababbabbababbababbabbababbababbabbababbabbababbababbabbababbabbababbababbabbababbababbabbababbabbababbababbabbababbabaabaabbabaaaabbabaaabbaababaababbabaaabbbbabbbaaaabababbbaaabaababb",
"output": "1054380"
},
{
"input": "bababbababbabbababbababbabbababbabbababbababbabbababbababbabbababbabbababbababbabbababbabbababbababbabbababbababbabbababbabbababbababbabbababbababbabbababbabbababbababbabbababbabbababbababbabbababbababbabbababbabbababbababbabbababbabbababbababbabbababbababbabbababbabbababbababbabbababbababbabbababbabbababbababbabbababbabbababbababbabbababbababbabbababbabbababbababbabbababbabbbbbbbaaabbaabbbabaabab",
"output": "2840036"
},
{
"input": "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",
"output": "3678765"
}
] | 1,682,117,977
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 1
| 92
| 0
|
import sys
input = lambda: sys.stdin.readline().rstrip()
S = input()
N = len(S)
def deal(S):
ls = [1]*N
for i in range(N):
l,r=i-1,i+1
while l>=0 and r<N and S[l]==S[r]:
ls[r]+=1
l-=1
r+=1
for i in range(1,N):
if S[i-1]==S[i]:
l,r=i-2,i+1
while l>=0 and r<N and S[l]==S[r]:
ls[r]+=1
l-=1
r+=1
return ls
ls = deal(S)
rs = deal(S[::-1])[::-1]
# print(ls)
# print(rs)
ans = 0
b = sum(rs)
for i in range(N-1):
b-=rs[i]
ans+=ls[i]*b
print(ans)
|
Title: Palindrome pairs
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a non-empty string *s* consisting of lowercase letters. Find the number of pairs of non-overlapping palindromic substrings of this string.
In a more formal way, you have to find the quantity of tuples (*a*,<=*b*,<=*x*,<=*y*) such that 1<=≤<=*a*<=≤<=*b*<=<<=*x*<=≤<=*y*<=≤<=|*s*| and substrings *s*[*a*... *b*], *s*[*x*... *y*] are palindromes.
A palindrome is a string that can be read the same way from left to right and from right to left. For example, "abacaba", "z", "abba" are palindromes.
A substring *s*[*i*... *j*] (1<=≤<=*i*<=≤<=*j*<=≤<=|*s*|) of string *s* = *s*1*s*2... *s*|*s*| is a string *s**i**s**i*<=+<=1... *s**j*. For example, substring *s*[2...4] of string *s* = "abacaba" equals "bac".
Input Specification:
The first line of input contains a non-empty string *s* which consists of lowercase letters ('a'...'z'), *s* contains at most 2000 characters.
Output Specification:
Output a single number — the quantity of pairs of non-overlapping palindromic substrings of *s*.
Please do not use the %lld format specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d format specifier.
Demo Input:
['aa\n', 'aaa\n', 'abacaba\n']
Demo Output:
['1\n', '5\n', '36\n']
Note:
none
|
```python
import sys
input = lambda: sys.stdin.readline().rstrip()
S = input()
N = len(S)
def deal(S):
ls = [1]*N
for i in range(N):
l,r=i-1,i+1
while l>=0 and r<N and S[l]==S[r]:
ls[r]+=1
l-=1
r+=1
for i in range(1,N):
if S[i-1]==S[i]:
l,r=i-2,i+1
while l>=0 and r<N and S[l]==S[r]:
ls[r]+=1
l-=1
r+=1
return ls
ls = deal(S)
rs = deal(S[::-1])[::-1]
# print(ls)
# print(rs)
ans = 0
b = sum(rs)
for i in range(N-1):
b-=rs[i]
ans+=ls[i]*b
print(ans)
```
| 0
|
|
0
|
none
|
none
|
none
| 0
|
[
"none"
] | 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 — 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".
| 0
|
[
{
"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"
}
] | 1,437,993,588
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 82
| 2,000
| 1,024,000
|
a = input()
b = input()
def equal_str(a, b):
ma = len(a)
mb = len(b)
if ma & 1:
return a == b;
ma >>= 1
mb >>= 1
if (equal_str(a[0 : ma], b[0 : mb]) and equal_str(a[ma:], b[mb:])) or (equal_str(a[0 : ma], b[mb:]) and equal_str(a[ma:], b[0 : mb])):
return True
return False
if (equal_str(a, b)):
print("YES")
else:
print("NO")
|
Title: none
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 — 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
a = input()
b = input()
def equal_str(a, b):
ma = len(a)
mb = len(b)
if ma & 1:
return a == b;
ma >>= 1
mb >>= 1
if (equal_str(a[0 : ma], b[0 : mb]) and equal_str(a[ma:], b[mb:])) or (equal_str(a[0 : ma], b[mb:]) and equal_str(a[ma:], b[0 : mb])):
return True
return False
if (equal_str(a, b)):
print("YES")
else:
print("NO")
```
| 0
|
|
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<=≤<=*n*<=≤<=100) — 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* = 5 there exists *j* = 3 such that *s*<sub class="lower-index">*i*</sub> = *s*<sub class="lower-index">*j*</sub> and *j* < *i*, which means that answer for *i* = 5 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"
}
] | 1,689,167,877
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 55
| 46
| 0
|
def check_diary_possession(n, names):
possession = set()
for name in names:
if name in possession:
print("YES")
else:
print("NO")
possession.add(name)
# Read input
n = int(input())
names = []
for _ in range(n):
names.append(input().lower())
# Check diary possession
check_diary_possession(n, names)
|
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<=≤<=*n*<=≤<=100) — 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* = 5 there exists *j* = 3 such that *s*<sub class="lower-index">*i*</sub> = *s*<sub class="lower-index">*j*</sub> and *j* < *i*, which means that answer for *i* = 5 is "YES".
|
```python
def check_diary_possession(n, names):
possession = set()
for name in names:
if name in possession:
print("YES")
else:
print("NO")
possession.add(name)
# Read input
n = int(input())
names = []
for _ in range(n):
names.append(input().lower())
# Check diary possession
check_diary_possession(n, names)
```
| 3
|
|
764
|
A
|
Taymyr is calling you
|
PROGRAMMING
| 800
|
[
"brute force",
"implementation",
"math"
] | null | null |
Comrade Dujikov is busy choosing artists for Timofey's birthday and is recieving calls from Taymyr from Ilia-alpinist.
Ilia-alpinist calls every *n* minutes, i.e. in minutes *n*, 2*n*, 3*n* and so on. Artists come to the comrade every *m* minutes, i.e. in minutes *m*, 2*m*, 3*m* and so on. The day is *z* minutes long, i.e. the day consists of minutes 1,<=2,<=...,<=*z*. How many artists should be killed so that there are no artists in the room when Ilia calls? Consider that a call and a talk with an artist take exactly one minute.
|
The only string contains three integers — *n*, *m* and *z* (1<=≤<=*n*,<=*m*,<=*z*<=≤<=104).
|
Print single integer — the minimum number of artists that should be killed so that there are no artists in the room when Ilia calls.
|
[
"1 1 10\n",
"1 2 5\n",
"2 3 9\n"
] |
[
"10\n",
"2\n",
"1\n"
] |
Taymyr is a place in the north of Russia.
In the first test the artists come each minute, as well as the calls, so we need to kill all of them.
In the second test we need to kill artists which come on the second and the fourth minutes.
In the third test — only the artist which comes on the sixth minute.
| 500
|
[
{
"input": "1 1 10",
"output": "10"
},
{
"input": "1 2 5",
"output": "2"
},
{
"input": "2 3 9",
"output": "1"
},
{
"input": "4 8 9",
"output": "1"
},
{
"input": "7 9 2",
"output": "0"
},
{
"input": "10000 10000 10000",
"output": "1"
},
{
"input": "24 22 9235",
"output": "34"
},
{
"input": "74 8 417",
"output": "1"
},
{
"input": "972 1 203",
"output": "0"
},
{
"input": "550 1 754",
"output": "1"
},
{
"input": "860 1 884",
"output": "1"
},
{
"input": "358 2 809",
"output": "2"
},
{
"input": "33 27 216",
"output": "0"
},
{
"input": "2940 1 9311",
"output": "3"
},
{
"input": "4624 1 1953",
"output": "0"
},
{
"input": "2696 2 7345",
"output": "2"
},
{
"input": "3443 2 6701",
"output": "0"
},
{
"input": "3 613 2275",
"output": "1"
},
{
"input": "1 10000 10000",
"output": "1"
},
{
"input": "10000 1 10000",
"output": "1"
},
{
"input": "1 1 1",
"output": "1"
},
{
"input": "1 1 10000",
"output": "10000"
},
{
"input": "34 27 10000",
"output": "10"
},
{
"input": "2 2 9999",
"output": "4999"
},
{
"input": "2 2 1",
"output": "0"
},
{
"input": "6 4 36",
"output": "3"
},
{
"input": "33 6 3005",
"output": "45"
},
{
"input": "5 1 20",
"output": "4"
},
{
"input": "1 2 10",
"output": "5"
},
{
"input": "2 1 100",
"output": "50"
},
{
"input": "10 20 10000",
"output": "500"
},
{
"input": "8 12 12",
"output": "0"
}
] | 1,699,889,930
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 32
| 576
| 512,000
|
a, b, c = map(int, input().split())
d = [0] * (c // a)
e = [0] * (c // b)
y = 0
f = 0
for i in range(a, c+1, a):
d[y] = i
y += 1
for i in range(b, c+1, b):
e[f] = i
f += 1
r = 0
for i in range(0, len(d)):
if d[i] in e:
r += 1
print(r)
|
Title: Taymyr is calling you
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Comrade Dujikov is busy choosing artists for Timofey's birthday and is recieving calls from Taymyr from Ilia-alpinist.
Ilia-alpinist calls every *n* minutes, i.e. in minutes *n*, 2*n*, 3*n* and so on. Artists come to the comrade every *m* minutes, i.e. in minutes *m*, 2*m*, 3*m* and so on. The day is *z* minutes long, i.e. the day consists of minutes 1,<=2,<=...,<=*z*. How many artists should be killed so that there are no artists in the room when Ilia calls? Consider that a call and a talk with an artist take exactly one minute.
Input Specification:
The only string contains three integers — *n*, *m* and *z* (1<=≤<=*n*,<=*m*,<=*z*<=≤<=104).
Output Specification:
Print single integer — the minimum number of artists that should be killed so that there are no artists in the room when Ilia calls.
Demo Input:
['1 1 10\n', '1 2 5\n', '2 3 9\n']
Demo Output:
['10\n', '2\n', '1\n']
Note:
Taymyr is a place in the north of Russia.
In the first test the artists come each minute, as well as the calls, so we need to kill all of them.
In the second test we need to kill artists which come on the second and the fourth minutes.
In the third test — only the artist which comes on the sixth minute.
|
```python
a, b, c = map(int, input().split())
d = [0] * (c // a)
e = [0] * (c // b)
y = 0
f = 0
for i in range(a, c+1, a):
d[y] = i
y += 1
for i in range(b, c+1, b):
e[f] = i
f += 1
r = 0
for i in range(0, len(d)):
if d[i] in e:
r += 1
print(r)
```
| 3
|
Subsets and Splits
Successful Python Submissions
Retrieves all records from the train dataset where the verdict is 'OK', providing basic filtering but limited analytical value.
SQL Console for MatrixStudio/Codeforces-Python-Submissions
Retrieves records of users with a rating of 1600 or higher and a verdict of 'OK', providing basic filtering but limited analytical value.
SQL Console for MatrixStudio/Codeforces-Python-Submissions
Counts the number of entries with a rating above 2000 and a verdict of 'OK', providing basic filtering but limited analytical value.
SQL Console for MatrixStudio/Codeforces-Python-Submissions
Counts the number of entries with a 'OK' verdict, providing a basic overview of a specific category within the dataset.