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3.99
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|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
793
|
A
|
Oleg and shares
|
PROGRAMMING
| 900
|
[
"implementation",
"math"
] | null | null |
Oleg the bank client checks share prices every day. There are *n* share prices he is interested in. Today he observed that each second exactly one of these prices decreases by *k* rubles (note that each second exactly one price changes, but at different seconds different prices can change). Prices can become negative. Oleg found this process interesting, and he asked Igor the financial analyst, what is the minimum time needed for all *n* prices to become equal, or it is impossible at all? Igor is busy right now, so he asked you to help Oleg. Can you answer this question?
|
The first line contains two integers *n* and *k* (1<=β€<=*n*<=β€<=105,<=1<=β€<=*k*<=β€<=109)Β β the number of share prices, and the amount of rubles some price decreases each second.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=β€<=*a**i*<=β€<=109)Β β the initial prices.
|
Print the only line containing the minimum number of seconds needed for prices to become equal, of Β«-1Β» if it is impossible.
|
[
"3 3\n12 9 15\n",
"2 2\n10 9\n",
"4 1\n1 1000000000 1000000000 1000000000\n"
] |
[
"3",
"-1",
"2999999997"
] |
Consider the first example.
Suppose the third price decreases in the first second and become equal 12 rubles, then the first price decreases and becomes equal 9 rubles, and in the third second the third price decreases again and becomes equal 9 rubles. In this case all prices become equal 9 rubles in 3 seconds.
There could be other possibilities, but this minimizes the time needed for all prices to become equal. Thus the answer is 3.
In the second example we can notice that parity of first and second price is different and never changes within described process. Thus prices never can become equal.
In the third example following scenario can take place: firstly, the second price drops, then the third price, and then fourth price. It happens 999999999 times, and, since in one second only one price can drop, the whole process takes 999999999β*β3β=β2999999997 seconds. We can note that this is the minimum possible time.
| 500
|
[
{
"input": "3 3\n12 9 15",
"output": "3"
},
{
"input": "2 2\n10 9",
"output": "-1"
},
{
"input": "4 1\n1 1000000000 1000000000 1000000000",
"output": "2999999997"
},
{
"input": "1 11\n123",
"output": "0"
},
{
"input": "20 6\n38 86 86 50 98 62 32 2 14 62 98 50 2 50 32 38 62 62 8 14",
"output": "151"
},
{
"input": "20 5\n59 54 19 88 55 100 54 3 6 13 99 38 36 71 59 6 64 85 45 54",
"output": "-1"
},
{
"input": "100 10\n340 70 440 330 130 120 340 210 440 110 410 120 180 40 50 230 70 110 310 360 480 70 230 120 230 310 470 60 210 60 210 480 290 250 450 440 150 40 500 230 280 250 30 50 310 50 230 360 420 260 330 80 50 160 70 470 140 180 380 190 250 30 220 410 80 310 280 50 20 430 440 180 310 190 190 330 90 190 320 390 170 460 230 30 80 500 470 370 80 500 400 120 220 150 70 120 70 320 260 260",
"output": "2157"
},
{
"input": "100 18\n489 42 300 366 473 105 220 448 70 488 201 396 168 281 67 235 324 291 313 387 407 223 39 144 224 233 72 318 229 377 62 171 448 119 354 282 147 447 260 384 172 199 67 326 311 431 337 142 281 202 404 468 38 120 90 437 33 420 249 372 367 253 255 411 309 333 103 176 162 120 203 41 352 478 216 498 224 31 261 493 277 99 375 370 394 229 71 488 246 194 233 13 66 111 366 456 277 360 116 354",
"output": "-1"
},
{
"input": "4 2\n1 2 3 4",
"output": "-1"
},
{
"input": "3 4\n3 5 5",
"output": "-1"
},
{
"input": "3 2\n88888884 88888886 88888888",
"output": "3"
},
{
"input": "2 1\n1000000000 1000000000",
"output": "0"
},
{
"input": "4 2\n1000000000 100000000 100000000 100000000",
"output": "450000000"
},
{
"input": "2 2\n1000000000 1000000000",
"output": "0"
},
{
"input": "3 3\n3 2 1",
"output": "-1"
},
{
"input": "3 4\n3 5 3",
"output": "-1"
},
{
"input": "3 2\n1 2 2",
"output": "-1"
},
{
"input": "4 2\n2 3 3 2",
"output": "-1"
},
{
"input": "3 2\n1 2 4",
"output": "-1"
},
{
"input": "3 2\n3 4 4",
"output": "-1"
},
{
"input": "3 3\n4 7 10",
"output": "3"
},
{
"input": "4 3\n2 2 5 1",
"output": "-1"
},
{
"input": "3 3\n1 3 5",
"output": "-1"
},
{
"input": "2 5\n5 9",
"output": "-1"
},
{
"input": "2 3\n5 7",
"output": "-1"
},
{
"input": "3 137\n1000000000 1000000000 1000000000",
"output": "0"
},
{
"input": "5 1000000000\n1000000000 1000000000 1000000000 1000000000 1000000000",
"output": "0"
},
{
"input": "3 5\n1 2 5",
"output": "-1"
},
{
"input": "3 3\n1000000000 1000000000 999999997",
"output": "2"
},
{
"input": "2 4\n5 6",
"output": "-1"
},
{
"input": "4 1\n1000000000 1000000000 1000000000 1000000000",
"output": "0"
},
{
"input": "2 3\n5 8",
"output": "1"
},
{
"input": "2 6\n8 16",
"output": "-1"
},
{
"input": "5 3\n15 14 9 12 18",
"output": "-1"
},
{
"input": "3 3\n1 2 3",
"output": "-1"
},
{
"input": "3 3\n3 4 5",
"output": "-1"
},
{
"input": "2 5\n8 17",
"output": "-1"
},
{
"input": "2 1\n1 2",
"output": "1"
},
{
"input": "1 1\n1000000000",
"output": "0"
},
{
"input": "3 3\n5 3 4",
"output": "-1"
},
{
"input": "3 6\n10 14 12",
"output": "-1"
},
{
"input": "2 2\n3 5",
"output": "1"
},
{
"input": "3 5\n1 3 4",
"output": "-1"
},
{
"input": "4 3\n1 6 6 6",
"output": "-1"
},
{
"input": "2 3\n1 8",
"output": "-1"
},
{
"input": "3 5\n6 11 17",
"output": "-1"
},
{
"input": "2 2\n1 4",
"output": "-1"
},
{
"input": "2 4\n6 8",
"output": "-1"
},
{
"input": "2 1\n2 3",
"output": "1"
},
{
"input": "4 4\n1 5 8 14",
"output": "-1"
},
{
"input": "3 3\n1 5 3",
"output": "-1"
},
{
"input": "4 3\n1 2 2 5",
"output": "-1"
},
{
"input": "3 2\n1 4 6",
"output": "-1"
},
{
"input": "2 3\n6 9",
"output": "1"
},
{
"input": "3 3\n2 3 4",
"output": "-1"
},
{
"input": "3 2\n9 10 10",
"output": "-1"
},
{
"input": "2 2\n9 12",
"output": "-1"
},
{
"input": "2 2\n100000003 100000005",
"output": "1"
},
{
"input": "2 3\n2 4",
"output": "-1"
},
{
"input": "3 2\n2 3 5",
"output": "-1"
},
{
"input": "3 3\n1 3 4",
"output": "-1"
},
{
"input": "10 2\n2 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000",
"output": "4499999991"
},
{
"input": "3 5\n2 4 5",
"output": "-1"
},
{
"input": "2 3\n7 10",
"output": "1"
},
{
"input": "3 10\n10 13 17",
"output": "-1"
},
{
"input": "2 3\n1 6",
"output": "-1"
},
{
"input": "1 7\n1000000000",
"output": "0"
},
{
"input": "2 4\n3 7",
"output": "1"
},
{
"input": "2 3\n2 5",
"output": "1"
},
{
"input": "20 1\n1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000",
"output": "0"
},
{
"input": "3 3\n7 8 8",
"output": "-1"
},
{
"input": "4 10\n1 11 100 11",
"output": "-1"
}
] | 1,611,815,204
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 88
| 202
| 10,956,800
|
import math
n,k=map(int,input().split())
a=list(map(int,input().split()))
a.sort()
r=a[0]%k;t=0
for i in range(1,n):
if a[i]%k!=r:
print(-1)
break
else:
t+=(a[i]-a[0])//k
else:
print(t)
|
Title: Oleg and shares
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Oleg the bank client checks share prices every day. There are *n* share prices he is interested in. Today he observed that each second exactly one of these prices decreases by *k* rubles (note that each second exactly one price changes, but at different seconds different prices can change). Prices can become negative. Oleg found this process interesting, and he asked Igor the financial analyst, what is the minimum time needed for all *n* prices to become equal, or it is impossible at all? Igor is busy right now, so he asked you to help Oleg. Can you answer this question?
Input Specification:
The first line contains two integers *n* and *k* (1<=β€<=*n*<=β€<=105,<=1<=β€<=*k*<=β€<=109)Β β the number of share prices, and the amount of rubles some price decreases each second.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=β€<=*a**i*<=β€<=109)Β β the initial prices.
Output Specification:
Print the only line containing the minimum number of seconds needed for prices to become equal, of Β«-1Β» if it is impossible.
Demo Input:
['3 3\n12 9 15\n', '2 2\n10 9\n', '4 1\n1 1000000000 1000000000 1000000000\n']
Demo Output:
['3', '-1', '2999999997']
Note:
Consider the first example.
Suppose the third price decreases in the first second and become equal 12 rubles, then the first price decreases and becomes equal 9 rubles, and in the third second the third price decreases again and becomes equal 9 rubles. In this case all prices become equal 9 rubles in 3 seconds.
There could be other possibilities, but this minimizes the time needed for all prices to become equal. Thus the answer is 3.
In the second example we can notice that parity of first and second price is different and never changes within described process. Thus prices never can become equal.
In the third example following scenario can take place: firstly, the second price drops, then the third price, and then fourth price. It happens 999999999 times, and, since in one second only one price can drop, the whole process takes 999999999β*β3β=β2999999997 seconds. We can note that this is the minimum possible time.
|
```python
import math
n,k=map(int,input().split())
a=list(map(int,input().split()))
a.sort()
r=a[0]%k;t=0
for i in range(1,n):
if a[i]%k!=r:
print(-1)
break
else:
t+=(a[i]-a[0])//k
else:
print(t)
```
| 3
|
|
265
|
A
|
Colorful Stones (Simplified Edition)
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
There is a sequence of colorful stones. The color of each stone is one of red, green, or blue. You are given a string *s*. The *i*-th (1-based) character of *s* represents the color of the *i*-th stone. If the character is "R", "G", or "B", the color of the corresponding stone is red, green, or blue, respectively.
Initially Squirrel Liss is standing on the first stone. You perform instructions one or more times.
Each instruction is one of the three types: "RED", "GREEN", or "BLUE". After an instruction *c*, if Liss is standing on a stone whose colors is *c*, Liss will move one stone forward, else she will not move.
You are given a string *t*. The number of instructions is equal to the length of *t*, and the *i*-th character of *t* represents the *i*-th instruction.
Calculate the final position of Liss (the number of the stone she is going to stand on in the end) after performing all the instructions, and print its 1-based position. It is guaranteed that Liss don't move out of the sequence.
|
The input contains two lines. The first line contains the string *s* (1<=β€<=|*s*|<=β€<=50). The second line contains the string *t* (1<=β€<=|*t*|<=β€<=50). The characters of each string will be one of "R", "G", or "B". It is guaranteed that Liss don't move out of the sequence.
|
Print the final 1-based position of Liss in a single line.
|
[
"RGB\nRRR\n",
"RRRBGBRBBB\nBBBRR\n",
"BRRBGBRGRBGRGRRGGBGBGBRGBRGRGGGRBRRRBRBBBGRRRGGBBB\nBBRBGGRGRGBBBRBGRBRBBBBRBRRRBGBBGBBRRBBGGRBRRBRGRB\n"
] |
[
"2\n",
"3\n",
"15\n"
] |
none
| 500
|
[
{
"input": "RGB\nRRR",
"output": "2"
},
{
"input": "RRRBGBRBBB\nBBBRR",
"output": "3"
},
{
"input": "BRRBGBRGRBGRGRRGGBGBGBRGBRGRGGGRBRRRBRBBBGRRRGGBBB\nBBRBGGRGRGBBBRBGRBRBBBBRBRRRBGBBGBBRRBBGGRBRRBRGRB",
"output": "15"
},
{
"input": "G\nRRBBRBRRBR",
"output": "1"
},
{
"input": "RRRRRBRRBRRGRBGGRRRGRBBRBBBBBRGRBGBRRGBBBRBBGBRGBB\nB",
"output": "1"
},
{
"input": "RRGGBRGRBG\nBRRGGBBGGR",
"output": "7"
},
{
"input": "BBRRGBGGRGBRGBRBRBGR\nGGGRBGGGBRRRRGRBGBGRGRRBGRBGBG",
"output": "15"
},
{
"input": "GBRRBGBGBBBBRRRGBGRRRGBGBBBRGR\nRRGBRRGRBBBBBBGRRBBR",
"output": "8"
},
{
"input": "BRGRRGRGRRGBBGBBBRRBBRRBGBBGRGBBGGRGBRBGGGRRRBGGBB\nRGBBGRRBBBRRGRRBRBBRGBBGGGRGBGRRRRBRBGGBRBGGGRGBRR",
"output": "16"
},
{
"input": "GGRGGBRRGRGBRRGGRBBGGRRGBBBGBBBGGRBGGBRBBRGBRRRBRG\nGGRGRRRRRRRRRGBBBBRGBRGRGRRGBBRGGBRBBRBGBRGRRRRGGR",
"output": "18"
},
{
"input": "RBBRBGBBGGGBRRBGBRGRRGGRBBBBGRBGGBRRBGBBRBRGBBGGGG\nBRRGRBGRBGBRGBRBGGBBBGBBRGRBGRGRBGRBGGRRBGBRGRRBBG",
"output": "23"
},
{
"input": "RRGGRRBRGRGBRGBRGBBRGGGBRBGBRGBRGGBRBRGBGRRBBGGRBR\nBRGRBBBGRBGRRGGRRRRRRRRGGGGGGGGBGGGGGGGRGRBBRGRGBR",
"output": "14"
},
{
"input": "RBRBRRRRRGGR\nBBBBRRGGBGRBRGBBRGGGRGGBRRBRBRBGGG",
"output": "12"
},
{
"input": "GGRGGGBRGRGRBBRGGGRBGRRR\nGBBBBGBGBGBBBRGGRGGBGRRGRRRGBGB",
"output": "12"
},
{
"input": "RGRBGBRBBGBRRGGGBBGBRRBGRGBBBBBGRRRGGRBB\nBGR",
"output": "2"
},
{
"input": "R\nB",
"output": "1"
},
{
"input": "RRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR\nRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR",
"output": "50"
},
{
"input": "RRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR\nGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG",
"output": "1"
},
{
"input": "R\nGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG",
"output": "1"
}
] | 1,677,513,194
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 19
| 92
| 0
|
input1=list(input())
instructions=list(input())
ans=1
counter1=0
counter2=0
for i in range(len(instructions)):
if instructions[counter1]==input1[counter2]:
ans+=1
counter2+=1
counter1+=1
print(ans)
|
Title: Colorful Stones (Simplified Edition)
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There is a sequence of colorful stones. The color of each stone is one of red, green, or blue. You are given a string *s*. The *i*-th (1-based) character of *s* represents the color of the *i*-th stone. If the character is "R", "G", or "B", the color of the corresponding stone is red, green, or blue, respectively.
Initially Squirrel Liss is standing on the first stone. You perform instructions one or more times.
Each instruction is one of the three types: "RED", "GREEN", or "BLUE". After an instruction *c*, if Liss is standing on a stone whose colors is *c*, Liss will move one stone forward, else she will not move.
You are given a string *t*. The number of instructions is equal to the length of *t*, and the *i*-th character of *t* represents the *i*-th instruction.
Calculate the final position of Liss (the number of the stone she is going to stand on in the end) after performing all the instructions, and print its 1-based position. It is guaranteed that Liss don't move out of the sequence.
Input Specification:
The input contains two lines. The first line contains the string *s* (1<=β€<=|*s*|<=β€<=50). The second line contains the string *t* (1<=β€<=|*t*|<=β€<=50). The characters of each string will be one of "R", "G", or "B". It is guaranteed that Liss don't move out of the sequence.
Output Specification:
Print the final 1-based position of Liss in a single line.
Demo Input:
['RGB\nRRR\n', 'RRRBGBRBBB\nBBBRR\n', 'BRRBGBRGRBGRGRRGGBGBGBRGBRGRGGGRBRRRBRBBBGRRRGGBBB\nBBRBGGRGRGBBBRBGRBRBBBBRBRRRBGBBGBBRRBBGGRBRRBRGRB\n']
Demo Output:
['2\n', '3\n', '15\n']
Note:
none
|
```python
input1=list(input())
instructions=list(input())
ans=1
counter1=0
counter2=0
for i in range(len(instructions)):
if instructions[counter1]==input1[counter2]:
ans+=1
counter2+=1
counter1+=1
print(ans)
```
| 3
|
|
552
|
A
|
Vanya and Table
|
PROGRAMMING
| 1,000
|
[
"implementation",
"math"
] | null | null |
Vanya has a table consisting of 100 rows, each row contains 100 cells. The rows are numbered by integers from 1 to 100 from bottom to top, the columns are numbered from 1 to 100 from left to right.
In this table, Vanya chose *n* rectangles with sides that go along borders of squares (some rectangles probably occur multiple times). After that for each cell of the table he counted the number of rectangles it belongs to and wrote this number into it. Now he wants to find the sum of values in all cells of the table and as the table is too large, he asks you to help him find the result.
|
The first line contains integer *n* (1<=β€<=*n*<=β€<=100) β the number of rectangles.
Each of the following *n* lines contains four integers *x*1,<=*y*1,<=*x*2,<=*y*2 (1<=β€<=*x*1<=β€<=*x*2<=β€<=100, 1<=β€<=*y*1<=β€<=*y*2<=β€<=100), where *x*1 and *y*1 are the number of the column and row of the lower left cell and *x*2 and *y*2 are the number of the column and row of the upper right cell of a rectangle.
|
In a single line print the sum of all values in the cells of the table.
|
[
"2\n1 1 2 3\n2 2 3 3\n",
"2\n1 1 3 3\n1 1 3 3\n"
] |
[
"10\n",
"18\n"
] |
Note to the first sample test:
Values of the table in the first three rows and columns will be as follows:
121
121
110
So, the sum of values will be equal to 10.
Note to the second sample test:
Values of the table in the first three rows and columns will be as follows:
222
222
222
So, the sum of values will be equal to 18.
| 500
|
[
{
"input": "2\n1 1 2 3\n2 2 3 3",
"output": "10"
},
{
"input": "2\n1 1 3 3\n1 1 3 3",
"output": "18"
},
{
"input": "5\n4 11 20 15\n7 5 12 20\n10 8 16 12\n7 5 12 15\n2 2 20 13",
"output": "510"
},
{
"input": "5\n4 11 20 20\n6 11 20 16\n5 2 19 15\n11 3 18 15\n3 2 14 11",
"output": "694"
},
{
"input": "5\n1 1 1 100\n1 1 1 100\n1 1 1 100\n1 1 1 100\n1 1 1 100",
"output": "500"
},
{
"input": "1\n1 1 1 1",
"output": "1"
},
{
"input": "1\n100 100 100 100",
"output": "1"
},
{
"input": "1\n1 1 1 100",
"output": "100"
},
{
"input": "3\n1 1 1 1\n1 2 1 2\n1 3 1 3",
"output": "3"
},
{
"input": "1\n1 1 100 100",
"output": "10000"
}
] | 1,695,059,582
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 26
| 77
| 0
|
def solve(x1, y1, x2, y2):
return sum((x2[i] - x1[i] + 1) * (y2[i] - y1[i] + 1) for i in range(len(x1)))
def main():
n = int(input())
x1 = []
y1 = []
x2 = []
y2 = []
for _ in range(n):
x1_i, y1_i, x2_i, y2_i = map(int, input().split())
x1.append(x1_i)
y1.append(y1_i)
x2.append(x2_i)
y2.append(y2_i)
result = solve(x1, y1, x2, y2)
print(result)
if __name__ == "__main__":
main()
|
Title: Vanya and Table
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vanya has a table consisting of 100 rows, each row contains 100 cells. The rows are numbered by integers from 1 to 100 from bottom to top, the columns are numbered from 1 to 100 from left to right.
In this table, Vanya chose *n* rectangles with sides that go along borders of squares (some rectangles probably occur multiple times). After that for each cell of the table he counted the number of rectangles it belongs to and wrote this number into it. Now he wants to find the sum of values in all cells of the table and as the table is too large, he asks you to help him find the result.
Input Specification:
The first line contains integer *n* (1<=β€<=*n*<=β€<=100) β the number of rectangles.
Each of the following *n* lines contains four integers *x*1,<=*y*1,<=*x*2,<=*y*2 (1<=β€<=*x*1<=β€<=*x*2<=β€<=100, 1<=β€<=*y*1<=β€<=*y*2<=β€<=100), where *x*1 and *y*1 are the number of the column and row of the lower left cell and *x*2 and *y*2 are the number of the column and row of the upper right cell of a rectangle.
Output Specification:
In a single line print the sum of all values in the cells of the table.
Demo Input:
['2\n1 1 2 3\n2 2 3 3\n', '2\n1 1 3 3\n1 1 3 3\n']
Demo Output:
['10\n', '18\n']
Note:
Note to the first sample test:
Values of the table in the first three rows and columns will be as follows:
121
121
110
So, the sum of values will be equal to 10.
Note to the second sample test:
Values of the table in the first three rows and columns will be as follows:
222
222
222
So, the sum of values will be equal to 18.
|
```python
def solve(x1, y1, x2, y2):
return sum((x2[i] - x1[i] + 1) * (y2[i] - y1[i] + 1) for i in range(len(x1)))
def main():
n = int(input())
x1 = []
y1 = []
x2 = []
y2 = []
for _ in range(n):
x1_i, y1_i, x2_i, y2_i = map(int, input().split())
x1.append(x1_i)
y1.append(y1_i)
x2.append(x2_i)
y2.append(y2_i)
result = solve(x1, y1, x2, y2)
print(result)
if __name__ == "__main__":
main()
```
| 3
|
|
155
|
A
|
I_love_\%username\%
|
PROGRAMMING
| 800
|
[
"brute force"
] | null | null |
Vasya adores sport programming. He can't write programs but he loves to watch the contests' progress. Vasya even has a favorite coder and Vasya pays special attention to him.
One day Vasya decided to collect the results of all contests where his favorite coder participated and track the progress of his coolness. For each contest where this coder participated, he wrote out a single non-negative number β the number of points his favorite coder earned in the contest. Vasya wrote out the points for the contest in the order, in which the contests run (naturally, no two contests ran simultaneously).
Vasya considers a coder's performance in a contest amazing in two situations: he can break either his best or his worst performance record. First, it is amazing if during the contest the coder earns strictly more points that he earned on each past contest. Second, it is amazing if during the contest the coder earns strictly less points that he earned on each past contest. A coder's first contest isn't considered amazing. Now he wants to count the number of amazing performances the coder had throughout his whole history of participating in contests. But the list of earned points turned out long and Vasya can't code... That's why he asks you to help him.
|
The first line contains the single integer *n* (1<=β€<=*n*<=β€<=1000) β the number of contests where the coder participated.
The next line contains *n* space-separated non-negative integer numbers β they are the points which the coder has earned. The points are given in the chronological order. All points do not exceed 10000.
|
Print the single number β the number of amazing performances the coder has had during his whole history of participating in the contests.
|
[
"5\n100 50 200 150 200\n",
"10\n4664 6496 5814 7010 5762 5736 6944 4850 3698 7242\n"
] |
[
"2\n",
"4\n"
] |
In the first sample the performances number 2 and 3 are amazing.
In the second sample the performances number 2, 4, 9 and 10 are amazing.
| 500
|
[
{
"input": "5\n100 50 200 150 200",
"output": "2"
},
{
"input": "10\n4664 6496 5814 7010 5762 5736 6944 4850 3698 7242",
"output": "4"
},
{
"input": "1\n6",
"output": "0"
},
{
"input": "2\n2 1",
"output": "1"
},
{
"input": "5\n100 36 53 7 81",
"output": "2"
},
{
"input": "5\n7 36 53 81 100",
"output": "4"
},
{
"input": "5\n100 81 53 36 7",
"output": "4"
},
{
"input": "10\n8 6 3 4 9 10 7 7 1 3",
"output": "5"
},
{
"input": "10\n1627 1675 1488 1390 1812 1137 1746 1324 1952 1862",
"output": "6"
},
{
"input": "10\n1 3 3 4 6 7 7 8 9 10",
"output": "7"
},
{
"input": "10\n1952 1862 1812 1746 1675 1627 1488 1390 1324 1137",
"output": "9"
},
{
"input": "25\n1448 4549 2310 2725 2091 3509 1565 2475 2232 3989 4231 779 2967 2702 608 3739 721 1552 2767 530 3114 665 1940 48 4198",
"output": "5"
},
{
"input": "33\n1097 1132 1091 1104 1049 1038 1023 1080 1104 1029 1035 1061 1049 1060 1088 1106 1105 1087 1063 1076 1054 1103 1047 1041 1028 1120 1126 1063 1117 1110 1044 1093 1101",
"output": "5"
},
{
"input": "34\n821 5536 2491 6074 7216 9885 764 1603 778 8736 8987 771 617 1587 8943 7922 439 7367 4115 8886 7878 6899 8811 5752 3184 3401 9760 9400 8995 4681 1323 6637 6554 6498",
"output": "7"
},
{
"input": "68\n6764 6877 6950 6768 6839 6755 6726 6778 6699 6805 6777 6985 6821 6801 6791 6805 6940 6761 6677 6999 6911 6699 6959 6933 6903 6843 6972 6717 6997 6756 6789 6668 6735 6852 6735 6880 6723 6834 6810 6694 6780 6679 6698 6857 6826 6896 6979 6968 6957 6988 6960 6700 6919 6892 6984 6685 6813 6678 6715 6857 6976 6902 6780 6686 6777 6686 6842 6679",
"output": "9"
},
{
"input": "60\n9000 9014 9034 9081 9131 9162 9174 9199 9202 9220 9221 9223 9229 9235 9251 9260 9268 9269 9270 9298 9307 9309 9313 9323 9386 9399 9407 9495 9497 9529 9531 9544 9614 9615 9627 9627 9643 9654 9656 9657 9685 9699 9701 9736 9745 9758 9799 9827 9843 9845 9854 9854 9885 9891 9896 9913 9942 9963 9986 9992",
"output": "57"
},
{
"input": "100\n7 61 12 52 41 16 34 99 30 44 48 89 31 54 21 1 48 52 61 15 35 87 21 76 64 92 44 81 16 93 84 92 32 15 68 76 53 39 26 4 11 26 7 4 99 99 61 65 55 85 65 67 47 39 2 74 63 49 98 87 5 94 22 30 25 42 31 84 49 23 89 60 16 26 92 27 9 57 75 61 94 35 83 47 99 100 63 24 91 88 79 10 15 45 22 64 3 11 89 83",
"output": "4"
},
{
"input": "100\n9999 9999 9999 9998 9998 9998 9997 9996 9996 9995 9993 9993 9991 9990 9989 9986 9984 9984 9983 9981 9981 9980 9980 9980 9979 9977 9977 9977 9977 9977 9976 9976 9975 9975 9973 9972 9972 9972 9972 9971 9969 9969 9969 9968 9967 9965 9962 9962 9962 9962 9960 9958 9958 9955 9954 9954 9954 9954 9950 9949 9949 9947 9943 9943 9940 9940 9939 9937 9937 9937 9936 9935 9932 9932 9931 9929 9927 9927 9926 9923 9919 9919 9918 9918 9918 9917 9917 9916 9916 9915 9914 9911 9910 9907 9907 9906 9905 9905 9904 9901",
"output": "57"
},
{
"input": "100\n158 159 157 156 155 154 160 153 161 152 162 163 151 164 150 165 149 148 147 166 146 167 145 144 143 142 168 141 169 170 140 139 138 137 171 136 135 134 172 173 174 133 175 132 177 178 131 179 180 130 129 181 128 127 182 126 183 125 124 184 123 122 121 120 119 118 117 185 186 187 188 189 116 190 115 191 192 193 114 113 112 111 110 109 108 194 195 107 106 105 196 197 104 198 199 103 102 200 101 100",
"output": "99"
},
{
"input": "2\n0 10000",
"output": "1"
},
{
"input": "2\n5 5",
"output": "0"
},
{
"input": "2\n1 1",
"output": "0"
},
{
"input": "2\n10 10",
"output": "0"
},
{
"input": "1\n0",
"output": "0"
}
] | 1,684,257,177
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 37
| 92
| 0
|
n = int(input())
m = list(map(int, input().split()))
hitler = 0
maxx = m[0]
minn = m[0]
for i in range(1, n):
if m[i] > maxx:
maxx = m[i]
hitler += 1
elif m[i] < minn:
minn = m[i]
hitler += 1
print(hitler)
|
Title: I_love_\%username\%
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya adores sport programming. He can't write programs but he loves to watch the contests' progress. Vasya even has a favorite coder and Vasya pays special attention to him.
One day Vasya decided to collect the results of all contests where his favorite coder participated and track the progress of his coolness. For each contest where this coder participated, he wrote out a single non-negative number β the number of points his favorite coder earned in the contest. Vasya wrote out the points for the contest in the order, in which the contests run (naturally, no two contests ran simultaneously).
Vasya considers a coder's performance in a contest amazing in two situations: he can break either his best or his worst performance record. First, it is amazing if during the contest the coder earns strictly more points that he earned on each past contest. Second, it is amazing if during the contest the coder earns strictly less points that he earned on each past contest. A coder's first contest isn't considered amazing. Now he wants to count the number of amazing performances the coder had throughout his whole history of participating in contests. But the list of earned points turned out long and Vasya can't code... That's why he asks you to help him.
Input Specification:
The first line contains the single integer *n* (1<=β€<=*n*<=β€<=1000) β the number of contests where the coder participated.
The next line contains *n* space-separated non-negative integer numbers β they are the points which the coder has earned. The points are given in the chronological order. All points do not exceed 10000.
Output Specification:
Print the single number β the number of amazing performances the coder has had during his whole history of participating in the contests.
Demo Input:
['5\n100 50 200 150 200\n', '10\n4664 6496 5814 7010 5762 5736 6944 4850 3698 7242\n']
Demo Output:
['2\n', '4\n']
Note:
In the first sample the performances number 2 and 3 are amazing.
In the second sample the performances number 2, 4, 9 and 10 are amazing.
|
```python
n = int(input())
m = list(map(int, input().split()))
hitler = 0
maxx = m[0]
minn = m[0]
for i in range(1, n):
if m[i] > maxx:
maxx = m[i]
hitler += 1
elif m[i] < minn:
minn = m[i]
hitler += 1
print(hitler)
```
| 3
|
|
581
|
A
|
Vasya the Hipster
|
PROGRAMMING
| 800
|
[
"implementation",
"math"
] | null | null |
One day Vasya the Hipster decided to count how many socks he had. It turned out that he had *a* red socks and *b* blue socks.
According to the latest fashion, hipsters should wear the socks of different colors: a red one on the left foot, a blue one on the right foot.
Every day Vasya puts on new socks in the morning and throws them away before going to bed as he doesn't want to wash them.
Vasya wonders, what is the maximum number of days when he can dress fashionable and wear different socks, and after that, for how many days he can then wear the same socks until he either runs out of socks or cannot make a single pair from the socks he's got.
Can you help him?
|
The single line of the input contains two positive integers *a* and *b* (1<=β€<=*a*,<=*b*<=β€<=100) β the number of red and blue socks that Vasya's got.
|
Print two space-separated integers β the maximum number of days when Vasya can wear different socks and the number of days when he can wear the same socks until he either runs out of socks or cannot make a single pair from the socks he's got.
Keep in mind that at the end of the day Vasya throws away the socks that he's been wearing on that day.
|
[
"3 1\n",
"2 3\n",
"7 3\n"
] |
[
"1 1\n",
"2 0\n",
"3 2\n"
] |
In the first sample Vasya can first put on one pair of different socks, after that he has two red socks left to wear on the second day.
| 500
|
[
{
"input": "3 1",
"output": "1 1"
},
{
"input": "2 3",
"output": "2 0"
},
{
"input": "7 3",
"output": "3 2"
},
{
"input": "100 100",
"output": "100 0"
},
{
"input": "4 10",
"output": "4 3"
},
{
"input": "6 10",
"output": "6 2"
},
{
"input": "6 11",
"output": "6 2"
},
{
"input": "10 40",
"output": "10 15"
},
{
"input": "11 56",
"output": "11 22"
},
{
"input": "34 30",
"output": "30 2"
},
{
"input": "33 33",
"output": "33 0"
},
{
"input": "100 45",
"output": "45 27"
},
{
"input": "100 23",
"output": "23 38"
},
{
"input": "45 12",
"output": "12 16"
},
{
"input": "1 1",
"output": "1 0"
},
{
"input": "1 100",
"output": "1 49"
},
{
"input": "100 1",
"output": "1 49"
},
{
"input": "68 59",
"output": "59 4"
},
{
"input": "45 99",
"output": "45 27"
},
{
"input": "99 100",
"output": "99 0"
},
{
"input": "100 98",
"output": "98 1"
},
{
"input": "59 12",
"output": "12 23"
},
{
"input": "86 4",
"output": "4 41"
},
{
"input": "68 21",
"output": "21 23"
},
{
"input": "100 11",
"output": "11 44"
},
{
"input": "100 10",
"output": "10 45"
},
{
"input": "15 45",
"output": "15 15"
},
{
"input": "11 32",
"output": "11 10"
},
{
"input": "34 96",
"output": "34 31"
},
{
"input": "89 89",
"output": "89 0"
}
] | 1,687,599,843
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 30
| 31
| 0
|
a, b = input().split()
a, b = int(a), int(b)
print(f"{min(a, b)} {abs(a-b)//2}")
|
Title: Vasya the Hipster
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Vasya the Hipster decided to count how many socks he had. It turned out that he had *a* red socks and *b* blue socks.
According to the latest fashion, hipsters should wear the socks of different colors: a red one on the left foot, a blue one on the right foot.
Every day Vasya puts on new socks in the morning and throws them away before going to bed as he doesn't want to wash them.
Vasya wonders, what is the maximum number of days when he can dress fashionable and wear different socks, and after that, for how many days he can then wear the same socks until he either runs out of socks or cannot make a single pair from the socks he's got.
Can you help him?
Input Specification:
The single line of the input contains two positive integers *a* and *b* (1<=β€<=*a*,<=*b*<=β€<=100) β the number of red and blue socks that Vasya's got.
Output Specification:
Print two space-separated integers β the maximum number of days when Vasya can wear different socks and the number of days when he can wear the same socks until he either runs out of socks or cannot make a single pair from the socks he's got.
Keep in mind that at the end of the day Vasya throws away the socks that he's been wearing on that day.
Demo Input:
['3 1\n', '2 3\n', '7 3\n']
Demo Output:
['1 1\n', '2 0\n', '3 2\n']
Note:
In the first sample Vasya can first put on one pair of different socks, after that he has two red socks left to wear on the second day.
|
```python
a, b = input().split()
a, b = int(a), int(b)
print(f"{min(a, b)} {abs(a-b)//2}")
```
| 3
|
|
202
|
A
|
LLPS
|
PROGRAMMING
| 800
|
[
"binary search",
"bitmasks",
"brute force",
"greedy",
"implementation",
"strings"
] | null | null |
This problem's actual name, "Lexicographically Largest Palindromic Subsequence" is too long to fit into the page headline.
You are given string *s* consisting of lowercase English letters only. Find its lexicographically largest palindromic subsequence.
We'll call a non-empty string *s*[*p*1*p*2... *p**k*] = *s**p*1*s**p*2... *s**p**k* (1 <=β€<= *p*1<=<<=*p*2<=<<=...<=<<=*p**k* <=β€<= |*s*|) a subsequence of string *s* = *s*1*s*2... *s*|*s*|, where |*s*| is the length of string *s*. For example, strings "abcb", "b" and "abacaba" are subsequences of string "abacaba".
String *x* = *x*1*x*2... *x*|*x*| is lexicographically larger than string *y* = *y*1*y*2... *y*|*y*| if either |*x*| > |*y*| and *x*1<==<=*y*1, *x*2<==<=*y*2, ...,<=*x*|*y*|<==<=*y*|*y*|, or there exists such number *r* (*r*<=<<=|*x*|, *r*<=<<=|*y*|) that *x*1<==<=*y*1, *x*2<==<=*y*2, ..., *x**r*<==<=*y**r* and *x**r*<=<=+<=<=1<=><=*y**r*<=<=+<=<=1. Characters in the strings are compared according to their ASCII codes. For example, string "ranger" is lexicographically larger than string "racecar" and string "poster" is lexicographically larger than string "post".
String *s* = *s*1*s*2... *s*|*s*| is a palindrome if it matches string *rev*(*s*) = *s*|*s*|*s*|*s*|<=-<=1... *s*1. In other words, a string is a palindrome if it reads the same way from left to right and from right to left. For example, palindromic strings are "racecar", "refer" and "z".
|
The only input line contains a non-empty string *s* consisting of lowercase English letters only. Its length does not exceed 10.
|
Print the lexicographically largest palindromic subsequence of string *s*.
|
[
"radar\n",
"bowwowwow\n",
"codeforces\n",
"mississipp\n"
] |
[
"rr\n",
"wwwww\n",
"s\n",
"ssss\n"
] |
Among all distinct subsequences of string "radar" the following ones are palindromes: "a", "d", "r", "aa", "rr", "ada", "rar", "rdr", "raar" and "radar". The lexicographically largest of them is "rr".
| 500
|
[
{
"input": "radar",
"output": "rr"
},
{
"input": "bowwowwow",
"output": "wwwww"
},
{
"input": "codeforces",
"output": "s"
},
{
"input": "mississipp",
"output": "ssss"
},
{
"input": "tourist",
"output": "u"
},
{
"input": "romka",
"output": "r"
},
{
"input": "helloworld",
"output": "w"
},
{
"input": "zzzzzzzazz",
"output": "zzzzzzzzz"
},
{
"input": "testcase",
"output": "tt"
},
{
"input": "hahahahaha",
"output": "hhhhh"
},
{
"input": "abbbbbbbbb",
"output": "bbbbbbbbb"
},
{
"input": "zaz",
"output": "zz"
},
{
"input": "aza",
"output": "z"
},
{
"input": "dcbaedcba",
"output": "e"
},
{
"input": "abcdeabcd",
"output": "e"
},
{
"input": "edcbabcde",
"output": "ee"
},
{
"input": "aaaaaaaaab",
"output": "b"
},
{
"input": "testzzzzzz",
"output": "zzzzzz"
},
{
"input": "zzzzzzwait",
"output": "zzzzzz"
},
{
"input": "rrrrrqponm",
"output": "rrrrr"
},
{
"input": "zzyzyy",
"output": "zzz"
},
{
"input": "aababb",
"output": "bbb"
},
{
"input": "zanzibar",
"output": "zz"
},
{
"input": "hhgfedcbaa",
"output": "hh"
},
{
"input": "aabcdefghh",
"output": "hh"
},
{
"input": "aruaru",
"output": "uu"
},
{
"input": "uraura",
"output": "uu"
},
{
"input": "aru",
"output": "u"
},
{
"input": "aburvabur",
"output": "v"
},
{
"input": "ura",
"output": "u"
},
{
"input": "eurottat",
"output": "u"
},
{
"input": "referee",
"output": "rr"
},
{
"input": "joking",
"output": "o"
},
{
"input": "seriously",
"output": "y"
},
{
"input": "sets",
"output": "t"
},
{
"input": "test",
"output": "tt"
},
{
"input": "klmgameklm",
"output": "mmm"
},
{
"input": "dfkjafdkdd",
"output": "kk"
},
{
"input": "zzzzzzzzzz",
"output": "zzzzzzzzzz"
},
{
"input": "aaaaaaaaaa",
"output": "aaaaaaaaaa"
},
{
"input": "buzz",
"output": "zz"
},
{
"input": "b",
"output": "b"
},
{
"input": "y",
"output": "y"
},
{
"input": "yy",
"output": "yy"
},
{
"input": "zz",
"output": "zz"
},
{
"input": "bb",
"output": "bb"
},
{
"input": "aa",
"output": "aa"
},
{
"input": "a",
"output": "a"
},
{
"input": "z",
"output": "z"
},
{
"input": "zz",
"output": "zz"
},
{
"input": "zzazazzzaz",
"output": "zzzzzzz"
},
{
"input": "hellhellhe",
"output": "llll"
},
{
"input": "hellohello",
"output": "oo"
},
{
"input": "refer",
"output": "rr"
}
] | 1,622,905,654
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 54
| 466
| 2,252,800
|
from sys import *
import sys
from math import *
from collections import *
import string
import re
from bisect import *
from functools import reduce
from itertools import permutations, combinations
# import numpy as np
# def arr(): return np.random.randint(1,50,5)
t=stdin.readline
R=range
p=stdout.write
mod = int(1e9)+7
MAX = 9223372036854775808
lower = string.ascii_lowercase
upper = string.ascii_uppercase
numbers = string.digits
def S(): return t().strip()
def I(): return int(t())
def GI(): return map(int, input().strip().split())
def GS(): return map(str, t().strip().split())
def IL(): return list(map(int, t().strip().split()))
def SL(): return list(map(str, t().strip().split()))
def mat(n): return [IL() for i in range(n)]
def sieve(n): return reduce(lambda r, x: r - set(range(x**2, n, x)) if x in r else r,
range(2, int(n**0.5) + 1), set(range(2,n)))
def iSq(x): return (ceil(float(sqrt(x))) == floor(float(sqrt(x))))
def ispow2(x): return ((x!=0) and (x&(x-1))==0)
def d2b(n):return bin(n).replace("0b", "")
def ls(x,y): return x<<y # adding '0' y bits from right and removing last y bits from left i.e floor(x/pow(2,y))
def rs(x,y): return x<<y # adding '0' y bits from left and removing first y bits from right i.e x*pow(2,y)
def powof2(x): return 1<<x
s=sorted(S())[::-1]
prev = s[0]
for i in s:
if prev==i: print(i,end='')
else: break
prev=i
|
Title: LLPS
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
This problem's actual name, "Lexicographically Largest Palindromic Subsequence" is too long to fit into the page headline.
You are given string *s* consisting of lowercase English letters only. Find its lexicographically largest palindromic subsequence.
We'll call a non-empty string *s*[*p*1*p*2... *p**k*] = *s**p*1*s**p*2... *s**p**k* (1 <=β€<= *p*1<=<<=*p*2<=<<=...<=<<=*p**k* <=β€<= |*s*|) a subsequence of string *s* = *s*1*s*2... *s*|*s*|, where |*s*| is the length of string *s*. For example, strings "abcb", "b" and "abacaba" are subsequences of string "abacaba".
String *x* = *x*1*x*2... *x*|*x*| is lexicographically larger than string *y* = *y*1*y*2... *y*|*y*| if either |*x*| > |*y*| and *x*1<==<=*y*1, *x*2<==<=*y*2, ...,<=*x*|*y*|<==<=*y*|*y*|, or there exists such number *r* (*r*<=<<=|*x*|, *r*<=<<=|*y*|) that *x*1<==<=*y*1, *x*2<==<=*y*2, ..., *x**r*<==<=*y**r* and *x**r*<=<=+<=<=1<=><=*y**r*<=<=+<=<=1. Characters in the strings are compared according to their ASCII codes. For example, string "ranger" is lexicographically larger than string "racecar" and string "poster" is lexicographically larger than string "post".
String *s* = *s*1*s*2... *s*|*s*| is a palindrome if it matches string *rev*(*s*) = *s*|*s*|*s*|*s*|<=-<=1... *s*1. In other words, a string is a palindrome if it reads the same way from left to right and from right to left. For example, palindromic strings are "racecar", "refer" and "z".
Input Specification:
The only input line contains a non-empty string *s* consisting of lowercase English letters only. Its length does not exceed 10.
Output Specification:
Print the lexicographically largest palindromic subsequence of string *s*.
Demo Input:
['radar\n', 'bowwowwow\n', 'codeforces\n', 'mississipp\n']
Demo Output:
['rr\n', 'wwwww\n', 's\n', 'ssss\n']
Note:
Among all distinct subsequences of string "radar" the following ones are palindromes: "a", "d", "r", "aa", "rr", "ada", "rar", "rdr", "raar" and "radar". The lexicographically largest of them is "rr".
|
```python
from sys import *
import sys
from math import *
from collections import *
import string
import re
from bisect import *
from functools import reduce
from itertools import permutations, combinations
# import numpy as np
# def arr(): return np.random.randint(1,50,5)
t=stdin.readline
R=range
p=stdout.write
mod = int(1e9)+7
MAX = 9223372036854775808
lower = string.ascii_lowercase
upper = string.ascii_uppercase
numbers = string.digits
def S(): return t().strip()
def I(): return int(t())
def GI(): return map(int, input().strip().split())
def GS(): return map(str, t().strip().split())
def IL(): return list(map(int, t().strip().split()))
def SL(): return list(map(str, t().strip().split()))
def mat(n): return [IL() for i in range(n)]
def sieve(n): return reduce(lambda r, x: r - set(range(x**2, n, x)) if x in r else r,
range(2, int(n**0.5) + 1), set(range(2,n)))
def iSq(x): return (ceil(float(sqrt(x))) == floor(float(sqrt(x))))
def ispow2(x): return ((x!=0) and (x&(x-1))==0)
def d2b(n):return bin(n).replace("0b", "")
def ls(x,y): return x<<y # adding '0' y bits from right and removing last y bits from left i.e floor(x/pow(2,y))
def rs(x,y): return x<<y # adding '0' y bits from left and removing first y bits from right i.e x*pow(2,y)
def powof2(x): return 1<<x
s=sorted(S())[::-1]
prev = s[0]
for i in s:
if prev==i: print(i,end='')
else: break
prev=i
```
| 3
|
|
339
|
A
|
Helpful Maths
|
PROGRAMMING
| 800
|
[
"greedy",
"implementation",
"sortings",
"strings"
] | null | null |
Xenia the beginner mathematician is a third year student at elementary school. She is now learning the addition operation.
The teacher has written down the sum of multiple numbers. Pupils should calculate the sum. To make the calculation easier, the sum only contains numbers 1, 2 and 3. Still, that isn't enough for Xenia. She is only beginning to count, so she can calculate a sum only if the summands follow in non-decreasing order. For example, she can't calculate sum 1+3+2+1 but she can calculate sums 1+1+2 and 3+3.
You've got the sum that was written on the board. Rearrange the summans and print the sum in such a way that Xenia can calculate the sum.
|
The first line contains a non-empty string *s* β the sum Xenia needs to count. String *s* contains no spaces. It only contains digits and characters "+". Besides, string *s* is a correct sum of numbers 1, 2 and 3. String *s* is at most 100 characters long.
|
Print the new sum that Xenia can count.
|
[
"3+2+1\n",
"1+1+3+1+3\n",
"2\n"
] |
[
"1+2+3\n",
"1+1+1+3+3\n",
"2\n"
] |
none
| 500
|
[
{
"input": "3+2+1",
"output": "1+2+3"
},
{
"input": "1+1+3+1+3",
"output": "1+1+1+3+3"
},
{
"input": "2",
"output": "2"
},
{
"input": "2+2+1+1+3",
"output": "1+1+2+2+3"
},
{
"input": "2+1+2+2+2+3+1+3+1+2",
"output": "1+1+1+2+2+2+2+2+3+3"
},
{
"input": "1+2+1+2+2+2+2+1+3+3",
"output": "1+1+1+2+2+2+2+2+3+3"
},
{
"input": "2+3+3+1+2+2+2+1+1+2+1+3+2+2+3+3+2+2+3+3+3+1+1+1+3+3+3+2+1+3+2+3+2+1+1+3+3+3+1+2+2+1+2+2+1+2+1+3+1+1",
"output": "1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+3+3+3+3+3+3+3+3+3+3+3+3+3+3+3+3+3"
},
{
"input": "1",
"output": "1"
},
{
"input": "2+1+2+2+1+3+2+3+1+1+2+1+2+2+3+1+1+3+3+3+2+2+3+2+2+2+1+2+1+2+3+2+2+2+1+3+1+3+3+3+1+2+1+2+2+2+2+3+1+1",
"output": "1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+3+3+3+3+3+3+3+3+3+3+3+3+3"
},
{
"input": "2+2+1+1+1+3+1+1+3+3+2+3+1+3+1+1+3+1+1+2+2+2+2+1+2+1+2+1+1+1+3+1+3+2+3+2+3+3+1+1+1+2+3+2+1+3+1+3+2+2",
"output": "1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+3+3+3+3+3+3+3+3+3+3+3+3+3+3"
},
{
"input": "3+2+3+3+2+2+1+2+1+2+3+1+2+3+2+3+2+1+2+2+1+1+2+2+3+2+1+3+1+1+3+2+2+2+2+3+3+2+2+3+3+1+1+2+3+3+2+3+3+3",
"output": "1+1+1+1+1+1+1+1+1+1+1+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+2+3+3+3+3+3+3+3+3+3+3+3+3+3+3+3+3+3+3"
},
{
"input": "3",
"output": "3"
},
{
"input": "1+1",
"output": "1+1"
},
{
"input": "1+2",
"output": "1+2"
},
{
"input": "1+3",
"output": "1+3"
},
{
"input": "2+1",
"output": "1+2"
},
{
"input": "2+2",
"output": "2+2"
},
{
"input": "2+3",
"output": "2+3"
},
{
"input": "3+1",
"output": "1+3"
},
{
"input": "3+2",
"output": "2+3"
},
{
"input": "3+3",
"output": "3+3"
}
] | 1,695,137,693
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 21
| 92
| 0
|
s = str(input())
s1 = s.split('+')
print('+'.join(sorted(s1)))
|
Title: Helpful Maths
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Xenia the beginner mathematician is a third year student at elementary school. She is now learning the addition operation.
The teacher has written down the sum of multiple numbers. Pupils should calculate the sum. To make the calculation easier, the sum only contains numbers 1, 2 and 3. Still, that isn't enough for Xenia. She is only beginning to count, so she can calculate a sum only if the summands follow in non-decreasing order. For example, she can't calculate sum 1+3+2+1 but she can calculate sums 1+1+2 and 3+3.
You've got the sum that was written on the board. Rearrange the summans and print the sum in such a way that Xenia can calculate the sum.
Input Specification:
The first line contains a non-empty string *s* β the sum Xenia needs to count. String *s* contains no spaces. It only contains digits and characters "+". Besides, string *s* is a correct sum of numbers 1, 2 and 3. String *s* is at most 100 characters long.
Output Specification:
Print the new sum that Xenia can count.
Demo Input:
['3+2+1\n', '1+1+3+1+3\n', '2\n']
Demo Output:
['1+2+3\n', '1+1+1+3+3\n', '2\n']
Note:
none
|
```python
s = str(input())
s1 = s.split('+')
print('+'.join(sorted(s1)))
```
| 3
|
|
231
|
A
|
Team
|
PROGRAMMING
| 800
|
[
"brute force",
"greedy"
] | null | null |
One day three best friends Petya, Vasya and Tonya decided to form a team and take part in programming contests. Participants are usually offered several problems during programming contests. Long before the start the friends decided that they will implement a problem if at least two of them are sure about the solution. Otherwise, the friends won't write the problem's solution.
This contest offers *n* problems to the participants. For each problem we know, which friend is sure about the solution. Help the friends find the number of problems for which they will write a solution.
|
The first input line contains a single integer *n* (1<=β€<=*n*<=β€<=1000) β the number of problems in the contest. Then *n* lines contain three integers each, each integer is either 0 or 1. If the first number in the line equals 1, then Petya is sure about the problem's solution, otherwise he isn't sure. The second number shows Vasya's view on the solution, the third number shows Tonya's view. The numbers on the lines are separated by spaces.
|
Print a single integer β the number of problems the friends will implement on the contest.
|
[
"3\n1 1 0\n1 1 1\n1 0 0\n",
"2\n1 0 0\n0 1 1\n"
] |
[
"2\n",
"1\n"
] |
In the first sample Petya and Vasya are sure that they know how to solve the first problem and all three of them know how to solve the second problem. That means that they will write solutions for these problems. Only Petya is sure about the solution for the third problem, but that isn't enough, so the friends won't take it.
In the second sample the friends will only implement the second problem, as Vasya and Tonya are sure about the solution.
| 500
|
[
{
"input": "3\n1 1 0\n1 1 1\n1 0 0",
"output": "2"
},
{
"input": "2\n1 0 0\n0 1 1",
"output": "1"
},
{
"input": "1\n1 0 0",
"output": "0"
},
{
"input": "2\n1 0 0\n1 1 1",
"output": "1"
},
{
"input": "5\n1 0 0\n0 1 0\n1 1 1\n0 0 1\n0 0 0",
"output": "1"
},
{
"input": "10\n0 1 0\n0 1 0\n1 1 0\n1 0 0\n0 0 1\n0 1 1\n1 1 1\n1 1 0\n0 0 0\n0 0 0",
"output": "4"
},
{
"input": "15\n0 1 0\n1 0 0\n1 1 0\n1 1 1\n0 1 0\n0 0 1\n1 0 1\n1 0 1\n1 0 1\n0 0 0\n1 1 1\n1 1 0\n0 1 1\n1 1 0\n1 1 1",
"output": "10"
},
{
"input": "50\n0 0 0\n0 1 1\n1 1 1\n0 1 0\n1 0 1\n1 1 1\n0 0 1\n1 0 0\n1 1 0\n1 0 1\n0 1 0\n0 0 1\n1 1 0\n0 1 0\n1 1 0\n0 0 0\n1 1 1\n1 0 1\n0 0 1\n1 1 0\n1 1 1\n0 1 1\n1 1 0\n0 0 0\n0 0 0\n1 1 1\n0 0 0\n1 1 1\n0 1 1\n0 0 1\n0 0 0\n0 0 0\n1 1 0\n1 1 0\n1 0 1\n1 0 0\n1 0 1\n1 0 1\n0 1 1\n1 1 0\n1 1 0\n0 1 0\n1 0 1\n0 0 0\n0 0 0\n0 0 0\n0 0 1\n1 1 1\n0 1 1\n1 0 1",
"output": "29"
},
{
"input": "1\n1 1 1",
"output": "1"
},
{
"input": "8\n0 0 0\n0 0 1\n0 0 0\n0 1 1\n1 0 0\n1 0 1\n1 1 0\n1 1 1",
"output": "4"
},
{
"input": "16\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1",
"output": "16"
}
] | 1,697,995,058
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 21
| 92
| 0
|
#!/usr/bin/python3
n = int(input())
votes = []
for i in range(n):
a, b, c = [int(char) for char in input().split(" ")]
if sum([a, b, c]) >= 2:
votes.append(1)
print(len(votes))
|
Title: Team
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day three best friends Petya, Vasya and Tonya decided to form a team and take part in programming contests. Participants are usually offered several problems during programming contests. Long before the start the friends decided that they will implement a problem if at least two of them are sure about the solution. Otherwise, the friends won't write the problem's solution.
This contest offers *n* problems to the participants. For each problem we know, which friend is sure about the solution. Help the friends find the number of problems for which they will write a solution.
Input Specification:
The first input line contains a single integer *n* (1<=β€<=*n*<=β€<=1000) β the number of problems in the contest. Then *n* lines contain three integers each, each integer is either 0 or 1. If the first number in the line equals 1, then Petya is sure about the problem's solution, otherwise he isn't sure. The second number shows Vasya's view on the solution, the third number shows Tonya's view. The numbers on the lines are separated by spaces.
Output Specification:
Print a single integer β the number of problems the friends will implement on the contest.
Demo Input:
['3\n1 1 0\n1 1 1\n1 0 0\n', '2\n1 0 0\n0 1 1\n']
Demo Output:
['2\n', '1\n']
Note:
In the first sample Petya and Vasya are sure that they know how to solve the first problem and all three of them know how to solve the second problem. That means that they will write solutions for these problems. Only Petya is sure about the solution for the third problem, but that isn't enough, so the friends won't take it.
In the second sample the friends will only implement the second problem, as Vasya and Tonya are sure about the solution.
|
```python
#!/usr/bin/python3
n = int(input())
votes = []
for i in range(n):
a, b, c = [int(char) for char in input().split(" ")]
if sum([a, b, c]) >= 2:
votes.append(1)
print(len(votes))
```
| 3
|
|
686
|
A
|
Free Ice Cream
|
PROGRAMMING
| 800
|
[
"constructive algorithms",
"implementation"
] | null | null |
After their adventure with the magic mirror Kay and Gerda have returned home and sometimes give free ice cream to kids in the summer.
At the start of the day they have *x* ice cream packs. Since the ice cream is free, people start standing in the queue before Kay and Gerda's house even in the night. Each person in the queue wants either to take several ice cream packs for himself and his friends or to give several ice cream packs to Kay and Gerda (carriers that bring ice cream have to stand in the same queue).
If a carrier with *d* ice cream packs comes to the house, then Kay and Gerda take all his packs. If a child who wants to take *d* ice cream packs comes to the house, then Kay and Gerda will give him *d* packs if they have enough ice cream, otherwise the child will get no ice cream at all and will leave in distress.
Kay wants to find the amount of ice cream they will have after all people will leave from the queue, and Gerda wants to find the number of distressed kids.
|
The first line contains two space-separated integers *n* and *x* (1<=β€<=*n*<=β€<=1000, 0<=β€<=*x*<=β€<=109).
Each of the next *n* lines contains a character '+' or '-', and an integer *d**i*, separated by a space (1<=β€<=*d**i*<=β€<=109). Record "+ *d**i*" in *i*-th line means that a carrier with *d**i* ice cream packs occupies *i*-th place from the start of the queue, and record "- *d**i*" means that a child who wants to take *d**i* packs stands in *i*-th place.
|
Print two space-separated integersΒ β number of ice cream packs left after all operations, and number of kids that left the house in distress.
|
[
"5 7\n+ 5\n- 10\n- 20\n+ 40\n- 20\n",
"5 17\n- 16\n- 2\n- 98\n+ 100\n- 98\n"
] |
[
"22 1\n",
"3 2\n"
] |
Consider the first sample.
1. Initially Kay and Gerda have 7 packs of ice cream. 1. Carrier brings 5 more, so now they have 12 packs. 1. A kid asks for 10 packs and receives them. There are only 2 packs remaining. 1. Another kid asks for 20 packs. Kay and Gerda do not have them, so the kid goes away distressed. 1. Carrier bring 40 packs, now Kay and Gerda have 42 packs. 1. Kid asks for 20 packs and receives them. There are 22 packs remaining.
| 500
|
[
{
"input": "5 7\n+ 5\n- 10\n- 20\n+ 40\n- 20",
"output": "22 1"
},
{
"input": "5 17\n- 16\n- 2\n- 98\n+ 100\n- 98",
"output": "3 2"
},
{
"input": "6 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000",
"output": "7000000000 0"
},
{
"input": "5 12\n- 12\n+ 7\n- 6\n- 1\n+ 46",
"output": "46 0"
},
{
"input": "11 1000\n- 100\n+ 100\n+ 100\n+ 100\n+ 100\n- 100\n- 100\n- 100\n- 100\n- 100\n- 100",
"output": "700 0"
},
{
"input": "1 0\n- 526403222",
"output": "0 1"
},
{
"input": "1 897986543\n- 371188251",
"output": "526798292 0"
},
{
"input": "1 0\n+ 1",
"output": "1 0"
},
{
"input": "1 0\n- 1",
"output": "0 1"
},
{
"input": "1 10\n+ 10",
"output": "20 0"
},
{
"input": "1 3\n- 5",
"output": "3 1"
},
{
"input": "1 0\n- 5",
"output": "0 1"
},
{
"input": "1 0\n+ 5",
"output": "5 0"
}
] | 1,675,424,104
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 34
| 46
| 0
|
x = list(map(int, input().split()))
s = 0
for i in range(x[0]):
y = list(map(str, input().split()))
y[1] = int(y[1])
if y[0] == '+':
x[1] = x[1] + y[1]
elif y[0] == '-':
if x[1] >= y[1]:
x[1] = x[1] - y[1]
else:
s = s+1
m = [str(x[1]),str(s)]
z = ' '.join(m)
print(z)
|
Title: Free Ice Cream
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
After their adventure with the magic mirror Kay and Gerda have returned home and sometimes give free ice cream to kids in the summer.
At the start of the day they have *x* ice cream packs. Since the ice cream is free, people start standing in the queue before Kay and Gerda's house even in the night. Each person in the queue wants either to take several ice cream packs for himself and his friends or to give several ice cream packs to Kay and Gerda (carriers that bring ice cream have to stand in the same queue).
If a carrier with *d* ice cream packs comes to the house, then Kay and Gerda take all his packs. If a child who wants to take *d* ice cream packs comes to the house, then Kay and Gerda will give him *d* packs if they have enough ice cream, otherwise the child will get no ice cream at all and will leave in distress.
Kay wants to find the amount of ice cream they will have after all people will leave from the queue, and Gerda wants to find the number of distressed kids.
Input Specification:
The first line contains two space-separated integers *n* and *x* (1<=β€<=*n*<=β€<=1000, 0<=β€<=*x*<=β€<=109).
Each of the next *n* lines contains a character '+' or '-', and an integer *d**i*, separated by a space (1<=β€<=*d**i*<=β€<=109). Record "+ *d**i*" in *i*-th line means that a carrier with *d**i* ice cream packs occupies *i*-th place from the start of the queue, and record "- *d**i*" means that a child who wants to take *d**i* packs stands in *i*-th place.
Output Specification:
Print two space-separated integersΒ β number of ice cream packs left after all operations, and number of kids that left the house in distress.
Demo Input:
['5 7\n+ 5\n- 10\n- 20\n+ 40\n- 20\n', '5 17\n- 16\n- 2\n- 98\n+ 100\n- 98\n']
Demo Output:
['22 1\n', '3 2\n']
Note:
Consider the first sample.
1. Initially Kay and Gerda have 7 packs of ice cream. 1. Carrier brings 5 more, so now they have 12 packs. 1. A kid asks for 10 packs and receives them. There are only 2 packs remaining. 1. Another kid asks for 20 packs. Kay and Gerda do not have them, so the kid goes away distressed. 1. Carrier bring 40 packs, now Kay and Gerda have 42 packs. 1. Kid asks for 20 packs and receives them. There are 22 packs remaining.
|
```python
x = list(map(int, input().split()))
s = 0
for i in range(x[0]):
y = list(map(str, input().split()))
y[1] = int(y[1])
if y[0] == '+':
x[1] = x[1] + y[1]
elif y[0] == '-':
if x[1] >= y[1]:
x[1] = x[1] - y[1]
else:
s = s+1
m = [str(x[1]),str(s)]
z = ' '.join(m)
print(z)
```
| 3
|
|
548
|
A
|
Mike and Fax
|
PROGRAMMING
| 1,100
|
[
"brute force",
"implementation",
"strings"
] | null | null |
While Mike was walking in the subway, all the stuff in his back-bag dropped on the ground. There were several fax messages among them. He concatenated these strings in some order and now he has string *s*.
He is not sure if this is his own back-bag or someone else's. He remembered that there were exactly *k* messages in his own bag, each was a palindrome string and all those strings had the same length.
He asked you to help him and tell him if he has worn his own back-bag. Check if the given string *s* is a concatenation of *k* palindromes of the same length.
|
The first line of input contains string *s* containing lowercase English letters (1<=β€<=|*s*|<=β€<=1000).
The second line contains integer *k* (1<=β€<=*k*<=β€<=1000).
|
Print "YES"(without quotes) if he has worn his own back-bag or "NO"(without quotes) otherwise.
|
[
"saba\n2\n",
"saddastavvat\n2\n"
] |
[
"NO\n",
"YES\n"
] |
Palindrome is a string reading the same forward and backward.
In the second sample, the faxes in his back-bag can be "saddas" and "tavvat".
| 500
|
[
{
"input": "saba\n2",
"output": "NO"
},
{
"input": "saddastavvat\n2",
"output": "YES"
},
{
"input": "aaaaaaaaaa\n3",
"output": "NO"
},
{
"input": "aaaaaa\n3",
"output": "YES"
},
{
"input": "abaacca\n2",
"output": "NO"
},
{
"input": "a\n1",
"output": "YES"
},
{
"input": "princeofpersia\n1",
"output": "NO"
},
{
"input": "xhwbdoryfiaxglripavycmxmcejbcpzidrqsqvikfzjyfnmedxrvlnusavyhillaxrblkynwdrlhthtqzjktzkullgrqsolqssocpfwcaizhovajlhmeibhiuwtxpljkyyiwykzpmazkkzampzkywiyykjlpxtwuihbiemhljavohziacwfpcossqlosqrgllukztkjzqththlrdwnyklbrxallihyvasunlvrxdemnfyjzfkivqsqrdizpcbjecmxmcyvapirlgxaifyrodbwhx\n1",
"output": "YES"
},
{
"input": "yfhqnbzaqeqmcvtsbcdn\n456",
"output": "NO"
},
{
"input": "lgsdfiforlqrohhjyzrigewkigiiffvbyrapzmjvtkklndeyuqpuukajgtguhlarjdqlxksyekbjgrmhuyiqdlzjqqzlxufffpelyptodwhvkfbalxbufrlcsjgxmfxeqsszqghcustqrqjljattgvzynyvfbjgbuynbcguqtyfowgtcbbaywvcrgzrulqpghwoflutswu\n584",
"output": "NO"
},
{
"input": "awlrhmxxivqbntvtapwkdkunamcqoerfncfmookhdnuxtttlxmejojpwbdyxirdsjippzjhdrpjepremruczbedxrjpodlyyldopjrxdebzcurmerpejprdhjzppijsdrixydbwpjojemxltttxundhkoomfcnfreoqcmanukdkwpatvtnbqvixxmhrlwa\n1",
"output": "YES"
},
{
"input": "kafzpsglcpzludxojtdhzynpbekzssvhzizfrboxbhqvojiqtjitrackqccxgenwwnegxccqkcartijtqijovqhbxobrfzizhvsszkebpnyzhdtjoxdulzpclgspzfakvcbbjejeubvrrzlvjjgrcprntbyuakoxowoybbxgdugjffgbtfwrfiobifrshyaqqayhsrfiboifrwftbgffjgudgxbbyowoxokauybtnrpcrgjjvlzrrvbuejejbbcv\n2",
"output": "YES"
},
{
"input": "zieqwmmbrtoxysvavwdemmdeatfrolsqvvlgphhhmojjfxfurtuiqdiilhlcwwqedlhblrzmvuoaczcwrqzyymiggpvbpkycibsvkhytrzhguksxyykkkvfljbbnjblylftmqxkojithwsegzsaexlpuicexbdzpwesrkzbqltxhifwqcehzsjgsqbwkujvjbjpqxdpmlimsusumizizpyigmkxwuberthdghnepyrxzvvidxeafwylegschhtywvqsxuqmsddhkzgkdiekodqpnftdyhnpicsnbhfxemxllvaurkmjvtrmqkulerxtaolmokiqqvqgechkqxmendpmgxwiaffcajmqjmvrwryzxujmiasuqtosuisiclnv\n8",
"output": "NO"
},
{
"input": "syghzncbi\n829",
"output": "NO"
},
{
"input": "ljpdpstntznciejqqtpysskztdfawuncqzwwfefrfsihyrdopwawowshquqnjhesxszuywezpebpzhtopgngrnqgwnoqhyrykojguybvdbjpfpmvkxscocywzsxcivysfrrzsonayztzzuybrkiombhqcfkszyscykzistiobrpavezedgobowjszfadcccmxyqehmkgywiwxffibzetb\n137",
"output": "NO"
},
{
"input": "eytuqriplfczwsqlsnjetfpzehzvzayickkbnfqddaisfpasvigwtnvbybwultsgrtjbaebktvubwofysgidpufzteuhuaaqkhmhguockoczlrmlrrzouvqtwbcchxxiydbohnvrmtqjzhkfmvdulojhdvgwudvidpausvfujkjprxsobliuauxleqvsmz\n253",
"output": "NO"
},
{
"input": "xkaqgwabuilhuqwhnrdtyattmqcjfbiqodjlwzgcyvghqncklbhnlmagvjvwysrfryrlmclninogumjfmyenkmydlmifxpkvlaapgnfarejaowftxxztshsesjtsgommaeslrhronruqdurvjesydrzmxirmxumrcqezznqltngsgdcthivdnjnshjfujtiqsltpttgbljfcbqsfwbzokciqlavrthgaqbzikpwwsebzwddlvdwrmztwmhcxdinwlbklwmteeybbdbzevfbsrtldapulwgusuvnreiflkytonzmervyrlbqhzapgxepwauaiwygpxarfeyqhimzlxntjuaaigeisgrvwgbhqemqetzyallzaoqprhzpjibkutgwrodruqu\n857",
"output": "NO"
},
{
"input": "rbehjxpblnzfgeebpkvzznwtzszghjuuxovreapmwehqyjymrkmksffbdpbdyegulabsmjiykeeqtuvqqyxlitpxjdpwmqtlmudqsksgwqekvwfjdsggzajcpsyserkctpbajgzdbiqaekfaepnecezdzamqszpwfvhlannszgaiewvcdnnvzhblmuzjtqeyjcqjqoxatavavokyxokuxwuqueskktxnxraihnqovrfykpzsyjmrhqsvbobzsnfqwvdwatvjxaubtiwhgvrqwjhgfnauqegqmcwnaruinplmtmxhvohhqgvcatsdqiwkvslhqzqevdgdhpijjevuyuitnixdjhwlmjsstmugownjdhvxueg\n677",
"output": "NO"
},
{
"input": "a\n3",
"output": "NO"
},
{
"input": "aa\n2",
"output": "YES"
},
{
"input": "abcaabca\n2",
"output": "NO"
},
{
"input": "aaaaaaaaab\n5",
"output": "NO"
},
{
"input": "aabbb\n3",
"output": "NO"
},
{
"input": "abcde\n5",
"output": "YES"
},
{
"input": "aabc\n2",
"output": "NO"
},
{
"input": "aabcaa\n3",
"output": "NO"
}
] | 1,600,822,233
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 56
| 140
| 0
|
n = input()
a = int(input())
if len(n)%a!=0:
print('NO')
else:
g = len(n)//a
p=0
if g%2==0:
for k in range(0,len(n),g):
u= n[k:k+g]
if u[:g//2]!=u[g//2:][::-1]:
print('NO')
p+=1
break
else:
for k in range(0,len(n),g):
u= n[k:k+g]
if u[:g//2]!=u[g//2+1:][::-1]:
print('NO')
p+=1
break
if p==0:
print('YES')
|
Title: Mike and Fax
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
While Mike was walking in the subway, all the stuff in his back-bag dropped on the ground. There were several fax messages among them. He concatenated these strings in some order and now he has string *s*.
He is not sure if this is his own back-bag or someone else's. He remembered that there were exactly *k* messages in his own bag, each was a palindrome string and all those strings had the same length.
He asked you to help him and tell him if he has worn his own back-bag. Check if the given string *s* is a concatenation of *k* palindromes of the same length.
Input Specification:
The first line of input contains string *s* containing lowercase English letters (1<=β€<=|*s*|<=β€<=1000).
The second line contains integer *k* (1<=β€<=*k*<=β€<=1000).
Output Specification:
Print "YES"(without quotes) if he has worn his own back-bag or "NO"(without quotes) otherwise.
Demo Input:
['saba\n2\n', 'saddastavvat\n2\n']
Demo Output:
['NO\n', 'YES\n']
Note:
Palindrome is a string reading the same forward and backward.
In the second sample, the faxes in his back-bag can be "saddas" and "tavvat".
|
```python
n = input()
a = int(input())
if len(n)%a!=0:
print('NO')
else:
g = len(n)//a
p=0
if g%2==0:
for k in range(0,len(n),g):
u= n[k:k+g]
if u[:g//2]!=u[g//2:][::-1]:
print('NO')
p+=1
break
else:
for k in range(0,len(n),g):
u= n[k:k+g]
if u[:g//2]!=u[g//2+1:][::-1]:
print('NO')
p+=1
break
if p==0:
print('YES')
```
| 3
|
|
259
|
A
|
Little Elephant and Chess
|
PROGRAMMING
| 1,000
|
[
"brute force",
"strings"
] | null | null |
The Little Elephant loves chess very much.
One day the Little Elephant and his friend decided to play chess. They've got the chess pieces but the board is a problem. They've got an 8<=Γ<=8 checkered board, each square is painted either black or white. The Little Elephant and his friend know that a proper chessboard doesn't have any side-adjacent cells with the same color and the upper left cell is white. To play chess, they want to make the board they have a proper chessboard. For that the friends can choose any row of the board and cyclically shift the cells of the chosen row, that is, put the last (rightmost) square on the first place in the row and shift the others one position to the right. You can run the described operation multiple times (or not run it at all).
For example, if the first line of the board looks like that "BBBBBBWW" (the white cells of the line are marked with character "W", the black cells are marked with character "B"), then after one cyclic shift it will look like that "WBBBBBBW".
Help the Little Elephant and his friend to find out whether they can use any number of the described operations to turn the board they have into a proper chessboard.
|
The input consists of exactly eight lines. Each line contains exactly eight characters "W" or "B" without any spaces: the *j*-th character in the *i*-th line stands for the color of the *j*-th cell of the *i*-th row of the elephants' board. Character "W" stands for the white color, character "B" stands for the black color.
Consider the rows of the board numbered from 1 to 8 from top to bottom, and the columns β from 1 to 8 from left to right. The given board can initially be a proper chessboard.
|
In a single line print "YES" (without the quotes), if we can make the board a proper chessboard and "NO" (without the quotes) otherwise.
|
[
"WBWBWBWB\nBWBWBWBW\nBWBWBWBW\nBWBWBWBW\nWBWBWBWB\nWBWBWBWB\nBWBWBWBW\nWBWBWBWB\n",
"WBWBWBWB\nWBWBWBWB\nBBWBWWWB\nBWBWBWBW\nBWBWBWBW\nBWBWBWWW\nBWBWBWBW\nBWBWBWBW\n"
] |
[
"YES\n",
"NO\n"
] |
In the first sample you should shift the following lines one position to the right: the 3-rd, the 6-th, the 7-th and the 8-th.
In the second sample there is no way you can achieve the goal.
| 500
|
[
{
"input": "WBWBWBWB\nBWBWBWBW\nBWBWBWBW\nBWBWBWBW\nWBWBWBWB\nWBWBWBWB\nBWBWBWBW\nWBWBWBWB",
"output": "YES"
},
{
"input": "WBWBWBWB\nWBWBWBWB\nBBWBWWWB\nBWBWBWBW\nBWBWBWBW\nBWBWBWWW\nBWBWBWBW\nBWBWBWBW",
"output": "NO"
},
{
"input": "BWBWBWBW\nWBWBWBWB\nBWBWBWBW\nBWBWBWBW\nWBWBWBWB\nWBWBWBWB\nWBWBWBWB\nWBWBWBWB",
"output": "YES"
},
{
"input": "BWBWBWBW\nWBWBWBWB\nBWBWBWBW\nWBWBWBWB\nBWBWBWBW\nWBWBWBWB\nWBWBWBWB\nWBWBWBWB",
"output": "YES"
},
{
"input": "WBWBWBWB\nBWBWBWBW\nBWBWBWBW\nWBWBWBWB\nBWBWBWBW\nBWBWBWBW\nBWBWBWBW\nBWBWBWBW",
"output": "YES"
},
{
"input": "WBWBWBWB\nWBWBWBWB\nBWBWBWBW\nWBWBWBWB\nWBWBWBWB\nWBWBWBWB\nWBWBWBWB\nBWWWBWBW",
"output": "NO"
},
{
"input": "BBBBBWWW\nWBBWBWWB\nWWWWWBWW\nBWBWWBWW\nBBBWWBWW\nBBBBBWBW\nWBBBWBWB\nWBWBWWWB",
"output": "NO"
},
{
"input": "BWBWBWBW\nBWBWBWBW\nBWWWWWBB\nBBWBWBWB\nWBWBWBWB\nWWBWWBWW\nBWBWBWBW\nWBWWBBBB",
"output": "NO"
},
{
"input": "WBWBWBWB\nWBWBWBWB\nWBWBWBWB\nBWBWBWBW\nWBWBWBWB\nBWBWBWBW\nWBWBWBWB\nWBWWBWBB",
"output": "NO"
},
{
"input": "WBWBWBWB\nBWBWBWBW\nWBWBWBWB\nBWBWBWBW\nWBWBWBWB\nWBWBWBWB\nBWBWBWBW\nBWBWBWBW",
"output": "YES"
},
{
"input": "WBWBWBWB\nWBWBWBWB\nBWBWBWBW\nWBWBWBWB\nBWBWBWBW\nWBWBWBWB\nBWBWBWBW\nBWBWBWBW",
"output": "YES"
},
{
"input": "WWWWBWWB\nBWBWBWBW\nBWBWBWBW\nWWBWBBBB\nBBWWBBBB\nBBBWWBBW\nBWWWWWWB\nBWWBBBWW",
"output": "NO"
},
{
"input": "WBBWWBWB\nBBWBWBWB\nBWBWBWBW\nBWBWBWBW\nWBWBWBBW\nWBWBBBBW\nBWWWWBWB\nBBBBBBBW",
"output": "NO"
},
{
"input": "BWBWBWBW\nBWBWBWBW\nBBWWWBBB\nWBBBBBWW\nWBBBBWBB\nWBWBWBWB\nWBWWBWWB\nWBBWBBWW",
"output": "NO"
},
{
"input": "WBBBBBWB\nBWBWBWBW\nBWBWBWBW\nWBWBWBWB\nWBWBWBWB\nBBBBBWBB\nWBBWWBWB\nBWBWBWBW",
"output": "NO"
},
{
"input": "BWBWBWBW\nBWBWBWBW\nBWBWBWBW\nWBWBWBWB\nWBWBWBWB\nBWBWBWBW\nBWBWBWBW\nWBBWWBWB",
"output": "NO"
},
{
"input": "BWBWBWBW\nWBWBWBWB\nBWBWBWBW\nBWWWBWBW\nWBWBWBWB\nWBWBWBWB\nBWBWBWBW\nWBWBWBBW",
"output": "NO"
},
{
"input": "WBWBWBWB\nWBWBWBWB\nBWBWBWBW\nBWBWBWBW\nBWBWBWBW\nBWBWBWBW\nWBWBWBWB\nBWBWBWBW",
"output": "YES"
},
{
"input": "BWBWBWBW\nWBWBWBWB\nBWBWBWBW\nBWBWBWBW\nBWBWBWBW\nBWBWBWBW\nWBWBWBWB\nBWBWBWBW",
"output": "YES"
},
{
"input": "BWBWBWBW\nBWBWBWBW\nWBWBWBWB\nBWBWBWBW\nWBWBWBWB\nBWBWBWBW\nWBWBWBWB\nBWBWBWBW",
"output": "YES"
},
{
"input": "WWBBWWBB\nBWWBBWWB\nBWBWBWBW\nWWBBWWWB\nWBWWWWBB\nWBWWBBWB\nBWBBWBWW\nBWBWWWWW",
"output": "NO"
},
{
"input": "WBWBWBWB\nWBWBWBWB\nWWBBWBBB\nWBWBWBWB\nWWWWBWWB\nWBBBBWWW\nBWBWWWBW\nWWWBWBBB",
"output": "NO"
},
{
"input": "WBWBWBWB\nBWWBWWWW\nWBWBWBWB\nBWBWBWBW\nWBWBWBWB\nWWBBBBBW\nWWWBWWBW\nWWBBBBWW",
"output": "NO"
},
{
"input": "BWBWBWBW\nBWBBBWWB\nWBWBWBWB\nBWBWBWBW\nBWBWBWBW\nBWBWBWBW\nWBWBWBWB\nBWBWBWBW",
"output": "NO"
},
{
"input": "BWBWBWBW\nBWBWBWBW\nWBWBWBWB\nWBWBWBWB\nWBWBWBWB\nBWBWBWBW\nWBWBWBWB\nBWBWBWBW",
"output": "YES"
},
{
"input": "BBBBBBBB\nBBBBBBBB\nBBBBBBBB\nBBBBBBBB\nWWWWWWWW\nWWWWWWWW\nWWWWWWWW\nWWWWWWWW",
"output": "NO"
},
{
"input": "BBBBBBBB\nBBBBBBBB\nBBBBBBBB\nBBBBBBBB\nBBBBBBBB\nBBBBBBBB\nBBBBBBBB\nBBBBBBBB",
"output": "NO"
},
{
"input": "BWBWBWBB\nBWBWBWBB\nBWBWBWBB\nBWBWBWBB\nBWBWBWBB\nBWBWBWBB\nBWBWBWBB\nBWBWBWBB",
"output": "NO"
},
{
"input": "WWBWWBWB\nBWBWBWBW\nWBWBWBWB\nBWBWBWBW\nWBWBWBWB\nBWBWBWBW\nWBWBWBWB\nBWBWBWBW",
"output": "NO"
},
{
"input": "WWWWWWWW\nBBBBBBBB\nWWWWWWWW\nBBBBBBBB\nWWWWWWWW\nBBBBBBBB\nWWWWWWWW\nBBBBBBBB",
"output": "NO"
},
{
"input": "BBBBBBBB\nBWBWBWBW\nBWBWBWBW\nBWBWBWBW\nWBWBWBWB\nWBWBWBWB\nBWBWBWBW\nWBWBWBWB",
"output": "NO"
},
{
"input": "BBBBBBBW\nBBBBBBBW\nBBBBBBBW\nBBBBBBBW\nBBBBBBBW\nBBBBBBBW\nBBBBBBBW\nBBBBBBBW",
"output": "NO"
},
{
"input": "BBBWWWWW\nWWWBBBBB\nBBBWWWWW\nWWWBBBBB\nBWBWBWBW\nWBWBWBWB\nBWBWBWBW\nWBWBWBWB",
"output": "NO"
}
] | 1,664,383,513
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 33
| 92
| 0
|
ans = 'YES'
for _ in range(8):
line = input()
for i in range(7):
if line[i] == line[i+1]:
ans = 'NO'
break
print(ans)
|
Title: Little Elephant and Chess
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Little Elephant loves chess very much.
One day the Little Elephant and his friend decided to play chess. They've got the chess pieces but the board is a problem. They've got an 8<=Γ<=8 checkered board, each square is painted either black or white. The Little Elephant and his friend know that a proper chessboard doesn't have any side-adjacent cells with the same color and the upper left cell is white. To play chess, they want to make the board they have a proper chessboard. For that the friends can choose any row of the board and cyclically shift the cells of the chosen row, that is, put the last (rightmost) square on the first place in the row and shift the others one position to the right. You can run the described operation multiple times (or not run it at all).
For example, if the first line of the board looks like that "BBBBBBWW" (the white cells of the line are marked with character "W", the black cells are marked with character "B"), then after one cyclic shift it will look like that "WBBBBBBW".
Help the Little Elephant and his friend to find out whether they can use any number of the described operations to turn the board they have into a proper chessboard.
Input Specification:
The input consists of exactly eight lines. Each line contains exactly eight characters "W" or "B" without any spaces: the *j*-th character in the *i*-th line stands for the color of the *j*-th cell of the *i*-th row of the elephants' board. Character "W" stands for the white color, character "B" stands for the black color.
Consider the rows of the board numbered from 1 to 8 from top to bottom, and the columns β from 1 to 8 from left to right. The given board can initially be a proper chessboard.
Output Specification:
In a single line print "YES" (without the quotes), if we can make the board a proper chessboard and "NO" (without the quotes) otherwise.
Demo Input:
['WBWBWBWB\nBWBWBWBW\nBWBWBWBW\nBWBWBWBW\nWBWBWBWB\nWBWBWBWB\nBWBWBWBW\nWBWBWBWB\n', 'WBWBWBWB\nWBWBWBWB\nBBWBWWWB\nBWBWBWBW\nBWBWBWBW\nBWBWBWWW\nBWBWBWBW\nBWBWBWBW\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first sample you should shift the following lines one position to the right: the 3-rd, the 6-th, the 7-th and the 8-th.
In the second sample there is no way you can achieve the goal.
|
```python
ans = 'YES'
for _ in range(8):
line = input()
for i in range(7):
if line[i] == line[i+1]:
ans = 'NO'
break
print(ans)
```
| 3
|
|
334
|
A
|
Candy Bags
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
Gerald has *n* younger brothers and their number happens to be even. One day he bought *n*2 candy bags. One bag has one candy, one bag has two candies, one bag has three candies and so on. In fact, for each integer *k* from 1 to *n*2 he has exactly one bag with *k* candies.
Help him give *n* bags of candies to each brother so that all brothers got the same number of candies.
|
The single line contains a single integer *n* (*n* is even, 2<=β€<=*n*<=β€<=100) β the number of Gerald's brothers.
|
Let's assume that Gerald indexes his brothers with numbers from 1 to *n*. You need to print *n* lines, on the *i*-th line print *n* integers β the numbers of candies in the bags for the *i*-th brother. Naturally, all these numbers should be distinct and be within limits from 1 to *n*2. You can print the numbers in the lines in any order.
It is guaranteed that the solution exists at the given limits.
|
[
"2\n"
] |
[
"1 4\n2 3\n"
] |
The sample shows Gerald's actions if he has two brothers. In this case, his bags contain 1, 2, 3 and 4 candies. He can give the bags with 1 and 4 candies to one brother and the bags with 2 and 3 to the other brother.
| 500
|
[
{
"input": "2",
"output": "1 4\n2 3"
},
{
"input": "4",
"output": "1 16 2 15\n3 14 4 13\n5 12 6 11\n7 10 8 9"
},
{
"input": "6",
"output": "1 36 2 35 3 34\n4 33 5 32 6 31\n7 30 8 29 9 28\n10 27 11 26 12 25\n13 24 14 23 15 22\n16 21 17 20 18 19"
},
{
"input": "8",
"output": "1 64 2 63 3 62 4 61\n5 60 6 59 7 58 8 57\n9 56 10 55 11 54 12 53\n13 52 14 51 15 50 16 49\n17 48 18 47 19 46 20 45\n21 44 22 43 23 42 24 41\n25 40 26 39 27 38 28 37\n29 36 30 35 31 34 32 33"
},
{
"input": "10",
"output": "1 100 2 99 3 98 4 97 5 96\n6 95 7 94 8 93 9 92 10 91\n11 90 12 89 13 88 14 87 15 86\n16 85 17 84 18 83 19 82 20 81\n21 80 22 79 23 78 24 77 25 76\n26 75 27 74 28 73 29 72 30 71\n31 70 32 69 33 68 34 67 35 66\n36 65 37 64 38 63 39 62 40 61\n41 60 42 59 43 58 44 57 45 56\n46 55 47 54 48 53 49 52 50 51"
},
{
"input": "100",
"output": "1 10000 2 9999 3 9998 4 9997 5 9996 6 9995 7 9994 8 9993 9 9992 10 9991 11 9990 12 9989 13 9988 14 9987 15 9986 16 9985 17 9984 18 9983 19 9982 20 9981 21 9980 22 9979 23 9978 24 9977 25 9976 26 9975 27 9974 28 9973 29 9972 30 9971 31 9970 32 9969 33 9968 34 9967 35 9966 36 9965 37 9964 38 9963 39 9962 40 9961 41 9960 42 9959 43 9958 44 9957 45 9956 46 9955 47 9954 48 9953 49 9952 50 9951\n51 9950 52 9949 53 9948 54 9947 55 9946 56 9945 57 9944 58 9943 59 9942 60 9941 61 9940 62 9939 63 9938 64 9937 65 993..."
},
{
"input": "62",
"output": "1 3844 2 3843 3 3842 4 3841 5 3840 6 3839 7 3838 8 3837 9 3836 10 3835 11 3834 12 3833 13 3832 14 3831 15 3830 16 3829 17 3828 18 3827 19 3826 20 3825 21 3824 22 3823 23 3822 24 3821 25 3820 26 3819 27 3818 28 3817 29 3816 30 3815 31 3814\n32 3813 33 3812 34 3811 35 3810 36 3809 37 3808 38 3807 39 3806 40 3805 41 3804 42 3803 43 3802 44 3801 45 3800 46 3799 47 3798 48 3797 49 3796 50 3795 51 3794 52 3793 53 3792 54 3791 55 3790 56 3789 57 3788 58 3787 59 3786 60 3785 61 3784 62 3783\n63 3782 64 3781 65 378..."
},
{
"input": "66",
"output": "1 4356 2 4355 3 4354 4 4353 5 4352 6 4351 7 4350 8 4349 9 4348 10 4347 11 4346 12 4345 13 4344 14 4343 15 4342 16 4341 17 4340 18 4339 19 4338 20 4337 21 4336 22 4335 23 4334 24 4333 25 4332 26 4331 27 4330 28 4329 29 4328 30 4327 31 4326 32 4325 33 4324\n34 4323 35 4322 36 4321 37 4320 38 4319 39 4318 40 4317 41 4316 42 4315 43 4314 44 4313 45 4312 46 4311 47 4310 48 4309 49 4308 50 4307 51 4306 52 4305 53 4304 54 4303 55 4302 56 4301 57 4300 58 4299 59 4298 60 4297 61 4296 62 4295 63 4294 64 4293 65 4292..."
},
{
"input": "18",
"output": "1 324 2 323 3 322 4 321 5 320 6 319 7 318 8 317 9 316\n10 315 11 314 12 313 13 312 14 311 15 310 16 309 17 308 18 307\n19 306 20 305 21 304 22 303 23 302 24 301 25 300 26 299 27 298\n28 297 29 296 30 295 31 294 32 293 33 292 34 291 35 290 36 289\n37 288 38 287 39 286 40 285 41 284 42 283 43 282 44 281 45 280\n46 279 47 278 48 277 49 276 50 275 51 274 52 273 53 272 54 271\n55 270 56 269 57 268 58 267 59 266 60 265 61 264 62 263 63 262\n64 261 65 260 66 259 67 258 68 257 69 256 70 255 71 254 72 253\n73 252 7..."
},
{
"input": "68",
"output": "1 4624 2 4623 3 4622 4 4621 5 4620 6 4619 7 4618 8 4617 9 4616 10 4615 11 4614 12 4613 13 4612 14 4611 15 4610 16 4609 17 4608 18 4607 19 4606 20 4605 21 4604 22 4603 23 4602 24 4601 25 4600 26 4599 27 4598 28 4597 29 4596 30 4595 31 4594 32 4593 33 4592 34 4591\n35 4590 36 4589 37 4588 38 4587 39 4586 40 4585 41 4584 42 4583 43 4582 44 4581 45 4580 46 4579 47 4578 48 4577 49 4576 50 4575 51 4574 52 4573 53 4572 54 4571 55 4570 56 4569 57 4568 58 4567 59 4566 60 4565 61 4564 62 4563 63 4562 64 4561 65 4560..."
},
{
"input": "86",
"output": "1 7396 2 7395 3 7394 4 7393 5 7392 6 7391 7 7390 8 7389 9 7388 10 7387 11 7386 12 7385 13 7384 14 7383 15 7382 16 7381 17 7380 18 7379 19 7378 20 7377 21 7376 22 7375 23 7374 24 7373 25 7372 26 7371 27 7370 28 7369 29 7368 30 7367 31 7366 32 7365 33 7364 34 7363 35 7362 36 7361 37 7360 38 7359 39 7358 40 7357 41 7356 42 7355 43 7354\n44 7353 45 7352 46 7351 47 7350 48 7349 49 7348 50 7347 51 7346 52 7345 53 7344 54 7343 55 7342 56 7341 57 7340 58 7339 59 7338 60 7337 61 7336 62 7335 63 7334 64 7333 65 7332..."
},
{
"input": "96",
"output": "1 9216 2 9215 3 9214 4 9213 5 9212 6 9211 7 9210 8 9209 9 9208 10 9207 11 9206 12 9205 13 9204 14 9203 15 9202 16 9201 17 9200 18 9199 19 9198 20 9197 21 9196 22 9195 23 9194 24 9193 25 9192 26 9191 27 9190 28 9189 29 9188 30 9187 31 9186 32 9185 33 9184 34 9183 35 9182 36 9181 37 9180 38 9179 39 9178 40 9177 41 9176 42 9175 43 9174 44 9173 45 9172 46 9171 47 9170 48 9169\n49 9168 50 9167 51 9166 52 9165 53 9164 54 9163 55 9162 56 9161 57 9160 58 9159 59 9158 60 9157 61 9156 62 9155 63 9154 64 9153 65 9152..."
},
{
"input": "12",
"output": "1 144 2 143 3 142 4 141 5 140 6 139\n7 138 8 137 9 136 10 135 11 134 12 133\n13 132 14 131 15 130 16 129 17 128 18 127\n19 126 20 125 21 124 22 123 23 122 24 121\n25 120 26 119 27 118 28 117 29 116 30 115\n31 114 32 113 33 112 34 111 35 110 36 109\n37 108 38 107 39 106 40 105 41 104 42 103\n43 102 44 101 45 100 46 99 47 98 48 97\n49 96 50 95 51 94 52 93 53 92 54 91\n55 90 56 89 57 88 58 87 59 86 60 85\n61 84 62 83 63 82 64 81 65 80 66 79\n67 78 68 77 69 76 70 75 71 74 72 73"
},
{
"input": "88",
"output": "1 7744 2 7743 3 7742 4 7741 5 7740 6 7739 7 7738 8 7737 9 7736 10 7735 11 7734 12 7733 13 7732 14 7731 15 7730 16 7729 17 7728 18 7727 19 7726 20 7725 21 7724 22 7723 23 7722 24 7721 25 7720 26 7719 27 7718 28 7717 29 7716 30 7715 31 7714 32 7713 33 7712 34 7711 35 7710 36 7709 37 7708 38 7707 39 7706 40 7705 41 7704 42 7703 43 7702 44 7701\n45 7700 46 7699 47 7698 48 7697 49 7696 50 7695 51 7694 52 7693 53 7692 54 7691 55 7690 56 7689 57 7688 58 7687 59 7686 60 7685 61 7684 62 7683 63 7682 64 7681 65 7680..."
},
{
"input": "28",
"output": "1 784 2 783 3 782 4 781 5 780 6 779 7 778 8 777 9 776 10 775 11 774 12 773 13 772 14 771\n15 770 16 769 17 768 18 767 19 766 20 765 21 764 22 763 23 762 24 761 25 760 26 759 27 758 28 757\n29 756 30 755 31 754 32 753 33 752 34 751 35 750 36 749 37 748 38 747 39 746 40 745 41 744 42 743\n43 742 44 741 45 740 46 739 47 738 48 737 49 736 50 735 51 734 52 733 53 732 54 731 55 730 56 729\n57 728 58 727 59 726 60 725 61 724 62 723 63 722 64 721 65 720 66 719 67 718 68 717 69 716 70 715\n71 714 72 713 73 712 74 7..."
},
{
"input": "80",
"output": "1 6400 2 6399 3 6398 4 6397 5 6396 6 6395 7 6394 8 6393 9 6392 10 6391 11 6390 12 6389 13 6388 14 6387 15 6386 16 6385 17 6384 18 6383 19 6382 20 6381 21 6380 22 6379 23 6378 24 6377 25 6376 26 6375 27 6374 28 6373 29 6372 30 6371 31 6370 32 6369 33 6368 34 6367 35 6366 36 6365 37 6364 38 6363 39 6362 40 6361\n41 6360 42 6359 43 6358 44 6357 45 6356 46 6355 47 6354 48 6353 49 6352 50 6351 51 6350 52 6349 53 6348 54 6347 55 6346 56 6345 57 6344 58 6343 59 6342 60 6341 61 6340 62 6339 63 6338 64 6337 65 6336..."
},
{
"input": "48",
"output": "1 2304 2 2303 3 2302 4 2301 5 2300 6 2299 7 2298 8 2297 9 2296 10 2295 11 2294 12 2293 13 2292 14 2291 15 2290 16 2289 17 2288 18 2287 19 2286 20 2285 21 2284 22 2283 23 2282 24 2281\n25 2280 26 2279 27 2278 28 2277 29 2276 30 2275 31 2274 32 2273 33 2272 34 2271 35 2270 36 2269 37 2268 38 2267 39 2266 40 2265 41 2264 42 2263 43 2262 44 2261 45 2260 46 2259 47 2258 48 2257\n49 2256 50 2255 51 2254 52 2253 53 2252 54 2251 55 2250 56 2249 57 2248 58 2247 59 2246 60 2245 61 2244 62 2243 63 2242 64 2241 65 224..."
},
{
"input": "54",
"output": "1 2916 2 2915 3 2914 4 2913 5 2912 6 2911 7 2910 8 2909 9 2908 10 2907 11 2906 12 2905 13 2904 14 2903 15 2902 16 2901 17 2900 18 2899 19 2898 20 2897 21 2896 22 2895 23 2894 24 2893 25 2892 26 2891 27 2890\n28 2889 29 2888 30 2887 31 2886 32 2885 33 2884 34 2883 35 2882 36 2881 37 2880 38 2879 39 2878 40 2877 41 2876 42 2875 43 2874 44 2873 45 2872 46 2871 47 2870 48 2869 49 2868 50 2867 51 2866 52 2865 53 2864 54 2863\n55 2862 56 2861 57 2860 58 2859 59 2858 60 2857 61 2856 62 2855 63 2854 64 2853 65 285..."
},
{
"input": "58",
"output": "1 3364 2 3363 3 3362 4 3361 5 3360 6 3359 7 3358 8 3357 9 3356 10 3355 11 3354 12 3353 13 3352 14 3351 15 3350 16 3349 17 3348 18 3347 19 3346 20 3345 21 3344 22 3343 23 3342 24 3341 25 3340 26 3339 27 3338 28 3337 29 3336\n30 3335 31 3334 32 3333 33 3332 34 3331 35 3330 36 3329 37 3328 38 3327 39 3326 40 3325 41 3324 42 3323 43 3322 44 3321 45 3320 46 3319 47 3318 48 3317 49 3316 50 3315 51 3314 52 3313 53 3312 54 3311 55 3310 56 3309 57 3308 58 3307\n59 3306 60 3305 61 3304 62 3303 63 3302 64 3301 65 330..."
},
{
"input": "64",
"output": "1 4096 2 4095 3 4094 4 4093 5 4092 6 4091 7 4090 8 4089 9 4088 10 4087 11 4086 12 4085 13 4084 14 4083 15 4082 16 4081 17 4080 18 4079 19 4078 20 4077 21 4076 22 4075 23 4074 24 4073 25 4072 26 4071 27 4070 28 4069 29 4068 30 4067 31 4066 32 4065\n33 4064 34 4063 35 4062 36 4061 37 4060 38 4059 39 4058 40 4057 41 4056 42 4055 43 4054 44 4053 45 4052 46 4051 47 4050 48 4049 49 4048 50 4047 51 4046 52 4045 53 4044 54 4043 55 4042 56 4041 57 4040 58 4039 59 4038 60 4037 61 4036 62 4035 63 4034 64 4033\n65 403..."
},
{
"input": "44",
"output": "1 1936 2 1935 3 1934 4 1933 5 1932 6 1931 7 1930 8 1929 9 1928 10 1927 11 1926 12 1925 13 1924 14 1923 15 1922 16 1921 17 1920 18 1919 19 1918 20 1917 21 1916 22 1915\n23 1914 24 1913 25 1912 26 1911 27 1910 28 1909 29 1908 30 1907 31 1906 32 1905 33 1904 34 1903 35 1902 36 1901 37 1900 38 1899 39 1898 40 1897 41 1896 42 1895 43 1894 44 1893\n45 1892 46 1891 47 1890 48 1889 49 1888 50 1887 51 1886 52 1885 53 1884 54 1883 55 1882 56 1881 57 1880 58 1879 59 1878 60 1877 61 1876 62 1875 63 1874 64 1873 65 187..."
}
] | 1,506,868,091
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 21
| 124
| 204,800
|
import math
n = int(input())
bags = int(math.pow(n,2))
#candies = math.factorial(bags)
n_bags = []
for j in range (n):
n_bags.append("")
for i in range(int(n/2)):
n_bags[j] += str((i*n)+j+1)
n_bags[j] += " "
n_bags[j] += str(bags-(j+(i*n)))
n_bags[j] += " "
#n_bags[j].append((i*n)+j+1)
#n_bags[j].append(bags-(j+(i*n)))
print (n_bags[j])
|
Title: Candy Bags
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Gerald has *n* younger brothers and their number happens to be even. One day he bought *n*2 candy bags. One bag has one candy, one bag has two candies, one bag has three candies and so on. In fact, for each integer *k* from 1 to *n*2 he has exactly one bag with *k* candies.
Help him give *n* bags of candies to each brother so that all brothers got the same number of candies.
Input Specification:
The single line contains a single integer *n* (*n* is even, 2<=β€<=*n*<=β€<=100) β the number of Gerald's brothers.
Output Specification:
Let's assume that Gerald indexes his brothers with numbers from 1 to *n*. You need to print *n* lines, on the *i*-th line print *n* integers β the numbers of candies in the bags for the *i*-th brother. Naturally, all these numbers should be distinct and be within limits from 1 to *n*2. You can print the numbers in the lines in any order.
It is guaranteed that the solution exists at the given limits.
Demo Input:
['2\n']
Demo Output:
['1 4\n2 3\n']
Note:
The sample shows Gerald's actions if he has two brothers. In this case, his bags contain 1, 2, 3 and 4 candies. He can give the bags with 1 and 4 candies to one brother and the bags with 2 and 3 to the other brother.
|
```python
import math
n = int(input())
bags = int(math.pow(n,2))
#candies = math.factorial(bags)
n_bags = []
for j in range (n):
n_bags.append("")
for i in range(int(n/2)):
n_bags[j] += str((i*n)+j+1)
n_bags[j] += " "
n_bags[j] += str(bags-(j+(i*n)))
n_bags[j] += " "
#n_bags[j].append((i*n)+j+1)
#n_bags[j].append(bags-(j+(i*n)))
print (n_bags[j])
```
| 3
|
|
899
|
A
|
Splitting in Teams
|
PROGRAMMING
| 800
|
[
"constructive algorithms",
"greedy",
"math"
] | null | null |
There were *n* groups of students which came to write a training contest. A group is either one person who can write the contest with anyone else, or two people who want to write the contest in the same team.
The coach decided to form teams of exactly three people for this training. Determine the maximum number of teams of three people he can form. It is possible that he can't use all groups to form teams. For groups of two, either both students should write the contest, or both should not. If two students from a group of two will write the contest, they should be in the same team.
|
The first line contains single integer *n* (2<=β€<=*n*<=β€<=2Β·105) β the number of groups.
The second line contains a sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (1<=β€<=*a**i*<=β€<=2), where *a**i* is the number of people in group *i*.
|
Print the maximum number of teams of three people the coach can form.
|
[
"4\n1 1 2 1\n",
"2\n2 2\n",
"7\n2 2 2 1 1 1 1\n",
"3\n1 1 1\n"
] |
[
"1\n",
"0\n",
"3\n",
"1\n"
] |
In the first example the coach can form one team. For example, he can take students from the first, second and fourth groups.
In the second example he can't make a single team.
In the third example the coach can form three teams. For example, he can do this in the following way:
- The first group (of two people) and the seventh group (of one person), - The second group (of two people) and the sixth group (of one person), - The third group (of two people) and the fourth group (of one person).
| 500
|
[
{
"input": "4\n1 1 2 1",
"output": "1"
},
{
"input": "2\n2 2",
"output": "0"
},
{
"input": "7\n2 2 2 1 1 1 1",
"output": "3"
},
{
"input": "3\n1 1 1",
"output": "1"
},
{
"input": "3\n2 2 2",
"output": "0"
},
{
"input": "3\n1 2 1",
"output": "1"
},
{
"input": "5\n2 2 1 1 1",
"output": "2"
},
{
"input": "7\n1 1 2 2 1 2 1",
"output": "3"
},
{
"input": "10\n1 2 2 1 2 2 1 2 1 1",
"output": "5"
},
{
"input": "5\n2 2 2 1 2",
"output": "1"
},
{
"input": "43\n1 2 2 2 1 1 2 2 1 1 2 2 2 2 1 2 2 2 2 2 1 2 1 2 1 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2",
"output": "10"
},
{
"input": "72\n1 2 1 2 2 1 2 1 1 1 1 2 2 1 2 1 2 1 2 2 2 2 1 2 2 2 2 1 2 1 1 2 2 1 1 2 2 2 2 2 1 1 1 1 2 2 1 1 2 1 1 1 1 2 2 1 2 2 1 2 1 1 2 1 2 2 1 1 1 2 2 2",
"output": "34"
},
{
"input": "64\n2 2 1 1 1 2 1 1 1 2 2 1 2 2 2 1 2 2 2 1 1 1 1 2 1 2 1 2 1 1 2 2 1 1 2 2 1 1 1 1 2 2 1 1 1 2 1 2 2 2 2 2 2 2 1 1 2 1 1 1 2 2 1 2",
"output": "32"
},
{
"input": "20\n1 1 1 1 2 1 2 2 2 1 2 1 2 1 2 1 1 2 1 2",
"output": "9"
},
{
"input": "23\n1 1 1 1 2 1 2 1 1 1 2 2 2 2 2 2 1 2 1 2 2 1 1",
"output": "11"
},
{
"input": "201\n1 1 2 2 2 2 1 1 1 2 2 1 2 1 2 1 2 2 2 1 1 2 1 1 1 2 1 2 1 1 1 2 1 1 2 1 2 2 1 1 1 1 2 1 1 2 1 1 1 2 2 2 2 1 2 1 2 2 2 2 2 2 1 1 1 2 2 1 1 1 1 2 2 1 2 1 1 2 2 1 1 2 2 2 1 1 1 2 1 1 2 1 2 2 1 2 2 2 2 1 1 1 2 1 2 2 2 2 2 1 2 1 1 1 2 2 2 2 2 1 2 1 1 2 2 2 1 1 2 2 1 2 2 2 1 1 1 2 1 1 1 2 1 1 2 2 2 1 2 1 1 1 2 2 1 1 2 2 2 2 2 2 1 2 2 1 2 2 2 1 1 2 2 1 1 2 1 1 1 1 2 1 1 1 2 2 1 2 1 1 2 2 1 1 2 1 2 1 1 1 2",
"output": "100"
},
{
"input": "247\n2 2 1 2 1 2 2 2 2 2 2 1 1 2 2 1 2 1 1 1 2 1 1 1 1 2 1 1 2 2 1 2 1 1 1 2 2 2 1 1 2 1 1 2 1 1 1 2 1 2 1 2 2 1 1 2 1 2 2 1 2 1 2 1 1 2 1 1 1 2 2 1 1 2 2 1 1 2 1 1 1 2 2 2 2 1 2 2 2 2 2 2 1 2 2 2 2 1 1 1 1 1 1 1 1 1 2 1 2 2 1 2 1 2 2 2 1 2 2 2 1 1 2 2 1 1 1 2 1 1 1 1 2 2 1 2 2 1 1 1 2 1 2 2 1 2 1 1 1 2 2 2 2 2 1 2 2 2 1 1 1 2 1 2 1 1 2 2 2 2 1 1 2 2 2 1 2 2 2 1 2 1 1 2 2 2 2 1 2 2 1 1 1 2 1 2 1 1 1 2 2 1 1 2 1 1 2 1 2 1 1 2 1 1 1 1 2 1 1 1 1 2 2 1 2 1 1 2 1 2 2 1 2 2 2 1 2 2 1 2 2 1 1 1 2 2 2",
"output": "123"
},
{
"input": "4\n2 2 2 2",
"output": "0"
},
{
"input": "4\n1 1 1 1",
"output": "1"
},
{
"input": "2\n1 1",
"output": "0"
},
{
"input": "2\n2 1",
"output": "1"
},
{
"input": "2\n1 2",
"output": "1"
},
{
"input": "3\n1 1 2",
"output": "1"
},
{
"input": "3\n1 2 2",
"output": "1"
},
{
"input": "3\n2 1 1",
"output": "1"
},
{
"input": "3\n2 1 2",
"output": "1"
},
{
"input": "3\n2 2 1",
"output": "1"
},
{
"input": "4\n1 1 1 2",
"output": "1"
},
{
"input": "4\n1 1 2 2",
"output": "2"
},
{
"input": "4\n1 2 2 2",
"output": "1"
},
{
"input": "4\n2 1 1 1",
"output": "1"
},
{
"input": "5\n1 1 1 1 2",
"output": "2"
},
{
"input": "14\n1 2 2 2 2 2 2 2 2 2 2 2 2 2",
"output": "1"
},
{
"input": "38\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 1",
"output": "1"
},
{
"input": "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 1",
"output": "1"
},
{
"input": "20\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1",
"output": "1"
},
{
"input": "26\n2 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",
"output": "1"
},
{
"input": "6\n1 1 1 2 2 2",
"output": "3"
},
{
"input": "5\n2 1 1 1 1",
"output": "2"
},
{
"input": "9\n1 1 1 1 1 1 2 2 2",
"output": "4"
},
{
"input": "10\n2 2 1 1 1 1 1 1 1 1",
"output": "4"
},
{
"input": "6\n1 1 1 1 1 1",
"output": "2"
}
] | 1,561,187,507
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 67
| 202
| 14,540,800
|
n = int(input())
l = list(map(int,input().split()))
k = l.count(1)
h = l.count(2)
if k>h:
print(h+(k-h)//3)
else:
print(k)
|
Title: Splitting in Teams
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There were *n* groups of students which came to write a training contest. A group is either one person who can write the contest with anyone else, or two people who want to write the contest in the same team.
The coach decided to form teams of exactly three people for this training. Determine the maximum number of teams of three people he can form. It is possible that he can't use all groups to form teams. For groups of two, either both students should write the contest, or both should not. If two students from a group of two will write the contest, they should be in the same team.
Input Specification:
The first line contains single integer *n* (2<=β€<=*n*<=β€<=2Β·105) β the number of groups.
The second line contains a sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (1<=β€<=*a**i*<=β€<=2), where *a**i* is the number of people in group *i*.
Output Specification:
Print the maximum number of teams of three people the coach can form.
Demo Input:
['4\n1 1 2 1\n', '2\n2 2\n', '7\n2 2 2 1 1 1 1\n', '3\n1 1 1\n']
Demo Output:
['1\n', '0\n', '3\n', '1\n']
Note:
In the first example the coach can form one team. For example, he can take students from the first, second and fourth groups.
In the second example he can't make a single team.
In the third example the coach can form three teams. For example, he can do this in the following way:
- The first group (of two people) and the seventh group (of one person), - The second group (of two people) and the sixth group (of one person), - The third group (of two people) and the fourth group (of one person).
|
```python
n = int(input())
l = list(map(int,input().split()))
k = l.count(1)
h = l.count(2)
if k>h:
print(h+(k-h)//3)
else:
print(k)
```
| 3
|
|
981
|
A
|
Antipalindrome
|
PROGRAMMING
| 900
|
[
"brute force",
"implementation",
"strings"
] | null | null |
A string is a palindrome if it reads the same from the left to the right and from the right to the left. For example, the strings "kek", "abacaba", "r" and "papicipap" are palindromes, while the strings "abb" and "iq" are not.
A substring $s[l \ldots r]$ ($1<=\leq<=l<=\leq<=r<=\leq<=|s|$) of a string $s<==<=s_{1}s_{2} \ldots s_{|s|}$ is the string $s_{l}s_{l<=+<=1} \ldots s_{r}$.
Anna does not like palindromes, so she makes her friends call her Ann. She also changes all the words she reads in a similar way. Namely, each word $s$ is changed into its longest substring that is not a palindrome. If all the substrings of $s$ are palindromes, she skips the word at all.
Some time ago Ann read the word $s$. What is the word she changed it into?
|
The first line contains a non-empty string $s$ with length at most $50$ characters, containing lowercase English letters only.
|
If there is such a substring in $s$ that is not a palindrome, print the maximum length of such a substring. Otherwise print $0$.
Note that there can be multiple longest substrings that are not palindromes, but their length is unique.
|
[
"mew\n",
"wuffuw\n",
"qqqqqqqq\n"
] |
[
"3\n",
"5\n",
"0\n"
] |
"mew" is not a palindrome, so the longest substring of it that is not a palindrome, is the string "mew" itself. Thus, the answer for the first example is $3$.
The string "uffuw" is one of the longest non-palindrome substrings (of length $5$) of the string "wuffuw", so the answer for the second example is $5$.
All substrings of the string "qqqqqqqq" consist of equal characters so they are palindromes. This way, there are no non-palindrome substrings. Thus, the answer for the third example is $0$.
| 500
|
[
{
"input": "mew",
"output": "3"
},
{
"input": "wuffuw",
"output": "5"
},
{
"input": "qqqqqqqq",
"output": "0"
},
{
"input": "ijvji",
"output": "4"
},
{
"input": "iiiiiii",
"output": "0"
},
{
"input": "wobervhvvkihcuyjtmqhaaigvvgiaahqmtjyuchikvvhvrebow",
"output": "49"
},
{
"input": "wwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww",
"output": "0"
},
{
"input": "wobervhvvkihcuyjtmqhaaigvahheoqleromusrartldojsjvy",
"output": "50"
},
{
"input": "ijvxljt",
"output": "7"
},
{
"input": "fyhcncnchyf",
"output": "10"
},
{
"input": "ffffffffffff",
"output": "0"
},
{
"input": "fyhcncfsepqj",
"output": "12"
},
{
"input": "ybejrrlbcinttnicblrrjeby",
"output": "23"
},
{
"input": "yyyyyyyyyyyyyyyyyyyyyyyyy",
"output": "0"
},
{
"input": "ybejrrlbcintahovgjddrqatv",
"output": "25"
},
{
"input": "oftmhcmclgyqaojljoaqyglcmchmtfo",
"output": "30"
},
{
"input": "oooooooooooooooooooooooooooooooo",
"output": "0"
},
{
"input": "oftmhcmclgyqaojllbotztajglsmcilv",
"output": "32"
},
{
"input": "gxandbtgpbknxvnkjaajknvxnkbpgtbdnaxg",
"output": "35"
},
{
"input": "gggggggggggggggggggggggggggggggggggg",
"output": "0"
},
{
"input": "gxandbtgpbknxvnkjaygommzqitqzjfalfkk",
"output": "36"
},
{
"input": "fcliblymyqckxvieotjooojtoeivxkcqymylbilcf",
"output": "40"
},
{
"input": "fffffffffffffffffffffffffffffffffffffffffff",
"output": "0"
},
{
"input": "fcliblymyqckxvieotjootiqwtyznhhvuhbaixwqnsy",
"output": "43"
},
{
"input": "rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr",
"output": "0"
},
{
"input": "rajccqwqnqmshmerpvjyfepxwpxyldzpzhctqjnstxyfmlhiy",
"output": "49"
},
{
"input": "a",
"output": "0"
},
{
"input": "abca",
"output": "4"
},
{
"input": "aaaaabaaaaa",
"output": "10"
},
{
"input": "aba",
"output": "2"
},
{
"input": "asaa",
"output": "4"
},
{
"input": "aabaa",
"output": "4"
},
{
"input": "aabbaa",
"output": "5"
},
{
"input": "abcdaaa",
"output": "7"
},
{
"input": "aaholaa",
"output": "7"
},
{
"input": "abcdefghijka",
"output": "12"
},
{
"input": "aaadcba",
"output": "7"
},
{
"input": "aaaabaaaa",
"output": "8"
},
{
"input": "abaa",
"output": "4"
},
{
"input": "abcbaa",
"output": "6"
},
{
"input": "ab",
"output": "2"
},
{
"input": "l",
"output": "0"
},
{
"input": "aaaabcaaaa",
"output": "10"
},
{
"input": "abbaaaaaabba",
"output": "11"
},
{
"input": "abaaa",
"output": "5"
},
{
"input": "baa",
"output": "3"
},
{
"input": "aaaaaaabbba",
"output": "11"
},
{
"input": "ccbcc",
"output": "4"
},
{
"input": "bbbaaab",
"output": "7"
},
{
"input": "abaaaaaaaa",
"output": "10"
},
{
"input": "abaaba",
"output": "5"
},
{
"input": "aabsdfaaaa",
"output": "10"
},
{
"input": "aaaba",
"output": "5"
},
{
"input": "aaabaaa",
"output": "6"
},
{
"input": "baaabbb",
"output": "7"
},
{
"input": "ccbbabbcc",
"output": "8"
},
{
"input": "cabc",
"output": "4"
},
{
"input": "aabcd",
"output": "5"
},
{
"input": "abcdea",
"output": "6"
},
{
"input": "bbabb",
"output": "4"
},
{
"input": "aaaaabababaaaaa",
"output": "14"
},
{
"input": "bbabbb",
"output": "6"
},
{
"input": "aababd",
"output": "6"
},
{
"input": "abaaaa",
"output": "6"
},
{
"input": "aaaaaaaabbba",
"output": "12"
},
{
"input": "aabca",
"output": "5"
},
{
"input": "aaabccbaaa",
"output": "9"
},
{
"input": "aaaaaaaaaaaaaaaaaaaab",
"output": "21"
},
{
"input": "babb",
"output": "4"
},
{
"input": "abcaa",
"output": "5"
},
{
"input": "qwqq",
"output": "4"
},
{
"input": "aaaaaaaaaaabbbbbbbbbbbbbbbaaaaaaaaaaaaaaaaaaaaaa",
"output": "48"
},
{
"input": "aaab",
"output": "4"
},
{
"input": "aaaaaabaaaaa",
"output": "12"
},
{
"input": "wwuww",
"output": "4"
},
{
"input": "aaaaabcbaaaaa",
"output": "12"
},
{
"input": "aaabbbaaa",
"output": "8"
},
{
"input": "aabcbaa",
"output": "6"
},
{
"input": "abccdefccba",
"output": "11"
},
{
"input": "aabbcbbaa",
"output": "8"
},
{
"input": "aaaabbaaaa",
"output": "9"
},
{
"input": "aabcda",
"output": "6"
},
{
"input": "abbca",
"output": "5"
},
{
"input": "aaaaaabbaaa",
"output": "11"
},
{
"input": "sssssspssssss",
"output": "12"
},
{
"input": "sdnmsdcs",
"output": "8"
},
{
"input": "aaabbbccbbbaaa",
"output": "13"
},
{
"input": "cbdbdc",
"output": "6"
},
{
"input": "abb",
"output": "3"
},
{
"input": "abcdefaaaa",
"output": "10"
},
{
"input": "abbbaaa",
"output": "7"
},
{
"input": "v",
"output": "0"
},
{
"input": "abccbba",
"output": "7"
},
{
"input": "axyza",
"output": "5"
},
{
"input": "abcdefgaaaa",
"output": "11"
},
{
"input": "aaabcdaaa",
"output": "9"
},
{
"input": "aaaacaaaa",
"output": "8"
},
{
"input": "aaaaaaaaaaaaaaaaaaaabaaaaaaaaaaaaaaaaaaaaa",
"output": "42"
},
{
"input": "abbbaa",
"output": "6"
},
{
"input": "abcdee",
"output": "6"
},
{
"input": "oom",
"output": "3"
},
{
"input": "aabcaa",
"output": "6"
},
{
"input": "abba",
"output": "3"
},
{
"input": "aaca",
"output": "4"
},
{
"input": "aacbca",
"output": "6"
},
{
"input": "ababa",
"output": "4"
},
{
"input": "abcda",
"output": "5"
},
{
"input": "cccaaccc",
"output": "7"
},
{
"input": "aaabcda",
"output": "7"
},
{
"input": "aa",
"output": "0"
},
{
"input": "aabaaaa",
"output": "7"
},
{
"input": "abbaaaa",
"output": "7"
},
{
"input": "aaabcbaaa",
"output": "8"
},
{
"input": "aabba",
"output": "5"
},
{
"input": "xyxx",
"output": "4"
},
{
"input": "aaaaaaaaaaaabc",
"output": "14"
},
{
"input": "bbaaaabb",
"output": "7"
},
{
"input": "aaabaa",
"output": "6"
},
{
"input": "sssssabsssss",
"output": "12"
},
{
"input": "bbbaaaabbb",
"output": "9"
},
{
"input": "abbbbaaaa",
"output": "9"
},
{
"input": "wwufuww",
"output": "6"
},
{
"input": "oowoo",
"output": "4"
},
{
"input": "cccaccc",
"output": "6"
},
{
"input": "aaa",
"output": "0"
},
{
"input": "bbbcc",
"output": "5"
},
{
"input": "abcdef",
"output": "6"
},
{
"input": "abbba",
"output": "4"
},
{
"input": "aab",
"output": "3"
},
{
"input": "aaba",
"output": "4"
},
{
"input": "azbyaaa",
"output": "7"
},
{
"input": "oooooiooooo",
"output": "10"
},
{
"input": "aabbbbbaaaaaa",
"output": "13"
}
] | 1,527,923,037
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 133
| 93
| 7,065,600
|
word = input()
if word == word[::-1]:
for i in range(len(word)):
slicedWord = word[i + 1:]
if slicedWord != slicedWord[::-1]:
print(len(word) - i - 1)
break
if len(slicedWord) == 0:
print(0)
break
else:
print(len(word))
|
Title: Antipalindrome
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A string is a palindrome if it reads the same from the left to the right and from the right to the left. For example, the strings "kek", "abacaba", "r" and "papicipap" are palindromes, while the strings "abb" and "iq" are not.
A substring $s[l \ldots r]$ ($1<=\leq<=l<=\leq<=r<=\leq<=|s|$) of a string $s<==<=s_{1}s_{2} \ldots s_{|s|}$ is the string $s_{l}s_{l<=+<=1} \ldots s_{r}$.
Anna does not like palindromes, so she makes her friends call her Ann. She also changes all the words she reads in a similar way. Namely, each word $s$ is changed into its longest substring that is not a palindrome. If all the substrings of $s$ are palindromes, she skips the word at all.
Some time ago Ann read the word $s$. What is the word she changed it into?
Input Specification:
The first line contains a non-empty string $s$ with length at most $50$ characters, containing lowercase English letters only.
Output Specification:
If there is such a substring in $s$ that is not a palindrome, print the maximum length of such a substring. Otherwise print $0$.
Note that there can be multiple longest substrings that are not palindromes, but their length is unique.
Demo Input:
['mew\n', 'wuffuw\n', 'qqqqqqqq\n']
Demo Output:
['3\n', '5\n', '0\n']
Note:
"mew" is not a palindrome, so the longest substring of it that is not a palindrome, is the string "mew" itself. Thus, the answer for the first example is $3$.
The string "uffuw" is one of the longest non-palindrome substrings (of length $5$) of the string "wuffuw", so the answer for the second example is $5$.
All substrings of the string "qqqqqqqq" consist of equal characters so they are palindromes. This way, there are no non-palindrome substrings. Thus, the answer for the third example is $0$.
|
```python
word = input()
if word == word[::-1]:
for i in range(len(word)):
slicedWord = word[i + 1:]
if slicedWord != slicedWord[::-1]:
print(len(word) - i - 1)
break
if len(slicedWord) == 0:
print(0)
break
else:
print(len(word))
```
| 3
|
|
443
|
A
|
Anton and Letters
|
PROGRAMMING
| 800
|
[
"constructive algorithms",
"implementation"
] | null | null |
Recently, Anton has found a set. The set consists of small English letters. Anton carefully wrote out all the letters from the set in one line, separated by a comma. He also added an opening curved bracket at the beginning of the line and a closing curved bracket at the end of the line.
Unfortunately, from time to time Anton would forget writing some letter and write it again. He asks you to count the total number of distinct letters in his set.
|
The first and the single line contains the set of letters. The length of the line doesn't exceed 1000. It is guaranteed that the line starts from an opening curved bracket and ends with a closing curved bracket. Between them, small English letters are listed, separated by a comma. Each comma is followed by a space.
|
Print a single number β the number of distinct letters in Anton's set.
|
[
"{a, b, c}\n",
"{b, a, b, a}\n",
"{}\n"
] |
[
"3\n",
"2\n",
"0\n"
] |
none
| 500
|
[
{
"input": "{a, b, c}",
"output": "3"
},
{
"input": "{b, a, b, a}",
"output": "2"
},
{
"input": "{}",
"output": "0"
},
{
"input": "{a, a, c, b, b, b, c, c, c, c}",
"output": "3"
},
{
"input": "{a, c, b, b}",
"output": "3"
},
{
"input": "{a, b}",
"output": "2"
},
{
"input": "{a}",
"output": "1"
},
{
"input": "{b, a, b, a, b, c, c, b, c, b}",
"output": "3"
},
{
"input": "{e, g, c, e}",
"output": "3"
},
{
"input": "{a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a}",
"output": "1"
},
{
"input": "{a, a, b}",
"output": "2"
},
{
"input": "{a, b, b, b, a, b, a, a, a, a, a, a, b, a, b, a, a, a, a, a, b, a, b, a}",
"output": "2"
},
{
"input": "{j, u, a, c, f, w, e, w, x, t, h, p, v, n, i, l, x, n, i, b, u, c, a, a}",
"output": "16"
},
{
"input": "{x, i, w, c, p, e, h, z, k, i}",
"output": "9"
},
{
"input": "{t, k, o, x, r, d, q, j, k, e, z, w, y, r, z, s, s, e, s, b, k, i}",
"output": "15"
},
{
"input": "{y}",
"output": "1"
},
{
"input": "{x}",
"output": "1"
},
{
"input": "{b, z, a, z}",
"output": "3"
},
{
"input": "{z}",
"output": "1"
},
{
"input": "{a, z}",
"output": "2"
},
{
"input": "{a, b, z}",
"output": "3"
},
{
"input": "{s, q, z, r, t, a, b, h, j, i, o, z, r, q}",
"output": "11"
}
] | 1,699,752,063
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 27
| 46
| 0
|
s = input()
if len(s) == 2:
print(0)
else:
print(len(set(s[1:-1].split(', '))))
|
Title: Anton and Letters
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Recently, Anton has found a set. The set consists of small English letters. Anton carefully wrote out all the letters from the set in one line, separated by a comma. He also added an opening curved bracket at the beginning of the line and a closing curved bracket at the end of the line.
Unfortunately, from time to time Anton would forget writing some letter and write it again. He asks you to count the total number of distinct letters in his set.
Input Specification:
The first and the single line contains the set of letters. The length of the line doesn't exceed 1000. It is guaranteed that the line starts from an opening curved bracket and ends with a closing curved bracket. Between them, small English letters are listed, separated by a comma. Each comma is followed by a space.
Output Specification:
Print a single number β the number of distinct letters in Anton's set.
Demo Input:
['{a, b, c}\n', '{b, a, b, a}\n', '{}\n']
Demo Output:
['3\n', '2\n', '0\n']
Note:
none
|
```python
s = input()
if len(s) == 2:
print(0)
else:
print(len(set(s[1:-1].split(', '))))
```
| 3
|
|
831
|
A
|
Unimodal Array
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
Array of integers is unimodal, if:
- it is strictly increasing in the beginning; - after that it is constant; - after that it is strictly decreasing.
The first block (increasing) and the last block (decreasing) may be absent. It is allowed that both of this blocks are absent.
For example, the following three arrays are unimodal: [5,<=7,<=11,<=11,<=2,<=1], [4,<=4,<=2], [7], but the following three are not unimodal: [5,<=5,<=6,<=6,<=1], [1,<=2,<=1,<=2], [4,<=5,<=5,<=6].
Write a program that checks if an array is unimodal.
|
The first line contains integer *n* (1<=β€<=*n*<=β€<=100) β the number of elements in the array.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=β€<=*a**i*<=β€<=1<=000) β the elements of the array.
|
Print "YES" if the given array is unimodal. Otherwise, print "NO".
You can output each letter in any case (upper or lower).
|
[
"6\n1 5 5 5 4 2\n",
"5\n10 20 30 20 10\n",
"4\n1 2 1 2\n",
"7\n3 3 3 3 3 3 3\n"
] |
[
"YES\n",
"YES\n",
"NO\n",
"YES\n"
] |
In the first example the array is unimodal, because it is strictly increasing in the beginning (from position 1 to position 2, inclusively), that it is constant (from position 2 to position 4, inclusively) and then it is strictly decreasing (from position 4 to position 6, inclusively).
| 500
|
[
{
"input": "6\n1 5 5 5 4 2",
"output": "YES"
},
{
"input": "5\n10 20 30 20 10",
"output": "YES"
},
{
"input": "4\n1 2 1 2",
"output": "NO"
},
{
"input": "7\n3 3 3 3 3 3 3",
"output": "YES"
},
{
"input": "6\n5 7 11 11 2 1",
"output": "YES"
},
{
"input": "1\n7",
"output": "YES"
},
{
"input": "100\n527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527",
"output": "YES"
},
{
"input": "5\n5 5 6 6 1",
"output": "NO"
},
{
"input": "3\n4 4 2",
"output": "YES"
},
{
"input": "4\n4 5 5 6",
"output": "NO"
},
{
"input": "3\n516 516 515",
"output": "YES"
},
{
"input": "5\n502 503 508 508 507",
"output": "YES"
},
{
"input": "10\n538 538 538 538 538 538 538 538 538 538",
"output": "YES"
},
{
"input": "15\n452 454 455 455 450 448 443 442 439 436 433 432 431 428 426",
"output": "YES"
},
{
"input": "20\n497 501 504 505 509 513 513 513 513 513 513 513 513 513 513 513 513 513 513 513",
"output": "YES"
},
{
"input": "50\n462 465 465 465 463 459 454 449 444 441 436 435 430 429 426 422 421 418 417 412 408 407 406 403 402 399 395 392 387 386 382 380 379 376 374 371 370 365 363 359 358 354 350 349 348 345 342 341 338 337",
"output": "YES"
},
{
"input": "70\n290 292 294 297 299 300 303 305 310 312 313 315 319 320 325 327 328 333 337 339 340 341 345 350 351 354 359 364 367 372 374 379 381 382 383 384 389 393 395 397 398 400 402 405 409 411 416 417 422 424 429 430 434 435 440 442 445 449 451 453 458 460 465 470 474 477 482 482 482 479",
"output": "YES"
},
{
"input": "99\n433 435 439 444 448 452 457 459 460 464 469 470 471 476 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 480 479 478 477 476 474 469 468 465 460 457 453 452 450 445 443 440 438 433 432 431 430 428 425 421 418 414 411 406 402 397 396 393",
"output": "YES"
},
{
"input": "100\n537 538 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543",
"output": "YES"
},
{
"input": "100\n524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 521",
"output": "YES"
},
{
"input": "100\n235 239 243 245 246 251 254 259 260 261 264 269 272 275 277 281 282 285 289 291 292 293 298 301 302 303 305 307 308 310 315 317 320 324 327 330 334 337 342 346 347 348 353 357 361 366 370 373 376 378 379 384 386 388 390 395 398 400 405 408 413 417 420 422 424 429 434 435 438 441 443 444 445 450 455 457 459 463 465 468 471 473 475 477 481 486 491 494 499 504 504 504 504 504 504 504 504 504 504 504",
"output": "YES"
},
{
"input": "100\n191 196 201 202 207 212 216 219 220 222 224 227 230 231 234 235 238 242 246 250 253 254 259 260 263 267 269 272 277 280 284 287 288 290 295 297 300 305 307 312 316 320 324 326 327 332 333 334 338 343 347 351 356 358 363 368 370 374 375 380 381 386 390 391 394 396 397 399 402 403 405 410 414 419 422 427 429 433 437 442 443 447 448 451 455 459 461 462 464 468 473 478 481 484 485 488 492 494 496 496",
"output": "YES"
},
{
"input": "100\n466 466 466 466 466 464 459 455 452 449 446 443 439 436 435 433 430 428 425 424 420 419 414 412 407 404 401 396 394 391 386 382 379 375 374 369 364 362 360 359 356 351 350 347 342 340 338 337 333 330 329 326 321 320 319 316 311 306 301 297 292 287 286 281 278 273 269 266 261 257 256 255 253 252 250 245 244 242 240 238 235 230 225 220 216 214 211 209 208 206 203 198 196 194 192 190 185 182 177 173",
"output": "YES"
},
{
"input": "100\n360 362 367 369 374 377 382 386 389 391 396 398 399 400 405 410 413 416 419 420 423 428 431 436 441 444 445 447 451 453 457 459 463 468 468 468 468 468 468 468 468 468 468 468 468 468 468 468 468 468 468 468 468 468 468 468 465 460 455 453 448 446 443 440 436 435 430 425 420 415 410 405 404 403 402 399 394 390 387 384 382 379 378 373 372 370 369 366 361 360 355 353 349 345 344 342 339 338 335 333",
"output": "YES"
},
{
"input": "1\n1000",
"output": "YES"
},
{
"input": "100\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"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": "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 1",
"output": "YES"
},
{
"input": "100\n1 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": "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 999 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\n998 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 999 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 999",
"output": "NO"
},
{
"input": "100\n537 538 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 691 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543 543",
"output": "NO"
},
{
"input": "100\n527 527 527 527 527 527 527 527 872 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527 527",
"output": "NO"
},
{
"input": "100\n524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 208 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 524 521",
"output": "NO"
},
{
"input": "100\n235 239 243 245 246 251 254 259 260 261 264 269 272 275 277 281 282 285 289 291 292 293 298 301 302 303 305 307 308 310 315 317 320 324 327 330 334 337 342 921 347 348 353 357 361 366 370 373 376 378 379 384 386 388 390 395 398 400 405 408 413 417 420 422 424 429 434 435 438 441 443 444 445 450 455 457 459 463 465 468 471 473 475 477 481 486 491 494 499 504 504 504 504 504 504 504 504 504 504 504",
"output": "NO"
},
{
"input": "100\n191 196 201 202 207 212 216 219 220 222 224 227 230 231 234 235 238 242 246 250 253 254 259 260 263 267 269 272 277 280 284 287 288 290 295 297 300 305 307 312 316 320 324 326 327 332 333 334 338 343 347 351 356 358 119 368 370 374 375 380 381 386 390 391 394 396 397 399 402 403 405 410 414 419 422 427 429 433 437 442 443 447 448 451 455 459 461 462 464 468 473 478 481 484 485 488 492 494 496 496",
"output": "NO"
},
{
"input": "100\n466 466 466 466 466 464 459 455 452 449 446 443 439 436 435 433 430 428 425 424 420 419 414 412 407 404 401 396 394 391 386 382 379 375 374 369 364 362 360 359 356 335 350 347 342 340 338 337 333 330 329 326 321 320 319 316 311 306 301 297 292 287 286 281 278 273 269 266 261 257 256 255 253 252 250 245 244 242 240 238 235 230 225 220 216 214 211 209 208 206 203 198 196 194 192 190 185 182 177 173",
"output": "NO"
},
{
"input": "100\n360 362 367 369 374 377 382 386 389 391 396 398 399 400 405 410 413 416 419 420 423 428 525 436 441 444 445 447 451 453 457 459 463 468 468 468 468 468 468 468 468 468 468 468 468 468 468 468 468 468 468 468 468 468 468 468 465 460 455 453 448 446 443 440 436 435 430 425 420 415 410 405 404 403 402 399 394 390 387 384 382 379 378 373 372 370 369 366 361 360 355 353 349 345 344 342 339 338 335 333",
"output": "NO"
},
{
"input": "3\n1 2 3",
"output": "YES"
},
{
"input": "3\n3 2 1",
"output": "YES"
},
{
"input": "3\n1 1 2",
"output": "NO"
},
{
"input": "3\n2 1 1",
"output": "NO"
},
{
"input": "3\n2 1 2",
"output": "NO"
},
{
"input": "3\n3 1 2",
"output": "NO"
},
{
"input": "3\n1 3 2",
"output": "YES"
},
{
"input": "100\n395 399 402 403 405 408 413 415 419 424 426 431 434 436 439 444 447 448 449 454 457 459 461 462 463 464 465 469 470 473 477 480 482 484 485 487 492 494 496 497 501 504 505 508 511 506 505 503 500 499 494 490 488 486 484 481 479 474 472 471 470 465 462 458 453 452 448 445 440 436 433 430 428 426 424 421 419 414 413 408 404 403 399 395 393 388 384 379 377 375 374 372 367 363 360 356 353 351 350 346",
"output": "YES"
},
{
"input": "100\n263 268 273 274 276 281 282 287 288 292 294 295 296 300 304 306 308 310 311 315 319 322 326 330 333 336 339 341 342 347 351 353 356 358 363 365 369 372 374 379 383 387 389 391 392 395 396 398 403 404 407 411 412 416 419 421 424 428 429 430 434 436 440 443 444 448 453 455 458 462 463 464 469 473 477 481 486 489 492 494 499 503 506 509 510 512 514 515 511 510 507 502 499 498 494 491 486 482 477 475",
"output": "YES"
},
{
"input": "100\n482 484 485 489 492 496 499 501 505 509 512 517 520 517 515 513 509 508 504 503 498 496 493 488 486 481 478 476 474 470 468 466 463 459 456 453 452 449 445 444 439 438 435 432 428 427 424 423 421 419 417 413 408 405 402 399 397 393 388 385 380 375 370 366 363 361 360 355 354 352 349 345 340 336 335 331 329 327 324 319 318 317 315 314 310 309 307 304 303 300 299 295 291 287 285 282 280 278 273 271",
"output": "YES"
},
{
"input": "100\n395 399 402 403 405 408 413 415 419 424 426 431 434 436 439 444 447 448 449 454 457 459 461 462 463 464 465 469 470 473 477 480 482 484 485 487 492 494 496 32 501 504 505 508 511 506 505 503 500 499 494 490 488 486 484 481 479 474 472 471 470 465 462 458 453 452 448 445 440 436 433 430 428 426 424 421 419 414 413 408 404 403 399 395 393 388 384 379 377 375 374 372 367 363 360 356 353 351 350 346",
"output": "NO"
},
{
"input": "100\n263 268 273 274 276 281 282 287 288 292 294 295 296 300 304 306 308 310 311 315 319 322 326 330 247 336 339 341 342 347 351 353 356 358 363 365 369 372 374 379 383 387 389 391 392 395 396 398 403 404 407 411 412 416 419 421 424 428 429 430 434 436 440 443 444 448 453 455 458 462 463 464 469 473 477 481 486 489 492 494 499 503 506 509 510 512 514 515 511 510 507 502 499 498 494 491 486 482 477 475",
"output": "NO"
},
{
"input": "100\n482 484 485 489 492 496 499 501 505 509 512 517 520 517 515 513 509 508 504 503 497 496 493 488 486 481 478 476 474 470 468 466 463 459 456 453 452 449 445 444 439 438 435 432 428 427 424 423 421 419 417 413 408 405 402 399 397 393 388 385 380 375 370 366 363 361 360 355 354 352 349 345 340 336 335 331 329 327 324 319 318 317 315 314 310 309 307 304 303 300 299 295 291 287 285 282 280 278 273 271",
"output": "YES"
},
{
"input": "2\n1 3",
"output": "YES"
},
{
"input": "2\n1 2",
"output": "YES"
},
{
"input": "5\n2 2 1 1 1",
"output": "NO"
},
{
"input": "4\n1 3 2 2",
"output": "NO"
},
{
"input": "6\n1 2 1 2 2 1",
"output": "NO"
},
{
"input": "2\n4 2",
"output": "YES"
},
{
"input": "3\n3 2 2",
"output": "NO"
},
{
"input": "9\n1 2 2 3 3 4 3 2 1",
"output": "NO"
},
{
"input": "4\n5 5 4 4",
"output": "NO"
},
{
"input": "2\n2 1",
"output": "YES"
},
{
"input": "5\n5 4 3 2 1",
"output": "YES"
},
{
"input": "7\n4 3 3 3 3 3 3",
"output": "NO"
},
{
"input": "5\n1 2 3 4 5",
"output": "YES"
},
{
"input": "3\n2 2 1",
"output": "YES"
},
{
"input": "3\n4 3 3",
"output": "NO"
},
{
"input": "7\n1 5 5 4 3 3 1",
"output": "NO"
},
{
"input": "6\n3 3 1 2 2 1",
"output": "NO"
},
{
"input": "5\n1 2 1 2 1",
"output": "NO"
},
{
"input": "2\n5 1",
"output": "YES"
},
{
"input": "9\n1 2 3 4 4 3 2 2 1",
"output": "NO"
},
{
"input": "3\n2 2 3",
"output": "NO"
},
{
"input": "2\n5 4",
"output": "YES"
},
{
"input": "5\n1 3 3 2 2",
"output": "NO"
},
{
"input": "10\n1 2 3 4 5 6 7 8 9 99",
"output": "YES"
},
{
"input": "4\n1 2 3 4",
"output": "YES"
},
{
"input": "3\n5 5 2",
"output": "YES"
},
{
"input": "4\n1 4 2 3",
"output": "NO"
},
{
"input": "2\n3 2",
"output": "YES"
},
{
"input": "5\n1 2 2 1 1",
"output": "NO"
},
{
"input": "4\n3 3 2 2",
"output": "NO"
},
{
"input": "5\n1 2 3 2 2",
"output": "NO"
},
{
"input": "5\n5 6 6 5 5",
"output": "NO"
},
{
"input": "4\n2 2 1 1",
"output": "NO"
},
{
"input": "5\n5 4 3 3 2",
"output": "NO"
},
{
"input": "7\n1 3 3 3 2 1 1",
"output": "NO"
},
{
"input": "9\n5 6 6 5 5 4 4 3 3",
"output": "NO"
},
{
"input": "6\n1 5 5 3 2 2",
"output": "NO"
},
{
"input": "5\n2 1 3 3 1",
"output": "NO"
},
{
"input": "2\n4 3",
"output": "YES"
},
{
"input": "5\n3 2 2 1 1",
"output": "NO"
},
{
"input": "4\n5 4 3 2",
"output": "YES"
},
{
"input": "4\n4 4 1 1",
"output": "NO"
},
{
"input": "4\n3 3 1 1",
"output": "NO"
},
{
"input": "4\n4 4 2 2",
"output": "NO"
},
{
"input": "5\n4 4 3 2 2",
"output": "NO"
},
{
"input": "8\n4 4 4 4 5 6 7 8",
"output": "NO"
},
{
"input": "5\n3 5 4 4 3",
"output": "NO"
},
{
"input": "6\n2 5 3 3 2 2",
"output": "NO"
},
{
"input": "4\n5 5 2 2",
"output": "NO"
},
{
"input": "5\n1 2 2 3 5",
"output": "NO"
}
] | 1,651,685,616
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 101
| 46
| 0
|
# -*- coding: utf-8 -*-
"""831.ipynb
Automatically generated by Colaboratory.
Original file is located at
https://colab.research.google.com/drive/1hYxPSks58iBj6lqO-ZFgsg_7bcRbm3l3
"""
#https://codeforces.com/contest/831/problem/A Unimodal Array
a=int(input())
b=list(map(int,input().split()))
i=1
while i<a and b[i-1]<b[i]:
i+=1
while i<a and b[i-1]==b[i]:
i+=1
while i<a and b[i-1]>b[i]:
i+=1
if i==a:
print("Yes")
else:
print("No")
|
Title: Unimodal Array
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Array of integers is unimodal, if:
- it is strictly increasing in the beginning; - after that it is constant; - after that it is strictly decreasing.
The first block (increasing) and the last block (decreasing) may be absent. It is allowed that both of this blocks are absent.
For example, the following three arrays are unimodal: [5,<=7,<=11,<=11,<=2,<=1], [4,<=4,<=2], [7], but the following three are not unimodal: [5,<=5,<=6,<=6,<=1], [1,<=2,<=1,<=2], [4,<=5,<=5,<=6].
Write a program that checks if an array is unimodal.
Input Specification:
The first line contains integer *n* (1<=β€<=*n*<=β€<=100) β the number of elements in the array.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=β€<=*a**i*<=β€<=1<=000) β the elements of the array.
Output Specification:
Print "YES" if the given array is unimodal. Otherwise, print "NO".
You can output each letter in any case (upper or lower).
Demo Input:
['6\n1 5 5 5 4 2\n', '5\n10 20 30 20 10\n', '4\n1 2 1 2\n', '7\n3 3 3 3 3 3 3\n']
Demo Output:
['YES\n', 'YES\n', 'NO\n', 'YES\n']
Note:
In the first example the array is unimodal, because it is strictly increasing in the beginning (from position 1 to position 2, inclusively), that it is constant (from position 2 to position 4, inclusively) and then it is strictly decreasing (from position 4 to position 6, inclusively).
|
```python
# -*- coding: utf-8 -*-
"""831.ipynb
Automatically generated by Colaboratory.
Original file is located at
https://colab.research.google.com/drive/1hYxPSks58iBj6lqO-ZFgsg_7bcRbm3l3
"""
#https://codeforces.com/contest/831/problem/A Unimodal Array
a=int(input())
b=list(map(int,input().split()))
i=1
while i<a and b[i-1]<b[i]:
i+=1
while i<a and b[i-1]==b[i]:
i+=1
while i<a and b[i-1]>b[i]:
i+=1
if i==a:
print("Yes")
else:
print("No")
```
| 3
|
|
262
|
A
|
Roma and Lucky Numbers
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Roma (a popular Russian name that means 'Roman') loves the Little Lvov Elephant's lucky numbers.
Let us remind you that lucky numbers are positive integers whose decimal representation only contains lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Roma's got *n* positive integers. He wonders, how many of those integers have not more than *k* lucky digits? Help him, write the program that solves the problem.
|
The first line contains two integers *n*, *k* (1<=β€<=*n*,<=*k*<=β€<=100). The second line contains *n* integers *a**i* (1<=β€<=*a**i*<=β€<=109) β the numbers that Roma has.
The numbers in the lines are separated by single spaces.
|
In a single line print a single integer β the answer to the problem.
|
[
"3 4\n1 2 4\n",
"3 2\n447 44 77\n"
] |
[
"3\n",
"2\n"
] |
In the first sample all numbers contain at most four lucky digits, so the answer is 3.
In the second sample number 447 doesn't fit in, as it contains more than two lucky digits. All other numbers are fine, so the answer is 2.
| 500
|
[
{
"input": "3 4\n1 2 4",
"output": "3"
},
{
"input": "3 2\n447 44 77",
"output": "2"
},
{
"input": "2 2\n507978501 180480073",
"output": "2"
},
{
"input": "9 6\n655243746 167613748 1470546 57644035 176077477 56984809 44677 215706823 369042089",
"output": "9"
},
{
"input": "6 100\n170427799 37215529 675016434 168544291 683447134 950090227",
"output": "6"
},
{
"input": "4 2\n194041605 706221269 69909135 257655784",
"output": "3"
},
{
"input": "4 2\n9581849 67346651 530497 272158241",
"output": "4"
},
{
"input": "3 47\n378261451 163985731 230342101",
"output": "3"
},
{
"input": "2 3\n247776868 480572137",
"output": "1"
},
{
"input": "7 77\n366496749 549646417 278840199 119255907 33557677 379268590 150378796",
"output": "7"
},
{
"input": "40 31\n32230963 709031779 144328646 513494529 36547831 416998222 84161665 318773941 170724397 553666286 368402971 48581613 31452501 368026285 47903381 939151438 204145360 189920160 288159400 133145006 314295423 450219949 160203213 358403181 478734385 29331901 31051111 110710191 567314089 139695685 111511396 87708701 317333277 103301481 110400517 634446253 481551313 39202255 105948 738066085",
"output": "40"
},
{
"input": "1 8\n55521105",
"output": "1"
},
{
"input": "49 3\n34644511 150953622 136135827 144208961 359490601 86708232 719413689 188605873 64330753 488776302 104482891 63360106 437791390 46521319 70778345 339141601 136198441 292941209 299339510 582531183 555958105 437904637 74219097 439816011 236010407 122674666 438442529 186501223 63932449 407678041 596993853 92223251 849265278 480265849 30983497 330283357 186901672 20271344 794252593 123774176 27851201 52717531 479907210 196833889 149331196 82147847 255966471 278600081 899317843",
"output": "44"
},
{
"input": "26 2\n330381357 185218042 850474297 483015466 296129476 1205865 538807493 103205601 160403321 694220263 416255901 7245756 507755361 88187633 91426751 1917161 58276681 59540376 576539745 595950717 390256887 105690055 607818885 28976353 488947089 50643601",
"output": "22"
},
{
"input": "38 1\n194481717 126247087 815196361 106258801 381703249 283859137 15290101 40086151 213688513 577996947 513899717 371428417 107799271 11136651 5615081 323386401 381128815 34217126 17709913 520702093 201694245 570931849 169037023 417019726 282437316 7417126 271667553 11375851 185087449 410130883 383045677 5764771 905017051 328584026 215330671 299553233 15838255 234532105",
"output": "20"
},
{
"input": "44 9\n683216389 250581469 130029957 467020047 188395565 206237982 63257361 68314981 732878407 563579660 199133851 53045209 665723851 16273169 10806790 556633156 350593410 474645249 478790761 708234243 71841230 18090541 19836685 146373571 17947452 534010506 46933264 377035021 311636557 75193963 54321761 12759959 71120181 548816939 23608621 31876417 107672995 72575155 369667956 20574379 210596751 532163173 75726739 853719629",
"output": "44"
},
{
"input": "8 6\n204157376 10514197 65483881 347219841 263304577 296402721 11739011 229776191",
"output": "8"
},
{
"input": "38 29\n333702889 680931737 61137217 203030505 68728281 11414209 642645708 590904616 3042901 607198177 189041074 700764043 813035201 198341461 126403544 401436841 420826465 45046581 20249976 46978855 46397957 706610773 24701041 57954481 51603266 593109701 385569073 178982291 582152863 287317968 1474090 34825141 432421977 130257781 151516903 540852403 548392 117246529",
"output": "38"
},
{
"input": "19 3\n562569697 549131571 50676718 84501863 74567295 702372009 365895280 451459937 40378543 167666701 158635641 53639293 442332661 825055617 100109161 326616021 862332843 533271196 4791547",
"output": "18"
},
{
"input": "1 1\n44",
"output": "0"
},
{
"input": "1 1\n4",
"output": "1"
},
{
"input": "10 3\n444 447 774 777 7777 4447 4 7 7 4",
"output": "8"
}
] | 1,589,023,440
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 34
| 186
| 6,656,000
|
[n, k] = [int(x) for x in input().split()]
L = [x for x in input().split()]
c = 0
for i in L:
if i.count('4') + i.count('7') <= k: c += 1
print(c)
|
Title: Roma and Lucky Numbers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Roma (a popular Russian name that means 'Roman') loves the Little Lvov Elephant's lucky numbers.
Let us remind you that lucky numbers are positive integers whose decimal representation only contains lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Roma's got *n* positive integers. He wonders, how many of those integers have not more than *k* lucky digits? Help him, write the program that solves the problem.
Input Specification:
The first line contains two integers *n*, *k* (1<=β€<=*n*,<=*k*<=β€<=100). The second line contains *n* integers *a**i* (1<=β€<=*a**i*<=β€<=109) β the numbers that Roma has.
The numbers in the lines are separated by single spaces.
Output Specification:
In a single line print a single integer β the answer to the problem.
Demo Input:
['3 4\n1 2 4\n', '3 2\n447 44 77\n']
Demo Output:
['3\n', '2\n']
Note:
In the first sample all numbers contain at most four lucky digits, so the answer is 3.
In the second sample number 447 doesn't fit in, as it contains more than two lucky digits. All other numbers are fine, so the answer is 2.
|
```python
[n, k] = [int(x) for x in input().split()]
L = [x for x in input().split()]
c = 0
for i in L:
if i.count('4') + i.count('7') <= k: c += 1
print(c)
```
| 3
|
|
427
|
A
|
Police Recruits
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
The police department of your city has just started its journey. Initially, they donβt have any manpower. So, they started hiring new recruits in groups.
Meanwhile, crimes keeps occurring within the city. One member of the police force can investigate only one crime during his/her lifetime.
If there is no police officer free (isn't busy with crime) during the occurrence of a crime, it will go untreated.
Given the chronological order of crime occurrences and recruit hirings, find the number of crimes which will go untreated.
|
The first line of input will contain an integer *n*Β (1<=β€<=*n*<=β€<=105), the number of events. The next line will contain *n* space-separated integers.
If the integer is -1 then it means a crime has occurred. Otherwise, the integer will be positive, the number of officers recruited together at that time. No more than 10 officers will be recruited at a time.
|
Print a single integer, the number of crimes which will go untreated.
|
[
"3\n-1 -1 1\n",
"8\n1 -1 1 -1 -1 1 1 1\n",
"11\n-1 -1 2 -1 -1 -1 -1 -1 -1 -1 -1\n"
] |
[
"2\n",
"1\n",
"8\n"
] |
Lets consider the second example:
1. Firstly one person is hired. 1. Then crime appears, the last hired person will investigate this crime. 1. One more person is hired. 1. One more crime appears, the last hired person will investigate this crime. 1. Crime appears. There is no free policeman at the time, so this crime will go untreated. 1. One more person is hired. 1. One more person is hired. 1. One more person is hired.
The answer is one, as one crime (on step 5) will go untreated.
| 500
|
[
{
"input": "3\n-1 -1 1",
"output": "2"
},
{
"input": "8\n1 -1 1 -1 -1 1 1 1",
"output": "1"
},
{
"input": "11\n-1 -1 2 -1 -1 -1 -1 -1 -1 -1 -1",
"output": "8"
},
{
"input": "7\n-1 -1 1 1 -1 -1 1",
"output": "2"
},
{
"input": "21\n-1 -1 -1 -1 -1 3 2 -1 6 -1 -1 2 1 -1 2 2 1 6 5 -1 5",
"output": "5"
},
{
"input": "98\n-1 -1 1 -1 -1 -1 -1 1 -1 -1 1 -1 -1 1 -1 1 1 1 -1 1 1 1 1 1 -1 1 -1 -1 -1 -1 1 -1 -1 1 1 -1 1 1 1 -1 -1 -1 -1 -1 -1 1 -1 -1 -1 1 -1 1 -1 1 -1 1 1 1 1 1 1 1 -1 -1 1 1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 1 -1 1 1 1 -1 1 1 -1 -1 -1 1 1 1 -1 -1 -1 1 -1 1 1",
"output": "13"
},
{
"input": "3\n-1 5 4",
"output": "1"
},
{
"input": "146\n4 -1 -1 -1 -1 -1 -1 -1 -1 -1 4 -1 3 -1 3 -1 -1 1 4 -1 2 -1 -1 3 -1 -1 -1 4 1 -1 -1 -1 4 -1 -1 -1 -1 -1 -1 3 2 -1 3 2 4 5 2 4 1 5 -1 -1 2 -1 -1 1 -1 5 3 -1 1 2 2 3 5 3 -1 -1 3 -1 -1 3 5 5 -1 -1 5 -1 4 4 1 -1 -1 -1 2 1 -1 -1 -1 2 5 3 -1 -1 -1 3 -1 5 4 -1 1 -1 -1 3 -1 -1 3 1 1 2 -1 -1 -1 1 3 1 -1 2 -1 -1 5 5 -1 -1 3 4 5 1 -1 2 -1 -1 -1 3 -1 5 3 2 -1 2 -1 -1 5 -1 3 -1",
"output": "5"
},
{
"input": "1\n2",
"output": "0"
},
{
"input": "1\n-1",
"output": "1"
},
{
"input": "2\n1 -1",
"output": "0"
},
{
"input": "2\n-1 1",
"output": "1"
},
{
"input": "2\n1 1",
"output": "0"
},
{
"input": "2\n-1 -1",
"output": "2"
},
{
"input": "4\n10 -1 -1 -1",
"output": "0"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "4\n-1 -1 1 1",
"output": "2"
}
] | 1,694,688,779
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 63
| 78
| 13,209,600
|
n = int(input())
li = list(map(int, input().split()))
c = 0
k = 0
for i in range(n):
if li[i] == -1 and c < 1:
k += 1
else:
c += li[i]
print(k)
"""
4
abacaba
acaba
abacaba
acab
"""
|
Title: Police Recruits
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The police department of your city has just started its journey. Initially, they donβt have any manpower. So, they started hiring new recruits in groups.
Meanwhile, crimes keeps occurring within the city. One member of the police force can investigate only one crime during his/her lifetime.
If there is no police officer free (isn't busy with crime) during the occurrence of a crime, it will go untreated.
Given the chronological order of crime occurrences and recruit hirings, find the number of crimes which will go untreated.
Input Specification:
The first line of input will contain an integer *n*Β (1<=β€<=*n*<=β€<=105), the number of events. The next line will contain *n* space-separated integers.
If the integer is -1 then it means a crime has occurred. Otherwise, the integer will be positive, the number of officers recruited together at that time. No more than 10 officers will be recruited at a time.
Output Specification:
Print a single integer, the number of crimes which will go untreated.
Demo Input:
['3\n-1 -1 1\n', '8\n1 -1 1 -1 -1 1 1 1\n', '11\n-1 -1 2 -1 -1 -1 -1 -1 -1 -1 -1\n']
Demo Output:
['2\n', '1\n', '8\n']
Note:
Lets consider the second example:
1. Firstly one person is hired. 1. Then crime appears, the last hired person will investigate this crime. 1. One more person is hired. 1. One more crime appears, the last hired person will investigate this crime. 1. Crime appears. There is no free policeman at the time, so this crime will go untreated. 1. One more person is hired. 1. One more person is hired. 1. One more person is hired.
The answer is one, as one crime (on step 5) will go untreated.
|
```python
n = int(input())
li = list(map(int, input().split()))
c = 0
k = 0
for i in range(n):
if li[i] == -1 and c < 1:
k += 1
else:
c += li[i]
print(k)
"""
4
abacaba
acaba
abacaba
acab
"""
```
| 3
|
|
801
|
A
|
Vicious Keyboard
|
PROGRAMMING
| 1,100
|
[
"brute force"
] | null | null |
Tonio has a keyboard with only two letters, "V" and "K".
One day, he has typed out a string *s* with only these two letters. He really likes it when the string "VK" appears, so he wishes to change at most one letter in the string (or do no changes) to maximize the number of occurrences of that string. Compute the maximum number of times "VK" can appear as a substring (i.Β e. a letter "K" right after a letter "V") in the resulting string.
|
The first line will contain a string *s* consisting only of uppercase English letters "V" and "K" with length not less than 1 and not greater than 100.
|
Output a single integer, the maximum number of times "VK" can appear as a substring of the given string after changing at most one character.
|
[
"VK\n",
"VV\n",
"V\n",
"VKKKKKKKKKVVVVVVVVVK\n",
"KVKV\n"
] |
[
"1\n",
"1\n",
"0\n",
"3\n",
"1\n"
] |
For the first case, we do not change any letters. "VK" appears once, which is the maximum number of times it could appear.
For the second case, we can change the second character from a "V" to a "K". This will give us the string "VK". This has one occurrence of the string "VK" as a substring.
For the fourth case, we can change the fourth character from a "K" to a "V". This will give us the string "VKKVKKKKKKVVVVVVVVVK". This has three occurrences of the string "VK" as a substring. We can check no other moves can give us strictly more occurrences.
| 500
|
[
{
"input": "VK",
"output": "1"
},
{
"input": "VV",
"output": "1"
},
{
"input": "V",
"output": "0"
},
{
"input": "VKKKKKKKKKVVVVVVVVVK",
"output": "3"
},
{
"input": "KVKV",
"output": "1"
},
{
"input": "VKKVVVKVKVK",
"output": "5"
},
{
"input": "VKVVKVKVVKVKKKKVVVVVVVVKVKVVVVVVKKVKKVKVVKVKKVVVVKV",
"output": "14"
},
{
"input": "VVKKVKKVVKKVKKVKVVKKVKKVVKKVKVVKKVKKVKVVKKVVKKVKVVKKVKVVKKVVKVVKKVKKVKKVKKVKKVKVVKKVKKVKKVKKVKKVVKVK",
"output": "32"
},
{
"input": "KVVKKVKVKVKVKVKKVKVKVVKVKVVKVVKVKKVKVKVKVKVKVKVKVKVKVKVKVKVKVKVVKVKVVKKVKVKK",
"output": "32"
},
{
"input": "KVVVVVKKVKVVKVVVKVVVKKKVKKKVVKVKKKVKKKKVKVVVVVKKKVVVVKKVVVVKKKVKVVVVVVVKKVKVKKKVVKVVVKVVKK",
"output": "21"
},
{
"input": "VVVVVKKVKVKVKVVKVVKKVVKVKKKKKKKVKKKVVVVVVKKVVVKVKVVKVKKVVKVVVKKKKKVVVVVKVVVVKVVVKKVKKVKKKVKKVKKVVKKV",
"output": "25"
},
{
"input": "KKVVKVVKVVKKVVKKVKVVKKV",
"output": "7"
},
{
"input": "KKVVKKVVVKKVKKVKKVVVKVVVKKVKKVVVKKVVVKVVVKVVVKKVVVKKVVVKVVVKKVVVKVVKKVVVKKVVVKKVVKVVVKKVVKKVKKVVVKKV",
"output": "24"
},
{
"input": "KVKVKVKVKVKVKVKVKVKVVKVKVKVKVKVKVKVVKVKVKKVKVKVKVKVVKVKVKVKVKVKVKVKVKKVKVKVV",
"output": "35"
},
{
"input": "VKVVVKKKVKVVKVKVKVKVKVV",
"output": "9"
},
{
"input": "KKKKVKKVKVKVKKKVVVVKK",
"output": "6"
},
{
"input": "KVKVKKVVVVVVKKKVKKKKVVVVKVKKVKVVK",
"output": "9"
},
{
"input": "KKVKKVKKKVKKKVKKKVKVVVKKVVVVKKKVKKVVKVKKVKVKVKVVVKKKVKKKKKVVKVVKVVVKKVVKVVKKKKKVK",
"output": "22"
},
{
"input": "VVVKVKVKVVVVVKVVVKKVVVKVVVVVKKVVKVVVKVVVKVKKKVVKVVVVVKVVVVKKVVKVKKVVKKKVKVVKVKKKKVVKVVVKKKVKVKKKKKK",
"output": "25"
},
{
"input": "VKVVKVVKKKVVKVKKKVVKKKVVKVVKVVKKVKKKVKVKKKVVKVKKKVVKVVKKKVVKKKVKKKVVKKVVKKKVKVKKKVKKKVKKKVKVKKKVVKVK",
"output": "29"
},
{
"input": "KKVKVVVKKVV",
"output": "3"
},
{
"input": "VKVKVKVKVKVKVKVKVKVKVVKVKVKVKVKVK",
"output": "16"
},
{
"input": "VVKKKVVKKKVVKKKVVKKKVVKKKVVKKKVVKKKVVKKKVVKKKVVKKKVVKKKVVKKKVV",
"output": "13"
},
{
"input": "VVKKVKVKKKVVVKVVVKVKKVKKKVVVKVVKVKKVKKVKVKVVKKVVKKVKVVKKKVVKKVVVKVKVVVKVKVVKVKKVKKV",
"output": "26"
},
{
"input": "VVKVKKVVKKVVKKVVKKVVKKVKKVVKVKKVVKKVVKKVVKKVVKKVVKVVKKVVKVVKKVVKVVKKVVKKVKKVVKVVKKVVKVVKKVV",
"output": "26"
},
{
"input": "K",
"output": "0"
},
{
"input": "VKVK",
"output": "2"
},
{
"input": "VKVV",
"output": "2"
},
{
"input": "KV",
"output": "0"
},
{
"input": "KK",
"output": "1"
},
{
"input": "KKVK",
"output": "2"
},
{
"input": "KKKK",
"output": "1"
},
{
"input": "KKV",
"output": "1"
},
{
"input": "KKVKVK",
"output": "3"
},
{
"input": "VKKVK",
"output": "2"
},
{
"input": "VKKK",
"output": "2"
},
{
"input": "KKK",
"output": "1"
},
{
"input": "KVV",
"output": "1"
},
{
"input": "KKVKV",
"output": "2"
},
{
"input": "VVK",
"output": "1"
},
{
"input": "VVVKVKVKVKVKVKVK",
"output": "8"
},
{
"input": "KVVVK",
"output": "2"
},
{
"input": "VVVKK",
"output": "2"
},
{
"input": "KKVV",
"output": "1"
},
{
"input": "KKKKKKK",
"output": "1"
},
{
"input": "VKKKVK",
"output": "3"
},
{
"input": "KKVVV",
"output": "1"
},
{
"input": "VVVVVV",
"output": "1"
},
{
"input": "KKKV",
"output": "1"
},
{
"input": "VVKVV",
"output": "2"
},
{
"input": "VKVKKK",
"output": "3"
},
{
"input": "VKKV",
"output": "1"
},
{
"input": "VKKVV",
"output": "2"
},
{
"input": "VVKKVV",
"output": "2"
},
{
"input": "KKVVKKV",
"output": "2"
},
{
"input": "KKKKK",
"output": "1"
},
{
"input": "VKVVKKVKKVVKVKKVKKKVKKVKVKK",
"output": "10"
},
{
"input": "VKVKVV",
"output": "3"
},
{
"input": "VKVVKVV",
"output": "3"
},
{
"input": "VVV",
"output": "1"
},
{
"input": "VVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVV",
"output": "1"
},
{
"input": "VVKKKKKKVKK",
"output": "3"
},
{
"input": "KVKVKVV",
"output": "3"
}
] | 1,492,357,420
| 520
|
Python 3
|
OK
|
TESTS
| 63
| 77
| 5,529,600
|
l=input()
s=[]
for i in l:
if(i=='V'):
s.append(1)
else:
s.append(0)
ls=len(s)
i=0
maxc = 0
c = 0
while i+1<ls:
if(s[i]==1 and s[i+1]==0):
c+=1
i+=1
maxc = max(maxc,c)
j=0
while j<ls:
if(s[j]==1):
s[j]=0
else:
s[j]=1
i=0
c=0
while i+1<ls:
if(s[i]==1 and s[i+1]==0):
c+=1
i+=1
maxc=max(c,maxc)
if(s[j]==1):
s[j]=0
else:
s[j]=1
j+=1
print(maxc)
|
Title: Vicious Keyboard
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Tonio has a keyboard with only two letters, "V" and "K".
One day, he has typed out a string *s* with only these two letters. He really likes it when the string "VK" appears, so he wishes to change at most one letter in the string (or do no changes) to maximize the number of occurrences of that string. Compute the maximum number of times "VK" can appear as a substring (i.Β e. a letter "K" right after a letter "V") in the resulting string.
Input Specification:
The first line will contain a string *s* consisting only of uppercase English letters "V" and "K" with length not less than 1 and not greater than 100.
Output Specification:
Output a single integer, the maximum number of times "VK" can appear as a substring of the given string after changing at most one character.
Demo Input:
['VK\n', 'VV\n', 'V\n', 'VKKKKKKKKKVVVVVVVVVK\n', 'KVKV\n']
Demo Output:
['1\n', '1\n', '0\n', '3\n', '1\n']
Note:
For the first case, we do not change any letters. "VK" appears once, which is the maximum number of times it could appear.
For the second case, we can change the second character from a "V" to a "K". This will give us the string "VK". This has one occurrence of the string "VK" as a substring.
For the fourth case, we can change the fourth character from a "K" to a "V". This will give us the string "VKKVKKKKKKVVVVVVVVVK". This has three occurrences of the string "VK" as a substring. We can check no other moves can give us strictly more occurrences.
|
```python
l=input()
s=[]
for i in l:
if(i=='V'):
s.append(1)
else:
s.append(0)
ls=len(s)
i=0
maxc = 0
c = 0
while i+1<ls:
if(s[i]==1 and s[i+1]==0):
c+=1
i+=1
maxc = max(maxc,c)
j=0
while j<ls:
if(s[j]==1):
s[j]=0
else:
s[j]=1
i=0
c=0
while i+1<ls:
if(s[i]==1 and s[i+1]==0):
c+=1
i+=1
maxc=max(c,maxc)
if(s[j]==1):
s[j]=0
else:
s[j]=1
j+=1
print(maxc)
```
| 3
|
|
637
|
B
|
Chat Order
|
PROGRAMMING
| 1,200
|
[
"*special",
"binary search",
"constructive algorithms",
"data structures",
"sortings"
] | null | null |
Polycarp is a big lover of killing time in social networks. A page with a chatlist in his favourite network is made so that when a message is sent to some friend, his friend's chat rises to the very top of the page. The relative order of the other chats doesn't change. If there was no chat with this friend before, then a new chat is simply inserted to the top of the list.
Assuming that the chat list is initially empty, given the sequence of Polycaprus' messages make a list of chats after all of his messages are processed. Assume that no friend wrote any message to Polycarpus.
|
The first line contains integer *n* (1<=β€<=*n*<=β€<=200<=000)Β β the number of Polycarpus' messages. Next *n* lines enlist the message recipients in the order in which the messages were sent. The name of each participant is a non-empty sequence of lowercase English letters of length at most 10.
|
Print all the recipients to who Polycarp talked to in the order of chats with them, from top to bottom.
|
[
"4\nalex\nivan\nroman\nivan\n",
"8\nalina\nmaria\nekaterina\ndarya\ndarya\nekaterina\nmaria\nalina\n"
] |
[
"ivan\nroman\nalex\n",
"alina\nmaria\nekaterina\ndarya\n"
] |
In the first test case Polycarpus first writes to friend by name "alex", and the list looks as follows:
1. alex
Then Polycarpus writes to friend by name "ivan" and the list looks as follows:
1. ivan 1. alex
Polycarpus writes the third message to friend by name "roman" and the list looks as follows:
1. roman 1. ivan 1. alex
Polycarpus writes the fourth message to friend by name "ivan", to who he has already sent a message, so the list of chats changes as follows:
1. ivan 1. roman 1. alex
| 1,000
|
[
{
"input": "4\nalex\nivan\nroman\nivan",
"output": "ivan\nroman\nalex"
},
{
"input": "8\nalina\nmaria\nekaterina\ndarya\ndarya\nekaterina\nmaria\nalina",
"output": "alina\nmaria\nekaterina\ndarya"
},
{
"input": "1\nwdi",
"output": "wdi"
},
{
"input": "2\nypg\nypg",
"output": "ypg"
},
{
"input": "3\nexhll\nexhll\narruapexj",
"output": "arruapexj\nexhll"
},
{
"input": "3\nfv\nle\nle",
"output": "le\nfv"
},
{
"input": "8\nm\nm\nm\nm\nm\nm\nm\nm",
"output": "m"
},
{
"input": "10\nr\nr\ni\nw\nk\nr\nb\nu\nu\nr",
"output": "r\nu\nb\nk\nw\ni"
},
{
"input": "7\ne\nfau\ncmk\nnzs\nby\nwx\ntjmok",
"output": "tjmok\nwx\nby\nnzs\ncmk\nfau\ne"
},
{
"input": "6\nklrj\nwe\nklrj\nwe\nwe\nwe",
"output": "we\nklrj"
},
{
"input": "8\nzncybqmh\naeebef\nzncybqmh\nn\naeebef\nzncybqmh\nzncybqmh\nzncybqmh",
"output": "zncybqmh\naeebef\nn"
},
{
"input": "30\nkqqcbs\nvap\nkymomn\nj\nkqqcbs\nfuzlzoum\nkymomn\ndbh\nfuzlzoum\nkymomn\nvap\nvlgzs\ndbh\nvlgzs\nbvy\ndbh\nkymomn\nkymomn\neoqql\nkymomn\nkymomn\nkqqcbs\nvlgzs\nkqqcbs\nkqqcbs\nfuzlzoum\nvlgzs\nrylgdoo\nvlgzs\nrylgdoo",
"output": "rylgdoo\nvlgzs\nfuzlzoum\nkqqcbs\nkymomn\neoqql\ndbh\nbvy\nvap\nj"
},
{
"input": "40\nji\nv\nv\nns\nji\nn\nji\nv\nfvy\nvje\nns\nvje\nv\nhas\nv\nusm\nhas\nfvy\nvje\nkdb\nn\nv\nji\nji\nn\nhas\nv\nji\nkdb\nr\nvje\nns\nv\nusm\nn\nvje\nhas\nns\nhas\nn",
"output": "n\nhas\nns\nvje\nusm\nv\nr\nkdb\nji\nfvy"
},
{
"input": "50\njcg\nvle\njopb\nepdb\nnkef\nfv\nxj\nufe\nfuy\noqta\ngbc\nyuz\nec\nyji\nkuux\ncwm\ntq\nnno\nhp\nzry\nxxpp\ntjvo\ngyz\nkwo\nvwqz\nyaqc\njnj\nwoav\nqcv\ndcu\ngc\nhovn\nop\nevy\ndc\ntrpu\nyb\nuzfa\npca\noq\nnhxy\nsiqu\nde\nhphy\nc\nwovu\nf\nbvv\ndsik\nlwyg",
"output": "lwyg\ndsik\nbvv\nf\nwovu\nc\nhphy\nde\nsiqu\nnhxy\noq\npca\nuzfa\nyb\ntrpu\ndc\nevy\nop\nhovn\ngc\ndcu\nqcv\nwoav\njnj\nyaqc\nvwqz\nkwo\ngyz\ntjvo\nxxpp\nzry\nhp\nnno\ntq\ncwm\nkuux\nyji\nec\nyuz\ngbc\noqta\nfuy\nufe\nxj\nfv\nnkef\nepdb\njopb\nvle\njcg"
},
{
"input": "100\nvhh\nvhh\nvhh\nfa\nfa\nvhh\nvhh\nvhh\nfa\nfa\nfa\nvhh\nfa\nvhh\nvhh\nvhh\nfa\nvhh\nvhh\nfa\nfa\nfa\nfa\nfa\nfa\nvhh\nfa\nfa\nvhh\nvhh\nvhh\nfa\nfa\nfa\nvhh\nfa\nvhh\nfa\nvhh\nvhh\nfa\nvhh\nfa\nvhh\nvhh\nvhh\nfa\nvhh\nfa\nfa\nvhh\nfa\nvhh\nvhh\nvhh\nvhh\nfa\nvhh\nvhh\nvhh\nvhh\nfa\nvhh\nvhh\nvhh\nvhh\nvhh\nfa\nvhh\nvhh\nfa\nfa\nfa\nvhh\nfa\nfa\nvhh\nfa\nvhh\nfa\nfa\nfa\nfa\nfa\nfa\nvhh\nvhh\nfa\nvhh\nfa\nfa\nvhh\nfa\nfa\nvhh\nfa\nvhh\nvhh\nfa\nvhh",
"output": "vhh\nfa"
},
{
"input": "2\naa\nbb",
"output": "bb\naa"
},
{
"input": "2\naa\na",
"output": "a\naa"
},
{
"input": "3\naa\naa\naa",
"output": "aa"
},
{
"input": "5\naa\na\naa\na\naa",
"output": "aa\na"
},
{
"input": "7\naaaa\naaaa\naaa\na\naa\naaaaaaa\naaa",
"output": "aaa\naaaaaaa\naa\na\naaaa"
},
{
"input": "5\na\naa\naaa\naaaa\na",
"output": "a\naaaa\naaa\naa"
},
{
"input": "12\naaaaa\naaaaaa\naaaa\naaaaaa\naa\naaaa\naaaa\naaaaaa\na\naaa\naaaaaaaa\naa",
"output": "aa\naaaaaaaa\naaa\na\naaaaaa\naaaa\naaaaa"
},
{
"input": "3\na\naa\naaa",
"output": "aaa\naa\na"
},
{
"input": "9\nzzz\nzzzzz\nzzz\nzzzz\nzz\nzzzz\nzzzzz\nzzzz\nzzzzzzz",
"output": "zzzzzzz\nzzzz\nzzzzz\nzz\nzzz"
}
] | 1,690,058,110
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 95
| 514
| 28,569,600
|
n = int(input())
v = []
for i in range(n):
name = input()
v.append(name)
# v contains all names in the initial order
# now we have to process the list in reverse order
s = set() # should use set() for set
for i in range(n-1, -1, -1): # start from the last index, dec by 1, stop at -1
name = v[i]
if name not in s:
print(name)
s.add(name)
|
Title: Chat Order
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp is a big lover of killing time in social networks. A page with a chatlist in his favourite network is made so that when a message is sent to some friend, his friend's chat rises to the very top of the page. The relative order of the other chats doesn't change. If there was no chat with this friend before, then a new chat is simply inserted to the top of the list.
Assuming that the chat list is initially empty, given the sequence of Polycaprus' messages make a list of chats after all of his messages are processed. Assume that no friend wrote any message to Polycarpus.
Input Specification:
The first line contains integer *n* (1<=β€<=*n*<=β€<=200<=000)Β β the number of Polycarpus' messages. Next *n* lines enlist the message recipients in the order in which the messages were sent. The name of each participant is a non-empty sequence of lowercase English letters of length at most 10.
Output Specification:
Print all the recipients to who Polycarp talked to in the order of chats with them, from top to bottom.
Demo Input:
['4\nalex\nivan\nroman\nivan\n', '8\nalina\nmaria\nekaterina\ndarya\ndarya\nekaterina\nmaria\nalina\n']
Demo Output:
['ivan\nroman\nalex\n', 'alina\nmaria\nekaterina\ndarya\n']
Note:
In the first test case Polycarpus first writes to friend by name "alex", and the list looks as follows:
1. alex
Then Polycarpus writes to friend by name "ivan" and the list looks as follows:
1. ivan 1. alex
Polycarpus writes the third message to friend by name "roman" and the list looks as follows:
1. roman 1. ivan 1. alex
Polycarpus writes the fourth message to friend by name "ivan", to who he has already sent a message, so the list of chats changes as follows:
1. ivan 1. roman 1. alex
|
```python
n = int(input())
v = []
for i in range(n):
name = input()
v.append(name)
# v contains all names in the initial order
# now we have to process the list in reverse order
s = set() # should use set() for set
for i in range(n-1, -1, -1): # start from the last index, dec by 1, stop at -1
name = v[i]
if name not in s:
print(name)
s.add(name)
```
| 3
|
|
385
|
A
|
Bear and Raspberry
|
PROGRAMMING
| 1,000
|
[
"brute force",
"greedy",
"implementation"
] | null | null |
The bear decided to store some raspberry for the winter. He cunningly found out the price for a barrel of honey in kilos of raspberry for each of the following *n* days. According to the bear's data, on the *i*-th (1<=β€<=*i*<=β€<=*n*) day, the price for one barrel of honey is going to is *x**i* kilos of raspberry.
Unfortunately, the bear has neither a honey barrel, nor the raspberry. At the same time, the bear's got a friend who is ready to lend him a barrel of honey for exactly one day for *c* kilograms of raspberry. That's why the bear came up with a smart plan. He wants to choose some day *d* (1<=β€<=*d*<=<<=*n*), lent a barrel of honey and immediately (on day *d*) sell it according to a daily exchange rate. The next day (*d*<=+<=1) the bear wants to buy a new barrel of honey according to a daily exchange rate (as he's got some raspberry left from selling the previous barrel) and immediately (on day *d*<=+<=1) give his friend the borrowed barrel of honey as well as *c* kilograms of raspberry for renting the barrel.
The bear wants to execute his plan at most once and then hibernate. What maximum number of kilograms of raspberry can he earn? Note that if at some point of the plan the bear runs out of the raspberry, then he won't execute such a plan.
|
The first line contains two space-separated integers, *n* and *c* (2<=β€<=*n*<=β€<=100,<=0<=β€<=*c*<=β€<=100), β the number of days and the number of kilos of raspberry that the bear should give for borrowing the barrel.
The second line contains *n* space-separated integers *x*1,<=*x*2,<=...,<=*x**n* (0<=β€<=*x**i*<=β€<=100), the price of a honey barrel on day *i*.
|
Print a single integer β the answer to the problem.
|
[
"5 1\n5 10 7 3 20\n",
"6 2\n100 1 10 40 10 40\n",
"3 0\n1 2 3\n"
] |
[
"3\n",
"97\n",
"0\n"
] |
In the first sample the bear will lend a honey barrel at day 3 and then sell it for 7. Then the bear will buy a barrel for 3 and return it to the friend. So, the profit is (7 - 3 - 1) = 3.
In the second sample bear will lend a honey barrel at day 1 and then sell it for 100. Then the bear buy the barrel for 1 at the day 2. So, the profit is (100 - 1 - 2) = 97.
| 500
|
[
{
"input": "5 1\n5 10 7 3 20",
"output": "3"
},
{
"input": "6 2\n100 1 10 40 10 40",
"output": "97"
},
{
"input": "3 0\n1 2 3",
"output": "0"
},
{
"input": "2 0\n2 1",
"output": "1"
},
{
"input": "10 5\n10 1 11 2 12 3 13 4 14 5",
"output": "4"
},
{
"input": "100 4\n2 57 70 8 44 10 88 67 50 44 93 79 72 50 69 19 21 9 71 47 95 13 46 10 68 72 54 40 15 83 57 92 58 25 4 22 84 9 8 55 87 0 16 46 86 58 5 21 32 28 10 46 11 29 13 33 37 34 78 33 33 21 46 70 77 51 45 97 6 21 68 61 87 54 8 91 37 12 76 61 57 9 100 45 44 88 5 71 98 98 26 45 37 87 34 50 33 60 64 77",
"output": "87"
},
{
"input": "100 5\n15 91 86 53 18 52 26 89 8 4 5 100 11 64 88 91 35 57 67 72 71 71 69 73 97 23 11 1 59 86 37 82 6 67 71 11 7 31 11 68 21 43 89 54 27 10 3 33 8 57 79 26 90 81 6 28 24 7 33 50 24 13 27 85 4 93 14 62 37 67 33 40 7 48 41 4 14 9 95 10 64 62 7 93 23 6 28 27 97 64 26 83 70 0 97 74 11 82 70 93",
"output": "84"
},
{
"input": "6 100\n10 9 8 7 6 5",
"output": "0"
},
{
"input": "100 9\n66 71 37 41 23 38 77 11 74 13 51 26 93 56 81 17 12 70 85 37 54 100 14 99 12 83 44 16 99 65 13 48 92 32 69 33 100 57 58 88 25 45 44 85 5 41 82 15 37 18 21 45 3 68 33 9 52 64 8 73 32 41 87 99 26 26 47 24 79 93 9 44 11 34 85 26 14 61 49 38 25 65 49 81 29 82 28 23 2 64 38 13 77 68 67 23 58 57 83 46",
"output": "78"
},
{
"input": "100 100\n9 72 46 37 26 94 80 1 43 85 26 53 58 18 24 19 67 2 100 52 61 81 48 15 73 41 97 93 45 1 73 54 75 51 28 79 0 14 41 42 24 50 70 18 96 100 67 1 68 48 44 39 63 77 78 18 10 51 32 53 26 60 1 13 66 39 55 27 23 71 75 0 27 88 73 31 16 95 87 84 86 71 37 40 66 70 65 83 19 4 81 99 26 51 67 63 80 54 23 44",
"output": "0"
},
{
"input": "43 65\n32 58 59 75 85 18 57 100 69 0 36 38 79 95 82 47 7 55 28 88 27 88 63 71 80 86 67 53 69 37 99 54 81 19 55 12 2 17 84 77 25 26 62",
"output": "4"
},
{
"input": "12 64\n14 87 40 24 32 36 4 41 38 77 68 71",
"output": "0"
},
{
"input": "75 94\n80 92 25 48 78 17 69 52 79 73 12 15 59 55 25 61 96 27 98 43 30 43 36 94 67 54 86 99 100 61 65 8 65 19 18 21 75 31 2 98 55 87 14 1 17 97 94 11 57 29 34 71 76 67 45 0 78 29 86 82 29 23 77 100 48 43 65 62 88 34 7 28 13 1 1",
"output": "0"
},
{
"input": "59 27\n76 61 24 66 48 18 69 84 21 8 64 90 19 71 36 90 9 36 30 37 99 37 100 56 9 79 55 37 54 63 11 11 49 71 91 70 14 100 10 44 52 23 21 19 96 13 93 66 52 79 76 5 62 6 90 35 94 7 27",
"output": "63"
},
{
"input": "86 54\n41 84 16 5 20 79 73 13 23 24 42 73 70 80 69 71 33 44 62 29 86 88 40 64 61 55 58 19 16 23 84 100 38 91 89 98 47 50 55 87 12 94 2 12 0 1 4 26 50 96 68 34 94 80 8 22 60 3 72 84 65 89 44 52 50 9 24 34 81 28 56 17 38 85 78 90 62 60 1 40 91 2 7 41 84 22",
"output": "38"
},
{
"input": "37 2\n65 36 92 92 92 76 63 56 15 95 75 26 15 4 73 50 41 92 26 20 19 100 63 55 25 75 61 96 35 0 14 6 96 3 28 41 83",
"output": "91"
},
{
"input": "19 4\n85 2 56 70 33 75 89 60 100 81 42 28 18 92 29 96 49 23 14",
"output": "79"
},
{
"input": "89 1\n50 53 97 41 68 27 53 66 93 19 11 78 46 49 38 69 96 9 43 16 1 63 95 64 96 6 34 34 45 40 19 4 53 8 11 18 95 25 50 16 64 33 97 49 23 81 63 10 30 73 76 55 7 70 9 98 6 36 75 78 3 92 85 75 40 75 55 71 9 91 15 17 47 55 44 35 55 88 53 87 61 22 100 56 14 87 36 84 24",
"output": "91"
},
{
"input": "67 0\n40 48 15 46 90 7 65 52 24 15 42 81 2 6 71 94 32 18 97 67 83 98 48 51 10 47 8 68 36 46 65 75 90 30 62 9 5 35 80 60 69 58 62 68 58 73 80 9 22 46 56 64 44 11 93 73 62 54 15 20 17 69 16 33 85 62 49",
"output": "83"
},
{
"input": "96 0\n38 97 82 43 80 40 1 99 50 94 81 63 92 13 57 24 4 10 25 32 79 56 96 19 25 14 69 56 66 22 23 78 87 76 37 30 75 77 61 64 35 64 62 32 44 62 6 84 91 44 99 5 71 19 17 12 35 52 1 14 35 18 8 36 54 42 4 67 80 11 88 44 34 35 12 38 66 42 4 90 45 10 1 44 37 96 23 28 100 90 75 17 27 67 51 70",
"output": "94"
},
{
"input": "14 14\n87 63 62 31 59 47 40 89 92 43 80 30 99 42",
"output": "43"
},
{
"input": "12 0\n100 1 100 2 100 3 100 4 100 5 100 0",
"output": "100"
},
{
"input": "3 1\n1 2 3",
"output": "0"
},
{
"input": "3 2\n3 3 3",
"output": "0"
},
{
"input": "3 3\n3 2 1",
"output": "0"
},
{
"input": "3 100\n1 2 3",
"output": "0"
},
{
"input": "2 100\n0 0",
"output": "0"
},
{
"input": "2 90\n10 5",
"output": "0"
},
{
"input": "2 5\n5 4",
"output": "0"
},
{
"input": "3 1\n19 20 1",
"output": "18"
},
{
"input": "5 1\n5 10 7 4 20",
"output": "2"
},
{
"input": "5 1\n1 2 3 4 5",
"output": "0"
}
] | 1,398,866,754
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 32
| 77
| 0
|
s = input().rstrip().split(' ')
n, c = int (s[0]), int(s[1])
x = input().rstrip().split(' ')
max = int (x[0]) - int (x[1]) - c
for i in range (1, n - 1):
if int (x[i]) - int (x[i + 1]) - c > max:
max = int (x[i]) - int (x[i+1]) - c
if max < 0:
max = 0
print (max)
|
Title: Bear and Raspberry
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The bear decided to store some raspberry for the winter. He cunningly found out the price for a barrel of honey in kilos of raspberry for each of the following *n* days. According to the bear's data, on the *i*-th (1<=β€<=*i*<=β€<=*n*) day, the price for one barrel of honey is going to is *x**i* kilos of raspberry.
Unfortunately, the bear has neither a honey barrel, nor the raspberry. At the same time, the bear's got a friend who is ready to lend him a barrel of honey for exactly one day for *c* kilograms of raspberry. That's why the bear came up with a smart plan. He wants to choose some day *d* (1<=β€<=*d*<=<<=*n*), lent a barrel of honey and immediately (on day *d*) sell it according to a daily exchange rate. The next day (*d*<=+<=1) the bear wants to buy a new barrel of honey according to a daily exchange rate (as he's got some raspberry left from selling the previous barrel) and immediately (on day *d*<=+<=1) give his friend the borrowed barrel of honey as well as *c* kilograms of raspberry for renting the barrel.
The bear wants to execute his plan at most once and then hibernate. What maximum number of kilograms of raspberry can he earn? Note that if at some point of the plan the bear runs out of the raspberry, then he won't execute such a plan.
Input Specification:
The first line contains two space-separated integers, *n* and *c* (2<=β€<=*n*<=β€<=100,<=0<=β€<=*c*<=β€<=100), β the number of days and the number of kilos of raspberry that the bear should give for borrowing the barrel.
The second line contains *n* space-separated integers *x*1,<=*x*2,<=...,<=*x**n* (0<=β€<=*x**i*<=β€<=100), the price of a honey barrel on day *i*.
Output Specification:
Print a single integer β the answer to the problem.
Demo Input:
['5 1\n5 10 7 3 20\n', '6 2\n100 1 10 40 10 40\n', '3 0\n1 2 3\n']
Demo Output:
['3\n', '97\n', '0\n']
Note:
In the first sample the bear will lend a honey barrel at day 3 and then sell it for 7. Then the bear will buy a barrel for 3 and return it to the friend. So, the profit is (7 - 3 - 1) = 3.
In the second sample bear will lend a honey barrel at day 1 and then sell it for 100. Then the bear buy the barrel for 1 at the day 2. So, the profit is (100 - 1 - 2) = 97.
|
```python
s = input().rstrip().split(' ')
n, c = int (s[0]), int(s[1])
x = input().rstrip().split(' ')
max = int (x[0]) - int (x[1]) - c
for i in range (1, n - 1):
if int (x[i]) - int (x[i + 1]) - c > max:
max = int (x[i]) - int (x[i+1]) - c
if max < 0:
max = 0
print (max)
```
| 3
|
|
977
|
A
|
Wrong Subtraction
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Little girl Tanya is learning how to decrease a number by one, but she does it wrong with a number consisting of two or more digits. Tanya subtracts one from a number by the following algorithm:
- if the last digit of the number is non-zero, she decreases the number by one; - if the last digit of the number is zero, she divides the number by 10 (i.e. removes the last digit).
You are given an integer number $n$. Tanya will subtract one from it $k$ times. Your task is to print the result after all $k$ subtractions.
It is guaranteed that the result will be positive integer number.
|
The first line of the input contains two integer numbers $n$ and $k$ ($2 \le n \le 10^9$, $1 \le k \le 50$) β the number from which Tanya will subtract and the number of subtractions correspondingly.
|
Print one integer number β the result of the decreasing $n$ by one $k$ times.
It is guaranteed that the result will be positive integer number.
|
[
"512 4\n",
"1000000000 9\n"
] |
[
"50\n",
"1\n"
] |
The first example corresponds to the following sequence: $512 \rightarrow 511 \rightarrow 510 \rightarrow 51 \rightarrow 50$.
| 0
|
[
{
"input": "512 4",
"output": "50"
},
{
"input": "1000000000 9",
"output": "1"
},
{
"input": "131203 11",
"output": "12"
},
{
"input": "999999999 50",
"output": "9999"
},
{
"input": "999999999 49",
"output": "99990"
},
{
"input": "131203 9",
"output": "130"
},
{
"input": "900000000 16",
"output": "1"
},
{
"input": "909090909 50",
"output": "3"
},
{
"input": "1001 2",
"output": "100"
},
{
"input": "5 2",
"output": "3"
},
{
"input": "2 1",
"output": "1"
}
] | 1,694,080,862
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 11
| 46
| 0
|
a=str(input(""))
b=0
f=-1
c=''
d=''
while a[f]!=" ":
d=a[f]+d
f=f-1
while a[b]!=" " :
c=c+a[b]
b=b+1
e=int(c)
g=int(d)
for i in range(0,g):
if e%10==0:
e=e/10
else:
e=e-1
print(int(e))
|
Title: Wrong Subtraction
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little girl Tanya is learning how to decrease a number by one, but she does it wrong with a number consisting of two or more digits. Tanya subtracts one from a number by the following algorithm:
- if the last digit of the number is non-zero, she decreases the number by one; - if the last digit of the number is zero, she divides the number by 10 (i.e. removes the last digit).
You are given an integer number $n$. Tanya will subtract one from it $k$ times. Your task is to print the result after all $k$ subtractions.
It is guaranteed that the result will be positive integer number.
Input Specification:
The first line of the input contains two integer numbers $n$ and $k$ ($2 \le n \le 10^9$, $1 \le k \le 50$) β the number from which Tanya will subtract and the number of subtractions correspondingly.
Output Specification:
Print one integer number β the result of the decreasing $n$ by one $k$ times.
It is guaranteed that the result will be positive integer number.
Demo Input:
['512 4\n', '1000000000 9\n']
Demo Output:
['50\n', '1\n']
Note:
The first example corresponds to the following sequence: $512 \rightarrow 511 \rightarrow 510 \rightarrow 51 \rightarrow 50$.
|
```python
a=str(input(""))
b=0
f=-1
c=''
d=''
while a[f]!=" ":
d=a[f]+d
f=f-1
while a[b]!=" " :
c=c+a[b]
b=b+1
e=int(c)
g=int(d)
for i in range(0,g):
if e%10==0:
e=e/10
else:
e=e-1
print(int(e))
```
| 3
|
|
276
|
B
|
Little Girl and Game
|
PROGRAMMING
| 1,300
|
[
"games",
"greedy"
] | null | null |
The Little Girl loves problems on games very much. Here's one of them.
Two players have got a string *s*, consisting of lowercase English letters. They play a game that is described by the following rules:
- The players move in turns; In one move the player can remove an arbitrary letter from string *s*. - If the player before his turn can reorder the letters in string *s* so as to get a palindrome, this player wins. A palindrome is a string that reads the same both ways (from left to right, and vice versa). For example, string "abba" is a palindrome and string "abc" isn't.
Determine which player will win, provided that both sides play optimally well β the one who moves first or the one who moves second.
|
The input contains a single line, containing string *s* (1<=β€<=|*s*|<=<=β€<=<=103). String *s* consists of lowercase English letters.
|
In a single line print word "First" if the first player wins (provided that both players play optimally well). Otherwise, print word "Second". Print the words without the quotes.
|
[
"aba\n",
"abca\n"
] |
[
"First\n",
"Second\n"
] |
none
| 1,000
|
[
{
"input": "aba",
"output": "First"
},
{
"input": "abca",
"output": "Second"
},
{
"input": "aabb",
"output": "First"
},
{
"input": "ctjxzuimsxnarlciuynqeoqmmbqtagszuo",
"output": "Second"
},
{
"input": "gevqgtaorjixsxnbcoybr",
"output": "First"
},
{
"input": "xvhtcbtouuddhylxhplgjxwlo",
"output": "First"
},
{
"input": "knaxhkbokmtfvnjvlsbrfoefpjpkqwlumeqqbeohodnwevhllkylposdpjuoizyunuxivzrjofiyxxiliuwhkjqpkqxukxroivfhikxjdtwcqngqswptdwrywxszxrqojjphzwzxqftnfhkapeejdgckfyrxtpuipfljsjwgpjfatmxpylpnerllshuvkbomlpghjrxcgxvktgeyuhrcwgvdmppqnkdmjtxukzlzqhfbgrishuhkyggkpstvqabpxoqjuovwjwcmazmvpfpnljdgpokpatjnvwacotkvxheorzbsrazldsquijzkmtmqahakjrjvzkquvayxpqrmqqcknilpqpjapagezonfpz",
"output": "Second"
},
{
"input": "desktciwoidfuswycratvovutcgjrcyzmilsmadzaegseetexygedzxdmorxzxgiqhcuppshcsjcozkopebegfmxzxxagzwoymlghgjexcgfojychyt",
"output": "First"
},
{
"input": "gfhuidxgxpxduqrfnqrnefgtyxgmrtehmddjkddwdiayyilaknxhlxszeslnsjpcrwnoqubmbpcehiftteirkfvbtfyibiikdaxmondnawtvqccctdxrjcfxqwqhvvrqmhqflbzskrayvruqvqijrmikucwzodxvufwxpxxjxlifdjzxrttjzatafkbzsjupsiefmipdufqltedjlytphzppoevxawjdhbxgennevbvdgpoeihasycctyddenzypoprchkoioouhcexjqwjflxvkgpgjatstlmledxasecfhwvabzwviywsiaryqrxyeceefblherqjevdzkfxslqiytwzz",
"output": "First"
},
{
"input": "fezzkpyctjvvqtncmmjsitrxaliyhirspnjjngvzdoudrkkvvdiwcwtcxobpobzukegtcrwsgxxzlcphdxkbxdximqbycaicfdeqlvzboptfimkzvjzdsvahorqqhcirpkhtwjkplitpacpkpbhnxtoxuoqsxcxnhtrmzvexmpvlethbkvmlzftimjnidrzvcunbpysvukzgwghjmwrvstsunaocnoqohcsggtrwxiworkliqejajewbrtdwgnyynpupbrrvtfqtlaaq",
"output": "Second"
},
{
"input": "tsvxmeixijyavdalmrvscwohzubhhgsocdvnjmjtctojbxxpezzbgfltixwgzmkfwdnlhidhrdgyajggmrvmwaoydodjmzqvgabyszfqcuhwdncyfqvmackvijgpjyiauxljvvwgiofdxccwmybdfcfcrqppbvbagmnvvvhngxauwbpourviyfokwjweypzzrrzjcmddnpoaqgqfgglssjnlshrerfffmrwhapzknxveiqixflykjbnpivogtdpyjakwrdoklsbvbkjhdojfnuwbpcfdycwxecysbyjfvoykxsxgg",
"output": "First"
},
{
"input": "upgqmhfmfnodsyosgqswugfvpdxhtkxvhlsxrjiqlojchoddxkpsamwmuvopdbncymcgrkurwlxerexgswricuqxhvqvgekeofkgqabypamozmyjyfvpifsaotnyzqydcenphcsmplekinwkmwzpjnlapfdbhxjdcnarlgkfgxzfbpgsuxqfyhnxjhtojrlnprnxprfbkkcyriqztjeeepkzgzcaiutvbqqofyhddfebozhvtvrigtidxqmydjxegxipakzjcnenjkdroyjmxugj",
"output": "Second"
},
{
"input": "aaaaaaaaaaaaaaaaaaaabbbbbbbbbbbbbbbbbbbbccccccccccccccccccccddddddddddeeeeeeeeeeffffgggghhhhiiiijjjjqqqqwwwweeeerrrrttttyyyyuuuuiiiiooooppppaaaassssddddffffgggghhhhjjjjkkkkllllzzzzxxxxccccvvvvbbbbnnnnmmmm",
"output": "First"
},
{
"input": "vnvtvnxjrtffdhrfvczzoyeokjabxcilmmsrhwuakghvuabcmfpmblyroodmhfivmhqoiqhapoglwaluewhqkunzitmvijaictjdncivccedfpaezcnpwemlohbhjjlqsonuclaumgbzjamsrhuzqdqtitygggsnruuccdtxkgbdd",
"output": "First"
},
{
"input": "vqdtkbvlbdyndheoiiwqhnvcmmhnhsmwwrvesnpdfxvprqbwzbodoihrywagphlsrcbtnvppjsquuuzkjazaenienjiyctyajsqdfsdiedzugkymgzllvpxfetkwfabbiotjcknzdwsvmbbuqrxrulvgljagvxdmfsqtcczhifhoghqgffkbviphbabwiaqburerfkbqfjbptkwlahysrrfwjbqfnrgnsnsukqqcxxwqtuhvdzqmpfwrbqzdwxcaifuyhvojgurmchh",
"output": "First"
},
{
"input": "hxueikegwnrctlciwguepdsgupguykrntbszeqzzbpdlouwnmqgzcxejidstxyxhdlnttnibxstduwiflouzfswfikdudkazoefawm",
"output": "Second"
},
{
"input": "ershkhsywqftixappwqzoojtnamvqjbyfauvuubwpctspioqusnnivwsiyszfhlrskbswaiaczurygcioonjcndntwvrlaejyrghfnecltqytfmkvjxuujifgtujrqsisdawpwgttxynewiqhdhronamabysvpxankxeybcjqttbqnciwuqiehzyfjoedaradqnfthuuwrezwrkjiytpgwfwbslawbiezdbdltenjlaygwaxddplgseiaojndqjcopvolqbvnacuvfvirzbrnlnyjixngeevcggmirzatenjihpgnyfjhgsjgzepohbyhmzbatfwuorwutavlqsogrvcjpqziuifrhurq",
"output": "First"
},
{
"input": "qilwpsuxogazrfgfznngwklnioueuccyjfatjoizcctgsweitzofwkyjustizbopzwtaqxbtovkdrxeplukrcuozhpymldstbbfynkgsmafigetvzkxloxqtphvtwkgfjkiczttcsxkjpsoutdpzxytrsqgjtbdljjrbmkudrkodfvcwkcuggbsthxdyogeeyfuyhmnwgyuatfkvchavpzadfacckdurlbqjkthqbnirzzbpusxcenkpgtizayjmsahvobobudfeaewcqmrlxxnocqzmkessnguxkiccrxyvnxxlqnqfwuzmupk",
"output": "First"
},
{
"input": "opfokvwzpllctflkphutcrkferbjyyrasqqkrcvoymyrxwaudgsugcqveccymdplxmtlzfoptmrapfeizpnnhbzlkuyznwacnswibxhqunazbhdvrlidghisuqunstbuevjzimvlfvopgqxrvahhngnaumgywscfrfwfpnfxwhfrelbunmedvkssykwjyartxjiplerntzkpiiaalijiwhyuhxlvhxpkgfypvrpqqsacuwocdampnnhvibsbolyduvscsjfayxpldyvqzjbqojjxdvxtctwtifcdfcaoonzbgegyllngwvjivtxeezoabhsihcuvgvdsgjtzbzwovjshvwrljkxavowovzrfpdufpogdtujkerdorpboufrxhyswirjehgsxthlvjufvahdpeajidqipwaxokbeycrievgcfkyuyavaq",
"output": "Second"
},
{
"input": "a",
"output": "First"
},
{
"input": "ab",
"output": "Second"
},
{
"input": "abacaba",
"output": "First"
},
{
"input": "abazaba",
"output": "First"
},
{
"input": "zz",
"output": "First"
},
{
"input": "aassddxyz",
"output": "First"
},
{
"input": "abcdefghijklmnopqrstuvwxyz",
"output": "Second"
},
{
"input": "aabc",
"output": "Second"
},
{
"input": "abcabc",
"output": "First"
},
{
"input": "aaabbbccdd",
"output": "Second"
},
{
"input": "aabbcccc",
"output": "First"
}
] | 1,696,898,167
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 51
| 154
| 0
|
import sys
input = sys.stdin.readline
from collections import Counter , defaultdict
def instr():
return input()[:-1]
import math
############################
w = instr()
c = Counter(w)
f = sum(x%2 != 0 for x in c.values())
if f == 0 :
print("First")
else :
print("First" if f%2 != 0 else "Second")
|
Title: Little Girl and Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Little Girl loves problems on games very much. Here's one of them.
Two players have got a string *s*, consisting of lowercase English letters. They play a game that is described by the following rules:
- The players move in turns; In one move the player can remove an arbitrary letter from string *s*. - If the player before his turn can reorder the letters in string *s* so as to get a palindrome, this player wins. A palindrome is a string that reads the same both ways (from left to right, and vice versa). For example, string "abba" is a palindrome and string "abc" isn't.
Determine which player will win, provided that both sides play optimally well β the one who moves first or the one who moves second.
Input Specification:
The input contains a single line, containing string *s* (1<=β€<=|*s*|<=<=β€<=<=103). String *s* consists of lowercase English letters.
Output Specification:
In a single line print word "First" if the first player wins (provided that both players play optimally well). Otherwise, print word "Second". Print the words without the quotes.
Demo Input:
['aba\n', 'abca\n']
Demo Output:
['First\n', 'Second\n']
Note:
none
|
```python
import sys
input = sys.stdin.readline
from collections import Counter , defaultdict
def instr():
return input()[:-1]
import math
############################
w = instr()
c = Counter(w)
f = sum(x%2 != 0 for x in c.values())
if f == 0 :
print("First")
else :
print("First" if f%2 != 0 else "Second")
```
| 3
|
|
897
|
B
|
Chtholly's request
|
PROGRAMMING
| 1,300
|
[
"brute force"
] | null | null |
β I experienced so many great things.
β You gave me memories like dreams... But I have to leave now...
β One last request, can you...
β Help me solve a Codeforces problem?
β ......
β What?
Chtholly has been thinking about a problem for days:
If a number is palindrome and length of its decimal representation without leading zeros is even, we call it a zcy number. A number is palindrome means when written in decimal representation, it contains no leading zeros and reads the same forwards and backwards. For example 12321 and 1221 are palindromes and 123 and 12451 are not. Moreover, 1221 is zcy number and 12321 is not.
Given integers *k* and *p*, calculate the sum of the *k* smallest zcy numbers and output this sum modulo *p*.
Unfortunately, Willem isn't good at solving this kind of problems, so he asks you for help!
|
The first line contains two integers *k* and *p* (1<=β€<=*k*<=β€<=105,<=1<=β€<=*p*<=β€<=109).
|
Output single integerΒ β answer to the problem.
|
[
"2 100\n",
"5 30\n"
] |
[
"33\n",
"15\n"
] |
In the first example, the smallest zcy number is 11, and the second smallest zcy number is 22.
In the second example, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/68fffad54395f7d920ad0384e07c6215ddc64141.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
| 1,000
|
[
{
"input": "2 100",
"output": "33"
},
{
"input": "5 30",
"output": "15"
},
{
"input": "42147 412393322",
"output": "251637727"
},
{
"input": "77809 868097296",
"output": "440411873"
},
{
"input": "5105 443422097",
"output": "363192634"
},
{
"input": "75615 376679484",
"output": "373089399"
},
{
"input": "22951 23793302",
"output": "1898631"
},
{
"input": "12785 993582106",
"output": "286204743"
},
{
"input": "60276 428978808",
"output": "376477293"
},
{
"input": "84776 104860385",
"output": "10209596"
},
{
"input": "41984 653766991",
"output": "17823101"
},
{
"input": "100000 1000000000",
"output": "495495496"
},
{
"input": "41163 472310076",
"output": "207304047"
},
{
"input": "6983 765352180",
"output": "586866999"
},
{
"input": "33493 967727004",
"output": "305705165"
},
{
"input": "90898 94010922",
"output": "65928728"
},
{
"input": "67298 349286579",
"output": "156435206"
},
{
"input": "92452 296773064",
"output": "229486976"
},
{
"input": "58832 563860457",
"output": "16775206"
},
{
"input": "90234 156145441",
"output": "44023160"
},
{
"input": "91454 977186148",
"output": "681779748"
},
{
"input": "11108 444095250",
"output": "188075844"
},
{
"input": "46304 584475527",
"output": "275627129"
},
{
"input": "1 1",
"output": "0"
},
{
"input": "1 1000000000",
"output": "11"
},
{
"input": "100000 1",
"output": "0"
}
] | 1,607,101,986
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 26
| 171
| 2,150,400
|
k, p = map(int, input().split())
res = 0
for i in range(1, k+1):
res += int(str(i)+(str(i)[::-1]))
res %= p
print(res)
|
Title: Chtholly's request
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
β I experienced so many great things.
β You gave me memories like dreams... But I have to leave now...
β One last request, can you...
β Help me solve a Codeforces problem?
β ......
β What?
Chtholly has been thinking about a problem for days:
If a number is palindrome and length of its decimal representation without leading zeros is even, we call it a zcy number. A number is palindrome means when written in decimal representation, it contains no leading zeros and reads the same forwards and backwards. For example 12321 and 1221 are palindromes and 123 and 12451 are not. Moreover, 1221 is zcy number and 12321 is not.
Given integers *k* and *p*, calculate the sum of the *k* smallest zcy numbers and output this sum modulo *p*.
Unfortunately, Willem isn't good at solving this kind of problems, so he asks you for help!
Input Specification:
The first line contains two integers *k* and *p* (1<=β€<=*k*<=β€<=105,<=1<=β€<=*p*<=β€<=109).
Output Specification:
Output single integerΒ β answer to the problem.
Demo Input:
['2 100\n', '5 30\n']
Demo Output:
['33\n', '15\n']
Note:
In the first example, the smallest zcy number is 11, and the second smallest zcy number is 22.
In the second example, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/68fffad54395f7d920ad0384e07c6215ddc64141.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
|
```python
k, p = map(int, input().split())
res = 0
for i in range(1, k+1):
res += int(str(i)+(str(i)[::-1]))
res %= p
print(res)
```
| 3
|
|
158
|
B
|
Taxi
|
PROGRAMMING
| 1,100
|
[
"*special",
"greedy",
"implementation"
] | null | null |
After the lessons *n* groups of schoolchildren went outside and decided to visit Polycarpus to celebrate his birthday. We know that the *i*-th group consists of *s**i* friends (1<=β€<=*s**i*<=β€<=4), and they want to go to Polycarpus together. They decided to get there by taxi. Each car can carry at most four passengers. What minimum number of cars will the children need if all members of each group should ride in the same taxi (but one taxi can take more than one group)?
|
The first line contains integer *n* (1<=β€<=*n*<=β€<=105) β the number of groups of schoolchildren. The second line contains a sequence of integers *s*1,<=*s*2,<=...,<=*s**n* (1<=β€<=*s**i*<=β€<=4). The integers are separated by a space, *s**i* is the number of children in the *i*-th group.
|
Print the single number β the minimum number of taxis necessary to drive all children to Polycarpus.
|
[
"5\n1 2 4 3 3\n",
"8\n2 3 4 4 2 1 3 1\n"
] |
[
"4\n",
"5\n"
] |
In the first test we can sort the children into four cars like this:
- the third group (consisting of four children), - the fourth group (consisting of three children), - the fifth group (consisting of three children), - the first and the second group (consisting of one and two children, correspondingly).
There are other ways to sort the groups into four cars.
| 1,000
|
[
{
"input": "5\n1 2 4 3 3",
"output": "4"
},
{
"input": "8\n2 3 4 4 2 1 3 1",
"output": "5"
},
{
"input": "5\n4 4 4 4 4",
"output": "5"
},
{
"input": "12\n1 1 1 1 1 1 1 1 1 1 1 1",
"output": "3"
},
{
"input": "2\n2 1",
"output": "1"
},
{
"input": "4\n3 2 1 3",
"output": "3"
},
{
"input": "4\n2 4 1 3",
"output": "3"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "1\n2",
"output": "1"
},
{
"input": "1\n3",
"output": "1"
},
{
"input": "1\n4",
"output": "1"
},
{
"input": "2\n1 1",
"output": "1"
},
{
"input": "2\n2 2",
"output": "1"
},
{
"input": "2\n3 3",
"output": "2"
},
{
"input": "2\n4 4",
"output": "2"
},
{
"input": "2\n2 1",
"output": "1"
},
{
"input": "2\n3 1",
"output": "1"
},
{
"input": "2\n4 1",
"output": "2"
},
{
"input": "2\n2 3",
"output": "2"
},
{
"input": "2\n4 2",
"output": "2"
},
{
"input": "2\n4 3",
"output": "2"
},
{
"input": "4\n2 2 1 1",
"output": "2"
},
{
"input": "4\n3 1 3 1",
"output": "2"
},
{
"input": "4\n1 4 1 4",
"output": "3"
},
{
"input": "4\n2 2 3 3",
"output": "3"
},
{
"input": "4\n2 4 4 2",
"output": "3"
},
{
"input": "4\n3 3 4 4",
"output": "4"
},
{
"input": "3\n1 1 2",
"output": "1"
},
{
"input": "3\n1 3 1",
"output": "2"
},
{
"input": "3\n4 1 1",
"output": "2"
},
{
"input": "3\n3 2 2",
"output": "2"
},
{
"input": "3\n2 4 2",
"output": "2"
},
{
"input": "3\n3 4 3",
"output": "3"
},
{
"input": "3\n2 2 1",
"output": "2"
},
{
"input": "3\n1 3 3",
"output": "2"
},
{
"input": "3\n4 4 1",
"output": "3"
},
{
"input": "3\n3 3 2",
"output": "3"
},
{
"input": "3\n4 2 4",
"output": "3"
},
{
"input": "3\n4 3 4",
"output": "3"
},
{
"input": "3\n4 3 2",
"output": "3"
},
{
"input": "3\n3 1 4",
"output": "2"
},
{
"input": "3\n2 1 4",
"output": "2"
},
{
"input": "3\n3 1 2",
"output": "2"
},
{
"input": "4\n4 4 3 2",
"output": "4"
},
{
"input": "4\n1 4 3 1",
"output": "3"
},
{
"input": "4\n2 2 4 1",
"output": "3"
},
{
"input": "4\n3 2 1 2",
"output": "2"
},
{
"input": "5\n2 4 2 3 4",
"output": "4"
},
{
"input": "5\n1 3 4 1 3",
"output": "3"
},
{
"input": "5\n1 1 2 4 2",
"output": "3"
},
{
"input": "5\n1 3 2 3 2",
"output": "3"
},
{
"input": "8\n1 1 2 1 1 1 3 2",
"output": "3"
},
{
"input": "78\n2 2 2 2 3 3 2 2 2 2 2 2 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 3 2 2 2 2 2 2 2 1 1 2 2 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",
"output": "39"
},
{
"input": "7\n2 2 2 1 2 1 2",
"output": "3"
},
{
"input": "9\n3 1 2 1 1 1 1 1 1",
"output": "3"
},
{
"input": "10\n3 1 2 2 2 2 2 2 1 2",
"output": "5"
},
{
"input": "18\n1 3 3 3 1 1 3 1 1 1 3 3 3 3 1 3 1 1",
"output": "9"
},
{
"input": "26\n3 1 3 3 1 3 2 3 1 3 3 2 1 2 3 2 2 1 2 1 2 1 1 3 2 1",
"output": "13"
}
] | 1,696,672,999
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 105
| 156
| 3,481,600
|
def min_taxis(groups):
counts = [groups.count(i) for i in range(1, 5)]
return counts[3] + counts[2] + (counts[1]*2 + max(0, counts[0] - counts[2]) + 3) // 4
# ΠΡΠΈΠΌΠ΅Ρ ΠΈΡΠΏΠΎΠ»ΡΠ·ΠΎΠ²Π°Π½ΠΈΡ
n = int(input())
groups = list(map(int, input().split()))
result = min_taxis(groups)
print(result)
|
Title: Taxi
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
After the lessons *n* groups of schoolchildren went outside and decided to visit Polycarpus to celebrate his birthday. We know that the *i*-th group consists of *s**i* friends (1<=β€<=*s**i*<=β€<=4), and they want to go to Polycarpus together. They decided to get there by taxi. Each car can carry at most four passengers. What minimum number of cars will the children need if all members of each group should ride in the same taxi (but one taxi can take more than one group)?
Input Specification:
The first line contains integer *n* (1<=β€<=*n*<=β€<=105) β the number of groups of schoolchildren. The second line contains a sequence of integers *s*1,<=*s*2,<=...,<=*s**n* (1<=β€<=*s**i*<=β€<=4). The integers are separated by a space, *s**i* is the number of children in the *i*-th group.
Output Specification:
Print the single number β the minimum number of taxis necessary to drive all children to Polycarpus.
Demo Input:
['5\n1 2 4 3 3\n', '8\n2 3 4 4 2 1 3 1\n']
Demo Output:
['4\n', '5\n']
Note:
In the first test we can sort the children into four cars like this:
- the third group (consisting of four children), - the fourth group (consisting of three children), - the fifth group (consisting of three children), - the first and the second group (consisting of one and two children, correspondingly).
There are other ways to sort the groups into four cars.
|
```python
def min_taxis(groups):
counts = [groups.count(i) for i in range(1, 5)]
return counts[3] + counts[2] + (counts[1]*2 + max(0, counts[0] - counts[2]) + 3) // 4
# ΠΡΠΈΠΌΠ΅Ρ ΠΈΡΠΏΠΎΠ»ΡΠ·ΠΎΠ²Π°Π½ΠΈΡ
n = int(input())
groups = list(map(int, input().split()))
result = min_taxis(groups)
print(result)
```
| 3
|
|
149
|
C
|
Division into Teams
|
PROGRAMMING
| 1,500
|
[
"greedy",
"math",
"sortings"
] | null | null |
Petya loves football very much, especially when his parents aren't home. Each morning he comes to the yard, gathers his friends and they play all day. From time to time they have a break to have some food or do some chores (for example, water the flowers).
The key in football is to divide into teams fairly before the game begins. There are *n* boys playing football in the yard (including Petya), each boy's football playing skill is expressed with a non-negative characteristic *a**i* (the larger it is, the better the boy plays).
Let's denote the number of players in the first team as *x*, the number of players in the second team as *y*, the individual numbers of boys who play for the first team as *p**i* and the individual numbers of boys who play for the second team as *q**i*. Division *n* boys into two teams is considered fair if three conditions are fulfilled:
- Each boy plays for exactly one team (*x*<=+<=*y*<==<=*n*). - The sizes of teams differ in no more than one (|*x*<=-<=*y*|<=β€<=1). - The total football playing skills for two teams differ in no more than by the value of skill the best player in the yard has. More formally:
Your task is to help guys divide into two teams fairly. It is guaranteed that a fair division into two teams always exists.
|
The first line contains the only integer *n* (2<=β€<=*n*<=β€<=105) which represents the number of guys in the yard. The next line contains *n* positive space-separated integers, *a**i* (1<=β€<=*a**i*<=β€<=104), the *i*-th number represents the *i*-th boy's playing skills.
|
On the first line print an integer *x* β the number of boys playing for the first team. On the second line print *x* integers β the individual numbers of boys playing for the first team. On the third line print an integer *y* β the number of boys playing for the second team, on the fourth line print *y* integers β the individual numbers of boys playing for the second team. Don't forget that you should fulfil all three conditions: *x*<=+<=*y*<==<=*n*, |*x*<=-<=*y*|<=β€<=1, and the condition that limits the total skills.
If there are multiple ways to solve the problem, print any of them.
The boys are numbered starting from one in the order in which their skills are given in the input data. You are allowed to print individual numbers of boys who belong to the same team in any order.
|
[
"3\n1 2 1\n",
"5\n2 3 3 1 1\n"
] |
[
"2\n1 2 \n1\n3 \n",
"3\n4 1 3 \n2\n5 2 \n"
] |
Let's consider the first sample test. There we send the first and the second boy to the first team and the third boy to the second team. Let's check all three conditions of a fair division. The first limitation is fulfilled (all boys play), the second limitation on the sizes of groups (|2β-β1|β=β1ββ€β1) is fulfilled, the third limitation on the difference in skills ((2β+β1)β-β(1)β=β2ββ€β2) is fulfilled.
| 1,500
|
[
{
"input": "3\n1 2 1",
"output": "2\n1 2 \n1\n3 "
},
{
"input": "5\n2 3 3 1 1",
"output": "3\n4 1 3 \n2\n5 2 "
},
{
"input": "10\n2 2 2 2 2 2 2 1 2 2",
"output": "5\n8 2 4 6 9 \n5\n1 3 5 7 10 "
},
{
"input": "10\n2 3 3 1 3 1 1 1 2 2",
"output": "5\n4 7 1 10 3 \n5\n6 8 9 2 5 "
},
{
"input": "10\n2 3 2 3 3 1 1 3 1 1",
"output": "5\n6 9 1 2 5 \n5\n7 10 3 4 8 "
},
{
"input": "11\n1 3 1 2 1 2 2 2 1 1 1",
"output": "6\n1 5 10 4 7 2 \n5\n3 9 11 6 8 "
},
{
"input": "11\n54 83 96 75 33 27 36 35 26 22 77",
"output": "6\n10 6 8 1 11 3 \n5\n9 5 7 4 2 "
},
{
"input": "11\n1 1 1 1 1 1 1 1 1 1 1",
"output": "6\n1 3 5 7 9 11 \n5\n2 4 6 8 10 "
},
{
"input": "2\n1 1",
"output": "1\n1 \n1\n2 "
},
{
"input": "2\n35 36",
"output": "1\n1 \n1\n2 "
},
{
"input": "25\n1 2 2 1 2 2 2 2 2 1 1 2 2 2 2 2 1 2 2 2 1 1 2 2 1",
"output": "13\n1 10 17 22 2 5 7 9 13 15 18 20 24 \n12\n4 11 21 25 3 6 8 12 14 16 19 23 "
},
{
"input": "27\n2 1 1 3 1 2 1 1 3 2 3 1 3 2 1 3 2 3 2 1 2 3 2 2 1 2 1",
"output": "14\n2 5 8 15 25 1 10 17 21 24 4 11 16 22 \n13\n3 7 12 20 27 6 14 19 23 26 9 13 18 "
},
{
"input": "30\n2 2 2 3 4 3 4 4 3 2 3 2 2 4 1 4 2 4 2 2 1 4 3 2 1 3 1 1 4 3",
"output": "15\n15 25 28 2 10 13 19 24 6 11 26 5 8 16 22 \n15\n21 27 1 3 12 17 20 4 9 23 30 7 14 18 29 "
},
{
"input": "100\n3 4 8 10 8 6 4 3 7 7 6 2 3 1 3 10 1 7 9 3 5 5 2 6 2 9 1 7 4 2 4 1 6 1 7 10 2 5 3 7 6 4 6 2 8 8 8 6 6 10 3 7 4 3 4 1 7 9 3 6 3 6 1 4 9 3 8 1 10 1 4 10 7 7 9 5 3 8 10 2 1 10 8 7 10 8 5 3 1 2 1 10 6 1 5 3 3 5 7 2",
"output": "50\n14 27 34 63 70 89 94 23 30 44 90 1 13 20 51 59 66 88 97 7 31 53 64 21 38 87 98 11 33 43 49 62 9 18 35 52 73 84 3 45 47 78 86 26 65 4 36 69 79 85 \n50\n17 32 56 68 81 91 12 25 37 80 100 8 15 39 54 61 77 96 2 29 42 55 71 22 76 95 6 24 41 48 60 93 10 28 40 57 74 99 5 46 67 83 19 58 75 16 50 72 82 92 "
},
{
"input": "100\n85 50 17 89 65 89 5 20 86 26 16 21 85 14 44 31 87 31 6 2 48 67 8 80 79 1 48 36 97 1 5 30 79 50 78 12 2 55 76 100 54 40 26 81 97 96 68 56 87 14 51 17 54 37 52 33 69 62 38 63 74 15 62 78 9 19 67 2 60 58 93 60 18 96 55 48 34 7 79 82 32 58 90 67 20 50 27 15 7 89 98 10 11 15 99 49 4 51 77 52",
"output": "50\n26 20 68 7 19 89 65 93 14 62 94 3 73 8 12 43 32 18 56 28 59 15 27 96 34 51 55 41 38 48 82 72 63 5 67 47 61 99 64 33 24 80 13 17 4 90 71 74 45 95 \n50\n30 37 97 31 78 23 92 36 50 88 11 52 66 85 10 87 16 81 77 54 42 21 76 2 86 98 100 53 75 70 69 58 60 22 84 57 39 35 25 79 44 1 9 49 6 83 46 29 91 40 "
},
{
"input": "100\n2382 7572 9578 1364 2325 2929 7670 5574 2836 2440 6553 1751 929 8785 6894 9373 9308 7338 6380 9541 9951 6785 8993 9942 5087 7544 6582 7139 8458 7424 9759 8199 9464 8817 7625 6200 4955 9373 9500 3062 849 4210 9337 5466 2190 8150 4971 3145 869 5675 1975 161 1998 378 5229 9000 8958 761 358 434 7636 8295 4406 73 375 812 2473 3652 9067 3052 5287 2850 6987 5442 2625 8894 8733 791 9763 5258 8259 9530 2050 7334 2118 2726 8221 5527 8827 1585 8334 8898 6399 6217 7400 2576 5164 9063 6247 9433",
"output": "50\n64 59 54 58 66 49 4 12 53 85 5 10 96 86 72 70 48 42 37 25 55 71 44 8 36 99 93 27 15 28 18 30 2 61 46 87 62 29 14 89 92 23 98 17 16 100 39 20 31 24 \n50\n52 65 60 78 41 13 90 51 83 45 1 67 75 9 6 40 68 63 47 97 80 74 88 50 94 19 11 22 73 84 95 26 35 7 32 81 91 77 34 76 57 56 69 43 38 33 82 3 79 21 "
},
{
"input": "3\n1 2 3",
"output": "2\n1 3 \n1\n2 "
},
{
"input": "3\n10 10 10",
"output": "2\n1 3 \n1\n2 "
},
{
"input": "3\n5 10 10",
"output": "2\n1 3 \n1\n2 "
},
{
"input": "5\n6 1 1 1 1",
"output": "3\n2 4 1 \n2\n3 5 "
},
{
"input": "5\n1 100 2 200 3",
"output": "3\n1 5 4 \n2\n3 2 "
}
] | 1,669,996,809
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 47
| 280
| 9,830,400
|
n = int(input())
a = list(map(int,input().split()))
b = list(range(1,n+1))
z = sorted(list(zip(a,b)))
x = []
y = []
sx = 0
sy = 0
for i,j in z:
if sx<=sy:
x.append(j)
sx+=i
else:
y.append(j)
sy+=i
print(len(x))
print(*x)
print(len(y))
print(*y)
|
Title: Division into Teams
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya loves football very much, especially when his parents aren't home. Each morning he comes to the yard, gathers his friends and they play all day. From time to time they have a break to have some food or do some chores (for example, water the flowers).
The key in football is to divide into teams fairly before the game begins. There are *n* boys playing football in the yard (including Petya), each boy's football playing skill is expressed with a non-negative characteristic *a**i* (the larger it is, the better the boy plays).
Let's denote the number of players in the first team as *x*, the number of players in the second team as *y*, the individual numbers of boys who play for the first team as *p**i* and the individual numbers of boys who play for the second team as *q**i*. Division *n* boys into two teams is considered fair if three conditions are fulfilled:
- Each boy plays for exactly one team (*x*<=+<=*y*<==<=*n*). - The sizes of teams differ in no more than one (|*x*<=-<=*y*|<=β€<=1). - The total football playing skills for two teams differ in no more than by the value of skill the best player in the yard has. More formally:
Your task is to help guys divide into two teams fairly. It is guaranteed that a fair division into two teams always exists.
Input Specification:
The first line contains the only integer *n* (2<=β€<=*n*<=β€<=105) which represents the number of guys in the yard. The next line contains *n* positive space-separated integers, *a**i* (1<=β€<=*a**i*<=β€<=104), the *i*-th number represents the *i*-th boy's playing skills.
Output Specification:
On the first line print an integer *x* β the number of boys playing for the first team. On the second line print *x* integers β the individual numbers of boys playing for the first team. On the third line print an integer *y* β the number of boys playing for the second team, on the fourth line print *y* integers β the individual numbers of boys playing for the second team. Don't forget that you should fulfil all three conditions: *x*<=+<=*y*<==<=*n*, |*x*<=-<=*y*|<=β€<=1, and the condition that limits the total skills.
If there are multiple ways to solve the problem, print any of them.
The boys are numbered starting from one in the order in which their skills are given in the input data. You are allowed to print individual numbers of boys who belong to the same team in any order.
Demo Input:
['3\n1 2 1\n', '5\n2 3 3 1 1\n']
Demo Output:
['2\n1 2 \n1\n3 \n', '3\n4 1 3 \n2\n5 2 \n']
Note:
Let's consider the first sample test. There we send the first and the second boy to the first team and the third boy to the second team. Let's check all three conditions of a fair division. The first limitation is fulfilled (all boys play), the second limitation on the sizes of groups (|2β-β1|β=β1ββ€β1) is fulfilled, the third limitation on the difference in skills ((2β+β1)β-β(1)β=β2ββ€β2) is fulfilled.
|
```python
n = int(input())
a = list(map(int,input().split()))
b = list(range(1,n+1))
z = sorted(list(zip(a,b)))
x = []
y = []
sx = 0
sy = 0
for i,j in z:
if sx<=sy:
x.append(j)
sx+=i
else:
y.append(j)
sy+=i
print(len(x))
print(*x)
print(len(y))
print(*y)
```
| 3
|
|
436
|
A
|
Feed with Candy
|
PROGRAMMING
| 1,500
|
[
"greedy"
] | null | null |
The hero of the Cut the Rope game is a little monster named Om Nom. He loves candies. And what a coincidence! He also is the hero of today's problem.
One day, Om Nom visited his friend Evan. Evan has *n* candies of two types (fruit drops and caramel drops), the *i*-th candy hangs at the height of *h**i* centimeters above the floor of the house, its mass is *m**i*. Om Nom wants to eat as many candies as possible. At the beginning Om Nom can make at most *x* centimeter high jumps. When Om Nom eats a candy of mass *y*, he gets stronger and the height of his jump increases by *y* centimeters.
What maximum number of candies can Om Nom eat if he never eats two candies of the same type in a row (Om Nom finds it too boring)?
|
The first line contains two integers, *n* and *x* (1<=β€<=*n*,<=*x*<=β€<=2000) β the number of sweets Evan has and the initial height of Om Nom's jump.
Each of the following *n* lines contains three integers *t**i*,<=*h**i*,<=*m**i* (0<=β€<=*t**i*<=β€<=1;Β 1<=β€<=*h**i*,<=*m**i*<=β€<=2000) β the type, height and the mass of the *i*-th candy. If number *t**i* equals 0, then the current candy is a caramel drop, otherwise it is a fruit drop.
|
Print a single integer β the maximum number of candies Om Nom can eat.
|
[
"5 3\n0 2 4\n1 3 1\n0 8 3\n0 20 10\n1 5 5\n"
] |
[
"4\n"
] |
One of the possible ways to eat 4 candies is to eat them in the order: 1, 5, 3, 2. Let's assume the following scenario:
1. Initially, the height of Om Nom's jump equals 3. He can reach candies 1 and 2. Let's assume that he eats candy 1. As the mass of this candy equals 4, the height of his jump will rise to 3β+β4β=β7. 1. Now Om Nom can reach candies 2 and 5. Let's assume that he eats candy 5. Then the height of his jump will be 7β+β5β=β12. 1. At this moment, Om Nom can reach two candies, 2 and 3. He won't eat candy 2 as its type matches the type of the previously eaten candy. Om Nom eats candy 3, the height of his jump is 12β+β3β=β15. 1. Om Nom eats candy 2, the height of his jump is 15β+β1β=β16. He cannot reach candy 4.
| 1,000
|
[
{
"input": "5 3\n0 2 4\n1 3 1\n0 8 3\n0 20 10\n1 5 5",
"output": "4"
},
{
"input": "5 2\n1 15 2\n1 11 2\n0 17 2\n0 16 1\n1 18 2",
"output": "0"
},
{
"input": "6 2\n1 17 3\n1 6 1\n0 4 2\n1 10 1\n1 7 3\n1 5 1",
"output": "0"
},
{
"input": "7 2\n1 14 1\n1 9 2\n0 6 3\n0 20 2\n0 4 2\n0 3 1\n0 9 2",
"output": "0"
},
{
"input": "8 2\n0 20 3\n1 5 2\n1 4 1\n1 7 1\n0 1 3\n1 5 3\n1 7 2\n1 3 1",
"output": "2"
},
{
"input": "9 2\n0 1 1\n1 8 2\n1 11 1\n0 9 1\n1 18 2\n1 7 3\n1 8 3\n0 16 1\n0 12 2",
"output": "1"
},
{
"input": "10 2\n0 2 3\n1 5 2\n0 7 3\n1 15 2\n0 14 3\n1 19 1\n1 5 3\n0 2 2\n0 10 2\n0 10 3",
"output": "9"
},
{
"input": "2 1\n0 1 1\n1 2 1",
"output": "2"
},
{
"input": "2 1\n1 1 1\n0 2 1",
"output": "2"
},
{
"input": "2 1\n0 1 1\n0 2 1",
"output": "1"
},
{
"input": "2 1\n1 1 1\n1 2 1",
"output": "1"
},
{
"input": "2 1\n0 1 1\n1 3 1",
"output": "1"
},
{
"input": "2 1\n1 1 1\n0 3 1",
"output": "1"
},
{
"input": "1 1\n1 2 1",
"output": "0"
},
{
"input": "3 4\n1 1 2\n1 4 100\n0 104 1",
"output": "3"
},
{
"input": "3 4\n1 1 100\n1 4 2\n0 104 1",
"output": "3"
},
{
"input": "3 100\n0 1 1\n1 1 1\n1 1 1",
"output": "3"
},
{
"input": "4 20\n0 10 10\n0 20 50\n1 40 1\n1 40 1",
"output": "4"
},
{
"input": "4 2\n0 1 1\n0 2 3\n1 4 1\n1 5 1",
"output": "4"
},
{
"input": "3 10\n0 9 1\n0 10 10\n1 20 1",
"output": "3"
},
{
"input": "3 5\n0 4 1\n0 5 10\n1 15 5",
"output": "3"
},
{
"input": "3 4\n0 2 1\n0 3 3\n1 6 5",
"output": "3"
},
{
"input": "3 3\n0 1 1\n0 2 100\n1 10 1",
"output": "3"
},
{
"input": "3 2\n0 1 1\n0 2 2\n1 4 4",
"output": "3"
},
{
"input": "5 3\n0 1 5\n0 1 5\n0 1 5\n1 1 10\n1 1 1",
"output": "5"
},
{
"input": "3 2\n0 1 1\n0 2 2\n1 4 2",
"output": "3"
},
{
"input": "4 10\n0 20 1\n1 1 9\n1 2 11\n1 3 8",
"output": "3"
},
{
"input": "7 1\n0 1 99\n1 100 1\n0 100 1\n0 101 1000\n1 1000 1\n0 1000 1\n1 1000 1",
"output": "7"
},
{
"input": "4 3\n0 1 1\n0 2 100\n0 3 1\n1 100 1",
"output": "3"
},
{
"input": "3 3\n0 1 100\n0 2 1\n1 100 100",
"output": "3"
},
{
"input": "3 2\n0 1 1\n0 2 100\n1 10 1",
"output": "3"
},
{
"input": "3 1\n0 1 1\n1 1 5\n0 7 1",
"output": "3"
},
{
"input": "3 5\n0 2 3\n1 9 10\n0 4 4",
"output": "3"
},
{
"input": "3 3\n0 2 1\n0 3 2\n1 5 10",
"output": "3"
}
] | 1,426,598,581
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 60
| 545
| 819,200
|
from copy import deepcopy
def getBetter(h, a):
maxi = -1
im = -1
for i in range(len(a)):
if (h >= a[i][0]):
if (maxi < a[i][1]):
im = i
maxi = a[i][1]
return(im, maxi)
n, h0 = map(int, input().split())
lolipops0 = [[], []]
for i in range(n):
t, h, m = map(int, input().split())
lolipops0[t].append((h, m))
lolipops0[1].sort()
lolipops0[0].sort()
lolipops1 = deepcopy(lolipops0)
lol0 = getBetter(h0, lolipops0[0])
t1 = 0
h1 = h0
lol1 = getBetter(h0, lolipops0[1])
t2 = 1 # ---- WARNING --------
h2 = h0
while(lol0[0] >= 0 or lol1[0] >=0):
if (lol0[0] > -1):
if (len(lolipops0[t1 % 2]) != 1 and lol0[0] != -1):
lolipops0[t1 % 2].pop(lol0[0])
else:
lolipops0[t1 % 2] = []
t1 += 1
h1 += lol0[1]
lol0 = getBetter(h1, lolipops0[t1 % 2])
if (lol1[0] > -1):
if (len(lolipops1[t2 % 2]) != 1 and lol1[0] != -1):
lolipops1[t2 % 2].pop(lol1[0])
else:
lolipops1[t2 % 2] = []
t2 += 1
h2 += lol1[1]
lol1 = getBetter(h2, lolipops1[t2 % 2])
print(max(t1, t2 - 1))
|
Title: Feed with Candy
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The hero of the Cut the Rope game is a little monster named Om Nom. He loves candies. And what a coincidence! He also is the hero of today's problem.
One day, Om Nom visited his friend Evan. Evan has *n* candies of two types (fruit drops and caramel drops), the *i*-th candy hangs at the height of *h**i* centimeters above the floor of the house, its mass is *m**i*. Om Nom wants to eat as many candies as possible. At the beginning Om Nom can make at most *x* centimeter high jumps. When Om Nom eats a candy of mass *y*, he gets stronger and the height of his jump increases by *y* centimeters.
What maximum number of candies can Om Nom eat if he never eats two candies of the same type in a row (Om Nom finds it too boring)?
Input Specification:
The first line contains two integers, *n* and *x* (1<=β€<=*n*,<=*x*<=β€<=2000) β the number of sweets Evan has and the initial height of Om Nom's jump.
Each of the following *n* lines contains three integers *t**i*,<=*h**i*,<=*m**i* (0<=β€<=*t**i*<=β€<=1;Β 1<=β€<=*h**i*,<=*m**i*<=β€<=2000) β the type, height and the mass of the *i*-th candy. If number *t**i* equals 0, then the current candy is a caramel drop, otherwise it is a fruit drop.
Output Specification:
Print a single integer β the maximum number of candies Om Nom can eat.
Demo Input:
['5 3\n0 2 4\n1 3 1\n0 8 3\n0 20 10\n1 5 5\n']
Demo Output:
['4\n']
Note:
One of the possible ways to eat 4 candies is to eat them in the order: 1, 5, 3, 2. Let's assume the following scenario:
1. Initially, the height of Om Nom's jump equals 3. He can reach candies 1 and 2. Let's assume that he eats candy 1. As the mass of this candy equals 4, the height of his jump will rise to 3β+β4β=β7. 1. Now Om Nom can reach candies 2 and 5. Let's assume that he eats candy 5. Then the height of his jump will be 7β+β5β=β12. 1. At this moment, Om Nom can reach two candies, 2 and 3. He won't eat candy 2 as its type matches the type of the previously eaten candy. Om Nom eats candy 3, the height of his jump is 12β+β3β=β15. 1. Om Nom eats candy 2, the height of his jump is 15β+β1β=β16. He cannot reach candy 4.
|
```python
from copy import deepcopy
def getBetter(h, a):
maxi = -1
im = -1
for i in range(len(a)):
if (h >= a[i][0]):
if (maxi < a[i][1]):
im = i
maxi = a[i][1]
return(im, maxi)
n, h0 = map(int, input().split())
lolipops0 = [[], []]
for i in range(n):
t, h, m = map(int, input().split())
lolipops0[t].append((h, m))
lolipops0[1].sort()
lolipops0[0].sort()
lolipops1 = deepcopy(lolipops0)
lol0 = getBetter(h0, lolipops0[0])
t1 = 0
h1 = h0
lol1 = getBetter(h0, lolipops0[1])
t2 = 1 # ---- WARNING --------
h2 = h0
while(lol0[0] >= 0 or lol1[0] >=0):
if (lol0[0] > -1):
if (len(lolipops0[t1 % 2]) != 1 and lol0[0] != -1):
lolipops0[t1 % 2].pop(lol0[0])
else:
lolipops0[t1 % 2] = []
t1 += 1
h1 += lol0[1]
lol0 = getBetter(h1, lolipops0[t1 % 2])
if (lol1[0] > -1):
if (len(lolipops1[t2 % 2]) != 1 and lol1[0] != -1):
lolipops1[t2 % 2].pop(lol1[0])
else:
lolipops1[t2 % 2] = []
t2 += 1
h2 += lol1[1]
lol1 = getBetter(h2, lolipops1[t2 % 2])
print(max(t1, t2 - 1))
```
| 3
|
|
34
|
B
|
Sale
|
PROGRAMMING
| 900
|
[
"greedy",
"sortings"
] |
B. Sale
|
2
|
256
|
Once Bob got to a sale of old TV sets. There were *n* TV sets at that sale. TV set with index *i* costs *a**i* bellars. Some TV sets have a negative price β their owners are ready to pay Bob if he buys their useless apparatus. Bob can Β«buyΒ» any TV sets he wants. Though he's very strong, Bob can carry at most *m* TV sets, and he has no desire to go to the sale for the second time. Please, help Bob find out the maximum sum of money that he can earn.
|
The first line contains two space-separated integers *n* and *m* (1<=β€<=*m*<=β€<=*n*<=β€<=100) β amount of TV sets at the sale, and amount of TV sets that Bob can carry. The following line contains *n* space-separated integers *a**i* (<=-<=1000<=β€<=*a**i*<=β€<=1000) β prices of the TV sets.
|
Output the only number β the maximum sum of money that Bob can earn, given that he can carry at most *m* TV sets.
|
[
"5 3\n-6 0 35 -2 4\n",
"4 2\n7 0 0 -7\n"
] |
[
"8\n",
"7\n"
] |
none
| 1,000
|
[
{
"input": "5 3\n-6 0 35 -2 4",
"output": "8"
},
{
"input": "4 2\n7 0 0 -7",
"output": "7"
},
{
"input": "6 6\n756 -611 251 -66 572 -818",
"output": "1495"
},
{
"input": "5 5\n976 437 937 788 518",
"output": "0"
},
{
"input": "5 3\n-2 -2 -2 -2 -2",
"output": "6"
},
{
"input": "5 1\n998 997 985 937 998",
"output": "0"
},
{
"input": "2 2\n-742 -187",
"output": "929"
},
{
"input": "3 3\n522 597 384",
"output": "0"
},
{
"input": "4 2\n-215 -620 192 647",
"output": "835"
},
{
"input": "10 6\n557 605 685 231 910 633 130 838 -564 -85",
"output": "649"
},
{
"input": "20 14\n932 442 960 943 624 624 955 998 631 910 850 517 715 123 1000 155 -10 961 966 59",
"output": "10"
},
{
"input": "30 5\n991 997 996 967 977 999 991 986 1000 965 984 997 998 1000 958 983 974 1000 991 999 1000 978 961 992 990 998 998 978 998 1000",
"output": "0"
},
{
"input": "50 20\n-815 -947 -946 -993 -992 -846 -884 -954 -963 -733 -940 -746 -766 -930 -821 -937 -937 -999 -914 -938 -936 -975 -939 -981 -977 -952 -925 -901 -952 -978 -994 -957 -946 -896 -905 -836 -994 -951 -887 -939 -859 -953 -985 -988 -946 -829 -956 -842 -799 -886",
"output": "19441"
},
{
"input": "88 64\n999 999 1000 1000 999 996 995 1000 1000 999 1000 997 998 1000 999 1000 997 1000 993 998 994 999 998 996 1000 997 1000 1000 1000 997 1000 998 997 1000 1000 998 1000 998 999 1000 996 999 999 999 996 995 999 1000 998 999 1000 999 999 1000 1000 1000 996 1000 1000 1000 997 1000 1000 997 999 1000 1000 1000 1000 1000 999 999 1000 1000 996 999 1000 1000 995 999 1000 996 1000 998 999 999 1000 999",
"output": "0"
},
{
"input": "99 17\n-993 -994 -959 -989 -991 -995 -976 -997 -990 -1000 -996 -994 -999 -995 -1000 -983 -979 -1000 -989 -968 -994 -992 -962 -993 -999 -983 -991 -979 -995 -993 -973 -999 -995 -995 -999 -993 -995 -992 -947 -1000 -999 -998 -982 -988 -979 -993 -963 -988 -980 -990 -979 -976 -995 -999 -981 -988 -998 -999 -970 -1000 -983 -994 -943 -975 -998 -977 -973 -997 -959 -999 -983 -985 -950 -977 -977 -991 -998 -973 -987 -985 -985 -986 -984 -994 -978 -998 -989 -989 -988 -970 -985 -974 -997 -981 -962 -972 -995 -988 -993",
"output": "16984"
},
{
"input": "100 37\n205 19 -501 404 912 -435 -322 -469 -655 880 -804 -470 793 312 -108 586 -642 -928 906 605 -353 -800 745 -440 -207 752 -50 -28 498 -800 -62 -195 602 -833 489 352 536 404 -775 23 145 -512 524 759 651 -461 -427 -557 684 -366 62 592 -563 -811 64 418 -881 -308 591 -318 -145 -261 -321 -216 -18 595 -202 960 -4 219 226 -238 -882 -963 425 970 -434 -160 243 -672 -4 873 8 -633 904 -298 -151 -377 -61 -72 -677 -66 197 -716 3 -870 -30 152 -469 981",
"output": "21743"
},
{
"input": "100 99\n-931 -806 -830 -828 -916 -962 -660 -867 -952 -966 -820 -906 -724 -982 -680 -717 -488 -741 -897 -613 -986 -797 -964 -939 -808 -932 -810 -860 -641 -916 -858 -628 -821 -929 -917 -976 -664 -985 -778 -665 -624 -928 -940 -958 -884 -757 -878 -896 -634 -526 -514 -873 -990 -919 -988 -878 -650 -973 -774 -783 -733 -648 -756 -895 -833 -974 -832 -725 -841 -748 -806 -613 -924 -867 -881 -943 -864 -991 -809 -926 -777 -817 -998 -682 -910 -996 -241 -722 -964 -904 -821 -920 -835 -699 -805 -632 -779 -317 -915 -654",
"output": "81283"
},
{
"input": "100 14\n995 994 745 684 510 737 984 690 979 977 542 933 871 603 758 653 962 997 747 974 773 766 975 770 527 960 841 989 963 865 974 967 950 984 757 685 986 809 982 959 931 880 978 867 805 562 970 900 834 782 616 885 910 608 974 918 576 700 871 980 656 941 978 759 767 840 573 859 841 928 693 853 716 927 976 851 962 962 627 797 707 873 869 988 993 533 665 887 962 880 929 980 877 887 572 790 721 883 848 782",
"output": "0"
},
{
"input": "100 84\n768 946 998 752 931 912 826 1000 991 910 875 962 901 952 958 733 959 908 872 840 923 826 952 980 974 980 947 955 959 822 997 963 966 933 829 923 971 999 926 932 865 984 974 858 994 855 949 941 992 861 951 949 991 711 763 728 935 485 716 907 869 952 960 859 909 963 978 942 968 933 923 909 997 962 687 764 924 774 875 1000 961 951 987 974 848 921 966 859 995 997 974 931 886 941 974 986 906 978 998 823",
"output": "0"
},
{
"input": "100 80\n-795 -994 -833 -930 -974 -980 -950 -940 -788 -927 -583 -956 -945 -949 -809 -974 -957 -736 -967 -908 -975 -961 -986 -983 -963 -771 -952 -847 -751 -741 -982 -959 -925 -931 -839 -937 -880 -914 -858 -998 -812 -911 -862 -965 -943 -984 -738 -920 -950 -998 -909 -998 -781 -901 -677 -940 -985 -951 -675 -952 -967 -949 -882 -641 -969 -937 -975 -993 -913 -941 -807 -851 -832 -960 -939 -943 -895 -929 -528 -880 -823 -930 -888 -862 -948 -966 -962 -857 -799 -969 -833 -998 -952 -878 -946 -971 -976 -974 -723 -992",
"output": "75068"
},
{
"input": "1 1\n0",
"output": "0"
},
{
"input": "1 1\n1",
"output": "0"
},
{
"input": "1 1\n555",
"output": "0"
},
{
"input": "1 1\n-1",
"output": "1"
},
{
"input": "1 1\n-24",
"output": "24"
}
] | 1,687,679,840
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 25
| 92
| 0
|
#F
n, m = [int(x) for x in input().split()]
total = int(0)
a = [int(x) for x in input().split()]
a.sort()
for i in range(m):
if (a[i]<=0):
total+=a[i]
else :
break
print(total*-1)
|
Title: Sale
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Once Bob got to a sale of old TV sets. There were *n* TV sets at that sale. TV set with index *i* costs *a**i* bellars. Some TV sets have a negative price β their owners are ready to pay Bob if he buys their useless apparatus. Bob can Β«buyΒ» any TV sets he wants. Though he's very strong, Bob can carry at most *m* TV sets, and he has no desire to go to the sale for the second time. Please, help Bob find out the maximum sum of money that he can earn.
Input Specification:
The first line contains two space-separated integers *n* and *m* (1<=β€<=*m*<=β€<=*n*<=β€<=100) β amount of TV sets at the sale, and amount of TV sets that Bob can carry. The following line contains *n* space-separated integers *a**i* (<=-<=1000<=β€<=*a**i*<=β€<=1000) β prices of the TV sets.
Output Specification:
Output the only number β the maximum sum of money that Bob can earn, given that he can carry at most *m* TV sets.
Demo Input:
['5 3\n-6 0 35 -2 4\n', '4 2\n7 0 0 -7\n']
Demo Output:
['8\n', '7\n']
Note:
none
|
```python
#F
n, m = [int(x) for x in input().split()]
total = int(0)
a = [int(x) for x in input().split()]
a.sort()
for i in range(m):
if (a[i]<=0):
total+=a[i]
else :
break
print(total*-1)
```
| 3.977
|
1
|
A
|
Theatre Square
|
PROGRAMMING
| 1,000
|
[
"math"
] |
A. Theatre Square
|
1
|
256
|
Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=Γ<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=Γ<=*a*.
What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square.
|
The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=β€<=<=*n*,<=*m*,<=*a*<=β€<=109).
|
Write the needed number of flagstones.
|
[
"6 6 4\n"
] |
[
"4\n"
] |
none
| 0
|
[
{
"input": "6 6 4",
"output": "4"
},
{
"input": "1 1 1",
"output": "1"
},
{
"input": "2 1 1",
"output": "2"
},
{
"input": "1 2 1",
"output": "2"
},
{
"input": "2 2 1",
"output": "4"
},
{
"input": "2 1 2",
"output": "1"
},
{
"input": "1 1 3",
"output": "1"
},
{
"input": "2 3 4",
"output": "1"
},
{
"input": "1000000000 1000000000 1",
"output": "1000000000000000000"
},
{
"input": "12 13 4",
"output": "12"
},
{
"input": "222 332 5",
"output": "3015"
},
{
"input": "1000 1000 10",
"output": "10000"
},
{
"input": "1001 1000 10",
"output": "10100"
},
{
"input": "100 10001 1000000000",
"output": "1"
},
{
"input": "1000000000 1000000000 1000000000",
"output": "1"
},
{
"input": "1000000000 1000000000 999999999",
"output": "4"
},
{
"input": "1000000000 1000000000 192",
"output": "27126743055556"
},
{
"input": "1000000000 987654321 1",
"output": "987654321000000000"
},
{
"input": "456784567 1000000000 51",
"output": "175618850864484"
},
{
"input": "39916800 134217728 40320",
"output": "3295710"
}
] | 1,695,700,468
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 20
| 62
| 0
|
n, m, a = map(int, input().split())
width_stones = -(-n // a)
length_stones = -(-m // a)
total_stones = width_stones * length_stones
print(total_stones)
|
Title: Theatre Square
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=Γ<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=Γ<=*a*.
What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square.
Input Specification:
The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=β€<=<=*n*,<=*m*,<=*a*<=β€<=109).
Output Specification:
Write the needed number of flagstones.
Demo Input:
['6 6 4\n']
Demo Output:
['4\n']
Note:
none
|
```python
n, m, a = map(int, input().split())
width_stones = -(-n // a)
length_stones = -(-m // a)
total_stones = width_stones * length_stones
print(total_stones)
```
| 3.969
|
37
|
A
|
Towers
|
PROGRAMMING
| 1,000
|
[
"sortings"
] |
A. Towers
|
2
|
256
|
Little Vasya has received a young builderβs kit. The kit consists of several wooden bars, the lengths of all of them are known. The bars can be put one on the top of the other if their lengths are the same.
Vasya wants to construct the minimal number of towers from the bars. Help Vasya to use the bars in the best way possible.
|
The first line contains an integer *N* (1<=β€<=*N*<=β€<=1000) β the number of bars at Vasyaβs disposal. The second line contains *N* space-separated integers *l**i* β the lengths of the bars. All the lengths are natural numbers not exceeding 1000.
|
In one line output two numbers β the height of the largest tower and their total number. Remember that Vasya should use all the bars.
|
[
"3\n1 2 3\n",
"4\n6 5 6 7\n"
] |
[
"1 3\n",
"2 3\n"
] |
none
| 500
|
[
{
"input": "3\n1 2 3",
"output": "1 3"
},
{
"input": "4\n6 5 6 7",
"output": "2 3"
},
{
"input": "4\n3 2 1 1",
"output": "2 3"
},
{
"input": "4\n1 2 3 3",
"output": "2 3"
},
{
"input": "3\n20 22 36",
"output": "1 3"
},
{
"input": "25\n47 30 94 41 45 20 96 51 110 129 24 116 9 47 32 82 105 114 116 75 154 151 70 42 162",
"output": "2 23"
},
{
"input": "45\n802 664 442 318 318 827 417 878 711 291 231 414 807 553 657 392 279 202 386 606 465 655 658 112 887 15 25 502 95 44 679 775 942 609 209 871 31 234 4 231 150 110 22 823 193",
"output": "2 43"
},
{
"input": "63\n93 180 116 7 8 179 268 279 136 94 221 153 264 190 278 19 19 63 153 26 158 225 25 49 89 218 111 149 255 225 197 122 243 80 3 224 107 178 202 17 53 92 69 42 228 24 81 205 95 8 265 82 228 156 127 241 172 159 106 60 67 155 111",
"output": "2 57"
},
{
"input": "83\n246 535 994 33 390 927 321 97 223 922 812 705 79 80 977 457 476 636 511 137 6 360 815 319 717 674 368 551 714 628 278 713 761 553 184 414 623 753 428 214 581 115 439 61 677 216 772 592 187 603 658 310 439 559 870 376 109 321 189 337 277 26 70 734 796 907 979 693 570 227 345 650 737 633 701 914 134 403 972 940 371 6 642",
"output": "2 80"
},
{
"input": "105\n246 57 12 204 165 123 246 68 191 310 3 152 386 333 374 257 158 104 333 50 80 290 8 340 101 76 221 316 388 289 138 359 316 26 93 290 105 178 81 195 41 196 218 180 244 292 187 97 315 323 174 119 248 239 92 312 31 2 101 180 307 170 338 314 163 281 217 31 142 238 280 190 190 156 70 74 329 113 151 8 141 313 366 40 253 116 168 124 135 230 294 266 353 389 371 359 195 200 183 237 93 102 315 118 188",
"output": "2 92"
},
{
"input": "123\n112 277 170 247 252 115 157 293 256 143 196 90 12 164 164 42 8 223 167 109 175 232 239 111 148 51 9 254 93 32 268 162 231 91 47 162 161 191 195 145 247 292 129 199 230 94 144 217 18 205 176 20 143 198 121 243 211 262 230 277 195 255 108 290 220 275 158 2 286 200 60 267 278 207 123 150 123 116 131 13 12 226 33 244 30 275 263 45 158 192 254 149 242 176 62 224 221 288 250 160 155 225 132 143 276 293 218 145 197 175 33 129 79 206 210 192 222 262 190 52 274 243 233",
"output": "3 101"
},
{
"input": "5\n5 5 5 5 5",
"output": "5 1"
},
{
"input": "3\n1000 1000 1000",
"output": "3 1"
},
{
"input": "1\n1000",
"output": "1 1"
},
{
"input": "1\n1",
"output": "1 1"
},
{
"input": "5\n1 1000 1000 1000 1000",
"output": "4 2"
},
{
"input": "5\n1000 1000 1000 8 7",
"output": "3 3"
}
] | 1,592,203,350
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 61
| 374
| 307,200
|
n=int(input())
lst=list(map(int,input().split(' ')))
lstc={}
for i in range(n):
if(lst[i] in lstc.keys()):
continue
else:
count=0
for j in range(n):
if(lst[i]==lst[j]):
count+=1
lstc[lst[i]]=count
height=max(lstc.values())
print(height,len(lstc.values()))
|
Title: Towers
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Little Vasya has received a young builderβs kit. The kit consists of several wooden bars, the lengths of all of them are known. The bars can be put one on the top of the other if their lengths are the same.
Vasya wants to construct the minimal number of towers from the bars. Help Vasya to use the bars in the best way possible.
Input Specification:
The first line contains an integer *N* (1<=β€<=*N*<=β€<=1000) β the number of bars at Vasyaβs disposal. The second line contains *N* space-separated integers *l**i* β the lengths of the bars. All the lengths are natural numbers not exceeding 1000.
Output Specification:
In one line output two numbers β the height of the largest tower and their total number. Remember that Vasya should use all the bars.
Demo Input:
['3\n1 2 3\n', '4\n6 5 6 7\n']
Demo Output:
['1 3\n', '2 3\n']
Note:
none
|
```python
n=int(input())
lst=list(map(int,input().split(' ')))
lstc={}
for i in range(n):
if(lst[i] in lstc.keys()):
continue
else:
count=0
for j in range(n):
if(lst[i]==lst[j]):
count+=1
lstc[lst[i]]=count
height=max(lstc.values())
print(height,len(lstc.values()))
```
| 3.905928
|
592
|
A
|
PawnChess
|
PROGRAMMING
| 1,200
|
[
"implementation"
] | null | null |
Galois is one of the strongest chess players of Byteforces. He has even invented a new variant of chess, which he named Β«PawnChessΒ».
This new game is played on a board consisting of 8 rows and 8 columns. At the beginning of every game some black and white pawns are placed on the board. The number of black pawns placed is not necessarily equal to the number of white pawns placed.
Lets enumerate rows and columns with integers from 1 to 8. Rows are numbered from top to bottom, while columns are numbered from left to right. Now we denote as (*r*,<=*c*) the cell located at the row *r* and at the column *c*.
There are always two players A and B playing the game. Player A plays with white pawns, while player B plays with black ones. The goal of player A is to put any of his pawns to the row 1, while player B tries to put any of his pawns to the row 8. As soon as any of the players completes his goal the game finishes immediately and the succeeded player is declared a winner.
Player A moves first and then they alternate turns. On his move player A must choose exactly one white pawn and move it one step upward and player B (at his turn) must choose exactly one black pawn and move it one step down. Any move is possible only if the targeted cell is empty. It's guaranteed that for any scenario of the game there will always be at least one move available for any of the players.
Moving upward means that the pawn located in (*r*,<=*c*) will go to the cell (*r*<=-<=1,<=*c*), while moving down means the pawn located in (*r*,<=*c*) will go to the cell (*r*<=+<=1,<=*c*). Again, the corresponding cell must be empty, i.e. not occupied by any other pawn of any color.
Given the initial disposition of the board, determine who wins the game if both players play optimally. Note that there will always be a winner due to the restriction that for any game scenario both players will have some moves available.
|
The input consists of the board description given in eight lines, each line contains eight characters. Character 'B' is used to denote a black pawn, and character 'W' represents a white pawn. Empty cell is marked with '.'.
It's guaranteed that there will not be white pawns on the first row neither black pawns on the last row.
|
Print 'A' if player A wins the game on the given board, and 'B' if player B will claim the victory. Again, it's guaranteed that there will always be a winner on the given board.
|
[
"........\n........\n.B....B.\n....W...\n........\n..W.....\n........\n........\n",
"..B.....\n..W.....\n......B.\n........\n.....W..\n......B.\n........\n........\n"
] |
[
"A\n",
"B\n"
] |
In the first sample player A is able to complete his goal in 3 steps by always moving a pawn initially located at (4,β5). Player B needs at least 5 steps for any of his pawns to reach the row 8. Hence, player A will be the winner.
| 500
|
[
{
"input": ".BB.B.B.\nB..B..B.\n.B.BB...\nBB.....B\nBBB....B\nB..BB...\nBB.B...B\n....WWW.",
"output": "B"
},
{
"input": "B.B.BB.B\nW.WWW.WW\n.WWWWW.W\nW.BB.WBW\n.W..BBWB\nBB.WWBBB\n.W.W.WWB\nWWW..WW.",
"output": "A"
},
{
"input": "BB..BB..\nBW.W.W.B\n..B.....\n.....BB.\n.B..B..B\n........\n...BB.B.\nW.WWWW.W",
"output": "A"
},
{
"input": "BB......\nW....BBW\n........\n.B.B.BBB\n....BB..\nB....BB.\n...WWWW.\n....WW..",
"output": "A"
},
{
"input": ".B.B..B.\nB.B....B\n...B.B.B\n..B.W..B\n.BBB.B.B\nB.BB.B.B\nBB..BBBB\nW.W.W.WW",
"output": "B"
},
{
"input": "..BB....\n.B.B.B.B\n..B.B...\n..B..B.B\nWWWBWWB.\n.BB...B.\n..BBB...\n......W.",
"output": "B"
},
{
"input": "..BB....\n.WBWBWBB\n.....BBB\n..WW....\n.W.W...W\nWWW...W.\n.W....W.\nW...W.W.",
"output": "A"
},
{
"input": "....BB..\nBB......\n.B.....B\nWW..WWW.\n...BB.B.\nB...BB..\n..W..WWW\n...W...W",
"output": "B"
},
{
"input": "B...BBBB\n...BBB..\nBBWBWW.W\n.B..BB.B\nW..W..WW\nW.WW....\n........\nWW.....W",
"output": "A"
},
{
"input": ".B......\n.B....B.\n...W....\n......W.\nW.WWWW.W\nW.WW....\n..WWW...\n..W...WW",
"output": "A"
},
{
"input": "B.......\nBBB.....\n.B....B.\n.W.BWB.W\n......B.\nW..WW...\n...W....\nW...W..W",
"output": "A"
},
{
"input": ".....B..\n........\n........\n.BB..B..\n..BB....\n........\n....WWW.\n......W.",
"output": "B"
},
{
"input": "B.B...B.\n...BBBBB\n....B...\n...B...B\nB.B.B..B\n........\n........\nWWW..WW.",
"output": "B"
},
{
"input": "B.B...B.\n........\n.......B\n.BB....B\n.....W..\n.W.WW.W.\n...W.WW.\nW..WW..W",
"output": "A"
},
{
"input": "......B.\nB....B..\n...B.BB.\n...B....\n........\n..W....W\nWW......\n.W....W.",
"output": "B"
},
{
"input": ".BBB....\nB.B.B...\nB.BB.B..\nB.BB.B.B\n........\n........\nW.....W.\n..WW..W.",
"output": "B"
},
{
"input": "..B..BBB\n........\n........\n........\n...W.W..\n...W..W.\nW.......\n..W...W.",
"output": "A"
},
{
"input": "........\n.B.B....\n...B..BB\n........\n........\nW...W...\nW...W...\nW.WW.W..",
"output": "A"
},
{
"input": "B....BB.\n...B...B\n.B......\n........\n........\n........\n........\n....W..W",
"output": "B"
},
{
"input": "...BB.BB\nBB...B..\n........\n........\n........\n........\n..W..W..\n......W.",
"output": "A"
},
{
"input": "...BB...\n........\n........\n........\n........\n........\n......W.\nWW...WW.",
"output": "A"
},
{
"input": "...B.B..\n........\n........\n........\n........\n........\n........\nWWW...WW",
"output": "A"
},
{
"input": "BBBBBBB.\n........\n........\n........\n........\n........\n........\n.WWWWWWW",
"output": "A"
},
{
"input": ".BBBBBB.\nB.......\n........\n........\n........\n........\n........\n.WWWWWWW",
"output": "B"
},
{
"input": ".BBBBBBB\n........\n........\n........\n........\n........\n........\nWWWWWWW.",
"output": "A"
},
{
"input": ".BBBBBB.\n.......B\n........\n........\n........\n........\n........\nWWWWWWW.",
"output": "B"
},
{
"input": "B..BB...\n..B...B.\n.WBB...B\nBW......\nW.B...W.\n..BBW.B.\nBW..BB..\n......W.",
"output": "B"
},
{
"input": "B.BBBBBB\nB..BBB.B\nW.BB.W.B\nB.BWBB.B\nBWBWBBBB\n...BBBBB\nB.B...BB\nWW..WW.W",
"output": "B"
},
{
"input": "BBBB.BBB\nBBBB.B.B\nB.B..BBB\nB.BB.BWW\nB.BB.BBB\nB.BB.BBB\n..BW.BB.\nW.WWWWWW",
"output": "B"
},
{
"input": "BBBB.BBB\n.B....WB\nBB.B...B\nWWWW.WWB\nBB...BWW\nWWW..BBB\nW.BW.BB.\nWWWWWWW.",
"output": "B"
},
{
"input": "B.BBBBBB\nW.WWBBBW\nW.BB.WBB\nW.W.BBBW\nW.BWW.WB\nB..B..BB\nB.B.W.BB\nWWWWW.WW",
"output": "B"
},
{
"input": "BBBBBB.B\n.BBWBB.B\nWWW..B.W\n..WW.W.W\nBWB..W.W\n..BW.B.W\nB..B....\nWWWW.WWW",
"output": "B"
},
{
"input": ".B...BB.\nWBB.BWBB\n.BWBW...\n..W...B.\nWB.BWW..\nWBW.....\n.W..W.B.\n.W.W.WW.",
"output": "A"
},
{
"input": ".B..BBBB\nBB...WWB\nB..B.W.B\nWB.W...B\n...W.WW.\nW.....W.\nWB.W.W.W\n.WW...WW",
"output": "A"
},
{
"input": "B.BBBBBB\nW.BB.W.B\nW.BBW...\n..WWWW.B\n....W..B\n.WW.W..W\n.W..WW.W\nW.W....W",
"output": "A"
},
{
"input": "........\n.B......\n.W......\n........\n....B...\n........\n........\n.......W",
"output": "B"
}
] | 1,489,420,890
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 56
| 109
| 23,142,400
|
grid = []
for k in range(8):
grid.append(input())
bbc = [] #blocked black columns
bwc = [] #blocked white columns
#first processing : downwards
white = 8
for k in range(8):
for j in range(8):
if grid[k][j] == "B":
bbc.append(j)
if grid[k][j] == "W" and not j in bbc and white == 8:
white = k
black = 8
for k in range(7,-1,-1):
for j in range(8):
if grid[k][j] == "W":
bwc.append(j)
if grid[k][j] == "B" and not j in bwc and black == 8:
black = 7-k
if white <= black:
print("A")
else:
print("B")
|
Title: PawnChess
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Galois is one of the strongest chess players of Byteforces. He has even invented a new variant of chess, which he named Β«PawnChessΒ».
This new game is played on a board consisting of 8 rows and 8 columns. At the beginning of every game some black and white pawns are placed on the board. The number of black pawns placed is not necessarily equal to the number of white pawns placed.
Lets enumerate rows and columns with integers from 1 to 8. Rows are numbered from top to bottom, while columns are numbered from left to right. Now we denote as (*r*,<=*c*) the cell located at the row *r* and at the column *c*.
There are always two players A and B playing the game. Player A plays with white pawns, while player B plays with black ones. The goal of player A is to put any of his pawns to the row 1, while player B tries to put any of his pawns to the row 8. As soon as any of the players completes his goal the game finishes immediately and the succeeded player is declared a winner.
Player A moves first and then they alternate turns. On his move player A must choose exactly one white pawn and move it one step upward and player B (at his turn) must choose exactly one black pawn and move it one step down. Any move is possible only if the targeted cell is empty. It's guaranteed that for any scenario of the game there will always be at least one move available for any of the players.
Moving upward means that the pawn located in (*r*,<=*c*) will go to the cell (*r*<=-<=1,<=*c*), while moving down means the pawn located in (*r*,<=*c*) will go to the cell (*r*<=+<=1,<=*c*). Again, the corresponding cell must be empty, i.e. not occupied by any other pawn of any color.
Given the initial disposition of the board, determine who wins the game if both players play optimally. Note that there will always be a winner due to the restriction that for any game scenario both players will have some moves available.
Input Specification:
The input consists of the board description given in eight lines, each line contains eight characters. Character 'B' is used to denote a black pawn, and character 'W' represents a white pawn. Empty cell is marked with '.'.
It's guaranteed that there will not be white pawns on the first row neither black pawns on the last row.
Output Specification:
Print 'A' if player A wins the game on the given board, and 'B' if player B will claim the victory. Again, it's guaranteed that there will always be a winner on the given board.
Demo Input:
['........\n........\n.B....B.\n....W...\n........\n..W.....\n........\n........\n', '..B.....\n..W.....\n......B.\n........\n.....W..\n......B.\n........\n........\n']
Demo Output:
['A\n', 'B\n']
Note:
In the first sample player A is able to complete his goal in 3 steps by always moving a pawn initially located at (4,β5). Player B needs at least 5 steps for any of his pawns to reach the row 8. Hence, player A will be the winner.
|
```python
grid = []
for k in range(8):
grid.append(input())
bbc = [] #blocked black columns
bwc = [] #blocked white columns
#first processing : downwards
white = 8
for k in range(8):
for j in range(8):
if grid[k][j] == "B":
bbc.append(j)
if grid[k][j] == "W" and not j in bbc and white == 8:
white = k
black = 8
for k in range(7,-1,-1):
for j in range(8):
if grid[k][j] == "W":
bwc.append(j)
if grid[k][j] == "B" and not j in bwc and black == 8:
black = 7-k
if white <= black:
print("A")
else:
print("B")
```
| 3
|
|
432
|
A
|
Choosing Teams
|
PROGRAMMING
| 800
|
[
"greedy",
"implementation",
"sortings"
] | null | null |
The Saratov State University Olympiad Programmers Training Center (SSU OPTC) has *n* students. For each student you know the number of times he/she has participated in the ACM ICPC world programming championship. According to the ACM ICPC rules, each person can participate in the world championship at most 5 times.
The head of the SSU OPTC is recently gathering teams to participate in the world championship. Each team must consist of exactly three people, at that, any person cannot be a member of two or more teams. What maximum number of teams can the head make if he wants each team to participate in the world championship with the same members at least *k* times?
|
The first line contains two integers, *n* and *k* (1<=β€<=*n*<=β€<=2000;Β 1<=β€<=*k*<=β€<=5). The next line contains *n* integers: *y*1,<=*y*2,<=...,<=*y**n* (0<=β€<=*y**i*<=β€<=5), where *y**i* shows the number of times the *i*-th person participated in the ACM ICPC world championship.
|
Print a single number β the answer to the problem.
|
[
"5 2\n0 4 5 1 0\n",
"6 4\n0 1 2 3 4 5\n",
"6 5\n0 0 0 0 0 0\n"
] |
[
"1\n",
"0\n",
"2\n"
] |
In the first sample only one team could be made: the first, the fourth and the fifth participants.
In the second sample no teams could be created.
In the third sample two teams could be created. Any partition into two teams fits.
| 500
|
[
{
"input": "5 2\n0 4 5 1 0",
"output": "1"
},
{
"input": "6 4\n0 1 2 3 4 5",
"output": "0"
},
{
"input": "6 5\n0 0 0 0 0 0",
"output": "2"
},
{
"input": "3 4\n0 1 0",
"output": "1"
},
{
"input": "3 4\n0 2 0",
"output": "0"
},
{
"input": "6 5\n0 0 0 0 0 0",
"output": "2"
},
{
"input": "12 2\n0 1 2 3 4 0 1 2 3 4 0 1",
"output": "3"
},
{
"input": "15 2\n0 1 2 3 4 0 1 2 3 4 0 1 2 3 4",
"output": "4"
},
{
"input": "13 1\n5 0 5 0 1 5 0 4 1 1 1 4 1",
"output": "3"
},
{
"input": "20 1\n5 0 4 2 2 3 2 1 2 4 1 3 5 5 5 4 4 1 3 0",
"output": "5"
},
{
"input": "20 2\n1 4 5 1 3 4 0 5 0 3 3 0 4 0 0 1 4 2 0 4",
"output": "4"
},
{
"input": "20 2\n1 4 5 1 3 4 0 5 0 3 3 0 4 0 0 1 4 2 0 4",
"output": "4"
},
{
"input": "20 2\n1 4 5 1 3 4 0 5 0 3 3 0 4 0 0 1 4 2 0 4",
"output": "4"
},
{
"input": "20 2\n1 4 5 1 3 4 0 5 0 3 3 0 4 0 0 1 4 2 0 4",
"output": "4"
},
{
"input": "6 3\n4 4 4 4 4 4",
"output": "0"
},
{
"input": "1 4\n1",
"output": "0"
},
{
"input": "1 1\n0",
"output": "0"
},
{
"input": "2 3\n0 0",
"output": "0"
}
] | 1,658,973,787
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 35
| 62
| 1,843,200
|
n,k = list(map(int,input().split()))
arr = list(map(int,input().split()))
for i in range(n):
arr[i] = (5-arr[i])-k
ans = 0
for i in range(n):
if arr[i] >=0:
ans += 1
print(ans//3)
|
Title: Choosing Teams
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Saratov State University Olympiad Programmers Training Center (SSU OPTC) has *n* students. For each student you know the number of times he/she has participated in the ACM ICPC world programming championship. According to the ACM ICPC rules, each person can participate in the world championship at most 5 times.
The head of the SSU OPTC is recently gathering teams to participate in the world championship. Each team must consist of exactly three people, at that, any person cannot be a member of two or more teams. What maximum number of teams can the head make if he wants each team to participate in the world championship with the same members at least *k* times?
Input Specification:
The first line contains two integers, *n* and *k* (1<=β€<=*n*<=β€<=2000;Β 1<=β€<=*k*<=β€<=5). The next line contains *n* integers: *y*1,<=*y*2,<=...,<=*y**n* (0<=β€<=*y**i*<=β€<=5), where *y**i* shows the number of times the *i*-th person participated in the ACM ICPC world championship.
Output Specification:
Print a single number β the answer to the problem.
Demo Input:
['5 2\n0 4 5 1 0\n', '6 4\n0 1 2 3 4 5\n', '6 5\n0 0 0 0 0 0\n']
Demo Output:
['1\n', '0\n', '2\n']
Note:
In the first sample only one team could be made: the first, the fourth and the fifth participants.
In the second sample no teams could be created.
In the third sample two teams could be created. Any partition into two teams fits.
|
```python
n,k = list(map(int,input().split()))
arr = list(map(int,input().split()))
for i in range(n):
arr[i] = (5-arr[i])-k
ans = 0
for i in range(n):
if arr[i] >=0:
ans += 1
print(ans//3)
```
| 3
|
|
595
|
A
|
Vitaly and Night
|
PROGRAMMING
| 800
|
[
"constructive algorithms",
"implementation"
] | null | null |
One day Vitaly was going home late at night and wondering: how many people aren't sleeping at that moment? To estimate, Vitaly decided to look which windows are lit in the house he was passing by at that moment.
Vitaly sees a building of *n* floors and 2Β·*m* windows on each floor. On each floor there are *m* flats numbered from 1 to *m*, and two consecutive windows correspond to each flat. If we number the windows from 1 to 2Β·*m* from left to right, then the *j*-th flat of the *i*-th floor has windows 2Β·*j*<=-<=1 and 2Β·*j* in the corresponding row of windows (as usual, floors are enumerated from the bottom). Vitaly thinks that people in the flat aren't sleeping at that moment if at least one of the windows corresponding to this flat has lights on.
Given the information about the windows of the given house, your task is to calculate the number of flats where, according to Vitaly, people aren't sleeping.
|
The first line of the input contains two integers *n* and *m* (1<=β€<=*n*,<=*m*<=β€<=100)Β β the number of floors in the house and the number of flats on each floor respectively.
Next *n* lines describe the floors from top to bottom and contain 2Β·*m* characters each. If the *i*-th window of the given floor has lights on, then the *i*-th character of this line is '1', otherwise it is '0'.
|
Print a single integerΒ β the number of flats that have lights on in at least one window, that is, the flats where, according to Vitaly, people aren't sleeping.
|
[
"2 2\n0 0 0 1\n1 0 1 1\n",
"1 3\n1 1 0 1 0 0\n"
] |
[
"3\n",
"2\n"
] |
In the first test case the house has two floors, two flats on each floor. That is, in total there are 4 flats. The light isn't on only on the second floor in the left flat. That is, in both rooms of the flat the light is off.
In the second test case the house has one floor and the first floor has three flats. The light is on in the leftmost flat (in both windows) and in the middle flat (in one window). In the right flat the light is off.
| 500
|
[
{
"input": "2 2\n0 0 0 1\n1 0 1 1",
"output": "3"
},
{
"input": "1 3\n1 1 0 1 0 0",
"output": "2"
},
{
"input": "3 3\n1 1 1 1 1 1\n1 1 0 1 1 0\n1 0 0 0 1 1",
"output": "8"
},
{
"input": "1 5\n1 0 1 1 1 0 1 1 1 1",
"output": "5"
},
{
"input": "1 100\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 0 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 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": "99"
},
{
"input": "1 100\n0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 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 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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 0 0 0 0 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 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0",
"output": "6"
},
{
"input": "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 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0",
"output": "0"
},
{
"input": "100 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 0\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n0 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 0\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 0\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n0 1\n1 1\n1 1\n1 0\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1",
"output": "100"
},
{
"input": "100 1\n0 0\n0 0\n0 1\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\n1 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 1\n0 0\n0 0\n0 0\n1 0\n0 0\n0 0\n0 0\n1 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 1\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 1\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\n1 0",
"output": "8"
},
{
"input": "100 1\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": "1 1\n0 0",
"output": "0"
},
{
"input": "1 1\n0 1",
"output": "1"
},
{
"input": "1 1\n1 0",
"output": "1"
},
{
"input": "1 1\n1 1",
"output": "1"
}
] | 1,529,501,681
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 36
| 155
| 614,400
|
n,m=list(map(int,input().split()))
c=0
for i in range(n):
l=list(map(int,input().split()))
for i in range(0,len(l),2):
if l[i]==0 and l[i+1]==1 or l[i]==1 and l[i+1]==1 or l[i]==1 and l[i+1]==0:
c+=1
i+=1
#i+=1
print(c)
|
Title: Vitaly and Night
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Vitaly was going home late at night and wondering: how many people aren't sleeping at that moment? To estimate, Vitaly decided to look which windows are lit in the house he was passing by at that moment.
Vitaly sees a building of *n* floors and 2Β·*m* windows on each floor. On each floor there are *m* flats numbered from 1 to *m*, and two consecutive windows correspond to each flat. If we number the windows from 1 to 2Β·*m* from left to right, then the *j*-th flat of the *i*-th floor has windows 2Β·*j*<=-<=1 and 2Β·*j* in the corresponding row of windows (as usual, floors are enumerated from the bottom). Vitaly thinks that people in the flat aren't sleeping at that moment if at least one of the windows corresponding to this flat has lights on.
Given the information about the windows of the given house, your task is to calculate the number of flats where, according to Vitaly, people aren't sleeping.
Input Specification:
The first line of the input contains two integers *n* and *m* (1<=β€<=*n*,<=*m*<=β€<=100)Β β the number of floors in the house and the number of flats on each floor respectively.
Next *n* lines describe the floors from top to bottom and contain 2Β·*m* characters each. If the *i*-th window of the given floor has lights on, then the *i*-th character of this line is '1', otherwise it is '0'.
Output Specification:
Print a single integerΒ β the number of flats that have lights on in at least one window, that is, the flats where, according to Vitaly, people aren't sleeping.
Demo Input:
['2 2\n0 0 0 1\n1 0 1 1\n', '1 3\n1 1 0 1 0 0\n']
Demo Output:
['3\n', '2\n']
Note:
In the first test case the house has two floors, two flats on each floor. That is, in total there are 4 flats. The light isn't on only on the second floor in the left flat. That is, in both rooms of the flat the light is off.
In the second test case the house has one floor and the first floor has three flats. The light is on in the leftmost flat (in both windows) and in the middle flat (in one window). In the right flat the light is off.
|
```python
n,m=list(map(int,input().split()))
c=0
for i in range(n):
l=list(map(int,input().split()))
for i in range(0,len(l),2):
if l[i]==0 and l[i+1]==1 or l[i]==1 and l[i+1]==1 or l[i]==1 and l[i+1]==0:
c+=1
i+=1
#i+=1
print(c)
```
| 3
|
|
44
|
C
|
Holidays
|
PROGRAMMING
| 1,300
|
[
"implementation"
] |
C. Holidays
|
2
|
256
|
School holidays come in Berland. The holidays are going to continue for *n* days. The students of school β*N* are having the time of their lives and the IT teacher Marina Sergeyevna, who has spent all the summer busy checking the BSE (Berland State Examination) results, has finally taken a vacation break! Some people are in charge of the daily watering of flowers in shifts according to the schedule. However when Marina Sergeyevna was making the schedule, she was so tired from work and so lost in dreams of the oncoming vacation that she perhaps made several mistakes. In fact, it is possible that according to the schedule, on some days during the holidays the flowers will not be watered or will be watered multiple times. Help Marina Sergeyevna to find a mistake.
|
The first input line contains two numbers *n* and *m* (1<=β€<=*n*,<=*m*<=β€<=100) β the number of days in Berland holidays and the number of people in charge of the watering respectively. The next *m* lines contain the description of the duty schedule. Each line contains two integers *a**i* and *b**i* (1<=β€<=*a**i*<=β€<=*b**i*<=β€<=*n*), meaning that the *i*-th person in charge should water the flowers from the *a**i*-th to the *b**i*-th day inclusively, once a day. The duty shifts are described sequentially, i.e. *b**i*<=β€<=*a**i*<=+<=1 for all *i* from 1 to *n*<=-<=1 inclusively.
|
Print "OK" (without quotes), if the schedule does not contain mistakes. Otherwise you have to find the minimal number of a day when the flowers will not be watered or will be watered multiple times, and output two integers β the day number and the number of times the flowers will be watered that day.
|
[
"10 5\n1 2\n3 3\n4 6\n7 7\n8 10\n",
"10 5\n1 2\n2 3\n4 5\n7 8\n9 10\n",
"10 5\n1 2\n3 3\n5 7\n7 7\n7 10\n"
] |
[
"OK\n",
"2 2\n",
"4 0\n"
] |
Keep in mind that in the second sample the mistake occurs not only on the second day, but also on the sixth day, when nobody waters the flowers. However, you have to print the second day, i.e. the day with the minimal number.
| 0
|
[
{
"input": "10 5\n1 2\n3 3\n4 6\n7 7\n8 10",
"output": "OK"
},
{
"input": "10 5\n1 2\n2 3\n4 5\n7 8\n9 10",
"output": "2 2"
},
{
"input": "10 5\n1 2\n3 3\n5 7\n7 7\n7 10",
"output": "4 0"
},
{
"input": "5 4\n1 1\n2 2\n3 3\n4 5",
"output": "OK"
},
{
"input": "100 50\n1 2\n3 3\n4 5\n6 8\n9 10\n11 11\n12 14\n15 15\n16 16\n17 17\n18 18\n19 19\n20 21\n22 23\n24 24\n25 26\n27 30\n31 34\n35 37\n38 38\n39 40\n41 43\n44 46\n47 53\n54 54\n55 55\n56 59\n60 60\n61 61\n62 64\n65 69\n70 72\n73 73\n74 74\n75 76\n77 79\n80 82\n83 83\n84 84\n85 85\n86 86\n87 88\n89 89\n90 90\n91 91\n92 92\n93 93\n94 97\n98 98\n99 100",
"output": "OK"
},
{
"input": "50 50\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\n21 21\n22 22\n23 23\n24 24\n25 25\n26 26\n27 27\n28 28\n29 29\n30 30\n31 31\n32 32\n33 33\n34 34\n35 35\n36 36\n37 37\n38 38\n39 39\n40 40\n41 41\n42 42\n43 43\n44 44\n45 45\n46 46\n47 47\n48 48\n49 49\n50 50",
"output": "OK"
},
{
"input": "5 1\n1 5",
"output": "OK"
},
{
"input": "6 2\n1 5\n6 6",
"output": "OK"
},
{
"input": "7 5\n1 1\n2 2\n3 3\n4 4\n5 7",
"output": "OK"
},
{
"input": "10 2\n1 2\n3 10",
"output": "OK"
},
{
"input": "21 15\n1 1\n2 2\n3 3\n4 5\n6 6\n7 7\n8 8\n9 9\n10 11\n12 12\n13 13\n14 14\n15 17\n18 19\n20 21",
"output": "OK"
},
{
"input": "100 7\n1 8\n9 26\n27 28\n29 30\n31 38\n39 95\n96 100",
"output": "OK"
},
{
"input": "100 13\n1 4\n5 11\n12 18\n19 24\n25 31\n32 38\n39 39\n40 45\n46 55\n56 69\n70 70\n71 75\n76 100",
"output": "OK"
},
{
"input": "100 50\n1 8\n9 12\n13 19\n20 22\n23 27\n28 31\n32 36\n36 40\n40 43\n47 47\n48 51\n51 55\n62 63\n69 77\n77 84\n85 90\n98 99\n100 100\n100 100\n100 100\n100 100\n100 100\n100 100\n100 100\n100 100\n100 100\n100 100\n100 100\n100 100\n100 100\n100 100\n100 100\n100 100\n100 100\n100 100\n100 100\n100 100\n100 100\n100 100\n100 100\n100 100\n100 100\n100 100\n100 100\n100 100\n100 100\n100 100\n100 100\n100 100\n100 100",
"output": "36 2"
},
{
"input": "1 1\n1 1",
"output": "OK"
},
{
"input": "10 1\n2 3",
"output": "1 0"
},
{
"input": "10 9\n1 1\n2 2\n3 4\n6 6\n8 8\n8 10\n10 10\n10 10\n10 10",
"output": "5 0"
},
{
"input": "27 10\n1 1\n2 3\n4 5\n6 7\n8 9\n10 11\n12 13\n14 15\n16 17\n17 18",
"output": "17 2"
},
{
"input": "67 15\n1 6\n7 14\n15 16\n17 23\n24 30\n31 34\n35 41\n42 48\n48 56\n56 62\n66 67\n67 67\n67 67\n67 67\n67 67",
"output": "48 2"
},
{
"input": "68 13\n1 2\n3 11\n12 21\n22 30\n31 38\n39 43\n44 44\n45 46\n47 50\n51 55\n64 68\n68 68\n68 68",
"output": "56 0"
},
{
"input": "47 45\n1 3\n4 7\n8 11\n12 15\n16 18\n19 23\n24 26\n27 28\n29 31\n32 33\n34 37\n37 40\n45 45\n46 46\n46 47\n47 47\n47 47\n47 47\n47 47\n47 47\n47 47\n47 47\n47 47\n47 47\n47 47\n47 47\n47 47\n47 47\n47 47\n47 47\n47 47\n47 47\n47 47\n47 47\n47 47\n47 47\n47 47\n47 47\n47 47\n47 47\n47 47\n47 47\n47 47\n47 47\n47 47",
"output": "37 2"
},
{
"input": "5 2\n1 1\n3 3",
"output": "2 0"
},
{
"input": "5 3\n1 2\n3 3\n3 5",
"output": "3 2"
},
{
"input": "5 4\n1 2\n3 4\n5 5\n5 5",
"output": "5 2"
},
{
"input": "10 5\n2 5\n5 6\n7 9\n9 9\n9 10",
"output": "1 0"
},
{
"input": "20 6\n1 1\n1 1\n1 3\n5 7\n7 13\n14 20",
"output": "1 3"
},
{
"input": "20 7\n1 3\n4 8\n8 8\n8 8\n8 9\n15 20\n20 20",
"output": "8 4"
},
{
"input": "20 7\n1 5\n6 8\n10 10\n12 15\n15 16\n16 16\n16 20",
"output": "9 0"
},
{
"input": "20 13\n1 2\n3 4\n5 7\n7 7\n7 7\n7 9\n10 11\n11 11\n11 12\n12 12\n12 13\n15 18\n19 20",
"output": "7 4"
},
{
"input": "20 7\n1 3\n4 5\n6 6\n7 11\n12 15\n16 17\n18 19",
"output": "20 0"
},
{
"input": "20 7\n1 6\n7 9\n10 11\n12 14\n15 19\n19 20\n20 20",
"output": "19 2"
}
] | 1,563,266,875
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 35
| 310
| 0
|
n,m = map(int,input().split())
Ln = [0]*n
while m!=0:
a,b = map(int,input().split())
for i in range(a,b+1):
Ln[i-1]+=1
m-=1
for i in range(0,len(Ln)):
if Ln[i]==0 :
print(i+1,0)
exit(0)
elif Ln[i]>1:
print(i+1,Ln[i])
exit(0)
print('OK')
exit(0)
|
Title: Holidays
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
School holidays come in Berland. The holidays are going to continue for *n* days. The students of school β*N* are having the time of their lives and the IT teacher Marina Sergeyevna, who has spent all the summer busy checking the BSE (Berland State Examination) results, has finally taken a vacation break! Some people are in charge of the daily watering of flowers in shifts according to the schedule. However when Marina Sergeyevna was making the schedule, she was so tired from work and so lost in dreams of the oncoming vacation that she perhaps made several mistakes. In fact, it is possible that according to the schedule, on some days during the holidays the flowers will not be watered or will be watered multiple times. Help Marina Sergeyevna to find a mistake.
Input Specification:
The first input line contains two numbers *n* and *m* (1<=β€<=*n*,<=*m*<=β€<=100) β the number of days in Berland holidays and the number of people in charge of the watering respectively. The next *m* lines contain the description of the duty schedule. Each line contains two integers *a**i* and *b**i* (1<=β€<=*a**i*<=β€<=*b**i*<=β€<=*n*), meaning that the *i*-th person in charge should water the flowers from the *a**i*-th to the *b**i*-th day inclusively, once a day. The duty shifts are described sequentially, i.e. *b**i*<=β€<=*a**i*<=+<=1 for all *i* from 1 to *n*<=-<=1 inclusively.
Output Specification:
Print "OK" (without quotes), if the schedule does not contain mistakes. Otherwise you have to find the minimal number of a day when the flowers will not be watered or will be watered multiple times, and output two integers β the day number and the number of times the flowers will be watered that day.
Demo Input:
['10 5\n1 2\n3 3\n4 6\n7 7\n8 10\n', '10 5\n1 2\n2 3\n4 5\n7 8\n9 10\n', '10 5\n1 2\n3 3\n5 7\n7 7\n7 10\n']
Demo Output:
['OK\n', '2 2\n', '4 0\n']
Note:
Keep in mind that in the second sample the mistake occurs not only on the second day, but also on the sixth day, when nobody waters the flowers. However, you have to print the second day, i.e. the day with the minimal number.
|
```python
n,m = map(int,input().split())
Ln = [0]*n
while m!=0:
a,b = map(int,input().split())
for i in range(a,b+1):
Ln[i-1]+=1
m-=1
for i in range(0,len(Ln)):
if Ln[i]==0 :
print(i+1,0)
exit(0)
elif Ln[i]>1:
print(i+1,Ln[i])
exit(0)
print('OK')
exit(0)
```
| 3.9225
|
732
|
A
|
Buy a Shovel
|
PROGRAMMING
| 800
|
[
"brute force",
"constructive algorithms",
"implementation",
"math"
] | null | null |
Polycarp urgently needs a shovel! He comes to the shop and chooses an appropriate one. The shovel that Policarp chooses is sold for *k* burles. Assume that there is an unlimited number of such shovels in the shop.
In his pocket Polycarp has an unlimited number of "10-burle coins" and exactly one coin of *r* burles (1<=β€<=*r*<=β€<=9).
What is the minimum number of shovels Polycarp has to buy so that he can pay for the purchase without any change? It is obvious that he can pay for 10 shovels without any change (by paying the requied amount of 10-burle coins and not using the coin of *r* burles). But perhaps he can buy fewer shovels and pay without any change. Note that Polycarp should buy at least one shovel.
|
The single line of input contains two integers *k* and *r* (1<=β€<=*k*<=β€<=1000, 1<=β€<=*r*<=β€<=9)Β β the price of one shovel and the denomination of the coin in Polycarp's pocket that is different from "10-burle coins".
Remember that he has an unlimited number of coins in the denomination of 10, that is, Polycarp has enough money to buy any number of shovels.
|
Print the required minimum number of shovels Polycarp has to buy so that he can pay for them without any change.
|
[
"117 3\n",
"237 7\n",
"15 2\n"
] |
[
"9\n",
"1\n",
"2\n"
] |
In the first example Polycarp can buy 9 shovels and pay 9Β·117β=β1053 burles. Indeed, he can pay this sum by using 10-burle coins and one 3-burle coin. He can't buy fewer shovels without any change.
In the second example it is enough for Polycarp to buy one shovel.
In the third example Polycarp should buy two shovels and pay 2Β·15β=β30 burles. It is obvious that he can pay this sum without any change.
| 500
|
[
{
"input": "117 3",
"output": "9"
},
{
"input": "237 7",
"output": "1"
},
{
"input": "15 2",
"output": "2"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "1 9",
"output": "9"
},
{
"input": "1000 3",
"output": "1"
},
{
"input": "1000 1",
"output": "1"
},
{
"input": "1000 9",
"output": "1"
},
{
"input": "1 2",
"output": "2"
},
{
"input": "999 9",
"output": "1"
},
{
"input": "999 8",
"output": "2"
},
{
"input": "105 6",
"output": "2"
},
{
"input": "403 9",
"output": "3"
},
{
"input": "546 4",
"output": "4"
},
{
"input": "228 9",
"output": "5"
},
{
"input": "57 2",
"output": "6"
},
{
"input": "437 9",
"output": "7"
},
{
"input": "997 6",
"output": "8"
},
{
"input": "109 1",
"output": "9"
},
{
"input": "998 9",
"output": "5"
},
{
"input": "4 2",
"output": "3"
},
{
"input": "9 3",
"output": "7"
},
{
"input": "8 2",
"output": "4"
},
{
"input": "1 3",
"output": "3"
},
{
"input": "1 4",
"output": "4"
},
{
"input": "1 5",
"output": "5"
},
{
"input": "1 6",
"output": "6"
},
{
"input": "1 7",
"output": "7"
},
{
"input": "1 8",
"output": "8"
},
{
"input": "100 3",
"output": "1"
},
{
"input": "1000 2",
"output": "1"
},
{
"input": "1000 4",
"output": "1"
},
{
"input": "1000 5",
"output": "1"
},
{
"input": "1000 6",
"output": "1"
},
{
"input": "1000 7",
"output": "1"
},
{
"input": "1000 8",
"output": "1"
},
{
"input": "23 4",
"output": "8"
},
{
"input": "33 1",
"output": "7"
},
{
"input": "33 2",
"output": "4"
},
{
"input": "666 5",
"output": "5"
},
{
"input": "2 3",
"output": "5"
},
{
"input": "5 5",
"output": "1"
},
{
"input": "3 6",
"output": "2"
},
{
"input": "12 4",
"output": "2"
},
{
"input": "15 5",
"output": "1"
},
{
"input": "2 5",
"output": "5"
},
{
"input": "25 5",
"output": "1"
},
{
"input": "2 9",
"output": "5"
},
{
"input": "6 7",
"output": "5"
},
{
"input": "8 9",
"output": "5"
},
{
"input": "2 7",
"output": "5"
},
{
"input": "4 7",
"output": "5"
},
{
"input": "2 1",
"output": "5"
},
{
"input": "261 1",
"output": "1"
}
] | 1,678,481,167
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 54
| 62
| 0
|
inp = input().split()
inp = [int(x) for x in inp]
a=0
for i in range(15):
if(i>0):
if(inp[0]%10==0 or (inp[0]-inp[1])%10==0 or ((inp[0]*i)-inp[1])%10==0 or (inp[0]*i)%10==0):# :
a=i
break
print(a)
|
Title: Buy a Shovel
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp urgently needs a shovel! He comes to the shop and chooses an appropriate one. The shovel that Policarp chooses is sold for *k* burles. Assume that there is an unlimited number of such shovels in the shop.
In his pocket Polycarp has an unlimited number of "10-burle coins" and exactly one coin of *r* burles (1<=β€<=*r*<=β€<=9).
What is the minimum number of shovels Polycarp has to buy so that he can pay for the purchase without any change? It is obvious that he can pay for 10 shovels without any change (by paying the requied amount of 10-burle coins and not using the coin of *r* burles). But perhaps he can buy fewer shovels and pay without any change. Note that Polycarp should buy at least one shovel.
Input Specification:
The single line of input contains two integers *k* and *r* (1<=β€<=*k*<=β€<=1000, 1<=β€<=*r*<=β€<=9)Β β the price of one shovel and the denomination of the coin in Polycarp's pocket that is different from "10-burle coins".
Remember that he has an unlimited number of coins in the denomination of 10, that is, Polycarp has enough money to buy any number of shovels.
Output Specification:
Print the required minimum number of shovels Polycarp has to buy so that he can pay for them without any change.
Demo Input:
['117 3\n', '237 7\n', '15 2\n']
Demo Output:
['9\n', '1\n', '2\n']
Note:
In the first example Polycarp can buy 9 shovels and pay 9Β·117β=β1053 burles. Indeed, he can pay this sum by using 10-burle coins and one 3-burle coin. He can't buy fewer shovels without any change.
In the second example it is enough for Polycarp to buy one shovel.
In the third example Polycarp should buy two shovels and pay 2Β·15β=β30 burles. It is obvious that he can pay this sum without any change.
|
```python
inp = input().split()
inp = [int(x) for x in inp]
a=0
for i in range(15):
if(i>0):
if(inp[0]%10==0 or (inp[0]-inp[1])%10==0 or ((inp[0]*i)-inp[1])%10==0 or (inp[0]*i)%10==0):# :
a=i
break
print(a)
```
| 3
|
|
41
|
A
|
Translation
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] |
A. Translation
|
2
|
256
|
The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the Β«translationΒ». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly.
|
The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols.
|
If the word *t* is a word *s*, written reversely, print YES, otherwise print NO.
|
[
"code\nedoc\n",
"abb\naba\n",
"code\ncode\n"
] |
[
"YES\n",
"NO\n",
"NO\n"
] |
none
| 500
|
[
{
"input": "code\nedoc",
"output": "YES"
},
{
"input": "abb\naba",
"output": "NO"
},
{
"input": "code\ncode",
"output": "NO"
},
{
"input": "abacaba\nabacaba",
"output": "YES"
},
{
"input": "q\nq",
"output": "YES"
},
{
"input": "asrgdfngfnmfgnhweratgjkk\nasrgdfngfnmfgnhweratgjkk",
"output": "NO"
},
{
"input": "z\na",
"output": "NO"
},
{
"input": "asd\ndsa",
"output": "YES"
},
{
"input": "abcdef\nfecdba",
"output": "NO"
},
{
"input": "ywjjbirapvskozubvxoemscfwl\ngnduubaogtfaiowjizlvjcu",
"output": "NO"
},
{
"input": "mfrmqxtzvgaeuleubcmcxcfqyruwzenguhgrmkuhdgnhgtgkdszwqyd\nmfxufheiperjnhyczclkmzyhcxntdfskzkzdwzzujdinf",
"output": "NO"
},
{
"input": "bnbnemvybqizywlnghlykniaxxxlkhftppbdeqpesrtgkcpoeqowjwhrylpsziiwcldodcoonpimudvrxejjo\ntiynnekmlalogyvrgptbinkoqdwzuiyjlrldxhzjmmp",
"output": "NO"
},
{
"input": "pwlpubwyhzqvcitemnhvvwkmwcaawjvdiwtoxyhbhbxerlypelevasmelpfqwjk\nstruuzebbcenziscuoecywugxncdwzyfozhljjyizpqcgkyonyetarcpwkqhuugsqjuixsxptmbnlfupdcfigacdhhrzb",
"output": "NO"
},
{
"input": "gdvqjoyxnkypfvdxssgrihnwxkeojmnpdeobpecytkbdwujqfjtxsqspxvxpqioyfagzjxupqqzpgnpnpxcuipweunqch\nkkqkiwwasbhezqcfeceyngcyuogrkhqecwsyerdniqiocjehrpkljiljophqhyaiefjpavoom",
"output": "NO"
},
{
"input": "umeszdawsvgkjhlqwzents\nhxqhdungbylhnikwviuh",
"output": "NO"
},
{
"input": "juotpscvyfmgntshcealgbsrwwksgrwnrrbyaqqsxdlzhkbugdyx\nibqvffmfktyipgiopznsqtrtxiijntdbgyy",
"output": "NO"
},
{
"input": "zbwueheveouatecaglziqmudxemhrsozmaujrwlqmppzoumxhamwugedikvkblvmxwuofmpafdprbcftew\nulczwrqhctbtbxrhhodwbcxwimncnexosksujlisgclllxokrsbnozthajnnlilyffmsyko",
"output": "NO"
},
{
"input": "nkgwuugukzcv\nqktnpxedwxpxkrxdvgmfgoxkdfpbzvwsduyiybynbkouonhvmzakeiruhfmvrktghadbfkmwxduoqv",
"output": "NO"
},
{
"input": "incenvizhqpcenhjhehvjvgbsnfixbatrrjstxjzhlmdmxijztphxbrldlqwdfimweepkggzcxsrwelodpnryntepioqpvk\ndhjbjjftlvnxibkklxquwmzhjfvnmwpapdrslioxisbyhhfymyiaqhlgecpxamqnocizwxniubrmpyubvpenoukhcobkdojlybxd",
"output": "NO"
},
{
"input": "w\nw",
"output": "YES"
},
{
"input": "vz\nzv",
"output": "YES"
},
{
"input": "ry\nyr",
"output": "YES"
},
{
"input": "xou\nuox",
"output": "YES"
},
{
"input": "axg\ngax",
"output": "NO"
},
{
"input": "zdsl\nlsdz",
"output": "YES"
},
{
"input": "kudl\nldku",
"output": "NO"
},
{
"input": "zzlzwnqlcl\nlclqnwzlzz",
"output": "YES"
},
{
"input": "vzzgicnzqooejpjzads\nsdazjpjeooqzncigzzv",
"output": "YES"
},
{
"input": "raqhmvmzuwaykjpyxsykr\nxkysrypjkyawuzmvmhqar",
"output": "NO"
},
{
"input": "ngedczubzdcqbxksnxuavdjaqtmdwncjnoaicvmodcqvhfezew\nwezefhvqcdomvciaonjcnwdmtqajdvauxnskxbqcdzbuzcdegn",
"output": "YES"
},
{
"input": "muooqttvrrljcxbroizkymuidvfmhhsjtumksdkcbwwpfqdyvxtrlymofendqvznzlmim\nmimlznzvqdnefomylrtxvydqfpwwbckdskmutjshhmfvdiumykziorbxcjlrrvttqooum",
"output": "YES"
},
{
"input": "vxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaivg\ngviayyikkitmuomcpiakhbxszgbnhvwyzkftwoagzixaearxpjacrnvpvbuzenvovehkmmxvblqyxvctroddksdsgebcmlluqpxv",
"output": "YES"
},
{
"input": "mnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfdc\ncdfmkdgrdptkpewbsqvszipgxvgvuiuzbkkwuowbafkikgvnqdkxnayzdjygvezmtsgywnupocdntipiyiorblqkrzjpzatxahnm",
"output": "NO"
},
{
"input": "dgxmzbqofstzcdgthbaewbwocowvhqpinehpjatnnbrijcolvsatbblsrxabzrpszoiecpwhfjmwuhqrapvtcgvikuxtzbftydkw\nwkdytfbztxukivgctvparqhuwmjfhwpceiozsprzbaxrslbbqasvlocjirbnntajphenipthvwocowbweabhtgdcztsfoqbzmxgd",
"output": "NO"
},
{
"input": "gxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwgeh\nhegwxvocotmzstqfbmpjvijgkcyodlxyjawrpkczpmdspsuhoiruavnnnuwvtwohglkdxjetshkboalvzqbgjgthoteceixioxg",
"output": "YES"
},
{
"input": "sihxuwvmaambplxvjfoskinghzicyfqebjtkysotattkahssumfcgrkheotdxwjckpvapbkaepqrxseyfrwtyaycmrzsrsngkh\nhkgnsrszrmcyaytwrfyesxrqpeakbpavpkcjwxdtoehkrgcfmusshakttatosyktjbeqfycizhgniksofjvxlpbmaamvwuxhis",
"output": "YES"
},
{
"input": "ycnahksbughnonldzrhkysujmylcgcfuludjvjiahtkyzqvkopzqcnwhltbzfugzojqkjjlggmvnultascmygelkiktmfieok\nkoeifmtkiklegkmcsatlunvmggkjjlqjozgufzbtlhwncqzpokvqzykthaijvjdulufcgclymjusyyhrzdlnonhgubskhancy",
"output": "NO"
},
{
"input": "wbqasaehtkfojruzyhrlgwmtyiovmzyfifslvlemhqheyaelzwnthrenjsbmntwaoryzwfbxmscmypvxlfmzpnkkjlvwvmtz\nztmvwvljkknpzmflxvpymcsmxbfwzyroawtnmbsjnerhtnwzleayehqhmelvlsfifyzmvoiytmwglrhyzurjofktheasaqbw",
"output": "YES"
},
{
"input": "imippqurprbhfugngtgifelytadegwrgaefnfhbjjnmzikvjaccotqzemufqieqldgnbmviisgkynzeldlhqxuqphjfmyij\njiymfjhpquxqhldleznykgsiivmbngdlqeiqfumezqtoccajvkizmnjjbhfnfeagrwgedatylefigtgngufhbrpruqppimi",
"output": "YES"
},
{
"input": "bikydffiuisckpvzqlteqfhegsagimodb\nbdomigasgehfqetlqzvpkcsiuiffdykib",
"output": "YES"
}
] | 1,570,801,198
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 40
| 218
| 0
|
s=input()
t=input()
x=0
if len(s)>len(t) or len(s)<len(t):
print('NO')
else:
for i in range(len(s)):
if s[i]==t[-i-1]:
x=x+1
if x==len(s):
print('YES')
else:
print('NO')
|
Title: Translation
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the Β«translationΒ». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly.
Input Specification:
The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols.
Output Specification:
If the word *t* is a word *s*, written reversely, print YES, otherwise print NO.
Demo Input:
['code\nedoc\n', 'abb\naba\n', 'code\ncode\n']
Demo Output:
['YES\n', 'NO\n', 'NO\n']
Note:
none
|
```python
s=input()
t=input()
x=0
if len(s)>len(t) or len(s)<len(t):
print('NO')
else:
for i in range(len(s)):
if s[i]==t[-i-1]:
x=x+1
if x==len(s):
print('YES')
else:
print('NO')
```
| 3.9455
|
43
|
A
|
Football
|
PROGRAMMING
| 1,000
|
[
"strings"
] |
A. Football
|
2
|
256
|
One day Vasya decided to have a look at the results of Berland 1910 Football Championshipβs finals. Unfortunately he didn't find the overall score of the match; however, he got hold of a profound description of the match's process. On the whole there are *n* lines in that description each of which described one goal. Every goal was marked with the name of the team that had scored it. Help Vasya, learn the name of the team that won the finals. It is guaranteed that the match did not end in a tie.
|
The first line contains an integer *n* (1<=β€<=*n*<=β€<=100) β the number of lines in the description. Then follow *n* lines β for each goal the names of the teams that scored it. The names are non-empty lines consisting of uppercase Latin letters whose lengths do not exceed 10 symbols. It is guaranteed that the match did not end in a tie and the description contains no more than two different teams.
|
Print the name of the winning team. We remind you that in football the team that scores more goals is considered the winner.
|
[
"1\nABC\n",
"5\nA\nABA\nABA\nA\nA\n"
] |
[
"ABC\n",
"A\n"
] |
none
| 500
|
[
{
"input": "1\nABC",
"output": "ABC"
},
{
"input": "5\nA\nABA\nABA\nA\nA",
"output": "A"
},
{
"input": "2\nXTSJEP\nXTSJEP",
"output": "XTSJEP"
},
{
"input": "3\nXZYDJAEDZ\nXZYDJAEDZ\nXZYDJAEDZ",
"output": "XZYDJAEDZ"
},
{
"input": "3\nQCCYXL\nQCCYXL\nAXGLFQDD",
"output": "QCCYXL"
},
{
"input": "3\nAZID\nEERWBC\nEERWBC",
"output": "EERWBC"
},
{
"input": "3\nHNCGYL\nHNCGYL\nHNCGYL",
"output": "HNCGYL"
},
{
"input": "4\nZZWZTG\nZZWZTG\nZZWZTG\nZZWZTG",
"output": "ZZWZTG"
},
{
"input": "4\nA\nA\nKUDLJMXCSE\nA",
"output": "A"
},
{
"input": "5\nPHBTW\nPHBTW\nPHBTW\nPHBTW\nPHBTW",
"output": "PHBTW"
},
{
"input": "5\nPKUZYTFYWN\nPKUZYTFYWN\nSTC\nPKUZYTFYWN\nPKUZYTFYWN",
"output": "PKUZYTFYWN"
},
{
"input": "5\nHH\nHH\nNTQWPA\nNTQWPA\nHH",
"output": "HH"
},
{
"input": "10\nW\nW\nW\nW\nW\nD\nW\nD\nD\nW",
"output": "W"
},
{
"input": "19\nXBCP\nTGACNIH\nXBCP\nXBCP\nXBCP\nXBCP\nXBCP\nTGACNIH\nXBCP\nXBCP\nXBCP\nXBCP\nXBCP\nTGACNIH\nXBCP\nXBCP\nTGACNIH\nTGACNIH\nXBCP",
"output": "XBCP"
},
{
"input": "33\nOWQWCKLLF\nOWQWCKLLF\nOWQWCKLLF\nPYPAS\nPYPAS\nPYPAS\nOWQWCKLLF\nPYPAS\nOWQWCKLLF\nPYPAS\nPYPAS\nOWQWCKLLF\nOWQWCKLLF\nOWQWCKLLF\nPYPAS\nOWQWCKLLF\nPYPAS\nPYPAS\nPYPAS\nPYPAS\nOWQWCKLLF\nPYPAS\nPYPAS\nOWQWCKLLF\nOWQWCKLLF\nPYPAS\nOWQWCKLLF\nOWQWCKLLF\nPYPAS\nPYPAS\nOWQWCKLLF\nPYPAS\nPYPAS",
"output": "PYPAS"
},
{
"input": "51\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC",
"output": "NC"
},
{
"input": "89\nH\nVOCI\nVOCI\nH\nVOCI\nH\nH\nVOCI\nVOCI\nVOCI\nH\nH\nH\nVOCI\nVOCI\nVOCI\nH\nVOCI\nVOCI\nH\nVOCI\nVOCI\nVOCI\nH\nVOCI\nH\nVOCI\nH\nVOCI\nH\nVOCI\nVOCI\nH\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nH\nVOCI\nVOCI\nVOCI\nVOCI\nH\nVOCI\nH\nH\nVOCI\nH\nVOCI\nH\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nH\nH\nVOCI\nH\nH\nVOCI\nH\nVOCI\nH\nVOCI\nVOCI\nH\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nH\nH\nH\nH\nH\nVOCI\nH\nVOCI\nH\nVOCI\nVOCI",
"output": "VOCI"
},
{
"input": "100\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nM\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA",
"output": "HA"
},
{
"input": "100\nG\nG\nS\nS\nG\nG\nS\nS\nG\nS\nS\nS\nG\nS\nG\nG\nS\nG\nS\nS\nG\nS\nS\nS\nS\nS\nG\nS\nG\nS\nS\nG\nG\nG\nS\nS\nS\nS\nG\nS\nS\nG\nG\nG\nG\nG\nS\nG\nG\nS\nS\nS\nS\nS\nG\nG\nS\nG\nG\nG\nG\nG\nS\nS\nG\nS\nS\nS\nS\nG\nS\nS\nG\nS\nG\nG\nG\nG\nG\nG\nG\nG\nG\nG\nG\nS\nS\nG\nS\nS\nS\nS\nG\nG\nG\nS\nG\nG\nG\nS",
"output": "G"
},
{
"input": "100\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nOBH\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL",
"output": "WL"
}
] | 1,684,592,469
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 34
| 186
| 0
|
n = int(input())
d = {}
ans = ""
m = 0
for i in range(n):
s = input()
if s not in d:
d[s] = 0
d[s] += 1
if d[s] > m:
m = d[s]
ans = s
print(ans)
|
Title: Football
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
One day Vasya decided to have a look at the results of Berland 1910 Football Championshipβs finals. Unfortunately he didn't find the overall score of the match; however, he got hold of a profound description of the match's process. On the whole there are *n* lines in that description each of which described one goal. Every goal was marked with the name of the team that had scored it. Help Vasya, learn the name of the team that won the finals. It is guaranteed that the match did not end in a tie.
Input Specification:
The first line contains an integer *n* (1<=β€<=*n*<=β€<=100) β the number of lines in the description. Then follow *n* lines β for each goal the names of the teams that scored it. The names are non-empty lines consisting of uppercase Latin letters whose lengths do not exceed 10 symbols. It is guaranteed that the match did not end in a tie and the description contains no more than two different teams.
Output Specification:
Print the name of the winning team. We remind you that in football the team that scores more goals is considered the winner.
Demo Input:
['1\nABC\n', '5\nA\nABA\nABA\nA\nA\n']
Demo Output:
['ABC\n', 'A\n']
Note:
none
|
```python
n = int(input())
d = {}
ans = ""
m = 0
for i in range(n):
s = input()
if s not in d:
d[s] = 0
d[s] += 1
if d[s] > m:
m = d[s]
ans = s
print(ans)
```
| 3.9535
|
127
|
B
|
Canvas Frames
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
Nicholas, a painter is going to paint several new canvases. Nicholas is sure that the canvases will turn out so great that each one will need framing and being hung on the wall. Frames are what Nicholas decided to begin with.
Nicholas has *n* sticks whose lengths equal *a*1,<=*a*2,<=... *a**n*. Nicholas does not want to break the sticks or glue them together. To make a *h*<=Γ<=*w*-sized frame, he needs two sticks whose lengths equal *h* and two sticks whose lengths equal *w*. Specifically, to make a square frame (when *h*<==<=*w*), he needs four sticks of the same length.
Now Nicholas wants to make from the sticks that he has as many frames as possible; to be able to paint as many canvases as possible to fill the frames. Help him in this uneasy task. Note that it is not necessary to use all the sticks Nicholas has.
|
The first line contains an integer *n* (1<=β€<=*n*<=β€<=100) β the number of sticks. The second line contains *n* space-separated integers. The *i*-th integer equals the length of the *i*-th stick *a**i* (1<=β€<=*a**i*<=β€<=100).
|
Print the single number β the maximum number of frames Nicholas can make for his future canvases.
|
[
"5\n2 4 3 2 3\n",
"13\n2 2 4 4 4 4 6 6 6 7 7 9 9\n",
"4\n3 3 3 5\n"
] |
[
"1",
"3",
"0"
] |
none
| 1,000
|
[
{
"input": "5\n2 4 3 2 3",
"output": "1"
},
{
"input": "13\n2 2 4 4 4 4 6 6 6 7 7 9 9",
"output": "3"
},
{
"input": "4\n3 3 3 5",
"output": "0"
},
{
"input": "2\n3 5",
"output": "0"
},
{
"input": "9\n1 2 3 4 5 6 7 8 9",
"output": "0"
},
{
"input": "14\n2 4 2 6 2 3 4 1 4 5 4 3 4 1",
"output": "2"
},
{
"input": "33\n1 2 2 6 10 10 33 11 17 32 25 6 7 29 11 32 33 8 13 17 17 6 11 11 11 8 10 26 29 26 32 33 36",
"output": "5"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "1\n10",
"output": "0"
},
{
"input": "2\n1 1",
"output": "0"
},
{
"input": "3\n1 1 1",
"output": "0"
},
{
"input": "3\n1 2 2",
"output": "0"
},
{
"input": "3\n3 2 1",
"output": "0"
},
{
"input": "4\n1 1 1 1",
"output": "1"
},
{
"input": "4\n1 2 1 2",
"output": "1"
},
{
"input": "4\n1 100 1 100",
"output": "1"
},
{
"input": "4\n10 100 100 10",
"output": "1"
},
{
"input": "4\n1 2 3 3",
"output": "0"
},
{
"input": "4\n8 5 9 13",
"output": "0"
},
{
"input": "4\n100 100 100 100",
"output": "1"
},
{
"input": "5\n1 1 1 1 1",
"output": "1"
},
{
"input": "5\n1 4 4 1 1",
"output": "1"
},
{
"input": "5\n1 100 1 1 100",
"output": "1"
},
{
"input": "5\n100 100 1 1 100",
"output": "1"
},
{
"input": "5\n100 1 100 100 100",
"output": "1"
},
{
"input": "5\n100 100 100 100 100",
"output": "1"
},
{
"input": "6\n1 1 1 1 1 1",
"output": "1"
},
{
"input": "6\n1 1 5 1 1 5",
"output": "1"
},
{
"input": "6\n1 100 100 1 1 1",
"output": "1"
},
{
"input": "6\n100 1 1 100 1 100",
"output": "1"
},
{
"input": "6\n1 2 3 2 3 1",
"output": "1"
},
{
"input": "6\n1 50 1 100 50 100",
"output": "1"
},
{
"input": "6\n10 10 10 12 13 14",
"output": "0"
},
{
"input": "7\n1 1 1 1 1 1 1",
"output": "1"
},
{
"input": "7\n1 2 1 1 1 1 1",
"output": "1"
},
{
"input": "7\n1 2 2 1 2 1 2",
"output": "1"
},
{
"input": "7\n1 1 2 2 1 2 3",
"output": "1"
},
{
"input": "7\n1 3 2 2 3 1 4",
"output": "1"
},
{
"input": "7\n1 3 4 3 5 4 6",
"output": "1"
},
{
"input": "7\n7 6 5 4 3 2 1",
"output": "0"
},
{
"input": "8\n1 2 1 2 2 2 2 2",
"output": "2"
},
{
"input": "8\n1 2 2 1 1 2 2 2",
"output": "1"
},
{
"input": "8\n1 2 2 2 3 1 1 3",
"output": "1"
},
{
"input": "8\n1 2 3 4 1 2 3 4",
"output": "2"
},
{
"input": "8\n1 1 1 1 2 3 2 3",
"output": "2"
},
{
"input": "8\n1 2 3 4 5 5 5 5",
"output": "1"
},
{
"input": "8\n1 2 1 3 4 1 5 6",
"output": "0"
},
{
"input": "8\n1 2 3 4 5 6 1 7",
"output": "0"
},
{
"input": "8\n8 6 3 4 5 2 1 7",
"output": "0"
},
{
"input": "8\n100 100 100 100 100 100 100 100",
"output": "2"
},
{
"input": "10\n1 1 1 1 1 1 1 1 1 1",
"output": "2"
},
{
"input": "10\n19 9 14 14 19 5 5 18 10 17",
"output": "1"
},
{
"input": "10\n72 86 73 25 84 29 33 34 20 29",
"output": "0"
},
{
"input": "10\n93 93 99 98 91 96 92 98 94 98",
"output": "1"
},
{
"input": "13\n35 6 21 30 67 55 70 39 75 72 11 13 69",
"output": "0"
},
{
"input": "17\n90 97 12 56 94 11 49 96 22 7 15 48 71 71 94 72 100",
"output": "1"
},
{
"input": "18\n39 72 67 28 69 41 43 51 66 99 4 57 68 93 28 27 37 27",
"output": "1"
},
{
"input": "23\n88 82 2 67 4 6 67 83 77 58 48 64 86 37 96 83 35 46 13 79 72 18 35",
"output": "1"
},
{
"input": "30\n43 34 38 50 47 24 26 20 7 5 26 29 98 87 90 46 10 53 88 61 90 39 78 81 65 13 72 95 53 27",
"output": "1"
},
{
"input": "33\n1 3 34 55 38 58 64 26 66 44 50 63 46 62 62 99 73 87 35 20 30 38 39 85 49 24 93 68 8 25 86 30 51",
"output": "1"
},
{
"input": "38\n65 69 80 93 28 36 40 81 53 75 55 50 82 95 8 51 66 65 50 4 40 92 18 70 38 68 42 100 34 57 98 79 95 84 82 35 100 89",
"output": "3"
},
{
"input": "40\n4 2 62 38 76 68 19 71 44 91 76 31 3 63 56 62 93 98 10 61 52 59 81 46 23 27 36 26 24 38 37 66 15 16 78 41 95 82 73 90",
"output": "1"
},
{
"input": "43\n62 31 14 43 67 2 60 77 64 70 91 9 3 43 76 7 56 84 5 20 88 50 47 42 7 39 8 56 71 24 49 59 70 61 81 17 76 44 80 61 77 5 96",
"output": "4"
},
{
"input": "49\n75 64 7 2 1 66 31 84 78 53 34 5 40 90 7 62 86 54 99 77 8 92 30 3 18 18 61 38 38 11 79 88 84 89 50 94 72 8 54 85 100 1 19 4 97 91 13 39 91",
"output": "4"
},
{
"input": "57\n83 94 42 57 19 9 40 25 56 92 9 38 58 66 43 19 50 10 100 3 49 96 77 36 20 3 48 15 38 19 99 100 66 14 52 13 16 73 65 99 29 85 75 18 97 64 57 82 70 19 16 25 40 11 9 22 89",
"output": "6"
},
{
"input": "67\n36 22 22 86 52 53 36 68 46 82 99 37 15 43 57 35 33 99 22 96 7 8 80 93 70 70 55 51 61 74 6 28 85 72 84 42 29 1 4 71 7 40 61 95 93 36 42 61 16 40 10 85 31 86 93 19 44 20 52 66 10 22 40 53 25 29 23",
"output": "8"
},
{
"input": "74\n90 26 58 69 87 23 44 9 32 25 33 13 79 84 52 90 4 7 93 77 29 85 22 1 96 69 98 16 76 87 57 16 44 41 57 28 18 70 77 83 37 17 59 87 27 19 89 63 14 84 77 40 46 77 82 73 86 73 30 58 6 30 70 36 31 12 43 50 93 3 3 57 38 91",
"output": "7"
},
{
"input": "87\n10 19 83 58 15 48 26 58 89 46 50 34 81 40 25 51 62 85 9 80 71 44 100 22 30 48 74 69 54 40 38 81 66 42 40 90 60 20 75 24 74 98 28 62 79 65 65 6 14 23 3 59 29 24 64 13 8 38 29 85 75 81 36 42 3 63 99 24 72 92 35 8 71 19 77 77 66 3 79 65 15 18 15 69 60 77 91",
"output": "11"
},
{
"input": "100\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "25"
},
{
"input": "100\n1 9 3 5 10 10 9 8 10 1 7 6 5 6 7 9 1 5 8 3 2 3 3 10 2 3 10 7 10 3 6 3 2 10 1 10 2 3 4 3 3 1 7 5 10 2 3 8 9 2 5 4 7 2 5 9 2 1 7 9 9 8 4 4 6 1 6 6 4 7 2 3 1 1 1 6 9 1 2 9 3 7 6 10 3 6 2 5 2 5 3 9 10 6 4 2 9 9 4 5",
"output": "23"
},
{
"input": "100\n70 70 75 70 74 70 70 73 72 73 74 75 70 74 73 70 70 74 72 72 75 70 73 72 70 75 73 70 74 70 73 75 71 74 70 71 75 74 75 71 74 70 73 73 70 75 71 73 73 74 73 74 71 73 73 71 72 71 70 75 74 74 72 72 71 72 75 75 70 73 71 73 72 71 70 75 71 75 73 75 73 72 75 71 73 71 72 74 75 70 70 74 75 73 70 73 73 75 71 74",
"output": "24"
},
{
"input": "100\n99 98 98 99 98 98 98 100 98 99 99 98 99 98 98 98 99 99 98 99 99 100 98 100 98 98 98 99 98 100 100 98 100 99 100 98 99 99 99 98 100 98 100 99 99 99 98 100 98 98 98 100 100 99 98 98 100 100 100 99 98 99 99 99 100 99 99 98 99 98 99 100 100 98 98 100 100 99 99 99 98 98 98 100 99 99 100 99 100 99 98 100 98 100 98 98 99 98 99 98",
"output": "24"
},
{
"input": "100\n94 87 92 91 94 89 93 94 87 93 93 94 89 91 87 87 92 91 87 94 90 89 92 92 87 88 90 90 90 89 90 92 91 91 89 88 93 89 88 94 91 89 88 87 92 89 91 87 88 90 88 92 90 87 93 94 94 92 92 87 90 88 88 91 94 93 87 94 93 93 87 90 92 92 90 88 88 90 92 91 90 88 89 91 91 88 90 93 90 94 94 93 90 91 91 93 94 94 92 93",
"output": "24"
},
{
"input": "100\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100",
"output": "25"
},
{
"input": "10\n1 1 1 1 1 1 1 1 1 1",
"output": "2"
},
{
"input": "7\n13 13 13 13 6 2 3",
"output": "1"
},
{
"input": "8\n1 1 1 1 1 1 1 1",
"output": "2"
},
{
"input": "5\n100 100 99 99 5",
"output": "1"
},
{
"input": "8\n2 2 2 2 2 2 2 2",
"output": "2"
},
{
"input": "8\n1 2 3 4 5 6 7 7",
"output": "0"
},
{
"input": "8\n4 4 4 4 4 4 4 4",
"output": "2"
},
{
"input": "10\n1 1 1 1 1 1 1 1 2 2",
"output": "2"
},
{
"input": "4\n100 100 100 99",
"output": "0"
},
{
"input": "4\n2 2 2 2",
"output": "1"
},
{
"input": "5\n100 100 99 99 2",
"output": "1"
},
{
"input": "9\n1 1 1 1 1 1 1 1 1",
"output": "2"
},
{
"input": "5\n2 2 3 4 4",
"output": "1"
},
{
"input": "100\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "25"
},
{
"input": "13\n1 2 3 4 5 6 7 8 9 10 11 12 13",
"output": "0"
},
{
"input": "20\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "5"
},
{
"input": "4\n4 4 4 4",
"output": "1"
},
{
"input": "5\n1 1 2 3 3",
"output": "1"
},
{
"input": "5\n30 30 30 1 1",
"output": "1"
}
] | 1,669,039,342
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 93
| 62
| 0
|
import math
n = int(input())
sticks = list(map(int,input().split()))
sticks.sort()
pair = 0
count = 1
for index in range(len(sticks)-1):
if sticks[index] == sticks[index+1]:
count = count + 1
if index == len(sticks)-2:
pair = pair + math.floor(count/2)
else:
pair = pair + math.floor(count/2)
count = 1
frames = math.floor(pair/2)
print(frames)
|
Title: Canvas Frames
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Nicholas, a painter is going to paint several new canvases. Nicholas is sure that the canvases will turn out so great that each one will need framing and being hung on the wall. Frames are what Nicholas decided to begin with.
Nicholas has *n* sticks whose lengths equal *a*1,<=*a*2,<=... *a**n*. Nicholas does not want to break the sticks or glue them together. To make a *h*<=Γ<=*w*-sized frame, he needs two sticks whose lengths equal *h* and two sticks whose lengths equal *w*. Specifically, to make a square frame (when *h*<==<=*w*), he needs four sticks of the same length.
Now Nicholas wants to make from the sticks that he has as many frames as possible; to be able to paint as many canvases as possible to fill the frames. Help him in this uneasy task. Note that it is not necessary to use all the sticks Nicholas has.
Input Specification:
The first line contains an integer *n* (1<=β€<=*n*<=β€<=100) β the number of sticks. The second line contains *n* space-separated integers. The *i*-th integer equals the length of the *i*-th stick *a**i* (1<=β€<=*a**i*<=β€<=100).
Output Specification:
Print the single number β the maximum number of frames Nicholas can make for his future canvases.
Demo Input:
['5\n2 4 3 2 3\n', '13\n2 2 4 4 4 4 6 6 6 7 7 9 9\n', '4\n3 3 3 5\n']
Demo Output:
['1', '3', '0']
Note:
none
|
```python
import math
n = int(input())
sticks = list(map(int,input().split()))
sticks.sort()
pair = 0
count = 1
for index in range(len(sticks)-1):
if sticks[index] == sticks[index+1]:
count = count + 1
if index == len(sticks)-2:
pair = pair + math.floor(count/2)
else:
pair = pair + math.floor(count/2)
count = 1
frames = math.floor(pair/2)
print(frames)
```
| 3
|
|
732
|
A
|
Buy a Shovel
|
PROGRAMMING
| 800
|
[
"brute force",
"constructive algorithms",
"implementation",
"math"
] | null | null |
Polycarp urgently needs a shovel! He comes to the shop and chooses an appropriate one. The shovel that Policarp chooses is sold for *k* burles. Assume that there is an unlimited number of such shovels in the shop.
In his pocket Polycarp has an unlimited number of "10-burle coins" and exactly one coin of *r* burles (1<=β€<=*r*<=β€<=9).
What is the minimum number of shovels Polycarp has to buy so that he can pay for the purchase without any change? It is obvious that he can pay for 10 shovels without any change (by paying the requied amount of 10-burle coins and not using the coin of *r* burles). But perhaps he can buy fewer shovels and pay without any change. Note that Polycarp should buy at least one shovel.
|
The single line of input contains two integers *k* and *r* (1<=β€<=*k*<=β€<=1000, 1<=β€<=*r*<=β€<=9)Β β the price of one shovel and the denomination of the coin in Polycarp's pocket that is different from "10-burle coins".
Remember that he has an unlimited number of coins in the denomination of 10, that is, Polycarp has enough money to buy any number of shovels.
|
Print the required minimum number of shovels Polycarp has to buy so that he can pay for them without any change.
|
[
"117 3\n",
"237 7\n",
"15 2\n"
] |
[
"9\n",
"1\n",
"2\n"
] |
In the first example Polycarp can buy 9 shovels and pay 9Β·117β=β1053 burles. Indeed, he can pay this sum by using 10-burle coins and one 3-burle coin. He can't buy fewer shovels without any change.
In the second example it is enough for Polycarp to buy one shovel.
In the third example Polycarp should buy two shovels and pay 2Β·15β=β30 burles. It is obvious that he can pay this sum without any change.
| 500
|
[
{
"input": "117 3",
"output": "9"
},
{
"input": "237 7",
"output": "1"
},
{
"input": "15 2",
"output": "2"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "1 9",
"output": "9"
},
{
"input": "1000 3",
"output": "1"
},
{
"input": "1000 1",
"output": "1"
},
{
"input": "1000 9",
"output": "1"
},
{
"input": "1 2",
"output": "2"
},
{
"input": "999 9",
"output": "1"
},
{
"input": "999 8",
"output": "2"
},
{
"input": "105 6",
"output": "2"
},
{
"input": "403 9",
"output": "3"
},
{
"input": "546 4",
"output": "4"
},
{
"input": "228 9",
"output": "5"
},
{
"input": "57 2",
"output": "6"
},
{
"input": "437 9",
"output": "7"
},
{
"input": "997 6",
"output": "8"
},
{
"input": "109 1",
"output": "9"
},
{
"input": "998 9",
"output": "5"
},
{
"input": "4 2",
"output": "3"
},
{
"input": "9 3",
"output": "7"
},
{
"input": "8 2",
"output": "4"
},
{
"input": "1 3",
"output": "3"
},
{
"input": "1 4",
"output": "4"
},
{
"input": "1 5",
"output": "5"
},
{
"input": "1 6",
"output": "6"
},
{
"input": "1 7",
"output": "7"
},
{
"input": "1 8",
"output": "8"
},
{
"input": "100 3",
"output": "1"
},
{
"input": "1000 2",
"output": "1"
},
{
"input": "1000 4",
"output": "1"
},
{
"input": "1000 5",
"output": "1"
},
{
"input": "1000 6",
"output": "1"
},
{
"input": "1000 7",
"output": "1"
},
{
"input": "1000 8",
"output": "1"
},
{
"input": "23 4",
"output": "8"
},
{
"input": "33 1",
"output": "7"
},
{
"input": "33 2",
"output": "4"
},
{
"input": "666 5",
"output": "5"
},
{
"input": "2 3",
"output": "5"
},
{
"input": "5 5",
"output": "1"
},
{
"input": "3 6",
"output": "2"
},
{
"input": "12 4",
"output": "2"
},
{
"input": "15 5",
"output": "1"
},
{
"input": "2 5",
"output": "5"
},
{
"input": "25 5",
"output": "1"
},
{
"input": "2 9",
"output": "5"
},
{
"input": "6 7",
"output": "5"
},
{
"input": "8 9",
"output": "5"
},
{
"input": "2 7",
"output": "5"
},
{
"input": "4 7",
"output": "5"
},
{
"input": "2 1",
"output": "5"
},
{
"input": "261 1",
"output": "1"
}
] | 1,684,598,231
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 54
| 46
| 0
|
l = input()
list1 = l.split(" ")
k = int(list1[0])
r = int(list1[1])
count =0
sum = 0
flag = 0
while flag == 0 :
sum += k
count += 1
if sum%10 ==0 :
flag = 1
print (count)
else :
sum -= r
if sum % 10 == 0:
flag = 1
print(count)
else :
sum += r
|
Title: Buy a Shovel
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp urgently needs a shovel! He comes to the shop and chooses an appropriate one. The shovel that Policarp chooses is sold for *k* burles. Assume that there is an unlimited number of such shovels in the shop.
In his pocket Polycarp has an unlimited number of "10-burle coins" and exactly one coin of *r* burles (1<=β€<=*r*<=β€<=9).
What is the minimum number of shovels Polycarp has to buy so that he can pay for the purchase without any change? It is obvious that he can pay for 10 shovels without any change (by paying the requied amount of 10-burle coins and not using the coin of *r* burles). But perhaps he can buy fewer shovels and pay without any change. Note that Polycarp should buy at least one shovel.
Input Specification:
The single line of input contains two integers *k* and *r* (1<=β€<=*k*<=β€<=1000, 1<=β€<=*r*<=β€<=9)Β β the price of one shovel and the denomination of the coin in Polycarp's pocket that is different from "10-burle coins".
Remember that he has an unlimited number of coins in the denomination of 10, that is, Polycarp has enough money to buy any number of shovels.
Output Specification:
Print the required minimum number of shovels Polycarp has to buy so that he can pay for them without any change.
Demo Input:
['117 3\n', '237 7\n', '15 2\n']
Demo Output:
['9\n', '1\n', '2\n']
Note:
In the first example Polycarp can buy 9 shovels and pay 9Β·117β=β1053 burles. Indeed, he can pay this sum by using 10-burle coins and one 3-burle coin. He can't buy fewer shovels without any change.
In the second example it is enough for Polycarp to buy one shovel.
In the third example Polycarp should buy two shovels and pay 2Β·15β=β30 burles. It is obvious that he can pay this sum without any change.
|
```python
l = input()
list1 = l.split(" ")
k = int(list1[0])
r = int(list1[1])
count =0
sum = 0
flag = 0
while flag == 0 :
sum += k
count += 1
if sum%10 ==0 :
flag = 1
print (count)
else :
sum -= r
if sum % 10 == 0:
flag = 1
print(count)
else :
sum += r
```
| 3
|
|
987
|
A
|
Infinity Gauntlet
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
You took a peek on Thanos wearing Infinity Gauntlet. In the Gauntlet there is a place for six Infinity Gems:
- the Power Gem of purple color, - the Time Gem of green color, - the Space Gem of blue color, - the Soul Gem of orange color, - the Reality Gem of red color, - the Mind Gem of yellow color.
Using colors of Gems you saw in the Gauntlet determine the names of absent Gems.
|
In the first line of input there is one integer $n$ ($0 \le n \le 6$)Β β the number of Gems in Infinity Gauntlet.
In next $n$ lines there are colors of Gems you saw. Words used for colors are: purple, green, blue, orange, red, yellow. It is guaranteed that all the colors are distinct. All colors are given in lowercase English letters.
|
In the first line output one integer $m$ ($0 \le m \le 6$)Β β the number of absent Gems.
Then in $m$ lines print the names of absent Gems, each on its own line. Words used for names are: Power, Time, Space, Soul, Reality, Mind. Names can be printed in any order. Keep the first letter uppercase, others lowercase.
|
[
"4\nred\npurple\nyellow\norange\n",
"0\n"
] |
[
"2\nSpace\nTime\n",
"6\nTime\nMind\nSoul\nPower\nReality\nSpace\n"
] |
In the first sample Thanos already has Reality, Power, Mind and Soul Gems, so he needs two more: Time and Space.
In the second sample Thanos doesn't have any Gems, so he needs all six.
| 500
|
[
{
"input": "4\nred\npurple\nyellow\norange",
"output": "2\nSpace\nTime"
},
{
"input": "0",
"output": "6\nMind\nSpace\nPower\nTime\nReality\nSoul"
},
{
"input": "6\npurple\nblue\nyellow\nred\ngreen\norange",
"output": "0"
},
{
"input": "1\npurple",
"output": "5\nTime\nReality\nSoul\nSpace\nMind"
},
{
"input": "3\nblue\norange\npurple",
"output": "3\nTime\nReality\nMind"
},
{
"input": "2\nyellow\nred",
"output": "4\nPower\nSoul\nSpace\nTime"
},
{
"input": "1\ngreen",
"output": "5\nReality\nSpace\nPower\nSoul\nMind"
},
{
"input": "2\npurple\ngreen",
"output": "4\nReality\nMind\nSpace\nSoul"
},
{
"input": "1\nblue",
"output": "5\nPower\nReality\nSoul\nTime\nMind"
},
{
"input": "2\npurple\nblue",
"output": "4\nMind\nSoul\nTime\nReality"
},
{
"input": "2\ngreen\nblue",
"output": "4\nReality\nMind\nPower\nSoul"
},
{
"input": "3\npurple\ngreen\nblue",
"output": "3\nMind\nReality\nSoul"
},
{
"input": "1\norange",
"output": "5\nReality\nTime\nPower\nSpace\nMind"
},
{
"input": "2\npurple\norange",
"output": "4\nReality\nMind\nTime\nSpace"
},
{
"input": "2\norange\ngreen",
"output": "4\nSpace\nMind\nReality\nPower"
},
{
"input": "3\norange\npurple\ngreen",
"output": "3\nReality\nSpace\nMind"
},
{
"input": "2\norange\nblue",
"output": "4\nTime\nMind\nReality\nPower"
},
{
"input": "3\nblue\ngreen\norange",
"output": "3\nPower\nMind\nReality"
},
{
"input": "4\nblue\norange\ngreen\npurple",
"output": "2\nMind\nReality"
},
{
"input": "1\nred",
"output": "5\nTime\nSoul\nMind\nPower\nSpace"
},
{
"input": "2\nred\npurple",
"output": "4\nMind\nSpace\nTime\nSoul"
},
{
"input": "2\nred\ngreen",
"output": "4\nMind\nSpace\nPower\nSoul"
},
{
"input": "3\nred\npurple\ngreen",
"output": "3\nSoul\nSpace\nMind"
},
{
"input": "2\nblue\nred",
"output": "4\nMind\nTime\nPower\nSoul"
},
{
"input": "3\nred\nblue\npurple",
"output": "3\nTime\nMind\nSoul"
},
{
"input": "3\nred\nblue\ngreen",
"output": "3\nSoul\nPower\nMind"
},
{
"input": "4\npurple\nblue\ngreen\nred",
"output": "2\nMind\nSoul"
},
{
"input": "2\norange\nred",
"output": "4\nPower\nMind\nTime\nSpace"
},
{
"input": "3\nred\norange\npurple",
"output": "3\nMind\nSpace\nTime"
},
{
"input": "3\nred\norange\ngreen",
"output": "3\nMind\nSpace\nPower"
},
{
"input": "4\nred\norange\ngreen\npurple",
"output": "2\nSpace\nMind"
},
{
"input": "3\nblue\norange\nred",
"output": "3\nPower\nMind\nTime"
},
{
"input": "4\norange\nblue\npurple\nred",
"output": "2\nTime\nMind"
},
{
"input": "4\ngreen\norange\nred\nblue",
"output": "2\nMind\nPower"
},
{
"input": "5\npurple\norange\nblue\nred\ngreen",
"output": "1\nMind"
},
{
"input": "1\nyellow",
"output": "5\nPower\nSoul\nReality\nSpace\nTime"
},
{
"input": "2\npurple\nyellow",
"output": "4\nTime\nReality\nSpace\nSoul"
},
{
"input": "2\ngreen\nyellow",
"output": "4\nSpace\nReality\nPower\nSoul"
},
{
"input": "3\npurple\nyellow\ngreen",
"output": "3\nSoul\nReality\nSpace"
},
{
"input": "2\nblue\nyellow",
"output": "4\nTime\nReality\nPower\nSoul"
},
{
"input": "3\nyellow\nblue\npurple",
"output": "3\nSoul\nReality\nTime"
},
{
"input": "3\ngreen\nyellow\nblue",
"output": "3\nSoul\nReality\nPower"
},
{
"input": "4\nyellow\nblue\ngreen\npurple",
"output": "2\nReality\nSoul"
},
{
"input": "2\nyellow\norange",
"output": "4\nTime\nSpace\nReality\nPower"
},
{
"input": "3\nyellow\npurple\norange",
"output": "3\nSpace\nReality\nTime"
},
{
"input": "3\norange\nyellow\ngreen",
"output": "3\nSpace\nReality\nPower"
},
{
"input": "4\ngreen\nyellow\norange\npurple",
"output": "2\nSpace\nReality"
},
{
"input": "3\nyellow\nblue\norange",
"output": "3\nTime\nReality\nPower"
},
{
"input": "4\norange\npurple\nblue\nyellow",
"output": "2\nReality\nTime"
},
{
"input": "4\nblue\norange\nyellow\ngreen",
"output": "2\nReality\nPower"
},
{
"input": "5\ngreen\nyellow\norange\nblue\npurple",
"output": "1\nReality"
},
{
"input": "3\nyellow\npurple\nred",
"output": "3\nTime\nSoul\nSpace"
},
{
"input": "3\nred\ngreen\nyellow",
"output": "3\nPower\nSoul\nSpace"
},
{
"input": "4\nred\npurple\ngreen\nyellow",
"output": "2\nSpace\nSoul"
},
{
"input": "3\nred\nyellow\nblue",
"output": "3\nPower\nSoul\nTime"
},
{
"input": "4\nblue\nyellow\nred\npurple",
"output": "2\nTime\nSoul"
},
{
"input": "4\nblue\nyellow\nred\ngreen",
"output": "2\nSoul\nPower"
},
{
"input": "5\nred\nyellow\ngreen\nblue\npurple",
"output": "1\nSoul"
},
{
"input": "3\nred\nyellow\norange",
"output": "3\nPower\nSpace\nTime"
},
{
"input": "4\norange\ngreen\nyellow\nred",
"output": "2\nPower\nSpace"
},
{
"input": "5\norange\nred\ngreen\nyellow\npurple",
"output": "1\nSpace"
},
{
"input": "4\nyellow\nred\norange\nblue",
"output": "2\nTime\nPower"
},
{
"input": "5\npurple\nblue\norange\nyellow\nred",
"output": "1\nTime"
},
{
"input": "5\norange\nblue\nyellow\nred\ngreen",
"output": "1\nPower"
}
] | 1,647,595,854
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 64
| 46
| 0
|
a=int(input())
gr=0
pu=0
bl=0
ore=0
r=0
y=0
p=0
for i in range(a):
v=input()
if v=='purple':
pu=v
p+=1
if v=='green':
gr=v
p+=1
if v=='blue':
bl=v
p+=1
if v=='orange':
ore=v
p+=1
if v=='red':
r=v
p+=1
if v=='yellow':
y=v
p+=1
print(6-p)
if pu==0:
print('Power')
if gr==0:
print('Time')
if bl==0:
print('Space')
if ore==0:
print('Soul')
if r==0:
print('Reality')
if y==0:
print('Mind')
|
Title: Infinity Gauntlet
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You took a peek on Thanos wearing Infinity Gauntlet. In the Gauntlet there is a place for six Infinity Gems:
- the Power Gem of purple color, - the Time Gem of green color, - the Space Gem of blue color, - the Soul Gem of orange color, - the Reality Gem of red color, - the Mind Gem of yellow color.
Using colors of Gems you saw in the Gauntlet determine the names of absent Gems.
Input Specification:
In the first line of input there is one integer $n$ ($0 \le n \le 6$)Β β the number of Gems in Infinity Gauntlet.
In next $n$ lines there are colors of Gems you saw. Words used for colors are: purple, green, blue, orange, red, yellow. It is guaranteed that all the colors are distinct. All colors are given in lowercase English letters.
Output Specification:
In the first line output one integer $m$ ($0 \le m \le 6$)Β β the number of absent Gems.
Then in $m$ lines print the names of absent Gems, each on its own line. Words used for names are: Power, Time, Space, Soul, Reality, Mind. Names can be printed in any order. Keep the first letter uppercase, others lowercase.
Demo Input:
['4\nred\npurple\nyellow\norange\n', '0\n']
Demo Output:
['2\nSpace\nTime\n', '6\nTime\nMind\nSoul\nPower\nReality\nSpace\n']
Note:
In the first sample Thanos already has Reality, Power, Mind and Soul Gems, so he needs two more: Time and Space.
In the second sample Thanos doesn't have any Gems, so he needs all six.
|
```python
a=int(input())
gr=0
pu=0
bl=0
ore=0
r=0
y=0
p=0
for i in range(a):
v=input()
if v=='purple':
pu=v
p+=1
if v=='green':
gr=v
p+=1
if v=='blue':
bl=v
p+=1
if v=='orange':
ore=v
p+=1
if v=='red':
r=v
p+=1
if v=='yellow':
y=v
p+=1
print(6-p)
if pu==0:
print('Power')
if gr==0:
print('Time')
if bl==0:
print('Space')
if ore==0:
print('Soul')
if r==0:
print('Reality')
if y==0:
print('Mind')
```
| 3
|
|
595
|
A
|
Vitaly and Night
|
PROGRAMMING
| 800
|
[
"constructive algorithms",
"implementation"
] | null | null |
One day Vitaly was going home late at night and wondering: how many people aren't sleeping at that moment? To estimate, Vitaly decided to look which windows are lit in the house he was passing by at that moment.
Vitaly sees a building of *n* floors and 2Β·*m* windows on each floor. On each floor there are *m* flats numbered from 1 to *m*, and two consecutive windows correspond to each flat. If we number the windows from 1 to 2Β·*m* from left to right, then the *j*-th flat of the *i*-th floor has windows 2Β·*j*<=-<=1 and 2Β·*j* in the corresponding row of windows (as usual, floors are enumerated from the bottom). Vitaly thinks that people in the flat aren't sleeping at that moment if at least one of the windows corresponding to this flat has lights on.
Given the information about the windows of the given house, your task is to calculate the number of flats where, according to Vitaly, people aren't sleeping.
|
The first line of the input contains two integers *n* and *m* (1<=β€<=*n*,<=*m*<=β€<=100)Β β the number of floors in the house and the number of flats on each floor respectively.
Next *n* lines describe the floors from top to bottom and contain 2Β·*m* characters each. If the *i*-th window of the given floor has lights on, then the *i*-th character of this line is '1', otherwise it is '0'.
|
Print a single integerΒ β the number of flats that have lights on in at least one window, that is, the flats where, according to Vitaly, people aren't sleeping.
|
[
"2 2\n0 0 0 1\n1 0 1 1\n",
"1 3\n1 1 0 1 0 0\n"
] |
[
"3\n",
"2\n"
] |
In the first test case the house has two floors, two flats on each floor. That is, in total there are 4 flats. The light isn't on only on the second floor in the left flat. That is, in both rooms of the flat the light is off.
In the second test case the house has one floor and the first floor has three flats. The light is on in the leftmost flat (in both windows) and in the middle flat (in one window). In the right flat the light is off.
| 500
|
[
{
"input": "2 2\n0 0 0 1\n1 0 1 1",
"output": "3"
},
{
"input": "1 3\n1 1 0 1 0 0",
"output": "2"
},
{
"input": "3 3\n1 1 1 1 1 1\n1 1 0 1 1 0\n1 0 0 0 1 1",
"output": "8"
},
{
"input": "1 5\n1 0 1 1 1 0 1 1 1 1",
"output": "5"
},
{
"input": "1 100\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 0 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 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": "99"
},
{
"input": "1 100\n0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 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 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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 0 0 0 0 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 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0",
"output": "6"
},
{
"input": "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 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0",
"output": "0"
},
{
"input": "100 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 0\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n0 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 0\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 0\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n0 1\n1 1\n1 1\n1 0\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1",
"output": "100"
},
{
"input": "100 1\n0 0\n0 0\n0 1\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\n1 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 1\n0 0\n0 0\n0 0\n1 0\n0 0\n0 0\n0 0\n1 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 1\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 1\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\n1 0",
"output": "8"
},
{
"input": "100 1\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": "1 1\n0 0",
"output": "0"
},
{
"input": "1 1\n0 1",
"output": "1"
},
{
"input": "1 1\n1 0",
"output": "1"
},
{
"input": "1 1\n1 1",
"output": "1"
}
] | 1,553,883,135
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 36
| 124
| 0
|
# import sys
# sys.stdin=open("input.in","r")
# sys.stdout=open("output.out","w")
l=list(map(int,input().split()))
n=l[0]
m=l[1]
count=0
for i in range(n):
p=list(map(int,input().split()))
for j in range(0,len(p)-1,2):
if(p[j]==1 or p[j+1]==1):
count+=1
print(count)
|
Title: Vitaly and Night
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Vitaly was going home late at night and wondering: how many people aren't sleeping at that moment? To estimate, Vitaly decided to look which windows are lit in the house he was passing by at that moment.
Vitaly sees a building of *n* floors and 2Β·*m* windows on each floor. On each floor there are *m* flats numbered from 1 to *m*, and two consecutive windows correspond to each flat. If we number the windows from 1 to 2Β·*m* from left to right, then the *j*-th flat of the *i*-th floor has windows 2Β·*j*<=-<=1 and 2Β·*j* in the corresponding row of windows (as usual, floors are enumerated from the bottom). Vitaly thinks that people in the flat aren't sleeping at that moment if at least one of the windows corresponding to this flat has lights on.
Given the information about the windows of the given house, your task is to calculate the number of flats where, according to Vitaly, people aren't sleeping.
Input Specification:
The first line of the input contains two integers *n* and *m* (1<=β€<=*n*,<=*m*<=β€<=100)Β β the number of floors in the house and the number of flats on each floor respectively.
Next *n* lines describe the floors from top to bottom and contain 2Β·*m* characters each. If the *i*-th window of the given floor has lights on, then the *i*-th character of this line is '1', otherwise it is '0'.
Output Specification:
Print a single integerΒ β the number of flats that have lights on in at least one window, that is, the flats where, according to Vitaly, people aren't sleeping.
Demo Input:
['2 2\n0 0 0 1\n1 0 1 1\n', '1 3\n1 1 0 1 0 0\n']
Demo Output:
['3\n', '2\n']
Note:
In the first test case the house has two floors, two flats on each floor. That is, in total there are 4 flats. The light isn't on only on the second floor in the left flat. That is, in both rooms of the flat the light is off.
In the second test case the house has one floor and the first floor has three flats. The light is on in the leftmost flat (in both windows) and in the middle flat (in one window). In the right flat the light is off.
|
```python
# import sys
# sys.stdin=open("input.in","r")
# sys.stdout=open("output.out","w")
l=list(map(int,input().split()))
n=l[0]
m=l[1]
count=0
for i in range(n):
p=list(map(int,input().split()))
for j in range(0,len(p)-1,2):
if(p[j]==1 or p[j+1]==1):
count+=1
print(count)
```
| 3
|
|
981
|
C
|
Useful Decomposition
|
PROGRAMMING
| 1,400
|
[
"implementation",
"trees"
] | null | null |
Ramesses knows a lot about problems involving trees (undirected connected graphs without cycles)!
He created a new useful tree decomposition, but he does not know how to construct it, so he asked you for help!
The decomposition is the splitting the edges of the tree in some simple paths in such a way that each two paths have at least one common vertex. Each edge of the tree should be in exactly one path.
Help Remesses, find such a decomposition of the tree or derermine that there is no such decomposition.
|
The first line contains a single integer $n$ ($2 \leq n \leq 10^{5}$) the number of nodes in the tree.
Each of the next $n<=-<=1$ lines contains two integers $a_i$ and $b_i$ ($1 \leq a_i, b_i \leq n$, $a_i \neq b_i$)Β β the edges of the tree. It is guaranteed that the given edges form a tree.
|
If there are no decompositions, print the only line containing "No".
Otherwise in the first line print "Yes", and in the second line print the number of paths in the decomposition $m$.
Each of the next $m$ lines should contain two integers $u_i$, $v_i$ ($1 \leq u_i, v_i \leq n$, $u_i \neq v_i$) denoting that one of the paths in the decomposition is the simple path between nodes $u_i$ and $v_i$.
Each pair of paths in the decomposition should have at least one common vertex, and each edge of the tree should be presented in exactly one path. You can print the paths and the ends of each path in arbitrary order.
If there are multiple decompositions, print any.
|
[
"4\n1 2\n2 3\n3 4\n",
"6\n1 2\n2 3\n3 4\n2 5\n3 6\n",
"5\n1 2\n1 3\n1 4\n1 5\n"
] |
[
"Yes\n1\n1 4\n",
"No\n",
"Yes\n4\n1 2\n1 3\n1 4\n1 5\n"
] |
The tree from the first example is shown on the picture below: <img class="tex-graphics" src="https://espresso.codeforces.com/9eb4b4c143d3ad267ae05d1e43341bd368b3088b.png" style="max-width: 100.0%;max-height: 100.0%;"/> The number next to each edge corresponds to the path number in the decomposition. It is easy to see that this decomposition suits the required conditions.
The tree from the second example is shown on the picture below: <img class="tex-graphics" src="https://espresso.codeforces.com/20704b97182d9bcde3321c00a16edcae4d772d93.png" style="max-width: 100.0%;max-height: 100.0%;"/> We can show that there are no valid decompositions of this tree.
The tree from the third example is shown on the picture below: <img class="tex-graphics" src="https://espresso.codeforces.com/357ff9496a4ed4746401160ee6ee63f5d57d81b9.png" style="max-width: 100.0%;max-height: 100.0%;"/> The number next to each edge corresponds to the path number in the decomposition. It is easy to see that this decomposition suits the required conditions.
| 1,250
|
[
{
"input": "4\n1 2\n2 3\n3 4",
"output": "Yes\n1\n1 4"
},
{
"input": "6\n1 2\n2 3\n3 4\n2 5\n3 6",
"output": "No"
},
{
"input": "5\n1 2\n1 3\n1 4\n1 5",
"output": "Yes\n4\n1 2\n1 3\n1 4\n1 5"
},
{
"input": "2\n1 2",
"output": "Yes\n1\n1 2"
},
{
"input": "8\n1 2\n1 3\n1 4\n1 8\n7 8\n6 8\n5 8",
"output": "No"
},
{
"input": "9\n1 2\n1 3\n1 4\n1 5\n1 6\n6 7\n7 8\n7 9",
"output": "No"
},
{
"input": "3\n2 3\n1 2",
"output": "Yes\n1\n1 3"
}
] | 1,527,435,141
| 2,541
|
PyPy 3
|
OK
|
TESTS
| 47
| 936
| 9,420,800
|
n = int(input())
orders = [0 for i in range(n)]
for i in range(n-1):
a, b = [int(j) -1 for j in input().split()]
orders[a] += 1
orders[b] += 1
roots = []
leafs = []
for i, x in enumerate(orders):
if x > 2:
roots.append(i)
elif x == 2:
pass
elif x == 1:
leafs.append(i)
else:
raise Exception('woww')
if len(roots) > 1:
print('No')
elif len(roots) == 0:
print('Yes')
print('1')
print(' '.join([str(l+1) for l in leafs]))
elif len(roots) == 1:
print('Yes')
print(str(len(leafs)))
root = str(roots[0] + 1)
for l in leafs:
print(root, str(l+1))
else:
raise Exception('woww')
|
Title: Useful Decomposition
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Ramesses knows a lot about problems involving trees (undirected connected graphs without cycles)!
He created a new useful tree decomposition, but he does not know how to construct it, so he asked you for help!
The decomposition is the splitting the edges of the tree in some simple paths in such a way that each two paths have at least one common vertex. Each edge of the tree should be in exactly one path.
Help Remesses, find such a decomposition of the tree or derermine that there is no such decomposition.
Input Specification:
The first line contains a single integer $n$ ($2 \leq n \leq 10^{5}$) the number of nodes in the tree.
Each of the next $n<=-<=1$ lines contains two integers $a_i$ and $b_i$ ($1 \leq a_i, b_i \leq n$, $a_i \neq b_i$)Β β the edges of the tree. It is guaranteed that the given edges form a tree.
Output Specification:
If there are no decompositions, print the only line containing "No".
Otherwise in the first line print "Yes", and in the second line print the number of paths in the decomposition $m$.
Each of the next $m$ lines should contain two integers $u_i$, $v_i$ ($1 \leq u_i, v_i \leq n$, $u_i \neq v_i$) denoting that one of the paths in the decomposition is the simple path between nodes $u_i$ and $v_i$.
Each pair of paths in the decomposition should have at least one common vertex, and each edge of the tree should be presented in exactly one path. You can print the paths and the ends of each path in arbitrary order.
If there are multiple decompositions, print any.
Demo Input:
['4\n1 2\n2 3\n3 4\n', '6\n1 2\n2 3\n3 4\n2 5\n3 6\n', '5\n1 2\n1 3\n1 4\n1 5\n']
Demo Output:
['Yes\n1\n1 4\n', 'No\n', 'Yes\n4\n1 2\n1 3\n1 4\n1 5\n']
Note:
The tree from the first example is shown on the picture below: <img class="tex-graphics" src="https://espresso.codeforces.com/9eb4b4c143d3ad267ae05d1e43341bd368b3088b.png" style="max-width: 100.0%;max-height: 100.0%;"/> The number next to each edge corresponds to the path number in the decomposition. It is easy to see that this decomposition suits the required conditions.
The tree from the second example is shown on the picture below: <img class="tex-graphics" src="https://espresso.codeforces.com/20704b97182d9bcde3321c00a16edcae4d772d93.png" style="max-width: 100.0%;max-height: 100.0%;"/> We can show that there are no valid decompositions of this tree.
The tree from the third example is shown on the picture below: <img class="tex-graphics" src="https://espresso.codeforces.com/357ff9496a4ed4746401160ee6ee63f5d57d81b9.png" style="max-width: 100.0%;max-height: 100.0%;"/> The number next to each edge corresponds to the path number in the decomposition. It is easy to see that this decomposition suits the required conditions.
|
```python
n = int(input())
orders = [0 for i in range(n)]
for i in range(n-1):
a, b = [int(j) -1 for j in input().split()]
orders[a] += 1
orders[b] += 1
roots = []
leafs = []
for i, x in enumerate(orders):
if x > 2:
roots.append(i)
elif x == 2:
pass
elif x == 1:
leafs.append(i)
else:
raise Exception('woww')
if len(roots) > 1:
print('No')
elif len(roots) == 0:
print('Yes')
print('1')
print(' '.join([str(l+1) for l in leafs]))
elif len(roots) == 1:
print('Yes')
print(str(len(leafs)))
root = str(roots[0] + 1)
for l in leafs:
print(root, str(l+1))
else:
raise Exception('woww')
```
| 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 β (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 β 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"
},
{
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},
{
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},
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},
{
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},
{
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"output": "1110100010111000101001001011101110011111100111000011011011"
},
{
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},
{
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"output": "111100101000000011101011011001110010101111000110010010000000"
},
{
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{
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{
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},
{
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},
{
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},
{
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},
{
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{
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"output": "00011111011111001000011100010011100011010100101011011000001001111110"
},
{
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"output": "001111000011001110100111010101111111011100110011001010010010000111011"
},
{
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"output": "0110001100110100010000110111000010011010011000011001010011010100010100"
},
{
"input": "00000111110010110001110110001010010101000111011001111111100110011110010\n00010111110100000100110101000010010001100001100011100000001100010100010",
"output": "00010000000110110101000011001000000100100110111010011111101010001010000"
},
{
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"output": "000100100000000110011100100001010110101001100101110010010011111001110111"
},
{
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"output": "1000111100010011010110011101000000101010101100011111100001101111001010010"
},
{
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},
{
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},
{
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"output": "1110111111110010111000001100101010101011010100101010100101100011001001111111"
},
{
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},
{
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"output": "110100100110000101010010011010011001100110000111010000010100001011110111111101"
},
{
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"output": "0001010101100110011000101011111000100100010100100010000000000001001100000100111"
},
{
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"output": "00001000101011011011110011001111010110100010101011000011110001101011110010001001"
},
{
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"output": "010111111011000000100001100111101000001010100010001001100101110101001010000111011"
},
{
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"output": "0100101010111101000000010111101001101101010010000110001100110111110001000100000101"
},
{
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},
{
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},
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{
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},
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},
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},
{
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"output": "010010111000010101000111111110111011001101010000000011010101010101000110111110101001010011001101011"
},
{
"input": "1010000011010110011101001101110001110010000111011101110010110110111111001001110100101100010101010001\n0111101000111100101100000101111010100100001000011101010100110011100011010011010101000100101011100011",
"output": "1101101011101010110001001000001011010110001111000000100110000101011100011010100001101000111110110010"
},
{
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"output": "1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111011"
},
{
"input": "0\n0",
"output": "0"
},
{
"input": "0\n1",
"output": "1"
}
] | 1,591,246,644
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 102
| 109
| 0
|
def ultra(n,m):
ans=''
for i in range(len(n)):
if(n[i]==m[i]):
ans+='0'
else:
ans+='1'
return ans
if __name__ == "__main__":
n=input()
m=input()
print(ultra(n,m))
|
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 β (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 β 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
def ultra(n,m):
ans=''
for i in range(len(n)):
if(n[i]==m[i]):
ans+='0'
else:
ans+='1'
return ans
if __name__ == "__main__":
n=input()
m=input()
print(ultra(n,m))
```
| 3.97275
|
41
|
A
|
Translation
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] |
A. Translation
|
2
|
256
|
The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the Β«translationΒ». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly.
|
The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols.
|
If the word *t* is a word *s*, written reversely, print YES, otherwise print NO.
|
[
"code\nedoc\n",
"abb\naba\n",
"code\ncode\n"
] |
[
"YES\n",
"NO\n",
"NO\n"
] |
none
| 500
|
[
{
"input": "code\nedoc",
"output": "YES"
},
{
"input": "abb\naba",
"output": "NO"
},
{
"input": "code\ncode",
"output": "NO"
},
{
"input": "abacaba\nabacaba",
"output": "YES"
},
{
"input": "q\nq",
"output": "YES"
},
{
"input": "asrgdfngfnmfgnhweratgjkk\nasrgdfngfnmfgnhweratgjkk",
"output": "NO"
},
{
"input": "z\na",
"output": "NO"
},
{
"input": "asd\ndsa",
"output": "YES"
},
{
"input": "abcdef\nfecdba",
"output": "NO"
},
{
"input": "ywjjbirapvskozubvxoemscfwl\ngnduubaogtfaiowjizlvjcu",
"output": "NO"
},
{
"input": "mfrmqxtzvgaeuleubcmcxcfqyruwzenguhgrmkuhdgnhgtgkdszwqyd\nmfxufheiperjnhyczclkmzyhcxntdfskzkzdwzzujdinf",
"output": "NO"
},
{
"input": "bnbnemvybqizywlnghlykniaxxxlkhftppbdeqpesrtgkcpoeqowjwhrylpsziiwcldodcoonpimudvrxejjo\ntiynnekmlalogyvrgptbinkoqdwzuiyjlrldxhzjmmp",
"output": "NO"
},
{
"input": "pwlpubwyhzqvcitemnhvvwkmwcaawjvdiwtoxyhbhbxerlypelevasmelpfqwjk\nstruuzebbcenziscuoecywugxncdwzyfozhljjyizpqcgkyonyetarcpwkqhuugsqjuixsxptmbnlfupdcfigacdhhrzb",
"output": "NO"
},
{
"input": "gdvqjoyxnkypfvdxssgrihnwxkeojmnpdeobpecytkbdwujqfjtxsqspxvxpqioyfagzjxupqqzpgnpnpxcuipweunqch\nkkqkiwwasbhezqcfeceyngcyuogrkhqecwsyerdniqiocjehrpkljiljophqhyaiefjpavoom",
"output": "NO"
},
{
"input": "umeszdawsvgkjhlqwzents\nhxqhdungbylhnikwviuh",
"output": "NO"
},
{
"input": "juotpscvyfmgntshcealgbsrwwksgrwnrrbyaqqsxdlzhkbugdyx\nibqvffmfktyipgiopznsqtrtxiijntdbgyy",
"output": "NO"
},
{
"input": "zbwueheveouatecaglziqmudxemhrsozmaujrwlqmppzoumxhamwugedikvkblvmxwuofmpafdprbcftew\nulczwrqhctbtbxrhhodwbcxwimncnexosksujlisgclllxokrsbnozthajnnlilyffmsyko",
"output": "NO"
},
{
"input": "nkgwuugukzcv\nqktnpxedwxpxkrxdvgmfgoxkdfpbzvwsduyiybynbkouonhvmzakeiruhfmvrktghadbfkmwxduoqv",
"output": "NO"
},
{
"input": "incenvizhqpcenhjhehvjvgbsnfixbatrrjstxjzhlmdmxijztphxbrldlqwdfimweepkggzcxsrwelodpnryntepioqpvk\ndhjbjjftlvnxibkklxquwmzhjfvnmwpapdrslioxisbyhhfymyiaqhlgecpxamqnocizwxniubrmpyubvpenoukhcobkdojlybxd",
"output": "NO"
},
{
"input": "w\nw",
"output": "YES"
},
{
"input": "vz\nzv",
"output": "YES"
},
{
"input": "ry\nyr",
"output": "YES"
},
{
"input": "xou\nuox",
"output": "YES"
},
{
"input": "axg\ngax",
"output": "NO"
},
{
"input": "zdsl\nlsdz",
"output": "YES"
},
{
"input": "kudl\nldku",
"output": "NO"
},
{
"input": "zzlzwnqlcl\nlclqnwzlzz",
"output": "YES"
},
{
"input": "vzzgicnzqooejpjzads\nsdazjpjeooqzncigzzv",
"output": "YES"
},
{
"input": "raqhmvmzuwaykjpyxsykr\nxkysrypjkyawuzmvmhqar",
"output": "NO"
},
{
"input": "ngedczubzdcqbxksnxuavdjaqtmdwncjnoaicvmodcqvhfezew\nwezefhvqcdomvciaonjcnwdmtqajdvauxnskxbqcdzbuzcdegn",
"output": "YES"
},
{
"input": "muooqttvrrljcxbroizkymuidvfmhhsjtumksdkcbwwpfqdyvxtrlymofendqvznzlmim\nmimlznzvqdnefomylrtxvydqfpwwbckdskmutjshhmfvdiumykziorbxcjlrrvttqooum",
"output": "YES"
},
{
"input": "vxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaivg\ngviayyikkitmuomcpiakhbxszgbnhvwyzkftwoagzixaearxpjacrnvpvbuzenvovehkmmxvblqyxvctroddksdsgebcmlluqpxv",
"output": "YES"
},
{
"input": "mnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfdc\ncdfmkdgrdptkpewbsqvszipgxvgvuiuzbkkwuowbafkikgvnqdkxnayzdjygvezmtsgywnupocdntipiyiorblqkrzjpzatxahnm",
"output": "NO"
},
{
"input": "dgxmzbqofstzcdgthbaewbwocowvhqpinehpjatnnbrijcolvsatbblsrxabzrpszoiecpwhfjmwuhqrapvtcgvikuxtzbftydkw\nwkdytfbztxukivgctvparqhuwmjfhwpceiozsprzbaxrslbbqasvlocjirbnntajphenipthvwocowbweabhtgdcztsfoqbzmxgd",
"output": "NO"
},
{
"input": "gxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwgeh\nhegwxvocotmzstqfbmpjvijgkcyodlxyjawrpkczpmdspsuhoiruavnnnuwvtwohglkdxjetshkboalvzqbgjgthoteceixioxg",
"output": "YES"
},
{
"input": "sihxuwvmaambplxvjfoskinghzicyfqebjtkysotattkahssumfcgrkheotdxwjckpvapbkaepqrxseyfrwtyaycmrzsrsngkh\nhkgnsrszrmcyaytwrfyesxrqpeakbpavpkcjwxdtoehkrgcfmusshakttatosyktjbeqfycizhgniksofjvxlpbmaamvwuxhis",
"output": "YES"
},
{
"input": "ycnahksbughnonldzrhkysujmylcgcfuludjvjiahtkyzqvkopzqcnwhltbzfugzojqkjjlggmvnultascmygelkiktmfieok\nkoeifmtkiklegkmcsatlunvmggkjjlqjozgufzbtlhwncqzpokvqzykthaijvjdulufcgclymjusyyhrzdlnonhgubskhancy",
"output": "NO"
},
{
"input": "wbqasaehtkfojruzyhrlgwmtyiovmzyfifslvlemhqheyaelzwnthrenjsbmntwaoryzwfbxmscmypvxlfmzpnkkjlvwvmtz\nztmvwvljkknpzmflxvpymcsmxbfwzyroawtnmbsjnerhtnwzleayehqhmelvlsfifyzmvoiytmwglrhyzurjofktheasaqbw",
"output": "YES"
},
{
"input": "imippqurprbhfugngtgifelytadegwrgaefnfhbjjnmzikvjaccotqzemufqieqldgnbmviisgkynzeldlhqxuqphjfmyij\njiymfjhpquxqhldleznykgsiivmbngdlqeiqfumezqtoccajvkizmnjjbhfnfeagrwgedatylefigtgngufhbrpruqppimi",
"output": "YES"
},
{
"input": "bikydffiuisckpvzqlteqfhegsagimodb\nbdomigasgehfqetlqzvpkcsiuiffdykib",
"output": "YES"
}
] | 1,674,719,877
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 40
| 92
| 0
|
t = list(input())
s = input()
i=0
j=len(t)-1
while(i<j):
t[i],t[j] = t[j],t[i]
i=i+1
j=j-1
x = "".join(t)
if(x==s):
print("YES")
else:
print("NO")
|
Title: Translation
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the Β«translationΒ». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly.
Input Specification:
The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols.
Output Specification:
If the word *t* is a word *s*, written reversely, print YES, otherwise print NO.
Demo Input:
['code\nedoc\n', 'abb\naba\n', 'code\ncode\n']
Demo Output:
['YES\n', 'NO\n', 'NO\n']
Note:
none
|
```python
t = list(input())
s = input()
i=0
j=len(t)-1
while(i<j):
t[i],t[j] = t[j],t[i]
i=i+1
j=j-1
x = "".join(t)
if(x==s):
print("YES")
else:
print("NO")
```
| 3.977
|
34
|
B
|
Sale
|
PROGRAMMING
| 900
|
[
"greedy",
"sortings"
] |
B. Sale
|
2
|
256
|
Once Bob got to a sale of old TV sets. There were *n* TV sets at that sale. TV set with index *i* costs *a**i* bellars. Some TV sets have a negative price β their owners are ready to pay Bob if he buys their useless apparatus. Bob can Β«buyΒ» any TV sets he wants. Though he's very strong, Bob can carry at most *m* TV sets, and he has no desire to go to the sale for the second time. Please, help Bob find out the maximum sum of money that he can earn.
|
The first line contains two space-separated integers *n* and *m* (1<=β€<=*m*<=β€<=*n*<=β€<=100) β amount of TV sets at the sale, and amount of TV sets that Bob can carry. The following line contains *n* space-separated integers *a**i* (<=-<=1000<=β€<=*a**i*<=β€<=1000) β prices of the TV sets.
|
Output the only number β the maximum sum of money that Bob can earn, given that he can carry at most *m* TV sets.
|
[
"5 3\n-6 0 35 -2 4\n",
"4 2\n7 0 0 -7\n"
] |
[
"8\n",
"7\n"
] |
none
| 1,000
|
[
{
"input": "5 3\n-6 0 35 -2 4",
"output": "8"
},
{
"input": "4 2\n7 0 0 -7",
"output": "7"
},
{
"input": "6 6\n756 -611 251 -66 572 -818",
"output": "1495"
},
{
"input": "5 5\n976 437 937 788 518",
"output": "0"
},
{
"input": "5 3\n-2 -2 -2 -2 -2",
"output": "6"
},
{
"input": "5 1\n998 997 985 937 998",
"output": "0"
},
{
"input": "2 2\n-742 -187",
"output": "929"
},
{
"input": "3 3\n522 597 384",
"output": "0"
},
{
"input": "4 2\n-215 -620 192 647",
"output": "835"
},
{
"input": "10 6\n557 605 685 231 910 633 130 838 -564 -85",
"output": "649"
},
{
"input": "20 14\n932 442 960 943 624 624 955 998 631 910 850 517 715 123 1000 155 -10 961 966 59",
"output": "10"
},
{
"input": "30 5\n991 997 996 967 977 999 991 986 1000 965 984 997 998 1000 958 983 974 1000 991 999 1000 978 961 992 990 998 998 978 998 1000",
"output": "0"
},
{
"input": "50 20\n-815 -947 -946 -993 -992 -846 -884 -954 -963 -733 -940 -746 -766 -930 -821 -937 -937 -999 -914 -938 -936 -975 -939 -981 -977 -952 -925 -901 -952 -978 -994 -957 -946 -896 -905 -836 -994 -951 -887 -939 -859 -953 -985 -988 -946 -829 -956 -842 -799 -886",
"output": "19441"
},
{
"input": "88 64\n999 999 1000 1000 999 996 995 1000 1000 999 1000 997 998 1000 999 1000 997 1000 993 998 994 999 998 996 1000 997 1000 1000 1000 997 1000 998 997 1000 1000 998 1000 998 999 1000 996 999 999 999 996 995 999 1000 998 999 1000 999 999 1000 1000 1000 996 1000 1000 1000 997 1000 1000 997 999 1000 1000 1000 1000 1000 999 999 1000 1000 996 999 1000 1000 995 999 1000 996 1000 998 999 999 1000 999",
"output": "0"
},
{
"input": "99 17\n-993 -994 -959 -989 -991 -995 -976 -997 -990 -1000 -996 -994 -999 -995 -1000 -983 -979 -1000 -989 -968 -994 -992 -962 -993 -999 -983 -991 -979 -995 -993 -973 -999 -995 -995 -999 -993 -995 -992 -947 -1000 -999 -998 -982 -988 -979 -993 -963 -988 -980 -990 -979 -976 -995 -999 -981 -988 -998 -999 -970 -1000 -983 -994 -943 -975 -998 -977 -973 -997 -959 -999 -983 -985 -950 -977 -977 -991 -998 -973 -987 -985 -985 -986 -984 -994 -978 -998 -989 -989 -988 -970 -985 -974 -997 -981 -962 -972 -995 -988 -993",
"output": "16984"
},
{
"input": "100 37\n205 19 -501 404 912 -435 -322 -469 -655 880 -804 -470 793 312 -108 586 -642 -928 906 605 -353 -800 745 -440 -207 752 -50 -28 498 -800 -62 -195 602 -833 489 352 536 404 -775 23 145 -512 524 759 651 -461 -427 -557 684 -366 62 592 -563 -811 64 418 -881 -308 591 -318 -145 -261 -321 -216 -18 595 -202 960 -4 219 226 -238 -882 -963 425 970 -434 -160 243 -672 -4 873 8 -633 904 -298 -151 -377 -61 -72 -677 -66 197 -716 3 -870 -30 152 -469 981",
"output": "21743"
},
{
"input": "100 99\n-931 -806 -830 -828 -916 -962 -660 -867 -952 -966 -820 -906 -724 -982 -680 -717 -488 -741 -897 -613 -986 -797 -964 -939 -808 -932 -810 -860 -641 -916 -858 -628 -821 -929 -917 -976 -664 -985 -778 -665 -624 -928 -940 -958 -884 -757 -878 -896 -634 -526 -514 -873 -990 -919 -988 -878 -650 -973 -774 -783 -733 -648 -756 -895 -833 -974 -832 -725 -841 -748 -806 -613 -924 -867 -881 -943 -864 -991 -809 -926 -777 -817 -998 -682 -910 -996 -241 -722 -964 -904 -821 -920 -835 -699 -805 -632 -779 -317 -915 -654",
"output": "81283"
},
{
"input": "100 14\n995 994 745 684 510 737 984 690 979 977 542 933 871 603 758 653 962 997 747 974 773 766 975 770 527 960 841 989 963 865 974 967 950 984 757 685 986 809 982 959 931 880 978 867 805 562 970 900 834 782 616 885 910 608 974 918 576 700 871 980 656 941 978 759 767 840 573 859 841 928 693 853 716 927 976 851 962 962 627 797 707 873 869 988 993 533 665 887 962 880 929 980 877 887 572 790 721 883 848 782",
"output": "0"
},
{
"input": "100 84\n768 946 998 752 931 912 826 1000 991 910 875 962 901 952 958 733 959 908 872 840 923 826 952 980 974 980 947 955 959 822 997 963 966 933 829 923 971 999 926 932 865 984 974 858 994 855 949 941 992 861 951 949 991 711 763 728 935 485 716 907 869 952 960 859 909 963 978 942 968 933 923 909 997 962 687 764 924 774 875 1000 961 951 987 974 848 921 966 859 995 997 974 931 886 941 974 986 906 978 998 823",
"output": "0"
},
{
"input": "100 80\n-795 -994 -833 -930 -974 -980 -950 -940 -788 -927 -583 -956 -945 -949 -809 -974 -957 -736 -967 -908 -975 -961 -986 -983 -963 -771 -952 -847 -751 -741 -982 -959 -925 -931 -839 -937 -880 -914 -858 -998 -812 -911 -862 -965 -943 -984 -738 -920 -950 -998 -909 -998 -781 -901 -677 -940 -985 -951 -675 -952 -967 -949 -882 -641 -969 -937 -975 -993 -913 -941 -807 -851 -832 -960 -939 -943 -895 -929 -528 -880 -823 -930 -888 -862 -948 -966 -962 -857 -799 -969 -833 -998 -952 -878 -946 -971 -976 -974 -723 -992",
"output": "75068"
},
{
"input": "1 1\n0",
"output": "0"
},
{
"input": "1 1\n1",
"output": "0"
},
{
"input": "1 1\n555",
"output": "0"
},
{
"input": "1 1\n-1",
"output": "1"
},
{
"input": "1 1\n-24",
"output": "24"
}
] | 1,663,424,934
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 25
| 92
| 0
|
n,m=list(map(int,input().split()))
lis=list(map(int,input().split()))
lis.sort()
sum=0
for i in range(m):
if(lis[i]<0):
sum=sum+lis[i]
print(abs(sum))
|
Title: Sale
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Once Bob got to a sale of old TV sets. There were *n* TV sets at that sale. TV set with index *i* costs *a**i* bellars. Some TV sets have a negative price β their owners are ready to pay Bob if he buys their useless apparatus. Bob can Β«buyΒ» any TV sets he wants. Though he's very strong, Bob can carry at most *m* TV sets, and he has no desire to go to the sale for the second time. Please, help Bob find out the maximum sum of money that he can earn.
Input Specification:
The first line contains two space-separated integers *n* and *m* (1<=β€<=*m*<=β€<=*n*<=β€<=100) β amount of TV sets at the sale, and amount of TV sets that Bob can carry. The following line contains *n* space-separated integers *a**i* (<=-<=1000<=β€<=*a**i*<=β€<=1000) β prices of the TV sets.
Output Specification:
Output the only number β the maximum sum of money that Bob can earn, given that he can carry at most *m* TV sets.
Demo Input:
['5 3\n-6 0 35 -2 4\n', '4 2\n7 0 0 -7\n']
Demo Output:
['8\n', '7\n']
Note:
none
|
```python
n,m=list(map(int,input().split()))
lis=list(map(int,input().split()))
lis.sort()
sum=0
for i in range(m):
if(lis[i]<0):
sum=sum+lis[i]
print(abs(sum))
```
| 3.977
|
71
|
A
|
Way Too Long Words
|
PROGRAMMING
| 800
|
[
"strings"
] |
A. Way Too Long Words
|
1
|
256
|
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalizationΒ» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
|
The first line contains an integer *n* (1<=β€<=*n*<=β€<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
|
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
|
[
"4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n"
] |
[
"word\nl10n\ni18n\np43s\n"
] |
none
| 500
|
[
{
"input": "4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis",
"output": "word\nl10n\ni18n\np43s"
},
{
"input": "5\nabcdefgh\nabcdefghi\nabcdefghij\nabcdefghijk\nabcdefghijklm",
"output": "abcdefgh\nabcdefghi\nabcdefghij\na9k\na11m"
},
{
"input": "3\nnjfngnrurunrgunrunvurn\njfvnjfdnvjdbfvsbdubruvbubvkdb\nksdnvidnviudbvibd",
"output": "n20n\nj27b\nk15d"
},
{
"input": "1\ntcyctkktcctrcyvbyiuhihhhgyvyvyvyvjvytchjckt",
"output": "t41t"
},
{
"input": "24\nyou\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nunofficially\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings",
"output": "you\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nu10y\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings"
},
{
"input": "1\na",
"output": "a"
},
{
"input": "26\na\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz",
"output": "a\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz"
},
{
"input": "1\nabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghij",
"output": "a98j"
},
{
"input": "10\ngyartjdxxlcl\nfzsck\nuidwu\nxbymclornemdmtj\nilppyoapitawgje\ncibzc\ndrgbeu\nhezplmsdekhhbo\nfeuzlrimbqbytdu\nkgdco",
"output": "g10l\nfzsck\nuidwu\nx13j\ni13e\ncibzc\ndrgbeu\nh12o\nf13u\nkgdco"
},
{
"input": "20\nlkpmx\nkovxmxorlgwaomlswjxlpnbvltfv\nhykasjxqyjrmybejnmeumzha\ntuevlumpqbbhbww\nqgqsphvrmupxxc\ntrissbaf\nqfgrlinkzvzqdryckaizutd\nzzqtoaxkvwoscyx\noswytrlnhpjvvnwookx\nlpuzqgec\ngyzqfwxggtvpjhzmzmdw\nrlxjgmvdftvrmvbdwudra\nvsntnjpepnvdaxiporggmglhagv\nxlvcqkqgcrbgtgglj\nlyxwxbiszyhlsrgzeedzprbmcpduvq\nyrmqqvrkqskqukzqrwukpsifgtdc\nxpuohcsjhhuhvr\nvvlfrlxpvqejngwrbfbpmqeirxlw\nsvmasocxdvadmaxtrpakysmeaympy\nyuflqboqfdt",
"output": "lkpmx\nk26v\nh22a\nt13w\nq12c\ntrissbaf\nq21d\nz13x\no17x\nlpuzqgec\ng18w\nr19a\nv25v\nx15j\nl28q\ny26c\nx12r\nv26w\ns27y\ny9t"
},
{
"input": "100\nm\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr",
"output": "m\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr"
}
] | 1,682,051,234
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 20
| 62
| 0
|
T = int(input())
i=0
while i<T:
word = input()
if(len(word)>10):
print(word[0],len(word)-2,word[-1],sep="")
else:
print(word)
i+=1
|
Title: Way Too Long Words
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalizationΒ» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
Input Specification:
The first line contains an integer *n* (1<=β€<=*n*<=β€<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
Output Specification:
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
Demo Input:
['4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n']
Demo Output:
['word\nl10n\ni18n\np43s\n']
Note:
none
|
```python
T = int(input())
i=0
while i<T:
word = input()
if(len(word)>10):
print(word[0],len(word)-2,word[-1],sep="")
else:
print(word)
i+=1
```
| 3.969
|
441
|
A
|
Valera and Antique Items
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
Valera is a collector. Once he wanted to expand his collection with exactly one antique item.
Valera knows *n* sellers of antiques, the *i*-th of them auctioned *k**i* items. Currently the auction price of the *j*-th object of the *i*-th seller is *s**ij*. Valera gets on well with each of the *n* sellers. He is perfectly sure that if he outbids the current price of one of the items in the auction (in other words, offers the seller the money that is strictly greater than the current price of the item at the auction), the seller of the object will immediately sign a contract with him.
Unfortunately, Valera has only *v* units of money. Help him to determine which of the *n* sellers he can make a deal with.
|
The first line contains two space-separated integers *n*,<=*v* (1<=β€<=*n*<=β€<=50;Β 104<=β€<=*v*<=β€<=106) β the number of sellers and the units of money the Valera has.
Then *n* lines follow. The *i*-th line first contains integer *k**i* (1<=β€<=*k**i*<=β€<=50) the number of items of the *i*-th seller. Then go *k**i* space-separated integers *s**i*1,<=*s**i*2,<=...,<=*s**ik**i* (104<=β€<=*s**ij*<=β€<=106) β the current prices of the items of the *i*-th seller.
|
In the first line, print integer *p* β the number of sellers with who Valera can make a deal.
In the second line print *p* space-separated integers *q*1,<=*q*2,<=...,<=*q**p* (1<=β€<=*q**i*<=β€<=*n*) β the numbers of the sellers with who Valera can make a deal. Print the numbers of the sellers in the increasing order.
|
[
"3 50000\n1 40000\n2 20000 60000\n3 10000 70000 190000\n",
"3 50000\n1 50000\n3 100000 120000 110000\n3 120000 110000 120000\n"
] |
[
"3\n1 2 3\n",
"0\n\n"
] |
In the first sample Valera can bargain with each of the sellers. He can outbid the following items: a 40000 item from the first seller, a 20000 item from the second seller, and a 10000 item from the third seller.
In the second sample Valera can not make a deal with any of the sellers, as the prices of all items in the auction too big for him.
| 500
|
[
{
"input": "3 50000\n1 40000\n2 20000 60000\n3 10000 70000 190000",
"output": "3\n1 2 3"
},
{
"input": "3 50000\n1 50000\n3 100000 120000 110000\n3 120000 110000 120000",
"output": "0"
},
{
"input": "2 100001\n1 895737\n1 541571",
"output": "0"
},
{
"input": "1 1000000\n1 1000000",
"output": "0"
},
{
"input": "1 1000000\n1 561774",
"output": "1\n1"
},
{
"input": "3 1000000\n5 1000000 568832 1000000 1000000 1000000\n13 1000000 1000000 1000000 596527 1000000 1000000 1000000 1000000 1000000 1000000 1000000 1000000 1000000\n20 1000000 1000000 1000000 1000000 1000000 1000000 1000000 1000000 1000000 1000000 1000000 1000000 1000000 1000000 1000000 1000000 1000000 1000000 1000000 1000000",
"output": "2\n1 2"
},
{
"input": "3 999999\n7 1000000 1000000 1000000 999999 1000000 999999 1000000\n6 999999 1000000 999999 1000000 999999 999999\n7 999999 1000000 1000000 999999 1000000 1000000 1000000",
"output": "0"
},
{
"input": "3 999999\n22 1000000 1000000 999999 999999 1000000 1000000 1000000 1000000 1000000 1000000 1000000 1000000 999999 1000000 1000000 999999 1000000 1000000 1000000 352800 999999 1000000\n14 999999 999999 999999 999999 999999 1000000 999999 999999 999999 999999 702638 999999 1000000 999999\n5 999999 1000000 1000000 999999 363236",
"output": "3\n1 2 3"
},
{
"input": "1 50001\n1 50000",
"output": "1\n1"
}
] | 1,585,640,870
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 26
| 109
| 0
|
a,b=map(int,input().split());k=[]
for i in range(1,a+1):
c,*d=map(int,input().split())
if min(d)<b:k+=[i]
print(len(k),"\n",*k)
|
Title: Valera and Antique Items
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Valera is a collector. Once he wanted to expand his collection with exactly one antique item.
Valera knows *n* sellers of antiques, the *i*-th of them auctioned *k**i* items. Currently the auction price of the *j*-th object of the *i*-th seller is *s**ij*. Valera gets on well with each of the *n* sellers. He is perfectly sure that if he outbids the current price of one of the items in the auction (in other words, offers the seller the money that is strictly greater than the current price of the item at the auction), the seller of the object will immediately sign a contract with him.
Unfortunately, Valera has only *v* units of money. Help him to determine which of the *n* sellers he can make a deal with.
Input Specification:
The first line contains two space-separated integers *n*,<=*v* (1<=β€<=*n*<=β€<=50;Β 104<=β€<=*v*<=β€<=106) β the number of sellers and the units of money the Valera has.
Then *n* lines follow. The *i*-th line first contains integer *k**i* (1<=β€<=*k**i*<=β€<=50) the number of items of the *i*-th seller. Then go *k**i* space-separated integers *s**i*1,<=*s**i*2,<=...,<=*s**ik**i* (104<=β€<=*s**ij*<=β€<=106) β the current prices of the items of the *i*-th seller.
Output Specification:
In the first line, print integer *p* β the number of sellers with who Valera can make a deal.
In the second line print *p* space-separated integers *q*1,<=*q*2,<=...,<=*q**p* (1<=β€<=*q**i*<=β€<=*n*) β the numbers of the sellers with who Valera can make a deal. Print the numbers of the sellers in the increasing order.
Demo Input:
['3 50000\n1 40000\n2 20000 60000\n3 10000 70000 190000\n', '3 50000\n1 50000\n3 100000 120000 110000\n3 120000 110000 120000\n']
Demo Output:
['3\n1 2 3\n', '0\n\n']
Note:
In the first sample Valera can bargain with each of the sellers. He can outbid the following items: a 40000 item from the first seller, a 20000 item from the second seller, and a 10000 item from the third seller.
In the second sample Valera can not make a deal with any of the sellers, as the prices of all items in the auction too big for him.
|
```python
a,b=map(int,input().split());k=[]
for i in range(1,a+1):
c,*d=map(int,input().split())
if min(d)<b:k+=[i]
print(len(k),"\n",*k)
```
| 3
|
|
10
|
A
|
Power Consumption Calculation
|
PROGRAMMING
| 900
|
[
"implementation"
] |
A. Power Consumption Calculation
|
1
|
256
|
Tom is interested in power consumption of his favourite laptop. His laptop has three modes. In normal mode laptop consumes *P*1 watt per minute. *T*1 minutes after Tom moved the mouse or touched the keyboard for the last time, a screensaver starts and power consumption changes to *P*2 watt per minute. Finally, after *T*2 minutes from the start of the screensaver, laptop switches to the "sleep" mode and consumes *P*3 watt per minute. If Tom moves the mouse or touches the keyboard when the laptop is in the second or in the third mode, it switches to the first (normal) mode. Tom's work with the laptop can be divided into *n* time periods [*l*1,<=*r*1],<=[*l*2,<=*r*2],<=...,<=[*l**n*,<=*r**n*]. During each interval Tom continuously moves the mouse and presses buttons on the keyboard. Between the periods Tom stays away from the laptop. Find out the total amount of power consumed by the laptop during the period [*l*1,<=*r**n*].
|
The first line contains 6 integer numbers *n*, *P*1, *P*2, *P*3, *T*1, *T*2 (1<=β€<=*n*<=β€<=100,<=0<=β€<=*P*1,<=*P*2,<=*P*3<=β€<=100,<=1<=β€<=*T*1,<=*T*2<=β€<=60). The following *n* lines contain description of Tom's work. Each *i*-th of these lines contains two space-separated integers *l**i* and *r**i* (0<=β€<=*l**i*<=<<=*r**i*<=β€<=1440, *r**i*<=<<=*l**i*<=+<=1 for *i*<=<<=*n*), which stand for the start and the end of the *i*-th period of work.
|
Output the answer to the problem.
|
[
"1 3 2 1 5 10\n0 10\n",
"2 8 4 2 5 10\n20 30\n50 100\n"
] |
[
"30",
"570"
] |
none
| 0
|
[
{
"input": "1 3 2 1 5 10\n0 10",
"output": "30"
},
{
"input": "2 8 4 2 5 10\n20 30\n50 100",
"output": "570"
},
{
"input": "3 15 9 95 39 19\n873 989\n1003 1137\n1172 1436",
"output": "8445"
},
{
"input": "4 73 2 53 58 16\n51 52\n209 242\n281 407\n904 945",
"output": "52870"
},
{
"input": "5 41 20 33 43 4\n46 465\n598 875\n967 980\n1135 1151\n1194 1245",
"output": "46995"
},
{
"input": "6 88 28 100 53 36\n440 445\n525 614\n644 844\n1238 1261\n1305 1307\n1425 1434",
"output": "85540"
},
{
"input": "7 46 61 55 28 59\n24 26\n31 61\n66 133\n161 612\n741 746\n771 849\n1345 1357",
"output": "67147"
},
{
"input": "8 83 18 30 28 5\n196 249\n313 544\n585 630\n718 843\n1040 1194\n1207 1246\n1268 1370\n1414 1422",
"output": "85876"
},
{
"input": "9 31 65 27 53 54\n164 176\n194 210\n485 538\n617 690\n875 886\n888 902\n955 957\n1020 1200\n1205 1282",
"output": "38570"
},
{
"input": "30 3 1 58 44 7\n11 13\n14 32\n37 50\n70 74\n101 106\n113 129\n184 195\n197 205\n213 228\n370 394\n443 446\n457 460\n461 492\n499 585\n602 627\n709 776\n812 818\n859 864\n910 913\n918 964\n1000 1010\n1051 1056\n1063 1075\n1106 1145\n1152 1189\n1211 1212\n1251 1259\n1272 1375\n1412 1417\n1430 1431",
"output": "11134"
},
{
"input": "30 42 3 76 28 26\n38 44\n55 66\n80 81\n84 283\n298 314\n331 345\n491 531\n569 579\n597 606\n612 617\n623 701\n723 740\n747 752\n766 791\n801 827\n842 846\n853 891\n915 934\n945 949\n955 964\n991 1026\n1051 1059\n1067 1179\n1181 1191\n1214 1226\n1228 1233\n1294 1306\n1321 1340\n1371 1374\n1375 1424",
"output": "59043"
},
{
"input": "30 46 5 93 20 46\n12 34\n40 41\n54 58\n100 121\n162 182\n220 349\n358 383\n390 398\n401 403\n408 409\n431 444\n466 470\n471 535\n556 568\n641 671\n699 709\n767 777\n786 859\n862 885\n912 978\n985 997\n1013 1017\n1032 1038\n1047 1048\n1062 1080\n1094 1097\n1102 1113\n1122 1181\n1239 1280\n1320 1369",
"output": "53608"
},
{
"input": "30 50 74 77 4 57\n17 23\n24 61\n67 68\n79 87\n93 101\n104 123\n150 192\n375 377\n398 414\n461 566\n600 633\n642 646\n657 701\n771 808\n812 819\n823 826\n827 833\n862 875\n880 891\n919 920\n928 959\n970 1038\n1057 1072\n1074 1130\n1165 1169\n1171 1230\n1265 1276\n1279 1302\n1313 1353\n1354 1438",
"output": "84067"
},
{
"input": "30 54 76 95 48 16\n9 11\n23 97\n112 116\n126 185\n214 223\n224 271\n278 282\n283 348\n359 368\n373 376\n452 463\n488 512\n532 552\n646 665\n681 685\n699 718\n735 736\n750 777\n791 810\n828 838\n841 858\n874 1079\n1136 1171\n1197 1203\n1210 1219\n1230 1248\n1280 1292\n1324 1374\n1397 1435\n1438 1439",
"output": "79844"
},
{
"input": "30 58 78 12 41 28\n20 26\n27 31\n35 36\n38 99\n103 104\n106 112\n133 143\n181 246\n248 251\n265 323\n350 357\n378 426\n430 443\n466 476\n510 515\n517 540\n542 554\n562 603\n664 810\n819 823\n826 845\n869 895\n921 973\n1002 1023\n1102 1136\n1143 1148\n1155 1288\n1316 1388\n1394 1403\n1434 1437",
"output": "82686"
},
{
"input": "30 62 80 97 25 47\n19 20\n43 75\n185 188\n199 242\n252 258\n277 310\n316 322\n336 357\n398 399\n404 436\n443 520\n549 617\n637 649\n679 694\n705 715\n725 730\n731 756\n768 793\n806 833\n834 967\n1003 1079\n1088 1097\n1100 1104\n1108 1121\n1127 1164\n1240 1263\n1274 1307\n1367 1407\n1419 1425\n1433 1437",
"output": "92356"
},
{
"input": "30 100 48 14 9 7\n26 55\n75 107\n145 146\n158 167\n197 199\n201 213\n222 238\n257 265\n271 338\n357 439\n454 462\n463 562\n633 660\n670 691\n696 829\n844 864\n880 888\n891 894\n895 900\n950 973\n1004 1007\n1018 1021\n1045 1049\n1062 1073\n1138 1184\n1240 1272\n1278 1315\n1403 1410\n1412 1418\n1426 1434",
"output": "107490"
},
{
"input": "30 3 50 32 1 18\n46 60\n66 94\n99 141\n162 239\n309 334\n400 550\n594 630\n659 663\n664 701\n735 736\n741 755\n756 801\n823 854\n865 915\n937 964\n984 986\n990 1007\n1008 1031\n1052 1064\n1073 1080\n1190 1191\n1192 1254\n1265 1266\n1272 1273\n1274 1277\n1284 1302\n1308 1312\n1358 1383\n1388 1389\n1390 1401",
"output": "25258"
},
{
"input": "30 7 52 49 46 37\n2 73\n129 130\n131 140\n143 150\n187 190\n217 238\n248 271\n375 426\n465 470\n475 503\n506 541\n569 571\n578 585\n680 683\n690 738\n742 761\n868 877\n890 892\n904 915\n935 961\n1076 1100\n1103 1114\n1232 1244\n1246 1247\n1257 1261\n1271 1276\n1280 1298\n1346 1360\n1373 1382\n1407 1419",
"output": "23992"
}
] | 1,674,683,181
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 30
| 124
| 0
|
n, P1, P2, P3, T1, T2=map(int,input().split())
s=0
for i in range(n):
l,r=map(int,input().split())
if i==0:
t=l
if T1>=(l-t):
s+=(l-t)*P1
else:
s+=T1*P1
if T2>=(l-T1-t):
s+=(l-T1-t)*P2
else:
s+=T2*P2
s+=(l-T1-T2-t)*P3
s+=(r-l)*P1
t=r
print(s)
|
Title: Power Consumption Calculation
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Tom is interested in power consumption of his favourite laptop. His laptop has three modes. In normal mode laptop consumes *P*1 watt per minute. *T*1 minutes after Tom moved the mouse or touched the keyboard for the last time, a screensaver starts and power consumption changes to *P*2 watt per minute. Finally, after *T*2 minutes from the start of the screensaver, laptop switches to the "sleep" mode and consumes *P*3 watt per minute. If Tom moves the mouse or touches the keyboard when the laptop is in the second or in the third mode, it switches to the first (normal) mode. Tom's work with the laptop can be divided into *n* time periods [*l*1,<=*r*1],<=[*l*2,<=*r*2],<=...,<=[*l**n*,<=*r**n*]. During each interval Tom continuously moves the mouse and presses buttons on the keyboard. Between the periods Tom stays away from the laptop. Find out the total amount of power consumed by the laptop during the period [*l*1,<=*r**n*].
Input Specification:
The first line contains 6 integer numbers *n*, *P*1, *P*2, *P*3, *T*1, *T*2 (1<=β€<=*n*<=β€<=100,<=0<=β€<=*P*1,<=*P*2,<=*P*3<=β€<=100,<=1<=β€<=*T*1,<=*T*2<=β€<=60). The following *n* lines contain description of Tom's work. Each *i*-th of these lines contains two space-separated integers *l**i* and *r**i* (0<=β€<=*l**i*<=<<=*r**i*<=β€<=1440, *r**i*<=<<=*l**i*<=+<=1 for *i*<=<<=*n*), which stand for the start and the end of the *i*-th period of work.
Output Specification:
Output the answer to the problem.
Demo Input:
['1 3 2 1 5 10\n0 10\n', '2 8 4 2 5 10\n20 30\n50 100\n']
Demo Output:
['30', '570']
Note:
none
|
```python
n, P1, P2, P3, T1, T2=map(int,input().split())
s=0
for i in range(n):
l,r=map(int,input().split())
if i==0:
t=l
if T1>=(l-t):
s+=(l-t)*P1
else:
s+=T1*P1
if T2>=(l-T1-t):
s+=(l-T1-t)*P2
else:
s+=T2*P2
s+=(l-T1-T2-t)*P3
s+=(r-l)*P1
t=r
print(s)
```
| 3.938
|
725
|
B
|
Food on the Plane
|
PROGRAMMING
| 1,200
|
[
"implementation",
"math"
] | null | null |
A new airplane SuperPuperJet has an infinite number of rows, numbered with positive integers starting with 1 from cockpit to tail. There are six seats in each row, denoted with letters from 'a' to 'f'. Seats 'a', 'b' and 'c' are located to the left of an aisle (if one looks in the direction of the cockpit), while seats 'd', 'e' and 'f' are located to the right. Seats 'a' and 'f' are located near the windows, while seats 'c' and 'd' are located near the aisle.
Β
It's lunch time and two flight attendants have just started to serve food. They move from the first rows to the tail, always maintaining a distance of two rows from each other because of the food trolley. Thus, at the beginning the first attendant serves row 1 while the second attendant serves row 3. When both rows are done they move one row forward: the first attendant serves row 2 while the second attendant serves row 4. Then they move three rows forward and the first attendant serves row 5 while the second attendant serves row 7. Then they move one row forward again and so on.
Flight attendants work with the same speed: it takes exactly 1 second to serve one passenger and 1 second to move one row forward. Each attendant first serves the passengers on the seats to the right of the aisle and then serves passengers on the seats to the left of the aisle (if one looks in the direction of the cockpit). Moreover, they always serve passengers in order from the window to the aisle. Thus, the first passenger to receive food in each row is located in seat 'f', and the last oneΒ β in seat 'c'. Assume that all seats are occupied.
Vasya has seat *s* in row *n* and wants to know how many seconds will pass before he gets his lunch.
|
The only line of input contains a description of Vasya's seat in the format *ns*, where *n* (1<=β€<=*n*<=β€<=1018) is the index of the row and *s* is the seat in this row, denoted as letter from 'a' to 'f'. The index of the row and the seat are not separated by a space.
|
Print one integerΒ β the number of seconds Vasya has to wait until he gets his lunch.
|
[
"1f\n",
"2d\n",
"4a\n",
"5e\n"
] |
[
"1\n",
"10\n",
"11\n",
"18\n"
] |
In the first sample, the first flight attendant serves Vasya first, so Vasya gets his lunch after 1 second.
In the second sample, the flight attendants will spend 6 seconds to serve everyone in the rows 1 and 3, then they will move one row forward in 1 second. As they first serve seats located to the right of the aisle in order from window to aisle, Vasya has to wait 3 more seconds. The total is 6β+β1β+β3β=β10.
| 1,000
|
[
{
"input": "1f",
"output": "1"
},
{
"input": "2d",
"output": "10"
},
{
"input": "4a",
"output": "11"
},
{
"input": "5e",
"output": "18"
},
{
"input": "2c",
"output": "13"
},
{
"input": "1b",
"output": "5"
},
{
"input": "1000000000000000000d",
"output": "3999999999999999994"
},
{
"input": "999999999999999997a",
"output": "3999999999999999988"
},
{
"input": "1c",
"output": "6"
},
{
"input": "1d",
"output": "3"
},
{
"input": "1e",
"output": "2"
},
{
"input": "1a",
"output": "4"
},
{
"input": "2a",
"output": "11"
},
{
"input": "2b",
"output": "12"
},
{
"input": "2e",
"output": "9"
},
{
"input": "2f",
"output": "8"
},
{
"input": "3a",
"output": "4"
},
{
"input": "3b",
"output": "5"
},
{
"input": "3c",
"output": "6"
},
{
"input": "3d",
"output": "3"
},
{
"input": "3e",
"output": "2"
},
{
"input": "3f",
"output": "1"
},
{
"input": "4b",
"output": "12"
},
{
"input": "4c",
"output": "13"
},
{
"input": "4d",
"output": "10"
},
{
"input": "4e",
"output": "9"
},
{
"input": "4f",
"output": "8"
},
{
"input": "999999997a",
"output": "3999999988"
},
{
"input": "999999997b",
"output": "3999999989"
},
{
"input": "999999997c",
"output": "3999999990"
},
{
"input": "999999997d",
"output": "3999999987"
},
{
"input": "999999997e",
"output": "3999999986"
},
{
"input": "999999997f",
"output": "3999999985"
},
{
"input": "999999998a",
"output": "3999999995"
},
{
"input": "999999998b",
"output": "3999999996"
},
{
"input": "999999998c",
"output": "3999999997"
},
{
"input": "999999998d",
"output": "3999999994"
},
{
"input": "999999998e",
"output": "3999999993"
},
{
"input": "999999998f",
"output": "3999999992"
},
{
"input": "999999999a",
"output": "3999999988"
},
{
"input": "999999999b",
"output": "3999999989"
},
{
"input": "999999999c",
"output": "3999999990"
},
{
"input": "999999999d",
"output": "3999999987"
},
{
"input": "999999999e",
"output": "3999999986"
},
{
"input": "999999999f",
"output": "3999999985"
},
{
"input": "1000000000a",
"output": "3999999995"
},
{
"input": "1000000000b",
"output": "3999999996"
},
{
"input": "1000000000c",
"output": "3999999997"
},
{
"input": "1000000000d",
"output": "3999999994"
},
{
"input": "1000000000e",
"output": "3999999993"
},
{
"input": "1000000000f",
"output": "3999999992"
},
{
"input": "100000b",
"output": "399996"
},
{
"input": "100000f",
"output": "399992"
},
{
"input": "100001d",
"output": "400003"
},
{
"input": "100001e",
"output": "400002"
},
{
"input": "100001f",
"output": "400001"
},
{
"input": "100002a",
"output": "400011"
},
{
"input": "100002b",
"output": "400012"
},
{
"input": "100002d",
"output": "400010"
},
{
"input": "1231273a",
"output": "4925092"
},
{
"input": "82784f",
"output": "331128"
},
{
"input": "88312c",
"output": "353245"
},
{
"input": "891237e",
"output": "3564946"
},
{
"input": "999999999999999997b",
"output": "3999999999999999989"
},
{
"input": "999999999999999997c",
"output": "3999999999999999990"
},
{
"input": "999999999999999997d",
"output": "3999999999999999987"
},
{
"input": "999999999999999997e",
"output": "3999999999999999986"
},
{
"input": "999999999999999997f",
"output": "3999999999999999985"
},
{
"input": "999999999999999998a",
"output": "3999999999999999995"
},
{
"input": "999999999999999998b",
"output": "3999999999999999996"
},
{
"input": "999999999999999998c",
"output": "3999999999999999997"
},
{
"input": "999999999999999998d",
"output": "3999999999999999994"
},
{
"input": "999999999999999998e",
"output": "3999999999999999993"
},
{
"input": "999999999999999998f",
"output": "3999999999999999992"
},
{
"input": "999999999999999999a",
"output": "3999999999999999988"
},
{
"input": "999999999999999999b",
"output": "3999999999999999989"
},
{
"input": "999999999999999999c",
"output": "3999999999999999990"
},
{
"input": "999999999999999999d",
"output": "3999999999999999987"
},
{
"input": "1000000000000000000a",
"output": "3999999999999999995"
},
{
"input": "1000000000000000000e",
"output": "3999999999999999993"
},
{
"input": "1000000000000000000f",
"output": "3999999999999999992"
},
{
"input": "1000000000000000000c",
"output": "3999999999999999997"
},
{
"input": "97a",
"output": "388"
},
{
"input": "6f",
"output": "24"
},
{
"input": "7f",
"output": "17"
},
{
"input": "7e",
"output": "18"
},
{
"input": "999999999999999992c",
"output": "3999999999999999965"
},
{
"input": "7a",
"output": "20"
},
{
"input": "8f",
"output": "24"
},
{
"input": "999999999999999992a",
"output": "3999999999999999963"
},
{
"input": "999999999999999992b",
"output": "3999999999999999964"
},
{
"input": "999999999999999992c",
"output": "3999999999999999965"
},
{
"input": "999999999999999992d",
"output": "3999999999999999962"
},
{
"input": "999999999999999992e",
"output": "3999999999999999961"
},
{
"input": "999999999999999992f",
"output": "3999999999999999960"
},
{
"input": "999999999999999993a",
"output": "3999999999999999972"
},
{
"input": "999999999999999993b",
"output": "3999999999999999973"
},
{
"input": "999999999999999993c",
"output": "3999999999999999974"
},
{
"input": "999999999999999993d",
"output": "3999999999999999971"
},
{
"input": "999999999999999993e",
"output": "3999999999999999970"
},
{
"input": "999999999999999993f",
"output": "3999999999999999969"
},
{
"input": "999999999999999994a",
"output": "3999999999999999979"
},
{
"input": "999999999999999994b",
"output": "3999999999999999980"
},
{
"input": "999999999999999994c",
"output": "3999999999999999981"
},
{
"input": "999999999999999994d",
"output": "3999999999999999978"
},
{
"input": "999999999999999994e",
"output": "3999999999999999977"
},
{
"input": "999999999999999994f",
"output": "3999999999999999976"
},
{
"input": "999999999999999995a",
"output": "3999999999999999972"
},
{
"input": "999999999999999995b",
"output": "3999999999999999973"
},
{
"input": "999999999999999995c",
"output": "3999999999999999974"
},
{
"input": "999999999999999995d",
"output": "3999999999999999971"
},
{
"input": "999999999999999995e",
"output": "3999999999999999970"
},
{
"input": "999999999999999995f",
"output": "3999999999999999969"
},
{
"input": "10a",
"output": "43"
},
{
"input": "11f",
"output": "33"
},
{
"input": "681572647b",
"output": "2726290581"
},
{
"input": "23f",
"output": "81"
},
{
"input": "123a",
"output": "484"
},
{
"input": "999999888888777777a",
"output": "3999999555555111108"
}
] | 1,610,378,357
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 119
| 77
| 0
|
c=input();n,s,d=int(c[:-1]),c[-1],{'a':4,'b':5,'c':6,'d':3,'e':2,'f':1}
x=(n//4*2+(n-1)%2-2*(n%4==0))*6+d[s]+n
print(x-1 if (n-1)%4<2 else x-3)
|
Title: Food on the Plane
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A new airplane SuperPuperJet has an infinite number of rows, numbered with positive integers starting with 1 from cockpit to tail. There are six seats in each row, denoted with letters from 'a' to 'f'. Seats 'a', 'b' and 'c' are located to the left of an aisle (if one looks in the direction of the cockpit), while seats 'd', 'e' and 'f' are located to the right. Seats 'a' and 'f' are located near the windows, while seats 'c' and 'd' are located near the aisle.
Β
It's lunch time and two flight attendants have just started to serve food. They move from the first rows to the tail, always maintaining a distance of two rows from each other because of the food trolley. Thus, at the beginning the first attendant serves row 1 while the second attendant serves row 3. When both rows are done they move one row forward: the first attendant serves row 2 while the second attendant serves row 4. Then they move three rows forward and the first attendant serves row 5 while the second attendant serves row 7. Then they move one row forward again and so on.
Flight attendants work with the same speed: it takes exactly 1 second to serve one passenger and 1 second to move one row forward. Each attendant first serves the passengers on the seats to the right of the aisle and then serves passengers on the seats to the left of the aisle (if one looks in the direction of the cockpit). Moreover, they always serve passengers in order from the window to the aisle. Thus, the first passenger to receive food in each row is located in seat 'f', and the last oneΒ β in seat 'c'. Assume that all seats are occupied.
Vasya has seat *s* in row *n* and wants to know how many seconds will pass before he gets his lunch.
Input Specification:
The only line of input contains a description of Vasya's seat in the format *ns*, where *n* (1<=β€<=*n*<=β€<=1018) is the index of the row and *s* is the seat in this row, denoted as letter from 'a' to 'f'. The index of the row and the seat are not separated by a space.
Output Specification:
Print one integerΒ β the number of seconds Vasya has to wait until he gets his lunch.
Demo Input:
['1f\n', '2d\n', '4a\n', '5e\n']
Demo Output:
['1\n', '10\n', '11\n', '18\n']
Note:
In the first sample, the first flight attendant serves Vasya first, so Vasya gets his lunch after 1 second.
In the second sample, the flight attendants will spend 6 seconds to serve everyone in the rows 1 and 3, then they will move one row forward in 1 second. As they first serve seats located to the right of the aisle in order from window to aisle, Vasya has to wait 3 more seconds. The total is 6β+β1β+β3β=β10.
|
```python
c=input();n,s,d=int(c[:-1]),c[-1],{'a':4,'b':5,'c':6,'d':3,'e':2,'f':1}
x=(n//4*2+(n-1)%2-2*(n%4==0))*6+d[s]+n
print(x-1 if (n-1)%4<2 else x-3)
```
| 3
|
|
574
|
A
|
Bear and Elections
|
PROGRAMMING
| 1,200
|
[
"greedy",
"implementation"
] | null | null |
Limak is a grizzly bear who desires power and adoration. He wants to win in upcoming elections and rule over the Bearland.
There are *n* candidates, including Limak. We know how many citizens are going to vote for each candidate. Now *i*-th candidate would get *a**i* votes. Limak is candidate number 1. To win in elections, he must get strictly more votes than any other candidate.
Victory is more important than everything else so Limak decided to cheat. He will steal votes from his opponents by bribing some citizens. To bribe a citizen, Limak must give him or her one candy - citizens are bears and bears like candies. Limak doesn't have many candies and wonders - how many citizens does he have to bribe?
|
The first line contains single integer *n* (2<=β€<=*n*<=β€<=100) - number of candidates.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=β€<=*a**i*<=β€<=1000) - number of votes for each candidate. Limak is candidate number 1.
Note that after bribing number of votes for some candidate might be zero or might be greater than 1000.
|
Print the minimum number of citizens Limak must bribe to have strictly more votes than any other candidate.
|
[
"5\n5 1 11 2 8\n",
"4\n1 8 8 8\n",
"2\n7 6\n"
] |
[
"4\n",
"6\n",
"0\n"
] |
In the first sample Limak has 5 votes. One of the ways to achieve victory is to bribe 4 citizens who want to vote for the third candidate. Then numbers of votes would be 9,β1,β7,β2,β8 (Limak would have 9 votes). Alternatively, Limak could steal only 3 votes from the third candidate and 1 vote from the second candidate to get situation 9,β0,β8,β2,β8.
In the second sample Limak will steal 2 votes from each candidate. Situation will be 7,β6,β6,β6.
In the third sample Limak is a winner without bribing any citizen.
| 500
|
[
{
"input": "5\n5 1 11 2 8",
"output": "4"
},
{
"input": "4\n1 8 8 8",
"output": "6"
},
{
"input": "2\n7 6",
"output": "0"
},
{
"input": "2\n1 1",
"output": "1"
},
{
"input": "10\n100 200 57 99 1 1000 200 200 200 500",
"output": "451"
},
{
"input": "16\n7 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000",
"output": "932"
},
{
"input": "100\n47 64 68 61 68 66 69 61 69 65 69 63 62 60 68 65 64 65 65 62 63 68 60 70 63 63 65 67 70 69 68 69 61 65 63 60 60 65 61 60 70 66 66 65 62 60 65 68 61 62 67 64 66 65 67 68 60 69 70 63 65 62 64 65 67 67 69 68 66 69 70 67 65 70 60 66 70 67 67 64 69 69 66 68 60 64 62 62 68 69 67 69 60 70 69 68 62 63 68 66",
"output": "23"
},
{
"input": "2\n96 97",
"output": "1"
},
{
"input": "2\n1000 1000",
"output": "1"
},
{
"input": "3\n999 1000 1000",
"output": "2"
},
{
"input": "3\n1 2 3",
"output": "2"
},
{
"input": "7\n10 940 926 990 946 980 985",
"output": "817"
},
{
"input": "10\n5 3 4 5 5 2 1 8 4 1",
"output": "2"
},
{
"input": "15\n17 15 17 16 13 17 13 16 14 14 17 17 13 15 17",
"output": "1"
},
{
"input": "20\n90 5 62 9 50 7 14 43 44 44 56 13 71 22 43 35 52 60 73 54",
"output": "0"
},
{
"input": "30\n27 85 49 7 77 38 4 68 23 28 81 100 40 9 78 38 1 60 60 49 98 44 45 92 46 39 98 24 37 39",
"output": "58"
},
{
"input": "51\n90 47 100 12 21 96 2 68 84 60 2 9 33 8 45 13 59 50 100 93 22 97 4 81 51 2 3 78 19 16 25 63 52 34 79 32 34 87 7 42 96 93 30 33 33 43 69 8 63 58 57",
"output": "8"
},
{
"input": "77\n1000 2 2 3 1 1 1 3 3 2 1 1 3 2 2 2 3 2 3 1 3 1 1 2 2 2 3 1 1 2 2 2 3 2 1 3 3 1 2 3 3 3 2 1 3 2 1 3 3 2 3 3 2 1 3 1 1 1 2 3 2 3 1 3 1 2 1 2 2 2 1 2 2 3 2 2 2",
"output": "0"
},
{
"input": "91\n3 92 89 83 85 80 91 94 95 82 92 95 80 88 90 85 81 90 87 86 94 88 90 87 88 82 95 84 84 93 83 95 91 85 89 88 88 85 87 90 93 80 89 95 94 92 93 86 83 82 86 84 91 80 90 95 84 86 84 85 84 92 82 84 83 91 87 95 94 95 90 95 86 92 86 80 95 86 88 80 82 87 84 83 91 93 81 81 91 89 88",
"output": "89"
},
{
"input": "100\n1 3 71 47 64 82 58 61 61 35 52 36 57 62 63 54 52 21 78 100 24 94 4 80 99 62 43 72 21 70 90 4 23 14 72 4 76 49 71 96 96 99 78 7 32 11 14 61 19 69 1 68 100 77 86 54 14 86 47 53 30 88 67 66 61 70 17 63 40 5 99 53 38 31 91 18 41 5 77 61 53 30 87 21 23 54 52 17 23 75 58 99 99 63 20 1 78 72 28 11",
"output": "90"
},
{
"input": "100\n1 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100",
"output": "99"
},
{
"input": "94\n3 100 100 99 99 99 100 99 99 99 99 99 100 99 100 100 99 100 99 99 100 99 100 99 100 100 100 99 100 99 100 99 100 99 99 99 100 99 99 99 99 99 100 99 100 100 99 100 99 99 99 99 100 99 100 99 99 99 100 100 99 100 100 99 99 100 100 100 99 100 99 99 99 99 99 100 100 100 100 100 100 100 100 100 99 99 99 99 100 99 100 99 100 100",
"output": "97"
},
{
"input": "97\n99 99 98 98 100 98 99 99 98 100 100 100 99 99 100 99 99 98 99 99 98 98 98 100 100 99 98 99 100 98 99 98 98 100 98 99 100 98 98 99 98 98 99 98 100 99 99 99 99 98 98 98 100 99 100 100 99 99 100 99 99 98 98 98 100 100 98 100 100 99 98 99 100 98 98 98 98 99 99 98 98 99 100 100 98 98 99 98 99 100 98 99 100 98 99 99 100",
"output": "2"
},
{
"input": "100\n100 55 70 81 73 51 6 75 45 85 33 61 98 63 11 59 1 8 14 28 78 74 44 80 7 69 7 5 90 73 43 78 64 64 43 92 59 70 80 19 33 39 31 70 38 85 24 23 86 79 98 56 92 63 92 4 36 8 79 74 2 81 54 13 69 44 49 63 17 76 78 99 42 36 47 71 19 90 9 58 83 53 27 2 35 51 65 59 90 51 74 87 84 48 98 44 84 100 84 93",
"output": "1"
},
{
"input": "100\n100 637 498 246 615 901 724 673 793 33 282 908 477 185 185 969 34 859 90 70 107 492 227 918 919 131 620 182 802 703 779 184 403 891 448 499 628 553 905 392 70 396 8 575 66 908 992 496 792 174 667 355 836 610 855 377 244 827 836 808 667 354 800 114 746 556 75 894 162 367 99 718 394 273 833 776 151 433 315 470 759 12 552 613 85 793 775 649 225 86 296 624 557 201 209 595 697 527 282 168",
"output": "749"
},
{
"input": "100\n107 172 549 883 564 56 399 970 173 990 224 217 601 381 948 631 159 958 512 136 61 584 633 202 652 355 26 723 663 237 410 721 688 552 699 24 748 186 461 88 34 243 872 205 471 298 654 693 244 33 359 533 471 116 386 653 654 887 531 303 335 829 319 340 827 89 602 191 422 289 361 200 593 421 592 402 256 813 606 589 741 9 148 893 3 142 50 169 219 360 642 45 810 818 507 624 561 743 303 111",
"output": "729"
},
{
"input": "90\n670 694 651 729 579 539 568 551 707 638 604 544 502 531 775 805 558 655 506 729 802 778 653 737 591 770 594 535 588 604 658 713 779 705 504 563 513 651 529 572 505 553 515 750 621 574 727 774 714 725 665 798 670 747 751 635 755 798 635 717 583 682 517 546 740 802 743 507 658 700 645 671 533 594 506 633 768 584 672 666 703 522 530 501 592 528 678 708 619 786",
"output": "111"
},
{
"input": "90\n10 265 429 431 343 305 806 746 284 313 503 221 594 351 83 653 232 431 427 610 458 88 255 215 529 205 492 549 55 694 535 104 45 327 816 432 595 549 454 141 216 557 250 415 531 494 190 749 718 380 78 447 784 347 196 814 16 780 262 462 776 315 160 307 593 694 692 41 528 725 376 777 337 44 438 630 345 502 384 184 742 429 570 361 394 267 820 778 662 377",
"output": "714"
},
{
"input": "95\n800 280 176 472 587 763 588 838 760 378 667 231 566 278 713 305 354 815 140 220 188 409 109 180 251 268 474 590 853 143 235 691 313 785 386 92 783 471 43 342 718 592 678 404 256 362 239 504 163 85 521 81 356 73 754 589 380 159 196 862 838 509 149 42 366 630 467 292 698 123 187 796 576 37 689 800 186 518 488 432 159 860 349 799 282 304 880 283 23 312 55 507 734 370 490",
"output": "52"
},
{
"input": "100\n95 88 84 85 74 97 100 52 91 94 62 66 90 56 86 66 95 73 79 68 54 67 99 52 82 62 81 71 93 85 72 72 95 52 72 63 57 90 92 89 88 77 84 78 95 59 72 86 98 64 89 64 80 70 54 93 88 86 79 78 94 64 89 66 50 90 54 82 52 96 99 54 81 66 83 79 69 80 51 73 81 69 93 82 76 52 58 87 93 92 52 67 78 63 63 87 77 95 58 78",
"output": "4"
},
{
"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": "1"
},
{
"input": "100\n999 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": "2"
},
{
"input": "100\n901 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": "99"
},
{
"input": "4\n8 5 11 3",
"output": "2"
},
{
"input": "3\n3 10 1",
"output": "4"
}
] | 1,665,167,303
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 35
| 62
| 1,740,800
|
n=int(input())
arr=list(map(int,input().split()))
t=ans=arr.pop(0)
mx=max(arr)
i=1
while (ans)<=mx:
ans+=1
arr.remove(mx)
arr.append(mx-1)
mx=max(arr)
print(ans-t)
|
Title: Bear and Elections
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Limak is a grizzly bear who desires power and adoration. He wants to win in upcoming elections and rule over the Bearland.
There are *n* candidates, including Limak. We know how many citizens are going to vote for each candidate. Now *i*-th candidate would get *a**i* votes. Limak is candidate number 1. To win in elections, he must get strictly more votes than any other candidate.
Victory is more important than everything else so Limak decided to cheat. He will steal votes from his opponents by bribing some citizens. To bribe a citizen, Limak must give him or her one candy - citizens are bears and bears like candies. Limak doesn't have many candies and wonders - how many citizens does he have to bribe?
Input Specification:
The first line contains single integer *n* (2<=β€<=*n*<=β€<=100) - number of candidates.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=β€<=*a**i*<=β€<=1000) - number of votes for each candidate. Limak is candidate number 1.
Note that after bribing number of votes for some candidate might be zero or might be greater than 1000.
Output Specification:
Print the minimum number of citizens Limak must bribe to have strictly more votes than any other candidate.
Demo Input:
['5\n5 1 11 2 8\n', '4\n1 8 8 8\n', '2\n7 6\n']
Demo Output:
['4\n', '6\n', '0\n']
Note:
In the first sample Limak has 5 votes. One of the ways to achieve victory is to bribe 4 citizens who want to vote for the third candidate. Then numbers of votes would be 9,β1,β7,β2,β8 (Limak would have 9 votes). Alternatively, Limak could steal only 3 votes from the third candidate and 1 vote from the second candidate to get situation 9,β0,β8,β2,β8.
In the second sample Limak will steal 2 votes from each candidate. Situation will be 7,β6,β6,β6.
In the third sample Limak is a winner without bribing any citizen.
|
```python
n=int(input())
arr=list(map(int,input().split()))
t=ans=arr.pop(0)
mx=max(arr)
i=1
while (ans)<=mx:
ans+=1
arr.remove(mx)
arr.append(mx-1)
mx=max(arr)
print(ans-t)
```
| 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<=β€<=*n*<=β€<=50) β 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 β 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"
}
] | 1,695,229,318
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 34
| 92
| 0
|
n = input()
n = int(n)
stones = input()
ans = 0
for i in range(1, n):
if stones[i] == stones[i-1]:
ans = ans + 1
print(ans)
|
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<=β€<=*n*<=β€<=50) β 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 β 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 = input()
n = int(n)
stones = input()
ans = 0
for i in range(1, n):
if stones[i] == stones[i-1]:
ans = ans + 1
print(ans)
```
| 3
|
|
792
|
B
|
Counting-out Rhyme
|
PROGRAMMING
| 1,300
|
[
"implementation"
] | null | null |
*n* children are standing in a circle and playing the counting-out game. Children are numbered clockwise from 1 to *n*. In the beginning, the first child is considered the leader. The game is played in *k* steps. In the *i*-th step the leader counts out *a**i* people in clockwise order, starting from the next person. The last one to be pointed at by the leader is eliminated, and the next player after him becomes the new leader.
For example, if there are children with numbers [8,<=10,<=13,<=14,<=16] currently in the circle, the leader is child 13 and *a**i*<==<=12, then counting-out rhyme ends on child 16, who is eliminated. Child 8 becomes the leader.
You have to write a program which prints the number of the child to be eliminated on every step.
|
The first line contains two integer numbers *n* and *k* (2<=β€<=*n*<=β€<=100, 1<=β€<=*k*<=β€<=*n*<=-<=1).
The next line contains *k* integer numbers *a*1,<=*a*2,<=...,<=*a**k* (1<=β€<=*a**i*<=β€<=109).
|
Print *k* numbers, the *i*-th one corresponds to the number of child to be eliminated at the *i*-th step.
|
[
"7 5\n10 4 11 4 1\n",
"3 2\n2 5\n"
] |
[
"4 2 5 6 1 \n",
"3 2 \n"
] |
Let's consider first example:
- In the first step child 4 is eliminated, child 5 becomes the leader. - In the second step child 2 is eliminated, child 3 becomes the leader. - In the third step child 5 is eliminated, child 6 becomes the leader. - In the fourth step child 6 is eliminated, child 7 becomes the leader. - In the final step child 1 is eliminated, child 3 becomes the leader.
| 0
|
[
{
"input": "7 5\n10 4 11 4 1",
"output": "4 2 5 6 1 "
},
{
"input": "3 2\n2 5",
"output": "3 2 "
},
{
"input": "2 1\n1",
"output": "2 "
},
{
"input": "2 1\n2",
"output": "1 "
},
{
"input": "2 1\n3",
"output": "2 "
},
{
"input": "10 7\n5 10 4 3 8 10 6",
"output": "6 8 3 9 2 4 10 "
},
{
"input": "10 8\n12 6 12 15 20 8 17 12",
"output": "3 10 6 8 2 9 4 5 "
},
{
"input": "12 10\n76 58 82 54 97 46 17 40 36 15",
"output": "5 9 12 1 3 10 8 11 2 4 "
},
{
"input": "12 6\n76 61 94 15 66 26",
"output": "5 12 6 2 7 3 "
},
{
"input": "90 10\n1045 8705 6077 3282 1459 9809 383 6206 2674 7274",
"output": "56 39 45 20 17 55 14 85 51 33 "
},
{
"input": "100 30\n601771 913885 829106 91674 465657 367068 142461 873149 294276 916519 720701 370006 551782 321506 68525 570684 81178 724855 564907 661130 10112 983124 799801 100639 766045 862312 513021 232094 979480 408554",
"output": "72 89 16 26 85 73 29 99 63 30 8 46 70 19 100 93 36 54 65 77 17 79 62 64 21 69 42 82 68 1 "
},
{
"input": "3 2\n20148340 81473314",
"output": "2 3 "
},
{
"input": "3 2\n301633543 643389490",
"output": "2 3 "
},
{
"input": "6 5\n532623340 628883728 583960589 690950241 488468353",
"output": "5 3 6 1 4 "
},
{
"input": "6 2\n458995521 294343587",
"output": "4 1 "
},
{
"input": "68 1\n5",
"output": "6 "
}
] | 1,637,303,391
| 291
|
PyPy 3-64
|
OK
|
TESTS
| 22
| 62
| 0
|
import sys
input = sys.stdin.readline
n, k = map(int, input().split())
l = sorted(list(range(1, n+1)))
c = 0
t = []
for i in [int(j) for j in input().split()]:
c = (c+i)%len(l)
t.append(l[c])
l = l[:c]+l[c+1:]
print(*t)
|
Title: Counting-out Rhyme
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
*n* children are standing in a circle and playing the counting-out game. Children are numbered clockwise from 1 to *n*. In the beginning, the first child is considered the leader. The game is played in *k* steps. In the *i*-th step the leader counts out *a**i* people in clockwise order, starting from the next person. The last one to be pointed at by the leader is eliminated, and the next player after him becomes the new leader.
For example, if there are children with numbers [8,<=10,<=13,<=14,<=16] currently in the circle, the leader is child 13 and *a**i*<==<=12, then counting-out rhyme ends on child 16, who is eliminated. Child 8 becomes the leader.
You have to write a program which prints the number of the child to be eliminated on every step.
Input Specification:
The first line contains two integer numbers *n* and *k* (2<=β€<=*n*<=β€<=100, 1<=β€<=*k*<=β€<=*n*<=-<=1).
The next line contains *k* integer numbers *a*1,<=*a*2,<=...,<=*a**k* (1<=β€<=*a**i*<=β€<=109).
Output Specification:
Print *k* numbers, the *i*-th one corresponds to the number of child to be eliminated at the *i*-th step.
Demo Input:
['7 5\n10 4 11 4 1\n', '3 2\n2 5\n']
Demo Output:
['4 2 5 6 1 \n', '3 2 \n']
Note:
Let's consider first example:
- In the first step child 4 is eliminated, child 5 becomes the leader. - In the second step child 2 is eliminated, child 3 becomes the leader. - In the third step child 5 is eliminated, child 6 becomes the leader. - In the fourth step child 6 is eliminated, child 7 becomes the leader. - In the final step child 1 is eliminated, child 3 becomes the leader.
|
```python
import sys
input = sys.stdin.readline
n, k = map(int, input().split())
l = sorted(list(range(1, n+1)))
c = 0
t = []
for i in [int(j) for j in input().split()]:
c = (c+i)%len(l)
t.append(l[c])
l = l[:c]+l[c+1:]
print(*t)
```
| 3
|
|
412
|
A
|
Poster
|
PROGRAMMING
| 900
|
[
"greedy",
"implementation"
] | null | null |
The R1 company has recently bought a high rise building in the centre of Moscow for its main office. It's time to decorate the new office, and the first thing to do is to write the company's slogan above the main entrance to the building.
The slogan of the company consists of *n* characters, so the decorators hung a large banner, *n* meters wide and 1 meter high, divided into *n* equal squares. The first character of the slogan must be in the first square (the leftmost) of the poster, the second character must be in the second square, and so on.
Of course, the R1 programmers want to write the slogan on the poster themselves. To do this, they have a large (and a very heavy) ladder which was put exactly opposite the *k*-th square of the poster. To draw the *i*-th character of the slogan on the poster, you need to climb the ladder, standing in front of the *i*-th square of the poster. This action (along with climbing up and down the ladder) takes one hour for a painter. The painter is not allowed to draw characters in the adjacent squares when the ladder is in front of the *i*-th square because the uncomfortable position of the ladder may make the characters untidy. Besides, the programmers can move the ladder. In one hour, they can move the ladder either a meter to the right or a meter to the left.
Drawing characters and moving the ladder is very tiring, so the programmers want to finish the job in as little time as possible. Develop for them an optimal poster painting plan!
|
The first line contains two integers, *n* and *k* (1<=β€<=*k*<=β€<=*n*<=β€<=100) β the number of characters in the slogan and the initial position of the ladder, correspondingly. The next line contains the slogan as *n* characters written without spaces. Each character of the slogan is either a large English letter, or digit, or one of the characters: '.', '!', ',', '?'.
|
In *t* lines, print the actions the programmers need to make. In the *i*-th line print:
- "LEFT" (without the quotes), if the *i*-th action was "move the ladder to the left"; - "RIGHT" (without the quotes), if the *i*-th action was "move the ladder to the right"; - "PRINT *x*" (without the quotes), if the *i*-th action was to "go up the ladder, paint character *x*, go down the ladder".
The painting time (variable *t*) must be minimum possible. If there are multiple optimal painting plans, you can print any of them.
|
[
"2 2\nR1\n",
"2 1\nR1\n",
"6 4\nGO?GO!\n"
] |
[
"PRINT 1\nLEFT\nPRINT R\n",
"PRINT R\nRIGHT\nPRINT 1\n",
"RIGHT\nRIGHT\nPRINT !\nLEFT\nPRINT O\nLEFT\nPRINT G\nLEFT\nPRINT ?\nLEFT\nPRINT O\nLEFT\nPRINT G\n"
] |
Note that the ladder cannot be shifted by less than one meter. The ladder can only stand in front of some square of the poster. For example, you cannot shift a ladder by half a meter and position it between two squares. Then go up and paint the first character and the second character.
| 500
|
[
{
"input": "2 2\nR1",
"output": "PRINT 1\nLEFT\nPRINT R"
},
{
"input": "2 1\nR1",
"output": "PRINT R\nRIGHT\nPRINT 1"
},
{
"input": "6 4\nGO?GO!",
"output": "RIGHT\nRIGHT\nPRINT !\nLEFT\nPRINT O\nLEFT\nPRINT G\nLEFT\nPRINT ?\nLEFT\nPRINT O\nLEFT\nPRINT G"
},
{
"input": "7 3\nME,YOU.",
"output": "LEFT\nLEFT\nPRINT M\nRIGHT\nPRINT E\nRIGHT\nPRINT ,\nRIGHT\nPRINT Y\nRIGHT\nPRINT O\nRIGHT\nPRINT U\nRIGHT\nPRINT ."
},
{
"input": "10 1\nEK5JQMS5QN",
"output": "PRINT E\nRIGHT\nPRINT K\nRIGHT\nPRINT 5\nRIGHT\nPRINT J\nRIGHT\nPRINT Q\nRIGHT\nPRINT M\nRIGHT\nPRINT S\nRIGHT\nPRINT 5\nRIGHT\nPRINT Q\nRIGHT\nPRINT N"
},
{
"input": "85 84\n73IW80UODC8B,UR7S8WMNATV0JSRF4W0B2VV8LCAX6SGCYY8?LHDKJEO29WXQWT9.WY1VY7408S1W04GNDZPK",
"output": "RIGHT\nPRINT K\nLEFT\nPRINT P\nLEFT\nPRINT Z\nLEFT\nPRINT D\nLEFT\nPRINT N\nLEFT\nPRINT G\nLEFT\nPRINT 4\nLEFT\nPRINT 0\nLEFT\nPRINT W\nLEFT\nPRINT 1\nLEFT\nPRINT S\nLEFT\nPRINT 8\nLEFT\nPRINT 0\nLEFT\nPRINT 4\nLEFT\nPRINT 7\nLEFT\nPRINT Y\nLEFT\nPRINT V\nLEFT\nPRINT 1\nLEFT\nPRINT Y\nLEFT\nPRINT W\nLEFT\nPRINT .\nLEFT\nPRINT 9\nLEFT\nPRINT T\nLEFT\nPRINT W\nLEFT\nPRINT Q\nLEFT\nPRINT X\nLEFT\nPRINT W\nLEFT\nPRINT 9\nLEFT\nPRINT 2\nLEFT\nPRINT O\nLEFT\nPRINT E\nLEFT\nPRINT J\nLEFT\nPRINT K\nLEFT\nPRINT D\n..."
},
{
"input": "59 53\n7NWD!9PC11C8S4TQABBTJO,?CO6YGOM!W0QR94CZJBD9U1YJY23YB354,8F",
"output": "RIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nPRINT F\nLEFT\nPRINT 8\nLEFT\nPRINT ,\nLEFT\nPRINT 4\nLEFT\nPRINT 5\nLEFT\nPRINT 3\nLEFT\nPRINT B\nLEFT\nPRINT Y\nLEFT\nPRINT 3\nLEFT\nPRINT 2\nLEFT\nPRINT Y\nLEFT\nPRINT J\nLEFT\nPRINT Y\nLEFT\nPRINT 1\nLEFT\nPRINT U\nLEFT\nPRINT 9\nLEFT\nPRINT D\nLEFT\nPRINT B\nLEFT\nPRINT J\nLEFT\nPRINT Z\nLEFT\nPRINT C\nLEFT\nPRINT 4\nLEFT\nPRINT 9\nLEFT\nPRINT R\nLEFT\nPRINT Q\nLEFT\nPRINT 0\nLEFT\nPRINT W\nLEFT\nPRINT !\nLEFT\nPRINT M\nLEFT\nPRINT O\nLEFT\nPRINT G\nLEFT\nPRIN..."
},
{
"input": "100 79\nF2.58O.L4A!QX!,.,YQUE.RZW.ENQCZKUFNG?.J6FT?L59BIHKFB?,44MAHSTD8?Z.UP3N!76YW6KVI?4AKWDPP0?3HPERM3PCUR",
"output": "RIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nPRINT R\nLEFT\nPRINT U\nLEFT\nPRINT C\nLEFT\nPRINT P\nLEFT\nPRINT 3\nLEFT\nPRINT M\nLEFT\nPRINT R\nLEFT\nPRINT E\nLEFT\nPRINT P\nLEFT\nPRINT H\nLEFT\nPRINT 3\nLEFT\nPRINT ?\nLEFT\nPRINT 0\nLEFT\nPRINT P\nLEFT\nPRINT P\nLEFT\nPRINT D\nLEFT\nPRINT W\nLEFT\nPRINT K\nLEFT\nPRINT A\nLEFT\nPRINT 4\nLEFT\nPRINT ?\nLEFT\nPRINT I\nLEFT\nPRINT V\nLEFT\nPRINT K\nLEFT\nPRIN..."
},
{
"input": "1 1\n!",
"output": "PRINT !"
},
{
"input": "34 20\n.C0QPPSWQKGBSH0,VGM!N,5SX.M9Q,D1DT",
"output": "RIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nPRINT T\nLEFT\nPRINT D\nLEFT\nPRINT 1\nLEFT\nPRINT D\nLEFT\nPRINT ,\nLEFT\nPRINT Q\nLEFT\nPRINT 9\nLEFT\nPRINT M\nLEFT\nPRINT .\nLEFT\nPRINT X\nLEFT\nPRINT S\nLEFT\nPRINT 5\nLEFT\nPRINT ,\nLEFT\nPRINT N\nLEFT\nPRINT !\nLEFT\nPRINT M\nLEFT\nPRINT G\nLEFT\nPRINT V\nLEFT\nPRINT ,\nLEFT\nPRINT 0\nLEFT\nPRINT H\nLEFT\nPRINT S\nLEFT\nPRINT B\nLEFT\nPRINT G\nLEFT\nPRINT K\nLEFT\nPRINT Q\nLEFT\nPRINT W\nLEFT\nPRINT S\n..."
},
{
"input": "99 98\nR8MZTEG240LNHY33H7.2CMWM73ZK,P5R,RGOA,KYKMIOG7CMPNHV3R2KM,N374IP8HN97XVMG.PSIPS8H3AXFGK0CJ76,EVKRZ9",
"output": "RIGHT\nPRINT 9\nLEFT\nPRINT Z\nLEFT\nPRINT R\nLEFT\nPRINT K\nLEFT\nPRINT V\nLEFT\nPRINT E\nLEFT\nPRINT ,\nLEFT\nPRINT 6\nLEFT\nPRINT 7\nLEFT\nPRINT J\nLEFT\nPRINT C\nLEFT\nPRINT 0\nLEFT\nPRINT K\nLEFT\nPRINT G\nLEFT\nPRINT F\nLEFT\nPRINT X\nLEFT\nPRINT A\nLEFT\nPRINT 3\nLEFT\nPRINT H\nLEFT\nPRINT 8\nLEFT\nPRINT S\nLEFT\nPRINT P\nLEFT\nPRINT I\nLEFT\nPRINT S\nLEFT\nPRINT P\nLEFT\nPRINT .\nLEFT\nPRINT G\nLEFT\nPRINT M\nLEFT\nPRINT V\nLEFT\nPRINT X\nLEFT\nPRINT 7\nLEFT\nPRINT 9\nLEFT\nPRINT N\nLEFT\nPRINT H\n..."
},
{
"input": "98 72\n.1?7CJ!EFZHO5WUKDZV,0EE92PTAGY078WKN!!41E,Q7381U60!9C,VONEZ6!SFFNDBI86MACX0?D?9!U2UV7S,977PNDSF0HY",
"output": "RIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nPRINT Y\nLEFT\nPRINT H\nLEFT\nPRINT 0\nLEFT\nPRINT F\nLEFT\nPRINT S\nLEFT\nPRINT D\nLEFT\nPRINT N\nLEFT\nPRINT P\nLEFT\nPRINT 7\nLEFT\nPRINT 7\nLEFT\nPRINT 9\nLEFT\nPRINT ,\nLEFT\nPRINT S\nLEFT\nPRINT 7\nLEFT\nPRINT V\nLEFT\nPRINT U\nLEFT\nPRINT 2\nLEFT\nPRINT U\nLEFT\nPRINT !\nLEFT\nPRINT 9\nLEFT\nPRINT ?\nLEFT\nPRINT D\nLEFT\n..."
},
{
"input": "97 41\nGQSPZGGRZ0KWUMI79GOXP7!RR9E?Z5YO?6WUL!I7GCXRS8T,PEFQM7CZOUG8HLC7198J1?C69JD00Q!QY1AK!27I?WB?UAUIG",
"output": "LEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nPRINT G\nRIGHT\nPRINT Q\nRIGHT\nPRINT S\nRIGHT\nPRINT P\nRIGHT\nPRINT Z\nRIGHT\nPRINT G\nRIGHT\nPRINT G\nRIGHT\nPRINT R\nRIGHT\nPRINT Z\nRIGHT\nPRINT 0\nRIGHT\nPRINT K\nRIGHT\nPRINT W\nRIGHT\nPRINT U\nRIGHT\nPRINT M\nRIGHT\nPRINT I\nRIGHT\nPRINT 7\nRIGHT\nPRINT 9\nRIGHT\n..."
},
{
"input": "96 28\nZCF!PLS27YGXHK8P46H,C.A7MW90ED,4BA!T0!XKIR2GE0HD..YZ0O20O8TA7E35G5YT3L4W5ESSYBHG8.TIQENS4I.R8WE,",
"output": "LEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nPRINT Z\nRIGHT\nPRINT C\nRIGHT\nPRINT F\nRIGHT\nPRINT !\nRIGHT\nPRINT P\nRIGHT\nPRINT L\nRIGHT\nPRINT S\nRIGHT\nPRINT 2\nRIGHT\nPRINT 7\nRIGHT\nPRINT Y\nRIGHT\nPRINT G\nRIGHT\nPRINT X\nRIGHT\nPRINT H\nRIGHT\nPRINT K\nRIGHT\nPRINT 8\nRIGHT\nPRINT P\nRIGHT\nPRINT 4\nRIGHT\nPRINT 6\nRIGHT\nPRINT H\nRIGHT\nPRINT ,\nRIGHT\nPRINT C\nRIGHT\nPRINT .\nRIGH..."
},
{
"input": "15 3\n!..!?!,!,..,?!.",
"output": "LEFT\nLEFT\nPRINT !\nRIGHT\nPRINT .\nRIGHT\nPRINT .\nRIGHT\nPRINT !\nRIGHT\nPRINT ?\nRIGHT\nPRINT !\nRIGHT\nPRINT ,\nRIGHT\nPRINT !\nRIGHT\nPRINT ,\nRIGHT\nPRINT .\nRIGHT\nPRINT .\nRIGHT\nPRINT ,\nRIGHT\nPRINT ?\nRIGHT\nPRINT !\nRIGHT\nPRINT ."
},
{
"input": "93 81\nGMIBVKYLURQLWHBGTFNJZZAZNUJJTPQKCPGDMGCDTTGXOANWKTDZSIYBUPFUXGQHCMVIEQCTINRTIUSPGMVZPGWBHPIXC",
"output": "RIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nPRINT C\nLEFT\nPRINT X\nLEFT\nPRINT I\nLEFT\nPRINT P\nLEFT\nPRINT H\nLEFT\nPRINT B\nLEFT\nPRINT W\nLEFT\nPRINT G\nLEFT\nPRINT P\nLEFT\nPRINT Z\nLEFT\nPRINT V\nLEFT\nPRINT M\nLEFT\nPRINT G\nLEFT\nPRINT P\nLEFT\nPRINT S\nLEFT\nPRINT U\nLEFT\nPRINT I\nLEFT\nPRINT T\nLEFT\nPRINT R\nLEFT\nPRINT N\nLEFT\nPRINT I\nLEFT\nPRINT T\nLEFT\nPRINT C\nLEFT\nPRINT Q\nLEFT\nPRINT E\nLEFT\nPRINT I\nLEFT\nPRINT V\nLEFT\nPRINT M\nLEFT\nPRINT C..."
},
{
"input": "88 30\n5847857685475132927321580125243001071762130696139249809763381765504146602574972381323476",
"output": "LEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nPRINT 5\nRIGHT\nPRINT 8\nRIGHT\nPRINT 4\nRIGHT\nPRINT 7\nRIGHT\nPRINT 8\nRIGHT\nPRINT 5\nRIGHT\nPRINT 7\nRIGHT\nPRINT 6\nRIGHT\nPRINT 8\nRIGHT\nPRINT 5\nRIGHT\nPRINT 4\nRIGHT\nPRINT 7\nRIGHT\nPRINT 5\nRIGHT\nPRINT 1\nRIGHT\nPRINT 3\nRIGHT\nPRINT 2\nRIGHT\nPRINT 9\nRIGHT\nPRINT 2\nRIGHT\nPRINT 7\nRIGHT\nPRINT 3\nRIGHT\nPRINT 2\nRIGHT\nP..."
},
{
"input": "100 50\n5B2N,CXCWOIWH71XV!HCFEUCN3U88JDRIFRO2VHY?!N.RGH.?W14X5S.Y00RIY6YA19BPD0T,WECXYI,O2RF1U4NX9,F5AVLPOYK",
"output": "LEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nPRINT 5\nRIGHT\nPRINT B\nRIGHT\nPRINT 2\nRIGHT\nPRINT N\nRIGHT\nPRINT ,\nRIGHT\nPRINT C\nRIGHT\nPRINT X\nRIGHT\nPRINT C\nRIGHT\nPRINT W\nRIGHT\nPRINT O\nRIGHT\nPRINT I\nRIGHT\nPRINT W\nRIGHT\nPRINT H\nRIGHT\nPRINT 7\n..."
},
{
"input": "100 51\n!X85PT!WJDNS9KA6D2SJBR,U,G7M914W07EK3EAJ4XG..UHA3KOOFYJ?M0MEFDC6KNCNGKS0A!S,C02H4TSZA1U7NDBTIY?,7XZ4",
"output": "RIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nPRINT 4\nLEFT\nPRINT Z\nLEFT\nPRINT X\nLEFT\nPRINT 7\nLEFT\nPRINT ,\nLEFT\nPRINT ?\nLEFT\nPRINT Y\nLEFT\nPRINT I\nLEFT\nPRINT T\nLEFT\nPRINT B\nLEFT\nPRINT D\nLEFT\nPRI..."
},
{
"input": "100 52\n!MLPE.0K72RW9XKHR60QE?69ILFSIKYSK5AG!TA5.02VG5OMY0967G2RI.62CNK9L8G!7IG9F0XNNCGSDOTFD?I,EBP31HRERZSX",
"output": "RIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nPRINT X\nLEFT\nPRINT S\nLEFT\nPRINT Z\nLEFT\nPRINT R\nLEFT\nPRINT E\nLEFT\nPRINT R\nLEFT\nPRINT H\nLEFT\nPRINT 1\nLEFT\nPRINT 3\nLEFT\nPRINT P\nLEFT\nPRINT B\nLEFT\nPRINT E\nL..."
},
{
"input": "100 49\n86C0NR7V,BE09,7,ER715OQ3GZ,P014H4BSQ5YS?OFNDD7YWI?S?UMKIWHSBDZ4398?SSDZLTDU1L?G4QVAB53HNDS!4PYW5C!VI",
"output": "LEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nPRINT 8\nRIGHT\nPRINT 6\nRIGHT\nPRINT C\nRIGHT\nPRINT 0\nRIGHT\nPRINT N\nRIGHT\nPRINT R\nRIGHT\nPRINT 7\nRIGHT\nPRINT V\nRIGHT\nPRINT ,\nRIGHT\nPRINT B\nRIGHT\nPRINT E\nRIGHT\nPRINT 0\nRIGHT\nPRINT 9\nRIGHT\nPRINT ,\nRIGHT\n..."
},
{
"input": "100 48\nFO,IYI4AAV?4?N5PWMZX1AINZLKAUJCKMDWU4CROT?.LYWYLYU5S80,15A6VGP!V0N,O.70CP?GEA52WG59UYWU1MMMU4BERVY.!",
"output": "LEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nPRINT F\nRIGHT\nPRINT O\nRIGHT\nPRINT ,\nRIGHT\nPRINT I\nRIGHT\nPRINT Y\nRIGHT\nPRINT I\nRIGHT\nPRINT 4\nRIGHT\nPRINT A\nRIGHT\nPRINT A\nRIGHT\nPRINT V\nRIGHT\nPRINT ?\nRIGHT\nPRINT 4\nRIGHT\nPRINT ?\nRIGHT\nPRINT N\nRIGHT\nPRINT..."
},
{
"input": "100 100\nE?F,W.,,O51!!G13ZWP?YHWRT69?RQPW7,V,EM3336F1YAIKJIME1M45?LJM42?45V7221?P.DIO9FK245LXKMR4ALKPDLA5YI2Y",
"output": "PRINT Y\nLEFT\nPRINT 2\nLEFT\nPRINT I\nLEFT\nPRINT Y\nLEFT\nPRINT 5\nLEFT\nPRINT A\nLEFT\nPRINT L\nLEFT\nPRINT D\nLEFT\nPRINT P\nLEFT\nPRINT K\nLEFT\nPRINT L\nLEFT\nPRINT A\nLEFT\nPRINT 4\nLEFT\nPRINT R\nLEFT\nPRINT M\nLEFT\nPRINT K\nLEFT\nPRINT X\nLEFT\nPRINT L\nLEFT\nPRINT 5\nLEFT\nPRINT 4\nLEFT\nPRINT 2\nLEFT\nPRINT K\nLEFT\nPRINT F\nLEFT\nPRINT 9\nLEFT\nPRINT O\nLEFT\nPRINT I\nLEFT\nPRINT D\nLEFT\nPRINT .\nLEFT\nPRINT P\nLEFT\nPRINT ?\nLEFT\nPRINT 1\nLEFT\nPRINT 2\nLEFT\nPRINT 2\nLEFT\nPRINT 7\nLEFT\nP..."
},
{
"input": "100 1\nJJ0ZOX4CY,SQ9L0K!2C9TM3C6K.6R21717I37VDSXGHBMR2!J820AI75D.O7NYMT6F.AGJ8R0RDETWOACK3P6UZAUYRKMKJ!G3WF",
"output": "PRINT J\nRIGHT\nPRINT J\nRIGHT\nPRINT 0\nRIGHT\nPRINT Z\nRIGHT\nPRINT O\nRIGHT\nPRINT X\nRIGHT\nPRINT 4\nRIGHT\nPRINT C\nRIGHT\nPRINT Y\nRIGHT\nPRINT ,\nRIGHT\nPRINT S\nRIGHT\nPRINT Q\nRIGHT\nPRINT 9\nRIGHT\nPRINT L\nRIGHT\nPRINT 0\nRIGHT\nPRINT K\nRIGHT\nPRINT !\nRIGHT\nPRINT 2\nRIGHT\nPRINT C\nRIGHT\nPRINT 9\nRIGHT\nPRINT T\nRIGHT\nPRINT M\nRIGHT\nPRINT 3\nRIGHT\nPRINT C\nRIGHT\nPRINT 6\nRIGHT\nPRINT K\nRIGHT\nPRINT .\nRIGHT\nPRINT 6\nRIGHT\nPRINT R\nRIGHT\nPRINT 2\nRIGHT\nPRINT 1\nRIGHT\nPRINT 7\nRIGHT\n..."
},
{
"input": "99 50\nLQJ!7GDFJ,SKQ8J2R?I4VA0K2.NDY.AZ?7K275NA81.YK!DO,PCQCJYL6BUU30XQ300FP0,LB!5TYTRSGOB4ELZ8IBKGVDNW8?B",
"output": "RIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nPRINT B\nLEFT\nPRINT ?\nLEFT\nPRINT 8\nLEFT\nPRINT W\nLEFT\nPRINT N\nLEFT\nPRINT D\nLEFT\nPRINT V\nLEFT\nPRINT G\nLEFT\nPRINT K\nLEFT\nPRINT B\nLEFT\nPRINT I\nLEFT\nPRI..."
},
{
"input": "99 51\nD9QHZXG46IWHHLTD2E,AZO0.M40R4B1WU6F,0QNZ37NQ0ACSU6!7Z?H02AD?0?9,5N5RG6PVOWIE6YA9QBCOHVNU??YT6,29SAC",
"output": "RIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nRIGHT\nPRINT C\nLEFT\nPRINT A\nLEFT\nPRINT S\nLEFT\nPRINT 9\nLEFT\nPRINT 2\nLEFT\nPRINT ,\nLEFT\nPRINT 6\nLEFT\nPRINT T\nLEFT\nPRINT Y\nLEFT\nPRINT ?\nLEFT\nPRINT ?\nLEFT\nPRINT U\nL..."
},
{
"input": "99 49\nOLUBX0Q3VPNSH,QCAWFVSKZA3NUURJ9PXBS3?72PMJ,27QTA7Z1N?6Q2CSJE,W0YX8XWS.W6B?K?M!PYAD30BX?8.VJCC,P8QL9",
"output": "LEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nPRINT O\nRIGHT\nPRINT L\nRIGHT\nPRINT U\nRIGHT\nPRINT B\nRIGHT\nPRINT X\nRIGHT\nPRINT 0\nRIGHT\nPRINT Q\nRIGHT\nPRINT 3\nRIGHT\nPRINT V\nRIGHT\nPRINT P\nRIGHT\nPRINT N\nRIGHT\nPRINT S\nRIGHT\nPRINT H\nRIGHT\nPRINT ,\nRIGHT\n..."
},
{
"input": "99 48\nW0GU5MNE5!JVIOO2SR5OO7RWLHDFH.HLCCX89O21SLD9!CU0MFG3RFZUFT!R0LWNVNSS.W54.67N4VAN1Q2J9NMO9Q6.UE8U6B8",
"output": "LEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nLEFT\nPRINT W\nRIGHT\nPRINT 0\nRIGHT\nPRINT G\nRIGHT\nPRINT U\nRIGHT\nPRINT 5\nRIGHT\nPRINT M\nRIGHT\nPRINT N\nRIGHT\nPRINT E\nRIGHT\nPRINT 5\nRIGHT\nPRINT !\nRIGHT\nPRINT J\nRIGHT\nPRINT V\nRIGHT\nPRINT I\nRIGHT\nPRINT O\nRIGHT\nPRINT..."
},
{
"input": "2 1\nOA",
"output": "PRINT O\nRIGHT\nPRINT A"
},
{
"input": "2 2\nGW",
"output": "PRINT W\nLEFT\nPRINT G"
},
{
"input": "3 1\n.VP",
"output": "PRINT .\nRIGHT\nPRINT V\nRIGHT\nPRINT P"
},
{
"input": "3 2\nUD0",
"output": "RIGHT\nPRINT 0\nLEFT\nPRINT D\nLEFT\nPRINT U"
},
{
"input": "3 3\nMYE",
"output": "PRINT E\nLEFT\nPRINT Y\nLEFT\nPRINT M"
},
{
"input": "4 1\nC5EJ",
"output": "PRINT C\nRIGHT\nPRINT 5\nRIGHT\nPRINT E\nRIGHT\nPRINT J"
},
{
"input": "4 2\n5QSW",
"output": "LEFT\nPRINT 5\nRIGHT\nPRINT Q\nRIGHT\nPRINT S\nRIGHT\nPRINT W"
},
{
"input": "4 3\n!F3D",
"output": "RIGHT\nPRINT D\nLEFT\nPRINT 3\nLEFT\nPRINT F\nLEFT\nPRINT !"
},
{
"input": "4 4\nS!?Y",
"output": "PRINT Y\nLEFT\nPRINT ?\nLEFT\nPRINT !\nLEFT\nPRINT S"
},
{
"input": "5 1\nIP4!N",
"output": "PRINT I\nRIGHT\nPRINT P\nRIGHT\nPRINT 4\nRIGHT\nPRINT !\nRIGHT\nPRINT N"
},
{
"input": "5 2\nI7??5",
"output": "LEFT\nPRINT I\nRIGHT\nPRINT 7\nRIGHT\nPRINT ?\nRIGHT\nPRINT ?\nRIGHT\nPRINT 5"
},
{
"input": "5 3\nASNUW",
"output": "RIGHT\nRIGHT\nPRINT W\nLEFT\nPRINT U\nLEFT\nPRINT N\nLEFT\nPRINT S\nLEFT\nPRINT A"
},
{
"input": "5 4\n3?,4L",
"output": "RIGHT\nPRINT L\nLEFT\nPRINT 4\nLEFT\nPRINT ,\nLEFT\nPRINT ?\nLEFT\nPRINT 3"
},
{
"input": "5 5\n30COA",
"output": "PRINT A\nLEFT\nPRINT O\nLEFT\nPRINT C\nLEFT\nPRINT 0\nLEFT\nPRINT 3"
},
{
"input": "4 2\nABCD",
"output": "LEFT\nPRINT A\nRIGHT\nPRINT B\nRIGHT\nPRINT C\nRIGHT\nPRINT D"
}
] | 1,397,838,629
| 1,229
|
Python 3
|
OK
|
TESTS
| 43
| 62
| 0
|
n,k=map(int,input('').split())
lozung=list(input())
if n-k<k:
for i in range(k,n):
print('RIGHT')
for i in reversed(range(0,n)):
print ('PRINT ',lozung[i])
if i!=0:print('LEFT')
else:
# print('2')
for i in reversed(range(0,k-1)):
print('LEFT')
for i in range(n):
print ('PRINT ',lozung[i])
if i!=n-1:print('RIGHT')
|
Title: Poster
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The R1 company has recently bought a high rise building in the centre of Moscow for its main office. It's time to decorate the new office, and the first thing to do is to write the company's slogan above the main entrance to the building.
The slogan of the company consists of *n* characters, so the decorators hung a large banner, *n* meters wide and 1 meter high, divided into *n* equal squares. The first character of the slogan must be in the first square (the leftmost) of the poster, the second character must be in the second square, and so on.
Of course, the R1 programmers want to write the slogan on the poster themselves. To do this, they have a large (and a very heavy) ladder which was put exactly opposite the *k*-th square of the poster. To draw the *i*-th character of the slogan on the poster, you need to climb the ladder, standing in front of the *i*-th square of the poster. This action (along with climbing up and down the ladder) takes one hour for a painter. The painter is not allowed to draw characters in the adjacent squares when the ladder is in front of the *i*-th square because the uncomfortable position of the ladder may make the characters untidy. Besides, the programmers can move the ladder. In one hour, they can move the ladder either a meter to the right or a meter to the left.
Drawing characters and moving the ladder is very tiring, so the programmers want to finish the job in as little time as possible. Develop for them an optimal poster painting plan!
Input Specification:
The first line contains two integers, *n* and *k* (1<=β€<=*k*<=β€<=*n*<=β€<=100) β the number of characters in the slogan and the initial position of the ladder, correspondingly. The next line contains the slogan as *n* characters written without spaces. Each character of the slogan is either a large English letter, or digit, or one of the characters: '.', '!', ',', '?'.
Output Specification:
In *t* lines, print the actions the programmers need to make. In the *i*-th line print:
- "LEFT" (without the quotes), if the *i*-th action was "move the ladder to the left"; - "RIGHT" (without the quotes), if the *i*-th action was "move the ladder to the right"; - "PRINT *x*" (without the quotes), if the *i*-th action was to "go up the ladder, paint character *x*, go down the ladder".
The painting time (variable *t*) must be minimum possible. If there are multiple optimal painting plans, you can print any of them.
Demo Input:
['2 2\nR1\n', '2 1\nR1\n', '6 4\nGO?GO!\n']
Demo Output:
['PRINT 1\nLEFT\nPRINT R\n', 'PRINT R\nRIGHT\nPRINT 1\n', 'RIGHT\nRIGHT\nPRINT !\nLEFT\nPRINT O\nLEFT\nPRINT G\nLEFT\nPRINT ?\nLEFT\nPRINT O\nLEFT\nPRINT G\n']
Note:
Note that the ladder cannot be shifted by less than one meter. The ladder can only stand in front of some square of the poster. For example, you cannot shift a ladder by half a meter and position it between two squares. Then go up and paint the first character and the second character.
|
```python
n,k=map(int,input('').split())
lozung=list(input())
if n-k<k:
for i in range(k,n):
print('RIGHT')
for i in reversed(range(0,n)):
print ('PRINT ',lozung[i])
if i!=0:print('LEFT')
else:
# print('2')
for i in reversed(range(0,k-1)):
print('LEFT')
for i in range(n):
print ('PRINT ',lozung[i])
if i!=n-1:print('RIGHT')
```
| 3
|
|
520
|
A
|
Pangram
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] | null | null |
A word or a sentence in some language is called a pangram if all the characters of the alphabet of this language appear in it at least once. Pangrams are often used to demonstrate fonts in printing or test the output devices.
You are given a string consisting of lowercase and uppercase Latin letters. Check whether this string is a pangram. We say that the string contains a letter of the Latin alphabet if this letter occurs in the string in uppercase or lowercase.
|
The first line contains a single integer *n* (1<=β€<=*n*<=β€<=100)Β β the number of characters in the string.
The second line contains the string. The string consists only of uppercase and lowercase Latin letters.
|
Output "YES", if the string is a pangram and "NO" otherwise.
|
[
"12\ntoosmallword\n",
"35\nTheQuickBrownFoxJumpsOverTheLazyDog\n"
] |
[
"NO\n",
"YES\n"
] |
none
| 500
|
[
{
"input": "12\ntoosmallword",
"output": "NO"
},
{
"input": "35\nTheQuickBrownFoxJumpsOverTheLazyDog",
"output": "YES"
},
{
"input": "1\na",
"output": "NO"
},
{
"input": "26\nqwertyuiopasdfghjklzxcvbnm",
"output": "YES"
},
{
"input": "26\nABCDEFGHIJKLMNOPQRSTUVWXYZ",
"output": "YES"
},
{
"input": "48\nthereisasyetinsufficientdataforameaningfulanswer",
"output": "NO"
},
{
"input": "30\nToBeOrNotToBeThatIsTheQuestion",
"output": "NO"
},
{
"input": "30\njackdawslovemybigsphinxofquarz",
"output": "NO"
},
{
"input": "31\nTHEFIVEBOXINGWIZARDSJUMPQUICKLY",
"output": "YES"
},
{
"input": "26\naaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "NO"
},
{
"input": "26\nMGJYIZDKsbhpVeNFlquRTcWoAx",
"output": "YES"
},
{
"input": "26\nfWMOhAPsbIVtyUEZrGNQXDklCJ",
"output": "YES"
},
{
"input": "26\nngPMVFSThiRCwLEuyOAbKxQzDJ",
"output": "YES"
},
{
"input": "25\nnxYTzLFwzNolAumjgcAboyxAj",
"output": "NO"
},
{
"input": "26\npRWdodGdxUESvcScPGbUoooZsC",
"output": "NO"
},
{
"input": "66\nBovdMlDzTaqKllZILFVfxbLGsRnzmtVVTmqiIDTYrossLEPlmsPrkUYtWEsGHVOnFj",
"output": "NO"
},
{
"input": "100\nmKtsiDRJypUieHIkvJaMFkwaKxcCIbBszZQLIyPpCDCjhNpAnYFngLjRpnKWpKWtGnwoSteeZXuFHWQxxxOpFlNeYTwKocsXuCoa",
"output": "YES"
},
{
"input": "26\nEoqxUbsLjPytUHMiFnvcGWZdRK",
"output": "NO"
},
{
"input": "26\nvCUFRKElZOnjmXGylWQaHDiPst",
"output": "NO"
},
{
"input": "26\nWtrPuaHdXLKJMsnvQfgOiJZBEY",
"output": "NO"
},
{
"input": "26\npGiFluRteQwkaVoPszJyNBChxM",
"output": "NO"
},
{
"input": "26\ncTUpqjPmANrdbzSFhlWIoKxgVY",
"output": "NO"
},
{
"input": "26\nLndjgvAEuICHKxPwqYztosrmBN",
"output": "NO"
},
{
"input": "26\nMdaXJrCipnOZLykfqHWEStevbU",
"output": "NO"
},
{
"input": "26\nEjDWsVxfKTqGXRnUMOLYcIzPba",
"output": "NO"
},
{
"input": "26\nxKwzRMpunYaqsdfaBgJcVElTHo",
"output": "NO"
},
{
"input": "26\nnRYUQsTwCPLZkgshfEXvBdoiMa",
"output": "NO"
},
{
"input": "26\nHNCQPfJutyAlDGsvRxZWMEbIdO",
"output": "NO"
},
{
"input": "26\nDaHJIpvKznQcmUyWsTGObXRFDe",
"output": "NO"
},
{
"input": "26\nkqvAnFAiRhzlJbtyuWedXSPcOG",
"output": "NO"
},
{
"input": "26\nhlrvgdwsIOyjcmUZXtAKEqoBpF",
"output": "NO"
},
{
"input": "26\njLfXXiMhBTcAwQVReGnpKzdsYu",
"output": "NO"
},
{
"input": "26\nlNMcVuwItjxRBGAekjhyDsQOzf",
"output": "NO"
},
{
"input": "26\nRkSwbNoYldUGtAZvpFMcxhIJFE",
"output": "NO"
},
{
"input": "26\nDqspXZJTuONYieKgaHLMBwfVSC",
"output": "NO"
},
{
"input": "26\necOyUkqNljFHRVXtIpWabGMLDz",
"output": "NO"
},
{
"input": "26\nEKAvqZhBnPmVCDRlgWJfOusxYI",
"output": "NO"
},
{
"input": "26\naLbgqeYchKdMrsZxIPFvTOWNjA",
"output": "NO"
},
{
"input": "26\nxfpBLsndiqtacOCHGmeWUjRkYz",
"output": "NO"
},
{
"input": "26\nXsbRKtqleZPNIVCdfUhyagAomJ",
"output": "NO"
},
{
"input": "26\nAmVtbrwquEthZcjKPLiyDgSoNF",
"output": "NO"
},
{
"input": "26\nOhvXDcwqAUmSEPRZGnjFLiKtNB",
"output": "NO"
},
{
"input": "26\nEKWJqCFLRmstxVBdYuinpbhaOg",
"output": "NO"
},
{
"input": "26\nmnbvcxxlkjhgfdsapoiuytrewq",
"output": "NO"
},
{
"input": "26\naAbcdefghijklmnopqrstuvwxy",
"output": "NO"
},
{
"input": "30\nABCDEFGHTYRIOPLabcdefghtyriopl",
"output": "NO"
},
{
"input": "25\nabcdefghijklmnopqrstuvwxy",
"output": "NO"
},
{
"input": "26\nabcdefhijklmnopqrstVxyzABC",
"output": "NO"
},
{
"input": "25\nqwertyuiopasdfghjklxcvbnm",
"output": "NO"
},
{
"input": "34\nTheQuickBrownFoxJumpsOverTheLayDog",
"output": "NO"
},
{
"input": "26\nabcdefghigklmnopqrstuvwxyz",
"output": "NO"
},
{
"input": "26\nabcdefghijklmnopqrstuvwxyA",
"output": "NO"
},
{
"input": "50\nqazwsxedcrfvtgbyhnujmikolQWERTYUIOASDFGHJKLZXCVBNM",
"output": "NO"
},
{
"input": "35\nTheQuickBrownFoxJumpsOverTheLasyDog",
"output": "NO"
},
{
"input": "25\nbcdefghijklmnopqrstuvwxyz",
"output": "NO"
},
{
"input": "38\nAbCdEfGhIjKlMnOpQrStVwXyZzzzzzzaaaaaaa",
"output": "NO"
},
{
"input": "26\nabcdefghiklmnopqrstvxyzABC",
"output": "NO"
},
{
"input": "26\nabcdefghijklmnopqrstuvwxzZ",
"output": "NO"
},
{
"input": "50\nabcdefghijklmnopqrstuvwxyABCDEFGHIJKLMNOPQRSTUVWXY",
"output": "NO"
}
] | 1,697,015,876
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 59
| 46
| 0
|
n=int(input())
ch=input()
m=0
letter='abcdefghijklmnopqrstuvwxyz'
for i in letter:
if i not in ch.lower():
m=1
if m==1:
print("No")
else:
print("Yes")
|
Title: Pangram
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A word or a sentence in some language is called a pangram if all the characters of the alphabet of this language appear in it at least once. Pangrams are often used to demonstrate fonts in printing or test the output devices.
You are given a string consisting of lowercase and uppercase Latin letters. Check whether this string is a pangram. We say that the string contains a letter of the Latin alphabet if this letter occurs in the string in uppercase or lowercase.
Input Specification:
The first line contains a single integer *n* (1<=β€<=*n*<=β€<=100)Β β the number of characters in the string.
The second line contains the string. The string consists only of uppercase and lowercase Latin letters.
Output Specification:
Output "YES", if the string is a pangram and "NO" otherwise.
Demo Input:
['12\ntoosmallword\n', '35\nTheQuickBrownFoxJumpsOverTheLazyDog\n']
Demo Output:
['NO\n', 'YES\n']
Note:
none
|
```python
n=int(input())
ch=input()
m=0
letter='abcdefghijklmnopqrstuvwxyz'
for i in letter:
if i not in ch.lower():
m=1
if m==1:
print("No")
else:
print("Yes")
```
| 3
|
|
146
|
A
|
Lucky Ticket
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya loves tickets very much. As we know, each ticket has a number that is a positive integer. Its length equals *n* (*n* is always even). Petya calls a ticket lucky if the ticket's number is a lucky number and the sum of digits in the first half (the sum of the first *n*<=/<=2 digits) equals the sum of digits in the second half (the sum of the last *n*<=/<=2 digits). Check if the given ticket is lucky.
|
The first line contains an even integer *n* (2<=β€<=*n*<=β€<=50) β the length of the ticket number that needs to be checked. The second line contains an integer whose length equals exactly *n* β the ticket number. The number may contain leading zeros.
|
On the first line print "YES" if the given ticket number is lucky. Otherwise, print "NO" (without the quotes).
|
[
"2\n47\n",
"4\n4738\n",
"4\n4774\n"
] |
[
"NO\n",
"NO\n",
"YES\n"
] |
In the first sample the sum of digits in the first half does not equal the sum of digits in the second half (4ββ β7).
In the second sample the ticket number is not the lucky number.
| 500
|
[
{
"input": "2\n47",
"output": "NO"
},
{
"input": "4\n4738",
"output": "NO"
},
{
"input": "4\n4774",
"output": "YES"
},
{
"input": "4\n4570",
"output": "NO"
},
{
"input": "6\n477477",
"output": "YES"
},
{
"input": "6\n777777",
"output": "YES"
},
{
"input": "20\n44444444444444444444",
"output": "YES"
},
{
"input": "2\n44",
"output": "YES"
},
{
"input": "10\n4745474547",
"output": "NO"
},
{
"input": "14\n77770004444444",
"output": "NO"
},
{
"input": "10\n4747777744",
"output": "YES"
},
{
"input": "10\n1234567890",
"output": "NO"
},
{
"input": "50\n44444444444444444444444444444444444444444444444444",
"output": "YES"
},
{
"input": "50\n44444444444444444444444444444444444444444444444447",
"output": "NO"
},
{
"input": "50\n74444444444444444444444444444444444444444444444444",
"output": "NO"
},
{
"input": "50\n07777777777777777777777777777777777777777777777770",
"output": "NO"
},
{
"input": "50\n77777777777777777777777777777777777777777777777777",
"output": "YES"
},
{
"input": "50\n44747747774474747747747447777447774747447477444474",
"output": "YES"
},
{
"input": "48\n447474444777444474747747744774447444747474774474",
"output": "YES"
},
{
"input": "32\n74474474777444474444747774474774",
"output": "YES"
},
{
"input": "40\n4747777444447747777447447747447474774777",
"output": "YES"
},
{
"input": "10\n4477477444",
"output": "YES"
},
{
"input": "18\n447747474447744747",
"output": "YES"
},
{
"input": "26\n44747744444774744774474447",
"output": "YES"
},
{
"input": "50\n44707747774474747747747447777447774747447477444474",
"output": "NO"
},
{
"input": "40\n4747777444447737777447447747447474774777",
"output": "NO"
},
{
"input": "36\n764477744747444444447747747474744444",
"output": "NO"
},
{
"input": "22\n4477407474777477744447",
"output": "NO"
},
{
"input": "32\n74274474777444474444747774474774",
"output": "NO"
},
{
"input": "16\n4744447974444747",
"output": "NO"
},
{
"input": "2\n11",
"output": "NO"
},
{
"input": "2\n22",
"output": "NO"
},
{
"input": "2\n33",
"output": "NO"
},
{
"input": "2\n74",
"output": "NO"
},
{
"input": "2\n55",
"output": "NO"
},
{
"input": "2\n66",
"output": "NO"
},
{
"input": "2\n77",
"output": "YES"
},
{
"input": "2\n88",
"output": "NO"
},
{
"input": "2\n99",
"output": "NO"
},
{
"input": "4\n4004",
"output": "NO"
},
{
"input": "14\n00077774444444",
"output": "NO"
},
{
"input": "6\n004444",
"output": "NO"
},
{
"input": "4\n0044",
"output": "NO"
},
{
"input": "14\n77771114444444",
"output": "NO"
},
{
"input": "14\n44444447777000",
"output": "NO"
},
{
"input": "6\n004774",
"output": "NO"
}
] | 1,568,028,194
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 46
| 218
| 0
|
import sys
n = int(input())
s = input()
i = 0
for ch in s:
if(ch != '4' and ch != '7'):
print('NO')
sys.exit(0)
n = int(n/2)
if(sum(map(int,s[:n])) == sum(map(int, s[-n:]))):
print("YES")
else:
print('NO')
|
Title: Lucky Ticket
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya loves tickets very much. As we know, each ticket has a number that is a positive integer. Its length equals *n* (*n* is always even). Petya calls a ticket lucky if the ticket's number is a lucky number and the sum of digits in the first half (the sum of the first *n*<=/<=2 digits) equals the sum of digits in the second half (the sum of the last *n*<=/<=2 digits). Check if the given ticket is lucky.
Input Specification:
The first line contains an even integer *n* (2<=β€<=*n*<=β€<=50) β the length of the ticket number that needs to be checked. The second line contains an integer whose length equals exactly *n* β the ticket number. The number may contain leading zeros.
Output Specification:
On the first line print "YES" if the given ticket number is lucky. Otherwise, print "NO" (without the quotes).
Demo Input:
['2\n47\n', '4\n4738\n', '4\n4774\n']
Demo Output:
['NO\n', 'NO\n', 'YES\n']
Note:
In the first sample the sum of digits in the first half does not equal the sum of digits in the second half (4ββ β7).
In the second sample the ticket number is not the lucky number.
|
```python
import sys
n = int(input())
s = input()
i = 0
for ch in s:
if(ch != '4' and ch != '7'):
print('NO')
sys.exit(0)
n = int(n/2)
if(sum(map(int,s[:n])) == sum(map(int, s[-n:]))):
print("YES")
else:
print('NO')
```
| 3
|
|
266
|
B
|
Queue at the School
|
PROGRAMMING
| 800
|
[
"constructive algorithms",
"graph matchings",
"implementation",
"shortest paths"
] | null | null |
During the break the schoolchildren, boys and girls, formed a queue of *n* people in the canteen. Initially the children stood in the order they entered the canteen. However, after a while the boys started feeling awkward for standing in front of the girls in the queue and they started letting the girls move forward each second.
Let's describe the process more precisely. Let's say that the positions in the queue are sequentially numbered by integers from 1 to *n*, at that the person in the position number 1 is served first. Then, if at time *x* a boy stands on the *i*-th position and a girl stands on the (*i*<=+<=1)-th position, then at time *x*<=+<=1 the *i*-th position will have a girl and the (*i*<=+<=1)-th position will have a boy. The time is given in seconds.
You've got the initial position of the children, at the initial moment of time. Determine the way the queue is going to look after *t* seconds.
|
The first line contains two integers *n* and *t* (1<=β€<=*n*,<=*t*<=β€<=50), which represent the number of children in the queue and the time after which the queue will transform into the arrangement you need to find.
The next line contains string *s*, which represents the schoolchildren's initial arrangement. If the *i*-th position in the queue contains a boy, then the *i*-th character of string *s* equals "B", otherwise the *i*-th character equals "G".
|
Print string *a*, which describes the arrangement after *t* seconds. If the *i*-th position has a boy after the needed time, then the *i*-th character *a* must equal "B", otherwise it must equal "G".
|
[
"5 1\nBGGBG\n",
"5 2\nBGGBG\n",
"4 1\nGGGB\n"
] |
[
"GBGGB\n",
"GGBGB\n",
"GGGB\n"
] |
none
| 500
|
[
{
"input": "5 1\nBGGBG",
"output": "GBGGB"
},
{
"input": "5 2\nBGGBG",
"output": "GGBGB"
},
{
"input": "4 1\nGGGB",
"output": "GGGB"
},
{
"input": "2 1\nBB",
"output": "BB"
},
{
"input": "2 1\nBG",
"output": "GB"
},
{
"input": "6 2\nBBGBBG",
"output": "GBBGBB"
},
{
"input": "8 3\nBBGBGBGB",
"output": "GGBGBBBB"
},
{
"input": "10 3\nBBGBBBBBBG",
"output": "GBBBBBGBBB"
},
{
"input": "22 7\nGBGGBGGGGGBBBGGBGBGBBB",
"output": "GGGGGGGGBGGBGGBBBBBBBB"
},
{
"input": "50 4\nGBBGBBBGGGGGBBGGBBBBGGGBBBGBBBGGBGGBGBBBGGBGGBGGBG",
"output": "GGBGBGBGBGBGGGBBGBGBGBGBBBGBGBGBGBGBGBGBGBGBGGBGBB"
},
{
"input": "50 8\nGGGGBGGBGGGBGBBBGGGGGGGGBBGBGBGBBGGBGGBGGGGGGGGBBG",
"output": "GGGGGGGGGGGGBGGBGBGBGBGBGGGGGGBGBGBGBGBGBGGBGGBGBB"
},
{
"input": "50 30\nBGGGGGGBGGBGBGGGGBGBBGBBBGGBBBGBGBGGGGGBGBBGBGBGGG",
"output": "GGGGGGGGGGGGGGGGGGGGGGGGGGGGGGBBBBBBBBBBBBBBBBBBBB"
},
{
"input": "20 20\nBBGGBGGGGBBBGBBGGGBB",
"output": "GGGGGGGGGGBBBBBBBBBB"
},
{
"input": "27 6\nGBGBGBGGGGGGBGGBGGBBGBBBGBB",
"output": "GGGGGGGBGBGBGGGGGBGBBBBBBBB"
},
{
"input": "46 11\nBGGGGGBGBGGBGGGBBGBBGBBGGBBGBBGBGGGGGGGBGBGBGB",
"output": "GGGGGGGGGGGBGGGGGBBGBGBGBGBGBGBGBGBGBGBGBBBBBB"
},
{
"input": "50 6\nBGGBBBBGGBBBBBBGGBGBGBBBBGBBBBBBGBBBBBBBBBBBBBBBBB",
"output": "GGGGBBBBBGBGBGBGBBBGBBBBBBGBBBBBBBBBBBBBBBBBBBBBBB"
},
{
"input": "50 10\nBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB",
"output": "BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB"
},
{
"input": "50 8\nGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG",
"output": "GGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG"
},
{
"input": "50 10\nBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBGB",
"output": "BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBGBBBBBBBBBBB"
},
{
"input": "50 13\nGGGBGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG",
"output": "GGGGGGGGGGGGGGGGBGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG"
},
{
"input": "1 1\nB",
"output": "B"
},
{
"input": "1 1\nG",
"output": "G"
},
{
"input": "1 50\nB",
"output": "B"
},
{
"input": "1 50\nG",
"output": "G"
},
{
"input": "50 50\nBBBBBBBBGGBBBBBBGBBBBBBBBBBBGBBBBBBBBBBBBBBGBBBBBB",
"output": "GGGGGBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB"
},
{
"input": "50 50\nGGBBGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGBBGGGGGGBG",
"output": "GGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGBBBBB"
},
{
"input": "6 3\nGGBBBG",
"output": "GGGBBB"
},
{
"input": "26 3\nGBBGBBBBBGGGBGBGGGBGBGGBBG",
"output": "GGBBBBGBGBGBGGGBGBGGGBGBBB"
},
{
"input": "46 3\nGGBBGGGGBBGBGBBBBBGGGBGGGBBGGGBBBGGBGGBBBGBGBB",
"output": "GGGGBGBGGGBBBBBGBGBGBGGGBGGBGBGBGBGBGBGBGBBBBB"
},
{
"input": "44 8\nBGBBBBBBBBBGGBBGBGBGGBBBBBGBBGBBBBBBBBBGBBGB",
"output": "GBBGBGBGBGBGBGBBBBGBBGBBBBBBBBBGBBGBBBBBBBBB"
},
{
"input": "20 20\nBBGGBGGGGBBBGBBGGGBB",
"output": "GGGGGGGGGGBBBBBBBBBB"
},
{
"input": "30 25\nBGGBBGBGGBGBGBBGBGGGGBGBGGBBBB",
"output": "GGGGGGGGGGGGGGGBBBBBBBBBBBBBBB"
},
{
"input": "17 42\nBBGBGBGGGGGGBBGGG",
"output": "GGGGGGGGGGGBBBBBB"
},
{
"input": "30 50\nBGGBBGGGGGGGGBBGGGBBGGBBBGBBGG",
"output": "GGGGGGGGGGGGGGGGGGBBBBBBBBBBBB"
},
{
"input": "31 33\nBBGGBBGGBGBBBBGGBGBBBGBGGBBGBBB",
"output": "GGGGGGGGGGGGBBBBBBBBBBBBBBBBBBB"
},
{
"input": "2 2\nBG",
"output": "GB"
},
{
"input": "8 8\nGGGBGGBB",
"output": "GGGGGBBB"
},
{
"input": "34 34\nBGGBBBBGGBBGBGGGGGGBBGGGGGBGGBGGGB",
"output": "GGGGGGGGGGGGGGGGGGGGGBBBBBBBBBBBBB"
},
{
"input": "34 20\nBBBBGBGGGBGGGBGBGGBBBBGGBBGGGBGBBG",
"output": "GGGGGGGGGGGGGGGGGBBBBBBBBBBBBBBBBB"
},
{
"input": "50 50\nBBGBBBBBBBBBBBBGBBBGBBBBBBBGBBBBBBGBBBGBBGBBGBBBGG",
"output": "GGGGGGGGGGBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB"
},
{
"input": "10 10\nGGGGGGGGGG",
"output": "GGGGGGGGGG"
},
{
"input": "10 10\nBBBBBBBBBB",
"output": "BBBBBBBBBB"
},
{
"input": "10 10\nBGBGBGBGBG",
"output": "GGGGGBBBBB"
},
{
"input": "1 1\nB",
"output": "B"
}
] | 1,698,452,541
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 44
| 92
| 0
|
q, r = map(int, input().split())
t1 = input()
t = []
for o in t1:
t.append(o)
for j in range(r):
i = 0
while i < len(t)-1:
if t[i] == 'B' and t[i + 1] == 'G':
t[i], t[i + 1] = t[i + 1], t[i]
i += 2
else:
i += 1
t1 = ''.join(t)
print(t1)
|
Title: Queue at the School
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
During the break the schoolchildren, boys and girls, formed a queue of *n* people in the canteen. Initially the children stood in the order they entered the canteen. However, after a while the boys started feeling awkward for standing in front of the girls in the queue and they started letting the girls move forward each second.
Let's describe the process more precisely. Let's say that the positions in the queue are sequentially numbered by integers from 1 to *n*, at that the person in the position number 1 is served first. Then, if at time *x* a boy stands on the *i*-th position and a girl stands on the (*i*<=+<=1)-th position, then at time *x*<=+<=1 the *i*-th position will have a girl and the (*i*<=+<=1)-th position will have a boy. The time is given in seconds.
You've got the initial position of the children, at the initial moment of time. Determine the way the queue is going to look after *t* seconds.
Input Specification:
The first line contains two integers *n* and *t* (1<=β€<=*n*,<=*t*<=β€<=50), which represent the number of children in the queue and the time after which the queue will transform into the arrangement you need to find.
The next line contains string *s*, which represents the schoolchildren's initial arrangement. If the *i*-th position in the queue contains a boy, then the *i*-th character of string *s* equals "B", otherwise the *i*-th character equals "G".
Output Specification:
Print string *a*, which describes the arrangement after *t* seconds. If the *i*-th position has a boy after the needed time, then the *i*-th character *a* must equal "B", otherwise it must equal "G".
Demo Input:
['5 1\nBGGBG\n', '5 2\nBGGBG\n', '4 1\nGGGB\n']
Demo Output:
['GBGGB\n', 'GGBGB\n', 'GGGB\n']
Note:
none
|
```python
q, r = map(int, input().split())
t1 = input()
t = []
for o in t1:
t.append(o)
for j in range(r):
i = 0
while i < len(t)-1:
if t[i] == 'B' and t[i + 1] == 'G':
t[i], t[i + 1] = t[i + 1], t[i]
i += 2
else:
i += 1
t1 = ''.join(t)
print(t1)
```
| 3
|
|
912
|
B
|
New Year's Eve
|
PROGRAMMING
| 1,300
|
[
"bitmasks",
"constructive algorithms",
"number theory"
] | null | null |
Since Grisha behaved well last year, at New Year's Eve he was visited by Ded Moroz who brought an enormous bag of gifts with him! The bag contains *n* sweet candies from the good ol' bakery, each labeled from 1 to *n* corresponding to its tastiness. No two candies have the same tastiness.
The choice of candies has a direct effect on Grisha's happiness. One can assume that he should take the tastiest onesΒ β but no, the holiday magic turns things upside down. It is the xor-sum of tastinesses that matters, not the ordinary sum!
A xor-sum of a sequence of integers *a*1,<=*a*2,<=...,<=*a**m* is defined as the bitwise XOR of all its elements: , here denotes the bitwise XOR operation; more about bitwise XOR can be found [here.](https://en.wikipedia.org/wiki/Bitwise_operation#XOR)
Ded Moroz warned Grisha he has more houses to visit, so Grisha can take no more than *k* candies from the bag. Help Grisha determine the largest xor-sum (largest xor-sum means maximum happiness!) he can obtain.
|
The sole string contains two integers *n* and *k* (1<=β€<=*k*<=β€<=*n*<=β€<=1018).
|
Output one numberΒ β the largest possible xor-sum.
|
[
"4 3\n",
"6 6\n"
] |
[
"7\n",
"7\n"
] |
In the first sample case, one optimal answer is 1, 2 and 4, giving the xor-sum of 7.
In the second sample case, one can, for example, take all six candies and obtain the xor-sum of 7.
| 1,000
|
[
{
"input": "4 3",
"output": "7"
},
{
"input": "6 6",
"output": "7"
},
{
"input": "2 2",
"output": "3"
},
{
"input": "1022 10",
"output": "1023"
},
{
"input": "415853337373441 52",
"output": "562949953421311"
},
{
"input": "75 12",
"output": "127"
},
{
"input": "1000000000000000000 1000000000000000000",
"output": "1152921504606846975"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "1000000000000000000 2",
"output": "1152921504606846975"
},
{
"input": "49194939 22",
"output": "67108863"
},
{
"input": "228104606 17",
"output": "268435455"
},
{
"input": "817034381 7",
"output": "1073741823"
},
{
"input": "700976748 4",
"output": "1073741823"
},
{
"input": "879886415 9",
"output": "1073741823"
},
{
"input": "18007336 10353515",
"output": "33554431"
},
{
"input": "196917003 154783328",
"output": "268435455"
},
{
"input": "785846777 496205300",
"output": "1073741823"
},
{
"input": "964756444 503568330",
"output": "1073741823"
},
{
"input": "848698811 317703059",
"output": "1073741823"
},
{
"input": "676400020444788 1",
"output": "676400020444788"
},
{
"input": "502643198528213 1",
"output": "502643198528213"
},
{
"input": "815936580997298686 684083143940282566",
"output": "1152921504606846975"
},
{
"input": "816762824175382110 752185261508428780",
"output": "1152921504606846975"
},
{
"input": "327942415253132295 222598158321260499",
"output": "576460752303423487"
},
{
"input": "328768654136248423 284493129147496637",
"output": "576460752303423487"
},
{
"input": "329594893019364551 25055600080496801",
"output": "576460752303423487"
},
{
"input": "921874985256864012 297786684518764536",
"output": "1152921504606846975"
},
{
"input": "922701224139980141 573634416190460758",
"output": "1152921504606846975"
},
{
"input": "433880815217730325 45629641110945892",
"output": "576460752303423487"
},
{
"input": "434707058395813749 215729375494216481",
"output": "576460752303423487"
},
{
"input": "435533301573897173 34078453236225189",
"output": "576460752303423487"
},
{
"input": "436359544751980597 199220719961060641",
"output": "576460752303423487"
},
{
"input": "437185783635096725 370972992240105630",
"output": "576460752303423487"
},
{
"input": "438012026813180149 111323110116193830",
"output": "576460752303423487"
},
{
"input": "438838269991263573 295468957052046146",
"output": "576460752303423487"
},
{
"input": "439664513169346997 46560240538186155",
"output": "576460752303423487"
},
{
"input": "440490752052463125 216165966013438147",
"output": "576460752303423487"
},
{
"input": "441316995230546549 401964286420555423",
"output": "576460752303423487"
},
{
"input": "952496582013329437 673506882352402278",
"output": "1152921504606846975"
},
{
"input": "1000000000000000000 1",
"output": "1000000000000000000"
},
{
"input": "2147483647 1",
"output": "2147483647"
},
{
"input": "2147483647 2",
"output": "2147483647"
},
{
"input": "2147483647 31",
"output": "2147483647"
},
{
"input": "8 2",
"output": "15"
},
{
"input": "3 3",
"output": "3"
},
{
"input": "4 1",
"output": "4"
},
{
"input": "10 2",
"output": "15"
},
{
"input": "288230376151711743 2",
"output": "288230376151711743"
},
{
"input": "5 2",
"output": "7"
},
{
"input": "576460752303423487 2",
"output": "576460752303423487"
},
{
"input": "36028797018963967 123",
"output": "36028797018963967"
},
{
"input": "1125899906842623 2",
"output": "1125899906842623"
},
{
"input": "576460752303423489 5",
"output": "1152921504606846975"
},
{
"input": "288230376151711743 3",
"output": "288230376151711743"
},
{
"input": "36028797018963967 345",
"output": "36028797018963967"
},
{
"input": "18014398509481984 30",
"output": "36028797018963967"
},
{
"input": "8 8",
"output": "15"
},
{
"input": "8 1",
"output": "8"
}
] | 1,516,725,943
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 58
| 77
| 5,632,000
|
n , k = map(int , input().split())
l = list(bin(n))[2:]
if k == 1: print(n)
else:
print ((1 << len(l)) -1)
|
Title: New Year's Eve
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Since Grisha behaved well last year, at New Year's Eve he was visited by Ded Moroz who brought an enormous bag of gifts with him! The bag contains *n* sweet candies from the good ol' bakery, each labeled from 1 to *n* corresponding to its tastiness. No two candies have the same tastiness.
The choice of candies has a direct effect on Grisha's happiness. One can assume that he should take the tastiest onesΒ β but no, the holiday magic turns things upside down. It is the xor-sum of tastinesses that matters, not the ordinary sum!
A xor-sum of a sequence of integers *a*1,<=*a*2,<=...,<=*a**m* is defined as the bitwise XOR of all its elements: , here denotes the bitwise XOR operation; more about bitwise XOR can be found [here.](https://en.wikipedia.org/wiki/Bitwise_operation#XOR)
Ded Moroz warned Grisha he has more houses to visit, so Grisha can take no more than *k* candies from the bag. Help Grisha determine the largest xor-sum (largest xor-sum means maximum happiness!) he can obtain.
Input Specification:
The sole string contains two integers *n* and *k* (1<=β€<=*k*<=β€<=*n*<=β€<=1018).
Output Specification:
Output one numberΒ β the largest possible xor-sum.
Demo Input:
['4 3\n', '6 6\n']
Demo Output:
['7\n', '7\n']
Note:
In the first sample case, one optimal answer is 1, 2 and 4, giving the xor-sum of 7.
In the second sample case, one can, for example, take all six candies and obtain the xor-sum of 7.
|
```python
n , k = map(int , input().split())
l = list(bin(n))[2:]
if k == 1: print(n)
else:
print ((1 << len(l)) -1)
```
| 3
|
|
268
|
A
|
Games
|
PROGRAMMING
| 800
|
[
"brute force"
] | null | null |
Manao works on a sports TV. He's spent much time watching the football games of some country. After a while he began to notice different patterns. For example, each team has two sets of uniforms: home uniform and guest uniform. When a team plays a game at home, the players put on the home uniform. When a team plays as a guest on somebody else's stadium, the players put on the guest uniform. The only exception to that rule is: when the home uniform color of the host team matches the guests' uniform, the host team puts on its guest uniform as well. For each team the color of the home and guest uniform is different.
There are *n* teams taking part in the national championship. The championship consists of *n*Β·(*n*<=-<=1) games: each team invites each other team to its stadium. At this point Manao wondered: how many times during the championship is a host team going to put on the guest uniform? Note that the order of the games does not affect this number.
You know the colors of the home and guest uniform for each team. For simplicity, the colors are numbered by integers in such a way that no two distinct colors have the same number. Help Manao find the answer to his question.
|
The first line contains an integer *n* (2<=β€<=*n*<=β€<=30). Each of the following *n* lines contains a pair of distinct space-separated integers *h**i*, *a**i* (1<=β€<=*h**i*,<=*a**i*<=β€<=100) β the colors of the *i*-th team's home and guest uniforms, respectively.
|
In a single line print the number of games where the host team is going to play in the guest uniform.
|
[
"3\n1 2\n2 4\n3 4\n",
"4\n100 42\n42 100\n5 42\n100 5\n",
"2\n1 2\n1 2\n"
] |
[
"1\n",
"5\n",
"0\n"
] |
In the first test case the championship consists of 6 games. The only game with the event in question is the game between teams 2 and 1 on the stadium of team 2.
In the second test sample the host team will have to wear guest uniform in the games between teams: 1 and 2, 2 and 1, 2 and 3, 3 and 4, 4 and 2 (the host team is written first).
| 500
|
[
{
"input": "3\n1 2\n2 4\n3 4",
"output": "1"
},
{
"input": "4\n100 42\n42 100\n5 42\n100 5",
"output": "5"
},
{
"input": "2\n1 2\n1 2",
"output": "0"
},
{
"input": "7\n4 7\n52 55\n16 4\n55 4\n20 99\n3 4\n7 52",
"output": "6"
},
{
"input": "10\n68 42\n1 35\n25 70\n59 79\n65 63\n46 6\n28 82\n92 62\n43 96\n37 28",
"output": "1"
},
{
"input": "30\n10 39\n89 1\n78 58\n75 99\n36 13\n77 50\n6 97\n79 28\n27 52\n56 5\n93 96\n40 21\n33 74\n26 37\n53 59\n98 56\n61 65\n42 57\n9 7\n25 63\n74 34\n96 84\n95 47\n12 23\n34 21\n71 6\n27 13\n15 47\n64 14\n12 77",
"output": "6"
},
{
"input": "30\n46 100\n87 53\n34 84\n44 66\n23 20\n50 34\n90 66\n17 39\n13 22\n94 33\n92 46\n63 78\n26 48\n44 61\n3 19\n41 84\n62 31\n65 89\n23 28\n58 57\n19 85\n26 60\n75 66\n69 67\n76 15\n64 15\n36 72\n90 89\n42 69\n45 35",
"output": "4"
},
{
"input": "2\n46 6\n6 46",
"output": "2"
},
{
"input": "29\n8 18\n33 75\n69 22\n97 95\n1 97\n78 10\n88 18\n13 3\n19 64\n98 12\n79 92\n41 72\n69 15\n98 31\n57 74\n15 56\n36 37\n15 66\n63 100\n16 42\n47 56\n6 4\n73 15\n30 24\n27 71\n12 19\n88 69\n85 6\n50 11",
"output": "10"
},
{
"input": "23\n43 78\n31 28\n58 80\n66 63\n20 4\n51 95\n40 20\n50 14\n5 34\n36 39\n77 42\n64 97\n62 89\n16 56\n8 34\n58 16\n37 35\n37 66\n8 54\n50 36\n24 8\n68 48\n85 33",
"output": "6"
},
{
"input": "13\n76 58\n32 85\n99 79\n23 58\n96 59\n72 35\n53 43\n96 55\n41 78\n75 10\n28 11\n72 7\n52 73",
"output": "0"
},
{
"input": "18\n6 90\n70 79\n26 52\n67 81\n29 95\n41 32\n94 88\n18 58\n59 65\n51 56\n64 68\n34 2\n6 98\n95 82\n34 2\n40 98\n83 78\n29 2",
"output": "1"
},
{
"input": "18\n6 90\n100 79\n26 100\n67 100\n29 100\n100 32\n94 88\n18 58\n59 65\n51 56\n64 68\n34 2\n6 98\n95 82\n34 2\n40 98\n83 78\n29 100",
"output": "8"
},
{
"input": "30\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1",
"output": "450"
},
{
"input": "30\n100 99\n58 59\n56 57\n54 55\n52 53\n50 51\n48 49\n46 47\n44 45\n42 43\n40 41\n38 39\n36 37\n34 35\n32 33\n30 31\n28 29\n26 27\n24 25\n22 23\n20 21\n18 19\n16 17\n14 15\n12 13\n10 11\n8 9\n6 7\n4 5\n2 3",
"output": "0"
},
{
"input": "15\n9 3\n2 6\n7 6\n5 10\n9 5\n8 1\n10 5\n2 8\n4 5\n9 8\n5 3\n3 8\n9 8\n4 10\n8 5",
"output": "20"
},
{
"input": "15\n2 1\n1 2\n1 2\n1 2\n2 1\n2 1\n2 1\n1 2\n2 1\n2 1\n2 1\n1 2\n2 1\n2 1\n1 2",
"output": "108"
},
{
"input": "25\n2 1\n1 2\n1 2\n1 2\n2 1\n1 2\n1 2\n1 2\n2 1\n2 1\n2 1\n1 2\n1 2\n1 2\n2 1\n2 1\n2 1\n1 2\n2 1\n1 2\n2 1\n2 1\n2 1\n2 1\n1 2",
"output": "312"
},
{
"input": "25\n91 57\n2 73\n54 57\n2 57\n23 57\n2 6\n57 54\n57 23\n91 54\n91 23\n57 23\n91 57\n54 2\n6 91\n57 54\n2 57\n57 91\n73 91\n57 23\n91 57\n2 73\n91 2\n23 6\n2 73\n23 6",
"output": "96"
},
{
"input": "28\n31 66\n31 91\n91 31\n97 66\n31 66\n31 66\n66 91\n91 31\n97 31\n91 97\n97 31\n66 31\n66 97\n91 31\n31 66\n31 66\n66 31\n31 97\n66 97\n97 31\n31 91\n66 91\n91 66\n31 66\n91 66\n66 31\n66 31\n91 97",
"output": "210"
},
{
"input": "29\n78 27\n50 68\n24 26\n68 43\n38 78\n26 38\n78 28\n28 26\n27 24\n23 38\n24 26\n24 43\n61 50\n38 78\n27 23\n61 26\n27 28\n43 23\n28 78\n43 27\n43 78\n27 61\n28 38\n61 78\n50 26\n43 27\n26 78\n28 50\n43 78",
"output": "73"
},
{
"input": "29\n80 27\n69 80\n27 80\n69 80\n80 27\n80 27\n80 27\n80 69\n27 69\n80 69\n80 27\n27 69\n69 27\n80 69\n27 69\n69 80\n27 69\n80 69\n80 27\n69 27\n27 69\n27 80\n80 27\n69 80\n27 69\n80 69\n69 80\n69 80\n27 80",
"output": "277"
},
{
"input": "30\n19 71\n7 89\n89 71\n21 7\n19 21\n7 89\n19 71\n89 8\n89 21\n19 8\n21 7\n8 89\n19 89\n7 21\n19 8\n19 7\n7 19\n8 21\n71 21\n71 89\n7 19\n7 19\n21 7\n21 19\n21 19\n71 8\n21 8\n71 19\n19 71\n8 21",
"output": "154"
},
{
"input": "30\n44 17\n44 17\n44 17\n17 44\n44 17\n44 17\n17 44\n17 44\n17 44\n44 17\n44 17\n44 17\n44 17\n44 17\n17 44\n17 44\n17 44\n44 17\n44 17\n17 44\n44 17\n44 17\n44 17\n17 44\n17 44\n44 17\n17 44\n44 17\n44 17\n44 17",
"output": "418"
},
{
"input": "22\n78 92\n15 92\n92 78\n78 80\n92 16\n24 80\n92 16\n16 92\n78 16\n24 78\n80 78\n92 80\n16 80\n80 78\n15 78\n92 16\n24 15\n24 80\n80 16\n16 80\n92 80\n24 80",
"output": "74"
},
{
"input": "24\n9 83\n90 31\n83 3\n83 3\n21 31\n83 3\n32 31\n12 21\n31 21\n90 32\n32 21\n12 9\n12 31\n9 83\n83 12\n32 3\n32 83\n90 31\n9 32\n31 21\n83 90\n32 21\n21 3\n32 9",
"output": "59"
},
{
"input": "30\n67 21\n85 39\n85 87\n21 39\n66 85\n10 95\n10 21\n87 85\n82 21\n67 21\n95 10\n21 39\n82 21\n21 66\n66 39\n95 30\n67 85\n66 82\n85 82\n21 66\n10 39\n67 10\n21 85\n10 82\n85 95\n10 85\n21 39\n85 39\n39 10\n95 67",
"output": "100"
},
{
"input": "4\n8 7\n8 7\n7 8\n7 8",
"output": "8"
},
{
"input": "6\n1 2\n1 2\n1 2\n1 2\n1 2\n2 1",
"output": "10"
},
{
"input": "12\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1",
"output": "72"
},
{
"input": "4\n1 2\n1 2\n2 1\n2 1",
"output": "8"
}
] | 1,673,285,663
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 31
| 92
| 0
|
teams = int(input())
homeKit = [None] * teams
guestKit = [None] * teams
for i in range(teams):
homeKit[i], guestKit[i] = map(int, input().split())
guestUni = 0
for i in range(teams):
for j in range(teams):
if i == j:
continue
if homeKit[i] == guestKit[j]:
guestUni += 1
print(guestUni)
|
Title: Games
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Manao works on a sports TV. He's spent much time watching the football games of some country. After a while he began to notice different patterns. For example, each team has two sets of uniforms: home uniform and guest uniform. When a team plays a game at home, the players put on the home uniform. When a team plays as a guest on somebody else's stadium, the players put on the guest uniform. The only exception to that rule is: when the home uniform color of the host team matches the guests' uniform, the host team puts on its guest uniform as well. For each team the color of the home and guest uniform is different.
There are *n* teams taking part in the national championship. The championship consists of *n*Β·(*n*<=-<=1) games: each team invites each other team to its stadium. At this point Manao wondered: how many times during the championship is a host team going to put on the guest uniform? Note that the order of the games does not affect this number.
You know the colors of the home and guest uniform for each team. For simplicity, the colors are numbered by integers in such a way that no two distinct colors have the same number. Help Manao find the answer to his question.
Input Specification:
The first line contains an integer *n* (2<=β€<=*n*<=β€<=30). Each of the following *n* lines contains a pair of distinct space-separated integers *h**i*, *a**i* (1<=β€<=*h**i*,<=*a**i*<=β€<=100) β the colors of the *i*-th team's home and guest uniforms, respectively.
Output Specification:
In a single line print the number of games where the host team is going to play in the guest uniform.
Demo Input:
['3\n1 2\n2 4\n3 4\n', '4\n100 42\n42 100\n5 42\n100 5\n', '2\n1 2\n1 2\n']
Demo Output:
['1\n', '5\n', '0\n']
Note:
In the first test case the championship consists of 6 games. The only game with the event in question is the game between teams 2 and 1 on the stadium of team 2.
In the second test sample the host team will have to wear guest uniform in the games between teams: 1 and 2, 2 and 1, 2 and 3, 3 and 4, 4 and 2 (the host team is written first).
|
```python
teams = int(input())
homeKit = [None] * teams
guestKit = [None] * teams
for i in range(teams):
homeKit[i], guestKit[i] = map(int, input().split())
guestUni = 0
for i in range(teams):
for j in range(teams):
if i == j:
continue
if homeKit[i] == guestKit[j]:
guestUni += 1
print(guestUni)
```
| 3
|
|
1,009
|
A
|
Game Shopping
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Maxim wants to buy some games at the local game shop. There are $n$ games in the shop, the $i$-th game costs $c_i$.
Maxim has a wallet which can be represented as an array of integers. His wallet contains $m$ bills, the $j$-th bill has value $a_j$.
Games in the shop are ordered from left to right, Maxim tries to buy every game in that order.
When Maxim stands at the position $i$ in the shop, he takes the first bill from his wallet (if his wallet is empty then he proceeds to the next position immediately) and tries to buy the $i$-th game using this bill. After Maxim tried to buy the $n$-th game, he leaves the shop.
Maxim buys the $i$-th game if and only if the value of the first bill (which he takes) from his wallet is greater or equal to the cost of the $i$-th game. If he successfully buys the $i$-th game, the first bill from his wallet disappears and the next bill becomes first. Otherwise Maxim leaves the first bill in his wallet (this bill still remains the first one) and proceeds to the next game.
For example, for array $c = [2, 4, 5, 2, 4]$ and array $a = [5, 3, 4, 6]$ the following process takes place: Maxim buys the first game using the first bill (its value is $5$), the bill disappears, after that the second bill (with value $3$) becomes the first one in Maxim's wallet, then Maxim doesn't buy the second game because $c_2 > a_2$, the same with the third game, then he buys the fourth game using the bill of value $a_2$ (the third bill becomes the first one in Maxim's wallet) and buys the fifth game using the bill of value $a_3$.
Your task is to get the number of games Maxim will buy.
|
The first line of the input contains two integers $n$ and $m$ ($1 \le n, m \le 1000$) β the number of games and the number of bills in Maxim's wallet.
The second line of the input contains $n$ integers $c_1, c_2, \dots, c_n$ ($1 \le c_i \le 1000$), where $c_i$ is the cost of the $i$-th game.
The third line of the input contains $m$ integers $a_1, a_2, \dots, a_m$ ($1 \le a_j \le 1000$), where $a_j$ is the value of the $j$-th bill from the Maxim's wallet.
|
Print a single integer β the number of games Maxim will buy.
|
[
"5 4\n2 4 5 2 4\n5 3 4 6\n",
"5 2\n20 40 50 20 40\n19 20\n",
"6 4\n4 8 15 16 23 42\n1000 1000 1000 1000\n"
] |
[
"3\n",
"0\n",
"4\n"
] |
The first example is described in the problem statement.
In the second example Maxim cannot buy any game because the value of the first bill in his wallet is smaller than the cost of any game in the shop.
In the third example the values of the bills in Maxim's wallet are large enough to buy any game he encounter until he runs out of bills in his wallet.
| 0
|
[
{
"input": "5 4\n2 4 5 2 4\n5 3 4 6",
"output": "3"
},
{
"input": "5 2\n20 40 50 20 40\n19 20",
"output": "0"
},
{
"input": "6 4\n4 8 15 16 23 42\n1000 1000 1000 1000",
"output": "4"
},
{
"input": "5 1\n1 1 1 1 1\n5",
"output": "1"
},
{
"input": "5 1\n10 1 1 1 1\n1000",
"output": "1"
},
{
"input": "5 1\n100 100 100 100 100\n100",
"output": "1"
},
{
"input": "2 1\n2 1\n1",
"output": "1"
},
{
"input": "2 3\n3 1\n2 4 2",
"output": "1"
},
{
"input": "1 5\n4\n1 4 3 3 2",
"output": "0"
},
{
"input": "5 3\n4 2 3 1 1\n2 1 3",
"output": "3"
},
{
"input": "3 5\n5 2 5\n1 4 1 4 2",
"output": "0"
},
{
"input": "7 3\n9 7 10 2 1 1 1\n8 9 6",
"output": "3"
},
{
"input": "5 3\n2 5 3 3 2\n2 5 3",
"output": "3"
}
] | 1,555,662,827
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 19
| 124
| 307,200
|
def main():
[n_games, n_bills] = [int(_) for _ in input().split()]
game_prices = [0] + [int(_) for _ in input().split()]
bills = [0] + [int(_) for _ in input().split()]
last_game = 0
last_bill = 0
count_games_bought = 0
while last_game < n_games and last_bill < n_bills:
last_bill += 1
bill = bills[last_bill]
try:
last_game = next(g for g in range(last_game + 1, n_games + 1) if game_prices[g] <= bill)
count_games_bought += 1
except StopIteration:
last_game = n_games
print(count_games_bought)
if __name__ == '__main__':
main()
|
Title: Game Shopping
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Maxim wants to buy some games at the local game shop. There are $n$ games in the shop, the $i$-th game costs $c_i$.
Maxim has a wallet which can be represented as an array of integers. His wallet contains $m$ bills, the $j$-th bill has value $a_j$.
Games in the shop are ordered from left to right, Maxim tries to buy every game in that order.
When Maxim stands at the position $i$ in the shop, he takes the first bill from his wallet (if his wallet is empty then he proceeds to the next position immediately) and tries to buy the $i$-th game using this bill. After Maxim tried to buy the $n$-th game, he leaves the shop.
Maxim buys the $i$-th game if and only if the value of the first bill (which he takes) from his wallet is greater or equal to the cost of the $i$-th game. If he successfully buys the $i$-th game, the first bill from his wallet disappears and the next bill becomes first. Otherwise Maxim leaves the first bill in his wallet (this bill still remains the first one) and proceeds to the next game.
For example, for array $c = [2, 4, 5, 2, 4]$ and array $a = [5, 3, 4, 6]$ the following process takes place: Maxim buys the first game using the first bill (its value is $5$), the bill disappears, after that the second bill (with value $3$) becomes the first one in Maxim's wallet, then Maxim doesn't buy the second game because $c_2 > a_2$, the same with the third game, then he buys the fourth game using the bill of value $a_2$ (the third bill becomes the first one in Maxim's wallet) and buys the fifth game using the bill of value $a_3$.
Your task is to get the number of games Maxim will buy.
Input Specification:
The first line of the input contains two integers $n$ and $m$ ($1 \le n, m \le 1000$) β the number of games and the number of bills in Maxim's wallet.
The second line of the input contains $n$ integers $c_1, c_2, \dots, c_n$ ($1 \le c_i \le 1000$), where $c_i$ is the cost of the $i$-th game.
The third line of the input contains $m$ integers $a_1, a_2, \dots, a_m$ ($1 \le a_j \le 1000$), where $a_j$ is the value of the $j$-th bill from the Maxim's wallet.
Output Specification:
Print a single integer β the number of games Maxim will buy.
Demo Input:
['5 4\n2 4 5 2 4\n5 3 4 6\n', '5 2\n20 40 50 20 40\n19 20\n', '6 4\n4 8 15 16 23 42\n1000 1000 1000 1000\n']
Demo Output:
['3\n', '0\n', '4\n']
Note:
The first example is described in the problem statement.
In the second example Maxim cannot buy any game because the value of the first bill in his wallet is smaller than the cost of any game in the shop.
In the third example the values of the bills in Maxim's wallet are large enough to buy any game he encounter until he runs out of bills in his wallet.
|
```python
def main():
[n_games, n_bills] = [int(_) for _ in input().split()]
game_prices = [0] + [int(_) for _ in input().split()]
bills = [0] + [int(_) for _ in input().split()]
last_game = 0
last_bill = 0
count_games_bought = 0
while last_game < n_games and last_bill < n_bills:
last_bill += 1
bill = bills[last_bill]
try:
last_game = next(g for g in range(last_game + 1, n_games + 1) if game_prices[g] <= bill)
count_games_bought += 1
except StopIteration:
last_game = n_games
print(count_games_bought)
if __name__ == '__main__':
main()
```
| 3
|
|
112
|
A
|
Petya and Strings
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] |
A. Petya and Strings
|
2
|
256
|
Little Petya loves presents. His mum bought him two strings of the same size for his birthday. The strings consist of uppercase and lowercase Latin letters. Now Petya wants to compare those two strings lexicographically. The letters' case does not matter, that is an uppercase letter is considered equivalent to the corresponding lowercase letter. Help Petya perform the comparison.
|
Each of the first two lines contains a bought string. The strings' lengths range from 1 to 100 inclusive. It is guaranteed that the strings are of the same length and also consist of uppercase and lowercase Latin letters.
|
If the first string is less than the second one, print "-1". If the second string is less than the first one, print "1". If the strings are equal, print "0". Note that the letters' case is not taken into consideration when the strings are compared.
|
[
"aaaa\naaaA\n",
"abs\nAbz\n",
"abcdefg\nAbCdEfF\n"
] |
[
"0\n",
"-1\n",
"1\n"
] |
If you want more formal information about the lexicographical order (also known as the "dictionary order" or "alphabetical order"), you can visit the following site:
- http://en.wikipedia.org/wiki/Lexicographical_order
| 500
|
[
{
"input": "aaaa\naaaA",
"output": "0"
},
{
"input": "abs\nAbz",
"output": "-1"
},
{
"input": "abcdefg\nAbCdEfF",
"output": "1"
},
{
"input": "asadasdasd\nasdwasdawd",
"output": "-1"
},
{
"input": "aslkjlkasdd\nasdlkjdajwi",
"output": "1"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "0"
},
{
"input": "aAaaaAAaAaaAzZsssSsdDfeEaeqZlpP\nAaaaAaaAaaAaZzSSSSsDdFeeAeQZLpp",
"output": "0"
},
{
"input": "bwuEhEveouaTECagLZiqmUdxEmhRSOzMauJRWLQMppZOumxhAmwuGeDIkvkBLvMXwUoFmpAfDprBcFtEwOULcZWRQhcTbTbX\nHhoDWbcxwiMnCNexOsKsujLiSGcLllXOkRSbnOzThAjnnliLYFFmsYkOfpTxRNEfBsoUHfoLTiqAINRPxWRqrTJhgfkKcDOH",
"output": "-1"
},
{
"input": "kGWUuguKzcvxqKTNpxeDWXpXkrXDvGMFGoXKDfPBZvWSDUyIYBynbKOUonHvmZaKeirUhfmVRKtGhAdBfKMWXDUoqvbfpfHYcg\ncvOULleuIIiYVVxcLZmHVpNGXuEpzcWZZWyMOwIwbpkKPwCfkVbKkUuosvxYCKjqfVmHfJKbdrsAcatPYgrCABaFcoBuOmMfFt",
"output": "1"
},
{
"input": "nCeNVIzHqPceNhjHeHvJvgBsNFiXBATRrjSTXJzhLMDMxiJztphxBRlDlqwDFImWeEPkggZCXSRwelOdpNrYnTepiOqpvkr\nHJbjJFtlvNxIbkKlxQUwmZHJFVNMwPAPDRslIoXISBYHHfymyIaQHLgECPxAmqnOCizwXnIUBRmpYUBVPenoUKhCobKdOjL",
"output": "1"
},
{
"input": "ttXjenUAlfixytHEOrPkgXmkKTSGYuyVXGIHYmWWYGlBYpHkujueqBSgjLguSgiMGJWATIGEUjjAjKXdMiVbHozZUmqQtFrT\nJziDBFBDmDJCcGqFsQwDFBYdOidLxxhBCtScznnDgnsiStlWFnEXQrJxqTXKPxZyIGfLIToETKWZBPUIBmLeImrlSBWCkTNo",
"output": "1"
},
{
"input": "AjQhPqSVhwQQjcgCycjKorWBgFCRuQBwgdVuAPSMJAvTyxGVuFHjfJzkKfsmfhFbKqFrFIohSZBbpjgEHebezmVlGLTPSCTMf\nXhxWuSnMmKFrCUOwkTUmvKAfbTbHWzzOTzxJatLLCdlGnHVaBUnxDlsqpvjLHMThOPAFBggVKDyKBrZAmjnjrhHlrnSkyzBja",
"output": "-1"
},
{
"input": "HCIgYtnqcMyjVngziNflxKHtdTmcRJhzMAjFAsNdWXFJYEhiTzsQUtFNkAbdrFBRmvLirkuirqTDvIpEfyiIqkrwsjvpPWTEdI\nErqiiWKsmIjyZuzgTlTqxYZwlrpvRyaVhRTOYUqtPMVGGtWOkDCOOQRKrkkRzPftyQCkYkzKkzTPqqXmeZhvvEEiEhkdOmoMvy",
"output": "1"
},
{
"input": "mtBeJYILXcECGyEVSyzLFdQJbiVnnfkbsYYsdUJSIRmyzLfTTtFwIBmRLVnwcewIqcuydkcLpflHAFyDaToLiFMgeHvQorTVbI\nClLvyejznjbRfCDcrCzkLvqQaGzTjwmWONBdCctJAPJBcQrcYvHaSLQgPIJbmkFBhFzuQLBiRzAdNHulCjIAkBvZxxlkdzUWLR",
"output": "1"
},
{
"input": "tjucSbGESVmVridTBjTmpVBCwwdWKBPeBvmgdxgIVLwQxveETnSdxkTVJpXoperWSgdpPMKNmwDiGeHfxnuqaDissgXPlMuNZIr\nHfjOOJhomqNIKHvqSgfySjlsWJQBuWYwhLQhlZYlpZwboMpoLoluGsBmhhlYgeIouwdkPfiaAIrkYRlxtiFazOPOllPsNZHcIZd",
"output": "1"
},
{
"input": "AanbDfbZNlUodtBQlvPMyomStKNhgvSGhSbTdabxGFGGXCdpsJDimsAykKjfBDPMulkhBMsqLmVKLDoesHZsRAEEdEzqigueXInY\ncwfyjoppiJNrjrOLNZkqcGimrpTsiyFBVgMWEPXsMrxLJDDbtYzerXiFGuLBcQYitLdqhGHBpdjRnkUegmnwhGHAKXGyFtscWDSI",
"output": "-1"
},
{
"input": "HRfxniwuJCaHOcaOVgjOGHXKrwxrDQxJpppeGDXnTAowyKbCsCQPbchCKeTWOcKbySSYnoaTJDnmRcyGPbfXJyZoPcARHBu\nxkLXvwkvGIWSQaFTznLOctUXNuzzBBOlqvzmVfTSejekTAlwidRrsxkbZTsGGeEWxCXHzqWVuLGoCyrGjKkQoHqduXwYQKC",
"output": "-1"
},
{
"input": "OjYwwNuPESIazoyLFREpObIaMKhCaKAMWMfRGgucEuyNYRantwdwQkmflzfqbcFRaXBnZoIUGsFqXZHGKwlaBUXABBcQEWWPvkjW\nRxLqGcTTpBwHrHltCOllnTpRKLDofBUqqHxnOtVWPgvGaeHIevgUSOeeDOJubfqonFpVNGVbHFcAhjnyFvrrqnRgKhkYqQZmRfUl",
"output": "-1"
},
{
"input": "tatuhQPIzjptlzzJpCAPXSRTKZRlwgfoCIsFjJquRoIDyZZYRSPdFUTjjUPhLBBfeEIfLQpygKXRcyQFiQsEtRtLnZErBqW\ntkHUjllbafLUWhVCnvblKjgYIEoHhsjVmrDBmAWbvtkHxDbRFvsXAjHIrujaDbYwOZmacknhZPeCcorbRgHjjgAgoJdjvLo",
"output": "-1"
},
{
"input": "cymCPGqdXKUdADEWDdUaLEEMHiXHsdAZuDnJDMUvxvrLRBrPSDpXPAgMRoGplLtniFRTomDTAHXWAdgUveTxaqKVSvnOyhOwiRN\nuhmyEWzapiRNPFDisvHTbenXMfeZaHqOFlKjrfQjUBwdFktNpeiRoDWuBftZLcCZZAVfioOihZVNqiNCNDIsUdIhvbcaxpTRWoV",
"output": "-1"
},
{
"input": "sSvpcITJAwghVfJaLKBmyjOkhltTGjYJVLWCYMFUomiJaKQYhXTajvZVHIMHbyckYROGQZzjWyWCcnmDmrkvTKfHSSzCIhsXgEZa\nvhCXkCwAmErGVBPBAnkSYEYvseFKbWSktoqaHYXUmYkHfOkRwuEyBRoGoBrOXBKVxXycjZGStuvDarnXMbZLWrbjrisDoJBdSvWJ",
"output": "-1"
},
{
"input": "hJDANKUNBisOOINDsTixJmYgHNogtpwswwcvVMptfGwIjvqgwTYFcqTdyAqaqlnhOCMtsnWXQqtjFwQlEcBtMFAtSqnqthVb\nrNquIcjNWESjpPVWmzUJFrelpUZeGDmSvCurCqVmKHKVAAPkaHksniOlzjiKYIJtvbuQWZRufMebpTFPqyxIWWjfPaWYiNlK",
"output": "-1"
},
{
"input": "ycLoapxsfsDTHMSfAAPIUpiEhQKUIXUcXEiopMBuuZLHtfPpLmCHwNMNQUwsEXxCEmKHTBSnKhtQhGWUvppUFZUgSpbeChX\ndCZhgVXofkGousCzObxZSJwXcHIaqUDSCPKzXntcVmPxtNcXmVcjsetZYxedmgQzXTZHMvzjoaXCMKsncGciSDqQWIIRlys",
"output": "1"
},
{
"input": "nvUbnrywIePXcoukIhwTfUVcHUEgXcsMyNQhmMlTltZiCooyZiIKRIGVHMCnTKgzXXIuvoNDEZswKoACOBGSyVNqTNQqMhAG\nplxuGSsyyJjdvpddrSebOARSAYcZKEaKjqbCwvjhNykuaECoQVHTVFMKXwvrQXRaqXsHsBaGVhCxGRxNyGUbMlxOarMZNXxy",
"output": "-1"
},
{
"input": "EncmXtAblQzcVRzMQqdDqXfAhXbtJKQwZVWyHoWUckohnZqfoCmNJDzexFgFJYrwNHGgzCJTzQQFnxGlhmvQTpicTkEeVICKac\nNIUNZoMLFMyAjVgQLITELJSodIXcGSDWfhFypRoGYuogJpnqGTotWxVqpvBHjFOWcDRDtARsaHarHaOkeNWEHGTaGOFCOFEwvK",
"output": "-1"
},
{
"input": "UG\nak",
"output": "1"
},
{
"input": "JZR\nVae",
"output": "-1"
},
{
"input": "a\nZ",
"output": "-1"
},
{
"input": "rk\nkv",
"output": "1"
},
{
"input": "RvuT\nbJzE",
"output": "1"
},
{
"input": "PPS\nydq",
"output": "-1"
},
{
"input": "q\nq",
"output": "0"
},
{
"input": "peOw\nIgSJ",
"output": "1"
},
{
"input": "PyK\noKN",
"output": "1"
},
{
"input": "O\ni",
"output": "1"
},
{
"input": "NmGY\npDlP",
"output": "-1"
},
{
"input": "nG\nZf",
"output": "-1"
},
{
"input": "m\na",
"output": "1"
},
{
"input": "MWyB\nWZEV",
"output": "-1"
},
{
"input": "Gre\nfxc",
"output": "1"
},
{
"input": "Ooq\nwap",
"output": "-1"
},
{
"input": "XId\nlbB",
"output": "1"
},
{
"input": "lfFpECEqUMEOJhipvkZjDPcpDNJedOVXiSMgBvBZbtfzIKekcvpWPCazKAhJyHircRtgcBIJwwstpHaLAgxFOngAWUZRgCef\nLfFPEcequmeojHIpVkzjDPcpdNJEDOVXiSmGBVBZBtfZikEKcvPwpCAzKAHJyHIrCRTgCbIJWwSTphALagXfOnGAwUzRGcEF",
"output": "0"
},
{
"input": "DQBdtSEDtFGiNRUeJNbOIfDZnsryUlzJHGTXGFXnwsVyxNtLgmklmFvRCzYETBVdmkpJJIvIOkMDgCFHZOTODiYrkwXd\nDQbDtsEdTFginRUEJNBOIfdZnsryulZJHGtxGFxnwSvYxnTLgmKlmFVRCzyEtBVdmKpJjiVioKMDgCFhzoTODiYrKwXD",
"output": "0"
},
{
"input": "tYWRijFQSzHBpCjUzqBtNvBKyzZRnIdWEuyqnORBQTLyOQglIGfYJIRjuxnbLvkqZakNqPiGDvgpWYkfxYNXsdoKXZtRkSasfa\nTYwRiJfqsZHBPcJuZQBTnVbkyZZRnidwEuYQnorbQTLYOqGligFyjirJUxnblVKqZaknQpigDVGPwyKfxyNXSDoKxztRKSaSFA",
"output": "0"
},
{
"input": "KhScXYiErQIUtmVhNTCXSLAviefIeHIIdiGhsYnPkSBaDTvMkyanfMLBOvDWgRybLtDqvXVdVjccNunDyijhhZEAKBrdz\nkHsCXyiErqIuTMVHNTCxSLaViEFIEhIIDiGHsYNpKsBAdTvMKyANFMLBovdwGRYbLtdQVxvDVJCcNUndYiJHhzeakBrdZ",
"output": "0"
},
{
"input": "cpPQMpjRQJKQVXjWDYECXbagSmNcVfOuBWNZxihdERraVuiOpSVDCPgTGuSQALNoVjySceHcKXwOEpSzXrEqWwwrYeppNiWhDVg\nCPPqmPjRqJkQvxJwdyECXBAGsMNcVfOuBWNzxIhderRavUiOpSvDCpGTgusqAlNovjyScEhCKXwoePSZxrEQwWwryEPPniWHDvG",
"output": "0"
},
{
"input": "SajcCGMepaLjZIWLRBGFcrZRCRvvoCsIyKsQerbrwsIamxxpRmQSZSalasJLVFbCHCuXJlubciQAvLxXYBazLsMKLHLdDQ\nsaJcCgmEpaLJziWlrBgFcRzrCrVVOcSIykSQerBrwSIamxxPrMqSzSalASjLVFbChCUxjLUbCIQAVlxxybAZLsmkLhLDdQ",
"output": "0"
},
{
"input": "kigPrWNTOUNDBskAfefjhHYZNYdnfZWuXWzHiBxFQryBbAkPtenFwWvCSTYGpzOntUNzNUhxRWjKmicTwLwJAnbAxj\nkigpRWntOUNdBsKaFEFjhhYZnYDNfzWuXwZhibxFQRybbakPteNfwwvcStyGPzoNTunznuHXrWjKMIctWLWJANBAxJ",
"output": "0"
},
{
"input": "nTomZZuTTRTAAPoUsySVFGElrpQRNLjqvFmcYytiheQnjUhPLnqNBiYtQkljbcvmjuNAVKbvQOWpqqFlQhAhULIhquoCnjUI\nntOmzZuttrtAAPOUSySVFgeLRPQrNLjQvfmCyYTiHEQnjuHPlNQNbIYtqKLJBCVmjunavkbvQOWPQQFlqHaHULIHQuOcnJUi",
"output": "0"
},
{
"input": "abac\nadaa",
"output": "-1"
},
{
"input": "Bbc\nabc",
"output": "1"
},
{
"input": "aaaba\naaaab",
"output": "1"
}
] | 1,695,174,378
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 53
| 92
| 0
|
a = input().lower()
b = input().lower()
abi = True
for i in range(len(a)):
if a[i] > b[i]:
print('1')
abi = False
break
elif a[i] < b[i]:
print('-1')
abi = False
break
if abi:
print('0')
|
Title: Petya and Strings
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Little Petya loves presents. His mum bought him two strings of the same size for his birthday. The strings consist of uppercase and lowercase Latin letters. Now Petya wants to compare those two strings lexicographically. The letters' case does not matter, that is an uppercase letter is considered equivalent to the corresponding lowercase letter. Help Petya perform the comparison.
Input Specification:
Each of the first two lines contains a bought string. The strings' lengths range from 1 to 100 inclusive. It is guaranteed that the strings are of the same length and also consist of uppercase and lowercase Latin letters.
Output Specification:
If the first string is less than the second one, print "-1". If the second string is less than the first one, print "1". If the strings are equal, print "0". Note that the letters' case is not taken into consideration when the strings are compared.
Demo Input:
['aaaa\naaaA\n', 'abs\nAbz\n', 'abcdefg\nAbCdEfF\n']
Demo Output:
['0\n', '-1\n', '1\n']
Note:
If you want more formal information about the lexicographical order (also known as the "dictionary order" or "alphabetical order"), you can visit the following site:
- http://en.wikipedia.org/wiki/Lexicographical_order
|
```python
a = input().lower()
b = input().lower()
abi = True
for i in range(len(a)):
if a[i] > b[i]:
print('1')
abi = False
break
elif a[i] < b[i]:
print('-1')
abi = False
break
if abi:
print('0')
```
| 3.977
|
847
|
M
|
Weather Tomorrow
|
PROGRAMMING
| 1,000
|
[
"implementation",
"math"
] | null | null |
Vasya came up with his own weather forecasting method. He knows the information about the average air temperature for each of the last *n* days. Assume that the average air temperature for each day is integral.
Vasya believes that if the average temperatures over the last *n* days form an arithmetic progression, where the first term equals to the average temperature on the first day, the second term equals to the average temperature on the second day and so on, then the average temperature of the next (*n*<=+<=1)-th day will be equal to the next term of the arithmetic progression. Otherwise, according to Vasya's method, the temperature of the (*n*<=+<=1)-th day will be equal to the temperature of the *n*-th day.
Your task is to help Vasya predict the average temperature for tomorrow, i. e. for the (*n*<=+<=1)-th day.
|
The first line contains a single integer *n* (2<=β€<=*n*<=β€<=100) β the number of days for which the average air temperature is known.
The second line contains a sequence of integers *t*1,<=*t*2,<=...,<=*t**n* (<=-<=1000<=β€<=*t**i*<=β€<=1000)Β β where *t**i* is the average temperature in the *i*-th day.
|
Print the average air temperature in the (*n*<=+<=1)-th day, which Vasya predicts according to his method. Note that the absolute value of the predicted temperature can exceed 1000.
|
[
"5\n10 5 0 -5 -10\n",
"4\n1 1 1 1\n",
"3\n5 1 -5\n",
"2\n900 1000\n"
] |
[
"-15\n",
"1\n",
"-5\n",
"1100\n"
] |
In the first example the sequence of the average temperatures is an arithmetic progression where the first term is 10 and each following terms decreases by 5. So the predicted average temperature for the sixth day is β-β10β-β5β=ββ-β15.
In the second example the sequence of the average temperatures is an arithmetic progression where the first term is 1 and each following terms equals to the previous one. So the predicted average temperature in the fifth day is 1.
In the third example the average temperatures do not form an arithmetic progression, so the average temperature of the fourth day equals to the temperature of the third day and equals to β-β5.
In the fourth example the sequence of the average temperatures is an arithmetic progression where the first term is 900 and each the following terms increase by 100. So predicted average temperature in the third day is 1000β+β100β=β1100.
| 0
|
[
{
"input": "5\n10 5 0 -5 -10",
"output": "-15"
},
{
"input": "4\n1 1 1 1",
"output": "1"
},
{
"input": "3\n5 1 -5",
"output": "-5"
},
{
"input": "2\n900 1000",
"output": "1100"
},
{
"input": "2\n1 2",
"output": "3"
},
{
"input": "3\n2 5 8",
"output": "11"
},
{
"input": "4\n4 1 -2 -5",
"output": "-8"
},
{
"input": "10\n-1000 -995 -990 -985 -980 -975 -970 -965 -960 -955",
"output": "-950"
},
{
"input": "11\n-1000 -800 -600 -400 -200 0 200 400 600 800 1000",
"output": "1200"
},
{
"input": "31\n1000 978 956 934 912 890 868 846 824 802 780 758 736 714 692 670 648 626 604 582 560 538 516 494 472 450 428 406 384 362 340",
"output": "318"
},
{
"input": "5\n1000 544 88 -368 -824",
"output": "-1280"
},
{
"input": "100\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0",
"output": "0"
},
{
"input": "33\n456 411 366 321 276 231 186 141 96 51 6 -39 -84 -129 -174 -219 -264 -309 -354 -399 -444 -489 -534 -579 -624 -669 -714 -759 -804 -849 -894 -939 -984",
"output": "-1029"
},
{
"input": "77\n-765 -742 -719 -696 -673 -650 -627 -604 -581 -558 -535 -512 -489 -466 -443 -420 -397 -374 -351 -328 -305 -282 -259 -236 -213 -190 -167 -144 -121 -98 -75 -52 -29 -6 17 40 63 86 109 132 155 178 201 224 247 270 293 316 339 362 385 408 431 454 477 500 523 546 569 592 615 638 661 684 707 730 753 776 799 822 845 868 891 914 937 960 983",
"output": "1006"
},
{
"input": "3\n2 4 8",
"output": "8"
},
{
"input": "4\n4 1 -3 -5",
"output": "-5"
},
{
"input": "10\n-1000 -995 -990 -984 -980 -975 -970 -965 -960 -955",
"output": "-955"
},
{
"input": "11\n-999 -800 -600 -400 -200 0 200 400 600 800 1000",
"output": "1000"
},
{
"input": "51\n-9 10 30 50 70 90 110 130 150 170 190 210 230 250 270 290 310 330 350 370 390 410 430 450 470 490 510 530 550 570 590 610 630 650 670 690 710 730 750 770 790 810 830 850 870 890 910 930 950 970 990",
"output": "990"
},
{
"input": "100\n10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50 52 54 56 58 60 62 64 66 68 70 72 74 76 78 80 82 84 86 88 90 92 94 96 98 100 102 104 106 108 110 112 114 116 118 120 122 124 126 128 130 132 134 136 138 140 142 144 146 148 150 152 154 156 158 160 162 164 166 168 170 172 174 176 178 180 182 184 186 188 190 192 194 196 198 200 202 204 206 207",
"output": "207"
},
{
"input": "2\n1000 1000",
"output": "1000"
},
{
"input": "2\n-1000 1000",
"output": "3000"
},
{
"input": "2\n1000 -1000",
"output": "-3000"
},
{
"input": "2\n-1000 -1000",
"output": "-1000"
},
{
"input": "100\n-85 -80 -76 -72 -68 -64 -60 -56 -52 -48 -44 -40 -36 -32 -28 -24 -20 -16 -12 -8 -4 0 4 8 12 16 20 24 28 32 36 40 44 48 52 56 60 64 68 72 76 80 84 88 92 96 100 104 108 112 116 120 124 128 132 136 140 144 148 152 156 160 164 168 172 176 180 184 188 192 196 200 204 208 212 216 220 224 228 232 236 240 244 248 252 256 260 264 268 272 276 280 284 288 292 296 300 304 308 312",
"output": "312"
},
{
"input": "4\n1 2 4 5",
"output": "5"
}
] | 1,680,878,020
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 26
| 46
| 0
|
n = int(input())
vals = list(map(int,(input().split())))
arith = True
diff = vals[0]-vals[1]
for i in range(n-1):
if vals[i]-vals[i+1]!=diff:
arith = False
break
if arith:
print(vals[-1]-diff)
else:
print(vals[-1])
|
Title: Weather Tomorrow
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya came up with his own weather forecasting method. He knows the information about the average air temperature for each of the last *n* days. Assume that the average air temperature for each day is integral.
Vasya believes that if the average temperatures over the last *n* days form an arithmetic progression, where the first term equals to the average temperature on the first day, the second term equals to the average temperature on the second day and so on, then the average temperature of the next (*n*<=+<=1)-th day will be equal to the next term of the arithmetic progression. Otherwise, according to Vasya's method, the temperature of the (*n*<=+<=1)-th day will be equal to the temperature of the *n*-th day.
Your task is to help Vasya predict the average temperature for tomorrow, i. e. for the (*n*<=+<=1)-th day.
Input Specification:
The first line contains a single integer *n* (2<=β€<=*n*<=β€<=100) β the number of days for which the average air temperature is known.
The second line contains a sequence of integers *t*1,<=*t*2,<=...,<=*t**n* (<=-<=1000<=β€<=*t**i*<=β€<=1000)Β β where *t**i* is the average temperature in the *i*-th day.
Output Specification:
Print the average air temperature in the (*n*<=+<=1)-th day, which Vasya predicts according to his method. Note that the absolute value of the predicted temperature can exceed 1000.
Demo Input:
['5\n10 5 0 -5 -10\n', '4\n1 1 1 1\n', '3\n5 1 -5\n', '2\n900 1000\n']
Demo Output:
['-15\n', '1\n', '-5\n', '1100\n']
Note:
In the first example the sequence of the average temperatures is an arithmetic progression where the first term is 10 and each following terms decreases by 5. So the predicted average temperature for the sixth day is β-β10β-β5β=ββ-β15.
In the second example the sequence of the average temperatures is an arithmetic progression where the first term is 1 and each following terms equals to the previous one. So the predicted average temperature in the fifth day is 1.
In the third example the average temperatures do not form an arithmetic progression, so the average temperature of the fourth day equals to the temperature of the third day and equals to β-β5.
In the fourth example the sequence of the average temperatures is an arithmetic progression where the first term is 900 and each the following terms increase by 100. So predicted average temperature in the third day is 1000β+β100β=β1100.
|
```python
n = int(input())
vals = list(map(int,(input().split())))
arith = True
diff = vals[0]-vals[1]
for i in range(n-1):
if vals[i]-vals[i+1]!=diff:
arith = False
break
if arith:
print(vals[-1]-diff)
else:
print(vals[-1])
```
| 3
|
|
129
|
A
|
Cookies
|
PROGRAMMING
| 900
|
[
"implementation"
] | null | null |
Olga came to visit the twins Anna and Maria and saw that they have many cookies. The cookies are distributed into bags. As there are many cookies, Olga decided that it's no big deal if she steals a bag. However, she doesn't want the sisters to quarrel because of nothing when they divide the cookies. That's why Olga wants to steal a bag with cookies so that the number of cookies in the remaining bags was even, that is, so that Anna and Maria could evenly divide it into two (even 0 remaining cookies will do, just as any other even number). How many ways there are to steal exactly one cookie bag so that the total number of cookies in the remaining bags was even?
|
The first line contains the only integer *n* (1<=β€<=*n*<=β€<=100) β the number of cookie bags Anna and Maria have. The second line contains *n* integers *a**i* (1<=β€<=*a**i*<=β€<=100) β the number of cookies in the *i*-th bag.
|
Print in the only line the only number β the sought number of ways. If there are no such ways print 0.
|
[
"1\n1\n",
"10\n1 2 2 3 4 4 4 2 2 2\n",
"11\n2 2 2 2 2 2 2 2 2 2 99\n"
] |
[
"1\n",
"8\n",
"1\n"
] |
In the first sample Olga should take the only bag so that the twins ended up with the even number of cookies.
In the second sample Olga can take any of five bags with two cookies or any of three bags with four cookies β 5β+β3β=β8 ways in total.
In the third sample, no matter which bag with two cookies Olga chooses, the twins are left with 2β*β9β+β99β=β117 cookies. Thus, Olga has only one option: to take the bag with 99 cookies.
| 500
|
[
{
"input": "1\n1",
"output": "1"
},
{
"input": "10\n1 2 2 3 4 4 4 2 2 2",
"output": "8"
},
{
"input": "11\n2 2 2 2 2 2 2 2 2 2 99",
"output": "1"
},
{
"input": "2\n1 1",
"output": "0"
},
{
"input": "2\n2 2",
"output": "2"
},
{
"input": "2\n1 2",
"output": "1"
},
{
"input": "7\n7 7 7 7 7 7 7",
"output": "7"
},
{
"input": "8\n1 2 3 4 5 6 7 8",
"output": "4"
},
{
"input": "100\n1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2",
"output": "50"
},
{
"input": "99\n99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99",
"output": "49"
},
{
"input": "82\n43 44 96 33 23 42 33 66 53 87 8 90 43 91 40 88 51 18 48 62 59 10 22 20 54 6 13 63 2 56 31 52 98 42 54 32 26 77 9 24 33 91 16 30 39 34 78 82 73 90 12 15 67 76 30 18 44 86 84 98 65 54 100 79 28 34 40 56 11 43 72 35 86 59 89 40 30 33 7 19 44 15",
"output": "50"
},
{
"input": "17\n50 14 17 77 74 74 38 76 41 27 45 29 66 98 38 73 38",
"output": "7"
},
{
"input": "94\n81 19 90 99 26 11 86 44 78 36 80 59 99 90 78 72 71 20 94 56 42 40 71 84 10 85 10 70 52 27 39 55 90 16 48 25 7 79 99 100 38 10 99 56 3 4 78 9 16 57 14 40 52 54 57 70 30 86 56 84 97 60 59 69 49 66 23 92 90 46 86 73 53 47 1 83 14 20 24 66 13 45 41 14 86 75 55 88 48 95 82 24 47 87",
"output": "39"
},
{
"input": "88\n64 95 12 90 40 65 98 45 52 54 79 7 81 25 98 19 68 82 41 53 35 50 5 22 32 21 8 39 8 6 72 27 81 30 12 79 21 42 60 2 66 87 46 93 62 78 52 71 76 32 78 94 86 85 55 15 34 76 41 20 32 26 94 81 89 45 74 49 11 40 40 39 49 46 80 85 90 23 80 40 86 58 70 26 48 93 23 53",
"output": "37"
},
{
"input": "84\n95 9 43 43 13 84 60 90 1 8 97 99 54 34 59 83 33 15 51 26 40 12 66 65 19 30 29 78 92 60 25 13 19 84 71 73 12 24 54 49 16 41 11 40 57 59 34 40 39 9 71 83 1 77 79 53 94 47 78 55 77 85 29 52 80 90 53 77 97 97 27 79 28 23 83 25 26 22 49 86 63 56 3 32",
"output": "51"
},
{
"input": "47\n61 97 76 94 91 22 2 68 62 73 90 47 16 79 44 71 98 68 43 6 53 52 40 27 68 67 43 96 14 91 60 61 96 24 97 13 32 65 85 96 81 77 34 18 23 14 80",
"output": "21"
},
{
"input": "69\n71 1 78 74 58 89 30 6 100 90 22 61 11 59 14 74 27 25 78 61 45 19 25 33 37 4 52 43 53 38 9 100 56 67 69 38 76 91 63 60 93 52 28 61 9 98 8 14 57 63 89 64 98 51 36 66 36 86 13 82 50 91 52 64 86 78 78 83 81",
"output": "37"
},
{
"input": "52\n38 78 36 75 19 3 56 1 39 97 24 79 84 16 93 55 96 64 12 24 1 86 80 29 12 32 36 36 73 39 76 65 53 98 30 20 28 8 86 43 70 22 75 69 62 65 81 25 53 40 71 59",
"output": "28"
},
{
"input": "74\n81 31 67 97 26 75 69 81 11 13 13 74 77 88 52 20 52 64 66 75 72 28 41 54 26 75 41 91 75 15 18 36 13 83 63 61 14 48 53 63 19 67 35 48 23 65 73 100 44 55 92 88 99 17 73 25 83 7 31 89 12 80 98 39 42 75 14 29 81 35 77 87 33 94",
"output": "47"
},
{
"input": "44\n46 56 31 31 37 71 94 2 14 100 45 72 36 72 80 3 38 54 42 98 50 32 31 42 62 31 45 50 95 100 18 17 64 22 18 25 52 56 70 57 43 40 81 28",
"output": "15"
},
{
"input": "22\n28 57 40 74 51 4 45 84 99 12 95 14 92 60 47 81 84 51 31 91 59 42",
"output": "11"
},
{
"input": "59\n73 45 94 76 41 49 65 13 74 66 36 25 47 75 40 23 92 72 11 32 32 8 81 26 68 56 41 8 76 47 96 55 70 11 84 14 83 18 70 22 30 39 28 100 48 11 92 45 78 69 86 1 54 90 98 91 13 17 35",
"output": "33"
},
{
"input": "63\n20 18 44 94 68 57 16 43 74 55 68 24 21 95 76 84 50 50 47 86 86 12 58 55 28 72 86 18 34 45 81 88 3 72 41 9 60 90 81 93 12 6 9 6 2 41 1 7 9 29 81 14 64 80 20 36 67 54 7 5 35 81 22",
"output": "37"
},
{
"input": "28\n49 84 48 19 44 91 11 82 96 95 88 90 71 82 87 25 31 23 18 13 98 45 26 65 35 12 31 14",
"output": "15"
},
{
"input": "61\n34 18 28 64 28 45 9 77 77 20 63 92 79 16 16 100 86 2 91 91 57 15 31 95 10 88 84 5 82 83 53 98 59 17 97 80 76 80 81 3 91 81 87 93 61 46 10 49 6 22 21 75 63 89 21 81 30 19 67 38 77",
"output": "35"
},
{
"input": "90\n41 90 43 1 28 75 90 50 3 70 76 64 81 63 25 69 83 82 29 91 59 66 21 61 7 55 72 49 38 69 72 20 64 58 30 81 61 29 96 14 39 5 100 20 29 98 75 29 44 78 97 45 26 77 73 59 22 99 41 6 3 96 71 20 9 18 96 18 90 62 34 78 54 5 41 6 73 33 2 54 26 21 18 6 45 57 43 73 95 75",
"output": "42"
},
{
"input": "45\n93 69 4 27 20 14 71 48 79 3 32 26 49 30 57 88 13 56 49 61 37 32 47 41 41 70 45 68 82 18 8 6 25 20 15 13 71 99 28 6 52 34 19 59 26",
"output": "23"
},
{
"input": "33\n29 95 48 49 91 10 83 71 47 25 66 36 51 12 34 10 54 74 41 96 89 26 89 1 42 33 1 62 9 32 49 65 78",
"output": "15"
},
{
"input": "34\n98 24 42 36 41 82 28 58 89 34 77 70 76 44 74 54 66 100 13 79 4 88 21 1 11 45 91 29 87 100 29 54 82 78",
"output": "13"
},
{
"input": "29\n91 84 26 84 9 63 52 9 65 56 90 2 36 7 67 33 91 14 65 38 53 36 81 83 85 14 33 95 51",
"output": "17"
},
{
"input": "100\n2 88 92 82 87 100 78 28 84 43 78 32 43 33 97 19 15 52 29 84 57 72 54 13 99 28 82 79 40 70 34 92 91 53 9 88 27 43 14 92 72 37 26 37 20 95 19 34 49 64 33 37 34 27 80 79 9 54 99 68 25 4 68 73 46 66 24 78 3 87 26 52 50 84 4 95 23 83 39 58 86 36 33 16 98 2 84 19 53 12 69 60 10 11 78 17 79 92 77 59",
"output": "45"
},
{
"input": "100\n2 95 45 73 9 54 20 97 57 82 88 26 18 71 25 27 75 54 31 11 58 85 69 75 72 91 76 5 25 80 45 49 4 73 8 81 81 38 5 12 53 77 7 96 90 35 28 80 73 94 19 69 96 17 94 49 69 9 32 19 5 12 46 29 26 40 59 59 6 95 82 50 72 2 45 69 12 5 72 29 39 72 23 96 81 28 28 56 68 58 37 41 30 1 90 84 15 24 96 43",
"output": "53"
},
{
"input": "100\n27 72 35 91 13 10 35 45 24 55 83 84 63 96 29 79 34 67 63 92 48 83 18 77 28 27 49 66 29 88 55 15 6 58 14 67 94 36 77 7 7 64 61 52 71 18 36 99 76 6 50 67 16 13 41 7 89 73 61 51 78 22 78 32 76 100 3 31 89 71 63 53 15 85 77 54 89 33 68 74 3 23 57 5 43 89 75 35 9 86 90 11 31 46 48 37 74 17 77 8",
"output": "40"
},
{
"input": "100\n69 98 69 88 11 49 55 8 25 91 17 81 47 26 15 73 96 71 18 42 42 61 48 14 92 78 35 72 4 27 62 75 83 79 17 16 46 80 96 90 82 54 37 69 85 21 67 70 96 10 46 63 21 59 56 92 54 88 77 30 75 45 44 29 86 100 51 11 65 69 66 56 82 63 27 1 51 51 13 10 3 55 26 85 34 16 87 72 13 100 81 71 90 95 86 50 83 55 55 54",
"output": "53"
},
{
"input": "100\n34 35 99 64 2 66 78 93 20 48 12 79 19 10 87 7 42 92 60 79 5 2 24 89 57 48 63 92 74 4 16 51 7 12 90 48 87 17 18 73 51 58 97 97 25 38 15 97 96 73 67 91 6 75 14 13 87 79 75 3 15 55 35 95 71 45 10 13 20 37 82 26 2 22 13 83 97 84 39 79 43 100 54 59 98 8 61 34 7 65 75 44 24 77 73 88 34 95 44 77",
"output": "55"
},
{
"input": "100\n15 86 3 1 51 26 74 85 37 87 64 58 10 6 57 26 30 47 85 65 24 72 50 40 12 35 91 47 91 60 47 87 95 34 80 91 26 3 36 39 14 86 28 70 51 44 28 21 72 79 57 61 16 71 100 94 57 67 36 74 24 21 89 85 25 2 97 67 76 53 76 80 97 64 35 13 8 32 21 52 62 61 67 14 74 73 66 44 55 76 24 3 43 42 99 61 36 80 38 66",
"output": "52"
},
{
"input": "100\n45 16 54 54 80 94 74 93 75 85 58 95 79 30 81 2 84 4 57 23 92 64 78 1 50 36 13 27 56 54 10 77 87 1 5 38 85 74 94 82 30 45 72 83 82 30 81 82 82 3 69 82 7 92 39 60 94 42 41 5 3 17 67 21 79 44 79 96 28 3 53 68 79 89 63 83 1 44 4 31 84 15 73 77 19 66 54 6 73 1 67 24 91 11 86 45 96 82 20 89",
"output": "51"
},
{
"input": "100\n84 23 50 32 90 71 92 43 58 70 6 82 7 55 85 19 70 89 12 26 29 56 74 30 2 27 4 39 63 67 91 81 11 33 75 10 82 88 39 43 43 80 68 35 55 67 53 62 73 65 86 74 43 51 14 48 42 92 83 57 22 33 24 99 5 27 78 96 7 28 11 15 8 38 85 67 5 92 24 96 57 59 14 95 91 4 9 18 45 33 74 83 64 85 14 51 51 94 29 2",
"output": "53"
},
{
"input": "100\n77 56 56 45 73 55 32 37 39 50 30 95 79 21 44 34 51 43 86 91 39 30 85 15 35 93 100 14 57 31 80 79 38 40 88 4 91 54 7 95 76 26 62 84 17 33 67 47 6 82 69 51 17 2 59 24 11 12 31 90 12 11 55 38 72 49 30 50 42 46 5 97 9 9 30 45 86 23 19 82 40 42 5 40 35 98 35 32 60 60 5 28 84 35 21 49 68 53 68 23",
"output": "48"
},
{
"input": "100\n78 38 79 61 45 86 83 83 86 90 74 69 2 84 73 39 2 5 20 71 24 80 54 89 58 34 77 40 39 62 2 47 28 53 97 75 88 98 94 96 33 71 44 90 47 36 19 89 87 98 90 87 5 85 34 79 82 3 42 88 89 63 35 7 89 30 40 48 12 41 56 76 83 60 80 80 39 56 77 4 72 96 30 55 57 51 7 19 11 1 66 1 91 87 11 62 95 85 79 25",
"output": "48"
},
{
"input": "100\n5 34 23 20 76 75 19 51 17 82 60 13 83 6 65 16 20 43 66 54 87 10 87 73 50 24 16 98 33 28 80 52 54 82 26 92 14 13 84 92 94 29 61 21 60 20 48 94 24 20 75 70 58 27 68 45 86 89 29 8 67 38 83 48 18 100 11 22 46 84 52 97 70 19 50 75 3 7 52 53 72 41 18 31 1 38 49 53 11 64 99 76 9 87 48 12 100 32 44 71",
"output": "58"
},
{
"input": "100\n76 89 68 78 24 72 73 95 98 72 58 15 2 5 56 32 9 65 50 70 94 31 29 54 89 52 31 93 43 56 26 35 72 95 51 55 78 70 11 92 17 5 54 94 81 31 78 95 73 91 95 37 59 9 53 48 65 55 84 8 45 97 64 37 96 34 36 53 66 17 72 48 99 23 27 18 92 84 44 73 60 78 53 29 68 99 19 39 61 40 69 6 77 12 47 29 15 4 8 45",
"output": "53"
},
{
"input": "100\n82 40 31 53 8 50 85 93 3 84 54 17 96 59 51 42 18 19 35 84 79 31 17 46 54 82 72 49 35 73 26 89 61 73 3 50 12 29 25 77 88 21 58 24 22 89 96 54 82 29 96 56 77 16 1 68 90 93 20 23 57 22 31 18 92 90 51 14 50 72 31 54 12 50 66 62 2 34 17 45 68 50 87 97 23 71 1 72 17 82 42 15 20 78 4 49 66 59 10 17",
"output": "54"
},
{
"input": "100\n32 82 82 24 39 53 48 5 29 24 9 37 91 37 91 95 1 97 84 52 12 56 93 47 22 20 14 17 40 22 79 34 24 2 69 30 69 29 3 89 21 46 60 92 39 29 18 24 49 18 40 22 60 13 77 50 39 64 50 70 99 8 66 31 90 38 20 54 7 21 5 56 41 68 69 20 54 89 69 62 9 53 43 89 81 97 15 2 52 78 89 65 16 61 59 42 56 25 32 52",
"output": "49"
},
{
"input": "100\n72 54 23 24 97 14 99 87 15 25 7 23 17 87 72 31 71 87 34 82 51 77 74 85 62 38 24 7 84 48 98 21 29 71 70 84 25 58 67 92 18 44 32 9 81 15 53 29 63 18 86 16 7 31 38 99 70 32 89 16 23 11 66 96 69 82 97 59 6 9 49 80 85 19 6 9 52 51 85 74 53 46 73 55 31 63 78 61 34 80 77 65 87 77 92 52 89 8 52 31",
"output": "44"
},
{
"input": "100\n56 88 8 19 7 15 11 54 35 50 19 57 63 72 51 43 50 19 57 90 40 100 8 92 11 96 30 32 59 65 93 47 62 3 50 41 30 50 72 83 61 46 83 60 20 46 33 1 5 18 83 22 34 16 41 95 63 63 7 59 55 95 91 29 64 60 64 81 45 45 10 9 88 37 69 85 21 82 41 76 42 34 47 78 51 83 65 100 13 22 59 76 63 1 26 86 36 94 99 74",
"output": "46"
},
{
"input": "100\n27 89 67 60 62 80 43 50 28 88 72 5 94 11 63 91 18 78 99 3 71 26 12 97 74 62 23 24 22 3 100 72 98 7 94 32 12 75 61 88 42 48 10 14 45 9 48 56 73 76 70 70 79 90 35 39 96 37 81 11 19 65 99 39 23 79 34 61 35 74 90 37 73 23 46 21 94 84 73 58 11 89 13 9 10 85 42 78 73 32 53 39 49 90 43 5 28 31 97 75",
"output": "53"
},
{
"input": "100\n33 24 97 96 1 14 99 51 13 65 67 20 46 88 42 44 20 49 5 89 98 83 15 40 74 83 58 3 10 79 34 2 69 28 37 100 55 52 14 8 44 94 97 89 6 42 11 28 30 33 55 56 20 57 52 25 75 1 87 42 62 41 37 12 54 85 95 80 42 36 94 96 28 76 54 36 4 17 26 24 62 15 17 79 84 36 92 78 74 91 96 77 54 92 81 91 62 98 37 37",
"output": "43"
},
{
"input": "100\n86 24 61 15 11 85 1 31 47 36 23 36 59 34 3 27 16 29 82 28 58 52 52 66 71 61 98 39 60 20 67 41 67 90 73 29 92 17 70 95 58 98 58 32 21 73 46 56 87 72 80 75 40 27 94 31 59 92 93 37 14 99 96 21 97 23 81 91 52 52 96 94 92 28 38 29 52 16 57 27 17 24 91 21 79 55 96 98 95 94 23 78 79 12 77 35 32 75 100 82",
"output": "51"
},
{
"input": "100\n88 85 41 37 69 21 7 69 36 5 92 26 64 75 22 46 67 20 70 22 62 66 38 24 47 49 68 30 90 31 67 86 86 82 9 51 43 45 48 42 73 44 31 94 45 60 54 66 20 87 11 94 34 32 87 66 56 28 75 39 37 90 72 93 55 72 31 42 30 71 87 61 4 12 12 81 23 61 56 98 71 32 30 33 96 63 92 16 8 78 47 91 47 54 49 3 81 82 41 5",
"output": "47"
},
{
"input": "1\n2",
"output": "1"
},
{
"input": "5\n1 1 3 2 2",
"output": "3"
}
] | 1,638,820,906
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 52
| 122
| 0
|
n=int(input())
l=list(map(int,input().split()))
Sum=sum(l)
if(Sum%2==0):
print(len([i for i in l if i%2==0]))
else:
print(len([i for i in l if(i%2==1)]))
|
Title: Cookies
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Olga came to visit the twins Anna and Maria and saw that they have many cookies. The cookies are distributed into bags. As there are many cookies, Olga decided that it's no big deal if she steals a bag. However, she doesn't want the sisters to quarrel because of nothing when they divide the cookies. That's why Olga wants to steal a bag with cookies so that the number of cookies in the remaining bags was even, that is, so that Anna and Maria could evenly divide it into two (even 0 remaining cookies will do, just as any other even number). How many ways there are to steal exactly one cookie bag so that the total number of cookies in the remaining bags was even?
Input Specification:
The first line contains the only integer *n* (1<=β€<=*n*<=β€<=100) β the number of cookie bags Anna and Maria have. The second line contains *n* integers *a**i* (1<=β€<=*a**i*<=β€<=100) β the number of cookies in the *i*-th bag.
Output Specification:
Print in the only line the only number β the sought number of ways. If there are no such ways print 0.
Demo Input:
['1\n1\n', '10\n1 2 2 3 4 4 4 2 2 2\n', '11\n2 2 2 2 2 2 2 2 2 2 99\n']
Demo Output:
['1\n', '8\n', '1\n']
Note:
In the first sample Olga should take the only bag so that the twins ended up with the even number of cookies.
In the second sample Olga can take any of five bags with two cookies or any of three bags with four cookies β 5β+β3β=β8 ways in total.
In the third sample, no matter which bag with two cookies Olga chooses, the twins are left with 2β*β9β+β99β=β117 cookies. Thus, Olga has only one option: to take the bag with 99 cookies.
|
```python
n=int(input())
l=list(map(int,input().split()))
Sum=sum(l)
if(Sum%2==0):
print(len([i for i in l if i%2==0]))
else:
print(len([i for i in l if(i%2==1)]))
```
| 3
|
|
656
|
A
|
Da Vinci Powers
|
PROGRAMMING
| 1,900
|
[
"*special"
] | null | null |
The input contains a single integer *a* (0<=β€<=*a*<=β€<=35).
Output a single integer.
|
The input contains a single integer *a* (0<=β€<=*a*<=β€<=35).
|
Output a single integer.
|
[
"3\n",
"10\n"
] |
[
"8\n",
"1024\n"
] |
none
| 0
|
[
{
"input": "3",
"output": "8"
},
{
"input": "10",
"output": "1024"
},
{
"input": "35",
"output": "33940307968"
},
{
"input": "0",
"output": "1"
},
{
"input": "1",
"output": "2"
},
{
"input": "2",
"output": "4"
},
{
"input": "4",
"output": "16"
},
{
"input": "5",
"output": "32"
},
{
"input": "6",
"output": "64"
},
{
"input": "7",
"output": "128"
},
{
"input": "8",
"output": "256"
},
{
"input": "9",
"output": "512"
},
{
"input": "11",
"output": "2048"
},
{
"input": "12",
"output": "4096"
},
{
"input": "13",
"output": "8092"
},
{
"input": "14",
"output": "16184"
},
{
"input": "15",
"output": "32368"
},
{
"input": "16",
"output": "64736"
},
{
"input": "17",
"output": "129472"
},
{
"input": "18",
"output": "258944"
},
{
"input": "19",
"output": "517888"
},
{
"input": "20",
"output": "1035776"
},
{
"input": "21",
"output": "2071552"
},
{
"input": "22",
"output": "4143104"
},
{
"input": "23",
"output": "8286208"
},
{
"input": "24",
"output": "16572416"
},
{
"input": "25",
"output": "33144832"
},
{
"input": "26",
"output": "66289664"
},
{
"input": "27",
"output": "132579328"
},
{
"input": "28",
"output": "265158656"
},
{
"input": "29",
"output": "530317312"
},
{
"input": "30",
"output": "1060634624"
},
{
"input": "31",
"output": "2121269248"
},
{
"input": "32",
"output": "4242538496"
},
{
"input": "33",
"output": "8485076992"
},
{
"input": "34",
"output": "16970153984"
}
] | 1,496,787,250
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 36
| 62
| 0
|
if __name__ == "__main__":
a = [1,
2,
4,
8,
16,
32,
64,
128,
256,
512,
1024,
2048,
4096,
8092,
16184,
32368,
64736,
129472,
258944,
517888,
1035776,
2071552,
4143104,
8286208,
16572416,
33144832,
66289664,
132579328,
265158656,
530317312,
1060634624,
2121269248,
4242538496,
8485076992,
16970153984,
33940307968]
print (a[int(input())])
|
Title: Da Vinci Powers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The input contains a single integer *a* (0<=β€<=*a*<=β€<=35).
Output a single integer.
Input Specification:
The input contains a single integer *a* (0<=β€<=*a*<=β€<=35).
Output Specification:
Output a single integer.
Demo Input:
['3\n', '10\n']
Demo Output:
['8\n', '1024\n']
Note:
none
|
```python
if __name__ == "__main__":
a = [1,
2,
4,
8,
16,
32,
64,
128,
256,
512,
1024,
2048,
4096,
8092,
16184,
32368,
64736,
129472,
258944,
517888,
1035776,
2071552,
4143104,
8286208,
16572416,
33144832,
66289664,
132579328,
265158656,
530317312,
1060634624,
2121269248,
4242538496,
8485076992,
16970153984,
33940307968]
print (a[int(input())])
```
| 3
|
|
976
|
A
|
Minimum Binary Number
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
String can be called correct if it consists of characters "0" and "1" and there are no redundant leading zeroes. Here are some examples: "0", "10", "1001".
You are given a correct string *s*.
You can perform two different operations on this string:
1. swap any pair of adjacent characters (for example, "101" "110"); 1. replace "11" with "1" (for example, "110" "10").
Let *val*(*s*) be such a number that *s* is its binary representation.
Correct string *a* is less than some other correct string *b* iff *val*(*a*)<=<<=*val*(*b*).
Your task is to find the minimum correct string that you can obtain from the given one using the operations described above. You can use these operations any number of times in any order (or even use no operations at all).
|
The first line contains integer number *n* (1<=β€<=*n*<=β€<=100) β the length of string *s*.
The second line contains the string *s* consisting of characters "0" and "1". It is guaranteed that the string *s* is correct.
|
Print one string β the minimum correct string that you can obtain from the given one.
|
[
"4\n1001\n",
"1\n1\n"
] |
[
"100\n",
"1\n"
] |
In the first example you can obtain the answer by the following sequence of operations: "1001" <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> "1010" <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> "1100" <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> "100".
In the second example you can't obtain smaller answer no matter what operations you use.
| 0
|
[
{
"input": "4\n1001",
"output": "100"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "100\n1110111100001111011111111010110011111111011110000111101101011100110110001011000000101010110101011100",
"output": "1000000000000000000000000000000000000000"
},
{
"input": "100\n1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "100\n1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111",
"output": "1"
},
{
"input": "100\n1111111111111111111111111111111111111111111111111111111110111111111111111111111111111111111111111111",
"output": "10"
},
{
"input": "1\n0",
"output": "0"
},
{
"input": "8\n10101010",
"output": "10000"
},
{
"input": "2\n10",
"output": "10"
},
{
"input": "3\n111",
"output": "1"
},
{
"input": "5\n11100",
"output": "100"
},
{
"input": "2\n11",
"output": "1"
},
{
"input": "3\n110",
"output": "10"
},
{
"input": "50\n10010010000000000000000000000000000000001000000000",
"output": "10000000000000000000000000000000000000000000000"
}
] | 1,694,614,508
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 14
| 46
| 0
|
def solve():
size = input()
s = input()
print('1' * ('1' in s) + '0' * s.count('0'))
if __name__ == "__main__":
solve()
|
Title: Minimum Binary Number
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
String can be called correct if it consists of characters "0" and "1" and there are no redundant leading zeroes. Here are some examples: "0", "10", "1001".
You are given a correct string *s*.
You can perform two different operations on this string:
1. swap any pair of adjacent characters (for example, "101" "110"); 1. replace "11" with "1" (for example, "110" "10").
Let *val*(*s*) be such a number that *s* is its binary representation.
Correct string *a* is less than some other correct string *b* iff *val*(*a*)<=<<=*val*(*b*).
Your task is to find the minimum correct string that you can obtain from the given one using the operations described above. You can use these operations any number of times in any order (or even use no operations at all).
Input Specification:
The first line contains integer number *n* (1<=β€<=*n*<=β€<=100) β the length of string *s*.
The second line contains the string *s* consisting of characters "0" and "1". It is guaranteed that the string *s* is correct.
Output Specification:
Print one string β the minimum correct string that you can obtain from the given one.
Demo Input:
['4\n1001\n', '1\n1\n']
Demo Output:
['100\n', '1\n']
Note:
In the first example you can obtain the answer by the following sequence of operations: "1001" <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> "1010" <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> "1100" <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> "100".
In the second example you can't obtain smaller answer no matter what operations you use.
|
```python
def solve():
size = input()
s = input()
print('1' * ('1' in s) + '0' * s.count('0'))
if __name__ == "__main__":
solve()
```
| 3
|
|
0
|
none
|
none
|
none
| 0
|
[
"none"
] | null | null |
In Absurdistan, there are *n* towns (numbered 1 through *n*) and *m* bidirectional railways. There is also an absurdly simple road networkΒ β 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 safetyΒ β 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<=β€<=*n*<=β€<=400, 0<=β€<=*m*<=β€<=*n*(*n*<=-<=1)<=/<=2)Β β 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<=β€<=*u*,<=*v*<=β€<=*n*, *u*<=β <=*v*).
You may assume that there is at most one railway connecting any two towns.
|
Output one integerΒ β 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.
| 0
|
[
{
"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"
}
] | 1,697,337,545
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 59
| 482
| 8,192,000
|
n, m = map(int, input().split())
a = [[0] * (n + 1) for _ in range(n + 1)]
for _ in range(m):
u, v = map(int, input().split())
a[u][v] = 1
a[v][u] = 1
def bfs(x):
fila = [(1, 0)]
visitados = [0] * (n + 1)
while fila:
u, v = fila.pop(0)
for i in range(1, n + 1):
if not visitados[i] and a[u][i] == x:
if i == n:
return v + 1
fila.append((i, v + 1))
visitados[i] = 1
return -1
if a[1][n]:
print(bfs(0))
else:
print(bfs(1))
|
Title: none
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 networkΒ β 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 safetyΒ β 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<=β€<=*n*<=β€<=400, 0<=β€<=*m*<=β€<=*n*(*n*<=-<=1)<=/<=2)Β β 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<=β€<=*u*,<=*v*<=β€<=*n*, *u*<=β <=*v*).
You may assume that there is at most one railway connecting any two towns.
Output Specification:
Output one integerΒ β 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
n, m = map(int, input().split())
a = [[0] * (n + 1) for _ in range(n + 1)]
for _ in range(m):
u, v = map(int, input().split())
a[u][v] = 1
a[v][u] = 1
def bfs(x):
fila = [(1, 0)]
visitados = [0] * (n + 1)
while fila:
u, v = fila.pop(0)
for i in range(1, n + 1):
if not visitados[i] and a[u][i] == x:
if i == n:
return v + 1
fila.append((i, v + 1))
visitados[i] = 1
return -1
if a[1][n]:
print(bfs(0))
else:
print(bfs(1))
```
| 3
|
|
507
|
B
|
Amr and Pins
|
PROGRAMMING
| 1,400
|
[
"geometry",
"math"
] | null | null |
Amr loves Geometry. One day he came up with a very interesting problem.
Amr has a circle of radius *r* and center in point (*x*,<=*y*). He wants the circle center to be in new position (*x*',<=*y*').
In one step Amr can put a pin to the border of the circle in a certain point, then rotate the circle around that pin by any angle and finally remove the pin.
Help Amr to achieve his goal in minimum number of steps.
|
Input consists of 5 space-separated integers *r*, *x*, *y*, *x*' *y*' (1<=β€<=*r*<=β€<=105, <=-<=105<=β€<=*x*,<=*y*,<=*x*',<=*y*'<=β€<=105), circle radius, coordinates of original center of the circle and coordinates of destination center of the circle respectively.
|
Output a single integer β minimum number of steps required to move the center of the circle to the destination point.
|
[
"2 0 0 0 4\n",
"1 1 1 4 4\n",
"4 5 6 5 6\n"
] |
[
"1\n",
"3\n",
"0\n"
] |
In the first sample test the optimal way is to put a pin at point (0,β2) and rotate the circle by 180 degrees counter-clockwise (or clockwise, no matter).
<img class="tex-graphics" src="https://espresso.codeforces.com/4e40fd4cc24a2050a0488aa131e6244369328039.png" style="max-width: 100.0%;max-height: 100.0%;"/>
| 1,000
|
[
{
"input": "2 0 0 0 4",
"output": "1"
},
{
"input": "1 1 1 4 4",
"output": "3"
},
{
"input": "4 5 6 5 6",
"output": "0"
},
{
"input": "10 20 0 40 0",
"output": "1"
},
{
"input": "9 20 0 40 0",
"output": "2"
},
{
"input": "5 -1 -6 -5 1",
"output": "1"
},
{
"input": "99125 26876 -21414 14176 17443",
"output": "1"
},
{
"input": "8066 7339 19155 -90534 -60666",
"output": "8"
},
{
"input": "100000 -100000 -100000 100000 100000",
"output": "2"
},
{
"input": "10 20 0 41 0",
"output": "2"
},
{
"input": "25 -64 -6 -56 64",
"output": "2"
},
{
"input": "125 455 450 439 721",
"output": "2"
},
{
"input": "5 6 3 7 2",
"output": "1"
},
{
"input": "24 130 14786 3147 2140",
"output": "271"
},
{
"input": "125 -363 176 93 330",
"output": "2"
},
{
"input": "1 14 30 30 14",
"output": "12"
},
{
"input": "25 96 13 7 2",
"output": "2"
},
{
"input": "4 100000 -100000 100000 -100000",
"output": "0"
},
{
"input": "1 3 4 2 5",
"output": "1"
},
{
"input": "1 -3 3 2 6",
"output": "3"
},
{
"input": "2 7 20 13 -5",
"output": "7"
},
{
"input": "1 1 1 1 4",
"output": "2"
},
{
"input": "249 -54242 -30537 -45023 -89682",
"output": "121"
},
{
"input": "4 100000 -100000 100000 -99999",
"output": "1"
},
{
"input": "97741 23818 78751 97583 26933",
"output": "1"
},
{
"input": "56767 -29030 51625 79823 -56297",
"output": "2"
},
{
"input": "98260 13729 74998 23701 9253",
"output": "1"
},
{
"input": "67377 -80131 -90254 -57320 14102",
"output": "1"
},
{
"input": "1 100000 100000 100000 -100000",
"output": "100000"
},
{
"input": "19312 19470 82059 58064 62231",
"output": "2"
},
{
"input": "67398 -68747 -79056 -34193 29400",
"output": "1"
},
{
"input": "91099 37184 -71137 75650 -3655",
"output": "1"
},
{
"input": "46456 -2621 -23623 -98302 -99305",
"output": "2"
},
{
"input": "100 100000 -100000 100000 -99999",
"output": "1"
},
{
"input": "1 100000 -100000 100000 -100000",
"output": "0"
},
{
"input": "8 0 0 0 32",
"output": "2"
},
{
"input": "100000 100000 1 -100000 0",
"output": "2"
}
] | 1,558,280,225
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 37
| 109
| 0
|
import math
r, x, y, x1, y1 = map(int, input().split())
print(math.ceil(math.sqrt((x-x1)**2 + (y-y1)**2)/(2*r)))
|
Title: Amr and Pins
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Amr loves Geometry. One day he came up with a very interesting problem.
Amr has a circle of radius *r* and center in point (*x*,<=*y*). He wants the circle center to be in new position (*x*',<=*y*').
In one step Amr can put a pin to the border of the circle in a certain point, then rotate the circle around that pin by any angle and finally remove the pin.
Help Amr to achieve his goal in minimum number of steps.
Input Specification:
Input consists of 5 space-separated integers *r*, *x*, *y*, *x*' *y*' (1<=β€<=*r*<=β€<=105, <=-<=105<=β€<=*x*,<=*y*,<=*x*',<=*y*'<=β€<=105), circle radius, coordinates of original center of the circle and coordinates of destination center of the circle respectively.
Output Specification:
Output a single integer β minimum number of steps required to move the center of the circle to the destination point.
Demo Input:
['2 0 0 0 4\n', '1 1 1 4 4\n', '4 5 6 5 6\n']
Demo Output:
['1\n', '3\n', '0\n']
Note:
In the first sample test the optimal way is to put a pin at point (0,β2) and rotate the circle by 180 degrees counter-clockwise (or clockwise, no matter).
<img class="tex-graphics" src="https://espresso.codeforces.com/4e40fd4cc24a2050a0488aa131e6244369328039.png" style="max-width: 100.0%;max-height: 100.0%;"/>
|
```python
import math
r, x, y, x1, y1 = map(int, input().split())
print(math.ceil(math.sqrt((x-x1)**2 + (y-y1)**2)/(2*r)))
```
| 3
|
|
686
|
A
|
Free Ice Cream
|
PROGRAMMING
| 800
|
[
"constructive algorithms",
"implementation"
] | null | null |
After their adventure with the magic mirror Kay and Gerda have returned home and sometimes give free ice cream to kids in the summer.
At the start of the day they have *x* ice cream packs. Since the ice cream is free, people start standing in the queue before Kay and Gerda's house even in the night. Each person in the queue wants either to take several ice cream packs for himself and his friends or to give several ice cream packs to Kay and Gerda (carriers that bring ice cream have to stand in the same queue).
If a carrier with *d* ice cream packs comes to the house, then Kay and Gerda take all his packs. If a child who wants to take *d* ice cream packs comes to the house, then Kay and Gerda will give him *d* packs if they have enough ice cream, otherwise the child will get no ice cream at all and will leave in distress.
Kay wants to find the amount of ice cream they will have after all people will leave from the queue, and Gerda wants to find the number of distressed kids.
|
The first line contains two space-separated integers *n* and *x* (1<=β€<=*n*<=β€<=1000, 0<=β€<=*x*<=β€<=109).
Each of the next *n* lines contains a character '+' or '-', and an integer *d**i*, separated by a space (1<=β€<=*d**i*<=β€<=109). Record "+ *d**i*" in *i*-th line means that a carrier with *d**i* ice cream packs occupies *i*-th place from the start of the queue, and record "- *d**i*" means that a child who wants to take *d**i* packs stands in *i*-th place.
|
Print two space-separated integersΒ β number of ice cream packs left after all operations, and number of kids that left the house in distress.
|
[
"5 7\n+ 5\n- 10\n- 20\n+ 40\n- 20\n",
"5 17\n- 16\n- 2\n- 98\n+ 100\n- 98\n"
] |
[
"22 1\n",
"3 2\n"
] |
Consider the first sample.
1. Initially Kay and Gerda have 7 packs of ice cream. 1. Carrier brings 5 more, so now they have 12 packs. 1. A kid asks for 10 packs and receives them. There are only 2 packs remaining. 1. Another kid asks for 20 packs. Kay and Gerda do not have them, so the kid goes away distressed. 1. Carrier bring 40 packs, now Kay and Gerda have 42 packs. 1. Kid asks for 20 packs and receives them. There are 22 packs remaining.
| 500
|
[
{
"input": "5 7\n+ 5\n- 10\n- 20\n+ 40\n- 20",
"output": "22 1"
},
{
"input": "5 17\n- 16\n- 2\n- 98\n+ 100\n- 98",
"output": "3 2"
},
{
"input": "6 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000",
"output": "7000000000 0"
},
{
"input": "5 12\n- 12\n+ 7\n- 6\n- 1\n+ 46",
"output": "46 0"
},
{
"input": "11 1000\n- 100\n+ 100\n+ 100\n+ 100\n+ 100\n- 100\n- 100\n- 100\n- 100\n- 100\n- 100",
"output": "700 0"
},
{
"input": "1 0\n- 526403222",
"output": "0 1"
},
{
"input": "1 897986543\n- 371188251",
"output": "526798292 0"
},
{
"input": "1 0\n+ 1",
"output": "1 0"
},
{
"input": "1 0\n- 1",
"output": "0 1"
},
{
"input": "1 10\n+ 10",
"output": "20 0"
},
{
"input": "1 3\n- 5",
"output": "3 1"
},
{
"input": "1 0\n- 5",
"output": "0 1"
},
{
"input": "1 0\n+ 5",
"output": "5 0"
}
] | 1,595,099,463
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 34
| 109
| 6,758,400
|
n,x = map(int,input().split())
count=0
for _ in range(n):
c,v = input().split()
v=int(v)
if(c=='+'):
x=x+v
if(c=='-'):
if(v<=x):
x=x-v
else:
count += 1
print(x,count)
|
Title: Free Ice Cream
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
After their adventure with the magic mirror Kay and Gerda have returned home and sometimes give free ice cream to kids in the summer.
At the start of the day they have *x* ice cream packs. Since the ice cream is free, people start standing in the queue before Kay and Gerda's house even in the night. Each person in the queue wants either to take several ice cream packs for himself and his friends or to give several ice cream packs to Kay and Gerda (carriers that bring ice cream have to stand in the same queue).
If a carrier with *d* ice cream packs comes to the house, then Kay and Gerda take all his packs. If a child who wants to take *d* ice cream packs comes to the house, then Kay and Gerda will give him *d* packs if they have enough ice cream, otherwise the child will get no ice cream at all and will leave in distress.
Kay wants to find the amount of ice cream they will have after all people will leave from the queue, and Gerda wants to find the number of distressed kids.
Input Specification:
The first line contains two space-separated integers *n* and *x* (1<=β€<=*n*<=β€<=1000, 0<=β€<=*x*<=β€<=109).
Each of the next *n* lines contains a character '+' or '-', and an integer *d**i*, separated by a space (1<=β€<=*d**i*<=β€<=109). Record "+ *d**i*" in *i*-th line means that a carrier with *d**i* ice cream packs occupies *i*-th place from the start of the queue, and record "- *d**i*" means that a child who wants to take *d**i* packs stands in *i*-th place.
Output Specification:
Print two space-separated integersΒ β number of ice cream packs left after all operations, and number of kids that left the house in distress.
Demo Input:
['5 7\n+ 5\n- 10\n- 20\n+ 40\n- 20\n', '5 17\n- 16\n- 2\n- 98\n+ 100\n- 98\n']
Demo Output:
['22 1\n', '3 2\n']
Note:
Consider the first sample.
1. Initially Kay and Gerda have 7 packs of ice cream. 1. Carrier brings 5 more, so now they have 12 packs. 1. A kid asks for 10 packs and receives them. There are only 2 packs remaining. 1. Another kid asks for 20 packs. Kay and Gerda do not have them, so the kid goes away distressed. 1. Carrier bring 40 packs, now Kay and Gerda have 42 packs. 1. Kid asks for 20 packs and receives them. There are 22 packs remaining.
|
```python
n,x = map(int,input().split())
count=0
for _ in range(n):
c,v = input().split()
v=int(v)
if(c=='+'):
x=x+v
if(c=='-'):
if(v<=x):
x=x-v
else:
count += 1
print(x,count)
```
| 3
|
|
231
|
A
|
Team
|
PROGRAMMING
| 800
|
[
"brute force",
"greedy"
] | null | null |
One day three best friends Petya, Vasya and Tonya decided to form a team and take part in programming contests. Participants are usually offered several problems during programming contests. Long before the start the friends decided that they will implement a problem if at least two of them are sure about the solution. Otherwise, the friends won't write the problem's solution.
This contest offers *n* problems to the participants. For each problem we know, which friend is sure about the solution. Help the friends find the number of problems for which they will write a solution.
|
The first input line contains a single integer *n* (1<=β€<=*n*<=β€<=1000) β the number of problems in the contest. Then *n* lines contain three integers each, each integer is either 0 or 1. If the first number in the line equals 1, then Petya is sure about the problem's solution, otherwise he isn't sure. The second number shows Vasya's view on the solution, the third number shows Tonya's view. The numbers on the lines are separated by spaces.
|
Print a single integer β the number of problems the friends will implement on the contest.
|
[
"3\n1 1 0\n1 1 1\n1 0 0\n",
"2\n1 0 0\n0 1 1\n"
] |
[
"2\n",
"1\n"
] |
In the first sample Petya and Vasya are sure that they know how to solve the first problem and all three of them know how to solve the second problem. That means that they will write solutions for these problems. Only Petya is sure about the solution for the third problem, but that isn't enough, so the friends won't take it.
In the second sample the friends will only implement the second problem, as Vasya and Tonya are sure about the solution.
| 500
|
[
{
"input": "3\n1 1 0\n1 1 1\n1 0 0",
"output": "2"
},
{
"input": "2\n1 0 0\n0 1 1",
"output": "1"
},
{
"input": "1\n1 0 0",
"output": "0"
},
{
"input": "2\n1 0 0\n1 1 1",
"output": "1"
},
{
"input": "5\n1 0 0\n0 1 0\n1 1 1\n0 0 1\n0 0 0",
"output": "1"
},
{
"input": "10\n0 1 0\n0 1 0\n1 1 0\n1 0 0\n0 0 1\n0 1 1\n1 1 1\n1 1 0\n0 0 0\n0 0 0",
"output": "4"
},
{
"input": "15\n0 1 0\n1 0 0\n1 1 0\n1 1 1\n0 1 0\n0 0 1\n1 0 1\n1 0 1\n1 0 1\n0 0 0\n1 1 1\n1 1 0\n0 1 1\n1 1 0\n1 1 1",
"output": "10"
},
{
"input": "50\n0 0 0\n0 1 1\n1 1 1\n0 1 0\n1 0 1\n1 1 1\n0 0 1\n1 0 0\n1 1 0\n1 0 1\n0 1 0\n0 0 1\n1 1 0\n0 1 0\n1 1 0\n0 0 0\n1 1 1\n1 0 1\n0 0 1\n1 1 0\n1 1 1\n0 1 1\n1 1 0\n0 0 0\n0 0 0\n1 1 1\n0 0 0\n1 1 1\n0 1 1\n0 0 1\n0 0 0\n0 0 0\n1 1 0\n1 1 0\n1 0 1\n1 0 0\n1 0 1\n1 0 1\n0 1 1\n1 1 0\n1 1 0\n0 1 0\n1 0 1\n0 0 0\n0 0 0\n0 0 0\n0 0 1\n1 1 1\n0 1 1\n1 0 1",
"output": "29"
},
{
"input": "1\n1 1 1",
"output": "1"
},
{
"input": "8\n0 0 0\n0 0 1\n0 0 0\n0 1 1\n1 0 0\n1 0 1\n1 1 0\n1 1 1",
"output": "4"
},
{
"input": "16\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1",
"output": "16"
}
] | 1,697,712,626
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 21
| 92
| 0
|
n = int(input())
count = 0
for i in range(n):
m = list(map(int,input().split()))
k = m.count(1)
if k >= 2:
count += 1
print(count)
|
Title: Team
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day three best friends Petya, Vasya and Tonya decided to form a team and take part in programming contests. Participants are usually offered several problems during programming contests. Long before the start the friends decided that they will implement a problem if at least two of them are sure about the solution. Otherwise, the friends won't write the problem's solution.
This contest offers *n* problems to the participants. For each problem we know, which friend is sure about the solution. Help the friends find the number of problems for which they will write a solution.
Input Specification:
The first input line contains a single integer *n* (1<=β€<=*n*<=β€<=1000) β the number of problems in the contest. Then *n* lines contain three integers each, each integer is either 0 or 1. If the first number in the line equals 1, then Petya is sure about the problem's solution, otherwise he isn't sure. The second number shows Vasya's view on the solution, the third number shows Tonya's view. The numbers on the lines are separated by spaces.
Output Specification:
Print a single integer β the number of problems the friends will implement on the contest.
Demo Input:
['3\n1 1 0\n1 1 1\n1 0 0\n', '2\n1 0 0\n0 1 1\n']
Demo Output:
['2\n', '1\n']
Note:
In the first sample Petya and Vasya are sure that they know how to solve the first problem and all three of them know how to solve the second problem. That means that they will write solutions for these problems. Only Petya is sure about the solution for the third problem, but that isn't enough, so the friends won't take it.
In the second sample the friends will only implement the second problem, as Vasya and Tonya are sure about the solution.
|
```python
n = int(input())
count = 0
for i in range(n):
m = list(map(int,input().split()))
k = m.count(1)
if k >= 2:
count += 1
print(count)
```
| 3
|
|
75
|
C
|
Modified GCD
|
PROGRAMMING
| 1,600
|
[
"binary search",
"number theory"
] |
C. Modified GCD
|
2
|
256
|
Well, here is another math class task. In mathematics, GCD is the greatest common divisor, and it's an easy task to calculate the GCD between two positive integers.
A common divisor for two positive numbers is a number which both numbers are divisible by.
But your teacher wants to give you a harder task, in this task you have to find the greatest common divisor *d* between two integers *a* and *b* that is in a given range from *low* to *high* (inclusive), i.e. *low*<=β€<=*d*<=β€<=*high*. It is possible that there is no common divisor in the given range.
You will be given the two integers *a* and *b*, then *n* queries. Each query is a range from *low* to *high* and you have to answer each query.
|
The first line contains two integers *a* and *b*, the two integers as described above (1<=β€<=*a*,<=*b*<=β€<=109). The second line contains one integer *n*, the number of queries (1<=β€<=*n*<=β€<=104). Then *n* lines follow, each line contains one query consisting of two integers, *low* and *high* (1<=β€<=*low*<=β€<=*high*<=β€<=109).
|
Print *n* lines. The *i*-th of them should contain the result of the *i*-th query in the input. If there is no common divisor in the given range for any query, you should print -1 as a result for this query.
|
[
"9 27\n3\n1 5\n10 11\n9 11\n"
] |
[
"3\n-1\n9\n"
] |
none
| 1,500
|
[
{
"input": "9 27\n3\n1 5\n10 11\n9 11",
"output": "3\n-1\n9"
},
{
"input": "48 72\n2\n8 29\n29 37",
"output": "24\n-1"
},
{
"input": "90 100\n10\n51 61\n6 72\n1 84\n33 63\n37 69\n18 21\n9 54\n49 90\n14 87\n37 90",
"output": "-1\n10\n10\n-1\n-1\n-1\n10\n-1\n-1\n-1"
},
{
"input": "84 36\n1\n18 32",
"output": "-1"
},
{
"input": "90 36\n16\n13 15\n5 28\n11 30\n26 35\n2 8\n19 36\n3 17\n5 14\n4 26\n22 33\n16 33\n18 27\n4 17\n1 2\n29 31\n18 36",
"output": "-1\n18\n18\n-1\n6\n-1\n9\n9\n18\n-1\n18\n18\n9\n2\n-1\n18"
},
{
"input": "84 90\n18\n10 75\n2 40\n30 56\n49 62\n19 33\n5 79\n61 83\n13 56\n73 78\n1 18\n23 35\n14 72\n22 33\n1 21\n8 38\n54 82\n6 80\n57 75",
"output": "-1\n6\n-1\n-1\n-1\n6\n-1\n-1\n-1\n6\n-1\n-1\n-1\n6\n-1\n-1\n6\n-1"
},
{
"input": "84 100\n16\n10 64\n3 61\n19 51\n42 67\n51 68\n12 40\n10 47\n52 53\n37 67\n2 26\n23 47\n17 75\n49 52\n3 83\n63 81\n8 43",
"output": "-1\n4\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n4\n-1\n-1\n-1\n4\n-1\n-1"
},
{
"input": "36 60\n2\n17 25\n16 20",
"output": "-1\n-1"
},
{
"input": "90 100\n8\n55 75\n46 68\n44 60\n32 71\n43 75\n23 79\n47 86\n11 57",
"output": "-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1"
},
{
"input": "90 36\n8\n1 19\n10 12\n14 28\n21 24\n8 8\n33 34\n10 26\n15 21",
"output": "18\n-1\n18\n-1\n-1\n-1\n18\n18"
},
{
"input": "48 80\n19\n1 1\n16 16\n1 16\n16 48\n16 80\n16 1000000000\n1000000000 1000000000\n1 1000000000\n500000000 1000000000\n15 17\n17 17\n15 15\n8 8\n8 15\n8 16\n8 17\n7 17\n7 15\n9 15",
"output": "1\n16\n16\n16\n16\n16\n-1\n16\n-1\n16\n-1\n-1\n8\n8\n16\n16\n16\n8\n-1"
},
{
"input": "31607 999002449\n18\n31607 31607\n31606 31608\n31607 31608\n31606 31607\n31606 31606\n31608 31608\n1 31607\n1 31606\n1 31608\n1 1000000000\n31607 1000000000\n31606 1000000000\n31608 1000000000\n1000000000 1000000000\n1 1\n2 31606\n2 31607\n2 31608",
"output": "31607\n31607\n31607\n31607\n-1\n-1\n31607\n1\n31607\n31607\n31607\n31607\n-1\n-1\n1\n-1\n31607\n31607"
},
{
"input": "999999937 999999929\n12\n999999929 999999937\n1 1\n1 1000000000\n2 1000000000\n1 2\n999999937 999999937\n999999929 999999929\n2 2\n3 3\n1 100\n1 999999937\n1 999999929",
"output": "-1\n1\n1\n-1\n1\n-1\n-1\n-1\n-1\n1\n1\n1"
}
] | 1,539,538,075
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 60
| 592
| 614,400
|
def gcd(a,b):
while a>0:
x=a
a=b%a
b=x
return(x)
def Bsearch1(L,x):
beg=0
last=len(L)-1
while beg<=last:
mid=(beg+last)//2
if x==L[mid]:
return(mid)
elif(x>L[mid]):
beg=mid+1
else:
last=mid-1
else:
return(beg)
def Bsearch2(L,x):
beg=0
last=len(L)-1
while beg<=last:
mid=(beg+last)//2
if x==L[mid]:
return(mid)
elif(x>L[mid]):
beg=mid+1
else:
last=mid-1
else:
return(last)
a,b=input().split()
a,b=[int(a),int(b)]
x=gcd(a,b)
from functools import reduce
def factors(n):
x=set(reduce(list.__add__,
([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0)))
return(list(x))
div=factors(x)
div.sort()
y=int(input())
for j in range(y):
c,d=input().split()
c,d=[int(c),int(d)]
f=Bsearch1(div,c)
g=Bsearch2(div,d)
if f>g:
print(-1)
else:
print(div[g])
|
Title: Modified GCD
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Well, here is another math class task. In mathematics, GCD is the greatest common divisor, and it's an easy task to calculate the GCD between two positive integers.
A common divisor for two positive numbers is a number which both numbers are divisible by.
But your teacher wants to give you a harder task, in this task you have to find the greatest common divisor *d* between two integers *a* and *b* that is in a given range from *low* to *high* (inclusive), i.e. *low*<=β€<=*d*<=β€<=*high*. It is possible that there is no common divisor in the given range.
You will be given the two integers *a* and *b*, then *n* queries. Each query is a range from *low* to *high* and you have to answer each query.
Input Specification:
The first line contains two integers *a* and *b*, the two integers as described above (1<=β€<=*a*,<=*b*<=β€<=109). The second line contains one integer *n*, the number of queries (1<=β€<=*n*<=β€<=104). Then *n* lines follow, each line contains one query consisting of two integers, *low* and *high* (1<=β€<=*low*<=β€<=*high*<=β€<=109).
Output Specification:
Print *n* lines. The *i*-th of them should contain the result of the *i*-th query in the input. If there is no common divisor in the given range for any query, you should print -1 as a result for this query.
Demo Input:
['9 27\n3\n1 5\n10 11\n9 11\n']
Demo Output:
['3\n-1\n9\n']
Note:
none
|
```python
def gcd(a,b):
while a>0:
x=a
a=b%a
b=x
return(x)
def Bsearch1(L,x):
beg=0
last=len(L)-1
while beg<=last:
mid=(beg+last)//2
if x==L[mid]:
return(mid)
elif(x>L[mid]):
beg=mid+1
else:
last=mid-1
else:
return(beg)
def Bsearch2(L,x):
beg=0
last=len(L)-1
while beg<=last:
mid=(beg+last)//2
if x==L[mid]:
return(mid)
elif(x>L[mid]):
beg=mid+1
else:
last=mid-1
else:
return(last)
a,b=input().split()
a,b=[int(a),int(b)]
x=gcd(a,b)
from functools import reduce
def factors(n):
x=set(reduce(list.__add__,
([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0)))
return(list(x))
div=factors(x)
div.sort()
y=int(input())
for j in range(y):
c,d=input().split()
c,d=[int(c),int(d)]
f=Bsearch1(div,c)
g=Bsearch2(div,d)
if f>g:
print(-1)
else:
print(div[g])
```
| 3.850856
|
58
|
A
|
Chat room
|
PROGRAMMING
| 1,000
|
[
"greedy",
"strings"
] |
A. Chat room
|
1
|
256
|
Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.
|
The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.
|
If Vasya managed to say hello, print "YES", otherwise print "NO".
|
[
"ahhellllloou\n",
"hlelo\n"
] |
[
"YES\n",
"NO\n"
] |
none
| 500
|
[
{
"input": "ahhellllloou",
"output": "YES"
},
{
"input": "hlelo",
"output": "NO"
},
{
"input": "helhcludoo",
"output": "YES"
},
{
"input": "hehwelloho",
"output": "YES"
},
{
"input": "pnnepelqomhhheollvlo",
"output": "YES"
},
{
"input": "tymbzjyqhymedasloqbq",
"output": "NO"
},
{
"input": "yehluhlkwo",
"output": "NO"
},
{
"input": "hatlevhhalrohairnolsvocafgueelrqmlqlleello",
"output": "YES"
},
{
"input": "hhhtehdbllnhwmbyhvelqqyoulretpbfokflhlhreeflxeftelziclrwllrpflflbdtotvlqgoaoqldlroovbfsq",
"output": "YES"
},
{
"input": "rzlvihhghnelqtwlexmvdjjrliqllolhyewgozkuovaiezgcilelqapuoeglnwmnlftxxiigzczlouooi",
"output": "YES"
},
{
"input": "pfhhwctyqdlkrwhebfqfelhyebwllhemtrmeblgrynmvyhioesqklclocxmlffuormljszllpoo",
"output": "YES"
},
{
"input": "lqllcolohwflhfhlnaow",
"output": "NO"
},
{
"input": "heheeellollvoo",
"output": "YES"
},
{
"input": "hellooo",
"output": "YES"
},
{
"input": "o",
"output": "NO"
},
{
"input": "hhqhzeclohlehljlhtesllylrolmomvuhcxsobtsckogdv",
"output": "YES"
},
{
"input": "yoegfuzhqsihygnhpnukluutocvvwuldiighpogsifealtgkfzqbwtmgghmythcxflebrkctlldlkzlagovwlstsghbouk",
"output": "YES"
},
{
"input": "uatqtgbvrnywfacwursctpagasnhydvmlinrcnqrry",
"output": "NO"
},
{
"input": "tndtbldbllnrwmbyhvqaqqyoudrstpbfokfoclnraefuxtftmgzicorwisrpfnfpbdtatvwqgyalqtdtrjqvbfsq",
"output": "NO"
},
{
"input": "rzlvirhgemelnzdawzpaoqtxmqucnahvqnwldklrmjiiyageraijfivigvozgwngiulttxxgzczptusoi",
"output": "YES"
},
{
"input": "kgyelmchocojsnaqdsyeqgnllytbqietpdlgknwwumqkxrexgdcnwoldicwzwofpmuesjuxzrasscvyuqwspm",
"output": "YES"
},
{
"input": "pnyvrcotjvgynbeldnxieghfltmexttuxzyac",
"output": "NO"
},
{
"input": "dtwhbqoumejligbenxvzhjlhosqojetcqsynlzyhfaevbdpekgbtjrbhlltbceobcok",
"output": "YES"
},
{
"input": "crrfpfftjwhhikwzeedrlwzblckkteseofjuxjrktcjfsylmlsvogvrcxbxtffujqshslemnixoeezivksouefeqlhhokwbqjz",
"output": "YES"
},
{
"input": "jhfbndhyzdvhbvhmhmefqllujdflwdpjbehedlsqfdsqlyelwjtyloxwsvasrbqosblzbowlqjmyeilcvotdlaouxhdpoeloaovb",
"output": "YES"
},
{
"input": "hwlghueoemiqtjhhpashjsouyegdlvoyzeunlroypoprnhlyiwiuxrghekaylndhrhllllwhbebezoglydcvykllotrlaqtvmlla",
"output": "YES"
},
{
"input": "wshiaunnqnqxodholbipwhhjmyeblhgpeleblklpzwhdunmpqkbuzloetmwwxmeltkrcomulxauzlwmlklldjodozxryghsnwgcz",
"output": "YES"
},
{
"input": "shvksednttggehroewuiptvvxtrzgidravtnjwuqrlnnkxbplctzkckinpkgjopjfoxdbojtcvsuvablcbkrzajrlhgobkcxeqti",
"output": "YES"
},
{
"input": "hyyhddqhxhekehkwfhlnlsihzefwchzerevcjtokefplholrbvxlltdlafjxrfhleglrvlolojoqaolagtbeyogxlbgfolllslli",
"output": "YES"
},
{
"input": "iaagrdhhelxpdegueiulflquvlzidoprzkehthkldaytrjwhyhprwjxlltinxvuilxohqgjqcvkkdcuoliekcnlilwgqlnlzouoo",
"output": "YES"
},
{
"input": "wfluaeseldgxyvxpwuhkptdmlflnlhktwxiabsvkolsquymrmhzczzoybvlilhmvsuunggvgxzgyyffk",
"output": "NO"
},
{
"input": "loee",
"output": "NO"
},
{
"input": "izdipwylefqmjbuoznfglgcfvedeouflkeehxbasaqmiooppfsqlhprospqgxvzrcpwlfdddngoqvpwvggsnvvxhmjoslwjudjlo",
"output": "NO"
},
{
"input": "pplynflbjlseflaegzqpdxswgunhnlglqhycrbcwdfhclmcrwekkzsoosmfblljeenwfjyifcvxihhhbyhmhlkvtskeglaovjnsi",
"output": "YES"
},
{
"input": "xqjqmenkodmlhzyzmmvofdngktygbbxbzpluzcohohmalkoeuwfikblltaaigv",
"output": "NO"
},
{
"input": "hello",
"output": "YES"
},
{
"input": "oohell",
"output": "NO"
},
{
"input": "hell",
"output": "NO"
},
{
"input": "eloellohoelo",
"output": "NO"
},
{
"input": "helllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllo",
"output": "YES"
}
] | 1,678,625,050
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 40
| 62
| 0
|
def stringcheck(s):
word="hello"
req="h"
c=0
for ch in s:
if c==5:
break
if ch==req:
c+=1
if c<5:
req=word[c]
if c==5:
print("YES")
else:
print("NO")
s=input()
stringcheck(s)
|
Title: Chat room
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.
Input Specification:
The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.
Output Specification:
If Vasya managed to say hello, print "YES", otherwise print "NO".
Demo Input:
['ahhellllloou\n', 'hlelo\n']
Demo Output:
['YES\n', 'NO\n']
Note:
none
|
```python
def stringcheck(s):
word="hello"
req="h"
c=0
for ch in s:
if c==5:
break
if ch==req:
c+=1
if c<5:
req=word[c]
if c==5:
print("YES")
else:
print("NO")
s=input()
stringcheck(s)
```
| 3.969
|
844
|
A
|
Diversity
|
PROGRAMMING
| 1,000
|
[
"greedy",
"implementation",
"strings"
] | null | null |
Calculate the minimum number of characters you need to change in the string *s*, so that it contains at least *k* different letters, or print that it is impossible.
String *s* consists only of lowercase Latin letters, and it is allowed to change characters only to lowercase Latin letters too.
|
First line of input contains string *s*, consisting only of lowercase Latin letters (1<=β€<=|*s*|<=β€<=1000, |*s*| denotes the length of *s*).
Second line of input contains integer *k* (1<=β€<=*k*<=β€<=26).
|
Print single line with a minimum number of necessary changes, or the word Β«impossibleΒ» (without quotes) if it is impossible.
|
[
"yandex\n6\n",
"yahoo\n5\n",
"google\n7\n"
] |
[
"0\n",
"1\n",
"impossible\n"
] |
In the first test case string contains 6 different letters, so we don't need to change anything.
In the second test case string contains 4 different letters: {'*a*',β'*h*',β'*o*',β'*y*'}. To get 5 different letters it is necessary to change one occurrence of '*o*' to some letter, which doesn't occur in the string, for example, {'*b*'}.
In the third test case, it is impossible to make 7 different letters because the length of the string is 6.
| 500
|
[
{
"input": "yandex\n6",
"output": "0"
},
{
"input": "yahoo\n5",
"output": "1"
},
{
"input": "google\n7",
"output": "impossible"
},
{
"input": "a\n1",
"output": "0"
},
{
"input": "z\n2",
"output": "impossible"
},
{
"input": "fwgfrwgkuwghfiruhewgirueguhergiqrbvgrgf\n26",
"output": "14"
},
{
"input": "nfevghreuoghrueighoqghbnebvnejbvnbgneluqe\n26",
"output": "12"
},
{
"input": "a\n3",
"output": "impossible"
},
{
"input": "smaxpqplaqqbxuqxalqmbmmgubbpspxhawbxsuqhhegpmmpebqmqpbbeplwaepxmsahuepuhuhwxeqmmlgqubuaxehwuwasgxpqmugbmuawuhwqlswllssueglbxepbmwgs\n1",
"output": "0"
},
{
"input": "cuguccgcugcugucgggggcgcgucgucugcuuuccccuugccg\n4",
"output": "1"
},
{
"input": "fcfccfcfccfcfcffcffffffcfccfccfcffccccfcffffccfccfcffcfcccccffcfffcccffcfccfffffcccfccffffffccfccccf\n20",
"output": "18"
},
{
"input": "swmkwaruyv\n5",
"output": "0"
},
{
"input": "tnbqpsuhkczmejirvyfdolxwga\n22",
"output": "0"
},
{
"input": "abcde\n3",
"output": "0"
},
{
"input": "abb\n1",
"output": "0"
},
{
"input": "aaaa\n1",
"output": "0"
},
{
"input": "abcde\n2",
"output": "0"
},
{
"input": "yandex\n4",
"output": "0"
},
{
"input": "aaabbbccc\n1",
"output": "0"
},
{
"input": "abcd\n2",
"output": "0"
},
{
"input": "asdfgh\n2",
"output": "0"
},
{
"input": "aab\n1",
"output": "0"
},
{
"input": "mynameissako\n5",
"output": "0"
},
{
"input": "abcde\n1",
"output": "0"
},
{
"input": "abcd\n3",
"output": "0"
},
{
"input": "abcdef\n2",
"output": "0"
},
{
"input": "abcdefg\n4",
"output": "0"
},
{
"input": "abc\n1",
"output": "0"
},
{
"input": "asdafjsgljdllgjdgkl\n5",
"output": "0"
},
{
"input": "yaay\n3",
"output": "1"
},
{
"input": "yaay\n4",
"output": "2"
},
{
"input": "zzzzzz\n2",
"output": "1"
}
] | 1,503,595,172
| 2,672
|
Python 3
|
OK
|
TESTS
| 41
| 77
| 0
|
s = input()
x = int(input())
if x > len(s):
print('impossible')
elif x < len(set(s)):
print(0)
else:
print(x - len(set(s)))
|
Title: Diversity
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Calculate the minimum number of characters you need to change in the string *s*, so that it contains at least *k* different letters, or print that it is impossible.
String *s* consists only of lowercase Latin letters, and it is allowed to change characters only to lowercase Latin letters too.
Input Specification:
First line of input contains string *s*, consisting only of lowercase Latin letters (1<=β€<=|*s*|<=β€<=1000, |*s*| denotes the length of *s*).
Second line of input contains integer *k* (1<=β€<=*k*<=β€<=26).
Output Specification:
Print single line with a minimum number of necessary changes, or the word Β«impossibleΒ» (without quotes) if it is impossible.
Demo Input:
['yandex\n6\n', 'yahoo\n5\n', 'google\n7\n']
Demo Output:
['0\n', '1\n', 'impossible\n']
Note:
In the first test case string contains 6 different letters, so we don't need to change anything.
In the second test case string contains 4 different letters: {'*a*',β'*h*',β'*o*',β'*y*'}. To get 5 different letters it is necessary to change one occurrence of '*o*' to some letter, which doesn't occur in the string, for example, {'*b*'}.
In the third test case, it is impossible to make 7 different letters because the length of the string is 6.
|
```python
s = input()
x = int(input())
if x > len(s):
print('impossible')
elif x < len(set(s)):
print(0)
else:
print(x - len(set(s)))
```
| 3
|
|
265
|
A
|
Colorful Stones (Simplified Edition)
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
There is a sequence of colorful stones. The color of each stone is one of red, green, or blue. You are given a string *s*. The *i*-th (1-based) character of *s* represents the color of the *i*-th stone. If the character is "R", "G", or "B", the color of the corresponding stone is red, green, or blue, respectively.
Initially Squirrel Liss is standing on the first stone. You perform instructions one or more times.
Each instruction is one of the three types: "RED", "GREEN", or "BLUE". After an instruction *c*, if Liss is standing on a stone whose colors is *c*, Liss will move one stone forward, else she will not move.
You are given a string *t*. The number of instructions is equal to the length of *t*, and the *i*-th character of *t* represents the *i*-th instruction.
Calculate the final position of Liss (the number of the stone she is going to stand on in the end) after performing all the instructions, and print its 1-based position. It is guaranteed that Liss don't move out of the sequence.
|
The input contains two lines. The first line contains the string *s* (1<=β€<=|*s*|<=β€<=50). The second line contains the string *t* (1<=β€<=|*t*|<=β€<=50). The characters of each string will be one of "R", "G", or "B". It is guaranteed that Liss don't move out of the sequence.
|
Print the final 1-based position of Liss in a single line.
|
[
"RGB\nRRR\n",
"RRRBGBRBBB\nBBBRR\n",
"BRRBGBRGRBGRGRRGGBGBGBRGBRGRGGGRBRRRBRBBBGRRRGGBBB\nBBRBGGRGRGBBBRBGRBRBBBBRBRRRBGBBGBBRRBBGGRBRRBRGRB\n"
] |
[
"2\n",
"3\n",
"15\n"
] |
none
| 500
|
[
{
"input": "RGB\nRRR",
"output": "2"
},
{
"input": "RRRBGBRBBB\nBBBRR",
"output": "3"
},
{
"input": "BRRBGBRGRBGRGRRGGBGBGBRGBRGRGGGRBRRRBRBBBGRRRGGBBB\nBBRBGGRGRGBBBRBGRBRBBBBRBRRRBGBBGBBRRBBGGRBRRBRGRB",
"output": "15"
},
{
"input": "G\nRRBBRBRRBR",
"output": "1"
},
{
"input": "RRRRRBRRBRRGRBGGRRRGRBBRBBBBBRGRBGBRRGBBBRBBGBRGBB\nB",
"output": "1"
},
{
"input": "RRGGBRGRBG\nBRRGGBBGGR",
"output": "7"
},
{
"input": "BBRRGBGGRGBRGBRBRBGR\nGGGRBGGGBRRRRGRBGBGRGRRBGRBGBG",
"output": "15"
},
{
"input": "GBRRBGBGBBBBRRRGBGRRRGBGBBBRGR\nRRGBRRGRBBBBBBGRRBBR",
"output": "8"
},
{
"input": "BRGRRGRGRRGBBGBBBRRBBRRBGBBGRGBBGGRGBRBGGGRRRBGGBB\nRGBBGRRBBBRRGRRBRBBRGBBGGGRGBGRRRRBRBGGBRBGGGRGBRR",
"output": "16"
},
{
"input": "GGRGGBRRGRGBRRGGRBBGGRRGBBBGBBBGGRBGGBRBBRGBRRRBRG\nGGRGRRRRRRRRRGBBBBRGBRGRGRRGBBRGGBRBBRBGBRGRRRRGGR",
"output": "18"
},
{
"input": "RBBRBGBBGGGBRRBGBRGRRGGRBBBBGRBGGBRRBGBBRBRGBBGGGG\nBRRGRBGRBGBRGBRBGGBBBGBBRGRBGRGRBGRBGGRRBGBRGRRBBG",
"output": "23"
},
{
"input": "RRGGRRBRGRGBRGBRGBBRGGGBRBGBRGBRGGBRBRGBGRRBBGGRBR\nBRGRBBBGRBGRRGGRRRRRRRRGGGGGGGGBGGGGGGGRGRBBRGRGBR",
"output": "14"
},
{
"input": "RBRBRRRRRGGR\nBBBBRRGGBGRBRGBBRGGGRGGBRRBRBRBGGG",
"output": "12"
},
{
"input": "GGRGGGBRGRGRBBRGGGRBGRRR\nGBBBBGBGBGBBBRGGRGGBGRRGRRRGBGB",
"output": "12"
},
{
"input": "RGRBGBRBBGBRRGGGBBGBRRBGRGBBBBBGRRRGGRBB\nBGR",
"output": "2"
},
{
"input": "R\nB",
"output": "1"
},
{
"input": "RRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR\nRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR",
"output": "50"
},
{
"input": "RRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR\nGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG",
"output": "1"
},
{
"input": "R\nGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG",
"output": "1"
}
] | 1,688,667,351
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 19
| 124
| 0
|
s = input()
t = input()
count=1
si=0
for i in range(len(t)):
if t[i] == s[si]:
count+=1
si+=1
print(count)
|
Title: Colorful Stones (Simplified Edition)
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There is a sequence of colorful stones. The color of each stone is one of red, green, or blue. You are given a string *s*. The *i*-th (1-based) character of *s* represents the color of the *i*-th stone. If the character is "R", "G", or "B", the color of the corresponding stone is red, green, or blue, respectively.
Initially Squirrel Liss is standing on the first stone. You perform instructions one or more times.
Each instruction is one of the three types: "RED", "GREEN", or "BLUE". After an instruction *c*, if Liss is standing on a stone whose colors is *c*, Liss will move one stone forward, else she will not move.
You are given a string *t*. The number of instructions is equal to the length of *t*, and the *i*-th character of *t* represents the *i*-th instruction.
Calculate the final position of Liss (the number of the stone she is going to stand on in the end) after performing all the instructions, and print its 1-based position. It is guaranteed that Liss don't move out of the sequence.
Input Specification:
The input contains two lines. The first line contains the string *s* (1<=β€<=|*s*|<=β€<=50). The second line contains the string *t* (1<=β€<=|*t*|<=β€<=50). The characters of each string will be one of "R", "G", or "B". It is guaranteed that Liss don't move out of the sequence.
Output Specification:
Print the final 1-based position of Liss in a single line.
Demo Input:
['RGB\nRRR\n', 'RRRBGBRBBB\nBBBRR\n', 'BRRBGBRGRBGRGRRGGBGBGBRGBRGRGGGRBRRRBRBBBGRRRGGBBB\nBBRBGGRGRGBBBRBGRBRBBBBRBRRRBGBBGBBRRBBGGRBRRBRGRB\n']
Demo Output:
['2\n', '3\n', '15\n']
Note:
none
|
```python
s = input()
t = input()
count=1
si=0
for i in range(len(t)):
if t[i] == s[si]:
count+=1
si+=1
print(count)
```
| 3
|
|
137
|
C
|
History
|
PROGRAMMING
| 1,500
|
[
"sortings"
] | null | null |
Polycarpus likes studying at school a lot and he is always diligent about his homework. Polycarpus has never had any problems with natural sciences as his great-great-grandfather was the great physicist Seinstein. On the other hand though, Polycarpus has never had an easy time with history.
Everybody knows that the World history encompasses exactly *n* events: the *i*-th event had continued from the year *a**i* to the year *b**i* inclusive (*a**i*<=<<=*b**i*). Polycarpus easily learned the dates when each of *n* events started and ended (Polycarpus inherited excellent memory from his great-great-granddad). But the teacher gave him a more complicated task: Polycaprus should know when all events began and ended and he should also find out for each event whether it includes another event. Polycarpus' teacher thinks that an event *j* includes an event *i* if *a**j*<=<<=*a**i* and *b**i*<=<<=*b**j*. Your task is simpler: find the number of events that are included in some other event.
|
The first input line contains integer *n* (1<=β€<=*n*<=β€<=105) which represents the number of events. Next *n* lines contain descriptions of the historical events, one event per line. The *i*<=+<=1 line contains two integers *a**i* and *b**i* (1<=β€<=*a**i*<=<<=*b**i*<=β€<=109) β the beginning and the end of the *i*-th event. No two events start or finish in the same year, that is, *a**i*<=β <=*a**j*,<=*a**i*<=β <=*b**j*,<=*b**i*<=β <=*a**j*,<=*b**i*<=β <=*b**j* for all *i*, *j* (where *i*<=β <=*j*). Events are given in arbitrary order.
|
Print the only integer β the answer to the problem.
|
[
"5\n1 10\n2 9\n3 8\n4 7\n5 6\n",
"5\n1 100\n2 50\n51 99\n52 98\n10 60\n",
"1\n1 1000000000\n"
] |
[
"4\n",
"4\n",
"0\n"
] |
In the first example the fifth event is contained in the fourth. Similarly, the fourth event is contained in the third, the third β in the second and the second β in the first.
In the second example all events except the first one are contained in the first.
In the third example only one event, so the answer is 0.
| 1,500
|
[
{
"input": "5\n1 10\n2 9\n3 8\n4 7\n5 6",
"output": "4"
},
{
"input": "5\n1 100\n2 50\n51 99\n52 98\n10 60",
"output": "4"
},
{
"input": "1\n1 1000000000",
"output": "0"
},
{
"input": "2\n100 1000\n500 1500",
"output": "0"
},
{
"input": "4\n1 100\n50 150\n120 200\n180 220",
"output": "0"
},
{
"input": "10\n1 20\n2 19\n3 18\n4 17\n5 16\n6 15\n7 14\n8 13\n9 12\n10 11",
"output": "9"
},
{
"input": "4\n1 5\n2 4\n10 15\n11 14",
"output": "2"
},
{
"input": "10\n1 2\n3 4\n5 6\n7 8\n9 10\n11 12\n13 14\n15 16\n17 18\n19 20",
"output": "0"
},
{
"input": "10\n58 22116\n112840 772160\n127432 399484\n182031 255417\n459340 548097\n603984 739189\n606173 663033\n280170 366159\n779239 906234\n874623 892907",
"output": "7"
},
{
"input": "2\n2 3\n1 4",
"output": "1"
},
{
"input": "3\n9 10\n1 2\n3 4",
"output": "0"
},
{
"input": "4\n1 2\n3 6\n4 5\n10 11",
"output": "1"
},
{
"input": "2\n5 6\n1 10",
"output": "1"
},
{
"input": "7\n1 10\n2 5\n3 9\n100 1000\n200 500\n1001 1002\n777 7777",
"output": "4"
}
] | 1,588,083,198
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 75
| 872
| 78,950,400
|
import sys
from math import log2,floor,ceil,sqrt,gcd
import bisect
# from collections import deque
sys.setrecursionlimit(10**5)
Ri = lambda : [int(x) for x in sys.stdin.readline().split()]
ri = lambda : sys.stdin.readline().strip()
def input(): return sys.stdin.readline().strip()
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
def list4d(a, b, c, d, e): return [[[[e] * d for j in range(c)] for j in range(b)] for i in range(a)]
def ceil(x, y=1): return int(-(-x // y))
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(N=None): return list(MAP()) if N is None else [INT() for i in range(N)]
def Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
INF = 10 ** 18
MOD = 1000000007
n =int(ri())
lis = []
for i in range(n):
temp = Ri()
lis.append(temp)
lis.sort(key = lambda x : x[0])
cnt = 0
right = lis[0][1]
for i in range(1,len(lis)):
if lis[i][1] < right:
cnt+=1
else:
right= lis[i][1]
print(cnt)
|
Title: History
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarpus likes studying at school a lot and he is always diligent about his homework. Polycarpus has never had any problems with natural sciences as his great-great-grandfather was the great physicist Seinstein. On the other hand though, Polycarpus has never had an easy time with history.
Everybody knows that the World history encompasses exactly *n* events: the *i*-th event had continued from the year *a**i* to the year *b**i* inclusive (*a**i*<=<<=*b**i*). Polycarpus easily learned the dates when each of *n* events started and ended (Polycarpus inherited excellent memory from his great-great-granddad). But the teacher gave him a more complicated task: Polycaprus should know when all events began and ended and he should also find out for each event whether it includes another event. Polycarpus' teacher thinks that an event *j* includes an event *i* if *a**j*<=<<=*a**i* and *b**i*<=<<=*b**j*. Your task is simpler: find the number of events that are included in some other event.
Input Specification:
The first input line contains integer *n* (1<=β€<=*n*<=β€<=105) which represents the number of events. Next *n* lines contain descriptions of the historical events, one event per line. The *i*<=+<=1 line contains two integers *a**i* and *b**i* (1<=β€<=*a**i*<=<<=*b**i*<=β€<=109) β the beginning and the end of the *i*-th event. No two events start or finish in the same year, that is, *a**i*<=β <=*a**j*,<=*a**i*<=β <=*b**j*,<=*b**i*<=β <=*a**j*,<=*b**i*<=β <=*b**j* for all *i*, *j* (where *i*<=β <=*j*). Events are given in arbitrary order.
Output Specification:
Print the only integer β the answer to the problem.
Demo Input:
['5\n1 10\n2 9\n3 8\n4 7\n5 6\n', '5\n1 100\n2 50\n51 99\n52 98\n10 60\n', '1\n1 1000000000\n']
Demo Output:
['4\n', '4\n', '0\n']
Note:
In the first example the fifth event is contained in the fourth. Similarly, the fourth event is contained in the third, the third β in the second and the second β in the first.
In the second example all events except the first one are contained in the first.
In the third example only one event, so the answer is 0.
|
```python
import sys
from math import log2,floor,ceil,sqrt,gcd
import bisect
# from collections import deque
sys.setrecursionlimit(10**5)
Ri = lambda : [int(x) for x in sys.stdin.readline().split()]
ri = lambda : sys.stdin.readline().strip()
def input(): return sys.stdin.readline().strip()
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
def list4d(a, b, c, d, e): return [[[[e] * d for j in range(c)] for j in range(b)] for i in range(a)]
def ceil(x, y=1): return int(-(-x // y))
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(N=None): return list(MAP()) if N is None else [INT() for i in range(N)]
def Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
INF = 10 ** 18
MOD = 1000000007
n =int(ri())
lis = []
for i in range(n):
temp = Ri()
lis.append(temp)
lis.sort(key = lambda x : x[0])
cnt = 0
right = lis[0][1]
for i in range(1,len(lis)):
if lis[i][1] < right:
cnt+=1
else:
right= lis[i][1]
print(cnt)
```
| 3
|
|
520
|
B
|
Two Buttons
|
PROGRAMMING
| 1,400
|
[
"dfs and similar",
"graphs",
"greedy",
"implementation",
"math",
"shortest paths"
] | null | null |
Vasya has found a strange device. On the front panel of a device there are: a red button, a blue button and a display showing some positive integer. After clicking the red button, device multiplies the displayed number by two. After clicking the blue button, device subtracts one from the number on the display. If at some point the number stops being positive, the device breaks down. The display can show arbitrarily large numbers. Initially, the display shows number *n*.
Bob wants to get number *m* on the display. What minimum number of clicks he has to make in order to achieve this result?
|
The first and the only line of the input contains two distinct integers *n* and *m* (1<=β€<=*n*,<=*m*<=β€<=104), separated by a space .
|
Print a single number β the minimum number of times one needs to push the button required to get the number *m* out of number *n*.
|
[
"4 6\n",
"10 1\n"
] |
[
"2\n",
"9\n"
] |
In the first example you need to push the blue button once, and then push the red button once.
In the second example, doubling the number is unnecessary, so we need to push the blue button nine times.
| 1,000
|
[
{
"input": "4 6",
"output": "2"
},
{
"input": "10 1",
"output": "9"
},
{
"input": "1 2",
"output": "1"
},
{
"input": "2 1",
"output": "1"
},
{
"input": "1 3",
"output": "3"
},
{
"input": "3 1",
"output": "2"
},
{
"input": "2 10",
"output": "5"
},
{
"input": "100 99",
"output": "1"
},
{
"input": "99 100",
"output": "50"
},
{
"input": "10 17",
"output": "3"
},
{
"input": "666 6666",
"output": "255"
},
{
"input": "6666 666",
"output": "6000"
},
{
"input": "1 8192",
"output": "13"
},
{
"input": "1 8193",
"output": "27"
},
{
"input": "9999 10000",
"output": "5000"
},
{
"input": "10000 9999",
"output": "1"
},
{
"input": "10000 1",
"output": "9999"
},
{
"input": "1234 5678",
"output": "528"
},
{
"input": "9102 9103",
"output": "4552"
},
{
"input": "8192 1",
"output": "8191"
},
{
"input": "9912 1023",
"output": "8889"
},
{
"input": "100 500",
"output": "41"
},
{
"input": "9997 9999",
"output": "4999"
},
{
"input": "5000 10000",
"output": "1"
},
{
"input": "4000 7997",
"output": "3"
},
{
"input": "1 10000",
"output": "20"
},
{
"input": "1 8191",
"output": "14"
},
{
"input": "9999 10000",
"output": "5000"
},
{
"input": "9097 9998",
"output": "4099"
},
{
"input": "886 9383",
"output": "305"
},
{
"input": "1 9",
"output": "7"
},
{
"input": "1918 10000",
"output": "671"
},
{
"input": "1 3",
"output": "3"
},
{
"input": "1 4",
"output": "2"
},
{
"input": "2 10000",
"output": "19"
},
{
"input": "3 10000",
"output": "17"
},
{
"input": "4 10000",
"output": "18"
},
{
"input": "9998 10000",
"output": "4999"
},
{
"input": "5000 10000",
"output": "1"
},
{
"input": "5001 10000",
"output": "2"
},
{
"input": "10000 1",
"output": "9999"
},
{
"input": "1 9999",
"output": "21"
},
{
"input": "7777 9999",
"output": "2779"
},
{
"input": "2 100",
"output": "9"
},
{
"input": "10 8722",
"output": "18"
},
{
"input": "848 4561",
"output": "283"
},
{
"input": "9967 9973",
"output": "4982"
},
{
"input": "5555 10000",
"output": "556"
},
{
"input": "999 10000",
"output": "378"
},
{
"input": "3 6",
"output": "1"
},
{
"input": "9 38",
"output": "8"
}
] | 1,698,290,071
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 51
| 61
| 1,228,800
|
from collections import deque
n, m = map(int, input().split())
queue = deque([(n, 0)])
visited = set()
while queue:
current, clicks = queue.popleft()
if current == m:
print(clicks)
break
if current * 2 not in visited and current * 2 <= 10**4:
queue.append((current * 2, clicks + 1))
visited.add(current * 2)
if current - 1 > 0 and current - 1 not in visited:
queue.append((current - 1, clicks + 1))
visited.add(current - 1)
|
Title: Two Buttons
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya has found a strange device. On the front panel of a device there are: a red button, a blue button and a display showing some positive integer. After clicking the red button, device multiplies the displayed number by two. After clicking the blue button, device subtracts one from the number on the display. If at some point the number stops being positive, the device breaks down. The display can show arbitrarily large numbers. Initially, the display shows number *n*.
Bob wants to get number *m* on the display. What minimum number of clicks he has to make in order to achieve this result?
Input Specification:
The first and the only line of the input contains two distinct integers *n* and *m* (1<=β€<=*n*,<=*m*<=β€<=104), separated by a space .
Output Specification:
Print a single number β the minimum number of times one needs to push the button required to get the number *m* out of number *n*.
Demo Input:
['4 6\n', '10 1\n']
Demo Output:
['2\n', '9\n']
Note:
In the first example you need to push the blue button once, and then push the red button once.
In the second example, doubling the number is unnecessary, so we need to push the blue button nine times.
|
```python
from collections import deque
n, m = map(int, input().split())
queue = deque([(n, 0)])
visited = set()
while queue:
current, clicks = queue.popleft()
if current == m:
print(clicks)
break
if current * 2 not in visited and current * 2 <= 10**4:
queue.append((current * 2, clicks + 1))
visited.add(current * 2)
if current - 1 > 0 and current - 1 not in visited:
queue.append((current - 1, clicks + 1))
visited.add(current - 1)
```
| 3
|
|
336
|
A
|
Vasily the Bear and Triangle
|
PROGRAMMING
| 1,000
|
[
"implementation",
"math"
] | null | null |
Vasily the bear has a favorite rectangle, it has one vertex at point (0,<=0), and the opposite vertex at point (*x*,<=*y*). Of course, the sides of Vasya's favorite rectangle are parallel to the coordinate axes.
Vasya also loves triangles, if the triangles have one vertex at point *B*<==<=(0,<=0). That's why today he asks you to find two points *A*<==<=(*x*1,<=*y*1) and *C*<==<=(*x*2,<=*y*2), such that the following conditions hold:
- the coordinates of points: *x*1, *x*2, *y*1, *y*2 are integers. Besides, the following inequation holds: *x*1<=<<=*x*2; - the triangle formed by point *A*, *B* and *C* is rectangular and isosceles ( is right); - all points of the favorite rectangle are located inside or on the border of triangle *ABC*; - the area of triangle *ABC* is as small as possible.
Help the bear, find the required points. It is not so hard to proof that these points are unique.
|
The first line contains two integers *x*,<=*y* (<=-<=109<=β€<=*x*,<=*y*<=β€<=109,<=*x*<=β <=0,<=*y*<=β <=0).
|
Print in the single line four integers *x*1,<=*y*1,<=*x*2,<=*y*2 β the coordinates of the required points.
|
[
"10 5\n",
"-10 5\n"
] |
[
"0 15 15 0\n",
"-15 0 0 15\n"
] |
<img class="tex-graphics" src="https://espresso.codeforces.com/a9ea2088c4294ce8f23801562fda36b830df2c3f.png" style="max-width: 100.0%;max-height: 100.0%;"/>
Figure to the first sample
| 500
|
[
{
"input": "10 5",
"output": "0 15 15 0"
},
{
"input": "-10 5",
"output": "-15 0 0 15"
},
{
"input": "20 -10",
"output": "0 -30 30 0"
},
{
"input": "-10 -1000000000",
"output": "-1000000010 0 0 -1000000010"
},
{
"input": "-1000000000 -1000000000",
"output": "-2000000000 0 0 -2000000000"
},
{
"input": "1000000000 1000000000",
"output": "0 2000000000 2000000000 0"
},
{
"input": "-123131 3123141",
"output": "-3246272 0 0 3246272"
},
{
"input": "-23423 -243242423",
"output": "-243265846 0 0 -243265846"
},
{
"input": "123112 4560954",
"output": "0 4684066 4684066 0"
},
{
"input": "1321 -23131",
"output": "0 -24452 24452 0"
},
{
"input": "1000000000 999999999",
"output": "0 1999999999 1999999999 0"
},
{
"input": "54543 432423",
"output": "0 486966 486966 0"
},
{
"input": "1 1",
"output": "0 2 2 0"
},
{
"input": "-1 -1",
"output": "-2 0 0 -2"
},
{
"input": "-1 1",
"output": "-2 0 0 2"
},
{
"input": "1 -1",
"output": "0 -2 2 0"
},
{
"input": "42 -2",
"output": "0 -44 44 0"
},
{
"input": "2 -435",
"output": "0 -437 437 0"
},
{
"input": "76 -76",
"output": "0 -152 152 0"
},
{
"input": "1000000000 1",
"output": "0 1000000001 1000000001 0"
},
{
"input": "1000000000 -1",
"output": "0 -1000000001 1000000001 0"
},
{
"input": "-1000000000 1",
"output": "-1000000001 0 0 1000000001"
},
{
"input": "-1000000000 -1",
"output": "-1000000001 0 0 -1000000001"
},
{
"input": "1000000000 -999999999",
"output": "0 -1999999999 1999999999 0"
},
{
"input": "-1000000000 999999999",
"output": "-1999999999 0 0 1999999999"
},
{
"input": "-1000000000 -999999999",
"output": "-1999999999 0 0 -1999999999"
},
{
"input": "999999999 1000000000",
"output": "0 1999999999 1999999999 0"
},
{
"input": "-999999999 1000000000",
"output": "-1999999999 0 0 1999999999"
},
{
"input": "999999999 -1000000000",
"output": "0 -1999999999 1999999999 0"
},
{
"input": "-999999999 -1000000000",
"output": "-1999999999 0 0 -1999999999"
}
] | 1,602,661,831
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 30
| 186
| 0
|
n=input().split()
x,y=int(n[0]),int(n[1])
if x>0 and y>0:
print(0,x+y,x+y,0)
elif x>0 and y<0:
print(0,y-x,x-y,0)
elif x<0 and y>0:
print(x-y,0,0,y-x)
else:
print(x+y,0,0,x+y)
|
Title: Vasily the Bear and Triangle
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasily the bear has a favorite rectangle, it has one vertex at point (0,<=0), and the opposite vertex at point (*x*,<=*y*). Of course, the sides of Vasya's favorite rectangle are parallel to the coordinate axes.
Vasya also loves triangles, if the triangles have one vertex at point *B*<==<=(0,<=0). That's why today he asks you to find two points *A*<==<=(*x*1,<=*y*1) and *C*<==<=(*x*2,<=*y*2), such that the following conditions hold:
- the coordinates of points: *x*1, *x*2, *y*1, *y*2 are integers. Besides, the following inequation holds: *x*1<=<<=*x*2; - the triangle formed by point *A*, *B* and *C* is rectangular and isosceles ( is right); - all points of the favorite rectangle are located inside or on the border of triangle *ABC*; - the area of triangle *ABC* is as small as possible.
Help the bear, find the required points. It is not so hard to proof that these points are unique.
Input Specification:
The first line contains two integers *x*,<=*y* (<=-<=109<=β€<=*x*,<=*y*<=β€<=109,<=*x*<=β <=0,<=*y*<=β <=0).
Output Specification:
Print in the single line four integers *x*1,<=*y*1,<=*x*2,<=*y*2 β the coordinates of the required points.
Demo Input:
['10 5\n', '-10 5\n']
Demo Output:
['0 15 15 0\n', '-15 0 0 15\n']
Note:
<img class="tex-graphics" src="https://espresso.codeforces.com/a9ea2088c4294ce8f23801562fda36b830df2c3f.png" style="max-width: 100.0%;max-height: 100.0%;"/>
Figure to the first sample
|
```python
n=input().split()
x,y=int(n[0]),int(n[1])
if x>0 and y>0:
print(0,x+y,x+y,0)
elif x>0 and y<0:
print(0,y-x,x-y,0)
elif x<0 and y>0:
print(x-y,0,0,y-x)
else:
print(x+y,0,0,x+y)
```
| 3
|
|
862
|
A
|
Mahmoud and Ehab and the MEX
|
PROGRAMMING
| 1,000
|
[
"greedy",
"implementation"
] | null | null |
Dr. Evil kidnapped Mahmoud and Ehab in the evil land because of their performance in the Evil Olympiad in Informatics (EOI). He decided to give them some problems to let them go.
Dr. Evil is interested in sets, He has a set of *n* integers. Dr. Evil calls a set of integers evil if the MEX of it is exactly *x*. the MEX of a set of integers is the minimum non-negative integer that doesn't exist in it. For example, the MEX of the set {0,<=2,<=4} is 1 and the MEX of the set {1,<=2,<=3} is 0 .
Dr. Evil is going to make his set evil. To do this he can perform some operations. During each operation he can add some non-negative integer to his set or erase some element from it. What is the minimal number of operations Dr. Evil has to perform to make his set evil?
|
The first line contains two integers *n* and *x* (1<=β€<=*n*<=β€<=100, 0<=β€<=*x*<=β€<=100)Β β the size of the set Dr. Evil owns, and the desired MEX.
The second line contains *n* distinct non-negative integers not exceeding 100 that represent the set.
|
The only line should contain one integerΒ β the minimal number of operations Dr. Evil should perform.
|
[
"5 3\n0 4 5 6 7\n",
"1 0\n0\n",
"5 0\n1 2 3 4 5\n"
] |
[
"2\n",
"1\n",
"0\n"
] |
For the first test case Dr. Evil should add 1 and 2 to the set performing 2 operations.
For the second test case Dr. Evil should erase 0 from the set. After that, the set becomes empty, so the MEX of it is 0.
In the third test case the set is already evil.
| 500
|
[
{
"input": "5 3\n0 4 5 6 7",
"output": "2"
},
{
"input": "1 0\n0",
"output": "1"
},
{
"input": "5 0\n1 2 3 4 5",
"output": "0"
},
{
"input": "10 5\n57 1 47 9 93 37 76 70 78 15",
"output": "4"
},
{
"input": "10 5\n99 98 93 97 95 100 92 94 91 96",
"output": "5"
},
{
"input": "10 5\n1 2 3 4 59 45 0 58 51 91",
"output": "0"
},
{
"input": "100 100\n79 13 21 11 3 87 28 40 29 4 96 34 8 78 61 46 33 45 99 30 92 67 22 97 39 86 73 31 74 44 62 55 57 2 54 63 80 69 25 48 77 98 17 93 15 16 89 12 43 23 37 95 14 38 83 90 49 56 72 10 20 0 50 71 70 88 19 1 76 81 52 41 82 68 85 47 6 7 35 60 18 64 75 84 27 9 65 91 94 42 53 24 66 26 59 36 51 32 5 58",
"output": "0"
},
{
"input": "100 50\n95 78 46 92 80 18 79 58 30 72 19 89 39 29 44 65 15 100 59 8 96 9 62 67 41 42 82 14 57 32 71 77 40 5 7 51 28 53 85 23 16 35 3 91 6 11 75 61 17 66 13 47 36 56 10 22 83 60 48 24 26 97 4 33 76 86 70 0 34 64 52 43 21 49 55 74 1 73 81 25 54 63 94 84 20 68 87 12 31 88 38 93 37 90 98 69 99 45 27 2",
"output": "0"
},
{
"input": "100 33\n28 11 79 92 88 62 77 72 7 41 96 97 67 84 44 8 81 35 38 1 64 68 46 17 98 83 31 12 74 21 2 22 47 6 36 75 65 61 37 26 25 45 59 48 100 51 93 76 78 49 3 57 16 4 87 29 55 82 70 39 53 0 60 15 24 71 58 20 66 89 95 42 13 43 63 90 85 52 50 30 54 40 56 23 27 34 32 18 10 19 69 9 99 73 91 14 5 80 94 86",
"output": "0"
},
{
"input": "99 33\n25 76 41 95 55 20 47 59 58 84 87 92 16 27 35 65 72 63 93 54 36 96 15 86 5 69 24 46 67 73 48 60 40 6 61 74 97 10 100 8 52 26 77 18 7 62 37 2 14 66 11 56 68 91 0 64 75 99 30 21 53 1 89 81 3 98 12 88 39 38 29 83 22 90 9 28 45 43 78 44 32 57 4 50 70 17 13 51 80 85 71 94 82 19 34 42 23 79 49",
"output": "1"
},
{
"input": "100 100\n65 56 84 46 44 33 99 74 62 72 93 67 43 92 75 88 38 34 66 12 55 76 58 90 78 8 14 45 97 59 48 32 64 18 39 89 31 51 54 81 29 36 70 77 40 22 49 27 3 1 73 13 98 42 87 37 2 57 4 6 50 25 23 79 28 86 68 61 80 17 19 10 15 63 52 11 35 60 21 16 24 85 30 91 7 5 69 20 71 82 53 94 41 95 96 9 26 83 0 47",
"output": "0"
},
{
"input": "100 100\n58 88 12 71 22 1 40 19 73 20 67 48 57 17 69 36 100 35 33 37 72 55 52 8 89 85 47 42 78 70 81 86 11 9 68 99 6 16 21 61 53 98 23 62 32 59 51 0 87 24 50 30 65 10 80 95 7 92 25 74 60 79 91 5 13 31 75 38 90 94 46 66 93 34 14 41 28 2 76 84 43 96 3 56 49 82 27 77 64 63 4 45 18 29 54 39 15 26 83 44",
"output": "2"
},
{
"input": "89 100\n58 96 17 41 86 34 28 84 18 40 8 77 87 89 68 79 33 35 53 49 0 6 22 12 72 90 48 55 21 50 56 62 75 2 37 95 69 74 14 20 44 46 27 32 31 59 63 60 10 85 71 70 38 52 94 30 61 51 80 26 36 23 39 47 76 45 100 57 15 78 97 66 54 13 99 16 93 73 24 4 83 5 98 81 92 25 29 88 65",
"output": "13"
},
{
"input": "100 50\n7 95 24 76 81 78 60 69 83 84 100 1 65 31 48 92 73 39 18 89 38 97 10 42 8 55 98 51 21 90 62 77 16 91 0 94 4 37 19 17 67 35 45 41 56 20 15 85 75 28 59 27 12 54 61 68 36 5 79 93 66 11 70 49 50 34 30 25 96 46 64 14 32 22 47 40 58 23 43 9 87 82 26 53 80 52 3 86 13 99 33 71 6 88 57 74 2 44 72 63",
"output": "2"
},
{
"input": "77 0\n27 8 20 92 21 41 53 98 17 65 67 35 81 11 55 49 61 44 2 66 51 89 40 28 52 62 86 91 64 24 18 5 94 82 96 99 71 6 39 83 26 29 16 30 45 97 80 90 69 12 13 33 76 73 46 19 78 56 88 38 42 34 57 77 47 4 59 58 7 100 95 72 9 74 15 43 54",
"output": "0"
},
{
"input": "100 50\n55 36 0 32 81 6 17 43 24 13 30 19 8 59 71 45 15 74 3 41 99 42 86 47 2 94 35 1 66 95 38 49 4 27 96 89 34 44 92 25 51 39 54 28 80 77 20 14 48 40 68 56 31 63 33 78 69 37 18 26 83 70 23 82 91 65 67 52 61 53 7 22 60 21 12 73 72 87 75 100 90 29 64 79 98 85 5 62 93 84 50 46 97 58 57 16 9 10 76 11",
"output": "1"
},
{
"input": "77 0\n12 8 19 87 9 54 55 86 97 7 27 85 25 48 94 73 26 1 13 57 72 69 76 39 38 91 75 40 42 28 93 21 70 84 65 11 60 90 20 95 66 89 59 47 34 99 6 61 52 100 50 3 77 81 82 53 15 24 0 45 44 14 68 96 58 5 18 35 10 98 29 74 92 49 83 71 17",
"output": "1"
},
{
"input": "100 70\n25 94 66 65 10 99 89 6 70 31 7 40 20 92 64 27 21 72 77 98 17 43 47 44 48 81 38 56 100 39 90 22 88 76 3 83 86 29 33 55 82 79 49 11 2 16 12 78 85 69 32 97 26 15 53 24 23 91 51 67 34 35 52 5 62 50 95 18 71 13 75 8 30 42 93 36 45 60 63 46 57 41 87 0 84 54 74 37 4 58 28 19 96 61 80 9 1 14 73 68",
"output": "2"
},
{
"input": "89 19\n14 77 85 81 79 38 91 45 55 51 50 11 62 67 73 76 2 27 16 23 3 29 65 98 78 17 4 58 22 20 34 66 64 31 72 5 32 44 12 75 80 47 18 25 99 0 61 56 71 84 48 88 10 7 86 8 49 24 43 21 37 28 33 54 46 57 40 89 36 97 6 96 39 95 26 74 1 69 9 100 52 30 83 87 68 60 92 90 35",
"output": "2"
},
{
"input": "89 100\n69 61 56 45 11 41 42 32 28 29 0 76 7 65 13 35 36 82 10 39 26 34 38 40 92 12 17 54 24 46 88 70 66 27 100 52 85 62 22 48 86 68 21 49 53 94 67 20 1 90 77 84 31 87 58 47 95 33 4 72 93 83 8 51 91 80 99 43 71 19 44 59 98 97 64 9 81 16 79 63 25 37 3 75 2 55 50 6 18",
"output": "13"
},
{
"input": "77 0\n38 76 24 74 42 88 29 75 96 46 90 32 59 97 98 60 41 57 80 37 100 49 25 63 95 31 61 68 53 78 27 66 84 48 94 83 30 26 36 99 71 62 45 47 70 28 35 54 34 85 79 43 91 72 86 33 67 92 77 65 69 52 82 55 87 64 56 40 50 44 51 73 89 81 58 93 39",
"output": "0"
},
{
"input": "89 100\n38 90 80 64 35 44 56 11 15 89 23 12 49 70 72 60 63 85 92 10 45 83 8 88 41 33 16 6 61 76 62 71 87 13 25 77 74 0 1 37 96 93 7 94 21 82 34 78 4 73 65 20 81 95 50 32 48 17 69 55 68 5 51 27 53 43 91 67 59 46 86 84 99 24 22 3 97 98 40 36 26 58 57 9 42 30 52 2 47",
"output": "11"
},
{
"input": "77 0\n55 71 78 86 68 35 53 10 59 32 81 19 74 97 62 61 93 87 96 44 25 18 43 82 84 16 34 48 92 39 64 36 49 91 45 76 95 31 57 29 75 79 13 2 14 24 52 23 33 20 47 99 63 15 5 80 58 67 12 3 85 6 1 27 73 90 4 42 37 70 8 11 89 77 9 22 94",
"output": "0"
},
{
"input": "77 0\n12 75 31 71 44 8 3 82 21 77 50 29 57 74 40 10 15 42 84 2 100 9 28 72 92 0 49 11 90 55 17 36 19 54 68 52 4 69 97 91 5 39 59 45 89 62 53 83 16 94 76 60 95 47 30 51 7 48 20 70 67 32 58 78 63 34 56 93 99 88 24 1 66 22 25 14 13",
"output": "1"
},
{
"input": "100 70\n91 82 8 85 26 25 95 97 40 87 81 93 7 73 38 94 64 96 74 18 90 19 65 68 72 61 23 43 36 41 60 88 30 33 71 24 52 39 15 3 16 89 86 79 55 4 9 58 67 44 46 29 6 48 84 69 27 21 78 54 51 57 80 53 76 50 47 77 45 12 34 10 100 0 17 31 56 99 98 11 92 5 2 42 32 59 66 62 37 63 28 75 35 1 22 13 83 49 20 14",
"output": "0"
},
{
"input": "77 0\n51 5 81 62 30 22 11 0 83 16 79 85 52 70 69 10 8 47 58 3 24 34 44 14 82 66 99 17 28 31 64 67 23 49 94 45 4 12 27 15 21 6 43 72 87 2 63 92 35 39 59 9 90 78 93 20 65 36 60 89 50 41 61 84 77 86 76 100 38 68 53 97 96 95 7 19 88",
"output": "1"
},
{
"input": "1 100\n0",
"output": "99"
},
{
"input": "1 0\n100",
"output": "0"
},
{
"input": "1 100\n100",
"output": "101"
},
{
"input": "2 100\n0 100",
"output": "100"
},
{
"input": "5 5\n1 2 3 4 5",
"output": "2"
},
{
"input": "5 3\n0 3 4 5 6",
"output": "3"
},
{
"input": "7 10\n0 1 2 3 4 5 10",
"output": "5"
},
{
"input": "2 2\n0 2",
"output": "2"
},
{
"input": "2 1\n1 2",
"output": "2"
},
{
"input": "5 1\n1 2 3 4 5",
"output": "2"
},
{
"input": "5 2\n1 2 3 4 5",
"output": "2"
},
{
"input": "5 6\n0 1 2 3 4",
"output": "1"
},
{
"input": "3 2\n3 4 5",
"output": "2"
}
] | 1,671,815,291
| 191
|
PyPy 3
|
OK
|
TESTS
| 39
| 93
| 0
|
import sys, os, io
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
n, x = map(int, input().split())
a = list(map(int, input().split()))
l = 105
cnt = [0] * l
for i in a:
cnt[i] += 1
ans = 0
for i in range(x):
if not cnt[i]:
ans += 1
ans += cnt[x]
print(ans)
|
Title: Mahmoud and Ehab and the MEX
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Dr. Evil kidnapped Mahmoud and Ehab in the evil land because of their performance in the Evil Olympiad in Informatics (EOI). He decided to give them some problems to let them go.
Dr. Evil is interested in sets, He has a set of *n* integers. Dr. Evil calls a set of integers evil if the MEX of it is exactly *x*. the MEX of a set of integers is the minimum non-negative integer that doesn't exist in it. For example, the MEX of the set {0,<=2,<=4} is 1 and the MEX of the set {1,<=2,<=3} is 0 .
Dr. Evil is going to make his set evil. To do this he can perform some operations. During each operation he can add some non-negative integer to his set or erase some element from it. What is the minimal number of operations Dr. Evil has to perform to make his set evil?
Input Specification:
The first line contains two integers *n* and *x* (1<=β€<=*n*<=β€<=100, 0<=β€<=*x*<=β€<=100)Β β the size of the set Dr. Evil owns, and the desired MEX.
The second line contains *n* distinct non-negative integers not exceeding 100 that represent the set.
Output Specification:
The only line should contain one integerΒ β the minimal number of operations Dr. Evil should perform.
Demo Input:
['5 3\n0 4 5 6 7\n', '1 0\n0\n', '5 0\n1 2 3 4 5\n']
Demo Output:
['2\n', '1\n', '0\n']
Note:
For the first test case Dr. Evil should add 1 and 2 to the set performing 2 operations.
For the second test case Dr. Evil should erase 0 from the set. After that, the set becomes empty, so the MEX of it is 0.
In the third test case the set is already evil.
|
```python
import sys, os, io
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
n, x = map(int, input().split())
a = list(map(int, input().split()))
l = 105
cnt = [0] * l
for i in a:
cnt[i] += 1
ans = 0
for i in range(x):
if not cnt[i]:
ans += 1
ans += cnt[x]
print(ans)
```
| 3
|
|
327
|
A
|
Flipping Game
|
PROGRAMMING
| 1,200
|
[
"brute force",
"dp",
"implementation"
] | null | null |
Iahub got bored, so he invented a game to be played on paper.
He writes *n* integers *a*1,<=*a*2,<=...,<=*a**n*. Each of those integers can be either 0 or 1. He's allowed to do exactly one move: he chooses two indices *i* and *j* (1<=β€<=*i*<=β€<=*j*<=β€<=*n*) and flips all values *a**k* for which their positions are in range [*i*,<=*j*] (that is *i*<=β€<=*k*<=β€<=*j*). Flip the value of *x* means to apply operation *x*<==<=1 - *x*.
The goal of the game is that after exactly one move to obtain the maximum number of ones. Write a program to solve the little game of Iahub.
|
The first line of the input contains an integer *n* (1<=β€<=*n*<=β€<=100). In the second line of the input there are *n* integers: *a*1,<=*a*2,<=...,<=*a**n*. It is guaranteed that each of those *n* values is either 0 or 1.
|
Print an integer β the maximal number of 1s that can be obtained after exactly one move.
|
[
"5\n1 0 0 1 0\n",
"4\n1 0 0 1\n"
] |
[
"4\n",
"4\n"
] |
In the first case, flip the segment from 2 to 5 (*i*β=β2,β*j*β=β5). That flip changes the sequence, it becomes: [1 1 1 0 1]. So, it contains four ones. There is no way to make the whole sequence equal to [1 1 1 1 1].
In the second case, flipping only the second and the third element (*i*β=β2,β*j*β=β3) will turn all numbers into 1.
| 500
|
[
{
"input": "5\n1 0 0 1 0",
"output": "4"
},
{
"input": "4\n1 0 0 1",
"output": "4"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "1\n0",
"output": "1"
},
{
"input": "8\n1 0 0 0 1 0 0 0",
"output": "7"
},
{
"input": "18\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0",
"output": "18"
},
{
"input": "23\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "22"
},
{
"input": "100\n0 1 0 1 1 1 0 1 0 1 0 0 1 1 1 1 0 0 1 1 1 1 1 1 1 0 0 1 1 1 0 1 1 0 0 0 1 1 1 1 0 0 1 1 1 0 0 1 1 0 1 1 1 0 0 0 1 0 0 0 0 0 1 1 0 0 1 1 1 1 1 1 1 1 0 1 1 1 0 1 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1",
"output": "70"
},
{
"input": "100\n0 1 1 0 1 0 0 1 0 0 0 1 1 0 0 0 1 1 1 0 1 0 0 0 0 0 1 0 1 0 1 0 1 0 1 0 0 0 1 0 0 0 0 0 1 1 0 1 0 1 0 1 1 1 0 1 0 1 1 0 0 1 1 0 0 1 1 1 0 0 1 0 0 1 1 0 1 0 0 1 1 0 1 0 0 1 1 0 0 0 0 1 0 0 0 0 1 1 1 1",
"output": "60"
},
{
"input": "18\n0 1 0 1 0 1 0 1 0 1 1 0 1 1 0 1 1 0",
"output": "11"
},
{
"input": "25\n0 1 0 0 0 0 0 1 0 1 0 1 0 0 0 0 1 1 1 0 0 1 1 0 1",
"output": "18"
},
{
"input": "55\n0 0 1 1 0 0 0 1 0 1 1 0 1 1 1 0 1 1 1 1 1 0 0 1 0 0 1 0 1 1 0 0 1 0 1 1 0 1 1 1 1 0 1 1 0 0 0 0 1 1 0 1 1 1 1",
"output": "36"
},
{
"input": "75\n1 1 0 1 0 1 1 0 0 0 0 0 1 1 1 1 1 0 1 0 1 0 0 0 0 1 1 1 0 1 0 0 1 1 0 1 0 0 1 1 0 1 0 1 0 1 0 0 0 0 1 0 0 1 1 1 0 0 1 0 1 1 0 0 0 0 1 1 0 0 0 1 0 0 0",
"output": "44"
},
{
"input": "100\n0 0 1 0 1 0 0 1 1 0 1 1 0 1 0 1 1 0 0 0 0 0 1 0 0 1 1 0 0 0 1 0 0 1 1 0 0 1 1 1 0 0 0 0 1 0 1 1 1 0 0 1 0 1 1 1 1 1 1 1 0 1 0 1 0 0 1 0 1 1 1 0 0 0 0 1 0 1 1 0 0 1 1 0 1 1 1 1 0 1 1 1 0 0 1 1 0 1 0 1",
"output": "61"
},
{
"input": "100\n0 0 0 1 0 0 0 1 0 1 1 0 1 1 1 1 1 0 1 0 1 1 0 0 1 1 0 1 0 1 0 1 0 1 1 0 1 1 0 0 0 1 1 1 1 0 1 1 0 1 1 1 1 0 1 0 0 1 0 1 0 0 0 0 1 1 0 0 1 0 0 1 1 0 1 1 0 1 0 0 1 1 0 1 1 1 1 0 1 0 0 1 0 1 1 1 0 1 1 0",
"output": "61"
},
{
"input": "100\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0",
"output": "100"
},
{
"input": "100\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "99"
},
{
"input": "100\n0 0 1 1 1 0 1 1 0 0 1 0 1 0 1 0 0 1 0 1 1 0 0 1 1 1 1 1 1 0 1 0 0 1 0 0 0 0 1 1 1 0 1 1 1 0 1 1 1 0 0 1 0 0 1 1 0 0 0 1 1 1 0 1 1 0 0 0 0 1 0 0 1 0 1 0 0 0 0 0 0 1 1 1 0 1 1 1 0 1 1 1 1 1 0 0 0 0 1 0",
"output": "61"
},
{
"input": "100\n0 1 1 1 1 1 1 0 0 0 1 1 0 0 0 1 0 0 0 0 0 0 1 1 1 0 0 0 1 1 1 0 1 0 0 1 1 0 0 0 1 1 1 1 1 1 1 1 1 1 0 0 1 0 1 0 0 0 0 1 0 1 0 1 0 0 1 0 1 0 1 1 0 0 0 0 0 1 1 1 0 1 1 0 1 1 0 0 0 1 0 1 1 0 0 0 0 0 0 1",
"output": "59"
},
{
"input": "99\n1 1 0 0 0 1 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 0 0 0 0 0 0 1 1 0 0 1 0 1 1 1 0 0 0 0 0 1 0 0 0 0 0 0 1 0 1 1 0 1 0 0 1 1 0 1 0 0 0 1 1 1 1 0 1 1 1 0 0 1 1 0 0 0 0 1 0 0 0 0 0 1 0 1 1 1",
"output": "61"
},
{
"input": "2\n1 1",
"output": "1"
}
] | 1,692,787,446
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 21
| 154
| 1,740,800
|
n = int(input())
a = list(map(int, input().split()))
max_ones = 0
for i in range(n):
for j in range(i, n):
temp = a.copy()
for k in range(i, j+1):
temp[k] = 1 - temp[k]
max_ones = max(max_ones, temp.count(1))
print(max_ones)
|
Title: Flipping Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Iahub got bored, so he invented a game to be played on paper.
He writes *n* integers *a*1,<=*a*2,<=...,<=*a**n*. Each of those integers can be either 0 or 1. He's allowed to do exactly one move: he chooses two indices *i* and *j* (1<=β€<=*i*<=β€<=*j*<=β€<=*n*) and flips all values *a**k* for which their positions are in range [*i*,<=*j*] (that is *i*<=β€<=*k*<=β€<=*j*). Flip the value of *x* means to apply operation *x*<==<=1 - *x*.
The goal of the game is that after exactly one move to obtain the maximum number of ones. Write a program to solve the little game of Iahub.
Input Specification:
The first line of the input contains an integer *n* (1<=β€<=*n*<=β€<=100). In the second line of the input there are *n* integers: *a*1,<=*a*2,<=...,<=*a**n*. It is guaranteed that each of those *n* values is either 0 or 1.
Output Specification:
Print an integer β the maximal number of 1s that can be obtained after exactly one move.
Demo Input:
['5\n1 0 0 1 0\n', '4\n1 0 0 1\n']
Demo Output:
['4\n', '4\n']
Note:
In the first case, flip the segment from 2 to 5 (*i*β=β2,β*j*β=β5). That flip changes the sequence, it becomes: [1 1 1 0 1]. So, it contains four ones. There is no way to make the whole sequence equal to [1 1 1 1 1].
In the second case, flipping only the second and the third element (*i*β=β2,β*j*β=β3) will turn all numbers into 1.
|
```python
n = int(input())
a = list(map(int, input().split()))
max_ones = 0
for i in range(n):
for j in range(i, n):
temp = a.copy()
for k in range(i, j+1):
temp[k] = 1 - temp[k]
max_ones = max(max_ones, temp.count(1))
print(max_ones)
```
| 3
|
|
272
|
A
|
Dima and Friends
|
PROGRAMMING
| 1,000
|
[
"implementation",
"math"
] | null | null |
Dima and his friends have been playing hide and seek at Dima's place all night. As a result, Dima's place got messy. In the morning they decided that they need to clean the place.
To decide who exactly would clean the apartment, the friends want to play a counting-out game. First, all the guys stand in a circle, and then each of them shows some number of fingers on one hand (one to five), and then the boys count in a circle, starting from Dima, the number of people, respective to the total number of fingers shown. The person on who the countdown stops will clean the apartment.
For example, if Dima and one of his friends played hide and seek, and 7 fingers were shown during the counting-out, then Dima would clean the place. If there were 2 or say, 8 fingers shown, then his friend would clean the place.
Dima knows how many fingers each of his friends will show during the counting-out. Now he is interested in the number of ways to show some number of fingers on one hand (one to five), so that he did not have to clean the place. Help Dima.
|
The first line contains integer *n* (1<=β€<=*n*<=β€<=100) β the number of Dima's friends. Dima himself isn't considered to be his own friend. The second line contains *n* positive integers, not exceeding 5, representing, how many fingers the Dima's friends will show.
The numbers in the lines are separated by a single space.
|
In a single line print the answer to the problem.
|
[
"1\n1\n",
"1\n2\n",
"2\n3 5\n"
] |
[
"3\n",
"2\n",
"3\n"
] |
In the first sample Dima can show 1, 3 or 5 fingers. If Dima shows 3 fingers, then the counting-out will go like that: Dima, his friend, Dima, his friend.
In the second sample Dima can show 2 or 4 fingers.
| 500
|
[
{
"input": "1\n1",
"output": "3"
},
{
"input": "1\n2",
"output": "2"
},
{
"input": "2\n3 5",
"output": "3"
},
{
"input": "2\n3 5",
"output": "3"
},
{
"input": "1\n5",
"output": "3"
},
{
"input": "5\n4 4 3 5 1",
"output": "4"
},
{
"input": "6\n2 3 2 2 1 3",
"output": "4"
},
{
"input": "8\n2 2 5 3 4 3 3 2",
"output": "4"
},
{
"input": "7\n4 1 3 2 2 4 5",
"output": "4"
},
{
"input": "3\n3 5 1",
"output": "4"
},
{
"input": "95\n4 2 3 4 4 5 2 2 4 4 3 5 3 3 3 5 4 2 5 4 2 1 1 3 4 2 1 3 5 4 2 1 1 5 1 1 2 2 4 4 5 4 5 5 2 1 2 2 2 4 5 5 2 4 3 4 4 3 5 2 4 1 5 4 5 1 3 2 4 2 2 1 5 3 1 5 3 4 3 3 2 1 2 2 1 3 1 5 2 3 1 1 2 5 2",
"output": "5"
},
{
"input": "31\n3 2 3 3 3 3 4 4 1 5 5 4 2 4 3 2 2 1 4 4 1 2 3 1 1 5 5 3 4 4 1",
"output": "4"
},
{
"input": "42\n3 1 2 2 5 1 2 2 4 5 4 5 2 5 4 5 4 4 1 4 3 3 4 4 4 4 3 2 1 3 4 5 5 2 1 2 1 5 5 2 4 4",
"output": "5"
},
{
"input": "25\n4 5 5 5 3 1 1 4 4 4 3 5 4 4 1 4 4 1 2 4 2 5 4 5 3",
"output": "5"
},
{
"input": "73\n3 4 3 4 5 1 3 4 2 1 4 2 2 3 5 3 1 4 2 3 2 1 4 5 3 5 2 2 4 3 2 2 5 3 2 3 5 1 3 1 1 4 5 2 4 2 5 1 4 3 1 3 1 4 2 3 3 3 3 5 5 2 5 2 5 4 3 1 1 5 5 2 3",
"output": "4"
},
{
"input": "46\n1 4 4 5 4 5 2 3 5 5 3 2 5 4 1 3 2 2 1 4 3 1 5 5 2 2 2 2 4 4 1 1 4 3 4 3 1 4 2 2 4 2 3 2 5 2",
"output": "4"
},
{
"input": "23\n5 2 1 1 4 2 5 5 3 5 4 5 5 1 1 5 2 4 5 3 4 4 3",
"output": "5"
},
{
"input": "6\n4 2 3 1 3 5",
"output": "4"
},
{
"input": "15\n5 5 5 3 5 4 1 3 3 4 3 4 1 4 4",
"output": "5"
},
{
"input": "93\n1 3 1 4 3 3 5 3 1 4 5 4 3 2 2 4 3 1 4 1 2 3 3 3 2 5 1 3 1 4 5 1 1 1 4 2 1 2 3 1 1 1 5 1 5 5 1 2 5 4 3 2 2 4 4 2 5 4 5 5 3 1 3 1 2 1 3 1 1 2 3 4 4 5 5 3 2 1 3 3 5 1 3 5 4 4 1 3 3 4 2 3 2",
"output": "5"
},
{
"input": "96\n1 5 1 3 2 1 2 2 2 2 3 4 1 1 5 4 4 1 2 3 5 1 4 4 4 1 3 3 1 4 5 4 1 3 5 3 4 4 3 2 1 1 4 4 5 1 1 2 5 1 2 3 1 4 1 2 2 2 3 2 3 3 2 5 2 2 3 3 3 3 2 1 2 4 5 5 1 5 3 2 1 4 3 5 5 5 3 3 5 3 4 3 4 2 1 3",
"output": "5"
},
{
"input": "49\n1 4 4 3 5 2 2 1 5 1 2 1 2 5 1 4 1 4 5 2 4 5 3 5 2 4 2 1 3 4 2 1 4 2 1 1 3 3 2 3 5 4 3 4 2 4 1 4 1",
"output": "5"
},
{
"input": "73\n4 1 3 3 3 1 5 2 1 4 1 1 3 5 1 1 4 5 2 1 5 4 1 5 3 1 5 2 4 5 1 4 3 3 5 2 2 3 3 2 5 1 4 5 2 3 1 4 4 3 5 2 3 5 1 4 3 5 1 2 4 1 3 3 5 4 2 4 2 4 1 2 5",
"output": "5"
},
{
"input": "41\n5 3 5 4 2 5 4 3 1 1 1 5 4 3 4 3 5 4 2 5 4 1 1 3 2 4 5 3 5 1 5 5 1 1 1 4 4 1 2 4 3",
"output": "5"
},
{
"input": "100\n3 3 1 4 2 4 4 3 1 5 1 1 4 4 3 4 4 3 5 4 5 2 4 3 4 1 2 4 5 4 2 1 5 4 1 1 4 3 2 4 1 2 1 4 4 5 5 4 4 5 3 2 5 1 4 2 2 1 1 2 5 2 5 1 5 3 1 4 3 2 4 3 2 2 4 5 5 1 2 3 1 4 1 2 2 2 5 5 2 3 2 4 3 1 1 2 1 2 1 2",
"output": "5"
},
{
"input": "100\n2 1 1 3 5 4 4 2 3 4 3 4 5 4 5 4 2 4 5 3 4 5 4 1 1 4 4 1 1 2 5 4 2 4 5 3 2 5 4 3 4 5 1 3 4 2 5 4 5 4 5 2 4 1 2 5 3 1 4 4 5 3 4 3 1 2 5 4 2 5 4 1 5 3 5 4 1 2 5 3 1 1 1 1 5 3 4 3 5 1 1 5 5 1 1 2 2 1 5 1",
"output": "5"
},
{
"input": "100\n4 4 3 3 2 5 4 4 2 1 4 4 4 5 4 1 2 1 5 2 4 3 4 1 4 1 2 5 1 4 5 4 2 1 2 5 3 4 5 5 2 1 2 2 2 2 2 3 2 5 1 2 2 3 2 5 5 1 3 4 5 2 1 3 4 2 2 4 4 3 3 3 2 3 2 1 5 5 5 2 1 4 2 3 5 1 4 4 2 3 2 5 5 4 3 5 1 3 5 5",
"output": "5"
},
{
"input": "100\n4 4 2 5 4 2 2 3 4 4 3 2 3 3 1 3 4 3 3 4 1 3 1 4 5 3 4 3 1 1 1 3 3 2 3 4 3 4 2 2 1 5 1 4 5 1 1 1 3 3 1 1 3 2 5 4 2 5 2 4 5 4 4 1 1 2 1 1 4 5 1 1 5 3 3 2 5 5 5 1 4 1 4 1 1 3 2 3 4 4 2 5 5 2 5 1 1 3 5 3",
"output": "5"
},
{
"input": "100\n4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4",
"output": "4"
},
{
"input": "100\n5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5",
"output": "5"
}
] | 1,591,859,419
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 30
| 280
| 0
|
n = int(input())
l = list(map(int,input().split()))
if n<3:
if n%2==0:
if sum(l)%2==0:
print(3)
else:
print(2)
else:
if sum(l)%2==0:
print(2)
else:
print(3)
else:
mod = sum(l)%(n+1)
mod = n+1-mod
if mod>=5:
print(5)
else:
print(4)
|
Title: Dima and Friends
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Dima and his friends have been playing hide and seek at Dima's place all night. As a result, Dima's place got messy. In the morning they decided that they need to clean the place.
To decide who exactly would clean the apartment, the friends want to play a counting-out game. First, all the guys stand in a circle, and then each of them shows some number of fingers on one hand (one to five), and then the boys count in a circle, starting from Dima, the number of people, respective to the total number of fingers shown. The person on who the countdown stops will clean the apartment.
For example, if Dima and one of his friends played hide and seek, and 7 fingers were shown during the counting-out, then Dima would clean the place. If there were 2 or say, 8 fingers shown, then his friend would clean the place.
Dima knows how many fingers each of his friends will show during the counting-out. Now he is interested in the number of ways to show some number of fingers on one hand (one to five), so that he did not have to clean the place. Help Dima.
Input Specification:
The first line contains integer *n* (1<=β€<=*n*<=β€<=100) β the number of Dima's friends. Dima himself isn't considered to be his own friend. The second line contains *n* positive integers, not exceeding 5, representing, how many fingers the Dima's friends will show.
The numbers in the lines are separated by a single space.
Output Specification:
In a single line print the answer to the problem.
Demo Input:
['1\n1\n', '1\n2\n', '2\n3 5\n']
Demo Output:
['3\n', '2\n', '3\n']
Note:
In the first sample Dima can show 1, 3 or 5 fingers. If Dima shows 3 fingers, then the counting-out will go like that: Dima, his friend, Dima, his friend.
In the second sample Dima can show 2 or 4 fingers.
|
```python
n = int(input())
l = list(map(int,input().split()))
if n<3:
if n%2==0:
if sum(l)%2==0:
print(3)
else:
print(2)
else:
if sum(l)%2==0:
print(2)
else:
print(3)
else:
mod = sum(l)%(n+1)
mod = n+1-mod
if mod>=5:
print(5)
else:
print(4)
```
| 3
|
|
69
|
A
|
Young Physicist
|
PROGRAMMING
| 1,000
|
[
"implementation",
"math"
] |
A. Young Physicist
|
2
|
256
|
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" β thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
|
The first line contains a positive integer *n* (1<=β€<=*n*<=β€<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=β€<=*x**i*,<=*y**i*,<=*z**i*<=β€<=100).
|
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
|
[
"3\n4 1 7\n-2 4 -1\n1 -5 -3\n",
"3\n3 -1 7\n-5 2 -4\n2 -1 -3\n"
] |
[
"NO",
"YES"
] |
none
| 500
|
[
{
"input": "3\n4 1 7\n-2 4 -1\n1 -5 -3",
"output": "NO"
},
{
"input": "3\n3 -1 7\n-5 2 -4\n2 -1 -3",
"output": "YES"
},
{
"input": "10\n21 32 -46\n43 -35 21\n42 2 -50\n22 40 20\n-27 -9 38\n-4 1 1\n-40 6 -31\n-13 -2 34\n-21 34 -12\n-32 -29 41",
"output": "NO"
},
{
"input": "10\n25 -33 43\n-27 -42 28\n-35 -20 19\n41 -42 -1\n49 -39 -4\n-49 -22 7\n-19 29 41\n8 -27 -43\n8 34 9\n-11 -3 33",
"output": "NO"
},
{
"input": "10\n-6 21 18\n20 -11 -8\n37 -11 41\n-5 8 33\n29 23 32\n30 -33 -11\n39 -49 -36\n28 34 -49\n22 29 -34\n-18 -6 7",
"output": "NO"
},
{
"input": "10\n47 -2 -27\n0 26 -14\n5 -12 33\n2 18 3\n45 -30 -49\n4 -18 8\n-46 -44 -41\n-22 -10 -40\n-35 -21 26\n33 20 38",
"output": "NO"
},
{
"input": "13\n-3 -36 -46\n-11 -50 37\n42 -11 -15\n9 42 44\n-29 -12 24\n3 9 -40\n-35 13 50\n14 43 18\n-13 8 24\n-48 -15 10\n50 9 -50\n21 0 -50\n0 0 -6",
"output": "YES"
},
{
"input": "14\n43 23 17\n4 17 44\n5 -5 -16\n-43 -7 -6\n47 -48 12\n50 47 -45\n2 14 43\n37 -30 15\n4 -17 -11\n17 9 -45\n-50 -3 -8\n-50 0 0\n-50 0 0\n-16 0 0",
"output": "YES"
},
{
"input": "13\n29 49 -11\n38 -11 -20\n25 1 -40\n-11 28 11\n23 -19 1\n45 -41 -17\n-3 0 -19\n-13 -33 49\n-30 0 28\n34 17 45\n-50 9 -27\n-50 0 0\n-37 0 0",
"output": "YES"
},
{
"input": "12\n3 28 -35\n-32 -44 -17\n9 -25 -6\n-42 -22 20\n-19 15 38\n-21 38 48\n-1 -37 -28\n-10 -13 -50\n-5 21 29\n34 28 50\n50 11 -49\n34 0 0",
"output": "YES"
},
{
"input": "37\n-64 -79 26\n-22 59 93\n-5 39 -12\n77 -9 76\n55 -86 57\n83 100 -97\n-70 94 84\n-14 46 -94\n26 72 35\n14 78 -62\n17 82 92\n-57 11 91\n23 15 92\n-80 -1 1\n12 39 18\n-23 -99 -75\n-34 50 19\n-39 84 -7\n45 -30 -39\n-60 49 37\n45 -16 -72\n33 -51 -56\n-48 28 5\n97 91 88\n45 -82 -11\n-21 -15 -90\n-53 73 -26\n-74 85 -90\n-40 23 38\n100 -13 49\n32 -100 -100\n0 -100 -70\n0 -100 0\n0 -100 0\n0 -100 0\n0 -100 0\n0 -37 0",
"output": "YES"
},
{
"input": "4\n68 3 100\n68 21 -100\n-100 -24 0\n-36 0 0",
"output": "YES"
},
{
"input": "33\n-1 -46 -12\n45 -16 -21\n-11 45 -21\n-60 -42 -93\n-22 -45 93\n37 96 85\n-76 26 83\n-4 9 55\n7 -52 -9\n66 8 -85\n-100 -54 11\n-29 59 74\n-24 12 2\n-56 81 85\n-92 69 -52\n-26 -97 91\n54 59 -51\n58 21 -57\n7 68 56\n-47 -20 -51\n-59 77 -13\n-85 27 91\n79 60 -56\n66 -80 5\n21 -99 42\n-31 -29 98\n66 93 76\n-49 45 61\n100 -100 -100\n100 -100 -100\n66 -75 -100\n0 0 -100\n0 0 -87",
"output": "YES"
},
{
"input": "3\n1 2 3\n3 2 1\n0 0 0",
"output": "NO"
},
{
"input": "2\n5 -23 12\n0 0 0",
"output": "NO"
},
{
"input": "1\n0 0 0",
"output": "YES"
},
{
"input": "1\n1 -2 0",
"output": "NO"
},
{
"input": "2\n-23 77 -86\n23 -77 86",
"output": "YES"
},
{
"input": "26\n86 7 20\n-57 -64 39\n-45 6 -93\n-44 -21 100\n-11 -49 21\n73 -71 -80\n-2 -89 56\n-65 -2 7\n5 14 84\n57 41 13\n-12 69 54\n40 -25 27\n-17 -59 0\n64 -91 -30\n-53 9 42\n-54 -8 14\n-35 82 27\n-48 -59 -80\n88 70 79\n94 57 97\n44 63 25\n84 -90 -40\n-100 100 -100\n-92 100 -100\n0 10 -100\n0 0 -82",
"output": "YES"
},
{
"input": "42\n11 27 92\n-18 -56 -57\n1 71 81\n33 -92 30\n82 83 49\n-87 -61 -1\n-49 45 49\n73 26 15\n-22 22 -77\n29 -93 87\n-68 44 -90\n-4 -84 20\n85 67 -6\n-39 26 77\n-28 -64 20\n65 -97 24\n-72 -39 51\n35 -75 -91\n39 -44 -8\n-25 -27 -57\n91 8 -46\n-98 -94 56\n94 -60 59\n-9 -95 18\n-53 -37 98\n-8 -94 -84\n-52 55 60\n15 -14 37\n65 -43 -25\n94 12 66\n-8 -19 -83\n29 81 -78\n-58 57 33\n24 86 -84\n-53 32 -88\n-14 7 3\n89 97 -53\n-5 -28 -91\n-100 100 -6\n-84 100 0\n0 100 0\n0 70 0",
"output": "YES"
},
{
"input": "3\n96 49 -12\n2 -66 28\n-98 17 -16",
"output": "YES"
},
{
"input": "5\n70 -46 86\n-100 94 24\n-27 63 -63\n57 -100 -47\n0 -11 0",
"output": "YES"
},
{
"input": "18\n-86 -28 70\n-31 -89 42\n31 -48 -55\n95 -17 -43\n24 -95 -85\n-21 -14 31\n68 -18 81\n13 31 60\n-15 28 99\n-42 15 9\n28 -61 -62\n-16 71 29\n-28 75 -48\n-77 -67 36\n-100 83 89\n100 100 -100\n57 34 -100\n0 0 -53",
"output": "YES"
},
{
"input": "44\n52 -54 -29\n-82 -5 -94\n-54 43 43\n91 16 71\n7 80 -91\n3 15 29\n-99 -6 -77\n-3 -77 -64\n73 67 34\n25 -10 -18\n-29 91 63\n-72 86 -16\n-68 85 -81\n-3 36 44\n-74 -14 -80\n34 -96 -97\n-76 -78 -33\n-24 44 -58\n98 12 77\n95 -63 -6\n-51 3 -90\n-92 -10 72\n7 3 -68\n57 -53 71\n29 57 -48\n35 -60 10\n79 -70 -61\n-20 77 55\n-86 -15 -35\n84 -88 -18\n100 -42 77\n-20 46 8\n-41 -43 -65\n38 -98 -23\n-100 65 45\n-7 -91 -63\n46 88 -85\n48 59 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 1",
"output": "YES"
},
{
"input": "18\n-14 -64 -91\n-8 -66 -86\n-23 92 -40\n6 -3 -53\n57 41 78\n-79 42 -22\n-88 -17 45\n4 -45 44\n83 -18 -25\n34 86 -92\n75 -30 12\n44 99 11\n-67 -13 72\n22 83 -56\n-37 71 72\n-9 -100 100\n0 -100 31\n0 -58 0",
"output": "YES"
},
{
"input": "23\n-70 37 78\n42 84 6\n28 -94 -24\n-49 76 95\n-67 18 84\n-53 78 -5\n65 -63 -64\n-66 -64 -69\n81 -93 95\n10 52 -79\n-89 -61 -64\n-64 -47 43\n-81 -35 55\n80 82 73\n-60 -81 -18\n49 -10 -19\n-58 70 12\n-24 -15 -93\n98 -93 -54\n-28 -75 11\n100 100 -63\n100 100 0\n56 34 0",
"output": "YES"
},
{
"input": "3\n0 2 -2\n1 -1 3\n-3 0 0",
"output": "NO"
}
] | 1,537,700,196
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 81
| 248
| 0
|
n = int(input())
xs, ys, zs = 0, 0, 0
for i in range(n):
l = [int(k) for k in input().split(' ')]
x, y, z = l[0], l[1], l[2]
xs, ys, zs = xs+x, ys+y, zs+z
if (xs==0) and (ys==0) and (zs==0):
print('YES')
else:
print('NO')
|
Title: Young Physicist
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" β thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
Input Specification:
The first line contains a positive integer *n* (1<=β€<=*n*<=β€<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=β€<=*x**i*,<=*y**i*,<=*z**i*<=β€<=100).
Output Specification:
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
Demo Input:
['3\n4 1 7\n-2 4 -1\n1 -5 -3\n', '3\n3 -1 7\n-5 2 -4\n2 -1 -3\n']
Demo Output:
['NO', 'YES']
Note:
none
|
```python
n = int(input())
xs, ys, zs = 0, 0, 0
for i in range(n):
l = [int(k) for k in input().split(' ')]
x, y, z = l[0], l[1], l[2]
xs, ys, zs = xs+x, ys+y, zs+z
if (xs==0) and (ys==0) and (zs==0):
print('YES')
else:
print('NO')
```
| 3.938
|
336
|
B
|
Vasily the Bear and Fly
|
PROGRAMMING
| 1,900
|
[
"math"
] | null | null |
One beautiful day Vasily the bear painted 2*m* circles of the same radius *R* on a coordinate plane. Circles with numbers from 1 to *m* had centers at points (2*R*<=-<=*R*,<=0), (4*R*<=-<=*R*,<=0), ..., (2*Rm*<=-<=*R*,<=0), respectively. Circles with numbers from *m*<=+<=1 to 2*m* had centers at points (2*R*<=-<=*R*,<=2*R*), (4*R*<=-<=*R*,<=2*R*), ..., (2*Rm*<=-<=*R*,<=2*R*), respectively.
Naturally, the bear painted the circles for a simple experiment with a fly. The experiment continued for *m*2 days. Each day of the experiment got its own unique number from 0 to *m*2<=-<=1, inclusive.
On the day number *i* the following things happened:
1. The fly arrived at the coordinate plane at the center of the circle with number ( is the result of dividing number *x* by number *y*, rounded down to an integer). 1. The fly went along the coordinate plane to the center of the circle number ( is the remainder after dividing number *x* by number *y*). The bear noticed that the fly went from the center of circle *v* to the center of circle *u* along the shortest path with all points lying on the border or inside at least one of the 2*m* circles. After the fly reached the center of circle *u*, it flew away in an unknown direction.
Help Vasily, count the average distance the fly went along the coordinate plane during each of these *m*2 days.
|
The first line contains two integers *m*,<=*R* (1<=β€<=*m*<=β€<=105, 1<=β€<=*R*<=β€<=10).
|
In a single line print a single real number β the answer to the problem. The answer will be considered correct if its absolute or relative error doesn't exceed 10<=-<=6.
|
[
"1 1\n",
"2 2\n"
] |
[
"2.0000000000\n",
"5.4142135624\n"
] |
<img class="tex-graphics" src="https://espresso.codeforces.com/9fe384073741e20965ddc4bf162afd3a604b6b39.png" style="max-width: 100.0%;max-height: 100.0%;"/>
Figure to the second sample
| 1,000
|
[
{
"input": "1 1",
"output": "2.0000000000"
},
{
"input": "2 2",
"output": "5.4142135624"
},
{
"input": "100000 3",
"output": "200002.4853316681"
},
{
"input": "2344 5",
"output": "7817.4790439982"
},
{
"input": "999 10",
"output": "6668.3010410807"
},
{
"input": "6 1",
"output": "5.0752966144"
},
{
"input": "43 4",
"output": "118.1337922078"
},
{
"input": "3333 3",
"output": "6668.4867900399"
},
{
"input": "4444 4",
"output": "11853.9818839104"
},
{
"input": "100000 10",
"output": "666674.9511055604"
},
{
"input": "3134 9",
"output": "18811.4606574435"
},
{
"input": "9 9",
"output": "63.0021484426"
},
{
"input": "32 9",
"output": "199.9170568378"
},
{
"input": "33333 9",
"output": "200005.4562967670"
},
{
"input": "99999 1",
"output": "66666.8284438896"
},
{
"input": "99999 9",
"output": "600001.4559950059"
},
{
"input": "99999 7",
"output": "466667.7991072268"
},
{
"input": "4234 4",
"output": "11293.9819587295"
},
{
"input": "66666 6",
"output": "266668.9707136318"
},
{
"input": "66666 9",
"output": "400003.4560704476"
},
{
"input": "67676 7",
"output": "315827.1324966100"
},
{
"input": "7777 7",
"output": "36298.4671653864"
},
{
"input": "7656 2",
"output": "10209.6572921612"
},
{
"input": "2 8",
"output": "21.6568542495"
},
{
"input": "2 10",
"output": "27.0710678119"
},
{
"input": "2 4",
"output": "10.8284271247"
},
{
"input": "8 1",
"output": "6.3530145174"
},
{
"input": "8 8",
"output": "50.8241161391"
},
{
"input": "92399 1",
"output": "61600.1617786019"
},
{
"input": "1 10",
"output": "20.0000000000"
},
{
"input": "3 1",
"output": "3.2570787221"
}
] | 1,377,735,208
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 31
| 686
| 0
|
import math
m,R = map (int,input().split())
D = math.sqrt (2) * R
result = 0
def sum_dist (n):
return n*(n+1)*R + 2*D*n
for i in range (1,m+1):
result += 2*R
if i-1 > 0: result += 2*R + D
if m-i > 0: result += 2*R + D
if i-2 > 0: result += sum_dist (i-2)
if m-i-1 > 0: result += sum_dist (m-i-1)
print (result / m / m)
|
Title: Vasily the Bear and Fly
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One beautiful day Vasily the bear painted 2*m* circles of the same radius *R* on a coordinate plane. Circles with numbers from 1 to *m* had centers at points (2*R*<=-<=*R*,<=0), (4*R*<=-<=*R*,<=0), ..., (2*Rm*<=-<=*R*,<=0), respectively. Circles with numbers from *m*<=+<=1 to 2*m* had centers at points (2*R*<=-<=*R*,<=2*R*), (4*R*<=-<=*R*,<=2*R*), ..., (2*Rm*<=-<=*R*,<=2*R*), respectively.
Naturally, the bear painted the circles for a simple experiment with a fly. The experiment continued for *m*2 days. Each day of the experiment got its own unique number from 0 to *m*2<=-<=1, inclusive.
On the day number *i* the following things happened:
1. The fly arrived at the coordinate plane at the center of the circle with number ( is the result of dividing number *x* by number *y*, rounded down to an integer). 1. The fly went along the coordinate plane to the center of the circle number ( is the remainder after dividing number *x* by number *y*). The bear noticed that the fly went from the center of circle *v* to the center of circle *u* along the shortest path with all points lying on the border or inside at least one of the 2*m* circles. After the fly reached the center of circle *u*, it flew away in an unknown direction.
Help Vasily, count the average distance the fly went along the coordinate plane during each of these *m*2 days.
Input Specification:
The first line contains two integers *m*,<=*R* (1<=β€<=*m*<=β€<=105, 1<=β€<=*R*<=β€<=10).
Output Specification:
In a single line print a single real number β the answer to the problem. The answer will be considered correct if its absolute or relative error doesn't exceed 10<=-<=6.
Demo Input:
['1 1\n', '2 2\n']
Demo Output:
['2.0000000000\n', '5.4142135624\n']
Note:
<img class="tex-graphics" src="https://espresso.codeforces.com/9fe384073741e20965ddc4bf162afd3a604b6b39.png" style="max-width: 100.0%;max-height: 100.0%;"/>
Figure to the second sample
|
```python
import math
m,R = map (int,input().split())
D = math.sqrt (2) * R
result = 0
def sum_dist (n):
return n*(n+1)*R + 2*D*n
for i in range (1,m+1):
result += 2*R
if i-1 > 0: result += 2*R + D
if m-i > 0: result += 2*R + D
if i-2 > 0: result += sum_dist (i-2)
if m-i-1 > 0: result += sum_dist (m-i-1)
print (result / m / m)
```
| 3
|
|
427
|
B
|
Prison Transfer
|
PROGRAMMING
| 1,100
|
[
"data structures",
"implementation"
] | null | null |
The prison of your city has *n* prisoners. As the prison can't accommodate all of them, the city mayor has decided to transfer *c* of the prisoners to a prison located in another city.
For this reason, he made the *n* prisoners to stand in a line, with a number written on their chests. The number is the severity of the crime he/she has committed. The greater the number, the more severe his/her crime was.
Then, the mayor told you to choose the *c* prisoners, who will be transferred to the other prison. He also imposed two conditions. They are,
- The chosen *c* prisoners has to form a contiguous segment of prisoners. - Any of the chosen prisoner's crime level should not be greater then *t*. Because, that will make the prisoner a severe criminal and the mayor doesn't want to take the risk of his running away during the transfer.
Find the number of ways you can choose the *c* prisoners.
|
The first line of input will contain three space separated integers *n*Β (1<=β€<=*n*<=β€<=2Β·105), *t*Β (0<=β€<=*t*<=β€<=109) and *c*Β (1<=β€<=*c*<=β€<=*n*). The next line will contain *n* space separated integers, the *i**th* integer is the severity *i**th* prisoner's crime. The value of crime severities will be non-negative and will not exceed 109.
|
Print a single integer β the number of ways you can choose the *c* prisoners.
|
[
"4 3 3\n2 3 1 1\n",
"1 1 1\n2\n",
"11 4 2\n2 2 0 7 3 2 2 4 9 1 4\n"
] |
[
"2\n",
"0\n",
"6\n"
] |
none
| 1,000
|
[
{
"input": "4 3 3\n2 3 1 1",
"output": "2"
},
{
"input": "1 1 1\n2",
"output": "0"
},
{
"input": "11 4 2\n2 2 0 7 3 2 2 4 9 1 4",
"output": "6"
},
{
"input": "57 2 10\n7 5 2 7 4 1 0 5 2 9 2 9 8 6 6 5 9 6 8 1 0 1 0 3 2 6 5 2 8 8 8 8 0 9 4 3 6 6 2 4 5 1 2 0 1 7 1 1 5 4 5 0 7 5 1 9 6",
"output": "0"
},
{
"input": "2 228885628 1\n90897004 258427916",
"output": "1"
},
{
"input": "3 1 1\n1 2 1",
"output": "2"
},
{
"input": "3 3 3\n3 2 3",
"output": "1"
},
{
"input": "4 2 2\n1 3 3 2",
"output": "0"
},
{
"input": "1 228 1\n1",
"output": "1"
}
] | 1,696,006,223
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 80
| 109
| 22,630,400
|
import sys
n, t, c = map(int, input().split())
vec = [-1]
for i, temp in enumerate(map(int, sys.stdin.readline().split())):
if temp > t:
vec.append(i)
vec.append(n)
ans = 0
for i in range(1, len(vec)):
diff = vec[i] - vec[i - 1] - 1
ans += max(diff - c + 1, 0)
sys.stdout.write(str(ans) + '\n')
|
Title: Prison Transfer
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The prison of your city has *n* prisoners. As the prison can't accommodate all of them, the city mayor has decided to transfer *c* of the prisoners to a prison located in another city.
For this reason, he made the *n* prisoners to stand in a line, with a number written on their chests. The number is the severity of the crime he/she has committed. The greater the number, the more severe his/her crime was.
Then, the mayor told you to choose the *c* prisoners, who will be transferred to the other prison. He also imposed two conditions. They are,
- The chosen *c* prisoners has to form a contiguous segment of prisoners. - Any of the chosen prisoner's crime level should not be greater then *t*. Because, that will make the prisoner a severe criminal and the mayor doesn't want to take the risk of his running away during the transfer.
Find the number of ways you can choose the *c* prisoners.
Input Specification:
The first line of input will contain three space separated integers *n*Β (1<=β€<=*n*<=β€<=2Β·105), *t*Β (0<=β€<=*t*<=β€<=109) and *c*Β (1<=β€<=*c*<=β€<=*n*). The next line will contain *n* space separated integers, the *i**th* integer is the severity *i**th* prisoner's crime. The value of crime severities will be non-negative and will not exceed 109.
Output Specification:
Print a single integer β the number of ways you can choose the *c* prisoners.
Demo Input:
['4 3 3\n2 3 1 1\n', '1 1 1\n2\n', '11 4 2\n2 2 0 7 3 2 2 4 9 1 4\n']
Demo Output:
['2\n', '0\n', '6\n']
Note:
none
|
```python
import sys
n, t, c = map(int, input().split())
vec = [-1]
for i, temp in enumerate(map(int, sys.stdin.readline().split())):
if temp > t:
vec.append(i)
vec.append(n)
ans = 0
for i in range(1, len(vec)):
diff = vec[i] - vec[i - 1] - 1
ans += max(diff - c + 1, 0)
sys.stdout.write(str(ans) + '\n')
```
| 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 β (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 β 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"
}
] | 1,635,769,339
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 102
| 46
| 4,300,800
|
a=input()
b=input()
res=[]
for i in range(len(a)):
if a[i]!=b[i]:
res.append('1')
else:
res.append('0')
print(*res,sep='')
|
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 β (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 β 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
a=input()
b=input()
res=[]
for i in range(len(a)):
if a[i]!=b[i]:
res.append('1')
else:
res.append('0')
print(*res,sep='')
```
| 3.980489
|
258
|
A
|
Little Elephant and Bits
|
PROGRAMMING
| 1,100
|
[
"greedy",
"math"
] | null | null |
The Little Elephant has an integer *a*, written in the binary notation. He wants to write this number on a piece of paper.
To make sure that the number *a* fits on the piece of paper, the Little Elephant ought to delete exactly one any digit from number *a* in the binary record. At that a new number appears. It consists of the remaining binary digits, written in the corresponding order (possible, with leading zeroes).
The Little Elephant wants the number he is going to write on the paper to be as large as possible. Help him find the maximum number that he can obtain after deleting exactly one binary digit and print it in the binary notation.
|
The single line contains integer *a*, written in the binary notation without leading zeroes. This number contains more than 1 and at most 105 digits.
|
In the single line print the number that is written without leading zeroes in the binary notation β the answer to the problem.
|
[
"101\n",
"110010\n"
] |
[
"11\n",
"11010\n"
] |
In the first sample the best strategy is to delete the second digit. That results in number 11<sub class="lower-index">2</sub>β=β3<sub class="lower-index">10</sub>.
In the second sample the best strategy is to delete the third or fourth digits β that results in number 11010<sub class="lower-index">2</sub>β=β26<sub class="lower-index">10</sub>.
| 500
|
[
{
"input": "101",
"output": "11"
},
{
"input": "110010",
"output": "11010"
},
{
"input": "10000",
"output": "1000"
},
{
"input": "1111111110",
"output": "111111111"
},
{
"input": "10100101011110101",
"output": "1100101011110101"
},
{
"input": "111010010111",
"output": "11110010111"
},
{
"input": "11110111011100000000",
"output": "1111111011100000000"
},
{
"input": "11110010010100001110110101110011110110100111101",
"output": "1111010010100001110110101110011110110100111101"
},
{
"input": "1001011111010010100111111",
"output": "101011111010010100111111"
},
{
"input": "1111111111",
"output": "111111111"
},
{
"input": "1111111111111111111100111101001110110111111000001111110101001101001110011000001011001111111000110101",
"output": "111111111111111111110111101001110110111111000001111110101001101001110011000001011001111111000110101"
},
{
"input": "11010110000100100101111110111001001010011000011011000010010100111010101000111010011101101111110001111000101000001100011101110100",
"output": "1110110000100100101111110111001001010011000011011000010010100111010101000111010011101101111110001111000101000001100011101110100"
},
{
"input": "11111111111111111111111110110111001101100111010010101101101001011100011011000111010011110010101100010001011101011010010100001000011100001101101001100010100001001010010100100001111110100110011000101100001111111011010111001011111110111101000100101001001011",
"output": "1111111111111111111111111110111001101100111010010101101101001011100011011000111010011110010101100010001011101011010010100001000011100001101101001100010100001001010010100100001111110100110011000101100001111111011010111001011111110111101000100101001001011"
},
{
"input": "11100010010010000110101101101100111111001010001101101001001111010110010111001011010000001100110101000101111000001111101111110010000010101110011110101101010110001100011101111011100010011101100111110010111111100110101000000111101000000000110100100101111101000110101010101101001110001110000101011010101100011100100111100010001011010010001100011111110010011010011000111000100111100010110100011010010101011011011111110100001110000011011",
"output": "1110010010010000110101101101100111111001010001101101001001111010110010111001011010000001100110101000101111000001111101111110010000010101110011110101101010110001100011101111011100010011101100111110010111111100110101000000111101000000000110100100101111101000110101010101101001110001110000101011010101100011100100111100010001011010010001100011111110010011010011000111000100111100010110100011010010101011011011111110100001110000011011"
},
{
"input": "11",
"output": "1"
},
{
"input": "111",
"output": "11"
},
{
"input": "111111",
"output": "11111"
},
{
"input": "11111",
"output": "1111"
},
{
"input": "1111",
"output": "111"
}
] | 1,618,995,477
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 37
| 218
| 2,355,200
|
s = input()
for idx in range(len(s)):
if s[idx] == "0":
break
print(s[:idx]+s[idx+1:])
|
Title: Little Elephant and Bits
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Little Elephant has an integer *a*, written in the binary notation. He wants to write this number on a piece of paper.
To make sure that the number *a* fits on the piece of paper, the Little Elephant ought to delete exactly one any digit from number *a* in the binary record. At that a new number appears. It consists of the remaining binary digits, written in the corresponding order (possible, with leading zeroes).
The Little Elephant wants the number he is going to write on the paper to be as large as possible. Help him find the maximum number that he can obtain after deleting exactly one binary digit and print it in the binary notation.
Input Specification:
The single line contains integer *a*, written in the binary notation without leading zeroes. This number contains more than 1 and at most 105 digits.
Output Specification:
In the single line print the number that is written without leading zeroes in the binary notation β the answer to the problem.
Demo Input:
['101\n', '110010\n']
Demo Output:
['11\n', '11010\n']
Note:
In the first sample the best strategy is to delete the second digit. That results in number 11<sub class="lower-index">2</sub>β=β3<sub class="lower-index">10</sub>.
In the second sample the best strategy is to delete the third or fourth digits β that results in number 11010<sub class="lower-index">2</sub>β=β26<sub class="lower-index">10</sub>.
|
```python
s = input()
for idx in range(len(s)):
if s[idx] == "0":
break
print(s[:idx]+s[idx+1:])
```
| 3
|
|
468
|
C
|
Hack it!
|
PROGRAMMING
| 2,500
|
[
"binary search",
"constructive algorithms",
"math"
] | null | null |
Little X has met the following problem recently.
Let's define *f*(*x*) as the sum of digits in decimal representation of number *x* (for example, *f*(1234)<==<=1<=+<=2<=+<=3<=+<=4). You are to calculate
Of course Little X has solved this problem quickly, has locked it, and then has tried to hack others. He has seen the following C++ code:
|
The first line contains a single integer *a*Β (1<=β€<=*a*<=β€<=1018).
|
Print two integers: *l*,<=*r*Β (1<=β€<=*l*<=β€<=*r*<=<<=10200) β the required test data. Leading zeros aren't allowed. It's guaranteed that the solution exists.
|
[
"46\n",
"126444381000032\n"
] |
[
"1 10\n",
"2333333 2333333333333\n"
] |
none
| 1,500
|
[
{
"input": "46",
"output": "1 10"
},
{
"input": "126444381000032",
"output": "2333333 2333333333333"
},
{
"input": "69645082595",
"output": "613752823618441225798858488535 713259406474207764329704856394"
},
{
"input": "70602205995",
"output": "11 249221334020432074498656960922"
},
{
"input": "33898130785",
"output": "9 558855506346909386939077840182"
},
{
"input": "58929554039",
"output": "22 855783114773435710171914224422"
},
{
"input": "81696185182",
"output": "499118531974994927425925323518 956291458400902769638235161661"
},
{
"input": "1",
"output": "149268802942315027273202513064 277551734280589260570057105889"
},
{
"input": "2",
"output": "119692200833686078608961312319 629363568954685219494592939495"
},
{
"input": "3",
"output": "2 302254410562920936884653943506"
},
{
"input": "4",
"output": "284378053387469023431537894255 317250990020830090421009164911"
},
{
"input": "5",
"output": "2 62668056583245293799710157951"
},
{
"input": "6",
"output": "3 93810188780011787541394067841"
},
{
"input": "7",
"output": "2 834286447477504059026206246185"
},
{
"input": "8",
"output": "3 257583347960907690857477857197"
},
{
"input": "10",
"output": "3 163048811987317819669274448265"
},
{
"input": "11",
"output": "3 919618203693907154039906935669"
},
{
"input": "12",
"output": "448221703341269567451520778454 698029790336105644790102859494"
},
{
"input": "43",
"output": "9 172412961300207091437973214327"
},
{
"input": "36",
"output": "8 619355518777647869838990701242"
},
{
"input": "65",
"output": "6 709024330418134127413755925068"
},
{
"input": "43",
"output": "7 669540448846929747909766131221"
},
{
"input": "23",
"output": "2 104579054315773428039906118259"
},
{
"input": "100",
"output": "15 324437778467489559125023403167"
},
{
"input": "10000",
"output": "2 936791129091842315790163514642"
},
{
"input": "1000000",
"output": "18 369591628030718549289473454545"
},
{
"input": "100000000",
"output": "7 870405265198051697453938746950"
},
{
"input": "10000000000",
"output": "20 972749766921651560604778558599"
},
{
"input": "1000000000000",
"output": "6 68997070398311657294228230677"
},
{
"input": "100000000000000",
"output": "249537318528661282822184562278 397003438246047829818181818181"
},
{
"input": "10000000000000000",
"output": "10 778165727326620883431915444624"
},
{
"input": "1000000000000000000",
"output": "408256298986776744812953390000 824018301451167837914299999999"
},
{
"input": "450000000000000000",
"output": "2 357722688084551093593033993033"
},
{
"input": "432022",
"output": "3 333556238531076799985515487090"
},
{
"input": "428033",
"output": "22 730314748425770554502599499142"
},
{
"input": "776930",
"output": "20 521232359366297130685112811874"
},
{
"input": "329824",
"output": "308969571112207311167474021348 745620588073413831210052337999"
},
{
"input": "85058261498638",
"output": "16 931187081941564769324316582547"
},
{
"input": "2130909834463",
"output": "21 895378349209612431051450316022"
},
{
"input": "3427089130241",
"output": "10 676758114393938690602742889714"
},
{
"input": "22881472397923",
"output": "174523915446146844994793303441 429392837423394397373605399524"
},
{
"input": "756499070280135900",
"output": "17 414254565210363110239866979636"
},
{
"input": "348919756917359012",
"output": "15 148711424846749040017925495268"
},
{
"input": "789136710974630947",
"output": "13 214740524663440316532540770911"
},
{
"input": "32851678509090784",
"output": "15 728787006985867306455350840047"
},
{
"input": "219760524792138559",
"output": "8 51530430692061277616906746431"
},
{
"input": "1000000007",
"output": "19 96093944063249823335491170999"
},
{
"input": "999999874000003969",
"output": "306259936121099507549028331741 983025349899931537481252156006"
},
{
"input": "999999999999999989",
"output": "140766388309929586494194507367 705096656792775954903252198639"
},
{
"input": "999999999999999983",
"output": "4 365946710276937286091309116340"
},
{
"input": "999999999999999999",
"output": "26 983966655749419565978291923474"
}
] | 1,424,004,438
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 51
| 62
| 0
|
a=int(input())
x=a-((10**20)*45*21+1)%a
print(1+x,10**21+x)
|
Title: Hack it!
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little X has met the following problem recently.
Let's define *f*(*x*) as the sum of digits in decimal representation of number *x* (for example, *f*(1234)<==<=1<=+<=2<=+<=3<=+<=4). You are to calculate
Of course Little X has solved this problem quickly, has locked it, and then has tried to hack others. He has seen the following C++ code:
Input Specification:
The first line contains a single integer *a*Β (1<=β€<=*a*<=β€<=1018).
Output Specification:
Print two integers: *l*,<=*r*Β (1<=β€<=*l*<=β€<=*r*<=<<=10200) β the required test data. Leading zeros aren't allowed. It's guaranteed that the solution exists.
Demo Input:
['46\n', '126444381000032\n']
Demo Output:
['1 10\n', '2333333 2333333333333\n']
Note:
none
|
```python
a=int(input())
x=a-((10**20)*45*21+1)%a
print(1+x,10**21+x)
```
| 3
|
|
71
|
A
|
Way Too Long Words
|
PROGRAMMING
| 800
|
[
"strings"
] |
A. Way Too Long Words
|
1
|
256
|
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalizationΒ» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
|
The first line contains an integer *n* (1<=β€<=*n*<=β€<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
|
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
|
[
"4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n"
] |
[
"word\nl10n\ni18n\np43s\n"
] |
none
| 500
|
[
{
"input": "4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis",
"output": "word\nl10n\ni18n\np43s"
},
{
"input": "5\nabcdefgh\nabcdefghi\nabcdefghij\nabcdefghijk\nabcdefghijklm",
"output": "abcdefgh\nabcdefghi\nabcdefghij\na9k\na11m"
},
{
"input": "3\nnjfngnrurunrgunrunvurn\njfvnjfdnvjdbfvsbdubruvbubvkdb\nksdnvidnviudbvibd",
"output": "n20n\nj27b\nk15d"
},
{
"input": "1\ntcyctkktcctrcyvbyiuhihhhgyvyvyvyvjvytchjckt",
"output": "t41t"
},
{
"input": "24\nyou\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nunofficially\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings",
"output": "you\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nu10y\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings"
},
{
"input": "1\na",
"output": "a"
},
{
"input": "26\na\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz",
"output": "a\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz"
},
{
"input": "1\nabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghij",
"output": "a98j"
},
{
"input": "10\ngyartjdxxlcl\nfzsck\nuidwu\nxbymclornemdmtj\nilppyoapitawgje\ncibzc\ndrgbeu\nhezplmsdekhhbo\nfeuzlrimbqbytdu\nkgdco",
"output": "g10l\nfzsck\nuidwu\nx13j\ni13e\ncibzc\ndrgbeu\nh12o\nf13u\nkgdco"
},
{
"input": "20\nlkpmx\nkovxmxorlgwaomlswjxlpnbvltfv\nhykasjxqyjrmybejnmeumzha\ntuevlumpqbbhbww\nqgqsphvrmupxxc\ntrissbaf\nqfgrlinkzvzqdryckaizutd\nzzqtoaxkvwoscyx\noswytrlnhpjvvnwookx\nlpuzqgec\ngyzqfwxggtvpjhzmzmdw\nrlxjgmvdftvrmvbdwudra\nvsntnjpepnvdaxiporggmglhagv\nxlvcqkqgcrbgtgglj\nlyxwxbiszyhlsrgzeedzprbmcpduvq\nyrmqqvrkqskqukzqrwukpsifgtdc\nxpuohcsjhhuhvr\nvvlfrlxpvqejngwrbfbpmqeirxlw\nsvmasocxdvadmaxtrpakysmeaympy\nyuflqboqfdt",
"output": "lkpmx\nk26v\nh22a\nt13w\nq12c\ntrissbaf\nq21d\nz13x\no17x\nlpuzqgec\ng18w\nr19a\nv25v\nx15j\nl28q\ny26c\nx12r\nv26w\ns27y\ny9t"
},
{
"input": "100\nm\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr",
"output": "m\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr"
}
] | 1,692,485,672
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 20
| 31
| 0
|
def way_too_long_words(txt_arr):
res = []
for input_txt in txt_arr:
n = len(input_txt)
if n <= 10:
res.append(input_txt)
else:
res.append(input_txt[0] + str(n-2) + input_txt[-1])
return res
if __name__ == "__main__":
n = int(input())
txt_arr = []
for _ in range(n):
txt_arr.append(input())
res_arr = way_too_long_words(txt_arr)
for res in res_arr:
print(res)
|
Title: Way Too Long Words
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalizationΒ» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
Input Specification:
The first line contains an integer *n* (1<=β€<=*n*<=β€<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
Output Specification:
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
Demo Input:
['4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n']
Demo Output:
['word\nl10n\ni18n\np43s\n']
Note:
none
|
```python
def way_too_long_words(txt_arr):
res = []
for input_txt in txt_arr:
n = len(input_txt)
if n <= 10:
res.append(input_txt)
else:
res.append(input_txt[0] + str(n-2) + input_txt[-1])
return res
if __name__ == "__main__":
n = int(input())
txt_arr = []
for _ in range(n):
txt_arr.append(input())
res_arr = way_too_long_words(txt_arr)
for res in res_arr:
print(res)
```
| 3.9845
|
962
|
A
|
Equator
|
PROGRAMMING
| 1,300
|
[
"implementation"
] | null | null |
Polycarp has created his own training plan to prepare for the programming contests. He will train for $n$ days, all days are numbered from $1$ to $n$, beginning from the first.
On the $i$-th day Polycarp will necessarily solve $a_i$ problems. One evening Polycarp plans to celebrate the equator. He will celebrate it on the first evening of such a day that from the beginning of the training and to this day inclusive he will solve half or more of all the problems.
Determine the index of day when Polycarp will celebrate the equator.
|
The first line contains a single integer $n$ ($1 \le n \le 200\,000$) β the number of days to prepare for the programming contests.
The second line contains a sequence $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10\,000$), where $a_i$ equals to the number of problems, which Polycarp will solve on the $i$-th day.
|
Print the index of the day when Polycarp will celebrate the equator.
|
[
"4\n1 3 2 1\n",
"6\n2 2 2 2 2 2\n"
] |
[
"2\n",
"3\n"
] |
In the first example Polycarp will celebrate the equator on the evening of the second day, because up to this day (inclusive) he will solve $4$ out of $7$ scheduled problems on four days of the training.
In the second example Polycarp will celebrate the equator on the evening of the third day, because up to this day (inclusive) he will solve $6$ out of $12$ scheduled problems on six days of the training.
| 0
|
[
{
"input": "4\n1 3 2 1",
"output": "2"
},
{
"input": "6\n2 2 2 2 2 2",
"output": "3"
},
{
"input": "1\n10000",
"output": "1"
},
{
"input": "3\n2 1 1",
"output": "1"
},
{
"input": "2\n1 3",
"output": "2"
},
{
"input": "4\n2 1 1 3",
"output": "3"
},
{
"input": "3\n1 1 3",
"output": "3"
},
{
"input": "3\n1 1 1",
"output": "2"
},
{
"input": "2\n1 2",
"output": "2"
},
{
"input": "3\n2 1 2",
"output": "2"
},
{
"input": "5\n1 2 4 3 5",
"output": "4"
},
{
"input": "5\n2 2 2 4 3",
"output": "4"
},
{
"input": "4\n1 2 3 1",
"output": "3"
},
{
"input": "6\n7 3 10 7 3 11",
"output": "4"
},
{
"input": "2\n3 4",
"output": "2"
},
{
"input": "5\n1 1 1 1 1",
"output": "3"
},
{
"input": "4\n1 3 2 3",
"output": "3"
},
{
"input": "2\n2 3",
"output": "2"
},
{
"input": "3\n32 10 23",
"output": "2"
},
{
"input": "7\n1 1 1 1 1 1 1",
"output": "4"
},
{
"input": "3\n1 2 4",
"output": "3"
},
{
"input": "6\n3 3 3 2 4 4",
"output": "4"
},
{
"input": "9\n1 1 1 1 1 1 1 1 1",
"output": "5"
},
{
"input": "5\n1 3 3 1 1",
"output": "3"
},
{
"input": "4\n1 1 1 2",
"output": "3"
},
{
"input": "4\n1 2 1 3",
"output": "3"
},
{
"input": "3\n2 2 1",
"output": "2"
},
{
"input": "4\n2 3 3 3",
"output": "3"
},
{
"input": "4\n3 2 3 3",
"output": "3"
},
{
"input": "4\n2 1 1 1",
"output": "2"
},
{
"input": "3\n2 1 4",
"output": "3"
},
{
"input": "2\n6 7",
"output": "2"
},
{
"input": "4\n3 3 4 3",
"output": "3"
},
{
"input": "4\n1 1 2 5",
"output": "4"
},
{
"input": "4\n1 8 7 3",
"output": "3"
},
{
"input": "6\n2 2 2 2 2 3",
"output": "4"
},
{
"input": "3\n2 2 5",
"output": "3"
},
{
"input": "4\n1 1 2 1",
"output": "3"
},
{
"input": "5\n1 1 2 2 3",
"output": "4"
},
{
"input": "5\n9 5 3 4 8",
"output": "3"
},
{
"input": "3\n3 3 1",
"output": "2"
},
{
"input": "4\n1 2 2 2",
"output": "3"
},
{
"input": "3\n1 3 5",
"output": "3"
},
{
"input": "4\n1 1 3 6",
"output": "4"
},
{
"input": "6\n1 2 1 1 1 1",
"output": "3"
},
{
"input": "3\n3 1 3",
"output": "2"
},
{
"input": "5\n3 4 5 1 2",
"output": "3"
},
{
"input": "11\n1 1 1 1 1 1 1 1 1 1 1",
"output": "6"
},
{
"input": "5\n3 1 2 5 2",
"output": "4"
},
{
"input": "4\n1 1 1 4",
"output": "4"
},
{
"input": "4\n2 6 1 10",
"output": "4"
},
{
"input": "4\n2 2 3 2",
"output": "3"
},
{
"input": "4\n4 2 2 1",
"output": "2"
},
{
"input": "6\n1 1 1 1 1 4",
"output": "5"
},
{
"input": "3\n3 2 2",
"output": "2"
},
{
"input": "6\n1 3 5 1 7 4",
"output": "5"
},
{
"input": "5\n1 2 4 8 16",
"output": "5"
},
{
"input": "5\n1 2 4 4 4",
"output": "4"
},
{
"input": "6\n4 2 1 2 3 1",
"output": "3"
},
{
"input": "4\n3 2 1 5",
"output": "3"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "3\n2 4 7",
"output": "3"
},
{
"input": "5\n1 1 1 1 3",
"output": "4"
},
{
"input": "3\n3 1 5",
"output": "3"
},
{
"input": "4\n1 2 3 7",
"output": "4"
},
{
"input": "3\n1 4 6",
"output": "3"
},
{
"input": "4\n2 1 2 2",
"output": "3"
},
{
"input": "2\n4 5",
"output": "2"
},
{
"input": "5\n1 2 1 2 1",
"output": "3"
},
{
"input": "3\n2 3 6",
"output": "3"
},
{
"input": "6\n1 1 4 1 1 5",
"output": "4"
},
{
"input": "5\n2 2 2 2 1",
"output": "3"
},
{
"input": "2\n5 6",
"output": "2"
},
{
"input": "4\n2 2 1 4",
"output": "3"
},
{
"input": "5\n2 2 3 4 4",
"output": "4"
},
{
"input": "4\n3 1 1 2",
"output": "2"
},
{
"input": "5\n3 4 1 4 5",
"output": "4"
},
{
"input": "4\n1 3 1 6",
"output": "4"
},
{
"input": "5\n1 1 1 2 2",
"output": "4"
},
{
"input": "4\n1 4 2 4",
"output": "3"
},
{
"input": "10\n1 1 1 1 1 1 1 1 1 8",
"output": "9"
},
{
"input": "4\n1 4 5 1",
"output": "3"
},
{
"input": "5\n1 1 1 1 5",
"output": "5"
},
{
"input": "4\n1 3 4 1",
"output": "3"
},
{
"input": "4\n2 2 2 3",
"output": "3"
},
{
"input": "4\n2 3 2 4",
"output": "3"
},
{
"input": "5\n2 2 1 2 2",
"output": "3"
},
{
"input": "3\n4 3 2",
"output": "2"
},
{
"input": "3\n6 5 2",
"output": "2"
},
{
"input": "69\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",
"output": "35"
},
{
"input": "6\n1 1 1 1 1 2",
"output": "4"
},
{
"input": "5\n1 2 5 4 5",
"output": "4"
},
{
"input": "2\n9 10",
"output": "2"
},
{
"input": "3\n1 1 5",
"output": "3"
},
{
"input": "4\n3 4 3 5",
"output": "3"
},
{
"input": "4\n1 4 3 3",
"output": "3"
},
{
"input": "4\n7 1 3 4",
"output": "2"
},
{
"input": "3\n100 100 1",
"output": "2"
},
{
"input": "4\n5 2 2 2",
"output": "2"
}
] | 1,524,523,715
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 106
| 171
| 20,582,400
|
#k=int(input())
#n,m=map(int,input().split())
import sys
#a=list(map(int,input().split()))
#b=list(map(int,input().split()))
import math
n=int(input())
a=list(map(int,input().split()))
ss=sum(a);
s=0
for i in range(n):
s+=a[i]
if(2*s>=ss):
print(i+1)
sys.exit()
|
Title: Equator
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp has created his own training plan to prepare for the programming contests. He will train for $n$ days, all days are numbered from $1$ to $n$, beginning from the first.
On the $i$-th day Polycarp will necessarily solve $a_i$ problems. One evening Polycarp plans to celebrate the equator. He will celebrate it on the first evening of such a day that from the beginning of the training and to this day inclusive he will solve half or more of all the problems.
Determine the index of day when Polycarp will celebrate the equator.
Input Specification:
The first line contains a single integer $n$ ($1 \le n \le 200\,000$) β the number of days to prepare for the programming contests.
The second line contains a sequence $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10\,000$), where $a_i$ equals to the number of problems, which Polycarp will solve on the $i$-th day.
Output Specification:
Print the index of the day when Polycarp will celebrate the equator.
Demo Input:
['4\n1 3 2 1\n', '6\n2 2 2 2 2 2\n']
Demo Output:
['2\n', '3\n']
Note:
In the first example Polycarp will celebrate the equator on the evening of the second day, because up to this day (inclusive) he will solve $4$ out of $7$ scheduled problems on four days of the training.
In the second example Polycarp will celebrate the equator on the evening of the third day, because up to this day (inclusive) he will solve $6$ out of $12$ scheduled problems on six days of the training.
|
```python
#k=int(input())
#n,m=map(int,input().split())
import sys
#a=list(map(int,input().split()))
#b=list(map(int,input().split()))
import math
n=int(input())
a=list(map(int,input().split()))
ss=sum(a);
s=0
for i in range(n):
s+=a[i]
if(2*s>=ss):
print(i+1)
sys.exit()
```
| 3
|
|
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<=β€<=*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<=β€<=*n*<=β€<=100, 1<=β€<=*m*<=β€<=*n*, 1<=β€<=*k*<=β€<=100)Β β 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<=β€<=*a**i*<=β€<=100)Β β 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 integerΒ β 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*β=β20 dollars, Zane can buy only house 5. The distance from house *m*β=β1 to house 5 is 10β+β10β+β10β+β10β=β40 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*β=β3 and house 6 are only 30 meters away, while house *m*β=β3 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"
}
] | 1,559,807,370
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 58
| 109
| 0
|
n,m,k=map(int,input().split())
m-=1
l=[int(x) for x in input().split()]
q=[]
for i in range(n):
if l[i]!=0 and l[i]<=k:
q.append((abs(i-m))*10)
print(min(q))
|
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<=β€<=*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<=β€<=*n*<=β€<=100, 1<=β€<=*m*<=β€<=*n*, 1<=β€<=*k*<=β€<=100)Β β 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<=β€<=*a**i*<=β€<=100)Β β 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 integerΒ β 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*β=β20 dollars, Zane can buy only house 5. The distance from house *m*β=β1 to house 5 is 10β+β10β+β10β+β10β=β40 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*β=β3 and house 6 are only 30 meters away, while house *m*β=β3 and house 7 are 40 meters away.
|
```python
n,m,k=map(int,input().split())
m-=1
l=[int(x) for x in input().split()]
q=[]
for i in range(n):
if l[i]!=0 and l[i]<=k:
q.append((abs(i-m))*10)
print(min(q))
```
| 3
|
|
584
|
A
|
Olesya and Rodion
|
PROGRAMMING
| 1,000
|
[
"math"
] | null | null |
Olesya loves numbers consisting of *n* digits, and Rodion only likes numbers that are divisible by *t*. Find some number that satisfies both of them.
Your task is: given the *n* and *t* print an integer strictly larger than zero consisting of *n* digits that is divisible by *t*. If such number doesn't exist, print <=-<=1.
|
The single line contains two numbers, *n* and *t* (1<=β€<=*n*<=β€<=100, 2<=β€<=*t*<=β€<=10) β the length of the number and the number it should be divisible by.
|
Print one such positive number without leading zeroes, β the answer to the problem, or <=-<=1, if such number doesn't exist. If there are multiple possible answers, you are allowed to print any of them.
|
[
"3 2\n"
] |
[
"712"
] |
none
| 500
|
[
{
"input": "3 2",
"output": "222"
},
{
"input": "2 2",
"output": "22"
},
{
"input": "4 3",
"output": "3333"
},
{
"input": "5 3",
"output": "33333"
},
{
"input": "10 7",
"output": "7777777777"
},
{
"input": "2 9",
"output": "99"
},
{
"input": "18 8",
"output": "888888888888888888"
},
{
"input": "1 5",
"output": "5"
},
{
"input": "1 10",
"output": "-1"
},
{
"input": "100 5",
"output": "5555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555"
},
{
"input": "10 2",
"output": "2222222222"
},
{
"input": "18 10",
"output": "111111111111111110"
},
{
"input": "1 9",
"output": "9"
},
{
"input": "7 6",
"output": "6666666"
},
{
"input": "4 4",
"output": "4444"
},
{
"input": "14 7",
"output": "77777777777777"
},
{
"input": "3 8",
"output": "888"
},
{
"input": "1 3",
"output": "3"
},
{
"input": "2 8",
"output": "88"
},
{
"input": "3 8",
"output": "888"
},
{
"input": "4 3",
"output": "3333"
},
{
"input": "5 9",
"output": "99999"
},
{
"input": "4 8",
"output": "8888"
},
{
"input": "3 4",
"output": "444"
},
{
"input": "9 4",
"output": "444444444"
},
{
"input": "8 10",
"output": "11111110"
},
{
"input": "1 6",
"output": "6"
},
{
"input": "20 3",
"output": "33333333333333333333"
},
{
"input": "15 10",
"output": "111111111111110"
},
{
"input": "31 4",
"output": "4444444444444444444444444444444"
},
{
"input": "18 9",
"output": "999999999999999999"
},
{
"input": "72 4",
"output": "444444444444444444444444444444444444444444444444444444444444444444444444"
},
{
"input": "76 8",
"output": "8888888888888888888888888888888888888888888888888888888888888888888888888888"
},
{
"input": "12 5",
"output": "555555555555"
},
{
"input": "54 5",
"output": "555555555555555555555555555555555555555555555555555555"
},
{
"input": "96 10",
"output": "111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111110"
},
{
"input": "15 9",
"output": "999999999999999"
},
{
"input": "100 2",
"output": "2222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222"
},
{
"input": "99 3",
"output": "333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333"
},
{
"input": "98 4",
"output": "44444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444"
},
{
"input": "97 5",
"output": "5555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555"
},
{
"input": "100 6",
"output": "6666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666"
},
{
"input": "99 7",
"output": "777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777"
},
{
"input": "98 8",
"output": "88888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888"
},
{
"input": "97 9",
"output": "9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999"
},
{
"input": "100 10",
"output": "1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111110"
},
{
"input": "3 10",
"output": "110"
},
{
"input": "2 4",
"output": "44"
},
{
"input": "2 10",
"output": "10"
},
{
"input": "10 10",
"output": "1111111110"
},
{
"input": "4 10",
"output": "1110"
},
{
"input": "25 10",
"output": "1111111111111111111111110"
},
{
"input": "50 10",
"output": "11111111111111111111111111111111111111111111111110"
},
{
"input": "5 10",
"output": "11110"
},
{
"input": "99 10",
"output": "111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111110"
},
{
"input": "100 10",
"output": "1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111110"
},
{
"input": "1 2",
"output": "2"
},
{
"input": "20 10",
"output": "11111111111111111110"
},
{
"input": "99 7",
"output": "777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777"
},
{
"input": "89 5",
"output": "55555555555555555555555555555555555555555555555555555555555555555555555555555555555555555"
},
{
"input": "6 10",
"output": "111110"
},
{
"input": "3 8",
"output": "888"
},
{
"input": "1 9",
"output": "9"
},
{
"input": "1 4",
"output": "4"
},
{
"input": "100 7",
"output": "7777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777"
}
] | 1,665,214,610
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 65
| 46
| 0
|
n,t = map(int,input().split())
s = '1'
i = 1
while i <= n-1:
s += '0'
i += 1
w = s + '0'
num1 = int(s)
num2 = int(w)
while num1 < num2:
if num1 % t == 0:
print(num1)
break
else:
num1 += 1
if num1 == num2:
print(-1)
|
Title: Olesya and Rodion
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Olesya loves numbers consisting of *n* digits, and Rodion only likes numbers that are divisible by *t*. Find some number that satisfies both of them.
Your task is: given the *n* and *t* print an integer strictly larger than zero consisting of *n* digits that is divisible by *t*. If such number doesn't exist, print <=-<=1.
Input Specification:
The single line contains two numbers, *n* and *t* (1<=β€<=*n*<=β€<=100, 2<=β€<=*t*<=β€<=10) β the length of the number and the number it should be divisible by.
Output Specification:
Print one such positive number without leading zeroes, β the answer to the problem, or <=-<=1, if such number doesn't exist. If there are multiple possible answers, you are allowed to print any of them.
Demo Input:
['3 2\n']
Demo Output:
['712']
Note:
none
|
```python
n,t = map(int,input().split())
s = '1'
i = 1
while i <= n-1:
s += '0'
i += 1
w = s + '0'
num1 = int(s)
num2 = int(w)
while num1 < num2:
if num1 % t == 0:
print(num1)
break
else:
num1 += 1
if num1 == num2:
print(-1)
```
| 3
|
|
801
|
A
|
Vicious Keyboard
|
PROGRAMMING
| 1,100
|
[
"brute force"
] | null | null |
Tonio has a keyboard with only two letters, "V" and "K".
One day, he has typed out a string *s* with only these two letters. He really likes it when the string "VK" appears, so he wishes to change at most one letter in the string (or do no changes) to maximize the number of occurrences of that string. Compute the maximum number of times "VK" can appear as a substring (i.Β e. a letter "K" right after a letter "V") in the resulting string.
|
The first line will contain a string *s* consisting only of uppercase English letters "V" and "K" with length not less than 1 and not greater than 100.
|
Output a single integer, the maximum number of times "VK" can appear as a substring of the given string after changing at most one character.
|
[
"VK\n",
"VV\n",
"V\n",
"VKKKKKKKKKVVVVVVVVVK\n",
"KVKV\n"
] |
[
"1\n",
"1\n",
"0\n",
"3\n",
"1\n"
] |
For the first case, we do not change any letters. "VK" appears once, which is the maximum number of times it could appear.
For the second case, we can change the second character from a "V" to a "K". This will give us the string "VK". This has one occurrence of the string "VK" as a substring.
For the fourth case, we can change the fourth character from a "K" to a "V". This will give us the string "VKKVKKKKKKVVVVVVVVVK". This has three occurrences of the string "VK" as a substring. We can check no other moves can give us strictly more occurrences.
| 500
|
[
{
"input": "VK",
"output": "1"
},
{
"input": "VV",
"output": "1"
},
{
"input": "V",
"output": "0"
},
{
"input": "VKKKKKKKKKVVVVVVVVVK",
"output": "3"
},
{
"input": "KVKV",
"output": "1"
},
{
"input": "VKKVVVKVKVK",
"output": "5"
},
{
"input": "VKVVKVKVVKVKKKKVVVVVVVVKVKVVVVVVKKVKKVKVVKVKKVVVVKV",
"output": "14"
},
{
"input": "VVKKVKKVVKKVKKVKVVKKVKKVVKKVKVVKKVKKVKVVKKVVKKVKVVKKVKVVKKVVKVVKKVKKVKKVKKVKKVKVVKKVKKVKKVKKVKKVVKVK",
"output": "32"
},
{
"input": "KVVKKVKVKVKVKVKKVKVKVVKVKVVKVVKVKKVKVKVKVKVKVKVKVKVKVKVKVKVKVKVVKVKVVKKVKVKK",
"output": "32"
},
{
"input": "KVVVVVKKVKVVKVVVKVVVKKKVKKKVVKVKKKVKKKKVKVVVVVKKKVVVVKKVVVVKKKVKVVVVVVVKKVKVKKKVVKVVVKVVKK",
"output": "21"
},
{
"input": "VVVVVKKVKVKVKVVKVVKKVVKVKKKKKKKVKKKVVVVVVKKVVVKVKVVKVKKVVKVVVKKKKKVVVVVKVVVVKVVVKKVKKVKKKVKKVKKVVKKV",
"output": "25"
},
{
"input": "KKVVKVVKVVKKVVKKVKVVKKV",
"output": "7"
},
{
"input": "KKVVKKVVVKKVKKVKKVVVKVVVKKVKKVVVKKVVVKVVVKVVVKKVVVKKVVVKVVVKKVVVKVVKKVVVKKVVVKKVVKVVVKKVVKKVKKVVVKKV",
"output": "24"
},
{
"input": "KVKVKVKVKVKVKVKVKVKVVKVKVKVKVKVKVKVVKVKVKKVKVKVKVKVVKVKVKVKVKVKVKVKVKKVKVKVV",
"output": "35"
},
{
"input": "VKVVVKKKVKVVKVKVKVKVKVV",
"output": "9"
},
{
"input": "KKKKVKKVKVKVKKKVVVVKK",
"output": "6"
},
{
"input": "KVKVKKVVVVVVKKKVKKKKVVVVKVKKVKVVK",
"output": "9"
},
{
"input": "KKVKKVKKKVKKKVKKKVKVVVKKVVVVKKKVKKVVKVKKVKVKVKVVVKKKVKKKKKVVKVVKVVVKKVVKVVKKKKKVK",
"output": "22"
},
{
"input": "VVVKVKVKVVVVVKVVVKKVVVKVVVVVKKVVKVVVKVVVKVKKKVVKVVVVVKVVVVKKVVKVKKVVKKKVKVVKVKKKKVVKVVVKKKVKVKKKKKK",
"output": "25"
},
{
"input": "VKVVKVVKKKVVKVKKKVVKKKVVKVVKVVKKVKKKVKVKKKVVKVKKKVVKVVKKKVVKKKVKKKVVKKVVKKKVKVKKKVKKKVKKKVKVKKKVVKVK",
"output": "29"
},
{
"input": "KKVKVVVKKVV",
"output": "3"
},
{
"input": "VKVKVKVKVKVKVKVKVKVKVVKVKVKVKVKVK",
"output": "16"
},
{
"input": "VVKKKVVKKKVVKKKVVKKKVVKKKVVKKKVVKKKVVKKKVVKKKVVKKKVVKKKVVKKKVV",
"output": "13"
},
{
"input": "VVKKVKVKKKVVVKVVVKVKKVKKKVVVKVVKVKKVKKVKVKVVKKVVKKVKVVKKKVVKKVVVKVKVVVKVKVVKVKKVKKV",
"output": "26"
},
{
"input": "VVKVKKVVKKVVKKVVKKVVKKVKKVVKVKKVVKKVVKKVVKKVVKKVVKVVKKVVKVVKKVVKVVKKVVKKVKKVVKVVKKVVKVVKKVV",
"output": "26"
},
{
"input": "K",
"output": "0"
},
{
"input": "VKVK",
"output": "2"
},
{
"input": "VKVV",
"output": "2"
},
{
"input": "KV",
"output": "0"
},
{
"input": "KK",
"output": "1"
},
{
"input": "KKVK",
"output": "2"
},
{
"input": "KKKK",
"output": "1"
},
{
"input": "KKV",
"output": "1"
},
{
"input": "KKVKVK",
"output": "3"
},
{
"input": "VKKVK",
"output": "2"
},
{
"input": "VKKK",
"output": "2"
},
{
"input": "KKK",
"output": "1"
},
{
"input": "KVV",
"output": "1"
},
{
"input": "KKVKV",
"output": "2"
},
{
"input": "VVK",
"output": "1"
},
{
"input": "VVVKVKVKVKVKVKVK",
"output": "8"
},
{
"input": "KVVVK",
"output": "2"
},
{
"input": "VVVKK",
"output": "2"
},
{
"input": "KKVV",
"output": "1"
},
{
"input": "KKKKKKK",
"output": "1"
},
{
"input": "VKKKVK",
"output": "3"
},
{
"input": "KKVVV",
"output": "1"
},
{
"input": "VVVVVV",
"output": "1"
},
{
"input": "KKKV",
"output": "1"
},
{
"input": "VVKVV",
"output": "2"
},
{
"input": "VKVKKK",
"output": "3"
},
{
"input": "VKKV",
"output": "1"
},
{
"input": "VKKVV",
"output": "2"
},
{
"input": "VVKKVV",
"output": "2"
},
{
"input": "KKVVKKV",
"output": "2"
},
{
"input": "KKKKK",
"output": "1"
},
{
"input": "VKVVKKVKKVVKVKKVKKKVKKVKVKK",
"output": "10"
},
{
"input": "VKVKVV",
"output": "3"
},
{
"input": "VKVVKVV",
"output": "3"
},
{
"input": "VVV",
"output": "1"
},
{
"input": "VVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVV",
"output": "1"
},
{
"input": "VVKKKKKKVKK",
"output": "3"
},
{
"input": "KVKVKVV",
"output": "3"
}
] | 1,492,357,293
| 393
|
Python 3
|
OK
|
TESTS
| 63
| 62
| 5,529,600
|
s = input().strip()
s = s.replace("VK", "*")
cnt = sum(map(lambda x: x=="*",s))
if "VV" in s or "KK" in s:
cnt +=1
print(cnt)
|
Title: Vicious Keyboard
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Tonio has a keyboard with only two letters, "V" and "K".
One day, he has typed out a string *s* with only these two letters. He really likes it when the string "VK" appears, so he wishes to change at most one letter in the string (or do no changes) to maximize the number of occurrences of that string. Compute the maximum number of times "VK" can appear as a substring (i.Β e. a letter "K" right after a letter "V") in the resulting string.
Input Specification:
The first line will contain a string *s* consisting only of uppercase English letters "V" and "K" with length not less than 1 and not greater than 100.
Output Specification:
Output a single integer, the maximum number of times "VK" can appear as a substring of the given string after changing at most one character.
Demo Input:
['VK\n', 'VV\n', 'V\n', 'VKKKKKKKKKVVVVVVVVVK\n', 'KVKV\n']
Demo Output:
['1\n', '1\n', '0\n', '3\n', '1\n']
Note:
For the first case, we do not change any letters. "VK" appears once, which is the maximum number of times it could appear.
For the second case, we can change the second character from a "V" to a "K". This will give us the string "VK". This has one occurrence of the string "VK" as a substring.
For the fourth case, we can change the fourth character from a "K" to a "V". This will give us the string "VKKVKKKKKKVVVVVVVVVK". This has three occurrences of the string "VK" as a substring. We can check no other moves can give us strictly more occurrences.
|
```python
s = input().strip()
s = s.replace("VK", "*")
cnt = sum(map(lambda x: x=="*",s))
if "VV" in s or "KK" in s:
cnt +=1
print(cnt)
```
| 3
|
|
120
|
A
|
Elevator
|
PROGRAMMING
| 1,000
|
[
"brute force",
"implementation",
"math"
] | null | null |
A sky scraper with 1000 floors has been built in the city of N. It has modern superfast elevators to help to travel from one floor to another. Each elevator has two doors, the front one and the back one. If one goes in through the front door, he goes out through the back one and vice versa. The elevator has two rails numbered with numbers 1 and 2. Rail 1 is located to the left of the entrance to the front door (or correspondingly, to the right of the entrance to the back door). Rail 2 is located opposite it, to the right of the entrance to the front door and to the left of the entrance to the back door. We know that each person in the city of N holds at a rail with the strongest hand.
One day a VIP person visited the city and of course, he took a look at the skyscraper and took a ride in the elevator. We know the door through which he entered and the rail he was holding at. Now we need to determine as soon as possible whether he is left-handed or right-handed.
|
The first line indicates the door through which the very important person entered the elevator. It contains "front" if the person enters the elevator through the front door and "back" if he entered the elevator through the back door. The second line contains integer *a* (1<=β€<=*a*<=β€<=2) which denotes the number of the rail at which the person was holding.
|
Print character "R" if the VIP is right-handed or "L" if he is left-handed.
|
[
"front\n1\n"
] |
[
"L\n"
] |
none
| 0
|
[
{
"input": "front\n1",
"output": "L"
},
{
"input": "back\n1",
"output": "R"
},
{
"input": "front\n2",
"output": "R"
},
{
"input": "back\n2",
"output": "L"
}
] | 1,447,199,712
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 4
| 92
| 0
|
fi = open('input.txt', 'r')
d, a = fi.readline().strip() == 'front', int(fi.readline()) == 1
print('L' if d == a else 'R', file=open('output.txt', 'w'))
|
Title: Elevator
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A sky scraper with 1000 floors has been built in the city of N. It has modern superfast elevators to help to travel from one floor to another. Each elevator has two doors, the front one and the back one. If one goes in through the front door, he goes out through the back one and vice versa. The elevator has two rails numbered with numbers 1 and 2. Rail 1 is located to the left of the entrance to the front door (or correspondingly, to the right of the entrance to the back door). Rail 2 is located opposite it, to the right of the entrance to the front door and to the left of the entrance to the back door. We know that each person in the city of N holds at a rail with the strongest hand.
One day a VIP person visited the city and of course, he took a look at the skyscraper and took a ride in the elevator. We know the door through which he entered and the rail he was holding at. Now we need to determine as soon as possible whether he is left-handed or right-handed.
Input Specification:
The first line indicates the door through which the very important person entered the elevator. It contains "front" if the person enters the elevator through the front door and "back" if he entered the elevator through the back door. The second line contains integer *a* (1<=β€<=*a*<=β€<=2) which denotes the number of the rail at which the person was holding.
Output Specification:
Print character "R" if the VIP is right-handed or "L" if he is left-handed.
Demo Input:
['front\n1\n']
Demo Output:
['L\n']
Note:
none
|
```python
fi = open('input.txt', 'r')
d, a = fi.readline().strip() == 'front', int(fi.readline()) == 1
print('L' if d == a else 'R', file=open('output.txt', 'w'))
```
| 3
|
|
263
|
A
|
Beautiful Matrix
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
You've got a 5<=Γ<=5 matrix, consisting of 24 zeroes and a single number one. Let's index the matrix rows by numbers from 1 to 5 from top to bottom, let's index the matrix columns by numbers from 1 to 5 from left to right. In one move, you are allowed to apply one of the two following transformations to the matrix:
1. Swap two neighboring matrix rows, that is, rows with indexes *i* and *i*<=+<=1 for some integer *i* (1<=β€<=*i*<=<<=5). 1. Swap two neighboring matrix columns, that is, columns with indexes *j* and *j*<=+<=1 for some integer *j* (1<=β€<=*j*<=<<=5).
You think that a matrix looks beautiful, if the single number one of the matrix is located in its middle (in the cell that is on the intersection of the third row and the third column). Count the minimum number of moves needed to make the matrix beautiful.
|
The input consists of five lines, each line contains five integers: the *j*-th integer in the *i*-th line of the input represents the element of the matrix that is located on the intersection of the *i*-th row and the *j*-th column. It is guaranteed that the matrix consists of 24 zeroes and a single number one.
|
Print a single integer β the minimum number of moves needed to make the matrix beautiful.
|
[
"0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n",
"0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n"
] |
[
"3\n",
"1\n"
] |
none
| 500
|
[
{
"input": "0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "1"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 1 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "0"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 1",
"output": "4"
},
{
"input": "0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "1 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "4"
},
{
"input": "0 0 1 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 1 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "4"
},
{
"input": "0 0 0 0 0\n1 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 1 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "1"
},
{
"input": "0 0 0 0 0\n0 0 0 1 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n1 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 1 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "1"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n1 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 1 0 0\n0 0 0 0 0",
"output": "1"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 1 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n1 0 0 0 0",
"output": "4"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 1 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 1 0",
"output": "3"
}
] | 1,698,415,903
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 25
| 92
| 0
|
l1=[]
for i in range(1,6):
l=list(map(int,input().split()))
l1.append(l)
for i in l1:
for j in i:
if j!=0:
c=i.index(j)
r=l1.index(i)
m=abs(r-2)
m=m+abs(c-2)
print(m)
|
Title: Beautiful Matrix
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You've got a 5<=Γ<=5 matrix, consisting of 24 zeroes and a single number one. Let's index the matrix rows by numbers from 1 to 5 from top to bottom, let's index the matrix columns by numbers from 1 to 5 from left to right. In one move, you are allowed to apply one of the two following transformations to the matrix:
1. Swap two neighboring matrix rows, that is, rows with indexes *i* and *i*<=+<=1 for some integer *i* (1<=β€<=*i*<=<<=5). 1. Swap two neighboring matrix columns, that is, columns with indexes *j* and *j*<=+<=1 for some integer *j* (1<=β€<=*j*<=<<=5).
You think that a matrix looks beautiful, if the single number one of the matrix is located in its middle (in the cell that is on the intersection of the third row and the third column). Count the minimum number of moves needed to make the matrix beautiful.
Input Specification:
The input consists of five lines, each line contains five integers: the *j*-th integer in the *i*-th line of the input represents the element of the matrix that is located on the intersection of the *i*-th row and the *j*-th column. It is guaranteed that the matrix consists of 24 zeroes and a single number one.
Output Specification:
Print a single integer β the minimum number of moves needed to make the matrix beautiful.
Demo Input:
['0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n', '0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n']
Demo Output:
['3\n', '1\n']
Note:
none
|
```python
l1=[]
for i in range(1,6):
l=list(map(int,input().split()))
l1.append(l)
for i in l1:
for j in i:
if j!=0:
c=i.index(j)
r=l1.index(i)
m=abs(r-2)
m=m+abs(c-2)
print(m)
```
| 3
|
|
69
|
A
|
Young Physicist
|
PROGRAMMING
| 1,000
|
[
"implementation",
"math"
] |
A. Young Physicist
|
2
|
256
|
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" β thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
|
The first line contains a positive integer *n* (1<=β€<=*n*<=β€<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=β€<=*x**i*,<=*y**i*,<=*z**i*<=β€<=100).
|
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
|
[
"3\n4 1 7\n-2 4 -1\n1 -5 -3\n",
"3\n3 -1 7\n-5 2 -4\n2 -1 -3\n"
] |
[
"NO",
"YES"
] |
none
| 500
|
[
{
"input": "3\n4 1 7\n-2 4 -1\n1 -5 -3",
"output": "NO"
},
{
"input": "3\n3 -1 7\n-5 2 -4\n2 -1 -3",
"output": "YES"
},
{
"input": "10\n21 32 -46\n43 -35 21\n42 2 -50\n22 40 20\n-27 -9 38\n-4 1 1\n-40 6 -31\n-13 -2 34\n-21 34 -12\n-32 -29 41",
"output": "NO"
},
{
"input": "10\n25 -33 43\n-27 -42 28\n-35 -20 19\n41 -42 -1\n49 -39 -4\n-49 -22 7\n-19 29 41\n8 -27 -43\n8 34 9\n-11 -3 33",
"output": "NO"
},
{
"input": "10\n-6 21 18\n20 -11 -8\n37 -11 41\n-5 8 33\n29 23 32\n30 -33 -11\n39 -49 -36\n28 34 -49\n22 29 -34\n-18 -6 7",
"output": "NO"
},
{
"input": "10\n47 -2 -27\n0 26 -14\n5 -12 33\n2 18 3\n45 -30 -49\n4 -18 8\n-46 -44 -41\n-22 -10 -40\n-35 -21 26\n33 20 38",
"output": "NO"
},
{
"input": "13\n-3 -36 -46\n-11 -50 37\n42 -11 -15\n9 42 44\n-29 -12 24\n3 9 -40\n-35 13 50\n14 43 18\n-13 8 24\n-48 -15 10\n50 9 -50\n21 0 -50\n0 0 -6",
"output": "YES"
},
{
"input": "14\n43 23 17\n4 17 44\n5 -5 -16\n-43 -7 -6\n47 -48 12\n50 47 -45\n2 14 43\n37 -30 15\n4 -17 -11\n17 9 -45\n-50 -3 -8\n-50 0 0\n-50 0 0\n-16 0 0",
"output": "YES"
},
{
"input": "13\n29 49 -11\n38 -11 -20\n25 1 -40\n-11 28 11\n23 -19 1\n45 -41 -17\n-3 0 -19\n-13 -33 49\n-30 0 28\n34 17 45\n-50 9 -27\n-50 0 0\n-37 0 0",
"output": "YES"
},
{
"input": "12\n3 28 -35\n-32 -44 -17\n9 -25 -6\n-42 -22 20\n-19 15 38\n-21 38 48\n-1 -37 -28\n-10 -13 -50\n-5 21 29\n34 28 50\n50 11 -49\n34 0 0",
"output": "YES"
},
{
"input": "37\n-64 -79 26\n-22 59 93\n-5 39 -12\n77 -9 76\n55 -86 57\n83 100 -97\n-70 94 84\n-14 46 -94\n26 72 35\n14 78 -62\n17 82 92\n-57 11 91\n23 15 92\n-80 -1 1\n12 39 18\n-23 -99 -75\n-34 50 19\n-39 84 -7\n45 -30 -39\n-60 49 37\n45 -16 -72\n33 -51 -56\n-48 28 5\n97 91 88\n45 -82 -11\n-21 -15 -90\n-53 73 -26\n-74 85 -90\n-40 23 38\n100 -13 49\n32 -100 -100\n0 -100 -70\n0 -100 0\n0 -100 0\n0 -100 0\n0 -100 0\n0 -37 0",
"output": "YES"
},
{
"input": "4\n68 3 100\n68 21 -100\n-100 -24 0\n-36 0 0",
"output": "YES"
},
{
"input": "33\n-1 -46 -12\n45 -16 -21\n-11 45 -21\n-60 -42 -93\n-22 -45 93\n37 96 85\n-76 26 83\n-4 9 55\n7 -52 -9\n66 8 -85\n-100 -54 11\n-29 59 74\n-24 12 2\n-56 81 85\n-92 69 -52\n-26 -97 91\n54 59 -51\n58 21 -57\n7 68 56\n-47 -20 -51\n-59 77 -13\n-85 27 91\n79 60 -56\n66 -80 5\n21 -99 42\n-31 -29 98\n66 93 76\n-49 45 61\n100 -100 -100\n100 -100 -100\n66 -75 -100\n0 0 -100\n0 0 -87",
"output": "YES"
},
{
"input": "3\n1 2 3\n3 2 1\n0 0 0",
"output": "NO"
},
{
"input": "2\n5 -23 12\n0 0 0",
"output": "NO"
},
{
"input": "1\n0 0 0",
"output": "YES"
},
{
"input": "1\n1 -2 0",
"output": "NO"
},
{
"input": "2\n-23 77 -86\n23 -77 86",
"output": "YES"
},
{
"input": "26\n86 7 20\n-57 -64 39\n-45 6 -93\n-44 -21 100\n-11 -49 21\n73 -71 -80\n-2 -89 56\n-65 -2 7\n5 14 84\n57 41 13\n-12 69 54\n40 -25 27\n-17 -59 0\n64 -91 -30\n-53 9 42\n-54 -8 14\n-35 82 27\n-48 -59 -80\n88 70 79\n94 57 97\n44 63 25\n84 -90 -40\n-100 100 -100\n-92 100 -100\n0 10 -100\n0 0 -82",
"output": "YES"
},
{
"input": "42\n11 27 92\n-18 -56 -57\n1 71 81\n33 -92 30\n82 83 49\n-87 -61 -1\n-49 45 49\n73 26 15\n-22 22 -77\n29 -93 87\n-68 44 -90\n-4 -84 20\n85 67 -6\n-39 26 77\n-28 -64 20\n65 -97 24\n-72 -39 51\n35 -75 -91\n39 -44 -8\n-25 -27 -57\n91 8 -46\n-98 -94 56\n94 -60 59\n-9 -95 18\n-53 -37 98\n-8 -94 -84\n-52 55 60\n15 -14 37\n65 -43 -25\n94 12 66\n-8 -19 -83\n29 81 -78\n-58 57 33\n24 86 -84\n-53 32 -88\n-14 7 3\n89 97 -53\n-5 -28 -91\n-100 100 -6\n-84 100 0\n0 100 0\n0 70 0",
"output": "YES"
},
{
"input": "3\n96 49 -12\n2 -66 28\n-98 17 -16",
"output": "YES"
},
{
"input": "5\n70 -46 86\n-100 94 24\n-27 63 -63\n57 -100 -47\n0 -11 0",
"output": "YES"
},
{
"input": "18\n-86 -28 70\n-31 -89 42\n31 -48 -55\n95 -17 -43\n24 -95 -85\n-21 -14 31\n68 -18 81\n13 31 60\n-15 28 99\n-42 15 9\n28 -61 -62\n-16 71 29\n-28 75 -48\n-77 -67 36\n-100 83 89\n100 100 -100\n57 34 -100\n0 0 -53",
"output": "YES"
},
{
"input": "44\n52 -54 -29\n-82 -5 -94\n-54 43 43\n91 16 71\n7 80 -91\n3 15 29\n-99 -6 -77\n-3 -77 -64\n73 67 34\n25 -10 -18\n-29 91 63\n-72 86 -16\n-68 85 -81\n-3 36 44\n-74 -14 -80\n34 -96 -97\n-76 -78 -33\n-24 44 -58\n98 12 77\n95 -63 -6\n-51 3 -90\n-92 -10 72\n7 3 -68\n57 -53 71\n29 57 -48\n35 -60 10\n79 -70 -61\n-20 77 55\n-86 -15 -35\n84 -88 -18\n100 -42 77\n-20 46 8\n-41 -43 -65\n38 -98 -23\n-100 65 45\n-7 -91 -63\n46 88 -85\n48 59 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 1",
"output": "YES"
},
{
"input": "18\n-14 -64 -91\n-8 -66 -86\n-23 92 -40\n6 -3 -53\n57 41 78\n-79 42 -22\n-88 -17 45\n4 -45 44\n83 -18 -25\n34 86 -92\n75 -30 12\n44 99 11\n-67 -13 72\n22 83 -56\n-37 71 72\n-9 -100 100\n0 -100 31\n0 -58 0",
"output": "YES"
},
{
"input": "23\n-70 37 78\n42 84 6\n28 -94 -24\n-49 76 95\n-67 18 84\n-53 78 -5\n65 -63 -64\n-66 -64 -69\n81 -93 95\n10 52 -79\n-89 -61 -64\n-64 -47 43\n-81 -35 55\n80 82 73\n-60 -81 -18\n49 -10 -19\n-58 70 12\n-24 -15 -93\n98 -93 -54\n-28 -75 11\n100 100 -63\n100 100 0\n56 34 0",
"output": "YES"
},
{
"input": "3\n0 2 -2\n1 -1 3\n-3 0 0",
"output": "NO"
}
] | 1,671,444,940
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 81
| 154
| 0
|
n = int(input())
x = []
for i in range(n):
a, b, c = map(int, input().split())
x.append([a,b,c])
s = 0
m = 0
l = 0
for i in range(n):
s += x[i][0]
m += x[i][1]
l += x[i][2]
if s == 0 and m == 0 and l == 0:
print("YES")
else:
print("NO")
|
Title: Young Physicist
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" β thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
Input Specification:
The first line contains a positive integer *n* (1<=β€<=*n*<=β€<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=β€<=*x**i*,<=*y**i*,<=*z**i*<=β€<=100).
Output Specification:
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
Demo Input:
['3\n4 1 7\n-2 4 -1\n1 -5 -3\n', '3\n3 -1 7\n-5 2 -4\n2 -1 -3\n']
Demo Output:
['NO', 'YES']
Note:
none
|
```python
n = int(input())
x = []
for i in range(n):
a, b, c = map(int, input().split())
x.append([a,b,c])
s = 0
m = 0
l = 0
for i in range(n):
s += x[i][0]
m += x[i][1]
l += x[i][2]
if s == 0 and m == 0 and l == 0:
print("YES")
else:
print("NO")
```
| 3.9615
|
981
|
A
|
Antipalindrome
|
PROGRAMMING
| 900
|
[
"brute force",
"implementation",
"strings"
] | null | null |
A string is a palindrome if it reads the same from the left to the right and from the right to the left. For example, the strings "kek", "abacaba", "r" and "papicipap" are palindromes, while the strings "abb" and "iq" are not.
A substring $s[l \ldots r]$ ($1<=\leq<=l<=\leq<=r<=\leq<=|s|$) of a string $s<==<=s_{1}s_{2} \ldots s_{|s|}$ is the string $s_{l}s_{l<=+<=1} \ldots s_{r}$.
Anna does not like palindromes, so she makes her friends call her Ann. She also changes all the words she reads in a similar way. Namely, each word $s$ is changed into its longest substring that is not a palindrome. If all the substrings of $s$ are palindromes, she skips the word at all.
Some time ago Ann read the word $s$. What is the word she changed it into?
|
The first line contains a non-empty string $s$ with length at most $50$ characters, containing lowercase English letters only.
|
If there is such a substring in $s$ that is not a palindrome, print the maximum length of such a substring. Otherwise print $0$.
Note that there can be multiple longest substrings that are not palindromes, but their length is unique.
|
[
"mew\n",
"wuffuw\n",
"qqqqqqqq\n"
] |
[
"3\n",
"5\n",
"0\n"
] |
"mew" is not a palindrome, so the longest substring of it that is not a palindrome, is the string "mew" itself. Thus, the answer for the first example is $3$.
The string "uffuw" is one of the longest non-palindrome substrings (of length $5$) of the string "wuffuw", so the answer for the second example is $5$.
All substrings of the string "qqqqqqqq" consist of equal characters so they are palindromes. This way, there are no non-palindrome substrings. Thus, the answer for the third example is $0$.
| 500
|
[
{
"input": "mew",
"output": "3"
},
{
"input": "wuffuw",
"output": "5"
},
{
"input": "qqqqqqqq",
"output": "0"
},
{
"input": "ijvji",
"output": "4"
},
{
"input": "iiiiiii",
"output": "0"
},
{
"input": "wobervhvvkihcuyjtmqhaaigvvgiaahqmtjyuchikvvhvrebow",
"output": "49"
},
{
"input": "wwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww",
"output": "0"
},
{
"input": "wobervhvvkihcuyjtmqhaaigvahheoqleromusrartldojsjvy",
"output": "50"
},
{
"input": "ijvxljt",
"output": "7"
},
{
"input": "fyhcncnchyf",
"output": "10"
},
{
"input": "ffffffffffff",
"output": "0"
},
{
"input": "fyhcncfsepqj",
"output": "12"
},
{
"input": "ybejrrlbcinttnicblrrjeby",
"output": "23"
},
{
"input": "yyyyyyyyyyyyyyyyyyyyyyyyy",
"output": "0"
},
{
"input": "ybejrrlbcintahovgjddrqatv",
"output": "25"
},
{
"input": "oftmhcmclgyqaojljoaqyglcmchmtfo",
"output": "30"
},
{
"input": "oooooooooooooooooooooooooooooooo",
"output": "0"
},
{
"input": "oftmhcmclgyqaojllbotztajglsmcilv",
"output": "32"
},
{
"input": "gxandbtgpbknxvnkjaajknvxnkbpgtbdnaxg",
"output": "35"
},
{
"input": "gggggggggggggggggggggggggggggggggggg",
"output": "0"
},
{
"input": "gxandbtgpbknxvnkjaygommzqitqzjfalfkk",
"output": "36"
},
{
"input": "fcliblymyqckxvieotjooojtoeivxkcqymylbilcf",
"output": "40"
},
{
"input": "fffffffffffffffffffffffffffffffffffffffffff",
"output": "0"
},
{
"input": "fcliblymyqckxvieotjootiqwtyznhhvuhbaixwqnsy",
"output": "43"
},
{
"input": "rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr",
"output": "0"
},
{
"input": "rajccqwqnqmshmerpvjyfepxwpxyldzpzhctqjnstxyfmlhiy",
"output": "49"
},
{
"input": "a",
"output": "0"
},
{
"input": "abca",
"output": "4"
},
{
"input": "aaaaabaaaaa",
"output": "10"
},
{
"input": "aba",
"output": "2"
},
{
"input": "asaa",
"output": "4"
},
{
"input": "aabaa",
"output": "4"
},
{
"input": "aabbaa",
"output": "5"
},
{
"input": "abcdaaa",
"output": "7"
},
{
"input": "aaholaa",
"output": "7"
},
{
"input": "abcdefghijka",
"output": "12"
},
{
"input": "aaadcba",
"output": "7"
},
{
"input": "aaaabaaaa",
"output": "8"
},
{
"input": "abaa",
"output": "4"
},
{
"input": "abcbaa",
"output": "6"
},
{
"input": "ab",
"output": "2"
},
{
"input": "l",
"output": "0"
},
{
"input": "aaaabcaaaa",
"output": "10"
},
{
"input": "abbaaaaaabba",
"output": "11"
},
{
"input": "abaaa",
"output": "5"
},
{
"input": "baa",
"output": "3"
},
{
"input": "aaaaaaabbba",
"output": "11"
},
{
"input": "ccbcc",
"output": "4"
},
{
"input": "bbbaaab",
"output": "7"
},
{
"input": "abaaaaaaaa",
"output": "10"
},
{
"input": "abaaba",
"output": "5"
},
{
"input": "aabsdfaaaa",
"output": "10"
},
{
"input": "aaaba",
"output": "5"
},
{
"input": "aaabaaa",
"output": "6"
},
{
"input": "baaabbb",
"output": "7"
},
{
"input": "ccbbabbcc",
"output": "8"
},
{
"input": "cabc",
"output": "4"
},
{
"input": "aabcd",
"output": "5"
},
{
"input": "abcdea",
"output": "6"
},
{
"input": "bbabb",
"output": "4"
},
{
"input": "aaaaabababaaaaa",
"output": "14"
},
{
"input": "bbabbb",
"output": "6"
},
{
"input": "aababd",
"output": "6"
},
{
"input": "abaaaa",
"output": "6"
},
{
"input": "aaaaaaaabbba",
"output": "12"
},
{
"input": "aabca",
"output": "5"
},
{
"input": "aaabccbaaa",
"output": "9"
},
{
"input": "aaaaaaaaaaaaaaaaaaaab",
"output": "21"
},
{
"input": "babb",
"output": "4"
},
{
"input": "abcaa",
"output": "5"
},
{
"input": "qwqq",
"output": "4"
},
{
"input": "aaaaaaaaaaabbbbbbbbbbbbbbbaaaaaaaaaaaaaaaaaaaaaa",
"output": "48"
},
{
"input": "aaab",
"output": "4"
},
{
"input": "aaaaaabaaaaa",
"output": "12"
},
{
"input": "wwuww",
"output": "4"
},
{
"input": "aaaaabcbaaaaa",
"output": "12"
},
{
"input": "aaabbbaaa",
"output": "8"
},
{
"input": "aabcbaa",
"output": "6"
},
{
"input": "abccdefccba",
"output": "11"
},
{
"input": "aabbcbbaa",
"output": "8"
},
{
"input": "aaaabbaaaa",
"output": "9"
},
{
"input": "aabcda",
"output": "6"
},
{
"input": "abbca",
"output": "5"
},
{
"input": "aaaaaabbaaa",
"output": "11"
},
{
"input": "sssssspssssss",
"output": "12"
},
{
"input": "sdnmsdcs",
"output": "8"
},
{
"input": "aaabbbccbbbaaa",
"output": "13"
},
{
"input": "cbdbdc",
"output": "6"
},
{
"input": "abb",
"output": "3"
},
{
"input": "abcdefaaaa",
"output": "10"
},
{
"input": "abbbaaa",
"output": "7"
},
{
"input": "v",
"output": "0"
},
{
"input": "abccbba",
"output": "7"
},
{
"input": "axyza",
"output": "5"
},
{
"input": "abcdefgaaaa",
"output": "11"
},
{
"input": "aaabcdaaa",
"output": "9"
},
{
"input": "aaaacaaaa",
"output": "8"
},
{
"input": "aaaaaaaaaaaaaaaaaaaabaaaaaaaaaaaaaaaaaaaaa",
"output": "42"
},
{
"input": "abbbaa",
"output": "6"
},
{
"input": "abcdee",
"output": "6"
},
{
"input": "oom",
"output": "3"
},
{
"input": "aabcaa",
"output": "6"
},
{
"input": "abba",
"output": "3"
},
{
"input": "aaca",
"output": "4"
},
{
"input": "aacbca",
"output": "6"
},
{
"input": "ababa",
"output": "4"
},
{
"input": "abcda",
"output": "5"
},
{
"input": "cccaaccc",
"output": "7"
},
{
"input": "aaabcda",
"output": "7"
},
{
"input": "aa",
"output": "0"
},
{
"input": "aabaaaa",
"output": "7"
},
{
"input": "abbaaaa",
"output": "7"
},
{
"input": "aaabcbaaa",
"output": "8"
},
{
"input": "aabba",
"output": "5"
},
{
"input": "xyxx",
"output": "4"
},
{
"input": "aaaaaaaaaaaabc",
"output": "14"
},
{
"input": "bbaaaabb",
"output": "7"
},
{
"input": "aaabaa",
"output": "6"
},
{
"input": "sssssabsssss",
"output": "12"
},
{
"input": "bbbaaaabbb",
"output": "9"
},
{
"input": "abbbbaaaa",
"output": "9"
},
{
"input": "wwufuww",
"output": "6"
},
{
"input": "oowoo",
"output": "4"
},
{
"input": "cccaccc",
"output": "6"
},
{
"input": "aaa",
"output": "0"
},
{
"input": "bbbcc",
"output": "5"
},
{
"input": "abcdef",
"output": "6"
},
{
"input": "abbba",
"output": "4"
},
{
"input": "aab",
"output": "3"
},
{
"input": "aaba",
"output": "4"
},
{
"input": "azbyaaa",
"output": "7"
},
{
"input": "oooooiooooo",
"output": "10"
},
{
"input": "aabbbbbaaaaaa",
"output": "13"
}
] | 1,638,956,940
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 133
| 109
| 512,000
|
"""https://codeforces.com/problemset/problem/981/A
"""
s = input()
for r in range(len(s), 1, -1):
if s[:r] != s[r - 1::-1]:
print(r)
break
else:
print(0)
|
Title: Antipalindrome
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A string is a palindrome if it reads the same from the left to the right and from the right to the left. For example, the strings "kek", "abacaba", "r" and "papicipap" are palindromes, while the strings "abb" and "iq" are not.
A substring $s[l \ldots r]$ ($1<=\leq<=l<=\leq<=r<=\leq<=|s|$) of a string $s<==<=s_{1}s_{2} \ldots s_{|s|}$ is the string $s_{l}s_{l<=+<=1} \ldots s_{r}$.
Anna does not like palindromes, so she makes her friends call her Ann. She also changes all the words she reads in a similar way. Namely, each word $s$ is changed into its longest substring that is not a palindrome. If all the substrings of $s$ are palindromes, she skips the word at all.
Some time ago Ann read the word $s$. What is the word she changed it into?
Input Specification:
The first line contains a non-empty string $s$ with length at most $50$ characters, containing lowercase English letters only.
Output Specification:
If there is such a substring in $s$ that is not a palindrome, print the maximum length of such a substring. Otherwise print $0$.
Note that there can be multiple longest substrings that are not palindromes, but their length is unique.
Demo Input:
['mew\n', 'wuffuw\n', 'qqqqqqqq\n']
Demo Output:
['3\n', '5\n', '0\n']
Note:
"mew" is not a palindrome, so the longest substring of it that is not a palindrome, is the string "mew" itself. Thus, the answer for the first example is $3$.
The string "uffuw" is one of the longest non-palindrome substrings (of length $5$) of the string "wuffuw", so the answer for the second example is $5$.
All substrings of the string "qqqqqqqq" consist of equal characters so they are palindromes. This way, there are no non-palindrome substrings. Thus, the answer for the third example is $0$.
|
```python
"""https://codeforces.com/problemset/problem/981/A
"""
s = input()
for r in range(len(s), 1, -1):
if s[:r] != s[r - 1::-1]:
print(r)
break
else:
print(0)
```
| 3
|
|
854
|
A
|
Fraction
|
PROGRAMMING
| 800
|
[
"brute force",
"constructive algorithms",
"math"
] | null | null |
Petya is a big fan of mathematics, especially its part related to fractions. Recently he learned that a fraction is called proper iff its numerator is smaller than its denominator (*a*<=<<=*b*) and that the fraction is called irreducible if its numerator and its denominator are coprime (they do not have positive common divisors except 1).
During his free time, Petya thinks about proper irreducible fractions and converts them to decimals using the calculator. One day he mistakenly pressed addition button (<=+<=) instead of division button (Γ·) and got sum of numerator and denominator that was equal to *n* instead of the expected decimal notation.
Petya wanted to restore the original fraction, but soon he realized that it might not be done uniquely. That's why he decided to determine maximum possible proper irreducible fraction such that sum of its numerator and denominator equals *n*. Help Petya deal with this problem.
|
In the only line of input there is an integer *n* (3<=β€<=*n*<=β€<=1000), the sum of numerator and denominator of the fraction.
|
Output two space-separated positive integers *a* and *b*, numerator and denominator of the maximum possible proper irreducible fraction satisfying the given sum.
|
[
"3\n",
"4\n",
"12\n"
] |
[
"1 2\n",
"1 3\n",
"5 7\n"
] |
none
| 500
|
[
{
"input": "3",
"output": "1 2"
},
{
"input": "4",
"output": "1 3"
},
{
"input": "12",
"output": "5 7"
},
{
"input": "34",
"output": "15 19"
},
{
"input": "13",
"output": "6 7"
},
{
"input": "11",
"output": "5 6"
},
{
"input": "24",
"output": "11 13"
},
{
"input": "17",
"output": "8 9"
},
{
"input": "10",
"output": "3 7"
},
{
"input": "69",
"output": "34 35"
},
{
"input": "100",
"output": "49 51"
},
{
"input": "57",
"output": "28 29"
},
{
"input": "1000",
"output": "499 501"
},
{
"input": "999",
"output": "499 500"
},
{
"input": "998",
"output": "497 501"
},
{
"input": "997",
"output": "498 499"
},
{
"input": "996",
"output": "497 499"
},
{
"input": "995",
"output": "497 498"
},
{
"input": "994",
"output": "495 499"
},
{
"input": "5",
"output": "2 3"
},
{
"input": "6",
"output": "1 5"
},
{
"input": "8",
"output": "3 5"
},
{
"input": "9",
"output": "4 5"
},
{
"input": "423",
"output": "211 212"
},
{
"input": "876",
"output": "437 439"
},
{
"input": "29",
"output": "14 15"
}
] | 1,613,645,596
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 26
| 62
| 0
|
n=int(input())
m=n//2
switcher={
0:[m-1,m+1],
1:[m,m+1],
2:[m-2,m+2],
3:[m,m+1],
}
ans=switcher.get(n%4)
print(ans[0],ans[1])
|
Title: Fraction
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya is a big fan of mathematics, especially its part related to fractions. Recently he learned that a fraction is called proper iff its numerator is smaller than its denominator (*a*<=<<=*b*) and that the fraction is called irreducible if its numerator and its denominator are coprime (they do not have positive common divisors except 1).
During his free time, Petya thinks about proper irreducible fractions and converts them to decimals using the calculator. One day he mistakenly pressed addition button (<=+<=) instead of division button (Γ·) and got sum of numerator and denominator that was equal to *n* instead of the expected decimal notation.
Petya wanted to restore the original fraction, but soon he realized that it might not be done uniquely. That's why he decided to determine maximum possible proper irreducible fraction such that sum of its numerator and denominator equals *n*. Help Petya deal with this problem.
Input Specification:
In the only line of input there is an integer *n* (3<=β€<=*n*<=β€<=1000), the sum of numerator and denominator of the fraction.
Output Specification:
Output two space-separated positive integers *a* and *b*, numerator and denominator of the maximum possible proper irreducible fraction satisfying the given sum.
Demo Input:
['3\n', '4\n', '12\n']
Demo Output:
['1 2\n', '1 3\n', '5 7\n']
Note:
none
|
```python
n=int(input())
m=n//2
switcher={
0:[m-1,m+1],
1:[m,m+1],
2:[m-2,m+2],
3:[m,m+1],
}
ans=switcher.get(n%4)
print(ans[0],ans[1])
```
| 3
|
|
82
|
A
|
Double Cola
|
PROGRAMMING
| 1,100
|
[
"implementation",
"math"
] |
A. Double Cola
|
1
|
256
|
Sheldon, Leonard, Penny, Rajesh and Howard are in the queue for a "Double Cola" drink vending machine; there are no other people in the queue. The first one in the queue (Sheldon) buys a can, drinks it and doubles! The resulting two Sheldons go to the end of the queue. Then the next in the queue (Leonard) buys a can, drinks it and gets to the end of the queue as two Leonards, and so on. This process continues ad infinitum.
For example, Penny drinks the third can of cola and the queue will look like this: Rajesh, Howard, Sheldon, Sheldon, Leonard, Leonard, Penny, Penny.
Write a program that will print the name of a man who will drink the *n*-th can.
Note that in the very beginning the queue looks like that: Sheldon, Leonard, Penny, Rajesh, Howard. The first person is Sheldon.
|
The input data consist of a single integer *n* (1<=β€<=*n*<=β€<=109).
It is guaranteed that the pretests check the spelling of all the five names, that is, that they contain all the five possible answers.
|
Print the single line β the name of the person who drinks the *n*-th can of cola. The cans are numbered starting from 1. Please note that you should spell the names like this: "Sheldon", "Leonard", "Penny", "Rajesh", "Howard" (without the quotes). In that order precisely the friends are in the queue initially.
|
[
"1\n",
"6\n",
"1802\n"
] |
[
"Sheldon\n",
"Sheldon\n",
"Penny\n"
] |
none
| 500
|
[
{
"input": "1",
"output": "Sheldon"
},
{
"input": "6",
"output": "Sheldon"
},
{
"input": "1802",
"output": "Penny"
},
{
"input": "1",
"output": "Sheldon"
},
{
"input": "2",
"output": "Leonard"
},
{
"input": "3",
"output": "Penny"
},
{
"input": "4",
"output": "Rajesh"
},
{
"input": "5",
"output": "Howard"
},
{
"input": "10",
"output": "Penny"
},
{
"input": "534",
"output": "Rajesh"
},
{
"input": "5033",
"output": "Howard"
},
{
"input": "10010",
"output": "Howard"
},
{
"input": "500000000",
"output": "Penny"
},
{
"input": "63",
"output": "Rajesh"
},
{
"input": "841",
"output": "Leonard"
},
{
"input": "3667",
"output": "Penny"
},
{
"input": "38614",
"output": "Howard"
},
{
"input": "282798",
"output": "Rajesh"
},
{
"input": "9266286",
"output": "Rajesh"
},
{
"input": "27385966",
"output": "Leonard"
},
{
"input": "121580142",
"output": "Penny"
},
{
"input": "5",
"output": "Howard"
},
{
"input": "300",
"output": "Howard"
},
{
"input": "1745",
"output": "Leonard"
},
{
"input": "8302",
"output": "Rajesh"
},
{
"input": "184518",
"output": "Sheldon"
},
{
"input": "1154414",
"output": "Rajesh"
},
{
"input": "28643950",
"output": "Leonard"
},
{
"input": "159222638",
"output": "Howard"
},
{
"input": "24",
"output": "Penny"
},
{
"input": "505",
"output": "Penny"
},
{
"input": "4425",
"output": "Rajesh"
},
{
"input": "12079",
"output": "Sheldon"
},
{
"input": "469726",
"output": "Penny"
},
{
"input": "3961838",
"output": "Penny"
},
{
"input": "57710446",
"output": "Leonard"
},
{
"input": "80719470",
"output": "Howard"
},
{
"input": "1000000000",
"output": "Penny"
},
{
"input": "999999999",
"output": "Penny"
},
{
"input": "999999998",
"output": "Penny"
},
{
"input": "5",
"output": "Howard"
}
] | 1,558,927,402
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 41
| 140
| 0
|
x = int(input())
n=1
lvl_arr = [0]
def LvL(n):
lvl_arr.append(2**(n-1)*5 + lvl_arr[n-1])
return lvl_arr[n]
while(x > LvL(n)):
n = n+1
p = ((x - lvl_arr[n-1] - 1) // 2**(n-1)) + 1
if(p == 1):
print("Sheldon")
elif(p == 2):
print("Leonard")
elif(p == 3):
print("Penny")
elif(p == 4):
print("Rajesh")
else:
print("Howard")
|
Title: Double Cola
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Sheldon, Leonard, Penny, Rajesh and Howard are in the queue for a "Double Cola" drink vending machine; there are no other people in the queue. The first one in the queue (Sheldon) buys a can, drinks it and doubles! The resulting two Sheldons go to the end of the queue. Then the next in the queue (Leonard) buys a can, drinks it and gets to the end of the queue as two Leonards, and so on. This process continues ad infinitum.
For example, Penny drinks the third can of cola and the queue will look like this: Rajesh, Howard, Sheldon, Sheldon, Leonard, Leonard, Penny, Penny.
Write a program that will print the name of a man who will drink the *n*-th can.
Note that in the very beginning the queue looks like that: Sheldon, Leonard, Penny, Rajesh, Howard. The first person is Sheldon.
Input Specification:
The input data consist of a single integer *n* (1<=β€<=*n*<=β€<=109).
It is guaranteed that the pretests check the spelling of all the five names, that is, that they contain all the five possible answers.
Output Specification:
Print the single line β the name of the person who drinks the *n*-th can of cola. The cans are numbered starting from 1. Please note that you should spell the names like this: "Sheldon", "Leonard", "Penny", "Rajesh", "Howard" (without the quotes). In that order precisely the friends are in the queue initially.
Demo Input:
['1\n', '6\n', '1802\n']
Demo Output:
['Sheldon\n', 'Sheldon\n', 'Penny\n']
Note:
none
|
```python
x = int(input())
n=1
lvl_arr = [0]
def LvL(n):
lvl_arr.append(2**(n-1)*5 + lvl_arr[n-1])
return lvl_arr[n]
while(x > LvL(n)):
n = n+1
p = ((x - lvl_arr[n-1] - 1) // 2**(n-1)) + 1
if(p == 1):
print("Sheldon")
elif(p == 2):
print("Leonard")
elif(p == 3):
print("Penny")
elif(p == 4):
print("Rajesh")
else:
print("Howard")
```
| 3.93
|
1
|
A
|
Theatre Square
|
PROGRAMMING
| 1,000
|
[
"math"
] |
A. Theatre Square
|
1
|
256
|
Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=Γ<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=Γ<=*a*.
What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square.
|
The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=β€<=<=*n*,<=*m*,<=*a*<=β€<=109).
|
Write the needed number of flagstones.
|
[
"6 6 4\n"
] |
[
"4\n"
] |
none
| 0
|
[
{
"input": "6 6 4",
"output": "4"
},
{
"input": "1 1 1",
"output": "1"
},
{
"input": "2 1 1",
"output": "2"
},
{
"input": "1 2 1",
"output": "2"
},
{
"input": "2 2 1",
"output": "4"
},
{
"input": "2 1 2",
"output": "1"
},
{
"input": "1 1 3",
"output": "1"
},
{
"input": "2 3 4",
"output": "1"
},
{
"input": "1000000000 1000000000 1",
"output": "1000000000000000000"
},
{
"input": "12 13 4",
"output": "12"
},
{
"input": "222 332 5",
"output": "3015"
},
{
"input": "1000 1000 10",
"output": "10000"
},
{
"input": "1001 1000 10",
"output": "10100"
},
{
"input": "100 10001 1000000000",
"output": "1"
},
{
"input": "1000000000 1000000000 1000000000",
"output": "1"
},
{
"input": "1000000000 1000000000 999999999",
"output": "4"
},
{
"input": "1000000000 1000000000 192",
"output": "27126743055556"
},
{
"input": "1000000000 987654321 1",
"output": "987654321000000000"
},
{
"input": "456784567 1000000000 51",
"output": "175618850864484"
},
{
"input": "39916800 134217728 40320",
"output": "3295710"
}
] | 1,683,039,348
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 20
| 46
| 0
|
n,m,a=map(int,input().split())
l1=n//a
l2=m//a
if n//a<n/a:
l1=n//a+1
if m//a<m/a:
l2=m//a +1
soni=l1*l2
print(soni)
|
Title: Theatre Square
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=Γ<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=Γ<=*a*.
What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square.
Input Specification:
The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=β€<=<=*n*,<=*m*,<=*a*<=β€<=109).
Output Specification:
Write the needed number of flagstones.
Demo Input:
['6 6 4\n']
Demo Output:
['4\n']
Note:
none
|
```python
n,m,a=map(int,input().split())
l1=n//a
l2=m//a
if n//a<n/a:
l1=n//a+1
if m//a<m/a:
l2=m//a +1
soni=l1*l2
print(soni)
```
| 3.977
|
111
|
A
|
Petya and Inequiations
|
PROGRAMMING
| 1,400
|
[
"greedy"
] |
A. Petya and Inequiations
|
2
|
256
|
Little Petya loves inequations. Help him find *n* positive integers *a*1,<=*a*2,<=...,<=*a**n*, such that the following two conditions are satisfied:
- *a*12<=+<=*a*22<=+<=...<=+<=*a**n*2<=β₯<=*x*- *a*1<=+<=*a*2<=+<=...<=+<=*a**n*<=β€<=*y*
|
The first line contains three space-separated integers *n*, *x* and *y* (1<=β€<=*n*<=β€<=105,<=1<=β€<=*x*<=β€<=1012,<=1<=β€<=*y*<=β€<=106).
Please do not use the %lld specificator to read or write 64-bit integers in Π‘++. It is recommended to use cin, cout streams or the %I64d specificator.
|
Print *n* positive integers that satisfy the conditions, one integer per line. If such numbers do not exist, print a single number "-1". If there are several solutions, print any of them.
|
[
"5 15 15\n",
"2 3 2\n",
"1 99 11\n"
] |
[
"4\n4\n1\n1\n2\n",
"-1\n",
"11\n"
] |
none
| 500
|
[
{
"input": "5 15 15",
"output": "11\n1\n1\n1\n1"
},
{
"input": "2 3 2",
"output": "-1"
},
{
"input": "1 99 11",
"output": "11"
},
{
"input": "100000 810000099998 1000000",
"output": "900001\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n..."
},
{
"input": "3 254 18",
"output": "16\n1\n1"
},
{
"input": "4 324 77",
"output": "74\n1\n1\n1"
},
{
"input": "5 315 90",
"output": "86\n1\n1\n1\n1"
},
{
"input": "6 225 59",
"output": "54\n1\n1\n1\n1\n1"
},
{
"input": "7 351 29",
"output": "23\n1\n1\n1\n1\n1\n1"
},
{
"input": "100 913723780421 955988",
"output": "-1"
},
{
"input": "200 894176381082 945808",
"output": "-1"
},
{
"input": "300 923251939897 961159",
"output": "960860\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n..."
},
{
"input": "1000 824905348050 909242",
"output": "-1"
},
{
"input": "10000 795416053320 901860",
"output": "891861\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n..."
},
{
"input": "31000 819461299082 936240",
"output": "-1"
},
{
"input": "44000 772772899626 923074",
"output": "-1"
},
{
"input": "45678 783917268558 931068",
"output": "885391\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n..."
},
{
"input": "99999 681508136225 925533",
"output": "-1"
},
{
"input": "99999 688345771552 929664",
"output": "829666\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n..."
},
{
"input": "99976 664640815001 915230",
"output": "-1"
},
{
"input": "100000 729199960625 953931",
"output": "-1"
},
{
"input": "50 890543266647 943735",
"output": "-1"
},
{
"input": "60 817630084499 904288",
"output": "904229\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1"
},
{
"input": "99999 716046078026 946193",
"output": "-1"
},
{
"input": "99998 729652614803 954194",
"output": "854197\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n..."
},
{
"input": "10000 950051796437 984705",
"output": "-1"
},
{
"input": "999 992972391401 997478",
"output": "-1"
},
{
"input": "99999 667887855532 917242",
"output": "817244\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n..."
},
{
"input": "12313 955817132591 989971",
"output": "977659\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n..."
},
{
"input": "1 983300308227 991615",
"output": "-1"
},
{
"input": "2 912219830404 955103",
"output": "955102\n1"
},
{
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"output": "-1"
},
{
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"output": "-1"
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{
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"output": "-1"
},
{
"input": "60 833021290059 912759",
"output": "912700\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1"
},
{
"input": "99999 715837929182 946070",
"output": "846072\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n..."
},
{
"input": "99998 714958284701 945549",
"output": "845552\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n..."
},
{
"input": "100000 730132752483 954477",
"output": "854478\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n..."
},
{
"input": "9999 850451005599 932197",
"output": "922199\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n..."
},
{
"input": "99999 741078317879 960857",
"output": "860859\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n..."
},
{
"input": "12313 873989408188 947186",
"output": "934874\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n..."
},
{
"input": "1 860113420929 927423",
"output": "927423"
},
{
"input": "2 933669982757 966267",
"output": "966266\n1"
},
{
"input": "3 933157932003 966003",
"output": "966001\n1\n1"
},
{
"input": "4 944626542564 971922",
"output": "971919\n1\n1\n1"
},
{
"input": "7 937519681542 968262",
"output": "968256\n1\n1\n1\n1\n1\n1"
},
{
"input": "100000 1000000000000 1000000",
"output": "-1"
},
{
"input": "99999 999999999999 999999",
"output": "-1"
},
{
"input": "100 1 1",
"output": "-1"
},
{
"input": "2 1 1",
"output": "-1"
},
{
"input": "11 10 10",
"output": "-1"
},
{
"input": "1 5 10",
"output": "10"
},
{
"input": "10 3 8",
"output": "-1"
},
{
"input": "5 37 10",
"output": "6\n1\n1\n1\n1"
},
{
"input": "5 1 4",
"output": "-1"
},
{
"input": "1 1 1",
"output": "1"
},
{
"input": "1 1000000000000 1",
"output": "-1"
},
{
"input": "1 1 1000000",
"output": "1000000"
},
{
"input": "100000 1 1",
"output": "-1"
},
{
"input": "100000 1 1000000",
"output": "900001\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n..."
},
{
"input": "100000 1000000000000 1",
"output": "-1"
},
{
"input": "1 1000000000000 1000000",
"output": "1000000"
}
] | 1,590,925,696
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 62
| 436
| 8,089,600
|
n,x,y=map(int,input().split())
import math
an=n-1
if(x<=n-1):
if(y>=n):
for i in range(n):
print(1)
else:
print(-1)
else:
me=math.sqrt(x-an)
if(me==int(me)):
me=int(me)
if(me+an<=y):
for i in range(n-1):
print(1)
print(me)
else:
print(-1)
else:
me=int(me)
me+=1
if(me+an<=y):
for i in range(n-1):
print(1)
print(me)
else:
print(-1)
|
Title: Petya and Inequiations
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Little Petya loves inequations. Help him find *n* positive integers *a*1,<=*a*2,<=...,<=*a**n*, such that the following two conditions are satisfied:
- *a*12<=+<=*a*22<=+<=...<=+<=*a**n*2<=β₯<=*x*- *a*1<=+<=*a*2<=+<=...<=+<=*a**n*<=β€<=*y*
Input Specification:
The first line contains three space-separated integers *n*, *x* and *y* (1<=β€<=*n*<=β€<=105,<=1<=β€<=*x*<=β€<=1012,<=1<=β€<=*y*<=β€<=106).
Please do not use the %lld specificator to read or write 64-bit integers in Π‘++. It is recommended to use cin, cout streams or the %I64d specificator.
Output Specification:
Print *n* positive integers that satisfy the conditions, one integer per line. If such numbers do not exist, print a single number "-1". If there are several solutions, print any of them.
Demo Input:
['5 15 15\n', '2 3 2\n', '1 99 11\n']
Demo Output:
['4\n4\n1\n1\n2\n', '-1\n', '11\n']
Note:
none
|
```python
n,x,y=map(int,input().split())
import math
an=n-1
if(x<=n-1):
if(y>=n):
for i in range(n):
print(1)
else:
print(-1)
else:
me=math.sqrt(x-an)
if(me==int(me)):
me=int(me)
if(me+an<=y):
for i in range(n-1):
print(1)
print(me)
else:
print(-1)
else:
me=int(me)
me+=1
if(me+an<=y):
for i in range(n-1):
print(1)
print(me)
else:
print(-1)
```
| 3.875932
|
157
|
B
|
Trace
|
PROGRAMMING
| 1,000
|
[
"geometry",
"sortings"
] | null | null |
One day, as Sherlock Holmes was tracking down one very important criminal, he found a wonderful painting on the wall. This wall could be represented as a plane. The painting had several concentric circles that divided the wall into several parts. Some parts were painted red and all the other were painted blue. Besides, any two neighboring parts were painted different colors, that is, the red and the blue color were alternating, i. e. followed one after the other. The outer area of the wall (the area that lied outside all circles) was painted blue. Help Sherlock Holmes determine the total area of red parts of the wall.
Let us remind you that two circles are called concentric if their centers coincide. Several circles are called concentric if any two of them are concentric.
|
The first line contains the single integer *n* (1<=β€<=*n*<=β€<=100). The second line contains *n* space-separated integers *r**i* (1<=β€<=*r**i*<=β€<=1000) β the circles' radii. It is guaranteed that all circles are different.
|
Print the single real number β total area of the part of the wall that is painted red. The answer is accepted if absolute or relative error doesn't exceed 10<=-<=4.
|
[
"1\n1\n",
"3\n1 4 2\n"
] |
[
"3.1415926536\n",
"40.8407044967\n"
] |
In the first sample the picture is just one circle of radius 1. Inner part of the circle is painted red. The area of the red part equals ΟβΓβ1<sup class="upper-index">2</sup>β=βΟ.
In the second sample there are three circles of radii 1, 4 and 2. Outside part of the second circle is painted blue. Part between the second and the third circles is painted red. Part between the first and the third is painted blue. And, finally, the inner part of the first circle is painted red. Overall there are two red parts: the ring between the second and the third circles and the inner part of the first circle. Total area of the red parts is equal (ΟβΓβ4<sup class="upper-index">2</sup>β-βΟβΓβ2<sup class="upper-index">2</sup>)β+βΟβΓβ1<sup class="upper-index">2</sup>β=βΟβΓβ12β+βΟβ=β13Ο
| 1,000
|
[
{
"input": "1\n1",
"output": "3.1415926536"
},
{
"input": "3\n1 4 2",
"output": "40.8407044967"
},
{
"input": "4\n4 1 3 2",
"output": "31.4159265359"
},
{
"input": "4\n100 10 2 1",
"output": "31111.1920484997"
},
{
"input": "10\n10 9 8 7 6 5 4 3 2 1",
"output": "172.7875959474"
},
{
"input": "1\n1000",
"output": "3141592.6535897931"
},
{
"input": "8\n8 1 7 2 6 3 5 4",
"output": "113.0973355292"
},
{
"input": "100\n1000 999 998 997 996 995 994 993 992 991 990 989 988 987 986 985 984 983 982 981 980 979 978 977 976 975 974 973 972 971 970 969 968 967 966 965 964 963 962 961 960 959 958 957 956 955 954 953 952 951 950 949 948 947 946 945 944 943 942 941 940 939 938 937 936 935 934 933 932 931 930 929 928 927 926 925 924 923 922 921 920 919 918 917 916 915 914 913 912 911 910 909 908 907 906 905 904 903 902 901",
"output": "298608.3817237098"
},
{
"input": "6\n109 683 214 392 678 10",
"output": "397266.9574170437"
},
{
"input": "2\n151 400",
"output": "431023.3704798660"
},
{
"input": "6\n258 877 696 425 663 934",
"output": "823521.3902487604"
},
{
"input": "9\n635 707 108 234 52 180 910 203 782",
"output": "1100144.9065826489"
},
{
"input": "8\n885 879 891 428 522 176 135 983",
"output": "895488.9947571954"
},
{
"input": "3\n269 918 721",
"output": "1241695.6467754442"
},
{
"input": "7\n920 570 681 428 866 935 795",
"output": "1469640.1849419588"
},
{
"input": "2\n517 331",
"output": "495517.1260654109"
},
{
"input": "2\n457 898",
"output": "1877274.3981158488"
},
{
"input": "8\n872 704 973 612 183 274 739 253",
"output": "1780774.0965755312"
},
{
"input": "74\n652 446 173 457 760 847 670 25 196 775 998 279 656 809 883 148 969 884 792 502 641 800 663 938 362 339 545 608 107 184 834 666 149 458 864 72 199 658 618 987 126 723 806 643 689 958 626 904 944 415 427 498 628 331 636 261 281 276 478 220 513 595 510 384 354 561 469 462 799 449 747 109 903 456",
"output": "1510006.5089479341"
},
{
"input": "76\n986 504 673 158 87 332 124 218 714 235 212 122 878 370 938 81 686 323 386 348 410 468 875 107 50 960 82 834 234 663 651 422 794 633 294 771 945 607 146 913 950 858 297 88 882 725 247 872 645 749 799 987 115 394 380 382 971 429 593 426 652 353 351 233 868 598 889 116 71 376 916 464 414 976 138 903",
"output": "1528494.7817143100"
},
{
"input": "70\n12 347 748 962 514 686 192 159 990 4 10 788 602 542 946 215 523 727 799 717 955 796 529 465 897 103 181 515 495 153 710 179 747 145 16 585 943 998 923 708 156 399 770 547 775 285 9 68 713 722 570 143 913 416 663 624 925 218 64 237 797 138 942 213 188 818 780 840 480 758",
"output": "1741821.4892636713"
},
{
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"output": "1818821.9252031571"
},
{
"input": "52\n270 658 808 249 293 707 700 78 791 167 92 772 807 502 830 991 945 102 968 376 556 578 326 980 688 368 280 853 646 256 666 638 424 737 321 996 925 405 199 680 953 541 716 481 727 143 577 919 892 355 346 298",
"output": "1272941.9273080483"
},
{
"input": "77\n482 532 200 748 692 697 171 863 586 547 301 149 326 812 147 698 303 691 527 805 681 387 619 947 598 453 167 799 840 508 893 688 643 974 998 341 804 230 538 669 271 404 477 759 943 596 949 235 880 160 151 660 832 82 969 539 708 889 258 81 224 655 790 144 462 582 646 256 445 52 456 920 67 819 631 484 534",
"output": "2045673.1891262225"
},
{
"input": "27\n167 464 924 575 775 97 944 390 297 315 668 296 533 829 851 406 702 366 848 512 71 197 321 900 544 529 116",
"output": "1573959.9105970615"
},
{
"input": "38\n488 830 887 566 720 267 583 102 65 200 884 220 263 858 510 481 316 804 754 568 412 166 374 869 356 977 145 421 500 58 664 252 745 70 381 927 670 772",
"output": "1479184.3434235646"
},
{
"input": "64\n591 387 732 260 840 397 563 136 571 876 831 953 799 493 579 13 559 872 53 678 256 232 969 993 847 14 837 365 547 997 604 199 834 529 306 443 739 49 19 276 343 835 904 588 900 870 439 576 975 955 518 117 131 347 800 83 432 882 869 709 32 950 314 450",
"output": "1258248.6984672088"
},
{
"input": "37\n280 281 169 68 249 389 977 101 360 43 448 447 368 496 125 507 747 392 338 270 916 150 929 428 118 266 589 470 774 852 263 644 187 817 808 58 637",
"output": "1495219.0323274869"
},
{
"input": "97\n768 569 306 968 437 779 227 561 412 60 44 807 234 645 169 858 580 396 343 145 842 723 416 80 456 247 81 150 297 116 760 964 312 558 101 850 549 650 299 868 121 435 579 705 118 424 302 812 970 397 659 565 916 183 933 459 6 593 518 717 326 305 744 470 75 981 824 221 294 324 194 293 251 446 481 215 338 861 528 829 921 945 540 89 450 178 24 460 990 392 148 219 934 615 932 340 937",
"output": "1577239.7333274092"
},
{
"input": "94\n145 703 874 425 277 652 239 496 458 658 339 842 564 699 893 352 625 980 432 121 798 872 499 859 850 721 414 825 543 843 304 111 342 45 219 311 50 748 465 902 781 822 504 985 919 656 280 310 917 438 464 527 491 713 906 329 635 777 223 810 501 535 156 252 806 112 971 719 103 443 165 98 579 554 244 996 221 560 301 51 977 422 314 858 528 772 448 626 185 194 536 66 577 677",
"output": "1624269.3753516484"
},
{
"input": "97\n976 166 649 81 611 927 480 231 998 711 874 91 969 521 531 414 993 790 317 981 9 261 437 332 173 573 904 777 882 990 658 878 965 64 870 896 271 732 431 53 761 943 418 602 708 949 930 130 512 240 363 458 673 319 131 784 224 48 919 126 208 212 911 59 677 535 450 273 479 423 79 807 336 18 72 290 724 28 123 605 287 228 350 897 250 392 885 655 746 417 643 114 813 378 355 635 905",
"output": "1615601.7212203942"
},
{
"input": "91\n493 996 842 9 748 178 1 807 841 519 796 998 84 670 778 143 707 208 165 893 154 943 336 150 761 881 434 112 833 55 412 682 552 945 758 189 209 600 354 325 440 844 410 20 136 665 88 791 688 17 539 821 133 236 94 606 483 446 429 60 960 476 915 134 137 852 754 908 276 482 117 252 297 903 981 203 829 811 471 135 188 667 710 393 370 302 874 872 551 457 692",
"output": "1806742.5014501044"
},
{
"input": "95\n936 736 17 967 229 607 589 291 242 244 29 698 800 566 630 667 90 416 11 94 812 838 668 520 678 111 490 823 199 973 681 676 683 721 262 896 682 713 402 691 874 44 95 704 56 322 822 887 639 433 406 35 988 61 176 496 501 947 440 384 372 959 577 370 754 802 1 945 427 116 746 408 308 391 397 730 493 183 203 871 831 862 461 565 310 344 504 378 785 137 279 123 475 138 415",
"output": "1611115.5269110680"
},
{
"input": "90\n643 197 42 218 582 27 66 704 195 445 641 675 285 639 503 686 242 327 57 955 848 287 819 992 756 749 363 48 648 736 580 117 752 921 923 372 114 313 202 337 64 497 399 25 883 331 24 871 917 8 517 486 323 529 325 92 891 406 864 402 263 773 931 253 625 31 17 271 140 131 232 586 893 525 846 54 294 562 600 801 214 55 768 683 389 738 314 284 328 804",
"output": "1569819.2914796301"
},
{
"input": "98\n29 211 984 75 333 96 840 21 352 168 332 433 130 944 215 210 620 442 363 877 91 491 513 955 53 82 351 19 998 706 702 738 770 453 344 117 893 590 723 662 757 16 87 546 312 669 568 931 224 374 927 225 751 962 651 587 361 250 256 240 282 600 95 64 384 589 813 783 39 918 412 648 506 283 886 926 443 173 946 241 310 33 622 565 261 360 547 339 943 367 354 25 479 743 385 485 896 741",
"output": "2042921.1539616778"
},
{
"input": "93\n957 395 826 67 185 4 455 880 683 654 463 84 258 878 553 592 124 585 9 133 20 609 43 452 725 125 801 537 700 685 771 155 566 376 19 690 383 352 174 208 177 416 304 1000 533 481 87 509 358 233 681 22 507 659 36 859 952 259 138 271 594 779 576 782 119 69 608 758 283 616 640 523 710 751 34 106 774 92 874 568 864 660 998 992 474 679 180 409 15 297 990 689 501",
"output": "1310703.8710041976"
},
{
"input": "97\n70 611 20 30 904 636 583 262 255 501 604 660 212 128 199 138 545 576 506 528 12 410 77 888 783 972 431 188 338 485 148 793 907 678 281 922 976 680 252 724 253 920 177 361 721 798 960 572 99 622 712 466 608 49 612 345 266 751 63 594 40 695 532 789 520 930 825 929 48 59 405 135 109 735 508 186 495 772 375 587 201 324 447 610 230 947 855 318 856 956 313 810 931 175 668 183 688",
"output": "1686117.9099228707"
},
{
"input": "96\n292 235 391 180 840 172 218 997 166 287 329 20 886 325 400 471 182 356 448 337 417 319 58 106 366 764 393 614 90 831 924 314 667 532 64 874 3 434 350 352 733 795 78 640 967 63 47 879 635 272 145 569 468 792 153 761 770 878 281 467 209 208 298 37 700 18 334 93 5 750 412 779 523 517 360 649 447 328 311 653 57 578 767 460 647 663 50 670 151 13 511 580 625 907 227 89",
"output": "1419726.5608617242"
},
{
"input": "100\n469 399 735 925 62 153 707 723 819 529 200 624 57 708 245 384 889 11 639 638 260 419 8 142 403 298 204 169 887 388 241 983 885 267 643 943 417 237 452 562 6 839 149 742 832 896 100 831 712 754 679 743 135 222 445 680 210 955 220 63 960 487 514 824 481 584 441 997 795 290 10 45 510 678 844 503 407 945 850 84 858 934 500 320 936 663 736 592 161 670 606 465 864 969 293 863 868 393 899 744",
"output": "1556458.0979239127"
},
{
"input": "100\n321 200 758 415 190 710 920 992 873 898 814 259 359 66 971 210 838 545 663 652 684 277 36 756 963 459 335 484 462 982 532 423 131 703 307 229 391 938 253 847 542 975 635 928 220 980 222 567 557 181 366 824 900 180 107 979 112 564 525 413 300 422 876 615 737 343 902 8 654 628 469 913 967 785 893 314 909 215 912 262 20 709 363 915 997 954 986 454 596 124 74 159 660 550 787 418 895 786 293 50",
"output": "1775109.8050211088"
},
{
"input": "100\n859 113 290 762 701 63 188 431 810 485 671 673 99 658 194 227 511 435 941 212 551 124 89 222 42 321 657 815 898 171 216 482 707 567 724 491 414 942 820 351 48 653 685 312 586 24 20 627 602 498 533 173 463 262 621 466 119 299 580 964 510 987 40 698 521 998 847 651 746 215 808 563 785 837 631 772 404 923 682 244 232 214 390 350 968 771 517 900 70 543 934 554 681 368 642 575 891 728 478 317",
"output": "1447969.4788174964"
},
{
"input": "100\n941 283 349 457 52 837 299 284 796 305 893 624 101 972 738 204 121 70 17 704 836 791 95 111 162 952 472 724 733 580 878 177 705 804 11 211 463 417 288 409 410 485 896 755 921 267 164 656 505 765 539 439 535 19 991 689 220 474 114 944 884 144 926 849 486 566 117 35 749 499 797 303 362 905 690 890 976 66 590 183 234 683 39 297 769 787 376 541 571 759 495 200 261 352 73 493 831 442 273 339",
"output": "1597889.4218394549"
},
{
"input": "100\n110 868 147 888 291 282 916 542 917 337 235 595 498 621 814 249 261 713 445 666 142 275 319 49 900 543 333 606 487 670 620 769 852 258 230 867 603 491 714 839 879 37 21 74 590 190 397 506 499 967 152 862 200 107 145 23 614 167 857 475 79 598 270 906 626 116 89 59 651 846 723 201 50 699 522 408 634 132 226 414 824 764 513 586 526 238 556 162 667 173 684 415 527 743 10 442 552 274 350 822",
"output": "1567230.6191330721"
},
{
"input": "100\n901 902 903 904 905 906 907 908 909 910 911 912 913 914 915 916 917 918 919 920 921 922 923 924 925 926 927 928 929 930 931 932 933 934 935 936 937 938 939 940 941 942 943 944 945 946 947 948 949 950 951 952 953 954 955 956 957 958 959 960 961 962 963 964 965 966 967 968 969 970 971 972 973 974 975 976 977 978 979 980 981 982 983 984 985 986 987 988 989 990 991 992 993 994 995 996 997 998 999 1000",
"output": "298608.3817237098"
}
] | 1,683,785,948
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 44
| 92
| 0
|
# Description of the problem can be found at http://codeforces.com/problemset/problem/157/B
import math
n = int(input())
l_s = list(map(int, input().split()))
l_s.sort(reverse = True)
t = 0
for index in range(n):
t += (-1 if index % 2 == 1 else 1) * l_s[index] ** 2
print(t * math.pi)
|
Title: Trace
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day, as Sherlock Holmes was tracking down one very important criminal, he found a wonderful painting on the wall. This wall could be represented as a plane. The painting had several concentric circles that divided the wall into several parts. Some parts were painted red and all the other were painted blue. Besides, any two neighboring parts were painted different colors, that is, the red and the blue color were alternating, i. e. followed one after the other. The outer area of the wall (the area that lied outside all circles) was painted blue. Help Sherlock Holmes determine the total area of red parts of the wall.
Let us remind you that two circles are called concentric if their centers coincide. Several circles are called concentric if any two of them are concentric.
Input Specification:
The first line contains the single integer *n* (1<=β€<=*n*<=β€<=100). The second line contains *n* space-separated integers *r**i* (1<=β€<=*r**i*<=β€<=1000) β the circles' radii. It is guaranteed that all circles are different.
Output Specification:
Print the single real number β total area of the part of the wall that is painted red. The answer is accepted if absolute or relative error doesn't exceed 10<=-<=4.
Demo Input:
['1\n1\n', '3\n1 4 2\n']
Demo Output:
['3.1415926536\n', '40.8407044967\n']
Note:
In the first sample the picture is just one circle of radius 1. Inner part of the circle is painted red. The area of the red part equals ΟβΓβ1<sup class="upper-index">2</sup>β=βΟ.
In the second sample there are three circles of radii 1, 4 and 2. Outside part of the second circle is painted blue. Part between the second and the third circles is painted red. Part between the first and the third is painted blue. And, finally, the inner part of the first circle is painted red. Overall there are two red parts: the ring between the second and the third circles and the inner part of the first circle. Total area of the red parts is equal (ΟβΓβ4<sup class="upper-index">2</sup>β-βΟβΓβ2<sup class="upper-index">2</sup>)β+βΟβΓβ1<sup class="upper-index">2</sup>β=βΟβΓβ12β+βΟβ=β13Ο
|
```python
# Description of the problem can be found at http://codeforces.com/problemset/problem/157/B
import math
n = int(input())
l_s = list(map(int, input().split()))
l_s.sort(reverse = True)
t = 0
for index in range(n):
t += (-1 if index % 2 == 1 else 1) * l_s[index] ** 2
print(t * math.pi)
```
| 3
|
|
863
|
A
|
Quasi-palindrome
|
PROGRAMMING
| 900
|
[
"brute force",
"implementation"
] | null | null |
Let quasi-palindromic number be such number that adding some leading zeros (possible none) to it produces a palindromic string.
String *t* is called a palindrome, if it reads the same from left to right and from right to left.
For example, numbers 131 and 2010200 are quasi-palindromic, they can be transformed to strings "131" and "002010200", respectively, which are palindromes.
You are given some integer number *x*. Check if it's a quasi-palindromic number.
|
The first line contains one integer number *x* (1<=β€<=*x*<=β€<=109). This number is given without any leading zeroes.
|
Print "YES" if number *x* is quasi-palindromic. Otherwise, print "NO" (without quotes).
|
[
"131\n",
"320\n",
"2010200\n"
] |
[
"YES\n",
"NO\n",
"YES\n"
] |
none
| 0
|
[
{
"input": "131",
"output": "YES"
},
{
"input": "320",
"output": "NO"
},
{
"input": "2010200",
"output": "YES"
},
{
"input": "1",
"output": "YES"
},
{
"input": "1000000000",
"output": "YES"
},
{
"input": "999999999",
"output": "YES"
},
{
"input": "999999998",
"output": "NO"
},
{
"input": "102000",
"output": "NO"
},
{
"input": "210000000",
"output": "NO"
},
{
"input": "213443120",
"output": "YES"
},
{
"input": "99",
"output": "YES"
},
{
"input": "22002",
"output": "NO"
},
{
"input": "1010",
"output": "YES"
},
{
"input": "1201",
"output": "NO"
},
{
"input": "6460046",
"output": "NO"
},
{
"input": "503435",
"output": "NO"
},
{
"input": "21002",
"output": "NO"
},
{
"input": "101001",
"output": "NO"
},
{
"input": "200102",
"output": "NO"
},
{
"input": "20010002",
"output": "NO"
},
{
"input": "33003",
"output": "NO"
},
{
"input": "100101",
"output": "NO"
},
{
"input": "1021",
"output": "NO"
},
{
"input": "1101",
"output": "NO"
},
{
"input": "10101100",
"output": "NO"
},
{
"input": "101",
"output": "YES"
},
{
"input": "1011",
"output": "NO"
},
{
"input": "11010",
"output": "NO"
},
{
"input": "10110",
"output": "NO"
},
{
"input": "110000",
"output": "YES"
},
{
"input": "2011",
"output": "NO"
},
{
"input": "10020001",
"output": "NO"
},
{
"input": "12505021",
"output": "NO"
},
{
"input": "12310",
"output": "NO"
},
{
"input": "100501",
"output": "NO"
},
{
"input": "11001",
"output": "NO"
},
{
"input": "20020002",
"output": "NO"
},
{
"input": "202002",
"output": "NO"
},
{
"input": "1001",
"output": "YES"
},
{
"input": "1020021",
"output": "NO"
},
{
"input": "60660",
"output": "NO"
},
{
"input": "98809",
"output": "NO"
},
{
"input": "11000000",
"output": "YES"
},
{
"input": "807008",
"output": "NO"
}
] | 1,691,031,915
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 44
| 62
| 0
|
x=list(input())
x.reverse()
g=False
while len(x)>1:
if x[0]=='0' and g==False:
x.pop(0)
else:
g=True
if x[0]!=x[-1]:
print('NO')
exit()
else:
x.pop(0)
x.pop()
print("YES")
|
Title: Quasi-palindrome
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Let quasi-palindromic number be such number that adding some leading zeros (possible none) to it produces a palindromic string.
String *t* is called a palindrome, if it reads the same from left to right and from right to left.
For example, numbers 131 and 2010200 are quasi-palindromic, they can be transformed to strings "131" and "002010200", respectively, which are palindromes.
You are given some integer number *x*. Check if it's a quasi-palindromic number.
Input Specification:
The first line contains one integer number *x* (1<=β€<=*x*<=β€<=109). This number is given without any leading zeroes.
Output Specification:
Print "YES" if number *x* is quasi-palindromic. Otherwise, print "NO" (without quotes).
Demo Input:
['131\n', '320\n', '2010200\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n']
Note:
none
|
```python
x=list(input())
x.reverse()
g=False
while len(x)>1:
if x[0]=='0' and g==False:
x.pop(0)
else:
g=True
if x[0]!=x[-1]:
print('NO')
exit()
else:
x.pop(0)
x.pop()
print("YES")
```
| 3
|
|
938
|
A
|
Word Correction
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Victor tries to write his own text editor, with word correction included. However, the rules of word correction are really strange.
Victor thinks that if a word contains two consecutive vowels, then it's kinda weird and it needs to be replaced. So the word corrector works in such a way: as long as there are two consecutive vowels in the word, it deletes the first vowel in a word such that there is another vowel right before it. If there are no two consecutive vowels in the word, it is considered to be correct.
You are given a word *s*. Can you predict what will it become after correction?
In this problem letters a, e, i, o, u and y are considered to be vowels.
|
The first line contains one integer *n* (1<=β€<=*n*<=β€<=100) β the number of letters in word *s* before the correction.
The second line contains a string *s* consisting of exactly *n* lowercase Latin letters β the word before the correction.
|
Output the word *s* after the correction.
|
[
"5\nweird\n",
"4\nword\n",
"5\naaeaa\n"
] |
[
"werd\n",
"word\n",
"a\n"
] |
Explanations of the examples:
1. There is only one replace: weird <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> werd;1. No replace needed since there are no two consecutive vowels;1. aaeaa <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> aeaa <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> aaa <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> aa <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> a.
| 0
|
[
{
"input": "5\nweird",
"output": "werd"
},
{
"input": "4\nword",
"output": "word"
},
{
"input": "5\naaeaa",
"output": "a"
},
{
"input": "100\naaaaabbbbboyoyoyoyoyacadabbbbbiuiufgiuiuaahjabbbklboyoyoyoyoyaaaaabbbbbiuiuiuiuiuaaaaabbbbbeyiyuyzyw",
"output": "abbbbbocadabbbbbifgihjabbbklbobbbbbibbbbbezyw"
},
{
"input": "69\nbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb",
"output": "bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb"
},
{
"input": "12\nmmmmmmmmmmmm",
"output": "mmmmmmmmmmmm"
},
{
"input": "18\nyaywptqwuyiqypwoyw",
"output": "ywptqwuqypwow"
},
{
"input": "85\nbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb",
"output": "bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb"
},
{
"input": "13\nmmmmmmmmmmmmm",
"output": "mmmmmmmmmmmmm"
},
{
"input": "10\nmmmmmmmmmm",
"output": "mmmmmmmmmm"
},
{
"input": "11\nmmmmmmmmmmm",
"output": "mmmmmmmmmmm"
},
{
"input": "15\nmmmmmmmmmmmmmmm",
"output": "mmmmmmmmmmmmmmm"
},
{
"input": "1\na",
"output": "a"
},
{
"input": "14\nmmmmmmmmmmmmmm",
"output": "mmmmmmmmmmmmmm"
},
{
"input": "33\nmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm",
"output": "mmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm"
},
{
"input": "79\nbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb",
"output": "bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb"
},
{
"input": "90\nbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb",
"output": "bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb"
},
{
"input": "2\naa",
"output": "a"
},
{
"input": "18\niuiuqpyyaoaetiwliu",
"output": "iqpytiwli"
},
{
"input": "5\nxxxxx",
"output": "xxxxx"
},
{
"input": "6\nxxxahg",
"output": "xxxahg"
},
{
"input": "3\nzcv",
"output": "zcv"
},
{
"input": "4\naepo",
"output": "apo"
},
{
"input": "5\nqqqqq",
"output": "qqqqq"
},
{
"input": "6\naaaaaa",
"output": "a"
},
{
"input": "4\naeta",
"output": "ata"
},
{
"input": "20\nttyttlwaoieulyiluuri",
"output": "ttyttlwalyluri"
},
{
"input": "1\nb",
"output": "b"
},
{
"input": "3\nanc",
"output": "anc"
},
{
"input": "1\ne",
"output": "e"
},
{
"input": "3\naie",
"output": "a"
},
{
"input": "3\nvio",
"output": "vi"
},
{
"input": "2\nea",
"output": "e"
},
{
"input": "3\nuas",
"output": "us"
},
{
"input": "2\nba",
"output": "ba"
},
{
"input": "2\naq",
"output": "aq"
},
{
"input": "2\nya",
"output": "y"
},
{
"input": "2\nou",
"output": "o"
},
{
"input": "2\nbb",
"output": "bb"
},
{
"input": "7\nayylmao",
"output": "alma"
},
{
"input": "2\nab",
"output": "ab"
},
{
"input": "19\nyuouiyaoiiweqrryqqp",
"output": "yweqrryqqp"
},
{
"input": "25\niqypwqpriiioetiuqqqttouei",
"output": "iqypwqpritiqqqtto"
},
{
"input": "100\naaaaabbbbboyoyoyoyoyacadabbbbbiuiufgiuiuaahjabbbklboyoyoyoyoyaaaaabbbbbiuiuiuiuiuaaaaabbbbbeyiyuyzyz",
"output": "abbbbbocadabbbbbifgihjabbbklbobbbbbibbbbbezyz"
},
{
"input": "17\naccccccccccccccca",
"output": "accccccccccccccca"
},
{
"input": "5\nababa",
"output": "ababa"
},
{
"input": "10\naaaaaaaaaa",
"output": "a"
},
{
"input": "22\naaaaabbbbboyoyoyoyoyac",
"output": "abbbbboc"
},
{
"input": "7\nmahmoud",
"output": "mahmod"
}
] | 1,609,959,402
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 49
| 140
| 0
|
vowels = ['a', 'e', 'i', 'o', 'u', 'y']
def word_correction(s):
result = []
result.append(s[0])
left = 0
right = 1
while right < len(s):
if not s[left] in vowels or not s[right] in vowels:
result.append(s[right])
left = right
right += 1
return ''.join(result)
n = int(input())
s = input()
print(word_correction(s))
|
Title: Word Correction
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Victor tries to write his own text editor, with word correction included. However, the rules of word correction are really strange.
Victor thinks that if a word contains two consecutive vowels, then it's kinda weird and it needs to be replaced. So the word corrector works in such a way: as long as there are two consecutive vowels in the word, it deletes the first vowel in a word such that there is another vowel right before it. If there are no two consecutive vowels in the word, it is considered to be correct.
You are given a word *s*. Can you predict what will it become after correction?
In this problem letters a, e, i, o, u and y are considered to be vowels.
Input Specification:
The first line contains one integer *n* (1<=β€<=*n*<=β€<=100) β the number of letters in word *s* before the correction.
The second line contains a string *s* consisting of exactly *n* lowercase Latin letters β the word before the correction.
Output Specification:
Output the word *s* after the correction.
Demo Input:
['5\nweird\n', '4\nword\n', '5\naaeaa\n']
Demo Output:
['werd\n', 'word\n', 'a\n']
Note:
Explanations of the examples:
1. There is only one replace: weird <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> werd;1. No replace needed since there are no two consecutive vowels;1. aaeaa <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> aeaa <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> aaa <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> aa <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> a.
|
```python
vowels = ['a', 'e', 'i', 'o', 'u', 'y']
def word_correction(s):
result = []
result.append(s[0])
left = 0
right = 1
while right < len(s):
if not s[left] in vowels or not s[right] in vowels:
result.append(s[right])
left = right
right += 1
return ''.join(result)
n = int(input())
s = input()
print(word_correction(s))
```
| 3
|
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