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381
A
Sereja and Dima
PROGRAMMING
800
[ "greedy", "implementation", "two pointers" ]
null
null
Sereja and Dima play a game. The rules of the game are very simple. The players have *n* cards in a row. Each card contains a number, all numbers on the cards are distinct. The players take turns, Sereja moves first. During his turn a player can take one card: either the leftmost card in a row, or the rightmost one. The game ends when there is no more cards. The player who has the maximum sum of numbers on his cards by the end of the game, wins. Sereja and Dima are being greedy. Each of them chooses the card with the larger number during his move. Inna is a friend of Sereja and Dima. She knows which strategy the guys are using, so she wants to determine the final score, given the initial state of the game. Help her.
The first line contains integer *n* (1<=≤<=*n*<=≤<=1000) — the number of cards on the table. The second line contains space-separated numbers on the cards from left to right. The numbers on the cards are distinct integers from 1 to 1000.
On a single line, print two integers. The first number is the number of Sereja's points at the end of the game, the second number is the number of Dima's points at the end of the game.
[ "4\n4 1 2 10\n", "7\n1 2 3 4 5 6 7\n" ]
[ "12 5\n", "16 12\n" ]
In the first sample Sereja will take cards with numbers 10 and 2, so Sereja's sum is 12. Dima will take cards with numbers 4 and 1, so Dima's sum is 5.
500
[ { "input": "4\n4 1 2 10", "output": "12 5" }, { "input": "7\n1 2 3 4 5 6 7", "output": "16 12" }, { "input": "42\n15 29 37 22 16 5 26 31 6 32 19 3 45 36 33 14 25 20 48 7 42 11 24 28 9 18 8 21 47 17 38 40 44 4 35 1 43 39 41 27 12 13", "output": "613 418" }, { "input": "43\n32 1 15 48 38 26 25 14 20 44 11 30 3 42 49 19 18 46 5 45 10 23 34 9 29 41 2 52 6 17 35 4 50 22 33 51 7 28 47 13 39 37 24", "output": "644 500" }, { "input": "1\n3", "output": "3 0" }, { "input": "45\n553 40 94 225 415 471 126 190 647 394 515 303 189 159 308 6 139 132 326 78 455 75 85 295 135 613 360 614 351 228 578 259 258 591 444 29 33 463 561 174 368 183 140 168 646", "output": "6848 6568" }, { "input": "44\n849 373 112 307 479 608 856 769 526 82 168 143 573 762 115 501 688 36 214 450 396 496 236 309 287 786 397 43 811 141 745 846 350 270 276 677 420 459 403 722 267 54 394 727", "output": "9562 9561" }, { "input": "35\n10 15 18 1 28 16 2 33 6 22 23 4 9 25 35 8 7 26 3 20 30 14 31 19 27 32 11 5 29 24 21 34 13 17 12", "output": "315 315" }, { "input": "17\n580 376 191 496 73 44 520 357 483 149 81 178 514 300 216 598 304", "output": "3238 2222" }, { "input": "30\n334 443 223 424 168 549 189 303 429 559 516 220 459 134 344 346 316 446 209 148 487 526 69 286 102 366 518 280 392 325", "output": "5246 4864" }, { "input": "95\n122 29 188 265 292 287 183 225 222 187 155 256 64 148 173 278 218 136 290 17 31 130 2 87 57 283 255 280 68 166 174 142 102 39 116 206 288 154 26 78 296 172 184 232 77 91 277 8 249 186 94 93 207 251 257 195 101 299 193 124 293 65 58 35 24 302 220 189 252 125 27 284 247 182 141 103 198 97 234 83 281 216 85 180 267 236 109 143 149 239 79 300 191 244 71", "output": "8147 7807" }, { "input": "1\n1", "output": "1 0" } ]
1,675,747,396
2,147,483,647
PyPy 3
COMPILATION_ERROR
TESTS
0
0
0
#include <bits/stdc++.h> using namespace std; #define TURBO ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0); #define ar array #define ll long long #define ld long double #define sza(x) ((int)x.size()) #define all(a) (a).begin(), (a).end() void solve(){ int n; cin >> n; vector<int>v(n); int s = 0, d = 0; for(int i = 0; i < n; i++) cin >> v[i]; sort(v.begin(), v.end(), greater<int>()); for(int i = 0; i < n; i+=2) s += v[i]; for(int i = 1; i < n; i+=2) d += v[i]; cout << s << " " << d << "\n"; return; } int main() { // your code goes here TURBO; ll testcase = 1; // cin >> testcase; for(int tc = 1; tc <= testcase; tc++) { // cout << "Case #" << tc << ": "; solve(); } return 0; }
Title: Sereja and Dima Time Limit: None seconds Memory Limit: None megabytes Problem Description: Sereja and Dima play a game. The rules of the game are very simple. The players have *n* cards in a row. Each card contains a number, all numbers on the cards are distinct. The players take turns, Sereja moves first. During his turn a player can take one card: either the leftmost card in a row, or the rightmost one. The game ends when there is no more cards. The player who has the maximum sum of numbers on his cards by the end of the game, wins. Sereja and Dima are being greedy. Each of them chooses the card with the larger number during his move. Inna is a friend of Sereja and Dima. She knows which strategy the guys are using, so she wants to determine the final score, given the initial state of the game. Help her. Input Specification: The first line contains integer *n* (1<=≤<=*n*<=≤<=1000) — the number of cards on the table. The second line contains space-separated numbers on the cards from left to right. The numbers on the cards are distinct integers from 1 to 1000. Output Specification: On a single line, print two integers. The first number is the number of Sereja's points at the end of the game, the second number is the number of Dima's points at the end of the game. Demo Input: ['4\n4 1 2 10\n', '7\n1 2 3 4 5 6 7\n'] Demo Output: ['12 5\n', '16 12\n'] Note: In the first sample Sereja will take cards with numbers 10 and 2, so Sereja's sum is 12. Dima will take cards with numbers 4 and 1, so Dima's sum is 5.
```python #include <bits/stdc++.h> using namespace std; #define TURBO ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0); #define ar array #define ll long long #define ld long double #define sza(x) ((int)x.size()) #define all(a) (a).begin(), (a).end() void solve(){ int n; cin >> n; vector<int>v(n); int s = 0, d = 0; for(int i = 0; i < n; i++) cin >> v[i]; sort(v.begin(), v.end(), greater<int>()); for(int i = 0; i < n; i+=2) s += v[i]; for(int i = 1; i < n; i+=2) d += v[i]; cout << s << " " << d << "\n"; return; } int main() { // your code goes here TURBO; ll testcase = 1; // cin >> testcase; for(int tc = 1; tc <= testcase; tc++) { // cout << "Case #" << tc << ": "; solve(); } return 0; } ```
-1
413
A
Data Recovery
PROGRAMMING
1,200
[ "implementation" ]
null
null
Not so long ago company R2 bought company R1 and consequently, all its developments in the field of multicore processors. Now the R2 laboratory is testing one of the R1 processors. The testing goes in *n* steps, at each step the processor gets some instructions, and then its temperature is measured. The head engineer in R2 is keeping a report record on the work of the processor: he writes down the minimum and the maximum measured temperature in his notebook. His assistant had to write down all temperatures into his notebook, but (for unknown reasons) he recorded only *m*. The next day, the engineer's assistant filed in a report with all the *m* temperatures. However, the chief engineer doubts that the assistant wrote down everything correctly (naturally, the chief engineer doesn't doubt his notes). So he asked you to help him. Given numbers *n*, *m*, *min*, *max* and the list of *m* temperatures determine whether you can upgrade the set of *m* temperatures to the set of *n* temperatures (that is add *n*<=-<=*m* temperatures), so that the minimum temperature was *min* and the maximum one was *max*.
The first line contains four integers *n*,<=*m*,<=*min*,<=*max* (1<=≤<=*m*<=&lt;<=*n*<=≤<=100; 1<=≤<=*min*<=&lt;<=*max*<=≤<=100). The second line contains *m* space-separated integers *t**i* (1<=≤<=*t**i*<=≤<=100) — the temperatures reported by the assistant. Note, that the reported temperatures, and the temperatures you want to add can contain equal temperatures.
If the data is consistent, print 'Correct' (without the quotes). Otherwise, print 'Incorrect' (without the quotes).
[ "2 1 1 2\n1\n", "3 1 1 3\n2\n", "2 1 1 3\n2\n" ]
[ "Correct\n", "Correct\n", "Incorrect\n" ]
In the first test sample one of the possible initial configurations of temperatures is [1, 2]. In the second test sample one of the possible initial configurations of temperatures is [2, 1, 3]. In the third test sample it is impossible to add one temperature to obtain the minimum equal to 1 and the maximum equal to 3.
500
[ { "input": "2 1 1 2\n1", "output": "Correct" }, { "input": "3 1 1 3\n2", "output": "Correct" }, { "input": "2 1 1 3\n2", "output": "Incorrect" }, { "input": "3 1 1 5\n3", "output": "Correct" }, { "input": "3 2 1 5\n1 5", "output": "Correct" }, { "input": "3 2 1 5\n1 1", "output": "Correct" }, { "input": "3 2 1 5\n5 5", "output": "Correct" }, { "input": "3 2 1 5\n1 6", "output": "Incorrect" }, { "input": "3 2 5 10\n1 10", "output": "Incorrect" }, { "input": "6 5 3 6\n4 4 4 4 4", "output": "Incorrect" }, { "input": "100 50 68 97\n20 42 93 1 98 6 32 11 48 46 82 96 24 73 40 100 99 10 55 87 65 80 97 54 59 48 30 22 16 92 66 2 22 60 23 81 64 60 34 60 99 99 4 70 91 99 30 20 41 96", "output": "Incorrect" }, { "input": "100 50 1 2\n1 1 2 1 1 2 2 1 1 1 1 1 2 2 1 2 1 2 2 1 1 1 2 2 2 1 1 2 1 1 1 1 2 2 1 1 1 1 1 2 1 1 1 2 1 2 2 2 1 2", "output": "Correct" }, { "input": "100 99 1 2\n2 1 1 1 2 2 1 1 1 2 2 2 1 2 1 1 2 1 1 2 1 2 2 1 2 1 2 1 2 1 2 2 2 2 1 1 1 1 1 2 1 2 2 1 2 2 2 1 1 1 1 1 2 2 2 2 1 2 2 1 1 1 2 1 1 2 1 1 2 1 2 1 2 1 1 1 1 2 1 1 1 1 1 2 2 2 1 1 1 1 2 2 2 2 1 1 2 2 2", "output": "Correct" }, { "input": "3 2 2 100\n40 1", "output": "Incorrect" }, { "input": "3 2 2 3\n4 4", "output": "Incorrect" }, { "input": "5 2 2 4\n2 2", "output": "Correct" }, { "input": "5 1 1 4\n1", "output": "Correct" }, { "input": "9 7 1 4\n4 3 3 2 2 4 1", "output": "Correct" }, { "input": "9 5 2 3\n4 2 4 3 3", "output": "Incorrect" }, { "input": "6 3 1 3\n1 4 2", "output": "Incorrect" }, { "input": "3 2 1 99\n34 100", "output": "Incorrect" }, { "input": "4 2 1 99\n100 38", "output": "Incorrect" }, { "input": "5 2 1 99\n100 38", "output": "Incorrect" }, { "input": "4 2 1 99\n36 51", "output": "Correct" }, { "input": "7 6 3 10\n5 10 7 7 4 5", "output": "Correct" }, { "input": "8 6 3 10\n8 5 7 8 4 4", "output": "Correct" }, { "input": "9 6 3 10\n9 7 7 5 3 10", "output": "Correct" }, { "input": "16 15 30 40\n36 37 35 36 34 34 37 35 32 33 31 38 39 38 38", "output": "Incorrect" }, { "input": "17 15 30 40\n38 36 37 34 30 38 38 31 38 38 36 39 39 37 35", "output": "Correct" }, { "input": "18 15 30 40\n35 37 31 32 30 33 36 38 36 38 31 30 39 32 36", "output": "Correct" }, { "input": "17 16 30 40\n39 32 37 31 40 32 36 34 56 34 40 36 37 36 33 36", "output": "Incorrect" }, { "input": "18 16 30 40\n32 35 33 39 34 30 37 34 30 34 39 18 32 37 37 36", "output": "Incorrect" }, { "input": "19 16 30 40\n36 30 37 30 37 32 34 30 35 35 33 35 39 37 46 37", "output": "Incorrect" }, { "input": "2 1 2 100\n38", "output": "Incorrect" }, { "input": "3 1 2 100\n1", "output": "Incorrect" }, { "input": "4 1 2 100\n1", "output": "Incorrect" }, { "input": "91 38 1 3\n3 2 3 2 3 2 3 3 1 1 1 2 2 1 3 2 3 1 3 3 1 3 3 2 1 2 2 3 1 2 1 3 2 2 3 1 1 2", "output": "Correct" }, { "input": "4 3 2 10\n6 3 10", "output": "Correct" }, { "input": "41 6 4 10\n10 7 4 9 9 10", "output": "Correct" }, { "input": "21 1 1 9\n9", "output": "Correct" }, { "input": "2 1 9 10\n10", "output": "Correct" }, { "input": "2 1 2 9\n9", "output": "Correct" }, { "input": "8 7 5 9\n6 7 8 5 5 6 6", "output": "Correct" }, { "input": "3 2 2 8\n7 2", "output": "Correct" }, { "input": "71 36 1 10\n7 10 8 1 3 8 5 7 3 10 8 1 6 4 5 7 8 2 4 3 4 10 8 5 1 2 8 8 10 10 4 3 7 9 7 8", "output": "Correct" }, { "input": "85 3 4 9\n4 8 7", "output": "Correct" }, { "input": "4 3 4 10\n9 10 5", "output": "Correct" }, { "input": "2 1 1 5\n1", "output": "Correct" }, { "input": "91 75 1 10\n2 6 9 7 4 9 4 8 10 6 4 1 10 6 5 9 7 5 1 4 6 4 8 2 1 3 5 7 6 9 5 5 8 1 7 1 4 2 8 3 1 6 6 2 10 6 2 2 8 5 4 5 5 3 10 9 4 3 1 9 10 3 2 4 8 7 4 9 3 1 1 1 3 4 5", "output": "Correct" }, { "input": "10 4 1 8\n7 9 6 6", "output": "Incorrect" }, { "input": "18 1 3 10\n2", "output": "Incorrect" }, { "input": "6 2 4 8\n6 3", "output": "Incorrect" }, { "input": "17 6 2 8\n3 8 6 1 6 4", "output": "Incorrect" }, { "input": "21 1 5 8\n4", "output": "Incorrect" }, { "input": "2 1 1 10\n9", "output": "Incorrect" }, { "input": "2 1 4 8\n5", "output": "Incorrect" }, { "input": "2 1 1 7\n6", "output": "Incorrect" }, { "input": "2 1 4 9\n5", "output": "Incorrect" }, { "input": "2 1 3 8\n7", "output": "Incorrect" }, { "input": "2 1 5 9\n6", "output": "Incorrect" }, { "input": "3 2 1 10\n4 9", "output": "Incorrect" }, { "input": "2 1 4 10\n7", "output": "Incorrect" }, { "input": "2 1 2 9\n8", "output": "Incorrect" }, { "input": "2 1 3 9\n3", "output": "Correct" }, { "input": "3 2 6 7\n6 6", "output": "Correct" }, { "input": "6 4 1 10\n11 10 9 1", "output": "Incorrect" }, { "input": "7 6 3 8\n3 4 5 6 7 8", "output": "Correct" }, { "input": "5 3 1 5\n2 3 4", "output": "Correct" } ]
1,399,145,101
301
Python 3
WRONG_ANSWER
TESTS
7
62
0
n, m, min, max = map(int, input().split()) a = list(map(int, input().split())) lim = 0 lim1 = 0 for i in range(len(a)): if a[i] == min: lim = 1 elif a[i] == max: lim1 = 1 if lim == 0 and lim1 == 0: if n - m == 1: print('Incorrect') else: print('Correct') else: print('Correct')
Title: Data Recovery Time Limit: None seconds Memory Limit: None megabytes Problem Description: Not so long ago company R2 bought company R1 and consequently, all its developments in the field of multicore processors. Now the R2 laboratory is testing one of the R1 processors. The testing goes in *n* steps, at each step the processor gets some instructions, and then its temperature is measured. The head engineer in R2 is keeping a report record on the work of the processor: he writes down the minimum and the maximum measured temperature in his notebook. His assistant had to write down all temperatures into his notebook, but (for unknown reasons) he recorded only *m*. The next day, the engineer's assistant filed in a report with all the *m* temperatures. However, the chief engineer doubts that the assistant wrote down everything correctly (naturally, the chief engineer doesn't doubt his notes). So he asked you to help him. Given numbers *n*, *m*, *min*, *max* and the list of *m* temperatures determine whether you can upgrade the set of *m* temperatures to the set of *n* temperatures (that is add *n*<=-<=*m* temperatures), so that the minimum temperature was *min* and the maximum one was *max*. Input Specification: The first line contains four integers *n*,<=*m*,<=*min*,<=*max* (1<=≤<=*m*<=&lt;<=*n*<=≤<=100; 1<=≤<=*min*<=&lt;<=*max*<=≤<=100). The second line contains *m* space-separated integers *t**i* (1<=≤<=*t**i*<=≤<=100) — the temperatures reported by the assistant. Note, that the reported temperatures, and the temperatures you want to add can contain equal temperatures. Output Specification: If the data is consistent, print 'Correct' (without the quotes). Otherwise, print 'Incorrect' (without the quotes). Demo Input: ['2 1 1 2\n1\n', '3 1 1 3\n2\n', '2 1 1 3\n2\n'] Demo Output: ['Correct\n', 'Correct\n', 'Incorrect\n'] Note: In the first test sample one of the possible initial configurations of temperatures is [1, 2]. In the second test sample one of the possible initial configurations of temperatures is [2, 1, 3]. In the third test sample it is impossible to add one temperature to obtain the minimum equal to 1 and the maximum equal to 3.
```python n, m, min, max = map(int, input().split()) a = list(map(int, input().split())) lim = 0 lim1 = 0 for i in range(len(a)): if a[i] == min: lim = 1 elif a[i] == max: lim1 = 1 if lim == 0 and lim1 == 0: if n - m == 1: print('Incorrect') else: print('Correct') else: print('Correct') ```
0
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 2·*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 2·*m* from left to right, then the *j*-th flat of the *i*-th floor has windows 2·*j*<=-<=1 and 2·*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<=≤<=*n*,<=*m*<=≤<=100) — 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 2·*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 integer — 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
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1,689,179,548
2,147,483,647
Python 3
OK
TESTS
36
77
102,400
p, q = map(int, input().split()) tot = 0 for k in range(p): a = list(map(int, input().split())) for l in range(0, len(a), 2): if a[l] == 1 or a[l+1] == 1: tot += 1 else: continue print(tot)
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 2·*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 2·*m* from left to right, then the *j*-th flat of the *i*-th floor has windows 2·*j*<=-<=1 and 2·*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<=≤<=*n*,<=*m*<=≤<=100) — 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 2·*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 integer — 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 p, q = map(int, input().split()) tot = 0 for k in range(p): a = list(map(int, input().split())) for l in range(0, len(a), 2): if a[l] == 1 or a[l+1] == 1: tot += 1 else: continue print(tot) ```
3
169
A
Chores
PROGRAMMING
800
[ "sortings" ]
null
null
Petya and Vasya are brothers. Today is a special day for them as their parents left them home alone and commissioned them to do *n* chores. Each chore is characterized by a single parameter — its complexity. The complexity of the *i*-th chore equals *h**i*. As Petya is older, he wants to take the chores with complexity larger than some value *x* (*h**i*<=&gt;<=*x*) to leave to Vasya the chores with complexity less than or equal to *x* (*h**i*<=≤<=*x*). The brothers have already decided that Petya will do exactly *a* chores and Vasya will do exactly *b* chores (*a*<=+<=*b*<==<=*n*). In how many ways can they choose an integer *x* so that Petya got exactly *a* chores and Vasya got exactly *b* chores?
The first input line contains three integers *n*,<=*a* and *b* (2<=≤<=*n*<=≤<=2000; *a*,<=*b*<=≥<=1; *a*<=+<=*b*<==<=*n*) — the total number of chores, the number of Petya's chores and the number of Vasya's chores. The next line contains a sequence of integers *h*1,<=*h*2,<=...,<=*h**n* (1<=≤<=*h**i*<=≤<=109), *h**i* is the complexity of the *i*-th chore. The numbers in the given sequence are not necessarily different. All numbers on the lines are separated by single spaces.
Print the required number of ways to choose an integer value of *x*. If there are no such ways, print 0.
[ "5 2 3\n6 2 3 100 1\n", "7 3 4\n1 1 9 1 1 1 1\n" ]
[ "3\n", "0\n" ]
In the first sample the possible values of *x* are 3, 4 or 5. In the second sample it is impossible to find such *x*, that Petya got 3 chores and Vasya got 4.
500
[ { "input": "5 2 3\n6 2 3 100 1", "output": "3" }, { "input": "7 3 4\n1 1 9 1 1 1 1", "output": "0" }, { "input": "2 1 1\n10 2", "output": "8" }, { "input": "2 1 1\n7 7", "output": "0" }, { "input": "2 1 1\n1 1000000000", "output": "999999999" }, { "input": "3 1 2\n6 5 5", "output": "1" }, { "input": "3 2 1\n10 10 8", "output": "2" }, { "input": "8 3 5\n42 55 61 72 83 10 22 33", "output": "6" }, { "input": "10 5 5\n1 2 3 4 5 999999999 999999998 999999997 999999996 999999995", "output": "999999990" }, { "input": "4 1 3\n10 8 7 3", "output": "2" }, { "input": "4 2 2\n402 10 10 402", "output": "392" }, { "input": "4 1 3\n10 402 402 10", "output": "0" }, { "input": "4 3 1\n100 100 200 200", "output": "0" }, { "input": "150 10 140\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "0" }, { "input": "102 101 1\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "0" } ]
1,516,246,229
2,147,483,647
Python 3
OK
TESTS
29
62
5,632,000
def inp(): return map(int, input().split()) n,a,b = inp() h = list(inp()) h.sort() print(h[b] - h[b-1])
Title: Chores Time Limit: None seconds Memory Limit: None megabytes Problem Description: Petya and Vasya are brothers. Today is a special day for them as their parents left them home alone and commissioned them to do *n* chores. Each chore is characterized by a single parameter — its complexity. The complexity of the *i*-th chore equals *h**i*. As Petya is older, he wants to take the chores with complexity larger than some value *x* (*h**i*<=&gt;<=*x*) to leave to Vasya the chores with complexity less than or equal to *x* (*h**i*<=≤<=*x*). The brothers have already decided that Petya will do exactly *a* chores and Vasya will do exactly *b* chores (*a*<=+<=*b*<==<=*n*). In how many ways can they choose an integer *x* so that Petya got exactly *a* chores and Vasya got exactly *b* chores? Input Specification: The first input line contains three integers *n*,<=*a* and *b* (2<=≤<=*n*<=≤<=2000; *a*,<=*b*<=≥<=1; *a*<=+<=*b*<==<=*n*) — the total number of chores, the number of Petya's chores and the number of Vasya's chores. The next line contains a sequence of integers *h*1,<=*h*2,<=...,<=*h**n* (1<=≤<=*h**i*<=≤<=109), *h**i* is the complexity of the *i*-th chore. The numbers in the given sequence are not necessarily different. All numbers on the lines are separated by single spaces. Output Specification: Print the required number of ways to choose an integer value of *x*. If there are no such ways, print 0. Demo Input: ['5 2 3\n6 2 3 100 1\n', '7 3 4\n1 1 9 1 1 1 1\n'] Demo Output: ['3\n', '0\n'] Note: In the first sample the possible values of *x* are 3, 4 or 5. In the second sample it is impossible to find such *x*, that Petya got 3 chores and Vasya got 4.
```python def inp(): return map(int, input().split()) n,a,b = inp() h = list(inp()) h.sort() print(h[b] - h[b-1]) ```
3
118
A
String Task
PROGRAMMING
1,000
[ "implementation", "strings" ]
null
null
Petya started to attend programming lessons. On the first lesson his task was to write a simple program. The program was supposed to do the following: in the given string, consisting if uppercase and lowercase Latin letters, it: - deletes all the vowels, - inserts a character "." before each consonant, - replaces all uppercase consonants with corresponding lowercase ones. Vowels are letters "A", "O", "Y", "E", "U", "I", and the rest are consonants. The program's input is exactly one string, it should return the output as a single string, resulting after the program's processing the initial string. Help Petya cope with this easy task.
The first line represents input string of Petya's program. This string only consists of uppercase and lowercase Latin letters and its length is from 1 to 100, inclusive.
Print the resulting string. It is guaranteed that this string is not empty.
[ "tour\n", "Codeforces\n", "aBAcAba\n" ]
[ ".t.r\n", ".c.d.f.r.c.s\n", ".b.c.b\n" ]
none
500
[ { "input": "tour", "output": ".t.r" }, { "input": "Codeforces", "output": ".c.d.f.r.c.s" }, { "input": "aBAcAba", "output": ".b.c.b" }, { "input": "obn", "output": ".b.n" }, { "input": "wpwl", "output": ".w.p.w.l" }, { "input": "ggdvq", "output": ".g.g.d.v.q" }, { "input": "pumesz", "output": ".p.m.s.z" }, { "input": "g", "output": ".g" }, { "input": "zjuotps", "output": ".z.j.t.p.s" }, { "input": "jzbwuehe", "output": ".j.z.b.w.h" }, { "input": "tnkgwuugu", "output": ".t.n.k.g.w.g" }, { "input": "kincenvizh", "output": ".k.n.c.n.v.z.h" }, { "input": "xattxjenual", "output": ".x.t.t.x.j.n.l" }, { "input": "ktajqhpqsvhw", "output": ".k.t.j.q.h.p.q.s.v.h.w" }, { "input": "xnhcigytnqcmy", "output": ".x.n.h.c.g.t.n.q.c.m" }, { "input": "jfmtbejyilxcec", "output": ".j.f.m.t.b.j.l.x.c.c" }, { "input": "D", "output": ".d" }, { "input": "ab", "output": ".b" }, { "input": "Ab", "output": ".b" }, { "input": "aB", "output": ".b" }, { "input": "AB", "output": ".b" }, { "input": "ba", "output": ".b" }, { "input": "bA", "output": ".b" }, { "input": "Ba", "output": ".b" }, { "input": "BA", "output": ".b" }, { "input": "aab", "output": ".b" }, { "input": "baa", "output": ".b" }, { "input": "femOZeCArKCpUiHYnbBPTIOFmsHmcpObtPYcLCdjFrUMIyqYzAokKUiiKZRouZiNMoiOuGVoQzaaCAOkquRjmmKKElLNqCnhGdQM", "output": ".f.m.z.c.r.k.c.p.h.n.b.b.p.t.f.m.s.h.m.c.p.b.t.p.c.l.c.d.j.f.r.m.q.z.k.k.k.z.r.z.n.m.g.v.q.z.c.k.q.r.j.m.m.k.k.l.l.n.q.c.n.h.g.d.q.m" }, { "input": "VMBPMCmMDCLFELLIISUJDWQRXYRDGKMXJXJHXVZADRZWVWJRKFRRNSAWKKDPZZLFLNSGUNIVJFBEQsMDHSBJVDTOCSCgZWWKvZZN", "output": ".v.m.b.p.m.c.m.m.d.c.l.f.l.l.s.j.d.w.q.r.x.r.d.g.k.m.x.j.x.j.h.x.v.z.d.r.z.w.v.w.j.r.k.f.r.r.n.s.w.k.k.d.p.z.z.l.f.l.n.s.g.n.v.j.f.b.q.s.m.d.h.s.b.j.v.d.t.c.s.c.g.z.w.w.k.v.z.z.n" }, { "input": "MCGFQQJNUKuAEXrLXibVjClSHjSxmlkQGTKZrRaDNDomIPOmtSgjJAjNVIVLeUGUAOHNkCBwNObVCHOWvNkLFQQbFnugYVMkJruJ", "output": ".m.c.g.f.q.q.j.n.k.x.r.l.x.b.v.j.c.l.s.h.j.s.x.m.l.k.q.g.t.k.z.r.r.d.n.d.m.p.m.t.s.g.j.j.j.n.v.v.l.g.h.n.k.c.b.w.n.b.v.c.h.w.v.n.k.l.f.q.q.b.f.n.g.v.m.k.j.r.j" }, { "input": "iyaiuiwioOyzUaOtAeuEYcevvUyveuyioeeueoeiaoeiavizeeoeyYYaaAOuouueaUioueauayoiuuyiuovyOyiyoyioaoyuoyea", "output": ".w.z.t.c.v.v.v.v.z.v" }, { "input": "yjnckpfyLtzwjsgpcrgCfpljnjwqzgVcufnOvhxplvflxJzqxnhrwgfJmPzifgubvspffmqrwbzivatlmdiBaddiaktdsfPwsevl", "output": ".j.n.c.k.p.f.l.t.z.w.j.s.g.p.c.r.g.c.f.p.l.j.n.j.w.q.z.g.v.c.f.n.v.h.x.p.l.v.f.l.x.j.z.q.x.n.h.r.w.g.f.j.m.p.z.f.g.b.v.s.p.f.f.m.q.r.w.b.z.v.t.l.m.d.b.d.d.k.t.d.s.f.p.w.s.v.l" }, { "input": "RIIIUaAIYJOiuYIUWFPOOAIuaUEZeIooyUEUEAoIyIHYOEAlVAAIiLUAUAeiUIEiUMuuOiAgEUOIAoOUYYEYFEoOIIVeOOAOIIEg", "output": ".r.j.w.f.p.z.h.l.v.l.m.g.f.v.g" }, { "input": "VBKQCFBMQHDMGNSGBQVJTGQCNHHRJMNKGKDPPSQRRVQTZNKBZGSXBPBRXPMVFTXCHZMSJVBRNFNTHBHGJLMDZJSVPZZBCCZNVLMQ", "output": ".v.b.k.q.c.f.b.m.q.h.d.m.g.n.s.g.b.q.v.j.t.g.q.c.n.h.h.r.j.m.n.k.g.k.d.p.p.s.q.r.r.v.q.t.z.n.k.b.z.g.s.x.b.p.b.r.x.p.m.v.f.t.x.c.h.z.m.s.j.v.b.r.n.f.n.t.h.b.h.g.j.l.m.d.z.j.s.v.p.z.z.b.c.c.z.n.v.l.m.q" }, { "input": "iioyoaayeuyoolyiyoeuouiayiiuyTueyiaoiueyioiouyuauouayyiaeoeiiigmioiououeieeeyuyyaYyioiiooaiuouyoeoeg", "output": ".l.t.g.m.g" }, { "input": "ueyiuiauuyyeueykeioouiiauzoyoeyeuyiaoaiiaaoaueyaeydaoauexuueafouiyioueeaaeyoeuaueiyiuiaeeayaioeouiuy", "output": ".k.z.d.x.f" }, { "input": "FSNRBXLFQHZXGVMKLQDVHWLDSLKGKFMDRQWMWSSKPKKQBNDZRSCBLRSKCKKFFKRDMZFZGCNSMXNPMZVDLKXGNXGZQCLRTTDXLMXQ", "output": ".f.s.n.r.b.x.l.f.q.h.z.x.g.v.m.k.l.q.d.v.h.w.l.d.s.l.k.g.k.f.m.d.r.q.w.m.w.s.s.k.p.k.k.q.b.n.d.z.r.s.c.b.l.r.s.k.c.k.k.f.f.k.r.d.m.z.f.z.g.c.n.s.m.x.n.p.m.z.v.d.l.k.x.g.n.x.g.z.q.c.l.r.t.t.d.x.l.m.x.q" }, { "input": "EYAYAYIOIOYOOAUOEUEUOUUYIYUUMOEOIIIAOIUOAAOIYOIOEUIERCEYYAOIOIGYUIAOYUEOEUAEAYPOYEYUUAUOAOEIYIEYUEEY", "output": ".m.r.c.g.p" }, { "input": "jvvzcdcxjstbbksmqjsngxkgtttdxrljjxtwptgwwqzpvqchvgrkqlzxmptzblxhhsmrkmzzmgdfskhtmmnqzzflpmqdctvrfgtx", "output": ".j.v.v.z.c.d.c.x.j.s.t.b.b.k.s.m.q.j.s.n.g.x.k.g.t.t.t.d.x.r.l.j.j.x.t.w.p.t.g.w.w.q.z.p.v.q.c.h.v.g.r.k.q.l.z.x.m.p.t.z.b.l.x.h.h.s.m.r.k.m.z.z.m.g.d.f.s.k.h.t.m.m.n.q.z.z.f.l.p.m.q.d.c.t.v.r.f.g.t.x" }, { "input": "YB", "output": ".b" }, { "input": "fly", "output": ".f.l" }, { "input": "YyyYYYyyYxdwdawdDAWDdaddYYYY", "output": ".x.d.w.d.w.d.d.w.d.d.d.d" } ]
1,699,780,721
2,147,483,647
Python 3
OK
TESTS
42
92
0
string = list(input()) vowels = ["A", "E", "I", "O", "U", "Y"] for x in string: if x.upper() in vowels: string[string.index(x)] = "" else: string[string.index(x)] = f".{x}" string1 = "".join(string) string1 = string1.lower() print(string1)
Title: String Task Time Limit: None seconds Memory Limit: None megabytes Problem Description: Petya started to attend programming lessons. On the first lesson his task was to write a simple program. The program was supposed to do the following: in the given string, consisting if uppercase and lowercase Latin letters, it: - deletes all the vowels, - inserts a character "." before each consonant, - replaces all uppercase consonants with corresponding lowercase ones. Vowels are letters "A", "O", "Y", "E", "U", "I", and the rest are consonants. The program's input is exactly one string, it should return the output as a single string, resulting after the program's processing the initial string. Help Petya cope with this easy task. Input Specification: The first line represents input string of Petya's program. This string only consists of uppercase and lowercase Latin letters and its length is from 1 to 100, inclusive. Output Specification: Print the resulting string. It is guaranteed that this string is not empty. Demo Input: ['tour\n', 'Codeforces\n', 'aBAcAba\n'] Demo Output: ['.t.r\n', '.c.d.f.r.c.s\n', '.b.c.b\n'] Note: none
```python string = list(input()) vowels = ["A", "E", "I", "O", "U", "Y"] for x in string: if x.upper() in vowels: string[string.index(x)] = "" else: string[string.index(x)] = f".{x}" string1 = "".join(string) string1 = string1.lower() print(string1) ```
3
954
A
Diagonal Walking
PROGRAMMING
800
[ "implementation" ]
null
null
Mikhail walks on a 2D plane. He can go either up or right. You are given a sequence of Mikhail's moves. He thinks that this sequence is too long and he wants to make it as short as possible. In the given sequence moving up is described by character U and moving right is described by character R. Mikhail can replace any pair of consecutive moves RU or UR with a diagonal move (described as character D). After that, he can go on and do some other replacements, until there is no pair of consecutive moves RU or UR left. Your problem is to print the minimum possible length of the sequence of moves after the replacements.
The first line of the input contains one integer *n* (1<=≤<=*n*<=≤<=100) — the length of the sequence. The second line contains the sequence consisting of *n* characters U and R.
Print the minimum possible length of the sequence of moves after all replacements are done.
[ "5\nRUURU\n", "17\nUUURRRRRUUURURUUU\n" ]
[ "3\n", "13\n" ]
In the first test the shortened sequence of moves may be DUD (its length is 3). In the second test the shortened sequence of moves can be UUDRRRDUDDUUU (its length is 13).
0
[ { "input": "5\nRUURU", "output": "3" }, { "input": "17\nUUURRRRRUUURURUUU", "output": "13" }, { "input": "100\nUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUU", "output": "100" }, { "input": "100\nRRURRUUUURURRRURRRRURRRRRRURRUURRRUUURUURURRURUURUURRUURUURRURURUUUUURUUUUUURRUUURRRURRURRRUURRUUUUR", "output": "67" }, { "input": "100\nUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUURUUUUUUUUUUUUUUUUUUUUU", "output": "99" }, { "input": "3\nRUR", "output": "2" }, { "input": "1\nR", "output": "1" }, { "input": "5\nRURUU", "output": "3" }, { "input": "1\nU", "output": "1" }, { "input": "2\nUR", "output": "1" }, { "input": "23\nUUUUUUUUUUUUUUUUUUUUUUU", "output": "23" } ]
1,637,228,196
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
2
46
0
s= int(input()) n = input() j =n.count('RU' and 'UR') print((s-(j*2))+j-1)
Title: Diagonal Walking Time Limit: None seconds Memory Limit: None megabytes Problem Description: Mikhail walks on a 2D plane. He can go either up or right. You are given a sequence of Mikhail's moves. He thinks that this sequence is too long and he wants to make it as short as possible. In the given sequence moving up is described by character U and moving right is described by character R. Mikhail can replace any pair of consecutive moves RU or UR with a diagonal move (described as character D). After that, he can go on and do some other replacements, until there is no pair of consecutive moves RU or UR left. Your problem is to print the minimum possible length of the sequence of moves after the replacements. Input Specification: The first line of the input contains one integer *n* (1<=≤<=*n*<=≤<=100) — the length of the sequence. The second line contains the sequence consisting of *n* characters U and R. Output Specification: Print the minimum possible length of the sequence of moves after all replacements are done. Demo Input: ['5\nRUURU\n', '17\nUUURRRRRUUURURUUU\n'] Demo Output: ['3\n', '13\n'] Note: In the first test the shortened sequence of moves may be DUD (its length is 3). In the second test the shortened sequence of moves can be UUDRRRDUDDUUU (its length is 13).
```python s= int(input()) n = input() j =n.count('RU' and 'UR') print((s-(j*2))+j-1) ```
0
199
A
Hexadecimal's theorem
PROGRAMMING
900
[ "brute force", "constructive algorithms", "implementation", "number theory" ]
null
null
Recently, a chaotic virus Hexadecimal advanced a new theorem which will shake the Universe. She thinks that each Fibonacci number can be represented as sum of three not necessary different Fibonacci numbers. Let's remember how Fibonacci numbers can be calculated. *F*0<==<=0, *F*1<==<=1, and all the next numbers are *F**i*<==<=*F**i*<=-<=2<=+<=*F**i*<=-<=1. So, Fibonacci numbers make a sequence of numbers: 0, 1, 1, 2, 3, 5, 8, 13, ... If you haven't run away from the PC in fear, you have to help the virus. Your task is to divide given Fibonacci number *n* by three not necessary different Fibonacci numbers or say that it is impossible.
The input contains of a single integer *n* (0<=≤<=*n*<=&lt;<=109) — the number that should be represented by the rules described above. It is guaranteed that *n* is a Fibonacci number.
Output three required numbers: *a*, *b* and *c*. If there is no answer for the test you have to print "I'm too stupid to solve this problem" without the quotes. If there are multiple answers, print any of them.
[ "3\n", "13\n" ]
[ "1 1 1\n", "2 3 8\n" ]
none
500
[ { "input": "3", "output": "1 1 1" }, { "input": "13", "output": "2 3 8" }, { "input": "0", "output": "0 0 0" }, { "input": "1", "output": "1 0 0" }, { "input": "2", "output": "1 1 0" }, { "input": "1597", "output": "233 377 987" }, { "input": "0", "output": "0 0 0" }, { "input": "1", "output": "1 0 0" }, { "input": "1", "output": "1 0 0" }, { "input": "2", "output": "1 1 0" }, { "input": "3", "output": "1 1 1" }, { "input": "5", "output": "1 1 3" }, { "input": "8", "output": "1 2 5" }, { "input": "13", "output": "2 3 8" }, { "input": "21", "output": "3 5 13" }, { "input": "34", "output": "5 8 21" }, { "input": "55", "output": "8 13 34" }, { "input": "89", "output": "13 21 55" }, { "input": "144", "output": "21 34 89" }, { "input": "233", "output": "34 55 144" }, { "input": "377", "output": "55 89 233" }, { "input": "610", "output": "89 144 377" }, { "input": "987", "output": "144 233 610" }, { "input": "1597", "output": "233 377 987" }, { "input": "2584", "output": "377 610 1597" }, { "input": "4181", "output": "610 987 2584" }, { "input": "6765", "output": "987 1597 4181" }, { "input": "10946", "output": "1597 2584 6765" }, { "input": "17711", "output": "2584 4181 10946" }, { "input": "28657", "output": "4181 6765 17711" }, { "input": "46368", "output": "6765 10946 28657" }, { "input": "75025", "output": "10946 17711 46368" }, { "input": "121393", "output": "17711 28657 75025" }, { "input": "196418", "output": "28657 46368 121393" }, { "input": "317811", "output": "46368 75025 196418" }, { "input": "514229", "output": "75025 121393 317811" }, { "input": "832040", "output": "121393 196418 514229" }, { "input": "1346269", "output": "196418 317811 832040" }, { "input": "2178309", "output": "317811 514229 1346269" }, { "input": "3524578", "output": "514229 832040 2178309" }, { "input": "5702887", "output": "832040 1346269 3524578" }, { "input": "9227465", "output": "1346269 2178309 5702887" }, { "input": "14930352", "output": "2178309 3524578 9227465" }, { "input": "24157817", "output": "3524578 5702887 14930352" }, { "input": "39088169", "output": "5702887 9227465 24157817" }, { "input": "63245986", "output": "9227465 14930352 39088169" }, { "input": "102334155", "output": "14930352 24157817 63245986" }, { "input": "165580141", "output": "24157817 39088169 102334155" }, { "input": "267914296", "output": "39088169 63245986 165580141" }, { "input": "433494437", "output": "63245986 102334155 267914296" }, { "input": "701408733", "output": "102334155 165580141 433494437" }, { "input": "701408733", "output": "102334155 165580141 433494437" }, { "input": "102334155", "output": "14930352 24157817 63245986" }, { "input": "63245986", "output": "9227465 14930352 39088169" } ]
1,588,462,778
2,147,483,647
PyPy 3
WRONG_ANSWER
TESTS
2
248
0
# 199A z = int(input()) all_fib = [1, 1] while all_fib[-1] < z: all_fib.append(all_fib[-2] + all_fib[-1]) def check(num): result = [0, 0, 0] for i in all_fib: for j in all_fib: for k in all_fib: if i + j + k == z: return (i, j, k) return result x = check(z) if x == [0, 0, 0]: print("I'm too stupid to solve this problem") else: print(x[0], x[1], x[2])
Title: Hexadecimal's theorem Time Limit: None seconds Memory Limit: None megabytes Problem Description: Recently, a chaotic virus Hexadecimal advanced a new theorem which will shake the Universe. She thinks that each Fibonacci number can be represented as sum of three not necessary different Fibonacci numbers. Let's remember how Fibonacci numbers can be calculated. *F*0<==<=0, *F*1<==<=1, and all the next numbers are *F**i*<==<=*F**i*<=-<=2<=+<=*F**i*<=-<=1. So, Fibonacci numbers make a sequence of numbers: 0, 1, 1, 2, 3, 5, 8, 13, ... If you haven't run away from the PC in fear, you have to help the virus. Your task is to divide given Fibonacci number *n* by three not necessary different Fibonacci numbers or say that it is impossible. Input Specification: The input contains of a single integer *n* (0<=≤<=*n*<=&lt;<=109) — the number that should be represented by the rules described above. It is guaranteed that *n* is a Fibonacci number. Output Specification: Output three required numbers: *a*, *b* and *c*. If there is no answer for the test you have to print "I'm too stupid to solve this problem" without the quotes. If there are multiple answers, print any of them. Demo Input: ['3\n', '13\n'] Demo Output: ['1 1 1\n', '2 3 8\n'] Note: none
```python # 199A z = int(input()) all_fib = [1, 1] while all_fib[-1] < z: all_fib.append(all_fib[-2] + all_fib[-1]) def check(num): result = [0, 0, 0] for i in all_fib: for j in all_fib: for k in all_fib: if i + j + k == z: return (i, j, k) return result x = check(z) if x == [0, 0, 0]: print("I'm too stupid to solve this problem") else: print(x[0], x[1], x[2]) ```
0
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 2·*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 2·*m* from left to right, then the *j*-th flat of the *i*-th floor has windows 2·*j*<=-<=1 and 2·*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<=≤<=*n*,<=*m*<=≤<=100) — 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 2·*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 integer — 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" } ]
1,601,666,321
2,147,483,647
Python 3
OK
TESTS
36
109
0
n,m = list(map(int,input().split())) c =0 for i in range(n): s = list(map(int,input().split()))[:2*m] for j in range(1,2*m,2): if s[j-1] == 1 or s[j] == 1: c+=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 2·*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 2·*m* from left to right, then the *j*-th flat of the *i*-th floor has windows 2·*j*<=-<=1 and 2·*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<=≤<=*n*,<=*m*<=≤<=100) — 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 2·*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 integer — 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): s = list(map(int,input().split()))[:2*m] for j in range(1,2*m,2): if s[j-1] == 1 or s[j] == 1: c+=1 print(c) ```
3
445
A
DZY Loves Chessboard
PROGRAMMING
1,200
[ "dfs and similar", "implementation" ]
null
null
DZY loves chessboard, and he enjoys playing with it. He has a chessboard of *n* rows and *m* columns. Some cells of the chessboard are bad, others are good. For every good cell, DZY wants to put a chessman on it. Each chessman is either white or black. After putting all chessmen, DZY wants that no two chessmen with the same color are on two adjacent cells. Two cells are adjacent if and only if they share a common edge. You task is to find any suitable placement of chessmen on the given chessboard.
The first line contains two space-separated integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100). Each of the next *n* lines contains a string of *m* characters: the *j*-th character of the *i*-th string is either "." or "-". A "." means that the corresponding cell (in the *i*-th row and the *j*-th column) is good, while a "-" means it is bad.
Output must contain *n* lines, each line must contain a string of *m* characters. The *j*-th character of the *i*-th string should be either "W", "B" or "-". Character "W" means the chessman on the cell is white, "B" means it is black, "-" means the cell is a bad cell. If multiple answers exist, print any of them. It is guaranteed that at least one answer exists.
[ "1 1\n.\n", "2 2\n..\n..\n", "3 3\n.-.\n---\n--." ]
[ "B\n", "BW\nWB\n", "B-B\n---\n--B" ]
In the first sample, DZY puts a single black chessman. Of course putting a white one is also OK. In the second sample, all 4 cells are good. No two same chessmen share an edge in the sample output. In the third sample, no good cells are adjacent. So you can just put 3 chessmen, no matter what their colors are.
500
[ { "input": "1 1\n.", "output": "B" }, { "input": "2 2\n..\n..", "output": "BW\nWB" }, { "input": "3 3\n.-.\n---\n--.", "output": "B-B\n---\n--B" }, { "input": "3 1\n-\n.\n.", "output": "-\nW\nB" }, { "input": "11 11\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------", "output": "-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------\n-----------" }, { "input": "1 1\n-", "output": "-" }, { "input": "4 2\n..\n..\n..\n..", "output": "BW\nWB\nBW\nWB" }, { "input": "5 1\n.\n.\n.\n.\n.", "output": "B\nW\nB\nW\nB" } ]
1,644,715,113
2,147,483,647
Python 3
RUNTIME_ERROR
TESTS
4
31
512,000
def resolve(actual_pos , table , symbol, passed_path): x , y= actual_pos pos = str(x) + str(y) if (x < 0 or x == len(table[0]) or y < 0 or y == len(table) or table[y][x] == "B" or table[y][x] == "W"): # print(actual_pos , "retornou !!!") return if(pos in passed_path): return passed_path.append(pos) # print(actual_pos) if (table[y][x] == "."): table[y][x] = symbol if(symbol == "B"): symbol = "W" else: symbol = "B" resolve([x + 1, y ], table , symbol, passed_path) resolve([x - 1, y ], table , symbol, passed_path) resolve([x, y + 1 ], table , symbol, passed_path) resolve([x, y -1 ], table , symbol, passed_path) n , m= input().split() table =[] for i in range(int(n)): line = input() table.append([e for e in line]) resolve([0,0] , table , "B", []) saida = "" for l in table: saida += "".join(l) + "\n" print(saida)
Title: DZY Loves Chessboard Time Limit: None seconds Memory Limit: None megabytes Problem Description: DZY loves chessboard, and he enjoys playing with it. He has a chessboard of *n* rows and *m* columns. Some cells of the chessboard are bad, others are good. For every good cell, DZY wants to put a chessman on it. Each chessman is either white or black. After putting all chessmen, DZY wants that no two chessmen with the same color are on two adjacent cells. Two cells are adjacent if and only if they share a common edge. You task is to find any suitable placement of chessmen on the given chessboard. Input Specification: The first line contains two space-separated integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100). Each of the next *n* lines contains a string of *m* characters: the *j*-th character of the *i*-th string is either "." or "-". A "." means that the corresponding cell (in the *i*-th row and the *j*-th column) is good, while a "-" means it is bad. Output Specification: Output must contain *n* lines, each line must contain a string of *m* characters. The *j*-th character of the *i*-th string should be either "W", "B" or "-". Character "W" means the chessman on the cell is white, "B" means it is black, "-" means the cell is a bad cell. If multiple answers exist, print any of them. It is guaranteed that at least one answer exists. Demo Input: ['1 1\n.\n', '2 2\n..\n..\n', '3 3\n.-.\n---\n--.'] Demo Output: ['B\n', 'BW\nWB\n', 'B-B\n---\n--B'] Note: In the first sample, DZY puts a single black chessman. Of course putting a white one is also OK. In the second sample, all 4 cells are good. No two same chessmen share an edge in the sample output. In the third sample, no good cells are adjacent. So you can just put 3 chessmen, no matter what their colors are.
```python def resolve(actual_pos , table , symbol, passed_path): x , y= actual_pos pos = str(x) + str(y) if (x < 0 or x == len(table[0]) or y < 0 or y == len(table) or table[y][x] == "B" or table[y][x] == "W"): # print(actual_pos , "retornou !!!") return if(pos in passed_path): return passed_path.append(pos) # print(actual_pos) if (table[y][x] == "."): table[y][x] = symbol if(symbol == "B"): symbol = "W" else: symbol = "B" resolve([x + 1, y ], table , symbol, passed_path) resolve([x - 1, y ], table , symbol, passed_path) resolve([x, y + 1 ], table , symbol, passed_path) resolve([x, y -1 ], table , symbol, passed_path) n , m= input().split() table =[] for i in range(int(n)): line = input() table.append([e for e in line]) resolve([0,0] , table , "B", []) saida = "" for l in table: saida += "".join(l) + "\n" print(saida) ```
-1
318
A
Even Odds
PROGRAMMING
900
[ "math" ]
null
null
Being a nonconformist, Volodya is displeased with the current state of things, particularly with the order of natural numbers (natural number is positive integer number). He is determined to rearrange them. But there are too many natural numbers, so Volodya decided to start with the first *n*. He writes down the following sequence of numbers: firstly all odd integers from 1 to *n* (in ascending order), then all even integers from 1 to *n* (also in ascending order). Help our hero to find out which number will stand at the position number *k*.
The only line of input contains integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=1012). Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
Print the number that will stand at the position number *k* after Volodya's manipulations.
[ "10 3\n", "7 7\n" ]
[ "5", "6" ]
In the first sample Volodya's sequence will look like this: {1, 3, 5, 7, 9, 2, 4, 6, 8, 10}. The third place in the sequence is therefore occupied by the number 5.
500
[ { "input": "10 3", "output": "5" }, { "input": "7 7", "output": "6" }, { "input": "7 1", "output": "1" }, { "input": "7 2", "output": "3" }, { "input": "8 5", "output": "2" }, { "input": "8 3", "output": "5" }, { "input": "8 4", "output": "7" }, { "input": "1000000000000 500000000001", "output": "2" }, { "input": "999999999997 499999999999", "output": "999999999997" }, { "input": "999999999999 999999999999", "output": "999999999998" }, { "input": "1000000000000 1", "output": "1" }, { "input": "999999999999 1", "output": "1" }, { "input": "1 1", "output": "1" }, { "input": "1000000000000 1000000000000", "output": "1000000000000" }, { "input": "1000000000000 500000000000", "output": "999999999999" }, { "input": "1000000000000 499999999999", "output": "999999999997" }, { "input": "999999999997 499999999998", "output": "999999999995" }, { "input": "619234238 556154835", "output": "493075432" }, { "input": "38151981 36650624", "output": "35149266" }, { "input": "680402465 442571217", "output": "204739968" }, { "input": "109135284 9408714", "output": "18817427" }, { "input": "603701841 56038951", "output": "112077901" }, { "input": "356764822 321510177", "output": "286255532" }, { "input": "284911189 142190783", "output": "284381565" }, { "input": "91028405 61435545", "output": "31842684" } ]
1,696,351,744
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
4
62
0
k,n = input().split(" "); k = int(k); n = int(n) if n <= k//2 + k%2: print(n*2-1) else: print(n//2*2)
Title: Even Odds Time Limit: None seconds Memory Limit: None megabytes Problem Description: Being a nonconformist, Volodya is displeased with the current state of things, particularly with the order of natural numbers (natural number is positive integer number). He is determined to rearrange them. But there are too many natural numbers, so Volodya decided to start with the first *n*. He writes down the following sequence of numbers: firstly all odd integers from 1 to *n* (in ascending order), then all even integers from 1 to *n* (also in ascending order). Help our hero to find out which number will stand at the position number *k*. Input Specification: The only line of input contains integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=1012). Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier. Output Specification: Print the number that will stand at the position number *k* after Volodya's manipulations. Demo Input: ['10 3\n', '7 7\n'] Demo Output: ['5', '6'] Note: In the first sample Volodya's sequence will look like this: {1, 3, 5, 7, 9, 2, 4, 6, 8, 10}. The third place in the sequence is therefore occupied by the number 5.
```python k,n = input().split(" "); k = int(k); n = int(n) if n <= k//2 + k%2: print(n*2-1) else: print(n//2*2) ```
0
931
B
World Cup
PROGRAMMING
1,200
[ "constructive algorithms", "implementation" ]
null
null
The last stage of Football World Cup is played using the play-off system. There are *n* teams left in this stage, they are enumerated from 1 to *n*. Several rounds are held, in each round the remaining teams are sorted in the order of their ids, then the first in this order plays with the second, the third — with the fourth, the fifth — with the sixth, and so on. It is guaranteed that in each round there is even number of teams. The winner of each game advances to the next round, the loser is eliminated from the tournament, there are no draws. In the last round there is the only game with two remaining teams: the round is called the Final, the winner is called the champion, and the tournament is over. Arkady wants his two favorite teams to play in the Final. Unfortunately, the team ids are already determined, and it may happen that it is impossible for teams to meet in the Final, because they are to meet in some earlier stage, if they are strong enough. Determine, in which round the teams with ids *a* and *b* can meet.
The only line contains three integers *n*, *a* and *b* (2<=≤<=*n*<=≤<=256, 1<=≤<=*a*,<=*b*<=≤<=*n*) — the total number of teams, and the ids of the teams that Arkady is interested in. It is guaranteed that *n* is such that in each round an even number of team advance, and that *a* and *b* are not equal.
In the only line print "Final!" (without quotes), if teams *a* and *b* can meet in the Final. Otherwise, print a single integer — the number of the round in which teams *a* and *b* can meet. The round are enumerated from 1.
[ "4 1 2\n", "8 2 6\n", "8 7 5\n" ]
[ "1\n", "Final!\n", "2\n" ]
In the first example teams 1 and 2 meet in the first round. In the second example teams 2 and 6 can only meet in the third round, which is the Final, if they win all their opponents in earlier rounds. In the third example the teams with ids 7 and 5 can meet in the second round, if they win their opponents in the first round.
1,000
[ { "input": "4 1 2", "output": "1" }, { "input": "8 2 6", "output": "Final!" }, { "input": "8 7 5", "output": "2" }, { "input": "128 30 98", "output": "Final!" }, { "input": "256 128 256", "output": "Final!" }, { "input": "256 2 127", "output": "7" }, { "input": "2 1 2", "output": "Final!" }, { "input": "2 2 1", "output": "Final!" }, { "input": "4 1 3", "output": "Final!" }, { "input": "4 1 4", "output": "Final!" }, { "input": "4 2 1", "output": "1" }, { "input": "4 2 3", "output": "Final!" }, { "input": "4 2 4", "output": "Final!" }, { "input": "4 3 1", "output": "Final!" }, { "input": "4 3 2", "output": "Final!" }, { "input": "4 3 4", "output": "1" }, { "input": "4 4 1", "output": "Final!" }, { "input": "4 4 2", "output": "Final!" }, { "input": "4 4 3", "output": "1" }, { "input": "8 8 7", "output": "1" }, { "input": "8 8 5", "output": "2" }, { "input": "8 8 1", "output": "Final!" }, { "input": "16 4 3", "output": "1" }, { "input": "16 2 4", "output": "2" }, { "input": "16 14 11", "output": "3" }, { "input": "16 3 11", "output": "Final!" }, { "input": "32 10 9", "output": "1" }, { "input": "32 25 28", "output": "2" }, { "input": "32 22 18", "output": "3" }, { "input": "32 17 25", "output": "4" }, { "input": "32 18 3", "output": "Final!" }, { "input": "64 40 39", "output": "1" }, { "input": "64 60 58", "output": "2" }, { "input": "64 34 37", "output": "3" }, { "input": "64 26 24", "output": "4" }, { "input": "64 50 43", "output": "5" }, { "input": "64 17 42", "output": "Final!" }, { "input": "128 116 115", "output": "1" }, { "input": "128 35 33", "output": "2" }, { "input": "128 61 59", "output": "3" }, { "input": "128 116 123", "output": "4" }, { "input": "128 17 15", "output": "5" }, { "input": "128 124 77", "output": "6" }, { "input": "128 4 80", "output": "Final!" }, { "input": "256 224 223", "output": "1" }, { "input": "256 24 22", "output": "2" }, { "input": "256 199 196", "output": "3" }, { "input": "256 148 159", "output": "4" }, { "input": "256 178 166", "output": "5" }, { "input": "256 75 97", "output": "6" }, { "input": "256 185 200", "output": "7" }, { "input": "256 3 238", "output": "Final!" }, { "input": "256 128 129", "output": "Final!" }, { "input": "256 255 129", "output": "7" }, { "input": "256 255 128", "output": "Final!" }, { "input": "256 129 256", "output": "7" }, { "input": "128 98 69", "output": "6" }, { "input": "128 47 83", "output": "Final!" }, { "input": "16 2 3", "output": "2" }, { "input": "64 32 30", "output": "2" }, { "input": "8 4 5", "output": "Final!" }, { "input": "8 7 8", "output": "1" }, { "input": "8 2 3", "output": "2" }, { "input": "8 2 5", "output": "Final!" } ]
1,611,579,971
2,147,483,647
PyPy 3
OK
TESTS
64
108
0
v, f, s = map(int, input().split()) f, s = f-1, s-1 a = list(range(v)) x = [] i = 0 z = True g = 1 while z: for i in range(0, v, 2): if a[i+1] == f and a[i] == s or a[i+1] == s and a[i] == f: print([g, "Final!"][v == 2]) exit(0) elif a[i+1] != f and a[i] == s or a[i+1] != s and a[i] == f: x.append(a[i]) elif a[i+1] == f and a[i] != s or a[i+1] == s and a[i] != f: x.append(a[i+1]) else: x.append(a[i+1]) a = x v = v//2 x = [] g += 1
Title: World Cup Time Limit: None seconds Memory Limit: None megabytes Problem Description: The last stage of Football World Cup is played using the play-off system. There are *n* teams left in this stage, they are enumerated from 1 to *n*. Several rounds are held, in each round the remaining teams are sorted in the order of their ids, then the first in this order plays with the second, the third — with the fourth, the fifth — with the sixth, and so on. It is guaranteed that in each round there is even number of teams. The winner of each game advances to the next round, the loser is eliminated from the tournament, there are no draws. In the last round there is the only game with two remaining teams: the round is called the Final, the winner is called the champion, and the tournament is over. Arkady wants his two favorite teams to play in the Final. Unfortunately, the team ids are already determined, and it may happen that it is impossible for teams to meet in the Final, because they are to meet in some earlier stage, if they are strong enough. Determine, in which round the teams with ids *a* and *b* can meet. Input Specification: The only line contains three integers *n*, *a* and *b* (2<=≤<=*n*<=≤<=256, 1<=≤<=*a*,<=*b*<=≤<=*n*) — the total number of teams, and the ids of the teams that Arkady is interested in. It is guaranteed that *n* is such that in each round an even number of team advance, and that *a* and *b* are not equal. Output Specification: In the only line print "Final!" (without quotes), if teams *a* and *b* can meet in the Final. Otherwise, print a single integer — the number of the round in which teams *a* and *b* can meet. The round are enumerated from 1. Demo Input: ['4 1 2\n', '8 2 6\n', '8 7 5\n'] Demo Output: ['1\n', 'Final!\n', '2\n'] Note: In the first example teams 1 and 2 meet in the first round. In the second example teams 2 and 6 can only meet in the third round, which is the Final, if they win all their opponents in earlier rounds. In the third example the teams with ids 7 and 5 can meet in the second round, if they win their opponents in the first round.
```python v, f, s = map(int, input().split()) f, s = f-1, s-1 a = list(range(v)) x = [] i = 0 z = True g = 1 while z: for i in range(0, v, 2): if a[i+1] == f and a[i] == s or a[i+1] == s and a[i] == f: print([g, "Final!"][v == 2]) exit(0) elif a[i+1] != f and a[i] == s or a[i+1] != s and a[i] == f: x.append(a[i]) elif a[i+1] == f and a[i] != s or a[i+1] == s and a[i] != f: x.append(a[i+1]) else: x.append(a[i+1]) a = x v = v//2 x = [] g += 1 ```
3
1,000
B
Light It Up
PROGRAMMING
1,500
[ "greedy" ]
null
null
Recently, you bought a brand new smart lamp with programming features. At first, you set up a schedule to the lamp. Every day it will turn power on at moment $0$ and turn power off at moment $M$. Moreover, the lamp allows you to set a program of switching its state (states are "lights on" and "lights off"). Unfortunately, some program is already installed into the lamp. The lamp allows only good programs. Good program can be represented as a non-empty array $a$, where $0 &lt; a_1 &lt; a_2 &lt; \dots &lt; a_{|a|} &lt; M$. All $a_i$ must be integers. Of course, preinstalled program is a good program. The lamp follows program $a$ in next manner: at moment $0$ turns power and light on. Then at moment $a_i$ the lamp flips its state to opposite (if it was lit, it turns off, and vice versa). The state of the lamp flips instantly: for example, if you turn the light off at moment $1$ and then do nothing, the total time when the lamp is lit will be $1$. Finally, at moment $M$ the lamp is turning its power off regardless of its state. Since you are not among those people who read instructions, and you don't understand the language it's written in, you realize (after some testing) the only possible way to alter the preinstalled program. You can insert at most one element into the program $a$, so it still should be a good program after alteration. Insertion can be done between any pair of consecutive elements of $a$, or even at the begining or at the end of $a$. Find such a way to alter the program that the total time when the lamp is lit is maximum possible. Maybe you should leave program untouched. If the lamp is lit from $x$ till moment $y$, then its lit for $y - x$ units of time. Segments of time when the lamp is lit are summed up.
First line contains two space separated integers $n$ and $M$ ($1 \le n \le 10^5$, $2 \le M \le 10^9$) — the length of program $a$ and the moment when power turns off. Second line contains $n$ space separated integers $a_1, a_2, \dots, a_n$ ($0 &lt; a_1 &lt; a_2 &lt; \dots &lt; a_n &lt; M$) — initially installed program $a$.
Print the only integer — maximum possible total time when the lamp is lit.
[ "3 10\n4 6 7\n", "2 12\n1 10\n", "2 7\n3 4\n" ]
[ "8\n", "9\n", "6\n" ]
In the first example, one of possible optimal solutions is to insert value $x = 3$ before $a_1$, so program will be $[3, 4, 6, 7]$ and time of lamp being lit equals $(3 - 0) + (6 - 4) + (10 - 7) = 8$. Other possible solution is to insert $x = 5$ in appropriate place. In the second example, there is only one optimal solution: to insert $x = 2$ between $a_1$ and $a_2$. Program will become $[1, 2, 10]$, and answer will be $(1 - 0) + (10 - 2) = 9$. In the third example, optimal answer is to leave program untouched, so answer will be $(3 - 0) + (7 - 4) = 6$.
0
[ { "input": "3 10\n4 6 7", "output": "8" }, { "input": "2 12\n1 10", "output": "9" }, { "input": "2 7\n3 4", "output": "6" }, { "input": "1 2\n1", "output": "1" }, { "input": "5 10\n1 3 5 6 8", "output": "6" }, { "input": "7 1000000000\n1 10001 10011 20011 20021 40021 40031", "output": "999999969" }, { "input": "7 1000000000\n3 10001 10011 20011 20021 40021 40031", "output": "999999969" }, { "input": "1 10\n1", "output": "9" }, { "input": "1 10000000\n1", "output": "9999999" }, { "input": "1 8\n1", "output": "7" }, { "input": "7 17\n1 5 9 10 11 14 16", "output": "9" }, { "input": "4 17\n1 5 9 10", "output": "12" }, { "input": "5 12\n1 2 3 4 5", "output": "9" }, { "input": "2 1000000000\n100 111", "output": "999999989" }, { "input": "3 90591\n90579 90580 90581", "output": "90589" }, { "input": "5 16\n1 2 3 4 5", "output": "13" }, { "input": "6 981231233\n1 6 15 30 130 1875", "output": "981229468" }, { "input": "1 3\n1", "output": "2" }, { "input": "3 12\n4 7 11", "output": "8" }, { "input": "1 100000\n1", "output": "99999" }, { "input": "2 5\n1 3", "output": "3" }, { "input": "1 1000000000\n1", "output": "999999999" }, { "input": "1 1000000000\n999999999", "output": "999999999" }, { "input": "3 4\n1 2 3", "output": "2" }, { "input": "3 5\n1 2 3", "output": "3" }, { "input": "1 4\n2", "output": "3" }, { "input": "3 5\n1 3 4", "output": "3" }, { "input": "9 20\n5 9 11 12 14 15 16 17 19", "output": "12" }, { "input": "2 4\n1 2", "output": "3" }, { "input": "1 10\n2", "output": "9" }, { "input": "2 6\n2 4", "output": "4" }, { "input": "2 4\n1 3", "output": "2" }, { "input": "3 6\n1 2 4", "output": "4" }, { "input": "7 9863\n65 96 97 98 101 112 1115", "output": "9819" } ]
1,663,251,664
2,147,483,647
PyPy 3-64
WRONG_ANSWER
TESTS
0
46
0
n,m = map(int,input().split()) aa = [int(i) for i in input().split()] aa.append(m) aa.insert(0,0) bb = aa[:] d = 0 c = 0 for b in range(1,n + 2,2): c = c + bb[b] - bb[b - 1] d = c for x in range(1,n): bb = aa[:] bb.insert(x,aa[x] - 1) print(bb) c = 0 for b in range(1,n + 3,2): c = c + bb[b] - bb[b - 1] if d < c: d = c bb = aa[:] bb.insert(x + 1,aa[x] + 1) c = 0 for b in range(1,n + 3,2): c = c + bb[b] - bb[b - 1] if d < c: d = c print(d)
Title: Light It Up Time Limit: None seconds Memory Limit: None megabytes Problem Description: Recently, you bought a brand new smart lamp with programming features. At first, you set up a schedule to the lamp. Every day it will turn power on at moment $0$ and turn power off at moment $M$. Moreover, the lamp allows you to set a program of switching its state (states are "lights on" and "lights off"). Unfortunately, some program is already installed into the lamp. The lamp allows only good programs. Good program can be represented as a non-empty array $a$, where $0 &lt; a_1 &lt; a_2 &lt; \dots &lt; a_{|a|} &lt; M$. All $a_i$ must be integers. Of course, preinstalled program is a good program. The lamp follows program $a$ in next manner: at moment $0$ turns power and light on. Then at moment $a_i$ the lamp flips its state to opposite (if it was lit, it turns off, and vice versa). The state of the lamp flips instantly: for example, if you turn the light off at moment $1$ and then do nothing, the total time when the lamp is lit will be $1$. Finally, at moment $M$ the lamp is turning its power off regardless of its state. Since you are not among those people who read instructions, and you don't understand the language it's written in, you realize (after some testing) the only possible way to alter the preinstalled program. You can insert at most one element into the program $a$, so it still should be a good program after alteration. Insertion can be done between any pair of consecutive elements of $a$, or even at the begining or at the end of $a$. Find such a way to alter the program that the total time when the lamp is lit is maximum possible. Maybe you should leave program untouched. If the lamp is lit from $x$ till moment $y$, then its lit for $y - x$ units of time. Segments of time when the lamp is lit are summed up. Input Specification: First line contains two space separated integers $n$ and $M$ ($1 \le n \le 10^5$, $2 \le M \le 10^9$) — the length of program $a$ and the moment when power turns off. Second line contains $n$ space separated integers $a_1, a_2, \dots, a_n$ ($0 &lt; a_1 &lt; a_2 &lt; \dots &lt; a_n &lt; M$) — initially installed program $a$. Output Specification: Print the only integer — maximum possible total time when the lamp is lit. Demo Input: ['3 10\n4 6 7\n', '2 12\n1 10\n', '2 7\n3 4\n'] Demo Output: ['8\n', '9\n', '6\n'] Note: In the first example, one of possible optimal solutions is to insert value $x = 3$ before $a_1$, so program will be $[3, 4, 6, 7]$ and time of lamp being lit equals $(3 - 0) + (6 - 4) + (10 - 7) = 8$. Other possible solution is to insert $x = 5$ in appropriate place. In the second example, there is only one optimal solution: to insert $x = 2$ between $a_1$ and $a_2$. Program will become $[1, 2, 10]$, and answer will be $(1 - 0) + (10 - 2) = 9$. In the third example, optimal answer is to leave program untouched, so answer will be $(3 - 0) + (7 - 4) = 6$.
```python n,m = map(int,input().split()) aa = [int(i) for i in input().split()] aa.append(m) aa.insert(0,0) bb = aa[:] d = 0 c = 0 for b in range(1,n + 2,2): c = c + bb[b] - bb[b - 1] d = c for x in range(1,n): bb = aa[:] bb.insert(x,aa[x] - 1) print(bb) c = 0 for b in range(1,n + 3,2): c = c + bb[b] - bb[b - 1] if d < c: d = c bb = aa[:] bb.insert(x + 1,aa[x] + 1) c = 0 for b in range(1,n + 3,2): c = c + bb[b] - bb[b - 1] if d < c: d = c print(d) ```
0
472
A
Design Tutorial: Learn from Math
PROGRAMMING
800
[ "math", "number theory" ]
null
null
One way to create a task is to learn from math. You can generate some random math statement or modify some theorems to get something new and build a new task from that. For example, there is a statement called the "Goldbach's conjecture". It says: "each even number no less than four can be expressed as the sum of two primes". Let's modify it. How about a statement like that: "each integer no less than 12 can be expressed as the sum of two composite numbers." Not like the Goldbach's conjecture, I can prove this theorem. You are given an integer *n* no less than 12, express it as a sum of two composite numbers.
The only line contains an integer *n* (12<=≤<=*n*<=≤<=106).
Output two composite integers *x* and *y* (1<=&lt;<=*x*,<=*y*<=&lt;<=*n*) such that *x*<=+<=*y*<==<=*n*. If there are multiple solutions, you can output any of them.
[ "12\n", "15\n", "23\n", "1000000\n" ]
[ "4 8\n", "6 9\n", "8 15\n", "500000 500000\n" ]
In the first example, 12 = 4 + 8 and both 4, 8 are composite numbers. You can output "6 6" or "8 4" as well. In the second example, 15 = 6 + 9. Note that you can't output "1 14" because 1 is not a composite number.
500
[ { "input": "12", "output": "4 8" }, { "input": "15", "output": "6 9" }, { "input": "23", "output": "8 15" }, { "input": "1000000", "output": "500000 500000" }, { "input": "63874", "output": "4 63870" }, { "input": "14568", "output": "4 14564" }, { "input": "192", "output": "4 188" }, { "input": "86", "output": "4 82" }, { "input": "46220", "output": "4 46216" }, { "input": "57114", "output": "4 57110" }, { "input": "869", "output": "4 865" }, { "input": "738457", "output": "4 738453" }, { "input": "58113", "output": "6 58107" }, { "input": "4864", "output": "4 4860" }, { "input": "15", "output": "6 9" }, { "input": "74752", "output": "4 74748" }, { "input": "6073", "output": "4 6069" }, { "input": "1289", "output": "4 1285" }, { "input": "20", "output": "4 16" }, { "input": "58134", "output": "4 58130" }, { "input": "57756", "output": "4 57752" }, { "input": "765", "output": "6 759" }, { "input": "59", "output": "4 55" }, { "input": "991666", "output": "4 991662" }, { "input": "70761", "output": "4 70757" }, { "input": "13", "output": "4 9" }, { "input": "999999", "output": "4 999995" }, { "input": "17", "output": "8 9" }, { "input": "21", "output": "6 15" }, { "input": "19", "output": "4 15" }, { "input": "100007", "output": "6 100001" }, { "input": "999987", "output": "6 999981" }, { "input": "22", "output": "4 18" } ]
1,676,050,290
2,147,483,647
PyPy 3-64
OK
TESTS
33
62
0
n=int(input()) if n%2: print(9,n-9) else: print(8,n-8)
Title: Design Tutorial: Learn from Math Time Limit: None seconds Memory Limit: None megabytes Problem Description: One way to create a task is to learn from math. You can generate some random math statement or modify some theorems to get something new and build a new task from that. For example, there is a statement called the "Goldbach's conjecture". It says: "each even number no less than four can be expressed as the sum of two primes". Let's modify it. How about a statement like that: "each integer no less than 12 can be expressed as the sum of two composite numbers." Not like the Goldbach's conjecture, I can prove this theorem. You are given an integer *n* no less than 12, express it as a sum of two composite numbers. Input Specification: The only line contains an integer *n* (12<=≤<=*n*<=≤<=106). Output Specification: Output two composite integers *x* and *y* (1<=&lt;<=*x*,<=*y*<=&lt;<=*n*) such that *x*<=+<=*y*<==<=*n*. If there are multiple solutions, you can output any of them. Demo Input: ['12\n', '15\n', '23\n', '1000000\n'] Demo Output: ['4 8\n', '6 9\n', '8 15\n', '500000 500000\n'] Note: In the first example, 12 = 4 + 8 and both 4, 8 are composite numbers. You can output "6 6" or "8 4" as well. In the second example, 15 = 6 + 9. Note that you can't output "1 14" because 1 is not a composite number.
```python n=int(input()) if n%2: print(9,n-9) else: print(8,n-8) ```
3
383
C
Propagating tree
PROGRAMMING
2,000
[ "data structures", "dfs and similar", "trees" ]
null
null
Iahub likes trees very much. Recently he discovered an interesting tree named propagating tree. The tree consists of *n* nodes numbered from 1 to *n*, each node *i* having an initial value *a**i*. The root of the tree is node 1. This tree has a special property: when a value *val* is added to a value of node *i*, the value -*val* is added to values of all the children of node *i*. Note that when you add value -*val* to a child of node *i*, you also add -(-*val*) to all children of the child of node *i* and so on. Look an example explanation to understand better how it works. This tree supports two types of queries: - "1 *x* *val*" — *val* is added to the value of node *x*; - "2 *x*" — print the current value of node *x*. In order to help Iahub understand the tree better, you must answer *m* queries of the preceding type.
The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=200000). The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=1000). Each of the next *n*–1 lines contains two integers *v**i* and *u**i* (1<=≤<=*v**i*,<=*u**i*<=≤<=*n*), meaning that there is an edge between nodes *v**i* and *u**i*. Each of the next *m* lines contains a query in the format described above. It is guaranteed that the following constraints hold for all queries: 1<=≤<=*x*<=≤<=*n*,<=1<=≤<=*val*<=≤<=1000.
For each query of type two (print the value of node *x*) you must print the answer to the query on a separate line. The queries must be answered in the order given in the input.
[ "5 5\n1 2 1 1 2\n1 2\n1 3\n2 4\n2 5\n1 2 3\n1 1 2\n2 1\n2 2\n2 4\n" ]
[ "3\n3\n0\n" ]
The values of the nodes are [1, 2, 1, 1, 2] at the beginning. Then value 3 is added to node 2. It propagates and value -3 is added to it's sons, node 4 and node 5. Then it cannot propagate any more. So the values of the nodes are [1, 5, 1,  - 2,  - 1]. Then value 2 is added to node 1. It propagates and value -2 is added to it's sons, node 2 and node 3. From node 2 it propagates again, adding value 2 to it's sons, node 4 and node 5. Node 3 has no sons, so it cannot propagate from there. The values of the nodes are [3, 3,  - 1, 0, 1]. You can see all the definitions about the tree at the following link: http://en.wikipedia.org/wiki/Tree_(graph_theory)
1,500
[ { "input": "5 5\n1 2 1 1 2\n1 2\n1 3\n2 4\n2 5\n1 2 3\n1 1 2\n2 1\n2 2\n2 4", "output": "3\n3\n0" }, { "input": "10 10\n137 197 856 768 825 894 86 174 218 326\n7 8\n4 7\n8 9\n7 10\n1 2\n2 4\n3 6\n3 5\n2 3\n1 9 624\n2 1\n2 4\n1 6 505\n1 8 467\n1 3 643\n2 1\n1 8 631\n2 4\n1 7 244", "output": "137\n768\n137\n768" }, { "input": "10 10\n418 45 865 869 745 901 177 773 854 462\n4 8\n1 4\n3 6\n1 5\n1 10\n5 9\n1 2\n4 7\n1 3\n2 2\n1 6 246\n1 4 296\n1 2 378\n1 8 648\n2 6\n1 5 288\n1 6 981\n1 2 868\n2 7", "output": "45\n1147\n-119" } ]
1,577,379,139
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
2
93
307,200
n,m = map(int,input().split()) values = list(map(int,input().split())) rebra = [] for i in range(n-1): rebra.append(list(map(int,input().split()))) zapr = [] for i in range(m): z = list(map(int,input().split())) if z[0] == 1: v = -z[2] values[z[1]-1] -= v for j in range(z[1],n-1): for e in rebra: if e[0] == j: #2 values[e[1]-1] += v v = -v else: zapr.append(values[z[1]-1]) for i in zapr: print(i)
Title: Propagating tree Time Limit: None seconds Memory Limit: None megabytes Problem Description: Iahub likes trees very much. Recently he discovered an interesting tree named propagating tree. The tree consists of *n* nodes numbered from 1 to *n*, each node *i* having an initial value *a**i*. The root of the tree is node 1. This tree has a special property: when a value *val* is added to a value of node *i*, the value -*val* is added to values of all the children of node *i*. Note that when you add value -*val* to a child of node *i*, you also add -(-*val*) to all children of the child of node *i* and so on. Look an example explanation to understand better how it works. This tree supports two types of queries: - "1 *x* *val*" — *val* is added to the value of node *x*; - "2 *x*" — print the current value of node *x*. In order to help Iahub understand the tree better, you must answer *m* queries of the preceding type. Input Specification: The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=200000). The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=1000). Each of the next *n*–1 lines contains two integers *v**i* and *u**i* (1<=≤<=*v**i*,<=*u**i*<=≤<=*n*), meaning that there is an edge between nodes *v**i* and *u**i*. Each of the next *m* lines contains a query in the format described above. It is guaranteed that the following constraints hold for all queries: 1<=≤<=*x*<=≤<=*n*,<=1<=≤<=*val*<=≤<=1000. Output Specification: For each query of type two (print the value of node *x*) you must print the answer to the query on a separate line. The queries must be answered in the order given in the input. Demo Input: ['5 5\n1 2 1 1 2\n1 2\n1 3\n2 4\n2 5\n1 2 3\n1 1 2\n2 1\n2 2\n2 4\n'] Demo Output: ['3\n3\n0\n'] Note: The values of the nodes are [1, 2, 1, 1, 2] at the beginning. Then value 3 is added to node 2. It propagates and value -3 is added to it's sons, node 4 and node 5. Then it cannot propagate any more. So the values of the nodes are [1, 5, 1,  - 2,  - 1]. Then value 2 is added to node 1. It propagates and value -2 is added to it's sons, node 2 and node 3. From node 2 it propagates again, adding value 2 to it's sons, node 4 and node 5. Node 3 has no sons, so it cannot propagate from there. The values of the nodes are [3, 3,  - 1, 0, 1]. You can see all the definitions about the tree at the following link: http://en.wikipedia.org/wiki/Tree_(graph_theory)
```python n,m = map(int,input().split()) values = list(map(int,input().split())) rebra = [] for i in range(n-1): rebra.append(list(map(int,input().split()))) zapr = [] for i in range(m): z = list(map(int,input().split())) if z[0] == 1: v = -z[2] values[z[1]-1] -= v for j in range(z[1],n-1): for e in rebra: if e[0] == j: #2 values[e[1]-1] += v v = -v else: zapr.append(values[z[1]-1]) for i in zapr: print(i) ```
0
71
A
Way Too Long Words
PROGRAMMING
800
[ "strings" ]
A. Way Too Long Words
1
256
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome. Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation. This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes. Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n". You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
[ "4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n" ]
[ "word\nl10n\ni18n\np43s\n" ]
none
500
[ { "input": "4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis", "output": "word\nl10n\ni18n\np43s" }, { "input": "5\nabcdefgh\nabcdefghi\nabcdefghij\nabcdefghijk\nabcdefghijklm", "output": "abcdefgh\nabcdefghi\nabcdefghij\na9k\na11m" }, { "input": "3\nnjfngnrurunrgunrunvurn\njfvnjfdnvjdbfvsbdubruvbubvkdb\nksdnvidnviudbvibd", "output": "n20n\nj27b\nk15d" }, { "input": "1\ntcyctkktcctrcyvbyiuhihhhgyvyvyvyvjvytchjckt", "output": "t41t" }, { "input": "24\nyou\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nunofficially\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings", "output": "you\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nu10y\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings" }, { "input": "1\na", "output": "a" }, { "input": "26\na\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz", "output": "a\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz" }, { "input": "1\nabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghij", "output": "a98j" }, { "input": "10\ngyartjdxxlcl\nfzsck\nuidwu\nxbymclornemdmtj\nilppyoapitawgje\ncibzc\ndrgbeu\nhezplmsdekhhbo\nfeuzlrimbqbytdu\nkgdco", "output": "g10l\nfzsck\nuidwu\nx13j\ni13e\ncibzc\ndrgbeu\nh12o\nf13u\nkgdco" }, { "input": "20\nlkpmx\nkovxmxorlgwaomlswjxlpnbvltfv\nhykasjxqyjrmybejnmeumzha\ntuevlumpqbbhbww\nqgqsphvrmupxxc\ntrissbaf\nqfgrlinkzvzqdryckaizutd\nzzqtoaxkvwoscyx\noswytrlnhpjvvnwookx\nlpuzqgec\ngyzqfwxggtvpjhzmzmdw\nrlxjgmvdftvrmvbdwudra\nvsntnjpepnvdaxiporggmglhagv\nxlvcqkqgcrbgtgglj\nlyxwxbiszyhlsrgzeedzprbmcpduvq\nyrmqqvrkqskqukzqrwukpsifgtdc\nxpuohcsjhhuhvr\nvvlfrlxpvqejngwrbfbpmqeirxlw\nsvmasocxdvadmaxtrpakysmeaympy\nyuflqboqfdt", "output": "lkpmx\nk26v\nh22a\nt13w\nq12c\ntrissbaf\nq21d\nz13x\no17x\nlpuzqgec\ng18w\nr19a\nv25v\nx15j\nl28q\ny26c\nx12r\nv26w\ns27y\ny9t" }, { "input": "100\nm\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr", "output": "m\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr" } ]
1,688,709,057
2,147,483,647
Python 3
OK
TESTS
20
46
0
n = int(input()) output = "" for x in range(0, n): w = input() l = len(w) if(l > 10): output = output + w[0] + "%i" % (l-2) + w[l-1] + "\n" else: output = output + w + "\n" print(output)
Title: Way Too Long Words Time Limit: 1 seconds Memory Limit: 256 megabytes Problem Description: Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome. Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation. This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes. Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n". You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes. Input Specification: The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters. Output Specification: Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data. Demo Input: ['4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n'] Demo Output: ['word\nl10n\ni18n\np43s\n'] Note: none
```python n = int(input()) output = "" for x in range(0, n): w = input() l = len(w) if(l > 10): output = output + w[0] + "%i" % (l-2) + w[l-1] + "\n" else: output = output + w + "\n" print(output) ```
3.977
37
A
Towers
PROGRAMMING
1,000
[ "sortings" ]
A. Towers
2
256
Little Vasya has received a young builder’s 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<=≤<=*N*<=≤<=1000) — the number of bars at Vasya’s disposal. The second line contains *N* space-separated integers *l**i* — the lengths of the bars. All the lengths are natural numbers not exceeding 1000.
In one line output two numbers — 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" } ]
1,434,527,080
2,147,483,647
Python 3
OK
TESTS
61
124
409,600
from collections import Counter input() c = Counter(int(i) for i in input().split()) print(max(c.values()), len(c))
Title: Towers Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Little Vasya has received a young builder’s 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<=≤<=*N*<=≤<=1000) — the number of bars at Vasya’s disposal. The second line contains *N* space-separated integers *l**i* — the lengths of the bars. All the lengths are natural numbers not exceeding 1000. Output Specification: In one line output two numbers — 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 from collections import Counter input() c = Counter(int(i) for i in input().split()) print(max(c.values()), len(c)) ```
3.968237
1
A
Theatre Square
PROGRAMMING
1,000
[ "math" ]
A. Theatre Square
1
256
Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=×<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=×<=*a*. What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square.
The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109).
Write the needed number of flagstones.
[ "6 6 4\n" ]
[ "4\n" ]
none
0
[ { "input": "6 6 4", "output": "4" }, { "input": "1 1 1", "output": "1" }, { "input": "2 1 1", "output": "2" }, { "input": "1 2 1", "output": "2" }, { "input": "2 2 1", "output": "4" }, { "input": "2 1 2", "output": "1" }, { "input": "1 1 3", "output": "1" }, { "input": "2 3 4", "output": "1" }, { "input": "1000000000 1000000000 1", "output": "1000000000000000000" }, { "input": "12 13 4", "output": "12" }, { "input": "222 332 5", "output": "3015" }, { "input": "1000 1000 10", "output": "10000" }, { "input": "1001 1000 10", "output": "10100" }, { "input": "100 10001 1000000000", "output": "1" }, { "input": "1000000000 1000000000 1000000000", "output": "1" }, { "input": "1000000000 1000000000 999999999", "output": "4" }, { "input": "1000000000 1000000000 192", "output": "27126743055556" }, { "input": "1000000000 987654321 1", "output": "987654321000000000" }, { "input": "456784567 1000000000 51", "output": "175618850864484" }, { "input": "39916800 134217728 40320", "output": "3295710" } ]
1,692,974,886
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
1
15
0
(n,m,a) = map(int,list(input().split(' '))) def cau(x,a): if x%a ==0: ans = x/a else: ans = int(x/a) + 1 return ans ans_n = cau(n,a) ans_m = cau(m,a) print(ans_n*ans_m)
Title: Theatre Square Time Limit: 1 seconds Memory Limit: 256 megabytes Problem Description: Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=×<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=×<=*a*. What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square. Input Specification: The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109). Output Specification: Write the needed number of flagstones. Demo Input: ['6 6 4\n'] Demo Output: ['4\n'] Note: none
```python (n,m,a) = map(int,list(input().split(' '))) def cau(x,a): if x%a ==0: ans = x/a else: ans = int(x/a) + 1 return ans ans_n = cau(n,a) ans_m = cau(m,a) print(ans_n*ans_m) ```
0
509
A
Maximum in Table
PROGRAMMING
800
[ "brute force", "implementation" ]
null
null
An *n*<=×<=*n* table *a* is defined as follows: - The first row and the first column contain ones, that is: *a**i*,<=1<==<=*a*1,<=*i*<==<=1 for all *i*<==<=1,<=2,<=...,<=*n*. - Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula *a**i*,<=*j*<==<=*a**i*<=-<=1,<=*j*<=+<=*a**i*,<=*j*<=-<=1. These conditions define all the values in the table. You are given a number *n*. You need to determine the maximum value in the *n*<=×<=*n* table defined by the rules above.
The only line of input contains a positive integer *n* (1<=≤<=*n*<=≤<=10) — the number of rows and columns of the table.
Print a single line containing a positive integer *m* — the maximum value in the table.
[ "1\n", "5\n" ]
[ "1", "70" ]
In the second test the rows of the table look as follows:
0
[ { "input": "1", "output": "1" }, { "input": "5", "output": "70" }, { "input": "2", "output": "2" }, { "input": "3", "output": "6" }, { "input": "4", "output": "20" }, { "input": "6", "output": "252" }, { "input": "7", "output": "924" }, { "input": "9", "output": "12870" }, { "input": "10", "output": "48620" }, { "input": "8", "output": "3432" } ]
1,652,965,516
2,147,483,647
PyPy 3-64
OK
TESTS
10
93
37,376,000
def answer(r,c): if(r==1 or c==1): return 1 return answer(r-1,c)+answer(r,c-1) n=int(input()) print(answer(n,n))
Title: Maximum in Table Time Limit: None seconds Memory Limit: None megabytes Problem Description: An *n*<=×<=*n* table *a* is defined as follows: - The first row and the first column contain ones, that is: *a**i*,<=1<==<=*a*1,<=*i*<==<=1 for all *i*<==<=1,<=2,<=...,<=*n*. - Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula *a**i*,<=*j*<==<=*a**i*<=-<=1,<=*j*<=+<=*a**i*,<=*j*<=-<=1. These conditions define all the values in the table. You are given a number *n*. You need to determine the maximum value in the *n*<=×<=*n* table defined by the rules above. Input Specification: The only line of input contains a positive integer *n* (1<=≤<=*n*<=≤<=10) — the number of rows and columns of the table. Output Specification: Print a single line containing a positive integer *m* — the maximum value in the table. Demo Input: ['1\n', '5\n'] Demo Output: ['1', '70'] Note: In the second test the rows of the table look as follows:
```python def answer(r,c): if(r==1 or c==1): return 1 return answer(r-1,c)+answer(r,c-1) n=int(input()) print(answer(n,n)) ```
3
381
A
Sereja and Dima
PROGRAMMING
800
[ "greedy", "implementation", "two pointers" ]
null
null
Sereja and Dima play a game. The rules of the game are very simple. The players have *n* cards in a row. Each card contains a number, all numbers on the cards are distinct. The players take turns, Sereja moves first. During his turn a player can take one card: either the leftmost card in a row, or the rightmost one. The game ends when there is no more cards. The player who has the maximum sum of numbers on his cards by the end of the game, wins. Sereja and Dima are being greedy. Each of them chooses the card with the larger number during his move. Inna is a friend of Sereja and Dima. She knows which strategy the guys are using, so she wants to determine the final score, given the initial state of the game. Help her.
The first line contains integer *n* (1<=≤<=*n*<=≤<=1000) — the number of cards on the table. The second line contains space-separated numbers on the cards from left to right. The numbers on the cards are distinct integers from 1 to 1000.
On a single line, print two integers. The first number is the number of Sereja's points at the end of the game, the second number is the number of Dima's points at the end of the game.
[ "4\n4 1 2 10\n", "7\n1 2 3 4 5 6 7\n" ]
[ "12 5\n", "16 12\n" ]
In the first sample Sereja will take cards with numbers 10 and 2, so Sereja's sum is 12. Dima will take cards with numbers 4 and 1, so Dima's sum is 5.
500
[ { "input": "4\n4 1 2 10", "output": "12 5" }, { "input": "7\n1 2 3 4 5 6 7", "output": "16 12" }, { "input": "42\n15 29 37 22 16 5 26 31 6 32 19 3 45 36 33 14 25 20 48 7 42 11 24 28 9 18 8 21 47 17 38 40 44 4 35 1 43 39 41 27 12 13", "output": "613 418" }, { "input": "43\n32 1 15 48 38 26 25 14 20 44 11 30 3 42 49 19 18 46 5 45 10 23 34 9 29 41 2 52 6 17 35 4 50 22 33 51 7 28 47 13 39 37 24", "output": "644 500" }, { "input": "1\n3", "output": "3 0" }, { "input": "45\n553 40 94 225 415 471 126 190 647 394 515 303 189 159 308 6 139 132 326 78 455 75 85 295 135 613 360 614 351 228 578 259 258 591 444 29 33 463 561 174 368 183 140 168 646", "output": "6848 6568" }, { "input": "44\n849 373 112 307 479 608 856 769 526 82 168 143 573 762 115 501 688 36 214 450 396 496 236 309 287 786 397 43 811 141 745 846 350 270 276 677 420 459 403 722 267 54 394 727", "output": "9562 9561" }, { "input": "35\n10 15 18 1 28 16 2 33 6 22 23 4 9 25 35 8 7 26 3 20 30 14 31 19 27 32 11 5 29 24 21 34 13 17 12", "output": "315 315" }, { "input": "17\n580 376 191 496 73 44 520 357 483 149 81 178 514 300 216 598 304", "output": "3238 2222" }, { "input": "30\n334 443 223 424 168 549 189 303 429 559 516 220 459 134 344 346 316 446 209 148 487 526 69 286 102 366 518 280 392 325", "output": "5246 4864" }, { "input": "95\n122 29 188 265 292 287 183 225 222 187 155 256 64 148 173 278 218 136 290 17 31 130 2 87 57 283 255 280 68 166 174 142 102 39 116 206 288 154 26 78 296 172 184 232 77 91 277 8 249 186 94 93 207 251 257 195 101 299 193 124 293 65 58 35 24 302 220 189 252 125 27 284 247 182 141 103 198 97 234 83 281 216 85 180 267 236 109 143 149 239 79 300 191 244 71", "output": "8147 7807" }, { "input": "1\n1", "output": "1 0" } ]
1,671,684,203
2,147,483,647
Python 3
RUNTIME_ERROR
TESTS
0
46
0
li=[] sum=0 sum_1=0 n=int(input()) for i in range(0,n): li.append(int(input())) for i in range(0,n): if i%2==0: sum= sum+max(li) li.remove(max(li)) else: sum_1=sum_1+max(li) li.remove(max(li)) print(sum,sum_1)
Title: Sereja and Dima Time Limit: None seconds Memory Limit: None megabytes Problem Description: Sereja and Dima play a game. The rules of the game are very simple. The players have *n* cards in a row. Each card contains a number, all numbers on the cards are distinct. The players take turns, Sereja moves first. During his turn a player can take one card: either the leftmost card in a row, or the rightmost one. The game ends when there is no more cards. The player who has the maximum sum of numbers on his cards by the end of the game, wins. Sereja and Dima are being greedy. Each of them chooses the card with the larger number during his move. Inna is a friend of Sereja and Dima. She knows which strategy the guys are using, so she wants to determine the final score, given the initial state of the game. Help her. Input Specification: The first line contains integer *n* (1<=≤<=*n*<=≤<=1000) — the number of cards on the table. The second line contains space-separated numbers on the cards from left to right. The numbers on the cards are distinct integers from 1 to 1000. Output Specification: On a single line, print two integers. The first number is the number of Sereja's points at the end of the game, the second number is the number of Dima's points at the end of the game. Demo Input: ['4\n4 1 2 10\n', '7\n1 2 3 4 5 6 7\n'] Demo Output: ['12 5\n', '16 12\n'] Note: In the first sample Sereja will take cards with numbers 10 and 2, so Sereja's sum is 12. Dima will take cards with numbers 4 and 1, so Dima's sum is 5.
```python li=[] sum=0 sum_1=0 n=int(input()) for i in range(0,n): li.append(int(input())) for i in range(0,n): if i%2==0: sum= sum+max(li) li.remove(max(li)) else: sum_1=sum_1+max(li) li.remove(max(li)) print(sum,sum_1) ```
-1
550
B
Preparing Olympiad
PROGRAMMING
1,400
[ "bitmasks", "brute force" ]
null
null
You have *n* problems. You have estimated the difficulty of the *i*-th one as integer *c**i*. Now you want to prepare a problemset for a contest, using some of the problems you've made. A problemset for the contest must consist of at least two problems. You think that the total difficulty of the problems of the contest must be at least *l* and at most *r*. Also, you think that the difference between difficulties of the easiest and the hardest of the chosen problems must be at least *x*. Find the number of ways to choose a problemset for the contest.
The first line contains four integers *n*, *l*, *r*, *x* (1<=≤<=*n*<=≤<=15, 1<=≤<=*l*<=≤<=*r*<=≤<=109, 1<=≤<=*x*<=≤<=106) — the number of problems you have, the minimum and maximum value of total difficulty of the problemset and the minimum difference in difficulty between the hardest problem in the pack and the easiest one, respectively. The second line contains *n* integers *c*1,<=*c*2,<=...,<=*c**n* (1<=≤<=*c**i*<=≤<=106) — the difficulty of each problem.
Print the number of ways to choose a suitable problemset for the contest.
[ "3 5 6 1\n1 2 3\n", "4 40 50 10\n10 20 30 25\n", "5 25 35 10\n10 10 20 10 20\n" ]
[ "2\n", "2\n", "6\n" ]
In the first example two sets are suitable, one consisting of the second and third problem, another one consisting of all three problems. In the second example, two sets of problems are suitable — the set of problems with difficulties 10 and 30 as well as the set of problems with difficulties 20 and 30. In the third example any set consisting of one problem of difficulty 10 and one problem of difficulty 20 is suitable.
750
[ { "input": "3 5 6 1\n1 2 3", "output": "2" }, { "input": "4 40 50 10\n10 20 30 25", "output": "2" }, { "input": "5 25 35 10\n10 10 20 10 20", "output": "6" }, { "input": "4 15 60 10\n10 20 30 25", "output": "6" }, { "input": "1 10 20 1\n15", "output": "0" }, { "input": "10 626451 11471247 246428\n369649 684428 303821 287098 422756 301599 720377 177567 515216 750602", "output": "914" }, { "input": "15 1415849 15540979 356865\n8352 960238 276753 259695 712845 945369 60023 920446 181269 392011 318488 857649 30681 740872 115749", "output": "31485" }, { "input": "7 1000 2000 1\n10 20 30 40 50 60 70", "output": "0" }, { "input": "4 10 20 1\n4 6 4 6", "output": "9" }, { "input": "4 10 20 1\n5 15 13 7", "output": "4" }, { "input": "2 10 20 5\n5 10", "output": "1" }, { "input": "5 1098816 3969849 167639\n85627 615007 794045 530104 7091", "output": "15" }, { "input": "13 700147 8713522 390093\n996812 94040 954140 545670 369698 423872 365802 784830 700267 960664 949252 84637 257447", "output": "8026" }, { "input": "15 4531977 20754263 137419\n637830 85299 755530 64382 896833 879525 331501 148182 741013 192101 112217 52165 702790 988594 587499", "output": "6759" }, { "input": "15 2572491 5084070 823435\n570344 78552 775918 501843 844935 71141 331498 636557 435494 715447 992666 831188 28969 171046 989614", "output": "15078" }, { "input": "15 4789415 23152928 233992\n502422 273992 449428 947379 700461 681985 857134 243310 478052 77769 936151 642380 464695 281772 964693", "output": "10875" }, { "input": "3 390224 390224 1\n264237 125987 288891", "output": "1" }, { "input": "7 1652707 1652707 1\n492387 684636 235422 332532 924898 499872 192988", "output": "1" }, { "input": "10 501107 501107 1\n843967 30518 196518 619138 204862 690754 274071 550121 173607 359971", "output": "1" }, { "input": "15 6627289 6627289 1\n683844 183950 184972 764255 211665 842336 790234 815301 914823 513046 93547 713159 554415 200951 388028", "output": "1" }, { "input": "15 5083470 5083470 1\n978510 643688 591921 723137 573784 346171 920030 352119 528857 365128 627302 308557 716247 263519 654230", "output": "1" }, { "input": "15 6558665 6558665 1\n572491 435494 916457 775918 823435 78552 501843 331498 71141 844935 636557 992666 570344 831188 715447", "output": "1" }, { "input": "10 159699 10967276 3542\n998862 999751 995306 992648 992661 991407 997503 998809 999740 997669", "output": "942" }, { "input": "5 2815840 8479687 4082\n991137 992161 997887 998891 994990", "output": "14" }, { "input": "15 2898377 6694755 721\n992733 999159 990076 996808 990975 993338 993234 994757 997873 993303 994409 993801 998027 990495 999287", "output": "9819" }, { "input": "6 20 70 1\n10 10 20 20 30 30", "output": "35" }, { "input": "6 20 70 1\n10 10 10 10 10 10", "output": "0" }, { "input": "15 1 1000000000 1\n10 20 30 40 50 60 70 80 90 100 110 120 130 140 150", "output": "32752" }, { "input": "6 30 40 1\n19 20 21 14 15 16", "output": "13" }, { "input": "4 5 234 2\n10 9 12 11", "output": "8" } ]
1,671,972,358
2,147,483,647
PyPy 3-64
OK
TESTS
30
93
3,584,000
import sys from itertools import chain, combinations def get_ints(): return map(int, sys.stdin.readline().strip().split()) def powerset(iterable): s = list(iterable) return chain.from_iterable(combinations(s, r) for r in range(len(s)+1)) n, l, r, x = get_ints() c = list(get_ints()) def isValid(list, l, r, x): intervalSum = sum(list) if l <= intervalSum <= r and max(list) - min(list) >= x: return True return False ans = 0 for list in powerset(c): if isValid(list, l, r, x): ans += 1 print(ans)
Title: Preparing Olympiad Time Limit: None seconds Memory Limit: None megabytes Problem Description: You have *n* problems. You have estimated the difficulty of the *i*-th one as integer *c**i*. Now you want to prepare a problemset for a contest, using some of the problems you've made. A problemset for the contest must consist of at least two problems. You think that the total difficulty of the problems of the contest must be at least *l* and at most *r*. Also, you think that the difference between difficulties of the easiest and the hardest of the chosen problems must be at least *x*. Find the number of ways to choose a problemset for the contest. Input Specification: The first line contains four integers *n*, *l*, *r*, *x* (1<=≤<=*n*<=≤<=15, 1<=≤<=*l*<=≤<=*r*<=≤<=109, 1<=≤<=*x*<=≤<=106) — the number of problems you have, the minimum and maximum value of total difficulty of the problemset and the minimum difference in difficulty between the hardest problem in the pack and the easiest one, respectively. The second line contains *n* integers *c*1,<=*c*2,<=...,<=*c**n* (1<=≤<=*c**i*<=≤<=106) — the difficulty of each problem. Output Specification: Print the number of ways to choose a suitable problemset for the contest. Demo Input: ['3 5 6 1\n1 2 3\n', '4 40 50 10\n10 20 30 25\n', '5 25 35 10\n10 10 20 10 20\n'] Demo Output: ['2\n', '2\n', '6\n'] Note: In the first example two sets are suitable, one consisting of the second and third problem, another one consisting of all three problems. In the second example, two sets of problems are suitable — the set of problems with difficulties 10 and 30 as well as the set of problems with difficulties 20 and 30. In the third example any set consisting of one problem of difficulty 10 and one problem of difficulty 20 is suitable.
```python import sys from itertools import chain, combinations def get_ints(): return map(int, sys.stdin.readline().strip().split()) def powerset(iterable): s = list(iterable) return chain.from_iterable(combinations(s, r) for r in range(len(s)+1)) n, l, r, x = get_ints() c = list(get_ints()) def isValid(list, l, r, x): intervalSum = sum(list) if l <= intervalSum <= r and max(list) - min(list) >= x: return True return False ans = 0 for list in powerset(c): if isValid(list, l, r, x): ans += 1 print(ans) ```
3
778
A
String Game
PROGRAMMING
1,700
[ "binary search", "greedy", "strings" ]
null
null
Little Nastya has a hobby, she likes to remove some letters from word, to obtain another word. But it turns out to be pretty hard for her, because she is too young. Therefore, her brother Sergey always helps her. Sergey gives Nastya the word *t* and wants to get the word *p* out of it. Nastya removes letters in a certain order (one after another, in this order strictly), which is specified by permutation of letters' indices of the word *t*: *a*1... *a*|*t*|. We denote the length of word *x* as |*x*|. Note that after removing one letter, the indices of other letters don't change. For example, if *t*<==<="nastya" and *a*<==<=[4,<=1,<=5,<=3,<=2,<=6] then removals make the following sequence of words "nastya" "nastya" "nastya" "nastya" "nastya" "nastya" "nastya". Sergey knows this permutation. His goal is to stop his sister at some point and continue removing by himself to get the word *p*. Since Nastya likes this activity, Sergey wants to stop her as late as possible. Your task is to determine, how many letters Nastya can remove before she will be stopped by Sergey. It is guaranteed that the word *p* can be obtained by removing the letters from word *t*.
The first and second lines of the input contain the words *t* and *p*, respectively. Words are composed of lowercase letters of the Latin alphabet (1<=≤<=|*p*|<=&lt;<=|*t*|<=≤<=200<=000). It is guaranteed that the word *p* can be obtained by removing the letters from word *t*. Next line contains a permutation *a*1,<=*a*2,<=...,<=*a*|*t*| of letter indices that specifies the order in which Nastya removes letters of *t* (1<=≤<=*a**i*<=≤<=|*t*|, all *a**i* are distinct).
Print a single integer number, the maximum number of letters that Nastya can remove.
[ "ababcba\nabb\n5 3 4 1 7 6 2\n", "bbbabb\nbb\n1 6 3 4 2 5\n" ]
[ "3", "4" ]
In the first sample test sequence of removing made by Nastya looks like this: "ababcba" <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> "ababcba" <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> "ababcba" <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> "ababcba" Nastya can not continue, because it is impossible to get word "abb" from word "ababcba". So, Nastya will remove only three letters.
500
[ { "input": "ababcba\nabb\n5 3 4 1 7 6 2", "output": "3" }, { "input": "bbbabb\nbb\n1 6 3 4 2 5", "output": "4" }, { "input": "cacaccccccacccc\ncacc\n10 9 14 5 1 7 15 3 6 12 4 8 11 13 2", "output": "9" }, { "input": "aaaabaaabaabaaaaaaaa\naaaa\n18 5 4 6 13 9 1 3 7 8 16 10 12 19 17 15 14 11 20 2", "output": "16" }, { "input": "aaaaaaaadbaaabbbbbddaaabdadbbbbbdbbabbbabaabdbbdababbbddddbdaabbddbbbbabbbbbabadaadabaaaadbbabbbaddb\naaaaaaaaaaaaaa\n61 52 5 43 53 81 7 96 6 9 34 78 79 12 8 63 22 76 18 46 41 56 3 20 57 21 75 73 100 94 35 69 32 4 70 95 88 44 68 10 71 98 23 89 36 62 28 51 24 30 74 55 27 80 38 48 93 1 19 84 13 11 86 60 87 33 39 29 83 91 67 72 54 2 17 85 82 14 15 90 64 50 99 26 66 65 31 49 40 45 77 37 25 42 97 47 58 92 59 16", "output": "57" } ]
1,641,895,924
2,147,483,647
PyPy 3-64
RUNTIME_ERROR
TESTS
10
763
133,017,600
class DSU: def __init__(self, size): self.p = [i for i in range(size)] self.rang = [0 for j in range(size)] def root(self, x): if self.p[x] == x: return x self.p[x] = DSU.root(self, self.p[x]) return self.p[x] def check(self, x1, x2): return DSU.root(self, x1) == DSU.root(self, x2) def join(self, x1, x2): x1 = DSU.root(self, x1) x2 = DSU.root(self, x2) self.p[max(x1, x2)] = min(x2, x1) # n, m = map(int, input().split()) # dsu = DSU(n) # flag = False # for i in range(m): # x1, y1, x2, y2 = [i - 1 for i in map(int, input().split())] # dsu.join(x1 * n + y1, x2 * n + y2) # for j in range(n * (n - 1), n ** 2): # print(f'{j} - {dsu.root(j)}') # if dsu.root(j) < n: # print(i + 1) # flag = True # break # if flag: # break # for j in range(0, n ** 2, n): # print(*dsu.p[j: j + n]) # print() # else: # print(-1) # def check(a, w, l): # zero_count = 0 # for i in range(len(l)): # if not l[i]: # zero_count += 1 # else: # if (zero_count + a) % (r_b - l_b + 1) >= w: # print(l[i - zero_count + 1: (i - zero_count + 1 + a) % (r_b - l_b + 1)]) # return True # zero_count = 0 # zero_count = 0 # for i in range(len(l) - 1, -1, -1): # if not l[i]: # zero_count += 1 # else: # if (zero_count + a) % (r_b - l_b + 1) >= w: # return True # zero_count = 0 # return False # # # def binary_search(w, walls, check): # l, r = -1, r_b - l_b + 1 # while r - l > 1: # m = (l + r) // 2 # if check(m, w, fence): # r = m # else: # l = m # print(l, r) # # return l if check(l, w, fence) and l >= 0 else r if check(r, w, fence) else -1 class min_queue(list): def push(self, a): while self and self[-1] > a: self.pop() self.append(a) def pop_left(self, removed_el): if self and self[0] == removed_el: self.pop(0) def DFS(v:int, graph:list, used:list, tiout:list): global t used[v] = True t += 1 tiout[v][0] = t for e in graph[v]: if not used[e]: DFS(e, graph, used, tiout) t += 1 tiout[v][1] = t # # def remove_same(st, l): # n = 0 # for i in l[::-1]: # if i == st[-1][0]: # l.pop() # n += 1 # print(l, n) # # else: # break # if st[-1][1] + n <= 1: # return # p, ti = st[-1][0] + st[-1][1] + n - 1, st[-1][1] # # for j in range(ti): # st.pop() # if st and p != st[-1][0]: # st.append([p, 1]) # else: # st.append([p, st[-1][1] + 1 if st else 1]) # remove_same(st, l) # # # n, l = int(input()), list(map(int, input().split())) # ind, k = map(int, input().split()) # l.insert(ind, k) # print(l) # st = [] # res = 0 # for i in l: # if not st: # st.append([i, 1]) # else: # if i == st[-1][0]: # st.append([i, st[-1][1] + 1]) # else: # remove_same(st, l) # if st and i != st[-1][0]: # st.append([i, 1]) # else: # st.append([i, st[-1][1] + 1 if st else 1]) # print(st) # # print(st, res) # # if st and st[-1][1] >= 3: # res += st[-1][1] # print(res) from collections import deque def BFS(v, graph, used): q = deque() q.append(v) used[v] = True while q: next = q.popleft() used[next] = True for u in graph[next]: if not used[u]: q.append(u) used[u] = True # def Dejcstra(s, graph:list, used:list, dist:list): # dist[s] = 0 # for i in range(n): # mi_v = min([i for i in range(len(dist)) if not used[i]], key=lambda i: dist[i]) # print(mi_v) # used[mi_v] = True # for e in graph[mi_v]: # to, length = e[0], e[1] # dist[to] = min(dist[to], dist[mi_v] + length # if l < mi: # mi = l # l_p = c # fll = True # if l == mi: # if # mi = l # fll = True # # if r > ma: # ma = r # r_p = c # flr = True # if r == ma: # r_p = min(c, r_p) # flr = True def issub(s1, s2): ''' :param s1: большая строка :param s2: возможноя подпоследовательность :return: ''' if not s2: return True if not s1: return False if s2[-1] == s1[-1]: return issub(s1[:-1], s2[:-1]) return issub(s1[:-1], s2) def check(s1, s2, pos, t): if len(s1) - t < len(s2): return False ls1 = list(s1) for i in pos[:t]: ls1[i - 1] = '.' s = ''.join([i for i in ls1 if i != '.']) # print(s2, s, issub(s, s2)) return s2 in s or issub(s, s2) def bin_search(s1, s2, pos): l, r = 0, len(s1) while l + 1 < r: mid = (l + r) // 2 if check(s1, s2, pos, mid): l = mid else: r = mid # print(l, r) return r if check(s1, s2, pos, r) else l s1, s2 = input(), input() poses = list(map(int, input().split())) print(bin_search(s1, s2, poses))
Title: String Game Time Limit: None seconds Memory Limit: None megabytes Problem Description: Little Nastya has a hobby, she likes to remove some letters from word, to obtain another word. But it turns out to be pretty hard for her, because she is too young. Therefore, her brother Sergey always helps her. Sergey gives Nastya the word *t* and wants to get the word *p* out of it. Nastya removes letters in a certain order (one after another, in this order strictly), which is specified by permutation of letters' indices of the word *t*: *a*1... *a*|*t*|. We denote the length of word *x* as |*x*|. Note that after removing one letter, the indices of other letters don't change. For example, if *t*<==<="nastya" and *a*<==<=[4,<=1,<=5,<=3,<=2,<=6] then removals make the following sequence of words "nastya" "nastya" "nastya" "nastya" "nastya" "nastya" "nastya". Sergey knows this permutation. His goal is to stop his sister at some point and continue removing by himself to get the word *p*. Since Nastya likes this activity, Sergey wants to stop her as late as possible. Your task is to determine, how many letters Nastya can remove before she will be stopped by Sergey. It is guaranteed that the word *p* can be obtained by removing the letters from word *t*. Input Specification: The first and second lines of the input contain the words *t* and *p*, respectively. Words are composed of lowercase letters of the Latin alphabet (1<=≤<=|*p*|<=&lt;<=|*t*|<=≤<=200<=000). It is guaranteed that the word *p* can be obtained by removing the letters from word *t*. Next line contains a permutation *a*1,<=*a*2,<=...,<=*a*|*t*| of letter indices that specifies the order in which Nastya removes letters of *t* (1<=≤<=*a**i*<=≤<=|*t*|, all *a**i* are distinct). Output Specification: Print a single integer number, the maximum number of letters that Nastya can remove. Demo Input: ['ababcba\nabb\n5 3 4 1 7 6 2\n', 'bbbabb\nbb\n1 6 3 4 2 5\n'] Demo Output: ['3', '4'] Note: In the first sample test sequence of removing made by Nastya looks like this: "ababcba" <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> "ababcba" <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> "ababcba" <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> "ababcba" Nastya can not continue, because it is impossible to get word "abb" from word "ababcba". So, Nastya will remove only three letters.
```python class DSU: def __init__(self, size): self.p = [i for i in range(size)] self.rang = [0 for j in range(size)] def root(self, x): if self.p[x] == x: return x self.p[x] = DSU.root(self, self.p[x]) return self.p[x] def check(self, x1, x2): return DSU.root(self, x1) == DSU.root(self, x2) def join(self, x1, x2): x1 = DSU.root(self, x1) x2 = DSU.root(self, x2) self.p[max(x1, x2)] = min(x2, x1) # n, m = map(int, input().split()) # dsu = DSU(n) # flag = False # for i in range(m): # x1, y1, x2, y2 = [i - 1 for i in map(int, input().split())] # dsu.join(x1 * n + y1, x2 * n + y2) # for j in range(n * (n - 1), n ** 2): # print(f'{j} - {dsu.root(j)}') # if dsu.root(j) < n: # print(i + 1) # flag = True # break # if flag: # break # for j in range(0, n ** 2, n): # print(*dsu.p[j: j + n]) # print() # else: # print(-1) # def check(a, w, l): # zero_count = 0 # for i in range(len(l)): # if not l[i]: # zero_count += 1 # else: # if (zero_count + a) % (r_b - l_b + 1) >= w: # print(l[i - zero_count + 1: (i - zero_count + 1 + a) % (r_b - l_b + 1)]) # return True # zero_count = 0 # zero_count = 0 # for i in range(len(l) - 1, -1, -1): # if not l[i]: # zero_count += 1 # else: # if (zero_count + a) % (r_b - l_b + 1) >= w: # return True # zero_count = 0 # return False # # # def binary_search(w, walls, check): # l, r = -1, r_b - l_b + 1 # while r - l > 1: # m = (l + r) // 2 # if check(m, w, fence): # r = m # else: # l = m # print(l, r) # # return l if check(l, w, fence) and l >= 0 else r if check(r, w, fence) else -1 class min_queue(list): def push(self, a): while self and self[-1] > a: self.pop() self.append(a) def pop_left(self, removed_el): if self and self[0] == removed_el: self.pop(0) def DFS(v:int, graph:list, used:list, tiout:list): global t used[v] = True t += 1 tiout[v][0] = t for e in graph[v]: if not used[e]: DFS(e, graph, used, tiout) t += 1 tiout[v][1] = t # # def remove_same(st, l): # n = 0 # for i in l[::-1]: # if i == st[-1][0]: # l.pop() # n += 1 # print(l, n) # # else: # break # if st[-1][1] + n <= 1: # return # p, ti = st[-1][0] + st[-1][1] + n - 1, st[-1][1] # # for j in range(ti): # st.pop() # if st and p != st[-1][0]: # st.append([p, 1]) # else: # st.append([p, st[-1][1] + 1 if st else 1]) # remove_same(st, l) # # # n, l = int(input()), list(map(int, input().split())) # ind, k = map(int, input().split()) # l.insert(ind, k) # print(l) # st = [] # res = 0 # for i in l: # if not st: # st.append([i, 1]) # else: # if i == st[-1][0]: # st.append([i, st[-1][1] + 1]) # else: # remove_same(st, l) # if st and i != st[-1][0]: # st.append([i, 1]) # else: # st.append([i, st[-1][1] + 1 if st else 1]) # print(st) # # print(st, res) # # if st and st[-1][1] >= 3: # res += st[-1][1] # print(res) from collections import deque def BFS(v, graph, used): q = deque() q.append(v) used[v] = True while q: next = q.popleft() used[next] = True for u in graph[next]: if not used[u]: q.append(u) used[u] = True # def Dejcstra(s, graph:list, used:list, dist:list): # dist[s] = 0 # for i in range(n): # mi_v = min([i for i in range(len(dist)) if not used[i]], key=lambda i: dist[i]) # print(mi_v) # used[mi_v] = True # for e in graph[mi_v]: # to, length = e[0], e[1] # dist[to] = min(dist[to], dist[mi_v] + length # if l < mi: # mi = l # l_p = c # fll = True # if l == mi: # if # mi = l # fll = True # # if r > ma: # ma = r # r_p = c # flr = True # if r == ma: # r_p = min(c, r_p) # flr = True def issub(s1, s2): ''' :param s1: большая строка :param s2: возможноя подпоследовательность :return: ''' if not s2: return True if not s1: return False if s2[-1] == s1[-1]: return issub(s1[:-1], s2[:-1]) return issub(s1[:-1], s2) def check(s1, s2, pos, t): if len(s1) - t < len(s2): return False ls1 = list(s1) for i in pos[:t]: ls1[i - 1] = '.' s = ''.join([i for i in ls1 if i != '.']) # print(s2, s, issub(s, s2)) return s2 in s or issub(s, s2) def bin_search(s1, s2, pos): l, r = 0, len(s1) while l + 1 < r: mid = (l + r) // 2 if check(s1, s2, pos, mid): l = mid else: r = mid # print(l, r) return r if check(s1, s2, pos, r) else l s1, s2 = input(), input() poses = list(map(int, input().split())) print(bin_search(s1, s2, poses)) ```
-1
78
E
Evacuation
PROGRAMMING
2,300
[ "flows", "graphs", "shortest paths" ]
E. Evacuation
1
256
They've screwed something up yet again... In one nuclear reactor of a research station an uncontrolled reaction is in progress and explosion which will destroy the whole station will happen soon. The station is represented by a square *n*<=×<=*n* divided into 1<=×<=1 blocks. Each block is either a reactor or a laboratory. There can be several reactors and exactly one of them will explode soon. The reactors can be considered impassable blocks, but one can move through laboratories. Between any two laboratories, which are in adjacent blocks, there is a corridor. Blocks are considered adjacent if they have a common edge. In each laboratory there is some number of scientists and some number of rescue capsules. Once the scientist climbs into a capsule, he is considered to be saved. Each capsule has room for not more than one scientist. The reactor, which is about to explode, is damaged and a toxic coolant trickles from it into the neighboring blocks. The block, which contains the reactor, is considered infected. Every minute the coolant spreads over the laboratories through corridors. If at some moment one of the blocks is infected, then the next minute all the neighboring laboratories also become infected. Once a lab is infected, all the scientists there that are not in rescue capsules die. The coolant does not spread through reactor blocks. There are exactly *t* minutes to the explosion. Any scientist in a minute can move down the corridor to the next lab, if it is not infected. On any corridor an unlimited number of scientists can simultaneously move in both directions. It is believed that the scientists inside a lab moves without consuming time. Moreover, any scientist could get into the rescue capsule instantly. It is also believed that any scientist at any given moment always has the time to perform their actions (move from the given laboratory into the next one, or climb into the rescue capsule) before the laboratory will be infected. Find the maximum number of scientists who will be able to escape.
The first line contains two integers *n* and *t* (2<=≤<=*n*<=≤<=10, 1<=≤<=*t*<=≤<=60). Each of the next *n* lines contains *n* characters. These lines describe the scientists' locations. Then exactly one empty line follows. Each of the next *n* more lines contains *n* characters. These lines describe the rescue capsules' locations. In the description of the scientists' and the rescue capsules' locations the character "Y" stands for a properly functioning reactor, "Z" stands for the malfunctioning reactor. The reactors' positions in both descriptions coincide. There is exactly one malfunctioning reactor on the station. The digits "0" - "9" stand for the laboratories. In the description of the scientists' locations those numbers stand for the number of scientists in the corresponding laboratories. In the rescue capsules' descriptions they stand for the number of such capsules in each laboratory.
Print a single number — the maximum number of scientists who will manage to save themselves.
[ "3 3\n1YZ\n1YY\n100\n\n0YZ\n0YY\n003\n", "4 4\nY110\n1Y1Z\n1Y0Y\n0100\n\nY001\n0Y0Z\n0Y0Y\n0005\n" ]
[ "2", "3" ]
In the second sample the events could take place as follows:
2,500
[]
1,690,909,804
2,147,483,647
PyPy 3-64
WRONG_ANSWER
TESTS
0
46
0
print("_RANDOM_GUESS_1690909804.325201")# 1690909804.32522
Title: Evacuation Time Limit: 1 seconds Memory Limit: 256 megabytes Problem Description: They've screwed something up yet again... In one nuclear reactor of a research station an uncontrolled reaction is in progress and explosion which will destroy the whole station will happen soon. The station is represented by a square *n*<=×<=*n* divided into 1<=×<=1 blocks. Each block is either a reactor or a laboratory. There can be several reactors and exactly one of them will explode soon. The reactors can be considered impassable blocks, but one can move through laboratories. Between any two laboratories, which are in adjacent blocks, there is a corridor. Blocks are considered adjacent if they have a common edge. In each laboratory there is some number of scientists and some number of rescue capsules. Once the scientist climbs into a capsule, he is considered to be saved. Each capsule has room for not more than one scientist. The reactor, which is about to explode, is damaged and a toxic coolant trickles from it into the neighboring blocks. The block, which contains the reactor, is considered infected. Every minute the coolant spreads over the laboratories through corridors. If at some moment one of the blocks is infected, then the next minute all the neighboring laboratories also become infected. Once a lab is infected, all the scientists there that are not in rescue capsules die. The coolant does not spread through reactor blocks. There are exactly *t* minutes to the explosion. Any scientist in a minute can move down the corridor to the next lab, if it is not infected. On any corridor an unlimited number of scientists can simultaneously move in both directions. It is believed that the scientists inside a lab moves without consuming time. Moreover, any scientist could get into the rescue capsule instantly. It is also believed that any scientist at any given moment always has the time to perform their actions (move from the given laboratory into the next one, or climb into the rescue capsule) before the laboratory will be infected. Find the maximum number of scientists who will be able to escape. Input Specification: The first line contains two integers *n* and *t* (2<=≤<=*n*<=≤<=10, 1<=≤<=*t*<=≤<=60). Each of the next *n* lines contains *n* characters. These lines describe the scientists' locations. Then exactly one empty line follows. Each of the next *n* more lines contains *n* characters. These lines describe the rescue capsules' locations. In the description of the scientists' and the rescue capsules' locations the character "Y" stands for a properly functioning reactor, "Z" stands for the malfunctioning reactor. The reactors' positions in both descriptions coincide. There is exactly one malfunctioning reactor on the station. The digits "0" - "9" stand for the laboratories. In the description of the scientists' locations those numbers stand for the number of scientists in the corresponding laboratories. In the rescue capsules' descriptions they stand for the number of such capsules in each laboratory. Output Specification: Print a single number — the maximum number of scientists who will manage to save themselves. Demo Input: ['3 3\n1YZ\n1YY\n100\n\n0YZ\n0YY\n003\n', '4 4\nY110\n1Y1Z\n1Y0Y\n0100\n\nY001\n0Y0Z\n0Y0Y\n0005\n'] Demo Output: ['2', '3'] Note: In the second sample the events could take place as follows:
```python print("_RANDOM_GUESS_1690909804.325201")# 1690909804.32522 ```
0
803
B
Distances to Zero
PROGRAMMING
1,200
[ "constructive algorithms" ]
null
null
You are given the array of integer numbers *a*0,<=*a*1,<=...,<=*a**n*<=-<=1. For each element find the distance to the nearest zero (to the element which equals to zero). There is at least one zero element in the given array.
The first line contains integer *n* (1<=≤<=*n*<=≤<=2·105) — length of the array *a*. The second line contains integer elements of the array separated by single spaces (<=-<=109<=≤<=*a**i*<=≤<=109).
Print the sequence *d*0,<=*d*1,<=...,<=*d**n*<=-<=1, where *d**i* is the difference of indices between *i* and nearest *j* such that *a**j*<==<=0. It is possible that *i*<==<=*j*.
[ "9\n2 1 0 3 0 0 3 2 4\n", "5\n0 1 2 3 4\n", "7\n5 6 0 1 -2 3 4\n" ]
[ "2 1 0 1 0 0 1 2 3 ", "0 1 2 3 4 ", "2 1 0 1 2 3 4 " ]
none
0
[ { "input": "9\n2 1 0 3 0 0 3 2 4", "output": "2 1 0 1 0 0 1 2 3 " }, { "input": "5\n0 1 2 3 4", "output": "0 1 2 3 4 " }, { "input": "7\n5 6 0 1 -2 3 4", "output": "2 1 0 1 2 3 4 " }, { "input": "1\n0", "output": "0 " }, { "input": "2\n0 0", "output": "0 0 " }, { "input": "2\n0 1", "output": "0 1 " }, { "input": "2\n1 0", "output": "1 0 " }, { "input": "5\n0 1000000000 1000000000 1000000000 1000000000", "output": "0 1 2 3 4 " }, { "input": "5\n-1000000000 -1000000000 0 1000000000 1000000000", "output": "2 1 0 1 2 " }, { "input": "5\n-1000000000 1000000000 1000000000 1000000000 0", "output": "4 3 2 1 0 " }, { "input": "15\n1000000000 -1000000000 -1000000000 1000000000 -1000000000 -1000000000 -1000000000 1000000000 1000000000 -1000000000 -1000000000 -1000000000 -1000000000 1000000000 0", "output": "14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 " }, { "input": "15\n0 0 0 0 1000000000 -1000000000 -1000000000 -1000000000 -1000000000 1000000000 1000000000 1000000000 -1000000000 -1000000000 1000000000", "output": "0 0 0 0 1 2 3 4 5 6 7 8 9 10 11 " }, { "input": "15\n-1000000000 1000000000 1000000000 -1000000000 -1000000000 1000000000 0 -1000000000 -1000000000 0 0 1000000000 -1000000000 0 -1000000000", "output": "6 5 4 3 2 1 0 1 1 0 0 1 1 0 1 " }, { "input": "15\n-1000000000 -1000000000 1000000000 1000000000 -1000000000 1000000000 1000000000 -1000000000 1000000000 1000000000 1000000000 0 0 0 0", "output": "11 10 9 8 7 6 5 4 3 2 1 0 0 0 0 " }, { "input": "4\n0 0 2 0", "output": "0 0 1 0 " }, { "input": "15\n1 2 3 4 0 1 2 3 -5 -4 -3 -1 0 5 4", "output": "4 3 2 1 0 1 2 3 4 3 2 1 0 1 2 " }, { "input": "2\n0 -1", "output": "0 1 " }, { "input": "5\n0 -1 -1 -1 0", "output": "0 1 2 1 0 " }, { "input": "5\n0 0 0 -1 0", "output": "0 0 0 1 0 " }, { "input": "3\n0 0 -1", "output": "0 0 1 " }, { "input": "3\n0 -1 -1", "output": "0 1 2 " }, { "input": "12\n0 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 0", "output": "0 1 2 3 4 5 5 4 3 2 1 0 " }, { "input": "18\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -1", "output": "0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 " }, { "input": "30\n0 0 0 0 0 0 0 0 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1", "output": "0 0 0 0 0 0 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 " }, { "input": "1\n0", "output": "0 " }, { "input": "1\n0", "output": "0 " }, { "input": "1\n0", "output": "0 " }, { "input": "2\n0 -1000000000", "output": "0 1 " }, { "input": "2\n0 1000000000", "output": "0 1 " }, { "input": "2\n-1000000000 0", "output": "1 0 " }, { "input": "2\n0 0", "output": "0 0 " }, { "input": "2\n0 0", "output": "0 0 " }, { "input": "2\n0 0", "output": "0 0 " }, { "input": "3\n0 -1000000000 -1000000000", "output": "0 1 2 " }, { "input": "3\n0 1000000000 1000000000", "output": "0 1 2 " }, { "input": "3\n1000000000 1000000000 0", "output": "2 1 0 " }, { "input": "3\n0 0 -1000000000", "output": "0 0 1 " }, { "input": "3\n0 1000000000 0", "output": "0 1 0 " }, { "input": "3\n-1000000000 0 0", "output": "1 0 0 " }, { "input": "3\n0 0 0", "output": "0 0 0 " }, { "input": "3\n0 0 0", "output": "0 0 0 " }, { "input": "3\n0 0 0", "output": "0 0 0 " }, { "input": "4\n0 -1000000000 -1000000000 -1000000000", "output": "0 1 2 3 " }, { "input": "4\n1000000000 -1000000000 0 -1000000000", "output": "2 1 0 1 " }, { "input": "4\n1000000000 -1000000000 1000000000 0", "output": "3 2 1 0 " }, { "input": "4\n0 0 -1000000000 1000000000", "output": "0 0 1 2 " }, { "input": "4\n0 0 1000000000 -1000000000", "output": "0 0 1 2 " }, { "input": "4\n-1000000000 1000000000 0 0", "output": "2 1 0 0 " }, { "input": "4\n0 0 0 -1000000000", "output": "0 0 0 1 " }, { "input": "4\n1000000000 0 0 0", "output": "1 0 0 0 " }, { "input": "4\n1000000000 0 0 0", "output": "1 0 0 0 " }, { "input": "4\n0 0 0 0", "output": "0 0 0 0 " }, { "input": "4\n0 0 0 0", "output": "0 0 0 0 " }, { "input": "4\n0 0 0 0", "output": "0 0 0 0 " }, { "input": "5\n0 1000000000 1000000000 1000000000 1000000000", "output": "0 1 2 3 4 " }, { "input": "5\n1000000000 -1000000000 -1000000000 1000000000 0", "output": "4 3 2 1 0 " }, { "input": "5\n1000000000 -1000000000 1000000000 -1000000000 0", "output": "4 3 2 1 0 " }, { "input": "5\n0 0 -1000000000 -1000000000 -1000000000", "output": "0 0 1 2 3 " }, { "input": "5\n1000000000 0 -1000000000 0 -1000000000", "output": "1 0 1 0 1 " }, { "input": "5\n1000000000 1000000000 1000000000 0 0", "output": "3 2 1 0 0 " }, { "input": "5\n0 0 0 -1000000000 -1000000000", "output": "0 0 0 1 2 " }, { "input": "5\n-1000000000 1000000000 0 0 0", "output": "2 1 0 0 0 " }, { "input": "5\n1000000000 1000000000 0 0 0", "output": "2 1 0 0 0 " }, { "input": "5\n0 0 0 0 -1000000000", "output": "0 0 0 0 1 " }, { "input": "5\n0 0 1000000000 0 0", "output": "0 0 1 0 0 " }, { "input": "5\n1000000000 0 0 0 0", "output": "1 0 0 0 0 " }, { "input": "5\n0 0 0 0 0", "output": "0 0 0 0 0 " }, { "input": "5\n0 0 0 0 0", "output": "0 0 0 0 0 " }, { "input": "5\n0 0 0 0 0", "output": "0 0 0 0 0 " }, { "input": "6\n0 1000000000 -1000000000 1000000000 -1000000000 1000000000", "output": "0 1 2 3 4 5 " }, { "input": "6\n-1000000000 -1000000000 1000000000 1000000000 1000000000 0", "output": "5 4 3 2 1 0 " }, { "input": "6\n-1000000000 1000000000 -1000000000 1000000000 -1000000000 0", "output": "5 4 3 2 1 0 " }, { "input": "6\n0 0 1000000000 1000000000 -1000000000 -1000000000", "output": "0 0 1 2 3 4 " }, { "input": "6\n0 0 1000000000 1000000000 -1000000000 -1000000000", "output": "0 0 1 2 3 4 " }, { "input": "6\n-1000000000 1000000000 -1000000000 -1000000000 0 0", "output": "4 3 2 1 0 0 " }, { "input": "6\n0 0 0 -1000000000 1000000000 1000000000", "output": "0 0 0 1 2 3 " }, { "input": "6\n-1000000000 1000000000 -1000000000 0 0 0", "output": "3 2 1 0 0 0 " }, { "input": "6\n-1000000000 -1000000000 1000000000 0 0 0", "output": "3 2 1 0 0 0 " }, { "input": "6\n0 0 0 0 -1000000000 1000000000", "output": "0 0 0 0 1 2 " }, { "input": "6\n0 0 0 -1000000000 1000000000 0", "output": "0 0 0 1 1 0 " }, { "input": "6\n1000000000 1000000000 0 0 0 0", "output": "2 1 0 0 0 0 " }, { "input": "6\n0 0 0 0 0 -1000000000", "output": "0 0 0 0 0 1 " }, { "input": "6\n0 0 0 1000000000 0 0", "output": "0 0 0 1 0 0 " }, { "input": "6\n1000000000 0 0 0 0 0", "output": "1 0 0 0 0 0 " }, { "input": "6\n0 0 0 0 0 0", "output": "0 0 0 0 0 0 " }, { "input": "6\n0 0 0 0 0 0", "output": "0 0 0 0 0 0 " }, { "input": "6\n0 0 0 0 0 0", "output": "0 0 0 0 0 0 " }, { "input": "7\n0 -1000000000 1000000000 -1000000000 -1000000000 -1000000000 -1000000000", "output": "0 1 2 3 4 5 6 " }, { "input": "7\n1000000000 1000000000 -1000000000 0 -1000000000 1000000000 -1000000000", "output": "3 2 1 0 1 2 3 " }, { "input": "7\n1000000000 1000000000 -1000000000 1000000000 -1000000000 -1000000000 0", "output": "6 5 4 3 2 1 0 " }, { "input": "7\n0 0 1000000000 1000000000 1000000000 1000000000 -1000000000", "output": "0 0 1 2 3 4 5 " }, { "input": "7\n0 1000000000 1000000000 -1000000000 1000000000 1000000000 0", "output": "0 1 2 3 2 1 0 " }, { "input": "7\n1000000000 -1000000000 -1000000000 1000000000 -1000000000 0 0", "output": "5 4 3 2 1 0 0 " }, { "input": "7\n0 0 0 1000000000 -1000000000 -1000000000 1000000000", "output": "0 0 0 1 2 3 4 " }, { "input": "7\n-1000000000 0 0 -1000000000 0 -1000000000 1000000000", "output": "1 0 0 1 0 1 2 " }, { "input": "7\n1000000000 1000000000 1000000000 -1000000000 0 0 0", "output": "4 3 2 1 0 0 0 " }, { "input": "7\n0 0 0 0 -1000000000 -1000000000 1000000000", "output": "0 0 0 0 1 2 3 " }, { "input": "7\n0 -1000000000 0 0 0 -1000000000 1000000000", "output": "0 1 0 0 0 1 2 " }, { "input": "7\n1000000000 1000000000 1000000000 0 0 0 0", "output": "3 2 1 0 0 0 0 " }, { "input": "7\n0 0 0 0 0 -1000000000 1000000000", "output": "0 0 0 0 0 1 2 " }, { "input": "7\n0 -1000000000 0 0 0 0 -1000000000", "output": "0 1 0 0 0 0 1 " }, { "input": "7\n-1000000000 1000000000 0 0 0 0 0", "output": "2 1 0 0 0 0 0 " }, { "input": "7\n0 0 0 0 0 0 -1000000000", "output": "0 0 0 0 0 0 1 " }, { "input": "7\n0 0 0 0 0 1000000000 0", "output": "0 0 0 0 0 1 0 " }, { "input": "7\n1000000000 0 0 0 0 0 0", "output": "1 0 0 0 0 0 0 " }, { "input": "7\n0 0 0 0 0 0 0", "output": "0 0 0 0 0 0 0 " }, { "input": "7\n0 0 0 0 0 0 0", "output": "0 0 0 0 0 0 0 " }, { "input": "7\n0 0 0 0 0 0 0", "output": "0 0 0 0 0 0 0 " }, { "input": "8\n0 -1000000000 -1000000000 1000000000 1000000000 1000000000 1000000000 -1000000000", "output": "0 1 2 3 4 5 6 7 " }, { "input": "8\n0 -1000000000 1000000000 1000000000 1000000000 -1000000000 1000000000 1000000000", "output": "0 1 2 3 4 5 6 7 " }, { "input": "8\n1000000000 -1000000000 -1000000000 -1000000000 1000000000 1000000000 1000000000 0", "output": "7 6 5 4 3 2 1 0 " }, { "input": "8\n0 0 -1000000000 -1000000000 1000000000 1000000000 1000000000 -1000000000", "output": "0 0 1 2 3 4 5 6 " }, { "input": "8\n1000000000 0 0 -1000000000 -1000000000 1000000000 -1000000000 -1000000000", "output": "1 0 0 1 2 3 4 5 " }, { "input": "8\n1000000000 -1000000000 1000000000 -1000000000 -1000000000 -1000000000 0 0", "output": "6 5 4 3 2 1 0 0 " }, { "input": "8\n0 0 0 1000000000 1000000000 -1000000000 -1000000000 -1000000000", "output": "0 0 0 1 2 3 4 5 " }, { "input": "8\n-1000000000 0 0 1000000000 1000000000 0 -1000000000 1000000000", "output": "1 0 0 1 1 0 1 2 " }, { "input": "8\n1000000000 1000000000 1000000000 -1000000000 -1000000000 0 0 0", "output": "5 4 3 2 1 0 0 0 " }, { "input": "8\n0 0 0 0 1000000000 1000000000 1000000000 -1000000000", "output": "0 0 0 0 1 2 3 4 " }, { "input": "8\n1000000000 0 1000000000 -1000000000 0 -1000000000 0 0", "output": "1 0 1 1 0 1 0 0 " }, { "input": "8\n-1000000000 -1000000000 -1000000000 -1000000000 0 0 0 0", "output": "4 3 2 1 0 0 0 0 " }, { "input": "8\n0 0 0 0 0 1000000000 1000000000 -1000000000", "output": "0 0 0 0 0 1 2 3 " }, { "input": "8\n-1000000000 0 -1000000000 0 0 1000000000 0 0", "output": "1 0 1 0 0 1 0 0 " }, { "input": "8\n1000000000 1000000000 1000000000 0 0 0 0 0", "output": "3 2 1 0 0 0 0 0 " }, { "input": "8\n0 0 0 0 0 0 -1000000000 -1000000000", "output": "0 0 0 0 0 0 1 2 " }, { "input": "8\n0 0 0 1000000000 -1000000000 0 0 0", "output": "0 0 0 1 1 0 0 0 " }, { "input": "8\n1000000000 1000000000 0 0 0 0 0 0", "output": "2 1 0 0 0 0 0 0 " }, { "input": "8\n0 0 0 0 0 0 0 -1000000000", "output": "0 0 0 0 0 0 0 1 " }, { "input": "8\n0 1000000000 0 0 0 0 0 0", "output": "0 1 0 0 0 0 0 0 " }, { "input": "8\n1000000000 0 0 0 0 0 0 0", "output": "1 0 0 0 0 0 0 0 " }, { "input": "8\n0 0 0 0 0 0 0 0", "output": "0 0 0 0 0 0 0 0 " }, { "input": "8\n0 0 0 0 0 0 0 0", "output": "0 0 0 0 0 0 0 0 " }, { "input": "8\n0 0 0 0 0 0 0 0", "output": "0 0 0 0 0 0 0 0 " } ]
1,493,393,619
1,719
Python 3
OK
TESTS
164
639
23,244,800
n = int(input()) l = list(map(int,input().split())) zeroes = [] for i in range(len(l)): if l[i]==0: zeroes.append(i) zero = -1 for i in range(n): zeroes.append(zeroes[-1]) s = "" for i in range(len(l)): if i > zeroes[zero+1]: zero += 1 if l[i]==0: s +=" 0" else: try: s += " "+str(min(abs(i-zeroes[zero]),abs(i-zeroes[zero+1]),abs(i-zeroes[zero+2]))) except: s += " "+str(min(abs(zeroes[zero+1]-i),abs(i-zeroes[zero]))) print(s[1:])
Title: Distances to Zero Time Limit: None seconds Memory Limit: None megabytes Problem Description: You are given the array of integer numbers *a*0,<=*a*1,<=...,<=*a**n*<=-<=1. For each element find the distance to the nearest zero (to the element which equals to zero). There is at least one zero element in the given array. Input Specification: The first line contains integer *n* (1<=≤<=*n*<=≤<=2·105) — length of the array *a*. The second line contains integer elements of the array separated by single spaces (<=-<=109<=≤<=*a**i*<=≤<=109). Output Specification: Print the sequence *d*0,<=*d*1,<=...,<=*d**n*<=-<=1, where *d**i* is the difference of indices between *i* and nearest *j* such that *a**j*<==<=0. It is possible that *i*<==<=*j*. Demo Input: ['9\n2 1 0 3 0 0 3 2 4\n', '5\n0 1 2 3 4\n', '7\n5 6 0 1 -2 3 4\n'] Demo Output: ['2 1 0 1 0 0 1 2 3 ', '0 1 2 3 4 ', '2 1 0 1 2 3 4 '] Note: none
```python n = int(input()) l = list(map(int,input().split())) zeroes = [] for i in range(len(l)): if l[i]==0: zeroes.append(i) zero = -1 for i in range(n): zeroes.append(zeroes[-1]) s = "" for i in range(len(l)): if i > zeroes[zero+1]: zero += 1 if l[i]==0: s +=" 0" else: try: s += " "+str(min(abs(i-zeroes[zero]),abs(i-zeroes[zero+1]),abs(i-zeroes[zero+2]))) except: s += " "+str(min(abs(zeroes[zero+1]-i),abs(i-zeroes[zero]))) print(s[1:]) ```
3
891
A
Pride
PROGRAMMING
1,500
[ "brute force", "dp", "greedy", "math", "number theory" ]
null
null
You have an array *a* with length *n*, you can perform operations. Each operation is like this: choose two adjacent elements from *a*, say *x* and *y*, and replace one of them with *gcd*(*x*,<=*y*), where *gcd* denotes the [greatest common divisor](https://en.wikipedia.org/wiki/Greatest_common_divisor). What is the minimum number of operations you need to make all of the elements equal to 1?
The first line of the input contains one integer *n* (1<=≤<=*n*<=≤<=2000) — the number of elements in the array. The second line contains *n* space separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the elements of the array.
Print -1, if it is impossible to turn all numbers to 1. Otherwise, print the minimum number of operations needed to make all numbers equal to 1.
[ "5\n2 2 3 4 6\n", "4\n2 4 6 8\n", "3\n2 6 9\n" ]
[ "5\n", "-1\n", "4\n" ]
In the first sample you can turn all numbers to 1 using the following 5 moves: - [2, 2, 3, 4, 6]. - [2, 1, 3, 4, 6] - [2, 1, 3, 1, 6] - [2, 1, 1, 1, 6] - [1, 1, 1, 1, 6] - [1, 1, 1, 1, 1] We can prove that in this case it is not possible to make all numbers one using less than 5 moves.
500
[ { "input": "5\n2 2 3 4 6", "output": "5" }, { "input": "4\n2 4 6 8", "output": "-1" }, { "input": "3\n2 6 9", "output": "4" }, { "input": "15\n10 10 10 10 10 10 21 21 21 21 21 21 21 21 21", "output": "15" }, { "input": "12\n10 10 14 14 14 14 14 14 14 14 21 21", "output": "20" }, { "input": "5\n10 10 14 21 21", "output": "6" }, { "input": "9\n10 10 10 10 10 14 14 21 21", "output": "11" }, { "input": "9\n10 10 10 10 10 10 10 10 21", "output": "9" }, { "input": "13\n10 10 10 15 15 15 15 15 15 15 15 21 21", "output": "21" }, { "input": "15\n10 10 10 10 10 10 10 10 10 10 10 10 15 15 21", "output": "17" }, { "input": "4\n1 1 1 1", "output": "0" }, { "input": "1\n3", "output": "-1" }, { "input": "2\n1 1", "output": "0" }, { "input": "2\n1000000000 1000000000", "output": "-1" }, { "input": "1\n1000000000", "output": "-1" }, { "input": "1\n1", "output": "0" }, { "input": "3\n42 15 35", "output": "4" }, { "input": "3\n6 10 15", "output": "4" }, { "input": "4\n2 1 1 1", "output": "1" }, { "input": "5\n2 1 1 1 2", "output": "2" }, { "input": "3\n30 14 21", "output": "4" }, { "input": "3\n15 6 10", "output": "4" }, { "input": "4\n1 1 1 2", "output": "1" }, { "input": "5\n1 1 1 2 2", "output": "2" }, { "input": "4\n2 6 9 1", "output": "3" }, { "input": "6\n2 3 4 1 1 1", "output": "3" }, { "input": "15\n2 6 6 6 3 3 3 15 5 5 5 7 5 5 5", "output": "15" }, { "input": "5\n2 3 2 6 9", "output": "5" }, { "input": "6\n6 15 10 6 15 10", "output": "7" } ]
1,557,479,503
2,147,483,647
Python 3
OK
TESTS
52
1,263
716,800
from functools import reduce def gcd(a, b): if b == 0: return a return gcd(b, a % b) n = int(input()) a = list(map(int, input().split(" "))) if a.count(1) > 0: print(n - a.count(1)) exit(0) c = reduce(gcd, a) if (c != 1): print(-1) exit(0) c = n for i in range(0, n-1): g = a[i] for j in range(i+1, n): g = gcd(g, a[j]) if g == 1: c = min(c, j-i) break print(c+n-1)
Title: Pride Time Limit: None seconds Memory Limit: None megabytes Problem Description: You have an array *a* with length *n*, you can perform operations. Each operation is like this: choose two adjacent elements from *a*, say *x* and *y*, and replace one of them with *gcd*(*x*,<=*y*), where *gcd* denotes the [greatest common divisor](https://en.wikipedia.org/wiki/Greatest_common_divisor). What is the minimum number of operations you need to make all of the elements equal to 1? Input Specification: The first line of the input contains one integer *n* (1<=≤<=*n*<=≤<=2000) — the number of elements in the array. The second line contains *n* space separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the elements of the array. Output Specification: Print -1, if it is impossible to turn all numbers to 1. Otherwise, print the minimum number of operations needed to make all numbers equal to 1. Demo Input: ['5\n2 2 3 4 6\n', '4\n2 4 6 8\n', '3\n2 6 9\n'] Demo Output: ['5\n', '-1\n', '4\n'] Note: In the first sample you can turn all numbers to 1 using the following 5 moves: - [2, 2, 3, 4, 6]. - [2, 1, 3, 4, 6] - [2, 1, 3, 1, 6] - [2, 1, 1, 1, 6] - [1, 1, 1, 1, 6] - [1, 1, 1, 1, 1] We can prove that in this case it is not possible to make all numbers one using less than 5 moves.
```python from functools import reduce def gcd(a, b): if b == 0: return a return gcd(b, a % b) n = int(input()) a = list(map(int, input().split(" "))) if a.count(1) > 0: print(n - a.count(1)) exit(0) c = reduce(gcd, a) if (c != 1): print(-1) exit(0) c = n for i in range(0, n-1): g = a[i] for j in range(i+1, n): g = gcd(g, a[j]) if g == 1: c = min(c, j-i) break print(c+n-1) ```
3
591
A
Wizards' Duel
PROGRAMMING
900
[ "implementation", "math" ]
null
null
Harry Potter and He-Who-Must-Not-Be-Named engaged in a fight to the death once again. This time they are located at opposite ends of the corridor of length *l*. Two opponents simultaneously charge a deadly spell in the enemy. We know that the impulse of Harry's magic spell flies at a speed of *p* meters per second, and the impulse of You-Know-Who's magic spell flies at a speed of *q* meters per second. The impulses are moving through the corridor toward each other, and at the time of the collision they turn round and fly back to those who cast them without changing their original speeds. Then, as soon as the impulse gets back to it's caster, the wizard reflects it and sends again towards the enemy, without changing the original speed of the impulse. Since Harry has perfectly mastered the basics of magic, he knows that after the second collision both impulses will disappear, and a powerful explosion will occur exactly in the place of their collision. However, the young wizard isn't good at math, so he asks you to calculate the distance from his position to the place of the second meeting of the spell impulses, provided that the opponents do not change positions during the whole fight.
The first line of the input contains a single integer *l* (1<=≤<=*l*<=≤<=1<=000) — the length of the corridor where the fight takes place. The second line contains integer *p*, the third line contains integer *q* (1<=≤<=*p*,<=*q*<=≤<=500) — the speeds of magical impulses for Harry Potter and He-Who-Must-Not-Be-Named, respectively.
Print a single real number — the distance from the end of the corridor, where Harry is located, to the place of the second meeting of the spell impulses. Your answer will be considered correct if its absolute or relative error will not exceed 10<=-<=4. Namely: let's assume that your answer equals *a*, and the answer of the jury is *b*. The checker program will consider your answer correct if .
[ "100\n50\n50\n", "199\n60\n40\n" ]
[ "50\n", "119.4\n" ]
In the first sample the speeds of the impulses are equal, so both of their meetings occur exactly in the middle of the corridor.
500
[ { "input": "100\n50\n50", "output": "50" }, { "input": "199\n60\n40", "output": "119.4" }, { "input": "1\n1\n1", "output": "0.5" }, { "input": "1\n1\n500", "output": "0.001996007984" }, { "input": "1\n500\n1", "output": "0.998003992" }, { "input": "1\n500\n500", "output": "0.5" }, { "input": "1000\n1\n1", "output": "500" }, { "input": "1000\n1\n500", "output": "1.996007984" }, { "input": "1000\n500\n1", "output": "998.003992" }, { "input": "1000\n500\n500", "output": "500" }, { "input": "101\n11\n22", "output": "33.66666667" }, { "input": "987\n1\n3", "output": "246.75" }, { "input": "258\n25\n431", "output": "14.14473684" }, { "input": "979\n39\n60", "output": "385.6666667" }, { "input": "538\n479\n416", "output": "287.9351955" }, { "input": "583\n112\n248", "output": "181.3777778" }, { "input": "978\n467\n371", "output": "545.0190931" }, { "input": "980\n322\n193", "output": "612.7378641" }, { "input": "871\n401\n17", "output": "835.576555" }, { "input": "349\n478\n378", "output": "194.885514" }, { "input": "425\n458\n118", "output": "337.9340278" }, { "input": "919\n323\n458", "output": "380.0729834" }, { "input": "188\n59\n126", "output": "59.95675676" }, { "input": "644\n428\n484", "output": "302.2280702" }, { "input": "253\n80\n276", "output": "56.85393258" }, { "input": "745\n152\n417", "output": "199.0158172" }, { "input": "600\n221\n279", "output": "265.2" }, { "input": "690\n499\n430", "output": "370.6243272" }, { "input": "105\n68\n403", "output": "15.15923567" }, { "input": "762\n462\n371", "output": "422.6218487" }, { "input": "903\n460\n362", "output": "505.3284672" }, { "input": "886\n235\n95", "output": "630.9393939" }, { "input": "655\n203\n18", "output": "601.6515837" }, { "input": "718\n29\n375", "output": "51.53960396" }, { "input": "296\n467\n377", "output": "163.7819905" }, { "input": "539\n61\n56", "output": "281.017094" }, { "input": "133\n53\n124", "output": "39.82485876" }, { "input": "998\n224\n65", "output": "773.5363322" }, { "input": "961\n173\n47", "output": "755.6954545" }, { "input": "285\n468\n62", "output": "251.6603774" }, { "input": "496\n326\n429", "output": "214.1668874" }, { "input": "627\n150\n285", "output": "216.2068966" }, { "input": "961\n443\n50", "output": "863.535497" }, { "input": "623\n422\n217", "output": "411.4334898" }, { "input": "678\n295\n29", "output": "617.3148148" } ]
1,696,414,132
2,147,483,647
PyPy 3-64
OK
TESTS
45
93
0
l = int(input()) p = int(input()) q = int(input()) s = l/(p+q) print(p*s)
Title: Wizards' Duel Time Limit: None seconds Memory Limit: None megabytes Problem Description: Harry Potter and He-Who-Must-Not-Be-Named engaged in a fight to the death once again. This time they are located at opposite ends of the corridor of length *l*. Two opponents simultaneously charge a deadly spell in the enemy. We know that the impulse of Harry's magic spell flies at a speed of *p* meters per second, and the impulse of You-Know-Who's magic spell flies at a speed of *q* meters per second. The impulses are moving through the corridor toward each other, and at the time of the collision they turn round and fly back to those who cast them without changing their original speeds. Then, as soon as the impulse gets back to it's caster, the wizard reflects it and sends again towards the enemy, without changing the original speed of the impulse. Since Harry has perfectly mastered the basics of magic, he knows that after the second collision both impulses will disappear, and a powerful explosion will occur exactly in the place of their collision. However, the young wizard isn't good at math, so he asks you to calculate the distance from his position to the place of the second meeting of the spell impulses, provided that the opponents do not change positions during the whole fight. Input Specification: The first line of the input contains a single integer *l* (1<=≤<=*l*<=≤<=1<=000) — the length of the corridor where the fight takes place. The second line contains integer *p*, the third line contains integer *q* (1<=≤<=*p*,<=*q*<=≤<=500) — the speeds of magical impulses for Harry Potter and He-Who-Must-Not-Be-Named, respectively. Output Specification: Print a single real number — the distance from the end of the corridor, where Harry is located, to the place of the second meeting of the spell impulses. Your answer will be considered correct if its absolute or relative error will not exceed 10<=-<=4. Namely: let's assume that your answer equals *a*, and the answer of the jury is *b*. The checker program will consider your answer correct if . Demo Input: ['100\n50\n50\n', '199\n60\n40\n'] Demo Output: ['50\n', '119.4\n'] Note: In the first sample the speeds of the impulses are equal, so both of their meetings occur exactly in the middle of the corridor.
```python l = int(input()) p = int(input()) q = int(input()) s = l/(p+q) print(p*s) ```
3
805
A
Fake NP
PROGRAMMING
1,000
[ "greedy", "math" ]
null
null
Tavak and Seyyed are good friends. Seyyed is very funny and he told Tavak to solve the following problem instead of longest-path. You are given *l* and *r*. For all integers from *l* to *r*, inclusive, we wrote down all of their integer divisors except 1. Find the integer that we wrote down the maximum number of times. Solve the problem to show that it's not a NP problem.
The first line contains two integers *l* and *r* (2<=≤<=*l*<=≤<=*r*<=≤<=109).
Print single integer, the integer that appears maximum number of times in the divisors. If there are multiple answers, print any of them.
[ "19 29\n", "3 6\n" ]
[ "2\n", "3\n" ]
Definition of a divisor: [https://www.mathsisfun.com/definitions/divisor-of-an-integer-.html](https://www.mathsisfun.com/definitions/divisor-of-an-integer-.html) The first example: from 19 to 29 these numbers are divisible by 2: {20, 22, 24, 26, 28}. The second example: from 3 to 6 these numbers are divisible by 3: {3, 6}.
500
[ { "input": "19 29", "output": "2" }, { "input": "3 6", "output": "2" }, { "input": "39 91", "output": "2" }, { "input": "76 134", "output": "2" }, { "input": "93 95", "output": "2" }, { "input": "17 35", "output": "2" }, { "input": "94 95", "output": "2" }, { "input": "51 52", "output": "2" }, { "input": "47 52", "output": "2" }, { "input": "38 98", "output": "2" }, { "input": "30 37", "output": "2" }, { "input": "56 92", "output": "2" }, { "input": "900000000 1000000000", "output": "2" }, { "input": "37622224 162971117", "output": "2" }, { "input": "760632746 850720703", "output": "2" }, { "input": "908580370 968054552", "output": "2" }, { "input": "951594860 953554446", "output": "2" }, { "input": "347877978 913527175", "output": "2" }, { "input": "620769961 988145114", "output": "2" }, { "input": "820844234 892579936", "output": "2" }, { "input": "741254764 741254768", "output": "2" }, { "input": "80270976 80270977", "output": "2" }, { "input": "392602363 392602367", "output": "2" }, { "input": "519002744 519002744", "output": "519002744" }, { "input": "331900277 331900277", "output": "331900277" }, { "input": "419873015 419873018", "output": "2" }, { "input": "349533413 349533413", "output": "349533413" }, { "input": "28829775 28829776", "output": "2" }, { "input": "568814539 568814539", "output": "568814539" }, { "input": "720270740 720270743", "output": "2" }, { "input": "871232720 871232722", "output": "2" }, { "input": "305693653 305693653", "output": "305693653" }, { "input": "634097178 634097179", "output": "2" }, { "input": "450868287 450868290", "output": "2" }, { "input": "252662256 252662260", "output": "2" }, { "input": "575062045 575062049", "output": "2" }, { "input": "273072892 273072894", "output": "2" }, { "input": "770439256 770439256", "output": "770439256" }, { "input": "2 1000000000", "output": "2" }, { "input": "6 8", "output": "2" }, { "input": "2 879190747", "output": "2" }, { "input": "5 5", "output": "5" }, { "input": "999999937 999999937", "output": "999999937" }, { "input": "3 3", "output": "3" }, { "input": "5 100", "output": "2" }, { "input": "2 2", "output": "2" }, { "input": "3 18", "output": "2" }, { "input": "7 7", "output": "7" }, { "input": "39916801 39916801", "output": "39916801" }, { "input": "3 8", "output": "2" }, { "input": "13 13", "output": "13" }, { "input": "4 8", "output": "2" }, { "input": "3 12", "output": "2" }, { "input": "6 12", "output": "2" }, { "input": "999999103 999999103", "output": "999999103" }, { "input": "100000007 100000007", "output": "100000007" }, { "input": "3 99", "output": "2" }, { "input": "999999733 999999733", "output": "999999733" }, { "input": "5 10", "output": "2" }, { "input": "982451653 982451653", "output": "982451653" }, { "input": "999900001 1000000000", "output": "2" }, { "input": "999727999 999727999", "output": "999727999" }, { "input": "2 999999999", "output": "2" }, { "input": "242 244", "output": "2" }, { "input": "3 10", "output": "2" }, { "input": "15 27", "output": "2" }, { "input": "998244353 998244353", "output": "998244353" }, { "input": "5 15", "output": "2" }, { "input": "999999797 999999797", "output": "999999797" }, { "input": "2 3", "output": "2" }, { "input": "999999929 999999929", "output": "999999929" }, { "input": "3 111111", "output": "2" }, { "input": "12 18", "output": "2" }, { "input": "479001599 479001599", "output": "479001599" }, { "input": "10000019 10000019", "output": "10000019" }, { "input": "715827883 715827883", "output": "715827883" }, { "input": "999992977 999992977", "output": "999992977" }, { "input": "11 11", "output": "11" }, { "input": "29 29", "output": "29" }, { "input": "1000003 1000003", "output": "1000003" }, { "input": "6 15", "output": "2" }, { "input": "1200007 1200007", "output": "1200007" }, { "input": "3 1000000000", "output": "2" }, { "input": "990000023 990000023", "output": "990000023" }, { "input": "1717 1717", "output": "1717" }, { "input": "141650963 141650963", "output": "141650963" }, { "input": "1002523 1002523", "output": "1002523" }, { "input": "900000011 900000011", "output": "900000011" }, { "input": "104729 104729", "output": "104729" }, { "input": "4 12", "output": "2" }, { "input": "100003 100003", "output": "100003" }, { "input": "17 17", "output": "17" }, { "input": "10 100", "output": "2" } ]
1,659,509,153
2,147,483,647
PyPy 3-64
WRONG_ANSWER
TESTS
0
30
0
l, r = map(int, input().split()) if l - r <= 3: print(l) else: print(2)
Title: Fake NP Time Limit: None seconds Memory Limit: None megabytes Problem Description: Tavak and Seyyed are good friends. Seyyed is very funny and he told Tavak to solve the following problem instead of longest-path. You are given *l* and *r*. For all integers from *l* to *r*, inclusive, we wrote down all of their integer divisors except 1. Find the integer that we wrote down the maximum number of times. Solve the problem to show that it's not a NP problem. Input Specification: The first line contains two integers *l* and *r* (2<=≤<=*l*<=≤<=*r*<=≤<=109). Output Specification: Print single integer, the integer that appears maximum number of times in the divisors. If there are multiple answers, print any of them. Demo Input: ['19 29\n', '3 6\n'] Demo Output: ['2\n', '3\n'] Note: Definition of a divisor: [https://www.mathsisfun.com/definitions/divisor-of-an-integer-.html](https://www.mathsisfun.com/definitions/divisor-of-an-integer-.html) The first example: from 19 to 29 these numbers are divisible by 2: {20, 22, 24, 26, 28}. The second example: from 3 to 6 these numbers are divisible by 3: {3, 6}.
```python l, r = map(int, input().split()) if l - r <= 3: print(l) else: print(2) ```
0
557
B
Pasha and Tea
PROGRAMMING
1,500
[ "constructive algorithms", "implementation", "math", "sortings" ]
null
null
Pasha decided to invite his friends to a tea party. For that occasion, he has a large teapot with the capacity of *w* milliliters and 2*n* tea cups, each cup is for one of Pasha's friends. The *i*-th cup can hold at most *a**i* milliliters of water. It turned out that among Pasha's friends there are exactly *n* boys and exactly *n* girls and all of them are going to come to the tea party. To please everyone, Pasha decided to pour the water for the tea as follows: - Pasha can boil the teapot exactly once by pouring there at most *w* milliliters of water; - Pasha pours the same amount of water to each girl; - Pasha pours the same amount of water to each boy; - if each girl gets *x* milliliters of water, then each boy gets 2*x* milliliters of water. In the other words, each boy should get two times more water than each girl does. Pasha is very kind and polite, so he wants to maximize the total amount of the water that he pours to his friends. Your task is to help him and determine the optimum distribution of cups between Pasha's friends.
The first line of the input contains two integers, *n* and *w* (1<=≤<=*n*<=≤<=105, 1<=≤<=*w*<=≤<=109) — the number of Pasha's friends that are boys (equal to the number of Pasha's friends that are girls) and the capacity of Pasha's teapot in milliliters. The second line of the input contains the sequence of integers *a**i* (1<=≤<=*a**i*<=≤<=109, 1<=≤<=*i*<=≤<=2*n*) — the capacities of Pasha's tea cups in milliliters.
Print a single real number — the maximum total amount of water in milliliters that Pasha can pour to his friends without violating the given conditions. Your answer will be considered correct if its absolute or relative error doesn't exceed 10<=-<=6.
[ "2 4\n1 1 1 1\n", "3 18\n4 4 4 2 2 2\n", "1 5\n2 3\n" ]
[ "3", "18", "4.5" ]
Pasha also has candies that he is going to give to girls but that is another task...
1,000
[ { "input": "2 4\n1 1 1 1", "output": "3.0000000000" }, { "input": "3 18\n4 4 4 2 2 2", "output": "18.0000000000" }, { "input": "1 5\n2 3", "output": "4.5000000000" }, { "input": "1 1\n1000000000 1000000000", "output": "1.0000000000" }, { "input": "4 1000000000\n1 1 1 1 1 1 1 1", "output": "6.0000000000" }, { "input": "4 1000000000\n1 1 1 1 2 2 2 2", "output": "12.0000000000" }, { "input": "4 1\n3 3 3 3 4 4 4 4", "output": "1.0000000000" }, { "input": "2 19\n3 3 5 5", "output": "15.0000000000" }, { "input": "3 31\n3 3 3 5 5 5", "output": "22.5000000000" }, { "input": "5 15\n2 3 4 1 2 4 5 3 5 10", "output": "15.0000000000" }, { "input": "5 14\n2 3 4 1 2 4 5 3 5 10", "output": "14.0000000000" }, { "input": "5 16\n2 3 4 1 2 4 5 3 5 10", "output": "15.0000000000" }, { "input": "1 100\n1 200", "output": "3.0000000000" }, { "input": "1 1\n1 1", "output": "1.0000000000" }, { "input": "2 1000000000\n1 1 1 100", "output": "3.0000000000" }, { "input": "4 30\n3 3 3 3 4 5 6 7", "output": "24.0000000000" }, { "input": "2 100\n1 1 1 10", "output": "3.0000000000" }, { "input": "3 18\n1 1 1 1 1 5", "output": "4.5000000000" } ]
1,443,502,312
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
0
46
0
n, k = list(map(int, input().split())) a = list(map(int, input().split())) a = sorted(a) minG = a[0] minB = a[n] ans = min(k/(3.0*n), minG ) print (ans*n + ans*2*n)
Title: Pasha and Tea Time Limit: None seconds Memory Limit: None megabytes Problem Description: Pasha decided to invite his friends to a tea party. For that occasion, he has a large teapot with the capacity of *w* milliliters and 2*n* tea cups, each cup is for one of Pasha's friends. The *i*-th cup can hold at most *a**i* milliliters of water. It turned out that among Pasha's friends there are exactly *n* boys and exactly *n* girls and all of them are going to come to the tea party. To please everyone, Pasha decided to pour the water for the tea as follows: - Pasha can boil the teapot exactly once by pouring there at most *w* milliliters of water; - Pasha pours the same amount of water to each girl; - Pasha pours the same amount of water to each boy; - if each girl gets *x* milliliters of water, then each boy gets 2*x* milliliters of water. In the other words, each boy should get two times more water than each girl does. Pasha is very kind and polite, so he wants to maximize the total amount of the water that he pours to his friends. Your task is to help him and determine the optimum distribution of cups between Pasha's friends. Input Specification: The first line of the input contains two integers, *n* and *w* (1<=≤<=*n*<=≤<=105, 1<=≤<=*w*<=≤<=109) — the number of Pasha's friends that are boys (equal to the number of Pasha's friends that are girls) and the capacity of Pasha's teapot in milliliters. The second line of the input contains the sequence of integers *a**i* (1<=≤<=*a**i*<=≤<=109, 1<=≤<=*i*<=≤<=2*n*) — the capacities of Pasha's tea cups in milliliters. Output Specification: Print a single real number — the maximum total amount of water in milliliters that Pasha can pour to his friends without violating the given conditions. Your answer will be considered correct if its absolute or relative error doesn't exceed 10<=-<=6. Demo Input: ['2 4\n1 1 1 1\n', '3 18\n4 4 4 2 2 2\n', '1 5\n2 3\n'] Demo Output: ['3', '18', '4.5'] Note: Pasha also has candies that he is going to give to girls but that is another task...
```python n, k = list(map(int, input().split())) a = list(map(int, input().split())) a = sorted(a) minG = a[0] minB = a[n] ans = min(k/(3.0*n), minG ) print (ans*n + ans*2*n) ```
0
239
A
Two Bags of Potatoes
PROGRAMMING
1,200
[ "greedy", "implementation", "math" ]
null
null
Valera had two bags of potatoes, the first of these bags contains *x* (*x*<=≥<=1) potatoes, and the second — *y* (*y*<=≥<=1) potatoes. Valera — very scattered boy, so the first bag of potatoes (it contains *x* potatoes) Valera lost. Valera remembers that the total amount of potatoes (*x*<=+<=*y*) in the two bags, firstly, was not gerater than *n*, and, secondly, was divisible by *k*. Help Valera to determine how many potatoes could be in the first bag. Print all such possible numbers in ascending order.
The first line of input contains three integers *y*, *k*, *n* (1<=≤<=*y*,<=*k*,<=*n*<=≤<=109; <=≤<=105).
Print the list of whitespace-separated integers — all possible values of *x* in ascending order. You should print each possible value of *x* exactly once. If there are no such values of *x* print a single integer -1.
[ "10 1 10\n", "10 6 40\n" ]
[ "-1\n", "2 8 14 20 26 \n" ]
none
500
[ { "input": "10 1 10", "output": "-1" }, { "input": "10 6 40", "output": "2 8 14 20 26 " }, { "input": "10 1 20", "output": "1 2 3 4 5 6 7 8 9 10 " }, { "input": "1 10000 1000000000", "output": "9999 19999 29999 39999 49999 59999 69999 79999 89999 99999 109999 119999 129999 139999 149999 159999 169999 179999 189999 199999 209999 219999 229999 239999 249999 259999 269999 279999 289999 299999 309999 319999 329999 339999 349999 359999 369999 379999 389999 399999 409999 419999 429999 439999 449999 459999 469999 479999 489999 499999 509999 519999 529999 539999 549999 559999 569999 579999 589999 599999 609999 619999 629999 639999 649999 659999 669999 679999 689999 699999 709999 719999 729999 739999 7499..." }, { "input": "84817 1 33457", "output": "-1" }, { "input": "21 37 99", "output": "16 53 " }, { "input": "78 7 15", "output": "-1" }, { "input": "74 17 27", "output": "-1" }, { "input": "79 23 43", "output": "-1" }, { "input": "32 33 3", "output": "-1" }, { "input": "55 49 44", "output": "-1" }, { "input": "64 59 404", "output": "54 113 172 231 290 " }, { "input": "61 69 820", "output": "8 77 146 215 284 353 422 491 560 629 698 " }, { "input": "17 28 532", "output": "11 39 67 95 123 151 179 207 235 263 291 319 347 375 403 431 459 487 515 " }, { "input": "46592 52 232", "output": "-1" }, { "input": "1541 58 648", "output": "-1" }, { "input": "15946 76 360", "output": "-1" }, { "input": "30351 86 424", "output": "-1" }, { "input": "1 2 37493", "output": "1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53 55 57 59 61 63 65 67 69 71 73 75 77 79 81 83 85 87 89 91 93 95 97 99 101 103 105 107 109 111 113 115 117 119 121 123 125 127 129 131 133 135 137 139 141 143 145 147 149 151 153 155 157 159 161 163 165 167 169 171 173 175 177 179 181 183 185 187 189 191 193 195 197 199 201 203 205 207 209 211 213 215 217 219 221 223 225 227 229 231 233 235 237 239 241 243 245 247 249 251 253 255 257 259 261 263 265 267 269 271 273 275 277 279 281 28..." }, { "input": "1 3 27764", "output": "2 5 8 11 14 17 20 23 26 29 32 35 38 41 44 47 50 53 56 59 62 65 68 71 74 77 80 83 86 89 92 95 98 101 104 107 110 113 116 119 122 125 128 131 134 137 140 143 146 149 152 155 158 161 164 167 170 173 176 179 182 185 188 191 194 197 200 203 206 209 212 215 218 221 224 227 230 233 236 239 242 245 248 251 254 257 260 263 266 269 272 275 278 281 284 287 290 293 296 299 302 305 308 311 314 317 320 323 326 329 332 335 338 341 344 347 350 353 356 359 362 365 368 371 374 377 380 383 386 389 392 395 398 401 404 407 410..." }, { "input": "10 4 9174", "output": "2 6 10 14 18 22 26 30 34 38 42 46 50 54 58 62 66 70 74 78 82 86 90 94 98 102 106 110 114 118 122 126 130 134 138 142 146 150 154 158 162 166 170 174 178 182 186 190 194 198 202 206 210 214 218 222 226 230 234 238 242 246 250 254 258 262 266 270 274 278 282 286 290 294 298 302 306 310 314 318 322 326 330 334 338 342 346 350 354 358 362 366 370 374 378 382 386 390 394 398 402 406 410 414 418 422 426 430 434 438 442 446 450 454 458 462 466 470 474 478 482 486 490 494 498 502 506 510 514 518 522 526 530 534 53..." }, { "input": "33 7 4971", "output": "2 9 16 23 30 37 44 51 58 65 72 79 86 93 100 107 114 121 128 135 142 149 156 163 170 177 184 191 198 205 212 219 226 233 240 247 254 261 268 275 282 289 296 303 310 317 324 331 338 345 352 359 366 373 380 387 394 401 408 415 422 429 436 443 450 457 464 471 478 485 492 499 506 513 520 527 534 541 548 555 562 569 576 583 590 597 604 611 618 625 632 639 646 653 660 667 674 681 688 695 702 709 716 723 730 737 744 751 758 765 772 779 786 793 800 807 814 821 828 835 842 849 856 863 870 877 884 891 898 905 912 919..." }, { "input": "981 1 3387", "output": "1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 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 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155..." }, { "input": "386 1 2747", "output": "1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 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 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155..." }, { "input": "123 2 50000", "output": "1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53 55 57 59 61 63 65 67 69 71 73 75 77 79 81 83 85 87 89 91 93 95 97 99 101 103 105 107 109 111 113 115 117 119 121 123 125 127 129 131 133 135 137 139 141 143 145 147 149 151 153 155 157 159 161 163 165 167 169 171 173 175 177 179 181 183 185 187 189 191 193 195 197 199 201 203 205 207 209 211 213 215 217 219 221 223 225 227 229 231 233 235 237 239 241 243 245 247 249 251 253 255 257 259 261 263 265 267 269 271 273 275 277 279 281 28..." }, { "input": "3123 100 10000000", "output": "77 177 277 377 477 577 677 777 877 977 1077 1177 1277 1377 1477 1577 1677 1777 1877 1977 2077 2177 2277 2377 2477 2577 2677 2777 2877 2977 3077 3177 3277 3377 3477 3577 3677 3777 3877 3977 4077 4177 4277 4377 4477 4577 4677 4777 4877 4977 5077 5177 5277 5377 5477 5577 5677 5777 5877 5977 6077 6177 6277 6377 6477 6577 6677 6777 6877 6977 7077 7177 7277 7377 7477 7577 7677 7777 7877 7977 8077 8177 8277 8377 8477 8577 8677 8777 8877 8977 9077 9177 9277 9377 9477 9577 9677 9777 9877 9977 10077 10177 10277 1037..." }, { "input": "2 10000 1000000000", "output": "9998 19998 29998 39998 49998 59998 69998 79998 89998 99998 109998 119998 129998 139998 149998 159998 169998 179998 189998 199998 209998 219998 229998 239998 249998 259998 269998 279998 289998 299998 309998 319998 329998 339998 349998 359998 369998 379998 389998 399998 409998 419998 429998 439998 449998 459998 469998 479998 489998 499998 509998 519998 529998 539998 549998 559998 569998 579998 589998 599998 609998 619998 629998 639998 649998 659998 669998 679998 689998 699998 709998 719998 729998 739998 7499..." }, { "input": "3 10000 1000000000", "output": "9997 19997 29997 39997 49997 59997 69997 79997 89997 99997 109997 119997 129997 139997 149997 159997 169997 179997 189997 199997 209997 219997 229997 239997 249997 259997 269997 279997 289997 299997 309997 319997 329997 339997 349997 359997 369997 379997 389997 399997 409997 419997 429997 439997 449997 459997 469997 479997 489997 499997 509997 519997 529997 539997 549997 559997 569997 579997 589997 599997 609997 619997 629997 639997 649997 659997 669997 679997 689997 699997 709997 719997 729997 739997 7499..." }, { "input": "12312223 10000 1000000000", "output": "7777 17777 27777 37777 47777 57777 67777 77777 87777 97777 107777 117777 127777 137777 147777 157777 167777 177777 187777 197777 207777 217777 227777 237777 247777 257777 267777 277777 287777 297777 307777 317777 327777 337777 347777 357777 367777 377777 387777 397777 407777 417777 427777 437777 447777 457777 467777 477777 487777 497777 507777 517777 527777 537777 547777 557777 567777 577777 587777 597777 607777 617777 627777 637777 647777 657777 667777 677777 687777 697777 707777 717777 727777 737777 7477..." }, { "input": "500000000 1000000000 1000000000", "output": "500000000 " }, { "input": "1 1000000000 1000000000", "output": "999999999 " }, { "input": "10 6 11", "output": "-1" }, { "input": "2 100 10", "output": "-1" }, { "input": "1 100000007 1000000000", "output": "100000006 200000013 300000020 400000027 500000034 600000041 700000048 800000055 900000062 " }, { "input": "1 999999999 1000000000", "output": "999999998 " }, { "input": "100000000 1000000000 1000000000", "output": "900000000 " }, { "input": "11 2 12", "output": "1 " }, { "input": "31 10 39", "output": "-1" }, { "input": "48 6 50", "output": "-1" }, { "input": "500000000 500000000 1000000000", "output": "500000000 " }, { "input": "1 1000000000 999999999", "output": "-1" }, { "input": "4 2 10", "output": "2 4 6 " }, { "input": "1000000000 1 1", "output": "-1" }, { "input": "1000000000 1 100000", "output": "-1" }, { "input": "1000000000 1 10", "output": "-1" }, { "input": "10 5 14", "output": "-1" }, { "input": "500000000 499999999 1000000000", "output": "499999998 " }, { "input": "1 999999997 1000000000", "output": "999999996 " } ]
1,609,789,035
2,147,483,647
Python 3
OK
TESTS
48
654
6,656,000
def STR(): return list(input()) def INT(): return int(input()) def MAP(): return map(int, input().split()) def MAP2():return map(float,input().split()) def LIST(): return list(map(int, input().split())) def STRING(): return input() import string import sys from heapq import heappop , heappush from bisect import * from collections import deque , Counter from math import * from itertools import permutations , accumulate dx = [-1 , 1 , 0 , 0 ] dy = [0 , 0 , 1 , - 1] #visited = [[False for i in range(m)] for j in range(n)] #sys.stdin = open(r'input.txt' , 'r') #sys.stdout = open(r'output.txt' , 'w') #for tt in range(INT()): y , k , n = MAP() x = k d = [x] while x <= n: x+=k d.append(x) #print(x) #print(d) ans = [] for i in d : if i > y and i <= n : z = i - y ans.append(z) if len(ans) == 0 : print('-1') exit(0) ans.sort() print(*ans)
Title: Two Bags of Potatoes Time Limit: None seconds Memory Limit: None megabytes Problem Description: Valera had two bags of potatoes, the first of these bags contains *x* (*x*<=≥<=1) potatoes, and the second — *y* (*y*<=≥<=1) potatoes. Valera — very scattered boy, so the first bag of potatoes (it contains *x* potatoes) Valera lost. Valera remembers that the total amount of potatoes (*x*<=+<=*y*) in the two bags, firstly, was not gerater than *n*, and, secondly, was divisible by *k*. Help Valera to determine how many potatoes could be in the first bag. Print all such possible numbers in ascending order. Input Specification: The first line of input contains three integers *y*, *k*, *n* (1<=≤<=*y*,<=*k*,<=*n*<=≤<=109; <=≤<=105). Output Specification: Print the list of whitespace-separated integers — all possible values of *x* in ascending order. You should print each possible value of *x* exactly once. If there are no such values of *x* print a single integer -1. Demo Input: ['10 1 10\n', '10 6 40\n'] Demo Output: ['-1\n', '2 8 14 20 26 \n'] Note: none
```python def STR(): return list(input()) def INT(): return int(input()) def MAP(): return map(int, input().split()) def MAP2():return map(float,input().split()) def LIST(): return list(map(int, input().split())) def STRING(): return input() import string import sys from heapq import heappop , heappush from bisect import * from collections import deque , Counter from math import * from itertools import permutations , accumulate dx = [-1 , 1 , 0 , 0 ] dy = [0 , 0 , 1 , - 1] #visited = [[False for i in range(m)] for j in range(n)] #sys.stdin = open(r'input.txt' , 'r') #sys.stdout = open(r'output.txt' , 'w') #for tt in range(INT()): y , k , n = MAP() x = k d = [x] while x <= n: x+=k d.append(x) #print(x) #print(d) ans = [] for i in d : if i > y and i <= n : z = i - y ans.append(z) if len(ans) == 0 : print('-1') exit(0) ans.sort() print(*ans) ```
3
71
A
Way Too Long Words
PROGRAMMING
800
[ "strings" ]
A. Way Too Long Words
1
256
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome. Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation. This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes. Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n". You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
[ "4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n" ]
[ "word\nl10n\ni18n\np43s\n" ]
none
500
[ { "input": "4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis", "output": "word\nl10n\ni18n\np43s" }, { "input": "5\nabcdefgh\nabcdefghi\nabcdefghij\nabcdefghijk\nabcdefghijklm", "output": "abcdefgh\nabcdefghi\nabcdefghij\na9k\na11m" }, { "input": "3\nnjfngnrurunrgunrunvurn\njfvnjfdnvjdbfvsbdubruvbubvkdb\nksdnvidnviudbvibd", "output": "n20n\nj27b\nk15d" }, { "input": "1\ntcyctkktcctrcyvbyiuhihhhgyvyvyvyvjvytchjckt", "output": "t41t" }, { "input": "24\nyou\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nunofficially\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings", "output": "you\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nu10y\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings" }, { "input": "1\na", "output": "a" }, { "input": "26\na\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz", "output": "a\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz" }, { "input": "1\nabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghij", "output": "a98j" }, { "input": "10\ngyartjdxxlcl\nfzsck\nuidwu\nxbymclornemdmtj\nilppyoapitawgje\ncibzc\ndrgbeu\nhezplmsdekhhbo\nfeuzlrimbqbytdu\nkgdco", "output": "g10l\nfzsck\nuidwu\nx13j\ni13e\ncibzc\ndrgbeu\nh12o\nf13u\nkgdco" }, { "input": "20\nlkpmx\nkovxmxorlgwaomlswjxlpnbvltfv\nhykasjxqyjrmybejnmeumzha\ntuevlumpqbbhbww\nqgqsphvrmupxxc\ntrissbaf\nqfgrlinkzvzqdryckaizutd\nzzqtoaxkvwoscyx\noswytrlnhpjvvnwookx\nlpuzqgec\ngyzqfwxggtvpjhzmzmdw\nrlxjgmvdftvrmvbdwudra\nvsntnjpepnvdaxiporggmglhagv\nxlvcqkqgcrbgtgglj\nlyxwxbiszyhlsrgzeedzprbmcpduvq\nyrmqqvrkqskqukzqrwukpsifgtdc\nxpuohcsjhhuhvr\nvvlfrlxpvqejngwrbfbpmqeirxlw\nsvmasocxdvadmaxtrpakysmeaympy\nyuflqboqfdt", "output": "lkpmx\nk26v\nh22a\nt13w\nq12c\ntrissbaf\nq21d\nz13x\no17x\nlpuzqgec\ng18w\nr19a\nv25v\nx15j\nl28q\ny26c\nx12r\nv26w\ns27y\ny9t" }, { "input": "100\nm\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr", "output": "m\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr" } ]
1,673,018,530
2,147,483,647
PyPy 3-64
OK
TESTS
20
62
0
t=int(input()) for _ in range(t): s=input() n=len(s) if n>10: print(s[0]+str(n-2)+s[n-1]) else: print(s)
Title: Way Too Long Words Time Limit: 1 seconds Memory Limit: 256 megabytes Problem Description: Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome. Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation. This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes. Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n". You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes. Input Specification: The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters. Output Specification: Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data. Demo Input: ['4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n'] Demo Output: ['word\nl10n\ni18n\np43s\n'] Note: none
```python t=int(input()) for _ in range(t): s=input() n=len(s) if n>10: print(s[0]+str(n-2)+s[n-1]) else: print(s) ```
3.969
279
B
Books
PROGRAMMING
1,400
[ "binary search", "brute force", "implementation", "two pointers" ]
null
null
When Valera has got some free time, he goes to the library to read some books. Today he's got *t* free minutes to read. That's why Valera took *n* books in the library and for each book he estimated the time he is going to need to read it. Let's number the books by integers from 1 to *n*. Valera needs *a**i* minutes to read the *i*-th book. Valera decided to choose an arbitrary book with number *i* and read the books one by one, starting from this book. In other words, he will first read book number *i*, then book number *i*<=+<=1, then book number *i*<=+<=2 and so on. He continues the process until he either runs out of the free time or finishes reading the *n*-th book. Valera reads each book up to the end, that is, he doesn't start reading the book if he doesn't have enough free time to finish reading it. Print the maximum number of books Valera can read.
The first line contains two integers *n* and *t* (1<=≤<=*n*<=≤<=105; 1<=≤<=*t*<=≤<=109) — the number of books and the number of free minutes Valera's got. The second line contains a sequence of *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=104), where number *a**i* shows the number of minutes that the boy needs to read the *i*-th book.
Print a single integer — the maximum number of books Valera can read.
[ "4 5\n3 1 2 1\n", "3 3\n2 2 3\n" ]
[ "3\n", "1\n" ]
none
1,000
[ { "input": "4 5\n3 1 2 1", "output": "3" }, { "input": "3 3\n2 2 3", "output": "1" }, { "input": "1 3\n5", "output": "0" }, { "input": "1 10\n4", "output": "1" }, { "input": "2 10\n6 4", "output": "2" }, { "input": "6 10\n2 3 4 2 1 1", "output": "4" }, { "input": "7 13\n6 8 14 9 4 11 10", "output": "2" }, { "input": "10 15\n10 9 1 1 5 10 5 3 7 2", "output": "3" }, { "input": "20 30\n8 1 2 6 9 4 1 9 9 10 4 7 8 9 5 7 1 8 7 4", "output": "6" }, { "input": "30 60\n16 13 22 38 13 35 17 17 20 38 12 19 9 22 20 3 35 34 34 21 35 40 22 3 27 19 12 4 8 19", "output": "4" }, { "input": "100 100\n75 92 18 6 81 67 7 92 100 65 82 32 50 67 85 31 80 91 84 63 39 52 92 81 1 98 24 12 43 48 17 86 51 72 48 95 45 50 12 66 19 79 49 89 34 1 97 75 20 33 96 27 42 23 73 71 93 1 85 19 66 14 17 61 20 39 36 33 42 61 56 64 23 91 80 99 40 74 13 18 98 85 74 39 62 84 46 74 50 23 38 11 79 14 9 25 66 100 25 52", "output": "3" }, { "input": "10 1\n4418 7528 8170 1736 1317 3205 8183 4995 8039 4708", "output": "0" }, { "input": "50 2\n124 214 63 73 996 760 38 571 451 300 970 1 706 937 837 494 619 88 851 411 957 990 842 613 821 649 627 34 693 678 734 116 816 985 705 940 499 493 922 967 854 439 112 644 961 438 189 572 655 550", "output": "1" } ]
1,681,034,849
2,147,483,647
Python 3
OK
TESTS
38
312
6,451,200
n,k = map(int,input().split()) P = list(map(int,input().split())) L,R,Sum,ans =0,0,0,0 for i in range(0,len(P)): Sum+=P[R] while(Sum>k): Sum-=P[L] L+=1 ans = max(ans,R-L+1) R+=1 print(ans)
Title: Books Time Limit: None seconds Memory Limit: None megabytes Problem Description: When Valera has got some free time, he goes to the library to read some books. Today he's got *t* free minutes to read. That's why Valera took *n* books in the library and for each book he estimated the time he is going to need to read it. Let's number the books by integers from 1 to *n*. Valera needs *a**i* minutes to read the *i*-th book. Valera decided to choose an arbitrary book with number *i* and read the books one by one, starting from this book. In other words, he will first read book number *i*, then book number *i*<=+<=1, then book number *i*<=+<=2 and so on. He continues the process until he either runs out of the free time or finishes reading the *n*-th book. Valera reads each book up to the end, that is, he doesn't start reading the book if he doesn't have enough free time to finish reading it. Print the maximum number of books Valera can read. Input Specification: The first line contains two integers *n* and *t* (1<=≤<=*n*<=≤<=105; 1<=≤<=*t*<=≤<=109) — the number of books and the number of free minutes Valera's got. The second line contains a sequence of *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=104), where number *a**i* shows the number of minutes that the boy needs to read the *i*-th book. Output Specification: Print a single integer — the maximum number of books Valera can read. Demo Input: ['4 5\n3 1 2 1\n', '3 3\n2 2 3\n'] Demo Output: ['3\n', '1\n'] Note: none
```python n,k = map(int,input().split()) P = list(map(int,input().split())) L,R,Sum,ans =0,0,0,0 for i in range(0,len(P)): Sum+=P[R] while(Sum>k): Sum-=P[L] L+=1 ans = max(ans,R-L+1) R+=1 print(ans) ```
3
79
A
Bus Game
PROGRAMMING
1,200
[ "greedy" ]
A. Bus Game
2
256
After Fox Ciel won an onsite round of a programming contest, she took a bus to return to her castle. The fee of the bus was 220 yen. She met Rabbit Hanako in the bus. They decided to play the following game because they got bored in the bus. - Initially, there is a pile that contains *x* 100-yen coins and *y* 10-yen coins. - They take turns alternatively. Ciel takes the first turn. - In each turn, they must take exactly 220 yen from the pile. In Ciel's turn, if there are multiple ways to take 220 yen, she will choose the way that contains the maximal number of 100-yen coins. In Hanako's turn, if there are multiple ways to take 220 yen, she will choose the way that contains the maximal number of 10-yen coins. - If Ciel or Hanako can't take exactly 220 yen from the pile, she loses. Determine the winner of the game.
The first line contains two integers *x* (0<=≤<=*x*<=≤<=106) and *y* (0<=≤<=*y*<=≤<=106), separated by a single space.
If Ciel wins, print "Ciel". Otherwise, print "Hanako".
[ "2 2\n", "3 22\n" ]
[ "Ciel\n", "Hanako\n" ]
In the first turn (Ciel's turn), she will choose 2 100-yen coins and 2 10-yen coins. In the second turn (Hanako's turn), she will choose 1 100-yen coin and 12 10-yen coins. In the third turn (Ciel's turn), she can't pay exactly 220 yen, so Ciel will lose.
500
[ { "input": "2 2", "output": "Ciel" }, { "input": "3 22", "output": "Hanako" }, { "input": "0 22", "output": "Ciel" }, { "input": "1000 1000", "output": "Ciel" }, { "input": "0 0", "output": "Hanako" }, { "input": "0 21", "output": "Hanako" }, { "input": "1 11", "output": "Hanako" }, { "input": "1 12", "output": "Ciel" }, { "input": "2 1", "output": "Hanako" }, { "input": "2 23", "output": "Ciel" }, { "input": "2 24", "output": "Hanako" }, { "input": "3 1", "output": "Hanako" }, { "input": "3 2", "output": "Ciel" }, { "input": "3 13", "output": "Ciel" }, { "input": "3 14", "output": "Hanako" }, { "input": "4 1", "output": "Hanako" }, { "input": "4 2", "output": "Ciel" }, { "input": "4 25", "output": "Hanako" }, { "input": "4 26", "output": "Ciel" }, { "input": "5 1", "output": "Hanako" }, { "input": "5 2", "output": "Ciel" }, { "input": "5 15", "output": "Hanako" }, { "input": "5 16", "output": "Ciel" }, { "input": "5 23", "output": "Ciel" }, { "input": "5 24", "output": "Hanako" }, { "input": "6 1", "output": "Hanako" }, { "input": "6 2", "output": "Ciel" }, { "input": "6 13", "output": "Ciel" }, { "input": "6 14", "output": "Hanako" }, { "input": "6 23", "output": "Ciel" }, { "input": "6 24", "output": "Hanako" }, { "input": "7 1", "output": "Hanako" }, { "input": "7 2", "output": "Ciel" }, { "input": "7 13", "output": "Ciel" }, { "input": "7 14", "output": "Hanako" }, { "input": "7 25", "output": "Hanako" }, { "input": "7 26", "output": "Ciel" }, { "input": "8 1", "output": "Hanako" }, { "input": "8 2", "output": "Ciel" }, { "input": "8 15", "output": "Hanako" }, { "input": "8 16", "output": "Ciel" }, { "input": "8 25", "output": "Hanako" }, { "input": "8 26", "output": "Ciel" }, { "input": "9 1", "output": "Hanako" }, { "input": "9 2", "output": "Ciel" }, { "input": "9 15", "output": "Hanako" }, { "input": "9 16", "output": "Ciel" }, { "input": "9 23", "output": "Ciel" }, { "input": "9 24", "output": "Hanako" }, { "input": "10 12", "output": "Ciel" }, { "input": "10 13", "output": "Ciel" }, { "input": "10 22", "output": "Ciel" }, { "input": "10 23", "output": "Ciel" }, { "input": "11 12", "output": "Ciel" }, { "input": "11 13", "output": "Ciel" }, { "input": "11 24", "output": "Hanako" }, { "input": "11 25", "output": "Hanako" }, { "input": "12 14", "output": "Hanako" }, { "input": "12 15", "output": "Hanako" }, { "input": "12 24", "output": "Hanako" }, { "input": "12 25", "output": "Hanako" }, { "input": "0 1000000", "output": "Hanako" }, { "input": "1000000 0", "output": "Hanako" }, { "input": "1000000 1000000", "output": "Ciel" }, { "input": "178087 42116", "output": "Ciel" }, { "input": "378897 104123", "output": "Ciel" }, { "input": "61207 166129", "output": "Hanako" }, { "input": "743519 228136", "output": "Ciel" }, { "input": "425829 771644", "output": "Ciel" }, { "input": "626640 833651", "output": "Ciel" }, { "input": "308950 895657", "output": "Hanako" }, { "input": "991262 957664", "output": "Ciel" }, { "input": "192071 19670", "output": "Hanako" }, { "input": "874382 81677", "output": "Hanako" }, { "input": "202081 745873", "output": "Hanako" }, { "input": "233663 723781", "output": "Hanako" }, { "input": "783744 701689", "output": "Hanako" }, { "input": "333825 679597", "output": "Hanako" }, { "input": "365407 657504", "output": "Hanako" }, { "input": "915488 635412", "output": "Hanako" }, { "input": "947070 613320", "output": "Hanako" }, { "input": "497151 591228", "output": "Hanako" }, { "input": "528732 87635", "output": "Ciel" }, { "input": "78813 65543", "output": "Hanako" }, { "input": "6 4", "output": "Hanako" }, { "input": "3 5", "output": "Ciel" }, { "input": "4 24", "output": "Hanako" }, { "input": "2 14", "output": "Ciel" }, { "input": "2 26", "output": "Hanako" }, { "input": "5 26", "output": "Ciel" }, { "input": "5 5", "output": "Hanako" }, { "input": "359 479", "output": "Hanako" }, { "input": "1 20", "output": "Ciel" }, { "input": "11 0", "output": "Hanako" }, { "input": "3 0", "output": "Hanako" }, { "input": "0 30", "output": "Ciel" }, { "input": "0 35", "output": "Ciel" }, { "input": "100 99", "output": "Ciel" }, { "input": "3 4", "output": "Ciel" } ]
1,553,776,107
2,147,483,647
PyPy 3
OK
TESTS
99
280
0
hundred, ten = [int(x) for x in input().split()] total = 100*hundred + 10*ten #reduce by number of rounds where Ciel can take 2 100-yen and 2 10-yen, and Hanako can take 22 10-yen skipped = min(hundred//2,ten//24) hundred = hundred - 2*skipped ten = ten - 24*skipped while True: win = 0 if hundred >= 2 and ten >= 2: hundred = hundred - 2 ten = ten - 2 elif hundred >= 1 and ten >= 12: hundred = hundred - 1 ten = ten - 12 elif ten >= 22: ten = ten - 22 else: break win = 1 if ten >= 22: ten = ten - 22 elif hundred >= 1 and ten >= 12: hundred = hundred - 1 ten = ten - 12 elif hundred >= 2 and ten >= 2: hundred = hundred - 2 ten = ten - 2 else: break if win == 0: print ("Hanako") else: print ("Ciel")
Title: Bus Game Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: After Fox Ciel won an onsite round of a programming contest, she took a bus to return to her castle. The fee of the bus was 220 yen. She met Rabbit Hanako in the bus. They decided to play the following game because they got bored in the bus. - Initially, there is a pile that contains *x* 100-yen coins and *y* 10-yen coins. - They take turns alternatively. Ciel takes the first turn. - In each turn, they must take exactly 220 yen from the pile. In Ciel's turn, if there are multiple ways to take 220 yen, she will choose the way that contains the maximal number of 100-yen coins. In Hanako's turn, if there are multiple ways to take 220 yen, she will choose the way that contains the maximal number of 10-yen coins. - If Ciel or Hanako can't take exactly 220 yen from the pile, she loses. Determine the winner of the game. Input Specification: The first line contains two integers *x* (0<=≤<=*x*<=≤<=106) and *y* (0<=≤<=*y*<=≤<=106), separated by a single space. Output Specification: If Ciel wins, print "Ciel". Otherwise, print "Hanako". Demo Input: ['2 2\n', '3 22\n'] Demo Output: ['Ciel\n', 'Hanako\n'] Note: In the first turn (Ciel's turn), she will choose 2 100-yen coins and 2 10-yen coins. In the second turn (Hanako's turn), she will choose 1 100-yen coin and 12 10-yen coins. In the third turn (Ciel's turn), she can't pay exactly 220 yen, so Ciel will lose.
```python hundred, ten = [int(x) for x in input().split()] total = 100*hundred + 10*ten #reduce by number of rounds where Ciel can take 2 100-yen and 2 10-yen, and Hanako can take 22 10-yen skipped = min(hundred//2,ten//24) hundred = hundred - 2*skipped ten = ten - 24*skipped while True: win = 0 if hundred >= 2 and ten >= 2: hundred = hundred - 2 ten = ten - 2 elif hundred >= 1 and ten >= 12: hundred = hundred - 1 ten = ten - 12 elif ten >= 22: ten = ten - 22 else: break win = 1 if ten >= 22: ten = ten - 22 elif hundred >= 1 and ten >= 12: hundred = hundred - 1 ten = ten - 12 elif hundred >= 2 and ten >= 2: hundred = hundred - 2 ten = ten - 2 else: break if win == 0: print ("Hanako") else: print ("Ciel") ```
3.93
364
D
Ghd
PROGRAMMING
2,900
[ "brute force", "math", "probabilities" ]
null
null
John Doe offered his sister Jane Doe find the gcd of some set of numbers *a*. Gcd is a positive integer *g*, such that all number from the set are evenly divisible by *g* and there isn't such *g*' (*g*'<=&gt;<=*g*), that all numbers of the set are evenly divisible by *g*'. Unfortunately Jane couldn't cope with the task and John offered her to find the ghd of the same subset of numbers. Ghd is a positive integer *g*, such that at least half of numbers from the set are evenly divisible by *g* and there isn't such *g*' (*g*'<=&gt;<=*g*) that at least half of the numbers from the set are evenly divisible by *g*'. Jane coped with the task for two hours. Please try it, too.
The first line contains an integer *n* (1<=≤<=*n*<=≤<=106) showing how many numbers are in set *a*. The second line contains space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1012). Please note, that given set can contain equal numbers. Please, do not write the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the %I64d specifier.
Print a single integer *g* — the Ghd of set *a*.
[ "6\n6 2 3 4 5 6\n", "5\n5 5 6 10 15\n" ]
[ "3\n", "5\n" ]
none
2,000
[ { "input": "6\n6 2 3 4 5 6", "output": "3" }, { "input": "5\n5 5 6 10 15", "output": "5" }, { "input": "100\n32 40 7 3 7560 21 7560 7560 10 12 3 7560 7560 7560 7560 5 7560 7560 6 7560 7560 7560 35 7560 18 7560 7560 7560 7560 7560 48 2 7 25 7560 2 2 49 7560 7560 15 16 7560 7560 2 7560 27 7560 7560 7560 7560 3 5 7560 8 7560 42 45 5 7560 5 7560 4 7 3 7560 7 3 7560 7 2 7560 7560 5 3 7560 7560 28 7560 7560 14 7560 5 7560 20 7560 24 7560 2 9 36 7 7560 7560 7560 7560 7560 30 7560 50", "output": "7560" }, { "input": "1\n3", "output": "3" }, { "input": "1\n7", "output": "7" }, { "input": "2\n1 7", "output": "7" }, { "input": "1\n1", "output": "1" } ]
1,581,768,288
2,147,483,647
PyPy 3
WRONG_ANSWER
TESTS
0
451
11,468,800
from random import randrange, randint from functools import reduce from typing import * from sys import stdin if not True: numb = 200560490130 n = 10 ** 6 arr = [randint(1, 4) * numb for i in range(n // 2)] arr += [randint(10 ** 5, 10 ** 12) for _ in range(n // 2)] def read_integers(): return list(map(int, stdin.readline().strip().split())) def decompose(n): return reduce(list.__add__, ([i, n // i] for i in range(1, int(n ** 0.5) + 1) if n % i == 0)) def ghd(arr: List[int]): n_trials = 10 sample_size = 1000 cache = {} sample_arr = [arr[randrange(0, len(arr))] for _ in range(n_trials)] factors = set(reduce(list.__add__, (decompose(x) for x in sample_arr))) # type: Set[int] for f in factors: cnt = sum(x % f == 0 for x in [arr[randrange(0, len(arr))] for _ in range(sample_size)]) if cnt * 2 >= sample_size: cache[f] = cnt return max(cache.keys()) if True: _, = read_integers() arr = read_integers() print(ghd(arr))
Title: Ghd Time Limit: None seconds Memory Limit: None megabytes Problem Description: John Doe offered his sister Jane Doe find the gcd of some set of numbers *a*. Gcd is a positive integer *g*, such that all number from the set are evenly divisible by *g* and there isn't such *g*' (*g*'<=&gt;<=*g*), that all numbers of the set are evenly divisible by *g*'. Unfortunately Jane couldn't cope with the task and John offered her to find the ghd of the same subset of numbers. Ghd is a positive integer *g*, such that at least half of numbers from the set are evenly divisible by *g* and there isn't such *g*' (*g*'<=&gt;<=*g*) that at least half of the numbers from the set are evenly divisible by *g*'. Jane coped with the task for two hours. Please try it, too. Input Specification: The first line contains an integer *n* (1<=≤<=*n*<=≤<=106) showing how many numbers are in set *a*. The second line contains space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1012). Please note, that given set can contain equal numbers. Please, do not write the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the %I64d specifier. Output Specification: Print a single integer *g* — the Ghd of set *a*. Demo Input: ['6\n6 2 3 4 5 6\n', '5\n5 5 6 10 15\n'] Demo Output: ['3\n', '5\n'] Note: none
```python from random import randrange, randint from functools import reduce from typing import * from sys import stdin if not True: numb = 200560490130 n = 10 ** 6 arr = [randint(1, 4) * numb for i in range(n // 2)] arr += [randint(10 ** 5, 10 ** 12) for _ in range(n // 2)] def read_integers(): return list(map(int, stdin.readline().strip().split())) def decompose(n): return reduce(list.__add__, ([i, n // i] for i in range(1, int(n ** 0.5) + 1) if n % i == 0)) def ghd(arr: List[int]): n_trials = 10 sample_size = 1000 cache = {} sample_arr = [arr[randrange(0, len(arr))] for _ in range(n_trials)] factors = set(reduce(list.__add__, (decompose(x) for x in sample_arr))) # type: Set[int] for f in factors: cnt = sum(x % f == 0 for x in [arr[randrange(0, len(arr))] for _ in range(sample_size)]) if cnt * 2 >= sample_size: cache[f] = cnt return max(cache.keys()) if True: _, = read_integers() arr = read_integers() print(ghd(arr)) ```
0
870
C
Maximum splitting
PROGRAMMING
1,300
[ "dp", "greedy", "math", "number theory" ]
null
null
You are given several queries. In the *i*-th query you are given a single positive integer *n**i*. You are to represent *n**i* as a sum of maximum possible number of composite summands and print this maximum number, or print -1, if there are no such splittings. An integer greater than 1 is composite, if it is not prime, i.e. if it has positive divisors not equal to 1 and the integer itself.
The first line contains single integer *q* (1<=≤<=*q*<=≤<=105) — the number of queries. *q* lines follow. The (*i*<=+<=1)-th line contains single integer *n**i* (1<=≤<=*n**i*<=≤<=109) — the *i*-th query.
For each query print the maximum possible number of summands in a valid splitting to composite summands, or -1, if there are no such splittings.
[ "1\n12\n", "2\n6\n8\n", "3\n1\n2\n3\n" ]
[ "3\n", "1\n2\n", "-1\n-1\n-1\n" ]
12 = 4 + 4 + 4 = 4 + 8 = 6 + 6 = 12, but the first splitting has the maximum possible number of summands. 8 = 4 + 4, 6 can't be split into several composite summands. 1, 2, 3 are less than any composite number, so they do not have valid splittings.
1,500
[ { "input": "1\n12", "output": "3" }, { "input": "2\n6\n8", "output": "1\n2" }, { "input": "3\n1\n2\n3", "output": "-1\n-1\n-1" }, { "input": "6\n1\n2\n3\n5\n7\n11", "output": "-1\n-1\n-1\n-1\n-1\n-1" }, { "input": "3\n4\n6\n9", "output": "1\n1\n1" }, { "input": "20\n8\n13\n20\n12\n9\n16\n4\n19\n7\n15\n10\n6\n14\n11\n3\n2\n5\n17\n18\n1", "output": "2\n2\n5\n3\n1\n4\n1\n3\n-1\n2\n2\n1\n3\n-1\n-1\n-1\n-1\n3\n4\n-1" }, { "input": "100\n611\n513\n544\n463\n38\n778\n347\n317\n848\n664\n382\n108\n718\n33\n334\n876\n234\n22\n944\n305\n159\n245\n513\n691\n639\n135\n308\n324\n813\n459\n304\n116\n331\n993\n184\n224\n853\n769\n121\n687\n93\n930\n751\n308\n485\n914\n400\n695\n95\n981\n175\n972\n121\n654\n242\n610\n617\n999\n237\n548\n742\n767\n613\n172\n223\n391\n102\n907\n673\n116\n230\n355\n189\n552\n399\n493\n903\n201\n985\n459\n776\n641\n693\n919\n253\n540\n427\n394\n655\n101\n461\n854\n417\n249\n66\n380\n213\n906\n212\n528", "output": "151\n127\n136\n114\n9\n194\n85\n78\n212\n166\n95\n27\n179\n7\n83\n219\n58\n5\n236\n75\n38\n60\n127\n171\n158\n32\n77\n81\n202\n113\n76\n29\n81\n247\n46\n56\n212\n191\n29\n170\n22\n232\n186\n77\n120\n228\n100\n172\n22\n244\n42\n243\n29\n163\n60\n152\n153\n248\n58\n137\n185\n190\n152\n43\n54\n96\n25\n225\n167\n29\n57\n87\n46\n138\n98\n122\n224\n49\n245\n113\n194\n159\n172\n228\n62\n135\n105\n98\n162\n24\n114\n213\n103\n61\n16\n95\n52\n226\n53\n132" }, { "input": "1\n10000001", "output": "2499999" } ]
1,601,560,275
2,147,483,647
PyPy 3
WRONG_ANSWER
TESTS
1
124
0
q = int(input()) for i in range(q): n = int(input()) if n % 2 == 0: if n == 4 or n > 6: print(n // 4) else: print(-1) else: if n < 9 or n == 11: print(-1) else: print(1 + (n-9) // 4)
Title: Maximum splitting Time Limit: None seconds Memory Limit: None megabytes Problem Description: You are given several queries. In the *i*-th query you are given a single positive integer *n**i*. You are to represent *n**i* as a sum of maximum possible number of composite summands and print this maximum number, or print -1, if there are no such splittings. An integer greater than 1 is composite, if it is not prime, i.e. if it has positive divisors not equal to 1 and the integer itself. Input Specification: The first line contains single integer *q* (1<=≤<=*q*<=≤<=105) — the number of queries. *q* lines follow. The (*i*<=+<=1)-th line contains single integer *n**i* (1<=≤<=*n**i*<=≤<=109) — the *i*-th query. Output Specification: For each query print the maximum possible number of summands in a valid splitting to composite summands, or -1, if there are no such splittings. Demo Input: ['1\n12\n', '2\n6\n8\n', '3\n1\n2\n3\n'] Demo Output: ['3\n', '1\n2\n', '-1\n-1\n-1\n'] Note: 12 = 4 + 4 + 4 = 4 + 8 = 6 + 6 = 12, but the first splitting has the maximum possible number of summands. 8 = 4 + 4, 6 can't be split into several composite summands. 1, 2, 3 are less than any composite number, so they do not have valid splittings.
```python q = int(input()) for i in range(q): n = int(input()) if n % 2 == 0: if n == 4 or n > 6: print(n // 4) else: print(-1) else: if n < 9 or n == 11: print(-1) else: print(1 + (n-9) // 4) ```
0
612
B
HDD is Outdated Technology
PROGRAMMING
1,200
[ "implementation", "math" ]
null
null
HDD hard drives group data by sectors. All files are split to fragments and each of them are written in some sector of hard drive. Note the fragments can be written in sectors in arbitrary order. One of the problems of HDD hard drives is the following: the magnetic head should move from one sector to another to read some file. Find the time need to read file split to *n* fragments. The *i*-th sector contains the *f**i*-th fragment of the file (1<=≤<=*f**i*<=≤<=*n*). Note different sectors contains the different fragments. At the start the magnetic head is in the position that contains the first fragment. The file are reading in the following manner: at first the first fragment is read, then the magnetic head moves to the sector that contains the second fragment, then the second fragment is read and so on until the *n*-th fragment is read. The fragments are read in the order from the first to the *n*-th. It takes |*a*<=-<=*b*| time units to move the magnetic head from the sector *a* to the sector *b*. Reading a fragment takes no time.
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=2·105) — the number of fragments. The second line contains *n* different integers *f**i* (1<=≤<=*f**i*<=≤<=*n*) — the number of the fragment written in the *i*-th sector.
Print the only integer — the number of time units needed to read the file.
[ "3\n3 1 2\n", "5\n1 3 5 4 2\n" ]
[ "3\n", "10\n" ]
In the second example the head moves in the following way: - 1-&gt;2 means movement from the sector 1 to the sector 5, i.e. it takes 4 time units - 2-&gt;3 means movement from the sector 5 to the sector 2, i.e. it takes 3 time units - 3-&gt;4 means movement from the sector 2 to the sector 4, i.e. it takes 2 time units - 4-&gt;5 means movement from the sector 4 to the sector 3, i.e. it takes 1 time units So the answer to the second example is 4 + 3 + 2 + 1 = 10.
0
[ { "input": "3\n3 1 2", "output": "3" }, { "input": "5\n1 3 5 4 2", "output": "10" }, { "input": "1\n1", "output": "0" }, { "input": "1\n1", "output": "0" }, { "input": "1\n1", "output": "0" }, { "input": "10\n8 2 10 3 4 6 1 7 9 5", "output": "40" }, { "input": "10\n1 2 3 4 5 6 7 8 9 10", "output": "9" }, { "input": "10\n1 3 5 7 9 10 8 6 4 2", "output": "45" }, { "input": "100\n11 9 35 34 51 74 16 67 26 21 14 80 84 79 7 61 28 3 53 43 42 5 56 36 69 30 22 88 1 27 65 91 46 31 59 50 17 96 25 18 64 55 78 2 63 24 95 48 93 13 38 76 89 94 15 90 45 81 52 87 83 73 44 49 23 82 85 75 86 33 47 19 58 97 37 20 40 10 92 4 6 68 77 54 71 12 62 60 100 39 41 99 72 29 57 8 70 32 66 98", "output": "3580" }, { "input": "100\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100", "output": "99" }, { "input": "100\n1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53 55 57 59 61 63 65 67 69 71 73 75 77 79 81 83 85 87 89 91 93 95 97 99 100 98 96 94 92 90 88 86 84 82 80 78 76 74 72 70 68 66 64 62 60 58 56 54 52 50 48 46 44 42 40 38 36 34 32 30 28 26 24 22 20 18 16 14 12 10 8 6 4 2", "output": "4950" } ]
1,636,027,470
2,147,483,647
Python 3
OK
TESTS
24
234
22,528,000
n = int(input()) values = list(map(int,input().split())) hm = {} for i, v in enumerate(values): hm[v] = i ans = 0 for i in range(1,n): ans += abs(hm[i+1]-hm[i]) print(ans)
Title: HDD is Outdated Technology Time Limit: None seconds Memory Limit: None megabytes Problem Description: HDD hard drives group data by sectors. All files are split to fragments and each of them are written in some sector of hard drive. Note the fragments can be written in sectors in arbitrary order. One of the problems of HDD hard drives is the following: the magnetic head should move from one sector to another to read some file. Find the time need to read file split to *n* fragments. The *i*-th sector contains the *f**i*-th fragment of the file (1<=≤<=*f**i*<=≤<=*n*). Note different sectors contains the different fragments. At the start the magnetic head is in the position that contains the first fragment. The file are reading in the following manner: at first the first fragment is read, then the magnetic head moves to the sector that contains the second fragment, then the second fragment is read and so on until the *n*-th fragment is read. The fragments are read in the order from the first to the *n*-th. It takes |*a*<=-<=*b*| time units to move the magnetic head from the sector *a* to the sector *b*. Reading a fragment takes no time. Input Specification: The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=2·105) — the number of fragments. The second line contains *n* different integers *f**i* (1<=≤<=*f**i*<=≤<=*n*) — the number of the fragment written in the *i*-th sector. Output Specification: Print the only integer — the number of time units needed to read the file. Demo Input: ['3\n3 1 2\n', '5\n1 3 5 4 2\n'] Demo Output: ['3\n', '10\n'] Note: In the second example the head moves in the following way: - 1-&gt;2 means movement from the sector 1 to the sector 5, i.e. it takes 4 time units - 2-&gt;3 means movement from the sector 5 to the sector 2, i.e. it takes 3 time units - 3-&gt;4 means movement from the sector 2 to the sector 4, i.e. it takes 2 time units - 4-&gt;5 means movement from the sector 4 to the sector 3, i.e. it takes 1 time units So the answer to the second example is 4 + 3 + 2 + 1 = 10.
```python n = int(input()) values = list(map(int,input().split())) hm = {} for i, v in enumerate(values): hm[v] = i ans = 0 for i in range(1,n): ans += abs(hm[i+1]-hm[i]) print(ans) ```
3
368
B
Sereja and Suffixes
PROGRAMMING
1,100
[ "data structures", "dp" ]
null
null
Sereja has an array *a*, consisting of *n* integers *a*1, *a*2, ..., *a**n*. The boy cannot sit and do nothing, he decided to study an array. Sereja took a piece of paper and wrote out *m* integers *l*1,<=*l*2,<=...,<=*l**m* (1<=≤<=*l**i*<=≤<=*n*). For each number *l**i* he wants to know how many distinct numbers are staying on the positions *l**i*, *l**i*<=+<=1, ..., *n*. Formally, he want to find the number of distinct numbers among *a**l**i*,<=*a**l**i*<=+<=1,<=...,<=*a**n*.? Sereja wrote out the necessary array elements but the array was so large and the boy was so pressed for time. Help him, find the answer for the described question for each *l**i*.
The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=105). The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=105) — the array elements. Next *m* lines contain integers *l*1,<=*l*2,<=...,<=*l**m*. The *i*-th line contains integer *l**i* (1<=≤<=*l**i*<=≤<=*n*).
Print *m* lines — on the *i*-th line print the answer to the number *l**i*.
[ "10 10\n1 2 3 4 1 2 3 4 100000 99999\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\n" ]
[ "6\n6\n6\n6\n6\n5\n4\n3\n2\n1\n" ]
none
1,000
[ { "input": "10 10\n1 2 3 4 1 2 3 4 100000 99999\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10", "output": "6\n6\n6\n6\n6\n5\n4\n3\n2\n1" }, { "input": "8 3\n8 6 4 3 4 2 4 8\n6\n4\n2", "output": "3\n4\n5" }, { "input": "7 10\n1 3 8 6 2 2 7\n4\n2\n6\n3\n4\n4\n6\n2\n7\n4", "output": "3\n5\n2\n4\n3\n3\n2\n5\n1\n3" }, { "input": "10 2\n2 6 5 7 2 2 3 2 4 8\n1\n2", "output": "7\n7" }, { "input": "7 1\n68346 10956 76708 23018 84063 34833 80407\n1", "output": "7" }, { "input": "2 2\n8 4\n1\n1", "output": "2\n2" }, { "input": "1 5\n5\n1\n1\n1\n1\n1", "output": "1\n1\n1\n1\n1" }, { "input": "4 7\n3 1 4 2\n4\n1\n2\n3\n2\n4\n4", "output": "1\n4\n3\n2\n3\n1\n1" }, { "input": "4 3\n9 1 7 1\n1\n4\n2", "output": "3\n1\n2" }, { "input": "8 3\n9280 6676 2720 6172 8329 10413 3975 1394\n5\n6\n7", "output": "4\n3\n2" }, { "input": "1 1\n1\n1", "output": "1" } ]
1,668,355,191
2,147,483,647
Python 3
TIME_LIMIT_EXCEEDED
TESTS
10
1,000
5,529,600
n, m = [int(x) for x in input().split()] arr = [int(x) for x in input().split()] for i in range(m): li = int(input()) arr_set = set(arr[li-1:]) print(len(arr_set))
Title: Sereja and Suffixes Time Limit: None seconds Memory Limit: None megabytes Problem Description: Sereja has an array *a*, consisting of *n* integers *a*1, *a*2, ..., *a**n*. The boy cannot sit and do nothing, he decided to study an array. Sereja took a piece of paper and wrote out *m* integers *l*1,<=*l*2,<=...,<=*l**m* (1<=≤<=*l**i*<=≤<=*n*). For each number *l**i* he wants to know how many distinct numbers are staying on the positions *l**i*, *l**i*<=+<=1, ..., *n*. Formally, he want to find the number of distinct numbers among *a**l**i*,<=*a**l**i*<=+<=1,<=...,<=*a**n*.? Sereja wrote out the necessary array elements but the array was so large and the boy was so pressed for time. Help him, find the answer for the described question for each *l**i*. Input Specification: The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=105). The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=105) — the array elements. Next *m* lines contain integers *l*1,<=*l*2,<=...,<=*l**m*. The *i*-th line contains integer *l**i* (1<=≤<=*l**i*<=≤<=*n*). Output Specification: Print *m* lines — on the *i*-th line print the answer to the number *l**i*. Demo Input: ['10 10\n1 2 3 4 1 2 3 4 100000 99999\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\n'] Demo Output: ['6\n6\n6\n6\n6\n5\n4\n3\n2\n1\n'] Note: none
```python n, m = [int(x) for x in input().split()] arr = [int(x) for x in input().split()] for i in range(m): li = int(input()) arr_set = set(arr[li-1:]) print(len(arr_set)) ```
0
409
H
A + B Strikes Back
PROGRAMMING
1,500
[ "*special", "brute force", "constructive algorithms", "dsu", "implementation" ]
null
null
A + B is often used as an example of the easiest problem possible to show some contest platform. However, some scientists have observed that sometimes this problem is not so easy to get accepted. Want to try?
The input contains two integers *a* and *b* (0<=≤<=*a*,<=*b*<=≤<=103), separated by a single space.
Output the sum of the given integers.
[ "5 14\n", "381 492\n" ]
[ "19\n", "873\n" ]
none
0
[ { "input": "5 14", "output": "19" }, { "input": "381 492", "output": "873" }, { "input": "536 298", "output": "834" }, { "input": "143 522", "output": "665" }, { "input": "433 126", "output": "559" }, { "input": "723 350", "output": "1073" }, { "input": "632 264", "output": "896" }, { "input": "923 488", "output": "1411" }, { "input": "522 92", "output": "614" }, { "input": "953 553", "output": "1506" }, { "input": "242 86", "output": "328" }, { "input": "151 691", "output": "842" }, { "input": "441 915", "output": "1356" }, { "input": "740 519", "output": "1259" }, { "input": "339 433", "output": "772" }, { "input": "629 657", "output": "1286" }, { "input": "538 261", "output": "799" }, { "input": "828 485", "output": "1313" }, { "input": "117 399", "output": "516" }, { "input": "399 280", "output": "679" } ]
1,658,573,100
2,147,483,647
Python 3
OK
TESTS
20
46
0
a, b = map(int, input().split(' ')) p = a + b print(p)
Title: A + B Strikes Back Time Limit: None seconds Memory Limit: None megabytes Problem Description: A + B is often used as an example of the easiest problem possible to show some contest platform. However, some scientists have observed that sometimes this problem is not so easy to get accepted. Want to try? Input Specification: The input contains two integers *a* and *b* (0<=≤<=*a*,<=*b*<=≤<=103), separated by a single space. Output Specification: Output the sum of the given integers. Demo Input: ['5 14\n', '381 492\n'] Demo Output: ['19\n', '873\n'] Note: none
```python a, b = map(int, input().split(' ')) p = a + b print(p) ```
3
41
A
Translation
PROGRAMMING
800
[ "implementation", "strings" ]
A. Translation
2
256
The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly.
The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols.
If the word *t* is a word *s*, written reversely, print YES, otherwise print NO.
[ "code\nedoc\n", "abb\naba\n", "code\ncode\n" ]
[ "YES\n", "NO\n", "NO\n" ]
none
500
[ { "input": "code\nedoc", "output": "YES" }, { "input": "abb\naba", "output": "NO" }, { "input": "code\ncode", "output": "NO" }, { "input": "abacaba\nabacaba", "output": "YES" }, { "input": "q\nq", "output": "YES" }, { "input": "asrgdfngfnmfgnhweratgjkk\nasrgdfngfnmfgnhweratgjkk", "output": "NO" }, { "input": "z\na", "output": "NO" }, { "input": "asd\ndsa", "output": "YES" }, { "input": "abcdef\nfecdba", "output": "NO" }, { "input": "ywjjbirapvskozubvxoemscfwl\ngnduubaogtfaiowjizlvjcu", "output": "NO" }, { "input": "mfrmqxtzvgaeuleubcmcxcfqyruwzenguhgrmkuhdgnhgtgkdszwqyd\nmfxufheiperjnhyczclkmzyhcxntdfskzkzdwzzujdinf", "output": "NO" }, { "input": "bnbnemvybqizywlnghlykniaxxxlkhftppbdeqpesrtgkcpoeqowjwhrylpsziiwcldodcoonpimudvrxejjo\ntiynnekmlalogyvrgptbinkoqdwzuiyjlrldxhzjmmp", "output": "NO" }, { "input": "pwlpubwyhzqvcitemnhvvwkmwcaawjvdiwtoxyhbhbxerlypelevasmelpfqwjk\nstruuzebbcenziscuoecywugxncdwzyfozhljjyizpqcgkyonyetarcpwkqhuugsqjuixsxptmbnlfupdcfigacdhhrzb", "output": "NO" }, { "input": "gdvqjoyxnkypfvdxssgrihnwxkeojmnpdeobpecytkbdwujqfjtxsqspxvxpqioyfagzjxupqqzpgnpnpxcuipweunqch\nkkqkiwwasbhezqcfeceyngcyuogrkhqecwsyerdniqiocjehrpkljiljophqhyaiefjpavoom", "output": "NO" }, { "input": "umeszdawsvgkjhlqwzents\nhxqhdungbylhnikwviuh", "output": "NO" }, { "input": "juotpscvyfmgntshcealgbsrwwksgrwnrrbyaqqsxdlzhkbugdyx\nibqvffmfktyipgiopznsqtrtxiijntdbgyy", "output": "NO" }, { "input": "zbwueheveouatecaglziqmudxemhrsozmaujrwlqmppzoumxhamwugedikvkblvmxwuofmpafdprbcftew\nulczwrqhctbtbxrhhodwbcxwimncnexosksujlisgclllxokrsbnozthajnnlilyffmsyko", "output": "NO" }, { "input": "nkgwuugukzcv\nqktnpxedwxpxkrxdvgmfgoxkdfpbzvwsduyiybynbkouonhvmzakeiruhfmvrktghadbfkmwxduoqv", "output": "NO" }, { "input": "incenvizhqpcenhjhehvjvgbsnfixbatrrjstxjzhlmdmxijztphxbrldlqwdfimweepkggzcxsrwelodpnryntepioqpvk\ndhjbjjftlvnxibkklxquwmzhjfvnmwpapdrslioxisbyhhfymyiaqhlgecpxamqnocizwxniubrmpyubvpenoukhcobkdojlybxd", "output": "NO" }, { "input": "w\nw", "output": "YES" }, { "input": "vz\nzv", "output": "YES" }, { "input": "ry\nyr", "output": "YES" }, { "input": "xou\nuox", "output": "YES" }, { "input": "axg\ngax", "output": "NO" }, { "input": "zdsl\nlsdz", "output": "YES" }, { "input": "kudl\nldku", "output": "NO" }, { "input": "zzlzwnqlcl\nlclqnwzlzz", "output": "YES" }, { "input": "vzzgicnzqooejpjzads\nsdazjpjeooqzncigzzv", "output": "YES" }, { "input": "raqhmvmzuwaykjpyxsykr\nxkysrypjkyawuzmvmhqar", "output": "NO" }, { "input": "ngedczubzdcqbxksnxuavdjaqtmdwncjnoaicvmodcqvhfezew\nwezefhvqcdomvciaonjcnwdmtqajdvauxnskxbqcdzbuzcdegn", "output": "YES" }, { "input": "muooqttvrrljcxbroizkymuidvfmhhsjtumksdkcbwwpfqdyvxtrlymofendqvznzlmim\nmimlznzvqdnefomylrtxvydqfpwwbckdskmutjshhmfvdiumykziorbxcjlrrvttqooum", "output": "YES" }, { "input": "vxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaivg\ngviayyikkitmuomcpiakhbxszgbnhvwyzkftwoagzixaearxpjacrnvpvbuzenvovehkmmxvblqyxvctroddksdsgebcmlluqpxv", "output": "YES" }, { "input": "mnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfdc\ncdfmkdgrdptkpewbsqvszipgxvgvuiuzbkkwuowbafkikgvnqdkxnayzdjygvezmtsgywnupocdntipiyiorblqkrzjpzatxahnm", "output": "NO" }, { "input": "dgxmzbqofstzcdgthbaewbwocowvhqpinehpjatnnbrijcolvsatbblsrxabzrpszoiecpwhfjmwuhqrapvtcgvikuxtzbftydkw\nwkdytfbztxukivgctvparqhuwmjfhwpceiozsprzbaxrslbbqasvlocjirbnntajphenipthvwocowbweabhtgdcztsfoqbzmxgd", "output": "NO" }, { "input": "gxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwgeh\nhegwxvocotmzstqfbmpjvijgkcyodlxyjawrpkczpmdspsuhoiruavnnnuwvtwohglkdxjetshkboalvzqbgjgthoteceixioxg", "output": "YES" }, { "input": "sihxuwvmaambplxvjfoskinghzicyfqebjtkysotattkahssumfcgrkheotdxwjckpvapbkaepqrxseyfrwtyaycmrzsrsngkh\nhkgnsrszrmcyaytwrfyesxrqpeakbpavpkcjwxdtoehkrgcfmusshakttatosyktjbeqfycizhgniksofjvxlpbmaamvwuxhis", "output": "YES" }, { "input": "ycnahksbughnonldzrhkysujmylcgcfuludjvjiahtkyzqvkopzqcnwhltbzfugzojqkjjlggmvnultascmygelkiktmfieok\nkoeifmtkiklegkmcsatlunvmggkjjlqjozgufzbtlhwncqzpokvqzykthaijvjdulufcgclymjusyyhrzdlnonhgubskhancy", "output": "NO" }, { "input": "wbqasaehtkfojruzyhrlgwmtyiovmzyfifslvlemhqheyaelzwnthrenjsbmntwaoryzwfbxmscmypvxlfmzpnkkjlvwvmtz\nztmvwvljkknpzmflxvpymcsmxbfwzyroawtnmbsjnerhtnwzleayehqhmelvlsfifyzmvoiytmwglrhyzurjofktheasaqbw", "output": "YES" }, { "input": "imippqurprbhfugngtgifelytadegwrgaefnfhbjjnmzikvjaccotqzemufqieqldgnbmviisgkynzeldlhqxuqphjfmyij\njiymfjhpquxqhldleznykgsiivmbngdlqeiqfumezqtoccajvkizmnjjbhfnfeagrwgedatylefigtgngufhbrpruqppimi", "output": "YES" }, { "input": "bikydffiuisckpvzqlteqfhegsagimodb\nbdomigasgehfqetlqzvpkcsiuiffdykib", "output": "YES" } ]
1,684,506,962
2,147,483,647
Python 3
OK
TESTS
40
92
0
a = input() b = input() q = 0 if len(a) == len(b): for i in range(len(a)): if a[i] != b[len(a) - 1 - i]: q += 1 break if q == 1: print("NO") else: print("YES") else: print("NO")
Title: Translation Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly. Input Specification: The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols. Output Specification: If the word *t* is a word *s*, written reversely, print YES, otherwise print NO. Demo Input: ['code\nedoc\n', 'abb\naba\n', 'code\ncode\n'] Demo Output: ['YES\n', 'NO\n', 'NO\n'] Note: none
```python a = input() b = input() q = 0 if len(a) == len(b): for i in range(len(a)): if a[i] != b[len(a) - 1 - i]: q += 1 break if q == 1: print("NO") else: print("YES") else: print("NO") ```
3.977
842
B
Gleb And Pizza
PROGRAMMING
1,100
[ "geometry" ]
null
null
Gleb ordered pizza home. When the courier delivered the pizza, he was very upset, because several pieces of sausage lay on the crust, and he does not really like the crust. The pizza is a circle of radius *r* and center at the origin. Pizza consists of the main part — circle of radius *r*<=-<=*d* with center at the origin, and crust around the main part of the width *d*. Pieces of sausage are also circles. The radius of the *i* -th piece of the sausage is *r**i*, and the center is given as a pair (*x**i*, *y**i*). Gleb asks you to help determine the number of pieces of sausage caught on the crust. A piece of sausage got on the crust, if it completely lies on the crust.
First string contains two integer numbers *r* and *d* (0<=≤<=*d*<=&lt;<=*r*<=≤<=500) — the radius of pizza and the width of crust. Next line contains one integer number *n* — the number of pieces of sausage (1<=≤<=*n*<=≤<=105). Each of next *n* lines contains three integer numbers *x**i*, *y**i* and *r**i* (<=-<=500<=≤<=*x**i*,<=*y**i*<=≤<=500, 0<=≤<=*r**i*<=≤<=500), where *x**i* and *y**i* are coordinates of the center of *i*-th peace of sausage, *r**i* — radius of *i*-th peace of sausage.
Output the number of pieces of sausage that lay on the crust.
[ "8 4\n7\n7 8 1\n-7 3 2\n0 2 1\n0 -2 2\n-3 -3 1\n0 6 2\n5 3 1\n", "10 8\n4\n0 0 9\n0 0 10\n1 0 1\n1 0 2\n" ]
[ "2\n", "0\n" ]
Below is a picture explaining the first example. Circles of green color denote pieces of sausage lying on the crust.
1,000
[ { "input": "8 4\n7\n7 8 1\n-7 3 2\n0 2 1\n0 -2 2\n-3 -3 1\n0 6 2\n5 3 1", "output": "2" }, { "input": "10 8\n4\n0 0 9\n0 0 10\n1 0 1\n1 0 2", "output": "0" }, { "input": "1 0\n1\n1 1 0", "output": "0" }, { "input": "3 0\n5\n3 0 0\n0 3 0\n-3 0 0\n0 -3 0\n3 0 1", "output": "4" }, { "input": "9 0\n5\n8 1 0\n8 2 0\n8 3 0\n-8 3 0\n-8 2 0", "output": "0" }, { "input": "10 2\n11\n1 1 0\n2 2 3\n3 3 0\n4 4 0\n5 5 0\n6 6 0\n7 7 4\n8 8 7\n9 9 3\n10 10 100\n9 0 1", "output": "2" }, { "input": "5 3\n1\n500 500 10", "output": "0" } ]
1,594,629,029
2,147,483,647
PyPy 3
OK
TESTS
34
920
34,406,400
from math import sqrt dist = lambda p: sqrt(sum([d*d for d in p])) inCrust = lambda l,r,rl,rr: (l-r)>=rl and (l+r)<=rr rr,d = list(map(int,input().split())) #500 xyrl = [list(map(int,input().split())) for _ in range(int(input()))] print(sum([inCrust(dist(xyr[:2]),xyr[2],rr-d,rr) for xyr in xyrl]))
Title: Gleb And Pizza Time Limit: None seconds Memory Limit: None megabytes Problem Description: Gleb ordered pizza home. When the courier delivered the pizza, he was very upset, because several pieces of sausage lay on the crust, and he does not really like the crust. The pizza is a circle of radius *r* and center at the origin. Pizza consists of the main part — circle of radius *r*<=-<=*d* with center at the origin, and crust around the main part of the width *d*. Pieces of sausage are also circles. The radius of the *i* -th piece of the sausage is *r**i*, and the center is given as a pair (*x**i*, *y**i*). Gleb asks you to help determine the number of pieces of sausage caught on the crust. A piece of sausage got on the crust, if it completely lies on the crust. Input Specification: First string contains two integer numbers *r* and *d* (0<=≤<=*d*<=&lt;<=*r*<=≤<=500) — the radius of pizza and the width of crust. Next line contains one integer number *n* — the number of pieces of sausage (1<=≤<=*n*<=≤<=105). Each of next *n* lines contains three integer numbers *x**i*, *y**i* and *r**i* (<=-<=500<=≤<=*x**i*,<=*y**i*<=≤<=500, 0<=≤<=*r**i*<=≤<=500), where *x**i* and *y**i* are coordinates of the center of *i*-th peace of sausage, *r**i* — radius of *i*-th peace of sausage. Output Specification: Output the number of pieces of sausage that lay on the crust. Demo Input: ['8 4\n7\n7 8 1\n-7 3 2\n0 2 1\n0 -2 2\n-3 -3 1\n0 6 2\n5 3 1\n', '10 8\n4\n0 0 9\n0 0 10\n1 0 1\n1 0 2\n'] Demo Output: ['2\n', '0\n'] Note: Below is a picture explaining the first example. Circles of green color denote pieces of sausage lying on the crust.
```python from math import sqrt dist = lambda p: sqrt(sum([d*d for d in p])) inCrust = lambda l,r,rl,rr: (l-r)>=rl and (l+r)<=rr rr,d = list(map(int,input().split())) #500 xyrl = [list(map(int,input().split())) for _ in range(int(input()))] print(sum([inCrust(dist(xyr[:2]),xyr[2],rr-d,rr) for xyr in xyrl])) ```
3
327
B
Hungry Sequence
PROGRAMMING
1,200
[ "math" ]
null
null
Iahub and Iahubina went to a date at a luxury restaurant. Everything went fine until paying for the food. Instead of money, the waiter wants Iahub to write a Hungry sequence consisting of *n* integers. A sequence *a*1, *a*2, ..., *a**n*, consisting of *n* integers, is Hungry if and only if: - Its elements are in increasing order. That is an inequality *a**i*<=&lt;<=*a**j* holds for any two indices *i*,<=*j* (*i*<=&lt;<=*j*). - For any two indices *i* and *j* (*i*<=&lt;<=*j*), *a**j* must not be divisible by *a**i*. Iahub is in trouble, so he asks you for help. Find a Hungry sequence with *n* elements.
The input contains a single integer: *n* (1<=≤<=*n*<=≤<=105).
Output a line that contains *n* space-separated integers *a*1 *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=107), representing a possible Hungry sequence. Note, that each *a**i* must not be greater than 10000000 (107) and less than 1. If there are multiple solutions you can output any one.
[ "3\n", "5\n" ]
[ "2 9 15\n", "11 14 20 27 31\n" ]
none
500
[ { "input": "3", "output": "2 9 15" }, { "input": "5", "output": "11 14 20 27 31" }, { "input": "1", "output": "3" }, { "input": "1000", "output": "3000 3001 3002 3003 3004 3005 3006 3007 3008 3009 3010 3011 3012 3013 3014 3015 3016 3017 3018 3019 3020 3021 3022 3023 3024 3025 3026 3027 3028 3029 3030 3031 3032 3033 3034 3035 3036 3037 3038 3039 3040 3041 3042 3043 3044 3045 3046 3047 3048 3049 3050 3051 3052 3053 3054 3055 3056 3057 3058 3059 3060 3061 3062 3063 3064 3065 3066 3067 3068 3069 3070 3071 3072 3073 3074 3075 3076 3077 3078 3079 3080 3081 3082 3083 3084 3085 3086 3087 3088 3089 3090 3091 3092 3093 3094 3095 3096 3097 3098 3099 3100 3101 3..." }, { "input": "100000", "output": "300000 300001 300002 300003 300004 300005 300006 300007 300008 300009 300010 300011 300012 300013 300014 300015 300016 300017 300018 300019 300020 300021 300022 300023 300024 300025 300026 300027 300028 300029 300030 300031 300032 300033 300034 300035 300036 300037 300038 300039 300040 300041 300042 300043 300044 300045 300046 300047 300048 300049 300050 300051 300052 300053 300054 300055 300056 300057 300058 300059 300060 300061 300062 300063 300064 300065 300066 300067 300068 300069 300070 300071 300072 ..." }, { "input": "46550", "output": "139650 139651 139652 139653 139654 139655 139656 139657 139658 139659 139660 139661 139662 139663 139664 139665 139666 139667 139668 139669 139670 139671 139672 139673 139674 139675 139676 139677 139678 139679 139680 139681 139682 139683 139684 139685 139686 139687 139688 139689 139690 139691 139692 139693 139694 139695 139696 139697 139698 139699 139700 139701 139702 139703 139704 139705 139706 139707 139708 139709 139710 139711 139712 139713 139714 139715 139716 139717 139718 139719 139720 139721 139722 ..." }, { "input": "61324", "output": "183972 183973 183974 183975 183976 183977 183978 183979 183980 183981 183982 183983 183984 183985 183986 183987 183988 183989 183990 183991 183992 183993 183994 183995 183996 183997 183998 183999 184000 184001 184002 184003 184004 184005 184006 184007 184008 184009 184010 184011 184012 184013 184014 184015 184016 184017 184018 184019 184020 184021 184022 184023 184024 184025 184026 184027 184028 184029 184030 184031 184032 184033 184034 184035 184036 184037 184038 184039 184040 184041 184042 184043 184044 ..." }, { "input": "13176", "output": "39528 39529 39530 39531 39532 39533 39534 39535 39536 39537 39538 39539 39540 39541 39542 39543 39544 39545 39546 39547 39548 39549 39550 39551 39552 39553 39554 39555 39556 39557 39558 39559 39560 39561 39562 39563 39564 39565 39566 39567 39568 39569 39570 39571 39572 39573 39574 39575 39576 39577 39578 39579 39580 39581 39582 39583 39584 39585 39586 39587 39588 39589 39590 39591 39592 39593 39594 39595 39596 39597 39598 39599 39600 39601 39602 39603 39604 39605 39606 39607 39608 39609 39610 39611 39612 3..." }, { "input": "73274", "output": "219822 219823 219824 219825 219826 219827 219828 219829 219830 219831 219832 219833 219834 219835 219836 219837 219838 219839 219840 219841 219842 219843 219844 219845 219846 219847 219848 219849 219850 219851 219852 219853 219854 219855 219856 219857 219858 219859 219860 219861 219862 219863 219864 219865 219866 219867 219868 219869 219870 219871 219872 219873 219874 219875 219876 219877 219878 219879 219880 219881 219882 219883 219884 219885 219886 219887 219888 219889 219890 219891 219892 219893 219894 ..." }, { "input": "86947", "output": "260841 260842 260843 260844 260845 260846 260847 260848 260849 260850 260851 260852 260853 260854 260855 260856 260857 260858 260859 260860 260861 260862 260863 260864 260865 260866 260867 260868 260869 260870 260871 260872 260873 260874 260875 260876 260877 260878 260879 260880 260881 260882 260883 260884 260885 260886 260887 260888 260889 260890 260891 260892 260893 260894 260895 260896 260897 260898 260899 260900 260901 260902 260903 260904 260905 260906 260907 260908 260909 260910 260911 260912 260913 ..." }, { "input": "26342", "output": "79026 79027 79028 79029 79030 79031 79032 79033 79034 79035 79036 79037 79038 79039 79040 79041 79042 79043 79044 79045 79046 79047 79048 79049 79050 79051 79052 79053 79054 79055 79056 79057 79058 79059 79060 79061 79062 79063 79064 79065 79066 79067 79068 79069 79070 79071 79072 79073 79074 79075 79076 79077 79078 79079 79080 79081 79082 79083 79084 79085 79086 79087 79088 79089 79090 79091 79092 79093 79094 79095 79096 79097 79098 79099 79100 79101 79102 79103 79104 79105 79106 79107 79108 79109 79110 7..." }, { "input": "22345", "output": "67035 67036 67037 67038 67039 67040 67041 67042 67043 67044 67045 67046 67047 67048 67049 67050 67051 67052 67053 67054 67055 67056 67057 67058 67059 67060 67061 67062 67063 67064 67065 67066 67067 67068 67069 67070 67071 67072 67073 67074 67075 67076 67077 67078 67079 67080 67081 67082 67083 67084 67085 67086 67087 67088 67089 67090 67091 67092 67093 67094 67095 67096 67097 67098 67099 67100 67101 67102 67103 67104 67105 67106 67107 67108 67109 67110 67111 67112 67113 67114 67115 67116 67117 67118 67119 6..." }, { "input": "19639", "output": "58917 58918 58919 58920 58921 58922 58923 58924 58925 58926 58927 58928 58929 58930 58931 58932 58933 58934 58935 58936 58937 58938 58939 58940 58941 58942 58943 58944 58945 58946 58947 58948 58949 58950 58951 58952 58953 58954 58955 58956 58957 58958 58959 58960 58961 58962 58963 58964 58965 58966 58967 58968 58969 58970 58971 58972 58973 58974 58975 58976 58977 58978 58979 58980 58981 58982 58983 58984 58985 58986 58987 58988 58989 58990 58991 58992 58993 58994 58995 58996 58997 58998 58999 59000 59001 5..." }, { "input": "12337", "output": "37011 37012 37013 37014 37015 37016 37017 37018 37019 37020 37021 37022 37023 37024 37025 37026 37027 37028 37029 37030 37031 37032 37033 37034 37035 37036 37037 37038 37039 37040 37041 37042 37043 37044 37045 37046 37047 37048 37049 37050 37051 37052 37053 37054 37055 37056 37057 37058 37059 37060 37061 37062 37063 37064 37065 37066 37067 37068 37069 37070 37071 37072 37073 37074 37075 37076 37077 37078 37079 37080 37081 37082 37083 37084 37085 37086 37087 37088 37089 37090 37091 37092 37093 37094 37095 3..." }, { "input": "67989", "output": "203967 203968 203969 203970 203971 203972 203973 203974 203975 203976 203977 203978 203979 203980 203981 203982 203983 203984 203985 203986 203987 203988 203989 203990 203991 203992 203993 203994 203995 203996 203997 203998 203999 204000 204001 204002 204003 204004 204005 204006 204007 204008 204009 204010 204011 204012 204013 204014 204015 204016 204017 204018 204019 204020 204021 204022 204023 204024 204025 204026 204027 204028 204029 204030 204031 204032 204033 204034 204035 204036 204037 204038 204039 ..." }, { "input": "57610", "output": "172830 172831 172832 172833 172834 172835 172836 172837 172838 172839 172840 172841 172842 172843 172844 172845 172846 172847 172848 172849 172850 172851 172852 172853 172854 172855 172856 172857 172858 172859 172860 172861 172862 172863 172864 172865 172866 172867 172868 172869 172870 172871 172872 172873 172874 172875 172876 172877 172878 172879 172880 172881 172882 172883 172884 172885 172886 172887 172888 172889 172890 172891 172892 172893 172894 172895 172896 172897 172898 172899 172900 172901 172902 ..." }, { "input": "63287", "output": "189861 189862 189863 189864 189865 189866 189867 189868 189869 189870 189871 189872 189873 189874 189875 189876 189877 189878 189879 189880 189881 189882 189883 189884 189885 189886 189887 189888 189889 189890 189891 189892 189893 189894 189895 189896 189897 189898 189899 189900 189901 189902 189903 189904 189905 189906 189907 189908 189909 189910 189911 189912 189913 189914 189915 189916 189917 189918 189919 189920 189921 189922 189923 189924 189925 189926 189927 189928 189929 189930 189931 189932 189933 ..." }, { "input": "952", "output": "2856 2857 2858 2859 2860 2861 2862 2863 2864 2865 2866 2867 2868 2869 2870 2871 2872 2873 2874 2875 2876 2877 2878 2879 2880 2881 2882 2883 2884 2885 2886 2887 2888 2889 2890 2891 2892 2893 2894 2895 2896 2897 2898 2899 2900 2901 2902 2903 2904 2905 2906 2907 2908 2909 2910 2911 2912 2913 2914 2915 2916 2917 2918 2919 2920 2921 2922 2923 2924 2925 2926 2927 2928 2929 2930 2931 2932 2933 2934 2935 2936 2937 2938 2939 2940 2941 2942 2943 2944 2945 2946 2947 2948 2949 2950 2951 2952 2953 2954 2955 2956 2957 2..." }, { "input": "77840", "output": "233520 233521 233522 233523 233524 233525 233526 233527 233528 233529 233530 233531 233532 233533 233534 233535 233536 233537 233538 233539 233540 233541 233542 233543 233544 233545 233546 233547 233548 233549 233550 233551 233552 233553 233554 233555 233556 233557 233558 233559 233560 233561 233562 233563 233564 233565 233566 233567 233568 233569 233570 233571 233572 233573 233574 233575 233576 233577 233578 233579 233580 233581 233582 233583 233584 233585 233586 233587 233588 233589 233590 233591 233592 ..." }, { "input": "42157", "output": "126471 126472 126473 126474 126475 126476 126477 126478 126479 126480 126481 126482 126483 126484 126485 126486 126487 126488 126489 126490 126491 126492 126493 126494 126495 126496 126497 126498 126499 126500 126501 126502 126503 126504 126505 126506 126507 126508 126509 126510 126511 126512 126513 126514 126515 126516 126517 126518 126519 126520 126521 126522 126523 126524 126525 126526 126527 126528 126529 126530 126531 126532 126533 126534 126535 126536 126537 126538 126539 126540 126541 126542 126543 ..." }, { "input": "46375", "output": "139125 139126 139127 139128 139129 139130 139131 139132 139133 139134 139135 139136 139137 139138 139139 139140 139141 139142 139143 139144 139145 139146 139147 139148 139149 139150 139151 139152 139153 139154 139155 139156 139157 139158 139159 139160 139161 139162 139163 139164 139165 139166 139167 139168 139169 139170 139171 139172 139173 139174 139175 139176 139177 139178 139179 139180 139181 139182 139183 139184 139185 139186 139187 139188 139189 139190 139191 139192 139193 139194 139195 139196 139197 ..." }, { "input": "55142", "output": "165426 165427 165428 165429 165430 165431 165432 165433 165434 165435 165436 165437 165438 165439 165440 165441 165442 165443 165444 165445 165446 165447 165448 165449 165450 165451 165452 165453 165454 165455 165456 165457 165458 165459 165460 165461 165462 165463 165464 165465 165466 165467 165468 165469 165470 165471 165472 165473 165474 165475 165476 165477 165478 165479 165480 165481 165482 165483 165484 165485 165486 165487 165488 165489 165490 165491 165492 165493 165494 165495 165496 165497 165498 ..." }, { "input": "60299", "output": "180897 180898 180899 180900 180901 180902 180903 180904 180905 180906 180907 180908 180909 180910 180911 180912 180913 180914 180915 180916 180917 180918 180919 180920 180921 180922 180923 180924 180925 180926 180927 180928 180929 180930 180931 180932 180933 180934 180935 180936 180937 180938 180939 180940 180941 180942 180943 180944 180945 180946 180947 180948 180949 180950 180951 180952 180953 180954 180955 180956 180957 180958 180959 180960 180961 180962 180963 180964 180965 180966 180967 180968 180969 ..." }, { "input": "28862", "output": "86586 86587 86588 86589 86590 86591 86592 86593 86594 86595 86596 86597 86598 86599 86600 86601 86602 86603 86604 86605 86606 86607 86608 86609 86610 86611 86612 86613 86614 86615 86616 86617 86618 86619 86620 86621 86622 86623 86624 86625 86626 86627 86628 86629 86630 86631 86632 86633 86634 86635 86636 86637 86638 86639 86640 86641 86642 86643 86644 86645 86646 86647 86648 86649 86650 86651 86652 86653 86654 86655 86656 86657 86658 86659 86660 86661 86662 86663 86664 86665 86666 86667 86668 86669 86670 8..." }, { "input": "48251", "output": "144753 144754 144755 144756 144757 144758 144759 144760 144761 144762 144763 144764 144765 144766 144767 144768 144769 144770 144771 144772 144773 144774 144775 144776 144777 144778 144779 144780 144781 144782 144783 144784 144785 144786 144787 144788 144789 144790 144791 144792 144793 144794 144795 144796 144797 144798 144799 144800 144801 144802 144803 144804 144805 144806 144807 144808 144809 144810 144811 144812 144813 144814 144815 144816 144817 144818 144819 144820 144821 144822 144823 144824 144825 ..." }, { "input": "2", "output": "6 7" }, { "input": "4", "output": "12 13 14 15" } ]
1,659,438,636
2,147,483,647
PyPy 3-64
COMPILATION_ERROR
TESTS
0
0
0
#include <bits/stdc++.h> using namespace std; typedef long long ll; typedef long double ld; typedef unsigned long long ull; typedef pair<int, int> ii; typedef vector<ii> vii; typedef vector<int> vi; #define rs resize #define all(x) (x).begin(), (x).end() #define rall(x) (x).rbegin(), (x).rend() #define pb push_back #define sz(x) (int)(x).size() #define X first #define Y second #define INF 1000000000 #define MOD 1000000007 #define rep(i, a, b) for(int i=a; i<b; i++) /// vector<bool> sieve(int n){ vector<bool> is_prime(n+1, true); for(int i=2; i<=n; i++){ if (is_prime[i]){ for (int j=2*i; j<=n; j+=i){ is_prime[j] = false; } } } return is_prime; } void solve(){ int n; cin >> n; vector<bool> primes=sieve(10000000); int have=0, i=2; while (have!=n){ if (primes[i]) { have++; cout << i<<' '; } i ++; } } int main() { ios::sync_with_stdio(false); cin.tie(0); cout.tie(0); int t; //cin >> t; t = 1; rep (_, 0, t) solve(); }
Title: Hungry Sequence Time Limit: None seconds Memory Limit: None megabytes Problem Description: Iahub and Iahubina went to a date at a luxury restaurant. Everything went fine until paying for the food. Instead of money, the waiter wants Iahub to write a Hungry sequence consisting of *n* integers. A sequence *a*1, *a*2, ..., *a**n*, consisting of *n* integers, is Hungry if and only if: - Its elements are in increasing order. That is an inequality *a**i*<=&lt;<=*a**j* holds for any two indices *i*,<=*j* (*i*<=&lt;<=*j*). - For any two indices *i* and *j* (*i*<=&lt;<=*j*), *a**j* must not be divisible by *a**i*. Iahub is in trouble, so he asks you for help. Find a Hungry sequence with *n* elements. Input Specification: The input contains a single integer: *n* (1<=≤<=*n*<=≤<=105). Output Specification: Output a line that contains *n* space-separated integers *a*1 *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=107), representing a possible Hungry sequence. Note, that each *a**i* must not be greater than 10000000 (107) and less than 1. If there are multiple solutions you can output any one. Demo Input: ['3\n', '5\n'] Demo Output: ['2 9 15\n', '11 14 20 27 31\n'] Note: none
```python #include <bits/stdc++.h> using namespace std; typedef long long ll; typedef long double ld; typedef unsigned long long ull; typedef pair<int, int> ii; typedef vector<ii> vii; typedef vector<int> vi; #define rs resize #define all(x) (x).begin(), (x).end() #define rall(x) (x).rbegin(), (x).rend() #define pb push_back #define sz(x) (int)(x).size() #define X first #define Y second #define INF 1000000000 #define MOD 1000000007 #define rep(i, a, b) for(int i=a; i<b; i++) /// vector<bool> sieve(int n){ vector<bool> is_prime(n+1, true); for(int i=2; i<=n; i++){ if (is_prime[i]){ for (int j=2*i; j<=n; j+=i){ is_prime[j] = false; } } } return is_prime; } void solve(){ int n; cin >> n; vector<bool> primes=sieve(10000000); int have=0, i=2; while (have!=n){ if (primes[i]) { have++; cout << i<<' '; } i ++; } } int main() { ios::sync_with_stdio(false); cin.tie(0); cout.tie(0); int t; //cin >> t; t = 1; rep (_, 0, t) solve(); } ```
-1
615
A
Bulbs
PROGRAMMING
800
[ "implementation" ]
null
null
Vasya wants to turn on Christmas lights consisting of *m* bulbs. Initially, all bulbs are turned off. There are *n* buttons, each of them is connected to some set of bulbs. Vasya can press any of these buttons. When the button is pressed, it turns on all the bulbs it's connected to. Can Vasya light up all the bulbs? If Vasya presses the button such that some bulbs connected to it are already turned on, they do not change their state, i.e. remain turned on.
The first line of the input contains integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of buttons and the number of bulbs respectively. Each of the next *n* lines contains *x**i* (0<=≤<=*x**i*<=≤<=*m*) — the number of bulbs that are turned on by the *i*-th button, and then *x**i* numbers *y**ij* (1<=≤<=*y**ij*<=≤<=*m*) — the numbers of these bulbs.
If it's possible to turn on all *m* bulbs print "YES", otherwise print "NO".
[ "3 4\n2 1 4\n3 1 3 1\n1 2\n", "3 3\n1 1\n1 2\n1 1\n" ]
[ "YES\n", "NO\n" ]
In the first sample you can press each button once and turn on all the bulbs. In the 2 sample it is impossible to turn on the 3-rd lamp.
500
[ { "input": "3 4\n2 1 4\n3 1 3 1\n1 2", "output": "YES" }, { "input": "3 3\n1 1\n1 2\n1 1", "output": "NO" }, { "input": "3 4\n1 1\n1 2\n1 3", "output": "NO" }, { "input": "1 5\n5 1 2 3 4 5", "output": "YES" }, { "input": "1 5\n5 4 4 1 2 3", "output": "NO" }, { "input": "1 5\n5 1 1 1 1 5", "output": "NO" }, { "input": "2 5\n4 3 1 4 2\n4 2 3 4 5", "output": "YES" }, { "input": "5 7\n2 6 7\n5 1 1 1 1 1\n3 6 5 4\n0\n4 4 3 2 1", "output": "YES" }, { "input": "100 100\n0\n0\n0\n1 53\n0\n0\n1 34\n1 54\n0\n1 14\n0\n1 33\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n1 82\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n1 34\n0\n0\n1 26\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n1 34\n0\n0\n0\n0\n0\n1 3\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n1 40\n0\n0\n0\n1 26\n0\n0\n0\n0\n0\n1 97\n0\n1 5\n0\n0\n0\n0\n0", "output": "NO" }, { "input": "100 100\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0", "output": "NO" }, { "input": "5 6\n3 1 2 6\n3 1 2 6\n1 1\n2 3 4\n3 1 5 6", "output": "YES" }, { "input": "5 2\n1 1\n1 1\n1 1\n1 1\n1 1", "output": "NO" }, { "input": "1 4\n3 1 2 3", "output": "NO" }, { "input": "1 4\n3 2 3 4", "output": "NO" }, { "input": "2 4\n3 2 3 4\n1 1", "output": "YES" }, { "input": "2 4\n3 1 2 3\n1 4", "output": "YES" }, { "input": "5 1\n0\n0\n0\n0\n0", "output": "NO" }, { "input": "1 1\n0", "output": "NO" }, { "input": "1 10\n10 1 2 3 4 5 6 7 8 9 10", "output": "YES" }, { "input": "1 1\n1 1", "output": "YES" }, { "input": "1 100\n99 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99", "output": "NO" }, { "input": "1 3\n3 1 2 1", "output": "NO" }, { "input": "1 100\n100 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 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": "YES" } ]
1,627,224,855
2,147,483,647
PyPy 3
WRONG_ANSWER
TESTS
4
93
20,172,800
nm = [int(x) for x in input().split()] freq = {} for i in range(nm[0]): for j in [int(x) for x in input().split()]: if j in freq: freq[j] += 1 else: freq[j] = 1 for i in range(1, nm[1] + 1): if i not in freq.keys(): print("NO") quit() print("YES")
Title: Bulbs Time Limit: None seconds Memory Limit: None megabytes Problem Description: Vasya wants to turn on Christmas lights consisting of *m* bulbs. Initially, all bulbs are turned off. There are *n* buttons, each of them is connected to some set of bulbs. Vasya can press any of these buttons. When the button is pressed, it turns on all the bulbs it's connected to. Can Vasya light up all the bulbs? If Vasya presses the button such that some bulbs connected to it are already turned on, they do not change their state, i.e. remain turned on. Input Specification: The first line of the input contains integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of buttons and the number of bulbs respectively. Each of the next *n* lines contains *x**i* (0<=≤<=*x**i*<=≤<=*m*) — the number of bulbs that are turned on by the *i*-th button, and then *x**i* numbers *y**ij* (1<=≤<=*y**ij*<=≤<=*m*) — the numbers of these bulbs. Output Specification: If it's possible to turn on all *m* bulbs print "YES", otherwise print "NO". Demo Input: ['3 4\n2 1 4\n3 1 3 1\n1 2\n', '3 3\n1 1\n1 2\n1 1\n'] Demo Output: ['YES\n', 'NO\n'] Note: In the first sample you can press each button once and turn on all the bulbs. In the 2 sample it is impossible to turn on the 3-rd lamp.
```python nm = [int(x) for x in input().split()] freq = {} for i in range(nm[0]): for j in [int(x) for x in input().split()]: if j in freq: freq[j] += 1 else: freq[j] = 1 for i in range(1, nm[1] + 1): if i not in freq.keys(): print("NO") quit() print("YES") ```
0
842
A
Kirill And The Game
PROGRAMMING
1,200
[ "brute force", "two pointers" ]
null
null
Kirill plays a new computer game. He came to the potion store where he can buy any potion. Each potion is characterized by two integers — amount of experience and cost. The efficiency of a potion is the ratio of the amount of experience to the cost. Efficiency may be a non-integer number. For each two integer numbers *a* and *b* such that *l*<=≤<=*a*<=≤<=*r* and *x*<=≤<=*b*<=≤<=*y* there is a potion with experience *a* and cost *b* in the store (that is, there are (*r*<=-<=*l*<=+<=1)·(*y*<=-<=*x*<=+<=1) potions). Kirill wants to buy a potion which has efficiency *k*. Will he be able to do this?
First string contains five integer numbers *l*, *r*, *x*, *y*, *k* (1<=≤<=*l*<=≤<=*r*<=≤<=107, 1<=≤<=*x*<=≤<=*y*<=≤<=107, 1<=≤<=*k*<=≤<=107).
Print "YES" without quotes if a potion with efficiency exactly *k* can be bought in the store and "NO" without quotes otherwise. You can output each of the letters in any register.
[ "1 10 1 10 1\n", "1 5 6 10 1\n" ]
[ "YES", "NO" ]
none
500
[ { "input": "1 10 1 10 1", "output": "YES" }, { "input": "1 5 6 10 1", "output": "NO" }, { "input": "1 1 1 1 1", "output": "YES" }, { "input": "1 1 1 1 2", "output": "NO" }, { "input": "1 100000 1 100000 100000", "output": "YES" }, { "input": "1 100000 1 100000 100001", "output": "NO" }, { "input": "25 10000 200 10000 5", "output": "YES" }, { "input": "1 100000 10 100000 50000", "output": "NO" }, { "input": "91939 94921 10197 89487 1", "output": "NO" }, { "input": "30518 58228 74071 77671 1", "output": "NO" }, { "input": "46646 79126 78816 91164 5", "output": "NO" }, { "input": "30070 83417 92074 99337 2", "output": "NO" }, { "input": "13494 17544 96820 99660 6", "output": "NO" }, { "input": "96918 97018 10077 86510 9", "output": "YES" }, { "input": "13046 45594 14823 52475 1", "output": "YES" }, { "input": "29174 40572 95377 97669 4", "output": "NO" }, { "input": "79894 92433 8634 86398 4", "output": "YES" }, { "input": "96022 98362 13380 94100 6", "output": "YES" }, { "input": "79446 95675 93934 96272 3", "output": "NO" }, { "input": "5440 46549 61481 99500 10", "output": "NO" }, { "input": "21569 53580 74739 87749 3", "output": "NO" }, { "input": "72289 78297 79484 98991 7", "output": "NO" }, { "input": "88417 96645 92742 98450 5", "output": "NO" }, { "input": "71841 96625 73295 77648 8", "output": "NO" }, { "input": "87969 99230 78041 94736 4", "output": "NO" }, { "input": "4 4 1 2 3", "output": "NO" }, { "input": "150 150 1 2 100", "output": "NO" }, { "input": "99 100 1 100 50", "output": "YES" }, { "input": "7 7 3 6 2", "output": "NO" }, { "input": "10 10 1 10 1", "output": "YES" }, { "input": "36 36 5 7 6", "output": "YES" }, { "input": "73 96 1 51 51", "output": "NO" }, { "input": "3 3 1 3 2", "output": "NO" }, { "input": "10000000 10000000 1 100000 10000000", "output": "YES" }, { "input": "9222174 9829060 9418763 9955619 9092468", "output": "NO" }, { "input": "70 70 1 2 50", "output": "NO" }, { "input": "100 200 1 20 5", "output": "YES" }, { "input": "1 200000 65536 65536 65537", "output": "NO" }, { "input": "15 15 1 100 1", "output": "YES" }, { "input": "10000000 10000000 1 10000000 100000", "output": "YES" }, { "input": "10 10 2 5 4", "output": "NO" }, { "input": "67 69 7 7 9", "output": "NO" }, { "input": "100000 10000000 1 10000000 100000", "output": "YES" }, { "input": "9 12 1 2 7", "output": "NO" }, { "input": "5426234 6375745 2636512 8492816 4409404", "output": "NO" }, { "input": "6134912 6134912 10000000 10000000 999869", "output": "NO" }, { "input": "3 3 1 100 1", "output": "YES" }, { "input": "10000000 10000000 10 10000000 100000", "output": "YES" }, { "input": "4 4 1 100 2", "output": "YES" }, { "input": "8 13 1 4 7", "output": "NO" }, { "input": "10 10 100000 10000000 10000000", "output": "NO" }, { "input": "5 6 1 4 2", "output": "YES" }, { "input": "1002 1003 1 2 1000", "output": "NO" }, { "input": "4 5 1 2 2", "output": "YES" }, { "input": "5 6 1 5 1", "output": "YES" }, { "input": "15 21 2 4 7", "output": "YES" }, { "input": "4 5 3 7 1", "output": "YES" }, { "input": "15 15 3 4 4", "output": "NO" }, { "input": "3 6 1 2 2", "output": "YES" }, { "input": "2 10 3 6 3", "output": "YES" }, { "input": "1 10000000 1 10000000 100000", "output": "YES" }, { "input": "8 13 1 2 7", "output": "NO" }, { "input": "98112 98112 100000 100000 128850", "output": "NO" }, { "input": "2 2 1 2 1", "output": "YES" }, { "input": "8 8 3 4 2", "output": "YES" }, { "input": "60 60 2 3 25", "output": "NO" }, { "input": "16 17 2 5 5", "output": "NO" }, { "input": "2 4 1 3 1", "output": "YES" }, { "input": "4 5 1 2 3", "output": "NO" }, { "input": "10 10 3 4 3", "output": "NO" }, { "input": "10 10000000 999999 10000000 300", "output": "NO" }, { "input": "100 120 9 11 10", "output": "YES" }, { "input": "8 20 1 3 4", "output": "YES" }, { "input": "10 14 2 3 4", "output": "YES" }, { "input": "2000 2001 1 3 1000", "output": "YES" }, { "input": "12 13 2 3 5", "output": "NO" }, { "input": "7 7 2 3 3", "output": "NO" }, { "input": "5 8 1 10000000 4", "output": "YES" }, { "input": "5 5 1 1 4", "output": "NO" }, { "input": "5 5 1 6 2", "output": "NO" }, { "input": "200 300 4000381 4000382 4000381", "output": "NO" }, { "input": "11 17 2 5 2", "output": "NO" }, { "input": "9999999 10000000 1 10000000 999997", "output": "NO" }, { "input": "7 8 2 3 3", "output": "NO" }, { "input": "7 7 3 3 2", "output": "NO" }, { "input": "15 15 2 3 7", "output": "NO" }, { "input": "65408 65408 859 859 10000000", "output": "NO" }, { "input": "1000000 10000000 1 100000 1", "output": "NO" }, { "input": "6 12 2 3 2", "output": "YES" }, { "input": "7 8 1 3 3", "output": "NO" }, { "input": "4 4 1 2 2", "output": "YES" }, { "input": "2 3 1 2 2", "output": "YES" }, { "input": "11 14 2 3 5", "output": "NO" }, { "input": "7 7 1 10 3", "output": "NO" }, { "input": "49 50 1 2 27", "output": "NO" }, { "input": "1 10000000 1 10000000 123456", "output": "YES" }, { "input": "100000 10000000 100 10000000 100000", "output": "YES" }, { "input": "17 19 2 3 8", "output": "NO" }, { "input": "4 6 3 9 1", "output": "YES" }, { "input": "19 20 6 7 3", "output": "NO" }, { "input": "5000000 10000000 1 4999999 1", "output": "NO" } ]
1,504,078,771
2,147,483,647
Python 3
TIME_LIMIT_EXCEEDED
TESTS
4
2,000
0
l, r, x, y, k = map(int, input().split()) ans = "NO" for i in range(l, r+1): for z in range(x, y+1): if i/z == k: ans = "YES" print(ans)
Title: Kirill And The Game Time Limit: None seconds Memory Limit: None megabytes Problem Description: Kirill plays a new computer game. He came to the potion store where he can buy any potion. Each potion is characterized by two integers — amount of experience and cost. The efficiency of a potion is the ratio of the amount of experience to the cost. Efficiency may be a non-integer number. For each two integer numbers *a* and *b* such that *l*<=≤<=*a*<=≤<=*r* and *x*<=≤<=*b*<=≤<=*y* there is a potion with experience *a* and cost *b* in the store (that is, there are (*r*<=-<=*l*<=+<=1)·(*y*<=-<=*x*<=+<=1) potions). Kirill wants to buy a potion which has efficiency *k*. Will he be able to do this? Input Specification: First string contains five integer numbers *l*, *r*, *x*, *y*, *k* (1<=≤<=*l*<=≤<=*r*<=≤<=107, 1<=≤<=*x*<=≤<=*y*<=≤<=107, 1<=≤<=*k*<=≤<=107). Output Specification: Print "YES" without quotes if a potion with efficiency exactly *k* can be bought in the store and "NO" without quotes otherwise. You can output each of the letters in any register. Demo Input: ['1 10 1 10 1\n', '1 5 6 10 1\n'] Demo Output: ['YES', 'NO'] Note: none
```python l, r, x, y, k = map(int, input().split()) ans = "NO" for i in range(l, r+1): for z in range(x, y+1): if i/z == k: ans = "YES" print(ans) ```
0
870
C
Maximum splitting
PROGRAMMING
1,300
[ "dp", "greedy", "math", "number theory" ]
null
null
You are given several queries. In the *i*-th query you are given a single positive integer *n**i*. You are to represent *n**i* as a sum of maximum possible number of composite summands and print this maximum number, or print -1, if there are no such splittings. An integer greater than 1 is composite, if it is not prime, i.e. if it has positive divisors not equal to 1 and the integer itself.
The first line contains single integer *q* (1<=≤<=*q*<=≤<=105) — the number of queries. *q* lines follow. The (*i*<=+<=1)-th line contains single integer *n**i* (1<=≤<=*n**i*<=≤<=109) — the *i*-th query.
For each query print the maximum possible number of summands in a valid splitting to composite summands, or -1, if there are no such splittings.
[ "1\n12\n", "2\n6\n8\n", "3\n1\n2\n3\n" ]
[ "3\n", "1\n2\n", "-1\n-1\n-1\n" ]
12 = 4 + 4 + 4 = 4 + 8 = 6 + 6 = 12, but the first splitting has the maximum possible number of summands. 8 = 4 + 4, 6 can't be split into several composite summands. 1, 2, 3 are less than any composite number, so they do not have valid splittings.
1,500
[ { "input": "1\n12", "output": "3" }, { "input": "2\n6\n8", "output": "1\n2" }, { "input": "3\n1\n2\n3", "output": "-1\n-1\n-1" }, { "input": "6\n1\n2\n3\n5\n7\n11", "output": "-1\n-1\n-1\n-1\n-1\n-1" }, { "input": "3\n4\n6\n9", "output": "1\n1\n1" }, { "input": "20\n8\n13\n20\n12\n9\n16\n4\n19\n7\n15\n10\n6\n14\n11\n3\n2\n5\n17\n18\n1", "output": "2\n2\n5\n3\n1\n4\n1\n3\n-1\n2\n2\n1\n3\n-1\n-1\n-1\n-1\n3\n4\n-1" }, { "input": "100\n611\n513\n544\n463\n38\n778\n347\n317\n848\n664\n382\n108\n718\n33\n334\n876\n234\n22\n944\n305\n159\n245\n513\n691\n639\n135\n308\n324\n813\n459\n304\n116\n331\n993\n184\n224\n853\n769\n121\n687\n93\n930\n751\n308\n485\n914\n400\n695\n95\n981\n175\n972\n121\n654\n242\n610\n617\n999\n237\n548\n742\n767\n613\n172\n223\n391\n102\n907\n673\n116\n230\n355\n189\n552\n399\n493\n903\n201\n985\n459\n776\n641\n693\n919\n253\n540\n427\n394\n655\n101\n461\n854\n417\n249\n66\n380\n213\n906\n212\n528", "output": "151\n127\n136\n114\n9\n194\n85\n78\n212\n166\n95\n27\n179\n7\n83\n219\n58\n5\n236\n75\n38\n60\n127\n171\n158\n32\n77\n81\n202\n113\n76\n29\n81\n247\n46\n56\n212\n191\n29\n170\n22\n232\n186\n77\n120\n228\n100\n172\n22\n244\n42\n243\n29\n163\n60\n152\n153\n248\n58\n137\n185\n190\n152\n43\n54\n96\n25\n225\n167\n29\n57\n87\n46\n138\n98\n122\n224\n49\n245\n113\n194\n159\n172\n228\n62\n135\n105\n98\n162\n24\n114\n213\n103\n61\n16\n95\n52\n226\n53\n132" }, { "input": "1\n10000001", "output": "2499999" } ]
1,597,052,736
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
5
93
6,963,200
def f(n): if n < 0: return -float('inf') if n == 0: return 0 if n % 4 == 1: return 1 + f(n - 9) if n % 4 == 2: return 1 + f(n - 6) if n % 4 == 3: return 1 + f(n - 13) return n // 4 q = int(input()) for _ in range(q): n = int(input()) c = f(n) if c < 0: c = -1 print(c)
Title: Maximum splitting Time Limit: None seconds Memory Limit: None megabytes Problem Description: You are given several queries. In the *i*-th query you are given a single positive integer *n**i*. You are to represent *n**i* as a sum of maximum possible number of composite summands and print this maximum number, or print -1, if there are no such splittings. An integer greater than 1 is composite, if it is not prime, i.e. if it has positive divisors not equal to 1 and the integer itself. Input Specification: The first line contains single integer *q* (1<=≤<=*q*<=≤<=105) — the number of queries. *q* lines follow. The (*i*<=+<=1)-th line contains single integer *n**i* (1<=≤<=*n**i*<=≤<=109) — the *i*-th query. Output Specification: For each query print the maximum possible number of summands in a valid splitting to composite summands, or -1, if there are no such splittings. Demo Input: ['1\n12\n', '2\n6\n8\n', '3\n1\n2\n3\n'] Demo Output: ['3\n', '1\n2\n', '-1\n-1\n-1\n'] Note: 12 = 4 + 4 + 4 = 4 + 8 = 6 + 6 = 12, but the first splitting has the maximum possible number of summands. 8 = 4 + 4, 6 can't be split into several composite summands. 1, 2, 3 are less than any composite number, so they do not have valid splittings.
```python def f(n): if n < 0: return -float('inf') if n == 0: return 0 if n % 4 == 1: return 1 + f(n - 9) if n % 4 == 2: return 1 + f(n - 6) if n % 4 == 3: return 1 + f(n - 13) return n // 4 q = int(input()) for _ in range(q): n = int(input()) c = f(n) if c < 0: c = -1 print(c) ```
0
4
A
Watermelon
PROGRAMMING
800
[ "brute force", "math" ]
A. Watermelon
1
64
One hot summer day Pete and his friend Billy decided to buy a watermelon. They chose the biggest and the ripest one, in their opinion. After that the watermelon was weighed, and the scales showed *w* kilos. They rushed home, dying of thirst, and decided to divide the berry, however they faced a hard problem. Pete and Billy are great fans of even numbers, that's why they want to divide the watermelon in such a way that each of the two parts weighs even number of kilos, at the same time it is not obligatory that the parts are equal. The boys are extremely tired and want to start their meal as soon as possible, that's why you should help them and find out, if they can divide the watermelon in the way they want. For sure, each of them should get a part of positive weight.
The first (and the only) input line contains integer number *w* (1<=≤<=*w*<=≤<=100) — the weight of the watermelon bought by the boys.
Print YES, if the boys can divide the watermelon into two parts, each of them weighing even number of kilos; and NO in the opposite case.
[ "8\n" ]
[ "YES\n" ]
For example, the boys can divide the watermelon into two parts of 2 and 6 kilos respectively (another variant — two parts of 4 and 4 kilos).
0
[ { "input": "8", "output": "YES" }, { "input": "5", "output": "NO" }, { "input": "4", "output": "YES" }, { "input": "3", "output": "NO" }, { "input": "2", "output": "NO" }, { "input": "1", "output": "NO" }, { "input": "7", "output": "NO" }, { "input": "6", "output": "YES" }, { "input": "10", "output": "YES" }, { "input": "9", "output": "NO" }, { "input": "53", "output": "NO" }, { "input": "77", "output": "NO" }, { "input": "32", "output": "YES" }, { "input": "44", "output": "YES" }, { "input": "98", "output": "YES" }, { "input": "99", "output": "NO" }, { "input": "90", "output": "YES" }, { "input": "67", "output": "NO" }, { "input": "100", "output": "YES" }, { "input": "88", "output": "YES" } ]
1,689,616,792
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
0
62
0
w = 5; if(w%2==0 and w>2): print ('Yes'); else: print ('No');
Title: Watermelon Time Limit: 1 seconds Memory Limit: 64 megabytes Problem Description: One hot summer day Pete and his friend Billy decided to buy a watermelon. They chose the biggest and the ripest one, in their opinion. After that the watermelon was weighed, and the scales showed *w* kilos. They rushed home, dying of thirst, and decided to divide the berry, however they faced a hard problem. Pete and Billy are great fans of even numbers, that's why they want to divide the watermelon in such a way that each of the two parts weighs even number of kilos, at the same time it is not obligatory that the parts are equal. The boys are extremely tired and want to start their meal as soon as possible, that's why you should help them and find out, if they can divide the watermelon in the way they want. For sure, each of them should get a part of positive weight. Input Specification: The first (and the only) input line contains integer number *w* (1<=≤<=*w*<=≤<=100) — the weight of the watermelon bought by the boys. Output Specification: Print YES, if the boys can divide the watermelon into two parts, each of them weighing even number of kilos; and NO in the opposite case. Demo Input: ['8\n'] Demo Output: ['YES\n'] Note: For example, the boys can divide the watermelon into two parts of 2 and 6 kilos respectively (another variant — two parts of 4 and 4 kilos).
```python w = 5; if(w%2==0 and w>2): print ('Yes'); else: print ('No'); ```
0
632
C
The Smallest String Concatenation
PROGRAMMING
1,700
[ "sortings", "strings" ]
null
null
You're given a list of *n* strings *a*1,<=*a*2,<=...,<=*a**n*. You'd like to concatenate them together in some order such that the resulting string would be lexicographically smallest. Given the list of strings, output the lexicographically smallest concatenation.
The first line contains integer *n* — the number of strings (1<=≤<=*n*<=≤<=5·104). Each of the next *n* lines contains one string *a**i* (1<=≤<=|*a**i*|<=≤<=50) consisting of only lowercase English letters. The sum of string lengths will not exceed 5·104.
Print the only string *a* — the lexicographically smallest string concatenation.
[ "4\nabba\nabacaba\nbcd\ner\n", "5\nx\nxx\nxxa\nxxaa\nxxaaa\n", "3\nc\ncb\ncba\n" ]
[ "abacabaabbabcder\n", "xxaaaxxaaxxaxxx\n", "cbacbc\n" ]
none
0
[ { "input": "4\nabba\nabacaba\nbcd\ner", "output": "abacabaabbabcder" }, { "input": "5\nx\nxx\nxxa\nxxaa\nxxaaa", "output": "xxaaaxxaaxxaxxx" }, { "input": "3\nc\ncb\ncba", "output": "cbacbc" }, { "input": "10\naba\nabaaca\naba\nacaaaabbac\nabaacac\nb\ncabbcccaab\nbaacbb\nbcab\ncc", "output": "abaabaabaacaabaacacacaaaabbacbaacbbbbcabcabbcccaabcc" }, { "input": "13\nclgknjjojyuvdtv\nclgknjjojyuvdtvzxz\nclgknjjojyuvdtvzxzxradqhm\ngvzpnckalbaubfviyhijosiixvxaydxagvymq\nclgknjjojyuvdtvjgcwarwvgilcctdiytmvmvfxxkezymt\nbvwzpuut\nclgknjjojyuvdtvjgcwarwvgilcctdiytmvmvfxxkezymth\nclgknjjojyuvdtvjgcwarwvgilcctdiytmvmvfxxkezymtlkdl\nrbwvaayhhunlvmwexvkmqbquypsnjbdkkeytthekttywypud\nqlvkfkshkxyrfefigtngifxsuhpjgkb\nlhxtkfjrrwcvzuamizotupjqckaq\nenngvhxvpaoiyhctddyscasyqtilvbvgfphayykjeoylcvmjz\nxkazwk", "output": "bvwzpuutclgknjjojyuvdtvclgknjjojyuvdtvjgcwarwvgilcctdiytmvmvfxxkezymtclgknjjojyuvdtvjgcwarwvgilcctdiytmvmvfxxkezymthclgknjjojyuvdtvjgcwarwvgilcctdiytmvmvfxxkezymtlkdlclgknjjojyuvdtvzxzclgknjjojyuvdtvzxzxradqhmenngvhxvpaoiyhctddyscasyqtilvbvgfphayykjeoylcvmjzgvzpnckalbaubfviyhijosiixvxaydxagvymqlhxtkfjrrwcvzuamizotupjqckaqqlvkfkshkxyrfefigtngifxsuhpjgkbrbwvaayhhunlvmwexvkmqbquypsnjbdkkeytthekttywypudxkazwk" } ]
1,617,019,816
2,147,483,647
PyPy 3
COMPILATION_ERROR
TESTS
0
0
0
#include <bits/stdc++.h> using namespace std; bool comp(string el1, string el2){ return el1 + el2 < el2 + el1; } int main(){ int n; cin >> n; vector<string> b(n); for (int i = 0; i < n; i ++){ cin >> b[i]; } sort(b.begin(), b.end(), comp); for (int i = 0; i < n; i ++){ cout << b[i]; } }
Title: The Smallest String Concatenation Time Limit: None seconds Memory Limit: None megabytes Problem Description: You're given a list of *n* strings *a*1,<=*a*2,<=...,<=*a**n*. You'd like to concatenate them together in some order such that the resulting string would be lexicographically smallest. Given the list of strings, output the lexicographically smallest concatenation. Input Specification: The first line contains integer *n* — the number of strings (1<=≤<=*n*<=≤<=5·104). Each of the next *n* lines contains one string *a**i* (1<=≤<=|*a**i*|<=≤<=50) consisting of only lowercase English letters. The sum of string lengths will not exceed 5·104. Output Specification: Print the only string *a* — the lexicographically smallest string concatenation. Demo Input: ['4\nabba\nabacaba\nbcd\ner\n', '5\nx\nxx\nxxa\nxxaa\nxxaaa\n', '3\nc\ncb\ncba\n'] Demo Output: ['abacabaabbabcder\n', 'xxaaaxxaaxxaxxx\n', 'cbacbc\n'] Note: none
```python #include <bits/stdc++.h> using namespace std; bool comp(string el1, string el2){ return el1 + el2 < el2 + el1; } int main(){ int n; cin >> n; vector<string> b(n); for (int i = 0; i < n; i ++){ cin >> b[i]; } sort(b.begin(), b.end(), comp); for (int i = 0; i < n; i ++){ cout << b[i]; } } ```
-1
228
A
Is your horseshoe on the other hoof?
PROGRAMMING
800
[ "implementation" ]
null
null
Valera the Horse is going to the party with friends. He has been following the fashion trends for a while, and he knows that it is very popular to wear all horseshoes of different color. Valera has got four horseshoes left from the last year, but maybe some of them have the same color. In this case he needs to go to the store and buy some few more horseshoes, not to lose face in front of his stylish comrades. Fortunately, the store sells horseshoes of all colors under the sun and Valera has enough money to buy any four of them. However, in order to save the money, he would like to spend as little money as possible, so you need to help Valera and determine what is the minimum number of horseshoes he needs to buy to wear four horseshoes of different colors to a party.
The first line contains four space-separated integers *s*1,<=*s*2,<=*s*3,<=*s*4 (1<=≤<=*s*1,<=*s*2,<=*s*3,<=*s*4<=≤<=109) — the colors of horseshoes Valera has. Consider all possible colors indexed with integers.
Print a single integer — the minimum number of horseshoes Valera needs to buy.
[ "1 7 3 3\n", "7 7 7 7\n" ]
[ "1\n", "3\n" ]
none
500
[ { "input": "1 7 3 3", "output": "1" }, { "input": "7 7 7 7", "output": "3" }, { "input": "81170865 673572653 756938629 995577259", "output": "0" }, { "input": "3491663 217797045 522540872 715355328", "output": "0" }, { "input": "251590420 586975278 916631563 586975278", "output": "1" }, { "input": "259504825 377489979 588153796 377489979", "output": "1" }, { "input": "652588203 931100304 931100304 652588203", "output": "2" }, { "input": "391958720 651507265 391958720 651507265", "output": "2" }, { "input": "90793237 90793237 90793237 90793237", "output": "3" }, { "input": "551651653 551651653 551651653 551651653", "output": "3" }, { "input": "156630260 609654355 668943582 973622757", "output": "0" }, { "input": "17061017 110313588 434481173 796661222", "output": "0" }, { "input": "24975422 256716298 337790533 690960249", "output": "0" }, { "input": "255635360 732742923 798648949 883146723", "output": "0" }, { "input": "133315691 265159773 734556507 265159773", "output": "1" }, { "input": "28442865 741657755 978106882 978106882", "output": "1" }, { "input": "131245479 174845575 497483467 131245479", "output": "1" }, { "input": "139159884 616215581 958341883 616215581", "output": "1" }, { "input": "147784432 947653080 947653080 947653080", "output": "2" }, { "input": "94055790 756126496 756126496 94055790", "output": "2" }, { "input": "240458500 511952208 240458500 511952208", "output": "2" }, { "input": "681828506 972810624 972810624 681828506", "output": "2" }, { "input": "454961014 454961014 454961014 454961014", "output": "3" }, { "input": "915819430 915819430 915819430 915819430", "output": "3" }, { "input": "671645142 671645142 671645142 671645142", "output": "3" }, { "input": "132503558 132503558 132503558 132503558", "output": "3" }, { "input": "5 5 999999 6", "output": "1" }, { "input": "1 1 2 5", "output": "1" }, { "input": "2 1 2 3", "output": "1" }, { "input": "1 1 3 5", "output": "1" }, { "input": "1 1 3 3", "output": "2" }, { "input": "2 2 2 1", "output": "2" }, { "input": "3 1 1 1", "output": "2" }, { "input": "1 2 2 2", "output": "2" } ]
1,687,319,744
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
2
62
0
a = "".join(input().split()) t = 0 for i in range(10): if a.count(str(i)) > 1: t = a.count(str(i))-1 print(t)
Title: Is your horseshoe on the other hoof? Time Limit: None seconds Memory Limit: None megabytes Problem Description: Valera the Horse is going to the party with friends. He has been following the fashion trends for a while, and he knows that it is very popular to wear all horseshoes of different color. Valera has got four horseshoes left from the last year, but maybe some of them have the same color. In this case he needs to go to the store and buy some few more horseshoes, not to lose face in front of his stylish comrades. Fortunately, the store sells horseshoes of all colors under the sun and Valera has enough money to buy any four of them. However, in order to save the money, he would like to spend as little money as possible, so you need to help Valera and determine what is the minimum number of horseshoes he needs to buy to wear four horseshoes of different colors to a party. Input Specification: The first line contains four space-separated integers *s*1,<=*s*2,<=*s*3,<=*s*4 (1<=≤<=*s*1,<=*s*2,<=*s*3,<=*s*4<=≤<=109) — the colors of horseshoes Valera has. Consider all possible colors indexed with integers. Output Specification: Print a single integer — the minimum number of horseshoes Valera needs to buy. Demo Input: ['1 7 3 3\n', '7 7 7 7\n'] Demo Output: ['1\n', '3\n'] Note: none
```python a = "".join(input().split()) t = 0 for i in range(10): if a.count(str(i)) > 1: t = a.count(str(i))-1 print(t) ```
0
478
B
Random Teams
PROGRAMMING
1,300
[ "combinatorics", "constructive algorithms", "greedy", "math" ]
null
null
*n* participants of the competition were split into *m* teams in some manner so that each team has at least one participant. After the competition each pair of participants from the same team became friends. Your task is to write a program that will find the minimum and the maximum number of pairs of friends that could have formed by the end of the competition.
The only line of input contains two integers *n* and *m*, separated by a single space (1<=≤<=*m*<=≤<=*n*<=≤<=109) — the number of participants and the number of teams respectively.
The only line of the output should contain two integers *k**min* and *k**max* — the minimum possible number of pairs of friends and the maximum possible number of pairs of friends respectively.
[ "5 1\n", "3 2\n", "6 3\n" ]
[ "10 10\n", "1 1\n", "3 6\n" ]
In the first sample all the participants get into one team, so there will be exactly ten pairs of friends. In the second sample at any possible arrangement one team will always have two participants and the other team will always have one participant. Thus, the number of pairs of friends will always be equal to one. In the third sample minimum number of newly formed friendships can be achieved if participants were split on teams consisting of 2 people, maximum number can be achieved if participants were split on teams of 1, 1 and 4 people.
1,000
[ { "input": "5 1", "output": "10 10" }, { "input": "3 2", "output": "1 1" }, { "input": "6 3", "output": "3 6" }, { "input": "5 3", "output": "2 3" }, { "input": "10 2", "output": "20 36" }, { "input": "10 6", "output": "4 10" }, { "input": "1000000000 1", "output": "499999999500000000 499999999500000000" }, { "input": "5000000 12", "output": "1041664166668 12499942500066" }, { "input": "1833 195", "output": "7722 1342341" }, { "input": "1000000000 1000000000", "output": "0 0" }, { "input": "1000000000 1000000", "output": "499500000000 499000500499500000" }, { "input": "1000000000 32170", "output": "15541930838100 499967831017438365" }, { "input": "1000000 1000", "output": "499500000 499000999500" }, { "input": "1234 1123", "output": "111 6216" }, { "input": "599222887 298488", "output": "601178656545 179355218158217800" }, { "input": "999999999 500000000", "output": "499999999 124999999750000000" }, { "input": "1000000000 384842119", "output": "845473643 189209609585784021" }, { "input": "1000000000 384842119", "output": "845473643 189209609585784021" }, { "input": "1000000000 2", "output": "249999999500000000 499999998500000001" }, { "input": "1000000000 999999999", "output": "1 1" }, { "input": "38447 383", "output": "1910550 724453080" }, { "input": "100000000 99999799", "output": "201 20301" }, { "input": "1 1", "output": "0 0" }, { "input": "2 1", "output": "1 1" }, { "input": "2 2", "output": "0 0" }, { "input": "10 10", "output": "0 0" } ]
1,651,936,565
2,147,483,647
PyPy 3-64
OK
TESTS
26
62
0
n, m = map(int, input().split()) q = n // m r = n % m mn = m * q * (q - 1) // 2 + r * q mx = (n - m + 1) * (n - m) // 2 print(mn, mx)
Title: Random Teams Time Limit: None seconds Memory Limit: None megabytes Problem Description: *n* participants of the competition were split into *m* teams in some manner so that each team has at least one participant. After the competition each pair of participants from the same team became friends. Your task is to write a program that will find the minimum and the maximum number of pairs of friends that could have formed by the end of the competition. Input Specification: The only line of input contains two integers *n* and *m*, separated by a single space (1<=≤<=*m*<=≤<=*n*<=≤<=109) — the number of participants and the number of teams respectively. Output Specification: The only line of the output should contain two integers *k**min* and *k**max* — the minimum possible number of pairs of friends and the maximum possible number of pairs of friends respectively. Demo Input: ['5 1\n', '3 2\n', '6 3\n'] Demo Output: ['10 10\n', '1 1\n', '3 6\n'] Note: In the first sample all the participants get into one team, so there will be exactly ten pairs of friends. In the second sample at any possible arrangement one team will always have two participants and the other team will always have one participant. Thus, the number of pairs of friends will always be equal to one. In the third sample minimum number of newly formed friendships can be achieved if participants were split on teams consisting of 2 people, maximum number can be achieved if participants were split on teams of 1, 1 and 4 people.
```python n, m = map(int, input().split()) q = n // m r = n % m mn = m * q * (q - 1) // 2 + r * q mx = (n - m + 1) * (n - m) // 2 print(mn, mx) ```
3
294
A
Shaass and Oskols
PROGRAMMING
800
[ "implementation", "math" ]
null
null
Shaass has decided to hunt some birds. There are *n* horizontal electricity wires aligned parallel to each other. Wires are numbered 1 to *n* from top to bottom. On each wire there are some oskols sitting next to each other. Oskol is the name of a delicious kind of birds in Shaass's territory. Supposed there are *a**i* oskols sitting on the *i*-th wire. Sometimes Shaass shots one of the birds and the bird dies (suppose that this bird sat at the *i*-th wire). Consequently all the birds on the *i*-th wire to the left of the dead bird get scared and jump up on the wire number *i*<=-<=1, if there exists no upper wire they fly away. Also all the birds to the right of the dead bird jump down on wire number *i*<=+<=1, if there exists no such wire they fly away. Shaass has shot *m* birds. You're given the initial number of birds on each wire, tell him how many birds are sitting on each wire after the shots.
The first line of the input contains an integer *n*, (1<=≤<=*n*<=≤<=100). The next line contains a list of space-separated integers *a*1,<=*a*2,<=...,<=*a**n*, (0<=≤<=*a**i*<=≤<=100). The third line contains an integer *m*, (0<=≤<=*m*<=≤<=100). Each of the next *m* lines contains two integers *x**i* and *y**i*. The integers mean that for the *i*-th time Shaass shoot the *y**i*-th (from left) bird on the *x**i*-th wire, (1<=≤<=*x**i*<=≤<=*n*,<=1<=≤<=*y**i*). It's guaranteed there will be at least *y**i* birds on the *x**i*-th wire at that moment.
On the *i*-th line of the output print the number of birds on the *i*-th wire.
[ "5\n10 10 10 10 10\n5\n2 5\n3 13\n2 12\n1 13\n4 6\n", "3\n2 4 1\n1\n2 2\n" ]
[ "0\n12\n5\n0\n16\n", "3\n0\n3\n" ]
none
500
[ { "input": "5\n10 10 10 10 10\n5\n2 5\n3 13\n2 12\n1 13\n4 6", "output": "0\n12\n5\n0\n16" }, { "input": "3\n2 4 1\n1\n2 2", "output": "3\n0\n3" }, { "input": "5\n58 51 45 27 48\n5\n4 9\n5 15\n4 5\n5 8\n1 43", "output": "0\n66\n57\n7\n0" }, { "input": "10\n48 53 10 28 91 56 81 2 67 52\n2\n2 40\n6 51", "output": "87\n0\n23\n28\n141\n0\n86\n2\n67\n52" }, { "input": "2\n72 45\n6\n1 69\n2 41\n1 19\n2 7\n1 5\n2 1", "output": "0\n0" }, { "input": "10\n95 54 36 39 98 30 19 24 14 12\n3\n9 5\n8 15\n7 5", "output": "95\n54\n36\n39\n98\n34\n0\n28\n13\n21" }, { "input": "100\n95 15 25 18 64 62 23 59 70 84 50 26 87 35 75 86 0 22 77 60 66 41 21 9 75 50 25 3 69 14 39 68 64 46 59 99 2 0 21 76 90 12 61 42 6 91 36 39 47 41 93 81 66 57 70 36 68 89 52 1 19 93 67 22 76 20 8 81 98 18 100 73 61 93 75 80 53 72 40 20 2 86 33 59 27 16 11 26 55 44 47 55 94 58 2 55 82 22 9 95\n0", "output": "95\n15\n25\n18\n64\n62\n23\n59\n70\n84\n50\n26\n87\n35\n75\n86\n0\n22\n77\n60\n66\n41\n21\n9\n75\n50\n25\n3\n69\n14\n39\n68\n64\n46\n59\n99\n2\n0\n21\n76\n90\n12\n61\n42\n6\n91\n36\n39\n47\n41\n93\n81\n66\n57\n70\n36\n68\n89\n52\n1\n19\n93\n67\n22\n76\n20\n8\n81\n98\n18\n100\n73\n61\n93\n75\n80\n53\n72\n40\n20\n2\n86\n33\n59\n27\n16\n11\n26\n55\n44\n47\n55\n94\n58\n2\n55\n82\n22\n9\n95" }, { "input": "1\n100\n1\n1 100", "output": "0" }, { "input": "1\n100\n1\n1 1", "output": "0" }, { "input": "1\n50\n1\n1 25", "output": "0" }, { "input": "2\n50 0\n1\n1 1", "output": "0\n49" }, { "input": "1\n10\n0", "output": "10" } ]
1,594,316,613
2,147,483,647
PyPy 3
RUNTIME_ERROR
TESTS
26
372
22,220,800
n=int(input()) l=list(map(int,input().split())) m=int(input()) for i in range(m): x,y=map(int,input().split()) if x==1: l[1]+=l[0]-y l[0]=0 elif x==n: l[x-2]+=y-1 l[x-1]=0 else: l[x]+=l[x-1]-y l[x-2]+=y-1 l[x-1]=0 for i in l: print(i)
Title: Shaass and Oskols Time Limit: None seconds Memory Limit: None megabytes Problem Description: Shaass has decided to hunt some birds. There are *n* horizontal electricity wires aligned parallel to each other. Wires are numbered 1 to *n* from top to bottom. On each wire there are some oskols sitting next to each other. Oskol is the name of a delicious kind of birds in Shaass's territory. Supposed there are *a**i* oskols sitting on the *i*-th wire. Sometimes Shaass shots one of the birds and the bird dies (suppose that this bird sat at the *i*-th wire). Consequently all the birds on the *i*-th wire to the left of the dead bird get scared and jump up on the wire number *i*<=-<=1, if there exists no upper wire they fly away. Also all the birds to the right of the dead bird jump down on wire number *i*<=+<=1, if there exists no such wire they fly away. Shaass has shot *m* birds. You're given the initial number of birds on each wire, tell him how many birds are sitting on each wire after the shots. Input Specification: The first line of the input contains an integer *n*, (1<=≤<=*n*<=≤<=100). The next line contains a list of space-separated integers *a*1,<=*a*2,<=...,<=*a**n*, (0<=≤<=*a**i*<=≤<=100). The third line contains an integer *m*, (0<=≤<=*m*<=≤<=100). Each of the next *m* lines contains two integers *x**i* and *y**i*. The integers mean that for the *i*-th time Shaass shoot the *y**i*-th (from left) bird on the *x**i*-th wire, (1<=≤<=*x**i*<=≤<=*n*,<=1<=≤<=*y**i*). It's guaranteed there will be at least *y**i* birds on the *x**i*-th wire at that moment. Output Specification: On the *i*-th line of the output print the number of birds on the *i*-th wire. Demo Input: ['5\n10 10 10 10 10\n5\n2 5\n3 13\n2 12\n1 13\n4 6\n', '3\n2 4 1\n1\n2 2\n'] Demo Output: ['0\n12\n5\n0\n16\n', '3\n0\n3\n'] Note: none
```python n=int(input()) l=list(map(int,input().split())) m=int(input()) for i in range(m): x,y=map(int,input().split()) if x==1: l[1]+=l[0]-y l[0]=0 elif x==n: l[x-2]+=y-1 l[x-1]=0 else: l[x]+=l[x-1]-y l[x-2]+=y-1 l[x-1]=0 for i in l: print(i) ```
-1
721
A
One-dimensional Japanese Crossword
PROGRAMMING
800
[ "implementation" ]
null
null
Recently Adaltik discovered japanese crosswords. Japanese crossword is a picture, represented as a table sized *a*<=×<=*b* squares, and each square is colored white or black. There are integers to the left of the rows and to the top of the columns, encrypting the corresponding row or column. The number of integers represents how many groups of black squares there are in corresponding row or column, and the integers themselves represents the number of consecutive black squares in corresponding group (you can find more detailed explanation in Wikipedia [https://en.wikipedia.org/wiki/Japanese_crossword](https://en.wikipedia.org/wiki/Japanese_crossword)). Adaltik decided that the general case of japanese crossword is too complicated and drew a row consisting of *n* squares (e.g. japanese crossword sized 1<=×<=*n*), which he wants to encrypt in the same way as in japanese crossword. Help Adaltik find the numbers encrypting the row he drew.
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the length of the row. The second line of the input contains a single string consisting of *n* characters 'B' or 'W', ('B' corresponds to black square, 'W' — to white square in the row that Adaltik drew).
The first line should contain a single integer *k* — the number of integers encrypting the row, e.g. the number of groups of black squares in the row. The second line should contain *k* integers, encrypting the row, e.g. corresponding to sizes of groups of consecutive black squares in the order from left to right.
[ "3\nBBW\n", "5\nBWBWB\n", "4\nWWWW\n", "4\nBBBB\n", "13\nWBBBBWWBWBBBW\n" ]
[ "1\n2 ", "3\n1 1 1 ", "0\n", "1\n4 ", "3\n4 1 3 " ]
The last sample case correspond to the picture in the statement.
500
[ { "input": "3\nBBW", "output": "1\n2 " }, { "input": "5\nBWBWB", "output": "3\n1 1 1 " }, { "input": "4\nWWWW", "output": "0" }, { "input": "4\nBBBB", "output": "1\n4 " }, { "input": "13\nWBBBBWWBWBBBW", "output": "3\n4 1 3 " }, { "input": "1\nB", "output": "1\n1 " }, { "input": "2\nBB", "output": "1\n2 " }, { "input": "100\nWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWB", "output": "50\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 " }, { "input": "1\nW", "output": "0" }, { "input": "2\nWW", "output": "0" }, { "input": "2\nWB", "output": "1\n1 " }, { "input": "2\nBW", "output": "1\n1 " }, { "input": "3\nBBB", "output": "1\n3 " }, { "input": "3\nBWB", "output": "2\n1 1 " }, { "input": "3\nWBB", "output": "1\n2 " }, { "input": "3\nWWB", "output": "1\n1 " }, { "input": "3\nWBW", "output": "1\n1 " }, { "input": "3\nBWW", "output": "1\n1 " }, { "input": "3\nWWW", "output": "0" }, { "input": "100\nBBBWWWWWWBBWWBBWWWBBWBBBBBBBBBBBWBBBWBBWWWBBWWBBBWBWWBBBWWBBBWBBBBBWWWBWWBBWWWWWWBWBBWWBWWWBWBWWWWWB", "output": "21\n3 2 2 2 11 3 2 2 3 1 3 3 5 1 2 1 2 1 1 1 1 " }, { "input": "5\nBBBWB", "output": "2\n3 1 " }, { "input": "5\nBWWWB", "output": "2\n1 1 " }, { "input": "5\nWWWWB", "output": "1\n1 " }, { "input": "5\nBWWWW", "output": "1\n1 " }, { "input": "5\nBBBWW", "output": "1\n3 " }, { "input": "5\nWWBBB", "output": "1\n3 " }, { "input": "10\nBBBBBWWBBB", "output": "2\n5 3 " }, { "input": "10\nBBBBWBBWBB", "output": "3\n4 2 2 " }, { "input": "20\nBBBBBWWBWBBWBWWBWBBB", "output": "6\n5 1 2 1 1 3 " }, { "input": "20\nBBBWWWWBBWWWBWBWWBBB", "output": "5\n3 2 1 1 3 " }, { "input": "20\nBBBBBBBBWBBBWBWBWBBB", "output": "5\n8 3 1 1 3 " }, { "input": "20\nBBBWBWBWWWBBWWWWBWBB", "output": "6\n3 1 1 2 1 2 " }, { "input": "40\nBBBBBBWWWWBWBWWWBWWWWWWWWWWWBBBBBBBBBBBB", "output": "5\n6 1 1 1 12 " }, { "input": "40\nBBBBBWBWWWBBWWWBWBWWBBBBWWWWBWBWBBBBBBBB", "output": "9\n5 1 2 1 1 4 1 1 8 " }, { "input": "50\nBBBBBBBBBBBWWWWBWBWWWWBBBBBBBBWWWWWWWBWWWWBWBBBBBB", "output": "7\n11 1 1 8 1 1 6 " }, { "input": "50\nWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWW", "output": "0" }, { "input": "50\nBBBBBWWWWWBWWWBWWWWWBWWWBWWWWWWBBWBBWWWWBWWWWWWWBW", "output": "9\n5 1 1 1 1 2 2 1 1 " }, { "input": "50\nWWWWBWWBWWWWWWWWWWWWWWWWWWWWWWWWWBWBWBWWWWWWWBBBBB", "output": "6\n1 1 1 1 1 5 " }, { "input": "50\nBBBBBWBWBWWBWBWWWWWWBWBWBWWWWWWWWWWWWWBWBWWWWBWWWB", "output": "12\n5 1 1 1 1 1 1 1 1 1 1 1 " }, { "input": "50\nBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB", "output": "1\n50 " }, { "input": "100\nBBBBBBBBBBBWBWWWWBWWBBWBBWWWWWWWWWWBWBWWBWWWWWWWWWWWBBBWWBBWWWWWBWBWWWWBWWWWWWWWWWWBWWWWWBBBBBBBBBBB", "output": "15\n11 1 1 2 2 1 1 1 3 2 1 1 1 1 11 " }, { "input": "100\nBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB", "output": "1\n100 " }, { "input": "100\nBBBBBBBBBBBBBBBBBBBBWBWBWWWWWBWWWWWWWWWWWWWWBBWWWBWWWWBWWBWWWWWWBWWWWWWWWWWWWWBWBBBBBBBBBBBBBBBBBBBB", "output": "11\n20 1 1 1 2 1 1 1 1 1 20 " }, { "input": "100\nBBBBWWWWWWWWWWWWWWWWWWWWWWWWWBWBWWWWWBWBWWWWWWBBWWWWWWWWWWWWBWWWWBWWWWWWWWWWWWBWWWWWWWBWWWWWWWBBBBBB", "output": "11\n4 1 1 1 1 2 1 1 1 1 6 " }, { "input": "5\nBWBWB", "output": "3\n1 1 1 " }, { "input": "10\nWWBWWWBWBB", "output": "3\n1 1 2 " }, { "input": "50\nBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB", "output": "1\n50 " }, { "input": "50\nBBBBBBBBBBBBBBBBBWWBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB", "output": "2\n17 31 " }, { "input": "100\nBBBBBBBBBBBBBBBBBBBBBBBBWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB", "output": "2\n24 42 " }, { "input": "90\nWWBWWBWBBWBBWWBWBWBBBWBWBBBWBWBWBWBWBWBWBWBBBBBWBBWWWWBWBBWBWWBBBWBWBWWBWBWBWBWWWWWWBWBBBB", "output": "30\n1 1 2 2 1 1 3 1 3 1 1 1 1 1 1 1 5 2 1 2 1 3 1 1 1 1 1 1 1 4 " }, { "input": "100\nBWWWBWBWBBBBBWBWWBWBWWWBWBWBWWBBWWBBBWBBBWWBWBWWBBBBWBWBBBWBWBBWWWWWWBWWBBBBWBWBWWBWBWWWBWBWWBWBWWWB", "output": "31\n1 1 1 5 1 1 1 1 1 1 2 3 3 1 1 4 1 3 1 2 1 4 1 1 1 1 1 1 1 1 1 " }, { "input": "90\nWBWBBBBBBWWWBBWWBWWWBBWWBWWWBWBBWBWBBWWWWBWBWBBWBBWBWWWBBWBBWWWWBWBBWWWBBBWBBWBWBBBBWWBWWB", "output": "25\n1 6 2 1 2 1 1 2 1 2 1 1 2 2 1 2 2 1 2 3 2 1 4 1 1 " }, { "input": "80\nBBWWBBBWBBWWWWBBWBWBBWWWWWBWBBWWBWBWBWBWBWWBWWBWWWBWBBWBBWBBWBBBWWBBBBBBBWBBBWBB", "output": "23\n2 3 2 2 1 2 1 2 1 1 1 1 1 1 1 1 2 2 2 3 7 3 2 " }, { "input": "65\nWWWWBWWWBBBBBWWWWWWBBBWWBBBBWWWWWWWWBBBWWWWBWBWWBBWWWWBWWWBBWBBBB", "output": "11\n1 5 3 4 3 1 1 2 1 2 4 " } ]
1,642,324,816
2,147,483,647
PyPy 3-64
OK
TESTS
54
93
0
n=int(input()) s=input().split('W') m=[] for i in range(len(s)): if len(s[i])>0: m.append(len(s[i])) print(len(m)) print(*m)
Title: One-dimensional Japanese Crossword Time Limit: None seconds Memory Limit: None megabytes Problem Description: Recently Adaltik discovered japanese crosswords. Japanese crossword is a picture, represented as a table sized *a*<=×<=*b* squares, and each square is colored white or black. There are integers to the left of the rows and to the top of the columns, encrypting the corresponding row or column. The number of integers represents how many groups of black squares there are in corresponding row or column, and the integers themselves represents the number of consecutive black squares in corresponding group (you can find more detailed explanation in Wikipedia [https://en.wikipedia.org/wiki/Japanese_crossword](https://en.wikipedia.org/wiki/Japanese_crossword)). Adaltik decided that the general case of japanese crossword is too complicated and drew a row consisting of *n* squares (e.g. japanese crossword sized 1<=×<=*n*), which he wants to encrypt in the same way as in japanese crossword. Help Adaltik find the numbers encrypting the row he drew. Input Specification: The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the length of the row. The second line of the input contains a single string consisting of *n* characters 'B' or 'W', ('B' corresponds to black square, 'W' — to white square in the row that Adaltik drew). Output Specification: The first line should contain a single integer *k* — the number of integers encrypting the row, e.g. the number of groups of black squares in the row. The second line should contain *k* integers, encrypting the row, e.g. corresponding to sizes of groups of consecutive black squares in the order from left to right. Demo Input: ['3\nBBW\n', '5\nBWBWB\n', '4\nWWWW\n', '4\nBBBB\n', '13\nWBBBBWWBWBBBW\n'] Demo Output: ['1\n2 ', '3\n1 1 1 ', '0\n', '1\n4 ', '3\n4 1 3 '] Note: The last sample case correspond to the picture in the statement.
```python n=int(input()) s=input().split('W') m=[] for i in range(len(s)): if len(s[i])>0: m.append(len(s[i])) print(len(m)) print(*m) ```
3
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" } ]
1,536,840,112
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
0
93
0
a=input() if 'hello' in a: print('Yes') else: print('No')
Title: Chat room Time Limit: 1 seconds Memory Limit: 256 megabytes Problem Description: Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*. Input Specification: The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters. Output Specification: If Vasya managed to say hello, print "YES", otherwise print "NO". Demo Input: ['ahhellllloou\n', 'hlelo\n'] Demo Output: ['YES\n', 'NO\n'] Note: none
```python a=input() if 'hello' in a: print('Yes') else: print('No') ```
0
433
A
Kitahara Haruki's Gift
PROGRAMMING
1,100
[ "brute force", "implementation" ]
null
null
Kitahara Haruki has bought *n* apples for Touma Kazusa and Ogiso Setsuna. Now he wants to divide all the apples between the friends. Each apple weights 100 grams or 200 grams. Of course Kitahara Haruki doesn't want to offend any of his friend. Therefore the total weight of the apples given to Touma Kazusa must be equal to the total weight of the apples given to Ogiso Setsuna. But unfortunately Kitahara Haruki doesn't have a knife right now, so he cannot split any apple into some parts. Please, tell him: is it possible to divide all the apples in a fair way between his friends?
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of apples. The second line contains *n* integers *w*1,<=*w*2,<=...,<=*w**n* (*w**i*<==<=100 or *w**i*<==<=200), where *w**i* is the weight of the *i*-th apple.
In a single line print "YES" (without the quotes) if it is possible to divide all the apples between his friends. Otherwise print "NO" (without the quotes).
[ "3\n100 200 100\n", "4\n100 100 100 200\n" ]
[ "YES\n", "NO\n" ]
In the first test sample Kitahara Haruki can give the first and the last apple to Ogiso Setsuna and the middle apple to Touma Kazusa.
500
[ { "input": "3\n100 200 100", "output": "YES" }, { "input": "4\n100 100 100 200", "output": "NO" }, { "input": "1\n100", "output": "NO" }, { "input": "1\n200", "output": "NO" }, { "input": "2\n100 100", "output": "YES" }, { "input": "2\n200 200", "output": "YES" }, { "input": "100\n200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200", "output": "YES" }, { "input": "100\n200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 100 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200", "output": "NO" }, { "input": "52\n200 200 100 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 100 200 200 200 200 200 200 200 200 200 200 200 200 200 200 100 200 200 200 200 100 200 100 200 200 200 100 200 200", "output": "YES" }, { "input": "2\n100 200", "output": "NO" }, { "input": "2\n200 100", "output": "NO" }, { "input": "3\n100 100 100", "output": "NO" }, { "input": "3\n200 200 200", "output": "NO" }, { "input": "3\n200 100 200", "output": "NO" }, { "input": "4\n100 100 100 100", "output": "YES" }, { "input": "4\n200 200 200 200", "output": "YES" }, { "input": "100\n200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 100 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 100 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200", "output": "YES" }, { "input": "100\n200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 100 200 100 200 100 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200", "output": "NO" }, { "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": "YES" }, { "input": "100\n100 100 200 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100", "output": "NO" }, { "input": "100\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 200 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 200 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100", "output": "YES" }, { "input": "100\n100 100 100 100 100 100 100 100 200 100 100 200 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 200 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100", "output": "NO" }, { "input": "99\n200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200", "output": "NO" }, { "input": "99\n200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 100 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200", "output": "NO" }, { "input": "99\n200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 100 200 200 200 200 200 100 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200", "output": "YES" }, { "input": "99\n200 200 200 200 200 200 200 200 200 200 200 100 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 100 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 100 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200", "output": "NO" }, { "input": "99\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", "output": "NO" }, { "input": "99\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 200 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100", "output": "YES" }, { "input": "99\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 200 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 200 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100", "output": "NO" }, { "input": "100\n100 100 200 100 100 200 200 200 200 100 200 100 100 100 200 100 100 100 100 200 100 100 100 100 100 100 200 100 100 200 200 100 100 100 200 200 200 100 200 200 100 200 100 100 200 100 200 200 100 200 200 100 100 200 200 100 200 200 100 100 200 100 200 100 200 200 200 200 200 100 200 200 200 200 200 200 100 100 200 200 200 100 100 100 200 100 100 200 100 100 100 200 200 100 100 200 200 200 200 100", "output": "YES" }, { "input": "100\n100 100 200 200 100 200 100 100 100 100 100 100 200 100 200 200 200 100 100 200 200 200 200 200 100 200 100 200 100 100 100 200 100 100 200 100 200 100 100 100 200 200 100 100 100 200 200 200 200 200 100 200 200 100 100 100 100 200 100 100 200 100 100 100 100 200 200 200 100 200 100 200 200 200 100 100 200 200 200 200 100 200 100 200 200 100 200 100 200 200 200 200 200 200 100 100 100 200 200 100", "output": "NO" }, { "input": "100\n100 200 100 100 200 200 200 200 100 200 200 200 200 200 200 200 200 200 100 100 100 200 200 200 200 200 100 200 200 200 200 100 200 200 100 100 200 100 100 100 200 100 100 100 200 100 200 100 200 200 200 100 100 200 100 200 100 200 100 100 100 200 100 200 100 100 100 100 200 200 200 200 100 200 200 100 200 100 100 100 200 100 100 100 100 100 200 100 100 100 200 200 200 100 200 100 100 100 200 200", "output": "YES" }, { "input": "99\n100 200 200 200 100 200 100 200 200 100 100 100 100 200 100 100 200 100 200 100 100 200 100 100 200 200 100 100 100 100 200 200 200 200 200 100 100 200 200 100 100 100 100 200 200 100 100 100 100 100 200 200 200 100 100 100 200 200 200 100 200 100 100 100 100 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 100 100 200 100 200 200 200 200 100 200 100 100 100 100 100 100 100 100 100", "output": "YES" }, { "input": "99\n100 200 100 100 100 100 200 200 100 200 100 100 200 100 100 100 100 100 100 200 100 100 100 100 100 100 100 200 100 200 100 100 100 100 100 100 100 200 200 200 200 200 200 200 100 200 100 200 100 200 100 200 100 100 200 200 200 100 200 200 200 200 100 200 100 200 200 200 200 100 200 100 200 200 100 200 200 200 200 200 100 100 200 100 100 100 100 200 200 200 100 100 200 200 200 200 200 200 200", "output": "NO" }, { "input": "99\n200 100 100 100 200 200 200 100 100 100 100 100 100 100 100 100 200 200 100 200 200 100 200 100 100 200 200 200 100 200 100 200 200 100 200 100 200 200 200 100 100 200 200 200 200 100 100 100 100 200 200 200 200 100 200 200 200 100 100 100 200 200 200 100 200 100 200 100 100 100 200 100 200 200 100 200 200 200 100 100 100 200 200 200 100 200 200 200 100 100 100 200 100 200 100 100 100 200 200", "output": "YES" }, { "input": "56\n100 200 200 200 200 200 100 200 100 100 200 100 100 100 100 100 200 200 200 100 200 100 100 200 200 200 100 200 100 200 200 100 100 100 100 100 200 100 200 100 200 200 200 100 100 200 200 200 200 200 200 200 200 200 200 100", "output": "YES" }, { "input": "72\n200 100 200 200 200 100 100 200 200 100 100 100 100 200 100 200 100 100 100 100 200 100 200 100 100 200 100 100 200 100 200 100 100 200 100 200 100 100 200 200 200 200 200 100 100 200 200 200 200 100 100 100 200 200 100 100 100 100 100 200 100 100 200 100 100 200 200 100 100 200 100 200", "output": "YES" }, { "input": "32\n200 200 200 100 100 100 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 100 200 200 200 200 200 200", "output": "YES" }, { "input": "48\n200 200 200 200 200 200 100 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 100 200 100 200 200 200 200 200 200", "output": "NO" }, { "input": "60\n100 100 200 200 100 200 100 200 100 100 100 100 100 100 200 100 100 100 200 100 200 100 100 100 100 100 200 100 200 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 200 100 100 100", "output": "YES" }, { "input": "24\n200 200 100 100 200 100 200 200 100 200 200 200 200 200 100 200 200 200 200 200 200 200 200 100", "output": "YES" }, { "input": "40\n100 100 200 200 200 200 100 100 100 200 100 100 200 200 100 100 100 100 100 200 100 200 200 100 200 200 200 100 100 100 100 100 200 200 100 200 100 100 200 100", "output": "NO" }, { "input": "5\n200 200 200 200 200", "output": "NO" }, { "input": "9\n100 100 100 200 100 100 200 100 200", "output": "YES" }, { "input": "1\n200", "output": "NO" }, { "input": "7\n200 200 200 100 200 200 200", "output": "NO" }, { "input": "4\n100 100 200 200", "output": "YES" }, { "input": "6\n100 100 100 200 200 200", "output": "NO" }, { "input": "4\n200 100 100 200", "output": "YES" }, { "input": "5\n100 100 100 100 200", "output": "YES" } ]
1,674,823,488
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
5
31
0
a=int(input());b=list(map(int, input().split()));print("YES" if (((b.count(100) % 2 == 0 or b.count(100) == 0) and (b.count(200) % 2 == 1 or b.count(200) == 0)) and a != 1) else "NO")
Title: Kitahara Haruki's Gift Time Limit: None seconds Memory Limit: None megabytes Problem Description: Kitahara Haruki has bought *n* apples for Touma Kazusa and Ogiso Setsuna. Now he wants to divide all the apples between the friends. Each apple weights 100 grams or 200 grams. Of course Kitahara Haruki doesn't want to offend any of his friend. Therefore the total weight of the apples given to Touma Kazusa must be equal to the total weight of the apples given to Ogiso Setsuna. But unfortunately Kitahara Haruki doesn't have a knife right now, so he cannot split any apple into some parts. Please, tell him: is it possible to divide all the apples in a fair way between his friends? Input Specification: The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of apples. The second line contains *n* integers *w*1,<=*w*2,<=...,<=*w**n* (*w**i*<==<=100 or *w**i*<==<=200), where *w**i* is the weight of the *i*-th apple. Output Specification: In a single line print "YES" (without the quotes) if it is possible to divide all the apples between his friends. Otherwise print "NO" (without the quotes). Demo Input: ['3\n100 200 100\n', '4\n100 100 100 200\n'] Demo Output: ['YES\n', 'NO\n'] Note: In the first test sample Kitahara Haruki can give the first and the last apple to Ogiso Setsuna and the middle apple to Touma Kazusa.
```python a=int(input());b=list(map(int, input().split()));print("YES" if (((b.count(100) % 2 == 0 or b.count(100) == 0) and (b.count(200) % 2 == 1 or b.count(200) == 0)) and a != 1) else "NO") ```
0
379
A
New Year Candles
PROGRAMMING
1,000
[ "implementation" ]
null
null
Vasily the Programmer loves romance, so this year he decided to illuminate his room with candles. Vasily has *a* candles.When Vasily lights up a new candle, it first burns for an hour and then it goes out. Vasily is smart, so he can make *b* went out candles into a new candle. As a result, this new candle can be used like any other new candle. Now Vasily wonders: for how many hours can his candles light up the room if he acts optimally well? Help him find this number.
The single line contains two integers, *a* and *b* (1<=≤<=*a*<=≤<=1000; 2<=≤<=*b*<=≤<=1000).
Print a single integer — the number of hours Vasily can light up the room for.
[ "4 2\n", "6 3\n" ]
[ "7\n", "8\n" ]
Consider the first sample. For the first four hours Vasily lights up new candles, then he uses four burned out candles to make two new ones and lights them up. When these candles go out (stop burning), Vasily can make another candle. Overall, Vasily can light up the room for 7 hours.
500
[ { "input": "4 2", "output": "7" }, { "input": "6 3", "output": "8" }, { "input": "1000 1000", "output": "1001" }, { "input": "123 5", "output": "153" }, { "input": "1000 2", "output": "1999" }, { "input": "1 2", "output": "1" }, { "input": "1 3", "output": "1" }, { "input": "1 4", "output": "1" }, { "input": "2 2", "output": "3" }, { "input": "3 2", "output": "5" }, { "input": "3 3", "output": "4" }, { "input": "999 2", "output": "1997" }, { "input": "1000 3", "output": "1499" }, { "input": "1000 4", "output": "1333" }, { "input": "1 1000", "output": "1" }, { "input": "80 970", "output": "80" }, { "input": "80 970", "output": "80" }, { "input": "80 970", "output": "80" }, { "input": "80 970", "output": "80" }, { "input": "80 970", "output": "80" }, { "input": "80 970", "output": "80" }, { "input": "10 4", "output": "13" }, { "input": "4 3", "output": "5" }, { "input": "91 5", "output": "113" }, { "input": "777 17", "output": "825" }, { "input": "100 4", "output": "133" }, { "input": "5 3", "output": "7" }, { "input": "6 4", "output": "7" }, { "input": "26 8", "output": "29" }, { "input": "9 4", "output": "11" }, { "input": "20 3", "output": "29" }, { "input": "17 3", "output": "25" } ]
1,688,903,450
2,147,483,647
Python 3
OK
TESTS
32
46
0
a,b = map(int,input().split()) time = 0 ost = 0 while a > 0 : time += a ost += a a = 0 a += ost // b ost = ost % b print(time)
Title: New Year Candles Time Limit: None seconds Memory Limit: None megabytes Problem Description: Vasily the Programmer loves romance, so this year he decided to illuminate his room with candles. Vasily has *a* candles.When Vasily lights up a new candle, it first burns for an hour and then it goes out. Vasily is smart, so he can make *b* went out candles into a new candle. As a result, this new candle can be used like any other new candle. Now Vasily wonders: for how many hours can his candles light up the room if he acts optimally well? Help him find this number. Input Specification: The single line contains two integers, *a* and *b* (1<=≤<=*a*<=≤<=1000; 2<=≤<=*b*<=≤<=1000). Output Specification: Print a single integer — the number of hours Vasily can light up the room for. Demo Input: ['4 2\n', '6 3\n'] Demo Output: ['7\n', '8\n'] Note: Consider the first sample. For the first four hours Vasily lights up new candles, then he uses four burned out candles to make two new ones and lights them up. When these candles go out (stop burning), Vasily can make another candle. Overall, Vasily can light up the room for 7 hours.
```python a,b = map(int,input().split()) time = 0 ost = 0 while a > 0 : time += a ost += a a = 0 a += ost // b ost = ost % b print(time) ```
3
633
A
Ebony and Ivory
PROGRAMMING
1,100
[ "brute force", "math", "number theory" ]
null
null
Dante is engaged in a fight with "The Savior". Before he can fight it with his sword, he needs to break its shields. He has two guns, Ebony and Ivory, each of them is able to perform any non-negative number of shots. For every bullet that hits the shield, Ebony deals *a* units of damage while Ivory deals *b* units of damage. In order to break the shield Dante has to deal exactly *c* units of damage. Find out if this is possible.
The first line of the input contains three integers *a*, *b*, *c* (1<=≤<=*a*,<=*b*<=≤<=100,<=1<=≤<=*c*<=≤<=10<=000) — the number of units of damage dealt by Ebony gun and Ivory gun, and the total number of damage required to break the shield, respectively.
Print "Yes" (without quotes) if Dante can deal exactly *c* damage to the shield and "No" (without quotes) otherwise.
[ "4 6 15\n", "3 2 7\n", "6 11 6\n" ]
[ "No\n", "Yes\n", "Yes\n" ]
In the second sample, Dante can fire 1 bullet from Ebony and 2 from Ivory to deal exactly 1·3 + 2·2 = 7 damage. In the third sample, Dante can fire 1 bullet from ebony and no bullets from ivory to do 1·6 + 0·11 = 6 damage.
250
[ { "input": "4 6 15", "output": "No" }, { "input": "3 2 7", "output": "Yes" }, { "input": "6 11 6", "output": "Yes" }, { "input": "3 12 15", "output": "Yes" }, { "input": "5 5 10", "output": "Yes" }, { "input": "6 6 7", "output": "No" }, { "input": "1 1 20", "output": "Yes" }, { "input": "12 14 19", "output": "No" }, { "input": "15 12 26", "output": "No" }, { "input": "2 4 8", "output": "Yes" }, { "input": "4 5 30", "output": "Yes" }, { "input": "4 5 48", "output": "Yes" }, { "input": "2 17 105", "output": "Yes" }, { "input": "10 25 282", "output": "No" }, { "input": "6 34 323", "output": "No" }, { "input": "2 47 464", "output": "Yes" }, { "input": "4 53 113", "output": "Yes" }, { "input": "6 64 546", "output": "Yes" }, { "input": "1 78 725", "output": "Yes" }, { "input": "1 84 811", "output": "Yes" }, { "input": "3 100 441", "output": "Yes" }, { "input": "20 5 57", "output": "No" }, { "input": "14 19 143", "output": "No" }, { "input": "17 23 248", "output": "No" }, { "input": "11 34 383", "output": "Yes" }, { "input": "20 47 568", "output": "Yes" }, { "input": "16 58 410", "output": "Yes" }, { "input": "11 70 1199", "output": "Yes" }, { "input": "16 78 712", "output": "Yes" }, { "input": "20 84 562", "output": "No" }, { "input": "19 100 836", "output": "Yes" }, { "input": "23 10 58", "output": "No" }, { "input": "25 17 448", "output": "Yes" }, { "input": "22 24 866", "output": "Yes" }, { "input": "24 35 67", "output": "No" }, { "input": "29 47 264", "output": "Yes" }, { "input": "23 56 45", "output": "No" }, { "input": "25 66 1183", "output": "Yes" }, { "input": "21 71 657", "output": "Yes" }, { "input": "29 81 629", "output": "No" }, { "input": "23 95 2226", "output": "Yes" }, { "input": "32 4 62", "output": "No" }, { "input": "37 15 789", "output": "Yes" }, { "input": "39 24 999", "output": "Yes" }, { "input": "38 32 865", "output": "No" }, { "input": "32 50 205", "output": "No" }, { "input": "31 57 1362", "output": "Yes" }, { "input": "38 68 1870", "output": "Yes" }, { "input": "36 76 549", "output": "No" }, { "input": "35 84 1257", "output": "No" }, { "input": "39 92 2753", "output": "Yes" }, { "input": "44 1 287", "output": "Yes" }, { "input": "42 12 830", "output": "No" }, { "input": "42 27 9", "output": "No" }, { "input": "49 40 1422", "output": "No" }, { "input": "44 42 2005", "output": "No" }, { "input": "50 55 2479", "output": "No" }, { "input": "48 65 917", "output": "No" }, { "input": "45 78 152", "output": "No" }, { "input": "43 90 4096", "output": "Yes" }, { "input": "43 94 4316", "output": "Yes" }, { "input": "60 7 526", "output": "Yes" }, { "input": "53 11 735", "output": "Yes" }, { "input": "52 27 609", "output": "Yes" }, { "input": "57 32 992", "output": "Yes" }, { "input": "52 49 421", "output": "No" }, { "input": "57 52 2634", "output": "Yes" }, { "input": "54 67 3181", "output": "Yes" }, { "input": "52 73 638", "output": "No" }, { "input": "57 84 3470", "output": "No" }, { "input": "52 100 5582", "output": "No" }, { "input": "62 1 501", "output": "Yes" }, { "input": "63 17 858", "output": "Yes" }, { "input": "70 24 1784", "output": "Yes" }, { "input": "65 32 1391", "output": "Yes" }, { "input": "62 50 2775", "output": "No" }, { "input": "62 58 88", "output": "No" }, { "input": "66 68 3112", "output": "Yes" }, { "input": "61 71 1643", "output": "No" }, { "input": "69 81 3880", "output": "No" }, { "input": "63 100 1960", "output": "Yes" }, { "input": "73 6 431", "output": "Yes" }, { "input": "75 19 736", "output": "Yes" }, { "input": "78 25 247", "output": "No" }, { "input": "79 36 2854", "output": "Yes" }, { "input": "80 43 1864", "output": "Yes" }, { "input": "76 55 2196", "output": "Yes" }, { "input": "76 69 4122", "output": "Yes" }, { "input": "76 76 4905", "output": "No" }, { "input": "75 89 3056", "output": "Yes" }, { "input": "73 100 3111", "output": "Yes" }, { "input": "84 9 530", "output": "No" }, { "input": "82 18 633", "output": "No" }, { "input": "85 29 2533", "output": "Yes" }, { "input": "89 38 2879", "output": "Yes" }, { "input": "89 49 2200", "output": "Yes" }, { "input": "88 60 4140", "output": "Yes" }, { "input": "82 68 1299", "output": "No" }, { "input": "90 76 2207", "output": "No" }, { "input": "83 84 4923", "output": "Yes" }, { "input": "89 99 7969", "output": "Yes" }, { "input": "94 9 168", "output": "No" }, { "input": "91 20 1009", "output": "No" }, { "input": "93 23 2872", "output": "Yes" }, { "input": "97 31 3761", "output": "Yes" }, { "input": "99 46 1341", "output": "Yes" }, { "input": "98 51 2845", "output": "No" }, { "input": "93 66 3412", "output": "No" }, { "input": "95 76 3724", "output": "Yes" }, { "input": "91 87 6237", "output": "Yes" }, { "input": "98 97 7886", "output": "Yes" }, { "input": "12 17 15", "output": "No" }, { "input": "93 94 95", "output": "No" }, { "input": "27 43 27", "output": "Yes" }, { "input": "17 43 68", "output": "Yes" }, { "input": "44 12 12", "output": "Yes" }, { "input": "44 50 150", "output": "Yes" }, { "input": "1 1 10000", "output": "Yes" }, { "input": "2 3 10000", "output": "Yes" }, { "input": "100 1 10", "output": "Yes" }, { "input": "3 2 1", "output": "No" }, { "input": "1 1 1", "output": "Yes" }, { "input": "9 9 10000", "output": "No" }, { "input": "2 3 9995", "output": "Yes" }, { "input": "3 5 4", "output": "No" }, { "input": "99 98 100", "output": "No" }, { "input": "6 10 2", "output": "No" }, { "input": "1 6 5", "output": "Yes" }, { "input": "1 4 3", "output": "Yes" }, { "input": "3 2 3", "output": "Yes" }, { "input": "1 7 6", "output": "Yes" }, { "input": "2 3 9871", "output": "Yes" }, { "input": "10 5 5", "output": "Yes" }, { "input": "10 8 2", "output": "No" } ]
1,573,662,975
2,147,483,647
PyPy 3
WRONG_ANSWER
TESTS
3
124
0
a,b,c=map(int,input().split()) flag=0 for i in range(c//a+2): t=c-a*i if t>=0 and t%b==0: flag=1 print('Yes') if flag==0: print('No')
Title: Ebony and Ivory Time Limit: None seconds Memory Limit: None megabytes Problem Description: Dante is engaged in a fight with "The Savior". Before he can fight it with his sword, he needs to break its shields. He has two guns, Ebony and Ivory, each of them is able to perform any non-negative number of shots. For every bullet that hits the shield, Ebony deals *a* units of damage while Ivory deals *b* units of damage. In order to break the shield Dante has to deal exactly *c* units of damage. Find out if this is possible. Input Specification: The first line of the input contains three integers *a*, *b*, *c* (1<=≤<=*a*,<=*b*<=≤<=100,<=1<=≤<=*c*<=≤<=10<=000) — the number of units of damage dealt by Ebony gun and Ivory gun, and the total number of damage required to break the shield, respectively. Output Specification: Print "Yes" (without quotes) if Dante can deal exactly *c* damage to the shield and "No" (without quotes) otherwise. Demo Input: ['4 6 15\n', '3 2 7\n', '6 11 6\n'] Demo Output: ['No\n', 'Yes\n', 'Yes\n'] Note: In the second sample, Dante can fire 1 bullet from Ebony and 2 from Ivory to deal exactly 1·3 + 2·2 = 7 damage. In the third sample, Dante can fire 1 bullet from ebony and no bullets from ivory to do 1·6 + 0·11 = 6 damage.
```python a,b,c=map(int,input().split()) flag=0 for i in range(c//a+2): t=c-a*i if t>=0 and t%b==0: flag=1 print('Yes') if flag==0: print('No') ```
0
702
A
Maximum Increase
PROGRAMMING
800
[ "dp", "greedy", "implementation" ]
null
null
You are given array consisting of *n* integers. Your task is to find the maximum length of an increasing subarray of the given array. A subarray is the sequence of consecutive elements of the array. Subarray is called increasing if each element of this subarray strictly greater than previous.
The first line contains single positive integer *n* (1<=≤<=*n*<=≤<=105) — the number of integers. The second line contains *n* positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109).
Print the maximum length of an increasing subarray of the given array.
[ "5\n1 7 2 11 15\n", "6\n100 100 100 100 100 100\n", "3\n1 2 3\n" ]
[ "3\n", "1\n", "3\n" ]
none
0
[ { "input": "5\n1 7 2 11 15", "output": "3" }, { "input": "6\n100 100 100 100 100 100", "output": "1" }, { "input": "3\n1 2 3", "output": "3" }, { "input": "1\n1000000000", "output": "1" }, { "input": "10\n802030518 598196518 640274071 983359971 71550121 96204862 799843967 446173607 796619138 402690754", "output": "3" }, { "input": "2\n2 1", "output": "1" }, { "input": "5\n1 2 3 3 4", "output": "3" }, { "input": "4\n1 2 2 3", "output": "2" }, { "input": "3\n2 1 1", "output": "1" }, { "input": "3\n1 2 1", "output": "2" }, { "input": "1\n1", "output": "1" }, { "input": "2\n1 2", "output": "2" }, { "input": "3\n1 1 2", "output": "2" }, { "input": "11\n1 2 3 1 2 3 2 1 2 3 4", "output": "4" }, { "input": "9\n1 2 3 4 5 6 7 8 9", "output": "9" }, { "input": "9\n1 1 1 1 1 1 1 1 1", "output": "1" }, { "input": "3\n3 2 1", "output": "1" }, { "input": "7\n1 2 3 4 5 6 7", "output": "7" }, { "input": "1\n1234394", "output": "1" } ]
1,654,747,261
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
0
31
0
n = int(input()) arr = list(map(int,input().split())) max_count = 0 index = 0 current_count = 11 while index+1 < n: if arr[index]<arr[index+1]: current_count +=1 else: current_count = 1 if current_count>max_count: max_count = current_count index+=1 print(max_count)
Title: Maximum Increase Time Limit: None seconds Memory Limit: None megabytes Problem Description: You are given array consisting of *n* integers. Your task is to find the maximum length of an increasing subarray of the given array. A subarray is the sequence of consecutive elements of the array. Subarray is called increasing if each element of this subarray strictly greater than previous. Input Specification: The first line contains single positive integer *n* (1<=≤<=*n*<=≤<=105) — the number of integers. The second line contains *n* positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109). Output Specification: Print the maximum length of an increasing subarray of the given array. Demo Input: ['5\n1 7 2 11 15\n', '6\n100 100 100 100 100 100\n', '3\n1 2 3\n'] Demo Output: ['3\n', '1\n', '3\n'] Note: none
```python n = int(input()) arr = list(map(int,input().split())) max_count = 0 index = 0 current_count = 11 while index+1 < n: if arr[index]<arr[index+1]: current_count +=1 else: current_count = 1 if current_count>max_count: max_count = current_count index+=1 print(max_count) ```
0
12
A
Super Agent
PROGRAMMING
800
[ "implementation" ]
A. Super Agent
2
256
There is a very secret base in Potatoland where potato mash is made according to a special recipe. The neighbours from Porridgia decided to seize this recipe and to sell it to Pilauland. For this mission they have been preparing special agent Pearlo for many years. When, finally, Pearlo learned all secrets of espionage, he penetrated into the Potatoland territory and reached the secret base. Now he is standing at the entrance, but to get inside he need to pass combination lock. Minute ago one of the workers entered the password on the terminal and opened the door. The terminal is a square digital keyboard 3<=×<=3 with digits from 1 to 9. Pearlo knows that the password consists from distinct digits and is probably symmetric with respect to the central button of the terminal. He has heat sensor which allowed him to detect the digits which the worker pressed. Now he wants to check whether the password entered by the worker is symmetric with respect to the central button of the terminal. This fact can Help Pearlo to reduce the number of different possible password combinations.
Input contains the matrix of three rows of three symbols each. Symbol «X» means that the corresponding button was pressed, and «.» means that is was not pressed. The matrix may contain no «X», also it may contain no «.».
Print YES if the password is symmetric with respect to the central button of the terminal and NO otherwise.
[ "XX.\n...\n.XX\n", "X.X\nX..\n...\n" ]
[ "YES\n", "NO\n" ]
If you are not familiar with the term «central symmetry», you may look into http://en.wikipedia.org/wiki/Central_symmetry
0
[ { "input": "XX.\n...\n.XX", "output": "YES" }, { "input": ".X.\n.X.\n.X.", "output": "YES" }, { "input": "XXX\nXXX\nXXX", "output": "YES" }, { "input": "XXX\nX.X\nXXX", "output": "YES" }, { "input": "X..\n.X.\n..X", "output": "YES" }, { "input": "...\nX.X\nX..", "output": "NO" }, { "input": ".X.\nX.X\n.X.", "output": "YES" }, { "input": "X.X\n.X.\nX.X", "output": "YES" }, { "input": "...\n...\n..X", "output": "NO" }, { "input": "XXX\n...\nXXX", "output": "YES" }, { "input": "..X\nX..\n..X", "output": "NO" }, { "input": ".X.\n...\nX.X", "output": "NO" }, { "input": "X.X\nX.X\nX.X", "output": "YES" }, { "input": ".X.\nX.X\nXX.", "output": "NO" }, { "input": "...\nXXX\nXXX", "output": "NO" }, { "input": "XXX\n..X\nXXX", "output": "NO" }, { "input": "X..\nX.X\n.X.", "output": "NO" }, { "input": "...\n..X\nXXX", "output": "NO" }, { "input": "..X\nX.X\nX..", "output": "YES" }, { "input": "..X\n..X\nXXX", "output": "NO" }, { "input": "X..\nX..\nX..", "output": "NO" }, { "input": "XXX\n.X.\nXXX", "output": "YES" }, { "input": "..X\n...\nX..", "output": "YES" }, { "input": "...\n...\nX..", "output": "NO" }, { "input": "...\n...\n.X.", "output": "NO" }, { "input": "...\n...\n..X", "output": "NO" }, { "input": "...\n.X.\nX..", "output": "NO" } ]
1,572,753,516
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
0
186
0
a=[] b=[] for i in range(3): n=input() n.split() for j in n: a.append(j) for k in range(8,0,-1): b.append(a[k]) b.append(a[0]) print(a,b) if a==b: print('YES') else: print('NO')
Title: Super Agent Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: There is a very secret base in Potatoland where potato mash is made according to a special recipe. The neighbours from Porridgia decided to seize this recipe and to sell it to Pilauland. For this mission they have been preparing special agent Pearlo for many years. When, finally, Pearlo learned all secrets of espionage, he penetrated into the Potatoland territory and reached the secret base. Now he is standing at the entrance, but to get inside he need to pass combination lock. Minute ago one of the workers entered the password on the terminal and opened the door. The terminal is a square digital keyboard 3<=×<=3 with digits from 1 to 9. Pearlo knows that the password consists from distinct digits and is probably symmetric with respect to the central button of the terminal. He has heat sensor which allowed him to detect the digits which the worker pressed. Now he wants to check whether the password entered by the worker is symmetric with respect to the central button of the terminal. This fact can Help Pearlo to reduce the number of different possible password combinations. Input Specification: Input contains the matrix of three rows of three symbols each. Symbol «X» means that the corresponding button was pressed, and «.» means that is was not pressed. The matrix may contain no «X», also it may contain no «.». Output Specification: Print YES if the password is symmetric with respect to the central button of the terminal and NO otherwise. Demo Input: ['XX.\n...\n.XX\n', 'X.X\nX..\n...\n'] Demo Output: ['YES\n', 'NO\n'] Note: If you are not familiar with the term «central symmetry», you may look into http://en.wikipedia.org/wiki/Central_symmetry
```python a=[] b=[] for i in range(3): n=input() n.split() for j in n: a.append(j) for k in range(8,0,-1): b.append(a[k]) b.append(a[0]) print(a,b) if a==b: print('YES') else: print('NO') ```
0
263
A
Beautiful Matrix
PROGRAMMING
800
[ "implementation" ]
null
null
You've got a 5<=×<=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<=≤<=*i*<=&lt;<=5). 1. Swap two neighboring matrix columns, that is, columns with indexes *j* and *j*<=+<=1 for some integer *j* (1<=≤<=*j*<=&lt;<=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 — 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" } ]
1,698,246,134
2,147,483,647
Python 3
OK
TESTS
25
92
0
# Read the input matrix matrix = [] for i in range(5): row = list(map(int, input().split())) matrix.append(row) # Find the row and column indices of the number 1 row_index = -1 col_index = -1 for i in range(5): for j in range(5): if matrix[i][j] == 1: row_index = i col_index = j break if row_index != -1: break # Calculate the number of moves needed to move the 1 to the center of the matrix moves = abs(row_index - 2) + abs(col_index - 2) # Output the number of moves print(moves)
Title: Beautiful Matrix Time Limit: None seconds Memory Limit: None megabytes Problem Description: You've got a 5<=×<=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<=≤<=*i*<=&lt;<=5). 1. Swap two neighboring matrix columns, that is, columns with indexes *j* and *j*<=+<=1 for some integer *j* (1<=≤<=*j*<=&lt;<=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 — 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 # Read the input matrix matrix = [] for i in range(5): row = list(map(int, input().split())) matrix.append(row) # Find the row and column indices of the number 1 row_index = -1 col_index = -1 for i in range(5): for j in range(5): if matrix[i][j] == 1: row_index = i col_index = j break if row_index != -1: break # Calculate the number of moves needed to move the 1 to the center of the matrix moves = abs(row_index - 2) + abs(col_index - 2) # Output the number of moves print(moves) ```
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<=≤<=*n*<=≤<=2000; 1<=≤<=*k*<=≤<=5). The next line contains *n* integers: *y*1,<=*y*2,<=...,<=*y**n* (0<=≤<=*y**i*<=≤<=5), where *y**i* shows the number of times the *i*-th person participated in the ACM ICPC world championship.
Print a single number — 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" } ]
1,668,081,378
2,147,483,647
PyPy 3
WRONG_ANSWER
TESTS
1
77
0
n,k=map(int,input().split()) y=list(map(int,input().split())) c= 0 for i in range (len(y)): z=y[i]>=k if (5)-z: c+=1 ans = c / 3 print(int(ans))
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<=≤<=*n*<=≤<=2000; 1<=≤<=*k*<=≤<=5). The next line contains *n* integers: *y*1,<=*y*2,<=...,<=*y**n* (0<=≤<=*y**i*<=≤<=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 — 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=map(int,input().split()) y=list(map(int,input().split())) c= 0 for i in range (len(y)): z=y[i]>=k if (5)-z: c+=1 ans = c / 3 print(int(ans)) ```
0
129
A
Cookies
PROGRAMMING
900
[ "implementation" ]
null
null
Olga came to visit the twins Anna and Maria and saw that they have many cookies. The cookies are distributed into bags. As there are many cookies, Olga decided that it's no big deal if she steals a bag. However, she doesn't want the sisters to quarrel because of nothing when they divide the cookies. That's why Olga wants to steal a bag with cookies so that the number of cookies in the remaining bags was even, that is, so that Anna and Maria could evenly divide it into two (even 0 remaining cookies will do, just as any other even number). How many ways there are to steal exactly one cookie bag so that the total number of cookies in the remaining bags was even?
The first line contains the only integer *n* (1<=≤<=*n*<=≤<=100) — the number of cookie bags Anna and Maria have. The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=100) — the number of cookies in the *i*-th bag.
Print in the only line the only number — the sought number of ways. If there are no such ways print 0.
[ "1\n1\n", "10\n1 2 2 3 4 4 4 2 2 2\n", "11\n2 2 2 2 2 2 2 2 2 2 99\n" ]
[ "1\n", "8\n", "1\n" ]
In the first sample Olga should take the only bag so that the twins ended up with the even number of cookies. In the second sample Olga can take any of five bags with two cookies or any of three bags with four cookies — 5 + 3 = 8 ways in total. In the third sample, no matter which bag with two cookies Olga chooses, the twins are left with 2 * 9 + 99 = 117 cookies. Thus, Olga has only one option: to take the bag with 99 cookies.
500
[ { "input": "1\n1", "output": "1" }, { "input": "10\n1 2 2 3 4 4 4 2 2 2", "output": "8" }, { "input": "11\n2 2 2 2 2 2 2 2 2 2 99", "output": "1" }, { "input": "2\n1 1", "output": "0" }, { "input": "2\n2 2", "output": "2" }, { "input": "2\n1 2", "output": "1" }, { "input": "7\n7 7 7 7 7 7 7", "output": "7" }, { "input": "8\n1 2 3 4 5 6 7 8", "output": "4" }, { "input": "100\n1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2", "output": "50" }, { "input": "99\n99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99", "output": "49" }, { "input": "82\n43 44 96 33 23 42 33 66 53 87 8 90 43 91 40 88 51 18 48 62 59 10 22 20 54 6 13 63 2 56 31 52 98 42 54 32 26 77 9 24 33 91 16 30 39 34 78 82 73 90 12 15 67 76 30 18 44 86 84 98 65 54 100 79 28 34 40 56 11 43 72 35 86 59 89 40 30 33 7 19 44 15", "output": "50" }, { "input": "17\n50 14 17 77 74 74 38 76 41 27 45 29 66 98 38 73 38", "output": "7" }, { "input": "94\n81 19 90 99 26 11 86 44 78 36 80 59 99 90 78 72 71 20 94 56 42 40 71 84 10 85 10 70 52 27 39 55 90 16 48 25 7 79 99 100 38 10 99 56 3 4 78 9 16 57 14 40 52 54 57 70 30 86 56 84 97 60 59 69 49 66 23 92 90 46 86 73 53 47 1 83 14 20 24 66 13 45 41 14 86 75 55 88 48 95 82 24 47 87", "output": "39" }, { "input": "88\n64 95 12 90 40 65 98 45 52 54 79 7 81 25 98 19 68 82 41 53 35 50 5 22 32 21 8 39 8 6 72 27 81 30 12 79 21 42 60 2 66 87 46 93 62 78 52 71 76 32 78 94 86 85 55 15 34 76 41 20 32 26 94 81 89 45 74 49 11 40 40 39 49 46 80 85 90 23 80 40 86 58 70 26 48 93 23 53", "output": "37" }, { "input": "84\n95 9 43 43 13 84 60 90 1 8 97 99 54 34 59 83 33 15 51 26 40 12 66 65 19 30 29 78 92 60 25 13 19 84 71 73 12 24 54 49 16 41 11 40 57 59 34 40 39 9 71 83 1 77 79 53 94 47 78 55 77 85 29 52 80 90 53 77 97 97 27 79 28 23 83 25 26 22 49 86 63 56 3 32", "output": "51" }, { "input": "47\n61 97 76 94 91 22 2 68 62 73 90 47 16 79 44 71 98 68 43 6 53 52 40 27 68 67 43 96 14 91 60 61 96 24 97 13 32 65 85 96 81 77 34 18 23 14 80", "output": "21" }, { "input": "69\n71 1 78 74 58 89 30 6 100 90 22 61 11 59 14 74 27 25 78 61 45 19 25 33 37 4 52 43 53 38 9 100 56 67 69 38 76 91 63 60 93 52 28 61 9 98 8 14 57 63 89 64 98 51 36 66 36 86 13 82 50 91 52 64 86 78 78 83 81", "output": "37" }, { "input": "52\n38 78 36 75 19 3 56 1 39 97 24 79 84 16 93 55 96 64 12 24 1 86 80 29 12 32 36 36 73 39 76 65 53 98 30 20 28 8 86 43 70 22 75 69 62 65 81 25 53 40 71 59", "output": "28" }, { "input": "74\n81 31 67 97 26 75 69 81 11 13 13 74 77 88 52 20 52 64 66 75 72 28 41 54 26 75 41 91 75 15 18 36 13 83 63 61 14 48 53 63 19 67 35 48 23 65 73 100 44 55 92 88 99 17 73 25 83 7 31 89 12 80 98 39 42 75 14 29 81 35 77 87 33 94", "output": "47" }, { "input": "44\n46 56 31 31 37 71 94 2 14 100 45 72 36 72 80 3 38 54 42 98 50 32 31 42 62 31 45 50 95 100 18 17 64 22 18 25 52 56 70 57 43 40 81 28", "output": "15" }, { "input": "22\n28 57 40 74 51 4 45 84 99 12 95 14 92 60 47 81 84 51 31 91 59 42", "output": "11" }, { "input": "59\n73 45 94 76 41 49 65 13 74 66 36 25 47 75 40 23 92 72 11 32 32 8 81 26 68 56 41 8 76 47 96 55 70 11 84 14 83 18 70 22 30 39 28 100 48 11 92 45 78 69 86 1 54 90 98 91 13 17 35", "output": "33" }, { "input": "63\n20 18 44 94 68 57 16 43 74 55 68 24 21 95 76 84 50 50 47 86 86 12 58 55 28 72 86 18 34 45 81 88 3 72 41 9 60 90 81 93 12 6 9 6 2 41 1 7 9 29 81 14 64 80 20 36 67 54 7 5 35 81 22", "output": "37" }, { "input": "28\n49 84 48 19 44 91 11 82 96 95 88 90 71 82 87 25 31 23 18 13 98 45 26 65 35 12 31 14", "output": "15" }, { "input": "61\n34 18 28 64 28 45 9 77 77 20 63 92 79 16 16 100 86 2 91 91 57 15 31 95 10 88 84 5 82 83 53 98 59 17 97 80 76 80 81 3 91 81 87 93 61 46 10 49 6 22 21 75 63 89 21 81 30 19 67 38 77", "output": "35" }, { "input": "90\n41 90 43 1 28 75 90 50 3 70 76 64 81 63 25 69 83 82 29 91 59 66 21 61 7 55 72 49 38 69 72 20 64 58 30 81 61 29 96 14 39 5 100 20 29 98 75 29 44 78 97 45 26 77 73 59 22 99 41 6 3 96 71 20 9 18 96 18 90 62 34 78 54 5 41 6 73 33 2 54 26 21 18 6 45 57 43 73 95 75", "output": "42" }, { "input": "45\n93 69 4 27 20 14 71 48 79 3 32 26 49 30 57 88 13 56 49 61 37 32 47 41 41 70 45 68 82 18 8 6 25 20 15 13 71 99 28 6 52 34 19 59 26", "output": "23" }, { "input": "33\n29 95 48 49 91 10 83 71 47 25 66 36 51 12 34 10 54 74 41 96 89 26 89 1 42 33 1 62 9 32 49 65 78", "output": "15" }, { "input": "34\n98 24 42 36 41 82 28 58 89 34 77 70 76 44 74 54 66 100 13 79 4 88 21 1 11 45 91 29 87 100 29 54 82 78", "output": "13" }, { "input": "29\n91 84 26 84 9 63 52 9 65 56 90 2 36 7 67 33 91 14 65 38 53 36 81 83 85 14 33 95 51", "output": "17" }, { "input": "100\n2 88 92 82 87 100 78 28 84 43 78 32 43 33 97 19 15 52 29 84 57 72 54 13 99 28 82 79 40 70 34 92 91 53 9 88 27 43 14 92 72 37 26 37 20 95 19 34 49 64 33 37 34 27 80 79 9 54 99 68 25 4 68 73 46 66 24 78 3 87 26 52 50 84 4 95 23 83 39 58 86 36 33 16 98 2 84 19 53 12 69 60 10 11 78 17 79 92 77 59", "output": "45" }, { "input": "100\n2 95 45 73 9 54 20 97 57 82 88 26 18 71 25 27 75 54 31 11 58 85 69 75 72 91 76 5 25 80 45 49 4 73 8 81 81 38 5 12 53 77 7 96 90 35 28 80 73 94 19 69 96 17 94 49 69 9 32 19 5 12 46 29 26 40 59 59 6 95 82 50 72 2 45 69 12 5 72 29 39 72 23 96 81 28 28 56 68 58 37 41 30 1 90 84 15 24 96 43", "output": "53" }, { "input": "100\n27 72 35 91 13 10 35 45 24 55 83 84 63 96 29 79 34 67 63 92 48 83 18 77 28 27 49 66 29 88 55 15 6 58 14 67 94 36 77 7 7 64 61 52 71 18 36 99 76 6 50 67 16 13 41 7 89 73 61 51 78 22 78 32 76 100 3 31 89 71 63 53 15 85 77 54 89 33 68 74 3 23 57 5 43 89 75 35 9 86 90 11 31 46 48 37 74 17 77 8", "output": "40" }, { "input": "100\n69 98 69 88 11 49 55 8 25 91 17 81 47 26 15 73 96 71 18 42 42 61 48 14 92 78 35 72 4 27 62 75 83 79 17 16 46 80 96 90 82 54 37 69 85 21 67 70 96 10 46 63 21 59 56 92 54 88 77 30 75 45 44 29 86 100 51 11 65 69 66 56 82 63 27 1 51 51 13 10 3 55 26 85 34 16 87 72 13 100 81 71 90 95 86 50 83 55 55 54", "output": "53" }, { "input": "100\n34 35 99 64 2 66 78 93 20 48 12 79 19 10 87 7 42 92 60 79 5 2 24 89 57 48 63 92 74 4 16 51 7 12 90 48 87 17 18 73 51 58 97 97 25 38 15 97 96 73 67 91 6 75 14 13 87 79 75 3 15 55 35 95 71 45 10 13 20 37 82 26 2 22 13 83 97 84 39 79 43 100 54 59 98 8 61 34 7 65 75 44 24 77 73 88 34 95 44 77", "output": "55" }, { "input": "100\n15 86 3 1 51 26 74 85 37 87 64 58 10 6 57 26 30 47 85 65 24 72 50 40 12 35 91 47 91 60 47 87 95 34 80 91 26 3 36 39 14 86 28 70 51 44 28 21 72 79 57 61 16 71 100 94 57 67 36 74 24 21 89 85 25 2 97 67 76 53 76 80 97 64 35 13 8 32 21 52 62 61 67 14 74 73 66 44 55 76 24 3 43 42 99 61 36 80 38 66", "output": "52" }, { "input": "100\n45 16 54 54 80 94 74 93 75 85 58 95 79 30 81 2 84 4 57 23 92 64 78 1 50 36 13 27 56 54 10 77 87 1 5 38 85 74 94 82 30 45 72 83 82 30 81 82 82 3 69 82 7 92 39 60 94 42 41 5 3 17 67 21 79 44 79 96 28 3 53 68 79 89 63 83 1 44 4 31 84 15 73 77 19 66 54 6 73 1 67 24 91 11 86 45 96 82 20 89", "output": "51" }, { "input": "100\n84 23 50 32 90 71 92 43 58 70 6 82 7 55 85 19 70 89 12 26 29 56 74 30 2 27 4 39 63 67 91 81 11 33 75 10 82 88 39 43 43 80 68 35 55 67 53 62 73 65 86 74 43 51 14 48 42 92 83 57 22 33 24 99 5 27 78 96 7 28 11 15 8 38 85 67 5 92 24 96 57 59 14 95 91 4 9 18 45 33 74 83 64 85 14 51 51 94 29 2", "output": "53" }, { "input": "100\n77 56 56 45 73 55 32 37 39 50 30 95 79 21 44 34 51 43 86 91 39 30 85 15 35 93 100 14 57 31 80 79 38 40 88 4 91 54 7 95 76 26 62 84 17 33 67 47 6 82 69 51 17 2 59 24 11 12 31 90 12 11 55 38 72 49 30 50 42 46 5 97 9 9 30 45 86 23 19 82 40 42 5 40 35 98 35 32 60 60 5 28 84 35 21 49 68 53 68 23", "output": "48" }, { "input": "100\n78 38 79 61 45 86 83 83 86 90 74 69 2 84 73 39 2 5 20 71 24 80 54 89 58 34 77 40 39 62 2 47 28 53 97 75 88 98 94 96 33 71 44 90 47 36 19 89 87 98 90 87 5 85 34 79 82 3 42 88 89 63 35 7 89 30 40 48 12 41 56 76 83 60 80 80 39 56 77 4 72 96 30 55 57 51 7 19 11 1 66 1 91 87 11 62 95 85 79 25", "output": "48" }, { "input": "100\n5 34 23 20 76 75 19 51 17 82 60 13 83 6 65 16 20 43 66 54 87 10 87 73 50 24 16 98 33 28 80 52 54 82 26 92 14 13 84 92 94 29 61 21 60 20 48 94 24 20 75 70 58 27 68 45 86 89 29 8 67 38 83 48 18 100 11 22 46 84 52 97 70 19 50 75 3 7 52 53 72 41 18 31 1 38 49 53 11 64 99 76 9 87 48 12 100 32 44 71", "output": "58" }, { "input": "100\n76 89 68 78 24 72 73 95 98 72 58 15 2 5 56 32 9 65 50 70 94 31 29 54 89 52 31 93 43 56 26 35 72 95 51 55 78 70 11 92 17 5 54 94 81 31 78 95 73 91 95 37 59 9 53 48 65 55 84 8 45 97 64 37 96 34 36 53 66 17 72 48 99 23 27 18 92 84 44 73 60 78 53 29 68 99 19 39 61 40 69 6 77 12 47 29 15 4 8 45", "output": "53" }, { "input": "100\n82 40 31 53 8 50 85 93 3 84 54 17 96 59 51 42 18 19 35 84 79 31 17 46 54 82 72 49 35 73 26 89 61 73 3 50 12 29 25 77 88 21 58 24 22 89 96 54 82 29 96 56 77 16 1 68 90 93 20 23 57 22 31 18 92 90 51 14 50 72 31 54 12 50 66 62 2 34 17 45 68 50 87 97 23 71 1 72 17 82 42 15 20 78 4 49 66 59 10 17", "output": "54" }, { "input": "100\n32 82 82 24 39 53 48 5 29 24 9 37 91 37 91 95 1 97 84 52 12 56 93 47 22 20 14 17 40 22 79 34 24 2 69 30 69 29 3 89 21 46 60 92 39 29 18 24 49 18 40 22 60 13 77 50 39 64 50 70 99 8 66 31 90 38 20 54 7 21 5 56 41 68 69 20 54 89 69 62 9 53 43 89 81 97 15 2 52 78 89 65 16 61 59 42 56 25 32 52", "output": "49" }, { "input": "100\n72 54 23 24 97 14 99 87 15 25 7 23 17 87 72 31 71 87 34 82 51 77 74 85 62 38 24 7 84 48 98 21 29 71 70 84 25 58 67 92 18 44 32 9 81 15 53 29 63 18 86 16 7 31 38 99 70 32 89 16 23 11 66 96 69 82 97 59 6 9 49 80 85 19 6 9 52 51 85 74 53 46 73 55 31 63 78 61 34 80 77 65 87 77 92 52 89 8 52 31", "output": "44" }, { "input": "100\n56 88 8 19 7 15 11 54 35 50 19 57 63 72 51 43 50 19 57 90 40 100 8 92 11 96 30 32 59 65 93 47 62 3 50 41 30 50 72 83 61 46 83 60 20 46 33 1 5 18 83 22 34 16 41 95 63 63 7 59 55 95 91 29 64 60 64 81 45 45 10 9 88 37 69 85 21 82 41 76 42 34 47 78 51 83 65 100 13 22 59 76 63 1 26 86 36 94 99 74", "output": "46" }, { "input": "100\n27 89 67 60 62 80 43 50 28 88 72 5 94 11 63 91 18 78 99 3 71 26 12 97 74 62 23 24 22 3 100 72 98 7 94 32 12 75 61 88 42 48 10 14 45 9 48 56 73 76 70 70 79 90 35 39 96 37 81 11 19 65 99 39 23 79 34 61 35 74 90 37 73 23 46 21 94 84 73 58 11 89 13 9 10 85 42 78 73 32 53 39 49 90 43 5 28 31 97 75", "output": "53" }, { "input": "100\n33 24 97 96 1 14 99 51 13 65 67 20 46 88 42 44 20 49 5 89 98 83 15 40 74 83 58 3 10 79 34 2 69 28 37 100 55 52 14 8 44 94 97 89 6 42 11 28 30 33 55 56 20 57 52 25 75 1 87 42 62 41 37 12 54 85 95 80 42 36 94 96 28 76 54 36 4 17 26 24 62 15 17 79 84 36 92 78 74 91 96 77 54 92 81 91 62 98 37 37", "output": "43" }, { "input": "100\n86 24 61 15 11 85 1 31 47 36 23 36 59 34 3 27 16 29 82 28 58 52 52 66 71 61 98 39 60 20 67 41 67 90 73 29 92 17 70 95 58 98 58 32 21 73 46 56 87 72 80 75 40 27 94 31 59 92 93 37 14 99 96 21 97 23 81 91 52 52 96 94 92 28 38 29 52 16 57 27 17 24 91 21 79 55 96 98 95 94 23 78 79 12 77 35 32 75 100 82", "output": "51" }, { "input": "100\n88 85 41 37 69 21 7 69 36 5 92 26 64 75 22 46 67 20 70 22 62 66 38 24 47 49 68 30 90 31 67 86 86 82 9 51 43 45 48 42 73 44 31 94 45 60 54 66 20 87 11 94 34 32 87 66 56 28 75 39 37 90 72 93 55 72 31 42 30 71 87 61 4 12 12 81 23 61 56 98 71 32 30 33 96 63 92 16 8 78 47 91 47 54 49 3 81 82 41 5", "output": "47" }, { "input": "1\n2", "output": "1" }, { "input": "5\n1 1 3 2 2", "output": "3" } ]
1,698,568,617
2,147,483,647
PyPy 3-64
WRONG_ANSWER
TESTS
6
92
0
n = int(input()) l = [int(x) for x in input().split()] odds = len([x for x in l if x%2]) evens = n-odds if n==odds: print(0 if n%2==0 else 1) exit() print(1 if 1 in (odds,evens) else max(odds,evens))
Title: Cookies Time Limit: None seconds Memory Limit: None megabytes Problem Description: Olga came to visit the twins Anna and Maria and saw that they have many cookies. The cookies are distributed into bags. As there are many cookies, Olga decided that it's no big deal if she steals a bag. However, she doesn't want the sisters to quarrel because of nothing when they divide the cookies. That's why Olga wants to steal a bag with cookies so that the number of cookies in the remaining bags was even, that is, so that Anna and Maria could evenly divide it into two (even 0 remaining cookies will do, just as any other even number). How many ways there are to steal exactly one cookie bag so that the total number of cookies in the remaining bags was even? Input Specification: The first line contains the only integer *n* (1<=≤<=*n*<=≤<=100) — the number of cookie bags Anna and Maria have. The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=100) — the number of cookies in the *i*-th bag. Output Specification: Print in the only line the only number — the sought number of ways. If there are no such ways print 0. Demo Input: ['1\n1\n', '10\n1 2 2 3 4 4 4 2 2 2\n', '11\n2 2 2 2 2 2 2 2 2 2 99\n'] Demo Output: ['1\n', '8\n', '1\n'] Note: In the first sample Olga should take the only bag so that the twins ended up with the even number of cookies. In the second sample Olga can take any of five bags with two cookies or any of three bags with four cookies — 5 + 3 = 8 ways in total. In the third sample, no matter which bag with two cookies Olga chooses, the twins are left with 2 * 9 + 99 = 117 cookies. Thus, Olga has only one option: to take the bag with 99 cookies.
```python n = int(input()) l = [int(x) for x in input().split()] odds = len([x for x in l if x%2]) evens = n-odds if n==odds: print(0 if n%2==0 else 1) exit() print(1 if 1 in (odds,evens) else max(odds,evens)) ```
0
139
A
Petr and Book
PROGRAMMING
1,000
[ "implementation" ]
null
null
One Sunday Petr went to a bookshop and bought a new book on sports programming. The book had exactly *n* pages. Petr decided to start reading it starting from the next day, that is, from Monday. Petr's got a very tight schedule and for each day of the week he knows how many pages he will be able to read on that day. Some days are so busy that Petr will have no time to read whatsoever. However, we know that he will be able to read at least one page a week. Assuming that Petr will not skip days and will read as much as he can every day, determine on which day of the week he will read the last page of the book.
The first input line contains the single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of pages in the book. The second line contains seven non-negative space-separated integers that do not exceed 1000 — those integers represent how many pages Petr can read on Monday, Tuesday, Wednesday, Thursday, Friday, Saturday and Sunday correspondingly. It is guaranteed that at least one of those numbers is larger than zero.
Print a single number — the number of the day of the week, when Petr will finish reading the book. The days of the week are numbered starting with one in the natural order: Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday.
[ "100\n15 20 20 15 10 30 45\n", "2\n1 0 0 0 0 0 0\n" ]
[ "6\n", "1\n" ]
Note to the first sample: By the end of Monday and therefore, by the beginning of Tuesday Petr has 85 pages left. He has 65 pages left by Wednesday, 45 by Thursday, 30 by Friday, 20 by Saturday and on Saturday Petr finishes reading the book (and he also has time to read 10 pages of something else). Note to the second sample: On Monday of the first week Petr will read the first page. On Monday of the second week Petr will read the second page and will finish reading the book.
500
[ { "input": "100\n15 20 20 15 10 30 45", "output": "6" }, { "input": "2\n1 0 0 0 0 0 0", "output": "1" }, { "input": "100\n100 200 100 200 300 400 500", "output": "1" }, { "input": "3\n1 1 1 1 1 1 1", "output": "3" }, { "input": "1\n1 1 1 1 1 1 1", "output": "1" }, { "input": "20\n5 3 7 2 1 6 4", "output": "6" }, { "input": "10\n5 1 1 1 1 1 5", "output": "6" }, { "input": "50\n10 1 10 1 10 1 10", "output": "1" }, { "input": "77\n11 11 11 11 11 11 10", "output": "1" }, { "input": "1\n1000 1000 1000 1000 1000 1000 1000", "output": "1" }, { "input": "1000\n100 100 100 100 100 100 100", "output": "3" }, { "input": "999\n10 20 10 20 30 20 10", "output": "3" }, { "input": "433\n109 58 77 10 39 125 15", "output": "7" }, { "input": "1\n0 0 0 0 0 0 1", "output": "7" }, { "input": "5\n1 0 1 0 1 0 1", "output": "1" }, { "input": "997\n1 1 0 0 1 0 1", "output": "1" }, { "input": "1000\n1 1 1 1 1 1 1", "output": "6" }, { "input": "1000\n1000 1000 1000 1000 1000 1000 1000", "output": "1" }, { "input": "1000\n1 0 0 0 0 0 0", "output": "1" }, { "input": "1000\n0 0 0 0 0 0 1", "output": "7" }, { "input": "1000\n1 0 0 1 0 0 1", "output": "1" }, { "input": "509\n105 23 98 0 7 0 155", "output": "2" }, { "input": "7\n1 1 1 1 1 1 1", "output": "7" }, { "input": "2\n1 1 0 0 0 0 0", "output": "2" }, { "input": "1\n0 0 0 0 0 1 0", "output": "6" }, { "input": "10\n0 0 0 0 0 0 1", "output": "7" }, { "input": "5\n0 0 0 0 0 6 0", "output": "6" }, { "input": "3\n0 1 0 0 0 0 0", "output": "2" }, { "input": "10\n0 0 0 0 0 0 10", "output": "7" }, { "input": "28\n1 2 3 4 5 6 7", "output": "7" }, { "input": "100\n5 5 5 5 5 5 5", "output": "6" }, { "input": "4\n1 0 0 0 0 0 1", "output": "7" }, { "input": "2\n0 0 0 0 0 0 1", "output": "7" }, { "input": "7\n0 0 0 0 0 0 7", "output": "7" }, { "input": "7\n2 1 1 1 1 1 0", "output": "6" }, { "input": "2\n0 0 1 1 0 0 0", "output": "4" }, { "input": "6\n1 1 1 1 1 1 0", "output": "6" }, { "input": "5\n1 1 1 0 0 1 1", "output": "7" }, { "input": "100\n10 20 30 10 10 10 10", "output": "7" }, { "input": "1\n0 0 0 1 0 0 0", "output": "4" }, { "input": "70\n10 10 10 10 10 10 10", "output": "7" }, { "input": "22\n1 2 3 4 5 6 10", "output": "7" }, { "input": "5\n0 0 0 1 0 0 0", "output": "4" }, { "input": "2\n0 0 0 1 0 0 0", "output": "4" }, { "input": "6\n1 0 0 0 0 0 2", "output": "7" }, { "input": "10\n1 2 2 1 2 1 1", "output": "7" }, { "input": "5\n0 0 0 0 0 0 10", "output": "7" }, { "input": "4\n0 1 1 0 0 0 0", "output": "3" }, { "input": "100\n0 0 0 0 0 1 0", "output": "6" } ]
1,644,554,143
2,147,483,647
Python 3
OK
TESTS
49
92
0
n = int(input()) day = -1 l = list(map(int,input().split())) while n>0: day = (day+1)%7 n-=l[day] print(day+1)
Title: Petr and Book Time Limit: None seconds Memory Limit: None megabytes Problem Description: One Sunday Petr went to a bookshop and bought a new book on sports programming. The book had exactly *n* pages. Petr decided to start reading it starting from the next day, that is, from Monday. Petr's got a very tight schedule and for each day of the week he knows how many pages he will be able to read on that day. Some days are so busy that Petr will have no time to read whatsoever. However, we know that he will be able to read at least one page a week. Assuming that Petr will not skip days and will read as much as he can every day, determine on which day of the week he will read the last page of the book. Input Specification: The first input line contains the single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of pages in the book. The second line contains seven non-negative space-separated integers that do not exceed 1000 — those integers represent how many pages Petr can read on Monday, Tuesday, Wednesday, Thursday, Friday, Saturday and Sunday correspondingly. It is guaranteed that at least one of those numbers is larger than zero. Output Specification: Print a single number — the number of the day of the week, when Petr will finish reading the book. The days of the week are numbered starting with one in the natural order: Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday. Demo Input: ['100\n15 20 20 15 10 30 45\n', '2\n1 0 0 0 0 0 0\n'] Demo Output: ['6\n', '1\n'] Note: Note to the first sample: By the end of Monday and therefore, by the beginning of Tuesday Petr has 85 pages left. He has 65 pages left by Wednesday, 45 by Thursday, 30 by Friday, 20 by Saturday and on Saturday Petr finishes reading the book (and he also has time to read 10 pages of something else). Note to the second sample: On Monday of the first week Petr will read the first page. On Monday of the second week Petr will read the second page and will finish reading the book.
```python n = int(input()) day = -1 l = list(map(int,input().split())) while n>0: day = (day+1)%7 n-=l[day] print(day+1) ```
3
486
A
Calculating Function
PROGRAMMING
800
[ "implementation", "math" ]
null
null
For a positive integer *n* let's define a function *f*: *f*(*n*)<==<=<=-<=1<=+<=2<=-<=3<=+<=..<=+<=(<=-<=1)*n**n* Your task is to calculate *f*(*n*) for a given integer *n*.
The single line contains the positive integer *n* (1<=≤<=*n*<=≤<=1015).
Print *f*(*n*) in a single line.
[ "4\n", "5\n" ]
[ "2\n", "-3\n" ]
*f*(4) =  - 1 + 2 - 3 + 4 = 2 *f*(5) =  - 1 + 2 - 3 + 4 - 5 =  - 3
500
[ { "input": "4", "output": "2" }, { "input": "5", "output": "-3" }, { "input": "1000000000", "output": "500000000" }, { "input": "1000000001", "output": "-500000001" }, { "input": "1000000000000000", "output": "500000000000000" }, { "input": "100", "output": "50" }, { "input": "101", "output": "-51" }, { "input": "102", "output": "51" }, { "input": "103", "output": "-52" }, { "input": "104", "output": "52" }, { "input": "105", "output": "-53" }, { "input": "106", "output": "53" }, { "input": "107", "output": "-54" }, { "input": "108", "output": "54" }, { "input": "109", "output": "-55" }, { "input": "208170109961052", "output": "104085054980526" }, { "input": "46017661651072", "output": "23008830825536" }, { "input": "4018154546667", "output": "-2009077273334" }, { "input": "288565475053", "output": "-144282737527" }, { "input": "3052460231", "output": "-1526230116" }, { "input": "29906716", "output": "14953358" }, { "input": "87897701693326", "output": "43948850846663" }, { "input": "8240", "output": "4120" }, { "input": "577935", "output": "-288968" }, { "input": "62", "output": "31" }, { "input": "1", "output": "-1" }, { "input": "2", "output": "1" }, { "input": "9999999999999", "output": "-5000000000000" }, { "input": "1000000000000", "output": "500000000000" }, { "input": "99999999999999", "output": "-50000000000000" }, { "input": "999999999999999", "output": "-500000000000000" }, { "input": "42191359342", "output": "21095679671" }, { "input": "100000000000000", "output": "50000000000000" }, { "input": "145645214654154", "output": "72822607327077" }, { "input": "4294967296", "output": "2147483648" }, { "input": "3037000499", "output": "-1518500250" }, { "input": "10000000000001", "output": "-5000000000001" }, { "input": "100000017040846", "output": "50000008520423" }, { "input": "98979894985999", "output": "-49489947493000" } ]
1,691,342,151
2,147,483,647
PyPy 3
RUNTIME_ERROR
TESTS
0
93
2,150,400
n = int(input()) if n%2==0: sum1 = sum(range(2,n+1,2)) sum2 = sum(1,n,2) else: sum1 = sum(range(2,n,2)) sum2 = sum(1,n+1,2) print(sum1-sum2)
Title: Calculating Function Time Limit: None seconds Memory Limit: None megabytes Problem Description: For a positive integer *n* let's define a function *f*: *f*(*n*)<==<=<=-<=1<=+<=2<=-<=3<=+<=..<=+<=(<=-<=1)*n**n* Your task is to calculate *f*(*n*) for a given integer *n*. Input Specification: The single line contains the positive integer *n* (1<=≤<=*n*<=≤<=1015). Output Specification: Print *f*(*n*) in a single line. Demo Input: ['4\n', '5\n'] Demo Output: ['2\n', '-3\n'] Note: *f*(4) =  - 1 + 2 - 3 + 4 = 2 *f*(5) =  - 1 + 2 - 3 + 4 - 5 =  - 3
```python n = int(input()) if n%2==0: sum1 = sum(range(2,n+1,2)) sum2 = sum(1,n,2) else: sum1 = sum(range(2,n,2)) sum2 = sum(1,n+1,2) print(sum1-sum2) ```
-1
71
A
Way Too Long Words
PROGRAMMING
800
[ "strings" ]
A. Way Too Long Words
1
256
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome. Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation. This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes. Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n". You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
[ "4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n" ]
[ "word\nl10n\ni18n\np43s\n" ]
none
500
[ { "input": "4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis", "output": "word\nl10n\ni18n\np43s" }, { "input": "5\nabcdefgh\nabcdefghi\nabcdefghij\nabcdefghijk\nabcdefghijklm", "output": "abcdefgh\nabcdefghi\nabcdefghij\na9k\na11m" }, { "input": "3\nnjfngnrurunrgunrunvurn\njfvnjfdnvjdbfvsbdubruvbubvkdb\nksdnvidnviudbvibd", "output": "n20n\nj27b\nk15d" }, { "input": "1\ntcyctkktcctrcyvbyiuhihhhgyvyvyvyvjvytchjckt", "output": "t41t" }, { "input": "24\nyou\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nunofficially\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings", "output": "you\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nu10y\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings" }, { "input": "1\na", "output": "a" }, { "input": "26\na\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz", "output": "a\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz" }, { "input": "1\nabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghij", "output": "a98j" }, { "input": "10\ngyartjdxxlcl\nfzsck\nuidwu\nxbymclornemdmtj\nilppyoapitawgje\ncibzc\ndrgbeu\nhezplmsdekhhbo\nfeuzlrimbqbytdu\nkgdco", "output": "g10l\nfzsck\nuidwu\nx13j\ni13e\ncibzc\ndrgbeu\nh12o\nf13u\nkgdco" }, { "input": "20\nlkpmx\nkovxmxorlgwaomlswjxlpnbvltfv\nhykasjxqyjrmybejnmeumzha\ntuevlumpqbbhbww\nqgqsphvrmupxxc\ntrissbaf\nqfgrlinkzvzqdryckaizutd\nzzqtoaxkvwoscyx\noswytrlnhpjvvnwookx\nlpuzqgec\ngyzqfwxggtvpjhzmzmdw\nrlxjgmvdftvrmvbdwudra\nvsntnjpepnvdaxiporggmglhagv\nxlvcqkqgcrbgtgglj\nlyxwxbiszyhlsrgzeedzprbmcpduvq\nyrmqqvrkqskqukzqrwukpsifgtdc\nxpuohcsjhhuhvr\nvvlfrlxpvqejngwrbfbpmqeirxlw\nsvmasocxdvadmaxtrpakysmeaympy\nyuflqboqfdt", "output": "lkpmx\nk26v\nh22a\nt13w\nq12c\ntrissbaf\nq21d\nz13x\no17x\nlpuzqgec\ng18w\nr19a\nv25v\nx15j\nl28q\ny26c\nx12r\nv26w\ns27y\ny9t" }, { "input": "100\nm\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr", "output": "m\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr" } ]
1,683,603,125
2,147,483,647
Python 3
RUNTIME_ERROR
TESTS
0
30
0
t=input() t=int(t) while t>0: string = input() l=len(string) if len(string)>10: print(string[0]+(l-2)+string[l-1]) else: print(string)
Title: Way Too Long Words Time Limit: 1 seconds Memory Limit: 256 megabytes Problem Description: Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome. Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation. This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes. Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n". You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes. Input Specification: The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters. Output Specification: Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data. Demo Input: ['4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n'] Demo Output: ['word\nl10n\ni18n\np43s\n'] Note: none
```python t=input() t=int(t) while t>0: string = input() l=len(string) if len(string)>10: print(string[0]+(l-2)+string[l-1]) else: print(string) ```
-1
264
A
Escape from Stones
PROGRAMMING
1,200
[ "constructive algorithms", "data structures", "implementation", "two pointers" ]
null
null
Squirrel Liss lived in a forest peacefully, but unexpected trouble happens. Stones fall from a mountain. Initially Squirrel Liss occupies an interval [0,<=1]. Next, *n* stones will fall and Liss will escape from the stones. The stones are numbered from 1 to *n* in order. The stones always fall to the center of Liss's interval. When Liss occupies the interval [*k*<=-<=*d*,<=*k*<=+<=*d*] and a stone falls to *k*, she will escape to the left or to the right. If she escapes to the left, her new interval will be [*k*<=-<=*d*,<=*k*]. If she escapes to the right, her new interval will be [*k*,<=*k*<=+<=*d*]. You are given a string *s* of length *n*. If the *i*-th character of *s* is "l" or "r", when the *i*-th stone falls Liss will escape to the left or to the right, respectively. Find the sequence of stones' numbers from left to right after all the *n* stones falls.
The input consists of only one line. The only line contains the string *s* (1<=≤<=|*s*|<=≤<=106). Each character in *s* will be either "l" or "r".
Output *n* lines — on the *i*-th line you should print the *i*-th stone's number from the left.
[ "llrlr\n", "rrlll\n", "lrlrr\n" ]
[ "3\n5\n4\n2\n1\n", "1\n2\n5\n4\n3\n", "2\n4\n5\n3\n1\n" ]
In the first example, the positions of stones 1, 2, 3, 4, 5 will be <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/58fdb5684df807bfcb705a9da9ce175613362b7d.png" style="max-width: 100.0%;max-height: 100.0%;"/>, respectively. So you should print the sequence: 3, 5, 4, 2, 1.
500
[ { "input": "llrlr", "output": "3\n5\n4\n2\n1" }, { "input": "rrlll", "output": "1\n2\n5\n4\n3" }, { "input": "lrlrr", "output": "2\n4\n5\n3\n1" }, { "input": "lllrlrllrl", "output": "4\n6\n9\n10\n8\n7\n5\n3\n2\n1" }, { "input": "llrlrrrlrr", "output": "3\n5\n6\n7\n9\n10\n8\n4\n2\n1" }, { "input": "rlrrrllrrr", "output": "1\n3\n4\n5\n8\n9\n10\n7\n6\n2" }, { "input": "lrrlrrllrrrrllllllrr", "output": "2\n3\n5\n6\n9\n10\n11\n12\n19\n20\n18\n17\n16\n15\n14\n13\n8\n7\n4\n1" }, { "input": "rlrrrlrrrllrrllrlrll", "output": "1\n3\n4\n5\n7\n8\n9\n12\n13\n16\n18\n20\n19\n17\n15\n14\n11\n10\n6\n2" }, { "input": "lllrrlrlrllrrrrrllrl", "output": "4\n5\n7\n9\n12\n13\n14\n15\n16\n19\n20\n18\n17\n11\n10\n8\n6\n3\n2\n1" }, { "input": "rrrllrrrlllrlllrlrrr", "output": "1\n2\n3\n6\n7\n8\n12\n16\n18\n19\n20\n17\n15\n14\n13\n11\n10\n9\n5\n4" }, { "input": "rrlllrrrlrrlrrrlllrlrlrrrlllrllrrllrllrrlrlrrllllrlrrrrlrlllrlrrrlrlrllrlrlrrlrrllrrrlrlrlllrrllllrl", "output": "1\n2\n6\n7\n8\n10\n11\n13\n14\n15\n19\n21\n23\n24\n25\n29\n32\n33\n36\n39\n40\n42\n44\n45\n50\n52\n53\n54\n55\n57\n61\n63\n64\n65\n67\n69\n72\n74\n76\n77\n79\n80\n83\n84\n85\n87\n89\n93\n94\n99\n100\n98\n97\n96\n95\n92\n91\n90\n88\n86\n82\n81\n78\n75\n73\n71\n70\n68\n66\n62\n60\n59\n58\n56\n51\n49\n48\n47\n46\n43\n41\n38\n37\n35\n34\n31\n30\n28\n27\n26\n22\n20\n18\n17\n16\n12\n9\n5\n4\n3" }, { "input": "llrlrlllrrllrllllrlrrlrlrrllrlrlrrlrrrrrrlllrrlrrrrrlrrrlrlrlrrlllllrrrrllrrlrlrrrllllrlrrlrrlrlrrll", "output": "3\n5\n9\n10\n13\n18\n20\n21\n23\n25\n26\n29\n31\n33\n34\n36\n37\n38\n39\n40\n41\n45\n46\n48\n49\n50\n51\n52\n54\n55\n56\n58\n60\n62\n63\n69\n70\n71\n72\n75\n76\n78\n80\n81\n82\n87\n89\n90\n92\n93\n95\n97\n98\n100\n99\n96\n94\n91\n88\n86\n85\n84\n83\n79\n77\n74\n73\n68\n67\n66\n65\n64\n61\n59\n57\n53\n47\n44\n43\n42\n35\n32\n30\n28\n27\n24\n22\n19\n17\n16\n15\n14\n12\n11\n8\n7\n6\n4\n2\n1" }, { "input": "llrrrrllrrlllrlrllrlrllllllrrrrrrrrllrrrrrrllrlrrrlllrrrrrrlllllllrrlrrllrrrllllrrlllrrrlrlrrlrlrllr", "output": "3\n4\n5\n6\n9\n10\n14\n16\n19\n21\n28\n29\n30\n31\n32\n33\n34\n35\n38\n39\n40\n41\n42\n43\n46\n48\n49\n50\n54\n55\n56\n57\n58\n59\n67\n68\n70\n71\n74\n75\n76\n81\n82\n86\n87\n88\n90\n92\n93\n95\n97\n100\n99\n98\n96\n94\n91\n89\n85\n84\n83\n80\n79\n78\n77\n73\n72\n69\n66\n65\n64\n63\n62\n61\n60\n53\n52\n51\n47\n45\n44\n37\n36\n27\n26\n25\n24\n23\n22\n20\n18\n17\n15\n13\n12\n11\n8\n7\n2\n1" }, { "input": "lllllrllrrlllrrrllrrrrlrrlrllllrrrrrllrlrllllllrrlrllrlrllrlrrlrlrrlrrrlrrrrllrlrrrrrrrllrllrrlrllrl", "output": "6\n9\n10\n14\n15\n16\n19\n20\n21\n22\n24\n25\n27\n32\n33\n34\n35\n36\n39\n41\n48\n49\n51\n54\n56\n59\n61\n62\n64\n66\n67\n69\n70\n71\n73\n74\n75\n76\n79\n81\n82\n83\n84\n85\n86\n87\n90\n93\n94\n96\n99\n100\n98\n97\n95\n92\n91\n89\n88\n80\n78\n77\n72\n68\n65\n63\n60\n58\n57\n55\n53\n52\n50\n47\n46\n45\n44\n43\n42\n40\n38\n37\n31\n30\n29\n28\n26\n23\n18\n17\n13\n12\n11\n8\n7\n5\n4\n3\n2\n1" }, { "input": "llrlrlrlrlrlrrlllllllrllllrllrrrlllrrllrllrrlllrrlllrlrrllllrrlllrrllrrllllrrlllrlllrrrllrrrrrrllrrl", "output": "3\n5\n7\n9\n11\n13\n14\n22\n27\n30\n31\n32\n36\n37\n40\n43\n44\n48\n49\n53\n55\n56\n61\n62\n66\n67\n70\n71\n76\n77\n81\n85\n86\n87\n90\n91\n92\n93\n94\n95\n98\n99\n100\n97\n96\n89\n88\n84\n83\n82\n80\n79\n78\n75\n74\n73\n72\n69\n68\n65\n64\n63\n60\n59\n58\n57\n54\n52\n51\n50\n47\n46\n45\n42\n41\n39\n38\n35\n34\n33\n29\n28\n26\n25\n24\n23\n21\n20\n19\n18\n17\n16\n15\n12\n10\n8\n6\n4\n2\n1" }, { "input": "l", "output": "1" }, { "input": "r", "output": "1" } ]
1,522,871,303
2,147,483,647
Python 3
COMPILATION_ERROR
TESTS
0
0
0
#include <bits/stdc++.h> using namespace std; #define MAX 13333333 #define EPS 1e-9 #define INF 1e7 #define MOD 1000000007 #define pb push_back #define mp make_pair #define fi first #define se second #define pi acos(-1) #define sf scanf #define pf printf #define SIZE(a) ((int)a.size()) #define All(S) S.begin(), S.end() #define Equal(a, b) (abs(a-b) < EPS) #define Greater(a, b) (a >= (b+EPS)) #define GreaterEqual(a, b) (a > (b-EPS)) #define fr(i, a, b) for(register int i = (a); i < (int)(b); i++) #define FastRead ios_base::sync_with_stdio(false); cin.tie(NULL); #define fileRead(S) freopen(S, "r", stdin); #define fileWrite(S) freopen(S, "w", stdout); #define Unique(X) X.erase(unique(X.begin(), X.end()), X.end()) #define error(args...) { string _s = #args; replace(_s.begin(), _s.end(), ',', ' '); stringstream _ss(_s); istream_iterator<string> _it(_ss); err(_it, args); } #define isOn(S, j) (S & (1 << j)) #define setBit(S, j) (S |= (1 << j)) #define clearBit(S, j) (S &= ~(1 << j)) #define toggleBit(S, j) (S ^= (1 << j)) #define lowBit(S) (S & (-S)) #define setAll(S, n) (S = (1 << n) - 1) typedef unsigned long long ull; typedef long long ll; typedef map<int, int> mii; typedef map<ll, ll>mll; typedef map<string, int> msi; typedef vector<int> vi; typedef vector<long long>vl; typedef pair<int, int> pii; typedef pair<ll, ll> pll; typedef vector<pair<int, int> > vii; typedef vector<pair<ll, ll> >vll; void err(istream_iterator<string> it) {} template<typename T, typename... Args> void err(istream_iterator<string> it, T a, Args... args) { // Debugger error(a, b, ....) cerr << *it << " = " << a << "\n"; err(++it, args...); } //int dx[] = {-1, 0, 1, 0}, dy[] = {0, 1, 0, -1}; //int dx[] = {-1, -1, -1, 0, 0, 1, 1, 1}, dy[] = {-1, 0, 1, -1, 1, -1, 0, 1}; //---------------------------------------------------------------------------------------------------------- int main() { FastRead; list<int>lst; string str; cin >> str; int i = 1; auto it = lst.begin(); for(auto elem : str) { lst.insert(it, i++); if(elem == 'l') it--; } for(auto it : lst) cout << it << endl; return 0; }
Title: Escape from Stones Time Limit: None seconds Memory Limit: None megabytes Problem Description: Squirrel Liss lived in a forest peacefully, but unexpected trouble happens. Stones fall from a mountain. Initially Squirrel Liss occupies an interval [0,<=1]. Next, *n* stones will fall and Liss will escape from the stones. The stones are numbered from 1 to *n* in order. The stones always fall to the center of Liss's interval. When Liss occupies the interval [*k*<=-<=*d*,<=*k*<=+<=*d*] and a stone falls to *k*, she will escape to the left or to the right. If she escapes to the left, her new interval will be [*k*<=-<=*d*,<=*k*]. If she escapes to the right, her new interval will be [*k*,<=*k*<=+<=*d*]. You are given a string *s* of length *n*. If the *i*-th character of *s* is "l" or "r", when the *i*-th stone falls Liss will escape to the left or to the right, respectively. Find the sequence of stones' numbers from left to right after all the *n* stones falls. Input Specification: The input consists of only one line. The only line contains the string *s* (1<=≤<=|*s*|<=≤<=106). Each character in *s* will be either "l" or "r". Output Specification: Output *n* lines — on the *i*-th line you should print the *i*-th stone's number from the left. Demo Input: ['llrlr\n', 'rrlll\n', 'lrlrr\n'] Demo Output: ['3\n5\n4\n2\n1\n', '1\n2\n5\n4\n3\n', '2\n4\n5\n3\n1\n'] Note: In the first example, the positions of stones 1, 2, 3, 4, 5 will be <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/58fdb5684df807bfcb705a9da9ce175613362b7d.png" style="max-width: 100.0%;max-height: 100.0%;"/>, respectively. So you should print the sequence: 3, 5, 4, 2, 1.
```python #include <bits/stdc++.h> using namespace std; #define MAX 13333333 #define EPS 1e-9 #define INF 1e7 #define MOD 1000000007 #define pb push_back #define mp make_pair #define fi first #define se second #define pi acos(-1) #define sf scanf #define pf printf #define SIZE(a) ((int)a.size()) #define All(S) S.begin(), S.end() #define Equal(a, b) (abs(a-b) < EPS) #define Greater(a, b) (a >= (b+EPS)) #define GreaterEqual(a, b) (a > (b-EPS)) #define fr(i, a, b) for(register int i = (a); i < (int)(b); i++) #define FastRead ios_base::sync_with_stdio(false); cin.tie(NULL); #define fileRead(S) freopen(S, "r", stdin); #define fileWrite(S) freopen(S, "w", stdout); #define Unique(X) X.erase(unique(X.begin(), X.end()), X.end()) #define error(args...) { string _s = #args; replace(_s.begin(), _s.end(), ',', ' '); stringstream _ss(_s); istream_iterator<string> _it(_ss); err(_it, args); } #define isOn(S, j) (S & (1 << j)) #define setBit(S, j) (S |= (1 << j)) #define clearBit(S, j) (S &= ~(1 << j)) #define toggleBit(S, j) (S ^= (1 << j)) #define lowBit(S) (S & (-S)) #define setAll(S, n) (S = (1 << n) - 1) typedef unsigned long long ull; typedef long long ll; typedef map<int, int> mii; typedef map<ll, ll>mll; typedef map<string, int> msi; typedef vector<int> vi; typedef vector<long long>vl; typedef pair<int, int> pii; typedef pair<ll, ll> pll; typedef vector<pair<int, int> > vii; typedef vector<pair<ll, ll> >vll; void err(istream_iterator<string> it) {} template<typename T, typename... Args> void err(istream_iterator<string> it, T a, Args... args) { // Debugger error(a, b, ....) cerr << *it << " = " << a << "\n"; err(++it, args...); } //int dx[] = {-1, 0, 1, 0}, dy[] = {0, 1, 0, -1}; //int dx[] = {-1, -1, -1, 0, 0, 1, 1, 1}, dy[] = {-1, 0, 1, -1, 1, -1, 0, 1}; //---------------------------------------------------------------------------------------------------------- int main() { FastRead; list<int>lst; string str; cin >> str; int i = 1; auto it = lst.begin(); for(auto elem : str) { lst.insert(it, i++); if(elem == 'l') it--; } for(auto it : lst) cout << it << endl; return 0; } ```
-1
263
A
Beautiful Matrix
PROGRAMMING
800
[ "implementation" ]
null
null
You've got a 5<=×<=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<=≤<=*i*<=&lt;<=5). 1. Swap two neighboring matrix columns, that is, columns with indexes *j* and *j*<=+<=1 for some integer *j* (1<=≤<=*j*<=&lt;<=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 — 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" } ]
1,696,325,325
2,147,483,647
PyPy 3-64
OK
TESTS
25
122
0
column = 0 row = 0 for i in range(5): new_str = input().split(" ") for j in range(len(new_str)): if new_str[j] == '1': column = j row = i break ans = abs(2 - row) + abs(2 - column) print(ans)
Title: Beautiful Matrix Time Limit: None seconds Memory Limit: None megabytes Problem Description: You've got a 5<=×<=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<=≤<=*i*<=&lt;<=5). 1. Swap two neighboring matrix columns, that is, columns with indexes *j* and *j*<=+<=1 for some integer *j* (1<=≤<=*j*<=&lt;<=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 — 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 column = 0 row = 0 for i in range(5): new_str = input().split(" ") for j in range(len(new_str)): if new_str[j] == '1': column = j row = i break ans = abs(2 - row) + abs(2 - column) print(ans) ```
3
38
A
Army
PROGRAMMING
800
[ "implementation" ]
A. Army
2
256
The Berland Armed Forces System consists of *n* ranks that are numbered using natural numbers from 1 to *n*, where 1 is the lowest rank and *n* is the highest rank. One needs exactly *d**i* years to rise from rank *i* to rank *i*<=+<=1. Reaching a certain rank *i* having not reached all the previous *i*<=-<=1 ranks is impossible. Vasya has just reached a new rank of *a*, but he dreams of holding the rank of *b*. Find for how many more years Vasya should serve in the army until he can finally realize his dream.
The first input line contains an integer *n* (2<=≤<=*n*<=≤<=100). The second line contains *n*<=-<=1 integers *d**i* (1<=≤<=*d**i*<=≤<=100). The third input line contains two integers *a* and *b* (1<=≤<=*a*<=&lt;<=*b*<=≤<=*n*). The numbers on the lines are space-separated.
Print the single number which is the number of years that Vasya needs to rise from rank *a* to rank *b*.
[ "3\n5 6\n1 2\n", "3\n5 6\n1 3\n" ]
[ "5\n", "11\n" ]
none
0
[ { "input": "3\n5 6\n1 2", "output": "5" }, { "input": "3\n5 6\n1 3", "output": "11" }, { "input": "2\n55\n1 2", "output": "55" }, { "input": "3\n85 78\n1 3", "output": "163" }, { "input": "4\n63 4 49\n2 3", "output": "4" }, { "input": "5\n93 83 42 56\n2 5", "output": "181" }, { "input": "6\n22 9 87 89 57\n1 6", "output": "264" }, { "input": "7\n52 36 31 23 74 78\n2 7", "output": "242" }, { "input": "8\n82 14 24 5 91 49 94\n3 8", "output": "263" }, { "input": "9\n12 40 69 39 59 21 59 5\n4 6", "output": "98" }, { "input": "10\n95 81 32 59 71 30 50 61 100\n1 6", "output": "338" }, { "input": "15\n89 55 94 4 15 69 19 60 91 77 3 94 91 62\n3 14", "output": "617" }, { "input": "20\n91 1 41 51 95 67 92 35 23 70 44 91 57 50 21 8 9 71 40\n8 17", "output": "399" }, { "input": "25\n70 95 21 84 97 39 12 98 53 24 78 29 84 65 70 22 100 17 69 27 62 48 35 80\n8 23", "output": "846" }, { "input": "30\n35 69 50 44 19 56 86 56 98 24 21 2 61 24 85 30 2 22 57 35 59 84 12 77 92 53 50 92 9\n1 16", "output": "730" }, { "input": "35\n2 34 47 15 27 61 6 88 67 20 53 65 29 68 77 5 78 86 44 98 32 81 91 79 54 84 95 23 65 97 22 33 42 87\n8 35", "output": "1663" }, { "input": "40\n32 88 59 36 95 45 28 78 73 30 97 13 13 47 48 100 43 21 22 45 88 25 15 13 63 25 72 92 29 5 25 11 50 5 54 51 48 84 23\n7 26", "output": "862" }, { "input": "45\n83 74 73 95 10 31 100 26 29 15 80 100 22 70 31 88 9 56 19 70 2 62 48 30 27 47 52 50 94 44 21 94 23 85 15 3 95 72 43 62 94 89 68 88\n17 40", "output": "1061" }, { "input": "50\n28 8 16 29 19 82 70 51 96 84 74 72 17 69 12 21 37 21 39 3 18 66 19 49 86 96 94 93 2 90 96 84 59 88 58 15 61 33 55 22 35 54 51 29 64 68 29 38 40\n23 28", "output": "344" }, { "input": "60\n24 28 25 21 43 71 64 73 71 90 51 83 69 43 75 43 78 72 56 61 99 7 23 86 9 16 16 94 23 74 18 56 20 72 13 31 75 34 35 86 61 49 4 72 84 7 65 70 66 52 21 38 6 43 69 40 73 46 5\n28 60", "output": "1502" }, { "input": "70\n69 95 34 14 67 61 6 95 94 44 28 94 73 66 39 13 19 71 73 71 28 48 26 22 32 88 38 95 43 59 88 77 80 55 17 95 40 83 67 1 38 95 58 63 56 98 49 2 41 4 73 8 78 41 64 71 60 71 41 61 67 4 4 19 97 14 39 20 27\n9 41", "output": "1767" }, { "input": "80\n65 15 43 6 43 98 100 16 69 98 4 54 25 40 2 35 12 23 38 29 10 89 30 6 4 8 7 96 64 43 11 49 89 38 20 59 54 85 46 16 16 89 60 54 28 37 32 34 67 9 78 30 50 87 58 53 99 48 77 3 5 6 19 99 16 20 31 10 80 76 82 56 56 83 72 81 84 60 28\n18 24", "output": "219" }, { "input": "90\n61 35 100 99 67 87 42 90 44 4 81 65 29 63 66 56 53 22 55 87 39 30 34 42 27 80 29 97 85 28 81 22 50 22 24 75 67 86 78 79 94 35 13 97 48 76 68 66 94 13 82 1 22 85 5 36 86 73 65 97 43 56 35 26 87 25 74 47 81 67 73 75 99 75 53 38 70 21 66 78 38 17 57 40 93 57 68 55 1\n12 44", "output": "1713" }, { "input": "95\n37 74 53 96 65 84 65 72 95 45 6 77 91 35 58 50 51 51 97 30 51 20 79 81 92 10 89 34 40 76 71 54 26 34 73 72 72 28 53 19 95 64 97 10 44 15 12 38 5 63 96 95 86 8 36 96 45 53 81 5 18 18 47 97 65 9 33 53 41 86 37 53 5 40 15 76 83 45 33 18 26 5 19 90 46 40 100 42 10 90 13 81 40 53\n6 15", "output": "570" }, { "input": "96\n51 32 95 75 23 54 70 89 67 3 1 51 4 100 97 30 9 35 56 38 54 77 56 98 43 17 60 43 72 46 87 61 100 65 81 22 74 38 16 96 5 10 54 22 23 22 10 91 9 54 49 82 29 73 33 98 75 8 4 26 24 90 71 42 90 24 94 74 94 10 41 98 56 63 18 43 56 21 26 64 74 33 22 38 67 66 38 60 64 76 53 10 4 65 76\n21 26", "output": "328" }, { "input": "97\n18 90 84 7 33 24 75 55 86 10 96 72 16 64 37 9 19 71 62 97 5 34 85 15 46 72 82 51 52 16 55 68 27 97 42 72 76 97 32 73 14 56 11 86 2 81 59 95 60 93 1 22 71 37 77 100 6 16 78 47 78 62 94 86 16 91 56 46 47 35 93 44 7 86 70 10 29 45 67 62 71 61 74 39 36 92 24 26 65 14 93 92 15 28 79 59\n6 68", "output": "3385" }, { "input": "98\n32 47 26 86 43 42 79 72 6 68 40 46 29 80 24 89 29 7 21 56 8 92 13 33 50 79 5 7 84 85 24 23 1 80 51 21 26 55 96 51 24 2 68 98 81 88 57 100 64 84 54 10 14 2 74 1 89 71 1 20 84 85 17 31 42 58 69 67 48 60 97 90 58 10 21 29 2 21 60 61 68 89 77 39 57 18 61 44 67 100 33 74 27 40 83 29 6\n8 77", "output": "3319" }, { "input": "99\n46 5 16 66 53 12 84 89 26 27 35 68 41 44 63 17 88 43 80 15 59 1 42 50 53 34 75 16 16 55 92 30 28 11 12 71 27 65 11 28 86 47 24 10 60 47 7 53 16 75 6 49 56 66 70 3 20 78 75 41 38 57 89 23 16 74 30 39 1 32 49 84 9 33 25 95 75 45 54 59 17 17 29 40 79 96 47 11 69 86 73 56 91 4 87 47 31 24\n23 36", "output": "514" }, { "input": "100\n63 65 21 41 95 23 3 4 12 23 95 50 75 63 58 34 71 27 75 31 23 94 96 74 69 34 43 25 25 55 44 19 43 86 68 17 52 65 36 29 72 96 84 25 84 23 71 54 6 7 71 7 21 100 99 58 93 35 62 47 36 70 68 9 75 13 35 70 76 36 62 22 52 51 2 87 66 41 54 35 78 62 30 35 65 44 74 93 78 37 96 70 26 32 71 27 85 85 63\n43 92", "output": "2599" }, { "input": "51\n85 38 22 38 42 36 55 24 36 80 49 15 66 91 88 61 46 82 1 61 89 92 6 56 28 8 46 80 56 90 91 38 38 17 69 64 57 68 13 44 45 38 8 72 61 39 87 2 73 88\n15 27", "output": "618" }, { "input": "2\n3\n1 2", "output": "3" }, { "input": "5\n6 8 22 22\n2 3", "output": "8" }, { "input": "6\n3 12 27 28 28\n3 4", "output": "27" }, { "input": "9\n1 2 2 2 2 3 3 5\n3 7", "output": "9" }, { "input": "10\n1 1 1 1 1 1 1 1 1\n6 8", "output": "2" }, { "input": "20\n1 1 1 1 1 1 1 1 2 2 2 2 2 3 3 3 3 3 3\n5 17", "output": "23" }, { "input": "25\n1 1 1 4 5 6 8 11 11 11 11 12 13 14 14 14 15 16 16 17 17 17 19 19\n4 8", "output": "23" }, { "input": "35\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2\n30 31", "output": "2" }, { "input": "45\n1 1 1 1 2 2 2 2 2 2 2 3 3 3 3 3 3 4 5 5 5 5 6 6 6 6 6 6 6 7 7 7 7 8 8 8 9 9 9 9 9 10 10 10\n42 45", "output": "30" }, { "input": "50\n1 8 8 13 14 15 15 16 19 21 22 24 26 31 32 37 45 47 47 47 50 50 51 54 55 56 58 61 61 61 63 63 64 66 66 67 67 70 71 80 83 84 85 92 92 94 95 95 100\n4 17", "output": "285" }, { "input": "60\n1 2 4 4 4 6 6 8 9 10 10 13 14 18 20 20 21 22 23 23 26 29 30 32 33 34 35 38 40 42 44 44 46 48 52 54 56 56 60 60 66 67 68 68 69 73 73 74 80 80 81 81 82 84 86 86 87 89 89\n56 58", "output": "173" }, { "input": "70\n1 2 3 3 4 5 5 7 7 7 8 8 8 8 9 9 10 12 12 12 12 13 16 16 16 16 16 16 17 17 18 18 20 20 21 23 24 25 25 26 29 29 29 29 31 32 32 34 35 36 36 37 37 38 39 39 40 40 40 40 41 41 42 43 44 44 44 45 45\n62 65", "output": "126" }, { "input": "80\n1 1 1 1 1 1 1 1 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 4 4 4 4 5 5 5 5 5 5 5 6 7 7 7 7 7 7 8 8 8 8 9 9 9 9 9 9 9 9 9 10 10 10 10 10 10 10 10 10 11 11 11 11 11 11 11 12 12 12 12 12 12 12 12\n17 65", "output": "326" }, { "input": "90\n1 1 3 5 8 9 10 11 11 11 11 12 13 14 15 15 15 16 16 19 19 20 22 23 24 25 25 28 29 29 30 31 33 34 35 37 37 38 41 43 43 44 45 47 51 54 55 56 58 58 59 59 60 62 66 67 67 67 68 68 69 70 71 72 73 73 76 77 77 78 78 78 79 79 79 82 83 84 85 85 87 87 89 93 93 93 95 99 99\n28 48", "output": "784" }, { "input": "95\n2 2 3 3 4 6 6 7 7 7 9 10 12 12 12 12 13 14 15 16 17 18 20 20 20 20 21 21 21 21 22 22 22 22 22 23 23 23 25 26 26 27 27 27 28 29 29 30 30 31 32 33 34 36 37 37 38 39 39 39 42 43 43 43 45 47 48 50 50 51 52 53 54 54 54 55 55 55 58 59 60 61 61 61 61 62 62 63 64 65 66 67 67 67\n64 93", "output": "1636" }, { "input": "96\n1 1 2 3 3 5 8 9 9 10 10 10 11 11 11 11 11 12 13 13 13 14 15 15 16 16 17 17 17 17 18 18 20 20 20 21 21 21 23 24 24 25 25 26 27 27 27 27 29 29 29 30 30 30 32 32 32 32 32 32 33 33 34 34 34 35 35 35 36 36 37 37 37 38 39 40 41 41 41 41 42 42 43 43 45 45 45 46 46 47 47 49 50 52 52\n76 96", "output": "898" }, { "input": "98\n2 3 4 4 5 7 8 10 10 10 11 11 12 12 12 12 13 14 15 15 16 16 18 19 19 20 21 21 21 21 22 23 24 25 26 26 27 27 27 27 29 29 30 30 31 31 37 40 40 40 41 41 41 42 43 44 44 44 46 46 47 49 49 50 50 50 51 53 55 55 56 56 56 56 56 57 57 58 59 60 60 60 62 62 63 64 64 64 65 66 66 67 68 70 70 71 71\n8 90", "output": "3016" }, { "input": "99\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n66 95", "output": "29" }, { "input": "100\n1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 3 3 3 4 4 4 4 4 4 4 4 4 4 5 5 5 5 5 5 6 6 6 6 6 6 6 6 6 6 6 6 7 7 7 7 7 7 8 8 8 8 9 9 9 9 10 10 10 10 11 11 11 11 12 12 12 13 13 13 13 13 13 13 13 13 13 14 14 14 14 14 14 15 15 15 15 15 15 16 16 16 17 17\n39 52", "output": "82" }, { "input": "51\n5 7 9 16 19 25 26 29 29 30 31 32 32 41 43 44 47 49 50 50 51 52 54 55 56 63 64 66 67 70 74 74 77 78 79 80 80 85 86 87 89 89 90 92 93 94 94 95 95 97\n3 44", "output": "2268" } ]
1,531,135,907
2,147,483,647
Python 3
OK
TESTS
50
218
0
n = int(input()) d = list(map(int,input().split())) d = [0]+d a,b = list(map(int,input().split())) dc = [sum(d[:i]) for i in range(1, n+1)] print(dc[b-1]-dc[a-1])
Title: Army Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: The Berland Armed Forces System consists of *n* ranks that are numbered using natural numbers from 1 to *n*, where 1 is the lowest rank and *n* is the highest rank. One needs exactly *d**i* years to rise from rank *i* to rank *i*<=+<=1. Reaching a certain rank *i* having not reached all the previous *i*<=-<=1 ranks is impossible. Vasya has just reached a new rank of *a*, but he dreams of holding the rank of *b*. Find for how many more years Vasya should serve in the army until he can finally realize his dream. Input Specification: The first input line contains an integer *n* (2<=≤<=*n*<=≤<=100). The second line contains *n*<=-<=1 integers *d**i* (1<=≤<=*d**i*<=≤<=100). The third input line contains two integers *a* and *b* (1<=≤<=*a*<=&lt;<=*b*<=≤<=*n*). The numbers on the lines are space-separated. Output Specification: Print the single number which is the number of years that Vasya needs to rise from rank *a* to rank *b*. Demo Input: ['3\n5 6\n1 2\n', '3\n5 6\n1 3\n'] Demo Output: ['5\n', '11\n'] Note: none
```python n = int(input()) d = list(map(int,input().split())) d = [0]+d a,b = list(map(int,input().split())) dc = [sum(d[:i]) for i in range(1, n+1)] print(dc[b-1]-dc[a-1]) ```
3.9455
467
A
George and Accommodation
PROGRAMMING
800
[ "implementation" ]
null
null
George has recently entered the BSUCP (Berland State University for Cool Programmers). George has a friend Alex who has also entered the university. Now they are moving into a dormitory. George and Alex want to live in the same room. The dormitory has *n* rooms in total. At the moment the *i*-th room has *p**i* people living in it and the room can accommodate *q**i* people in total (*p**i*<=≤<=*q**i*). Your task is to count how many rooms has free place for both George and Alex.
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of rooms. The *i*-th of the next *n* lines contains two integers *p**i* and *q**i* (0<=≤<=*p**i*<=≤<=*q**i*<=≤<=100) — the number of people who already live in the *i*-th room and the room's capacity.
Print a single integer — the number of rooms where George and Alex can move in.
[ "3\n1 1\n2 2\n3 3\n", "3\n1 10\n0 10\n10 10\n" ]
[ "0\n", "2\n" ]
none
500
[ { "input": "3\n1 1\n2 2\n3 3", "output": "0" }, { "input": "3\n1 10\n0 10\n10 10", "output": "2" }, { "input": "2\n36 67\n61 69", "output": "2" }, { "input": "3\n21 71\n10 88\n43 62", "output": "3" }, { "input": "3\n1 2\n2 3\n3 4", "output": "0" }, { "input": "10\n0 10\n0 20\n0 30\n0 40\n0 50\n0 60\n0 70\n0 80\n0 90\n0 100", "output": "10" }, { "input": "13\n14 16\n30 31\n45 46\n19 20\n15 17\n66 67\n75 76\n95 97\n29 30\n37 38\n0 2\n36 37\n8 9", "output": "4" }, { "input": "19\n66 67\n97 98\n89 91\n67 69\n67 68\n18 20\n72 74\n28 30\n91 92\n27 28\n75 77\n17 18\n74 75\n28 30\n16 18\n90 92\n9 11\n22 24\n52 54", "output": "12" }, { "input": "15\n55 57\n95 97\n57 59\n34 36\n50 52\n96 98\n39 40\n13 15\n13 14\n74 76\n47 48\n56 58\n24 25\n11 13\n67 68", "output": "10" }, { "input": "17\n68 69\n47 48\n30 31\n52 54\n41 43\n33 35\n38 40\n56 58\n45 46\n92 93\n73 74\n61 63\n65 66\n37 39\n67 68\n77 78\n28 30", "output": "8" }, { "input": "14\n64 66\n43 44\n10 12\n76 77\n11 12\n25 27\n87 88\n62 64\n39 41\n58 60\n10 11\n28 29\n57 58\n12 14", "output": "7" }, { "input": "38\n74 76\n52 54\n78 80\n48 49\n40 41\n64 65\n28 30\n6 8\n49 51\n68 70\n44 45\n57 59\n24 25\n46 48\n49 51\n4 6\n63 64\n76 78\n57 59\n18 20\n63 64\n71 73\n88 90\n21 22\n89 90\n65 66\n89 91\n96 98\n42 44\n1 1\n74 76\n72 74\n39 40\n75 76\n29 30\n48 49\n87 89\n27 28", "output": "22" }, { "input": "100\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0", "output": "0" }, { "input": "26\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\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2", "output": "0" }, { "input": "68\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2\n0 2", "output": "68" }, { "input": "7\n0 1\n1 5\n2 4\n3 5\n4 6\n5 6\n6 8", "output": "5" }, { "input": "1\n0 0", "output": "0" }, { "input": "1\n100 100", "output": "0" }, { "input": "44\n0 8\n1 11\n2 19\n3 5\n4 29\n5 45\n6 6\n7 40\n8 19\n9 22\n10 18\n11 26\n12 46\n13 13\n14 27\n15 48\n16 25\n17 20\n18 29\n19 27\n20 45\n21 39\n22 29\n23 39\n24 42\n25 37\n26 52\n27 36\n28 43\n29 35\n30 38\n31 70\n32 47\n33 38\n34 61\n35 71\n36 51\n37 71\n38 59\n39 77\n40 70\n41 80\n42 77\n43 73", "output": "42" }, { "input": "3\n1 3\n2 7\n8 9", "output": "2" }, { "input": "53\n0 1\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n7 8\n8 9\n9 10\n10 11\n11 12\n12 13\n13 14\n14 15\n15 16\n16 17\n17 18\n18 19\n19 20\n20 21\n21 22\n22 23\n23 24\n24 25\n25 26\n26 27\n27 28\n28 29\n29 30\n30 31\n31 32\n32 33\n33 34\n34 35\n35 36\n36 37\n37 38\n38 39\n39 40\n40 41\n41 42\n42 43\n43 44\n44 45\n45 46\n46 47\n47 48\n48 49\n49 50\n50 51\n51 52\n52 53", "output": "0" }, { "input": "55\n0 0\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\n51 51\n52 52\n53 53\n54 54", "output": "0" }, { "input": "51\n55 55\n55 55\n55 55\n55 55\n55 55\n55 55\n55 55\n55 55\n55 55\n55 62\n55 55\n55 55\n55 55\n55 55\n55 55\n55 55\n55 55\n55 55\n55 55\n55 55\n55 55\n55 73\n55 55\n55 55\n55 55\n55 55\n55 55\n55 55\n55 55\n55 55\n55 55\n55 82\n55 68\n55 70\n55 63\n55 55\n55 55\n55 55\n55 75\n55 75\n55 55\n55 55\n55 55\n55 55\n55 55\n55 55\n55 73\n55 55\n55 82\n55 99\n55 60", "output": "12" }, { "input": "14\n1 1\n1 1\n1 55\n1 16\n1 1\n1 1\n1 55\n1 62\n1 53\n1 26\n1 1\n1 36\n1 2\n1 3", "output": "8" } ]
1,691,685,322
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
1
46
0
n = input() n = int(n) count=0 for i in range(1,n+1): b,a=input().split() b=int(b) a=int(a) if ((b-a)>=2): count=count+1 print(count)
Title: George and Accommodation Time Limit: None seconds Memory Limit: None megabytes Problem Description: George has recently entered the BSUCP (Berland State University for Cool Programmers). George has a friend Alex who has also entered the university. Now they are moving into a dormitory. George and Alex want to live in the same room. The dormitory has *n* rooms in total. At the moment the *i*-th room has *p**i* people living in it and the room can accommodate *q**i* people in total (*p**i*<=≤<=*q**i*). Your task is to count how many rooms has free place for both George and Alex. Input Specification: The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of rooms. The *i*-th of the next *n* lines contains two integers *p**i* and *q**i* (0<=≤<=*p**i*<=≤<=*q**i*<=≤<=100) — the number of people who already live in the *i*-th room and the room's capacity. Output Specification: Print a single integer — the number of rooms where George and Alex can move in. Demo Input: ['3\n1 1\n2 2\n3 3\n', '3\n1 10\n0 10\n10 10\n'] Demo Output: ['0\n', '2\n'] Note: none
```python n = input() n = int(n) count=0 for i in range(1,n+1): b,a=input().split() b=int(b) a=int(a) if ((b-a)>=2): count=count+1 print(count) ```
0
743
A
Vladik and flights
PROGRAMMING
1,200
[ "constructive algorithms", "greedy", "implementation" ]
null
null
Vladik is a competitive programmer. This year he is going to win the International Olympiad in Informatics. But it is not as easy as it sounds: the question Vladik face now is to find the cheapest way to get to the olympiad. Vladik knows *n* airports. All the airports are located on a straight line. Each airport has unique id from 1 to *n*, Vladik's house is situated next to the airport with id *a*, and the place of the olympiad is situated next to the airport with id *b*. It is possible that Vladik's house and the place of the olympiad are located near the same airport. To get to the olympiad, Vladik can fly between any pair of airports any number of times, but he has to start his route at the airport *a* and finish it at the airport *b*. Each airport belongs to one of two companies. The cost of flight from the airport *i* to the airport *j* is zero if both airports belong to the same company, and |*i*<=-<=*j*| if they belong to different companies. Print the minimum cost Vladik has to pay to get to the olympiad.
The first line contains three integers *n*, *a*, and *b* (1<=≤<=*n*<=≤<=105, 1<=≤<=*a*,<=*b*<=≤<=*n*) — the number of airports, the id of the airport from which Vladik starts his route and the id of the airport which he has to reach. The second line contains a string with length *n*, which consists only of characters 0 and 1. If the *i*-th character in this string is 0, then *i*-th airport belongs to first company, otherwise it belongs to the second.
Print single integer — the minimum cost Vladik has to pay to get to the olympiad.
[ "4 1 4\n1010\n", "5 5 2\n10110\n" ]
[ "1", "0" ]
In the first example Vladik can fly to the airport 2 at first and pay |1 - 2| = 1 (because the airports belong to different companies), and then fly from the airport 2 to the airport 4 for free (because the airports belong to the same company). So the cost of the whole flight is equal to 1. It's impossible to get to the olympiad for free, so the answer is equal to 1. In the second example Vladik can fly directly from the airport 5 to the airport 2, because they belong to the same company.
500
[ { "input": "4 1 4\n1010", "output": "1" }, { "input": "5 5 2\n10110", "output": "0" }, { "input": "10 9 5\n1011111001", "output": "1" }, { "input": "7 3 7\n1110111", "output": "0" }, { "input": "1 1 1\n1", "output": "0" }, { "input": "10 3 3\n1001011011", "output": "0" }, { "input": "1 1 1\n0", "output": "0" }, { "input": "10 5 8\n1000001110", "output": "1" }, { "input": "10 1 10\n0000011111", "output": "1" }, { "input": "4 1 4\n0011", "output": "1" }, { "input": "10 3 7\n0000011111", "output": "1" }, { "input": "5 1 5\n11010", "output": "1" }, { "input": "6 1 6\n111000", "output": "1" }, { "input": "2 1 2\n01", "output": "1" }, { "input": "10 10 1\n0000011111", "output": "1" }, { "input": "6 1 6\n000111", "output": "1" }, { "input": "10 2 10\n0000011111", "output": "1" }, { "input": "8 1 8\n11110000", "output": "1" }, { "input": "6 1 5\n100000", "output": "1" }, { "input": "16 4 12\n0000000011111111", "output": "1" }, { "input": "6 1 5\n111000", "output": "1" }, { "input": "8 2 7\n11110000", "output": "1" }, { "input": "6 2 5\n111000", "output": "1" }, { "input": "9 9 1\n111000000", "output": "1" }, { "input": "2 2 1\n01", "output": "1" }, { "input": "5 2 5\n00001", "output": "1" }, { "input": "5 1 5\n10000", "output": "1" }, { "input": "6 1 6\n011111", "output": "1" }, { "input": "5 5 1\n11110", "output": "1" }, { "input": "2 1 2\n10", "output": "1" }, { "input": "4 2 4\n0001", "output": "1" }, { "input": "10 1 10\n1111100000", "output": "1" }, { "input": "8 4 5\n00001111", "output": "1" }, { "input": "4 4 1\n0111", "output": "1" }, { "input": "8 1 8\n00101001", "output": "1" } ]
1,594,120,645
2,147,483,647
PyPy 3
WRONG_ANSWER
TESTS
2
140
20,172,800
n,a,b=map(int,input().split()) s=input() s1=s[a-1] s2=s[b-1] m=10**9 for i in range(n): if s[i]==s1: m=min(m,abs(i+1-b)) print(m)
Title: Vladik and flights Time Limit: None seconds Memory Limit: None megabytes Problem Description: Vladik is a competitive programmer. This year he is going to win the International Olympiad in Informatics. But it is not as easy as it sounds: the question Vladik face now is to find the cheapest way to get to the olympiad. Vladik knows *n* airports. All the airports are located on a straight line. Each airport has unique id from 1 to *n*, Vladik's house is situated next to the airport with id *a*, and the place of the olympiad is situated next to the airport with id *b*. It is possible that Vladik's house and the place of the olympiad are located near the same airport. To get to the olympiad, Vladik can fly between any pair of airports any number of times, but he has to start his route at the airport *a* and finish it at the airport *b*. Each airport belongs to one of two companies. The cost of flight from the airport *i* to the airport *j* is zero if both airports belong to the same company, and |*i*<=-<=*j*| if they belong to different companies. Print the minimum cost Vladik has to pay to get to the olympiad. Input Specification: The first line contains three integers *n*, *a*, and *b* (1<=≤<=*n*<=≤<=105, 1<=≤<=*a*,<=*b*<=≤<=*n*) — the number of airports, the id of the airport from which Vladik starts his route and the id of the airport which he has to reach. The second line contains a string with length *n*, which consists only of characters 0 and 1. If the *i*-th character in this string is 0, then *i*-th airport belongs to first company, otherwise it belongs to the second. Output Specification: Print single integer — the minimum cost Vladik has to pay to get to the olympiad. Demo Input: ['4 1 4\n1010\n', '5 5 2\n10110\n'] Demo Output: ['1', '0'] Note: In the first example Vladik can fly to the airport 2 at first and pay |1 - 2| = 1 (because the airports belong to different companies), and then fly from the airport 2 to the airport 4 for free (because the airports belong to the same company). So the cost of the whole flight is equal to 1. It's impossible to get to the olympiad for free, so the answer is equal to 1. In the second example Vladik can fly directly from the airport 5 to the airport 2, because they belong to the same company.
```python n,a,b=map(int,input().split()) s=input() s1=s[a-1] s2=s[b-1] m=10**9 for i in range(n): if s[i]==s1: m=min(m,abs(i+1-b)) print(m) ```
0
88
A
Chord
PROGRAMMING
1,200
[ "brute force", "implementation" ]
A. Chord
2
256
Vasya studies music. He has learned lots of interesting stuff. For example, he knows that there are 12 notes: C, C#, D, D#, E, F, F#, G, G#, A, B, H. He also knows that the notes are repeated cyclically: after H goes C again, and before C stands H. We will consider the C note in the row's beginning and the C note after the H similar and we will identify them with each other. The distance between the notes along the musical scale is measured in tones: between two consecutive notes there's exactly one semitone, that is, 0.5 tone. The distance is taken from the lowest tone to the uppest one, that is, the distance between C and E is 4 semitones and between E and C is 8 semitones Vasya also knows what a chord is. A chord is an unordered set of no less than three notes. However, for now Vasya only works with triads, that is with the chords that consist of exactly three notes. He can already distinguish between two types of triads — major and minor. Let's define a major triad. Let the triad consist of notes *X*, *Y* and *Z*. If we can order the notes so as the distance along the musical scale between *X* and *Y* equals 4 semitones and the distance between *Y* and *Z* is 3 semitones, then the triad is major. The distance between *X* and *Z*, accordingly, equals 7 semitones. A minor triad is different in that the distance between *X* and *Y* should be 3 semitones and between *Y* and *Z* — 4 semitones. For example, the triad "C E G" is major: between C and E are 4 semitones, and between E and G are 3 semitones. And the triplet "C# B F" is minor, because if we order the notes as "B C# F", than between B and C# will be 3 semitones, and between C# and F — 4 semitones. Help Vasya classify the triad the teacher has given to him.
The only line contains 3 space-separated notes in the above-given notation.
Print "major" if the chord is major, "minor" if it is minor, and "strange" if the teacher gave Vasya some weird chord which is neither major nor minor. Vasya promises you that the answer will always be unambiguous. That is, there are no chords that are both major and minor simultaneously.
[ "C E G\n", "C# B F\n", "A B H\n" ]
[ "major\n", "minor\n", "strange\n" ]
none
500
[ { "input": "C E G", "output": "major" }, { "input": "C# B F", "output": "minor" }, { "input": "A B H", "output": "strange" }, { "input": "G H E", "output": "minor" }, { "input": "D# B G", "output": "major" }, { "input": "D# B F#", "output": "minor" }, { "input": "F H E", "output": "strange" }, { "input": "B F# G", "output": "strange" }, { "input": "F# H C", "output": "strange" }, { "input": "C# F C", "output": "strange" }, { "input": "G# C# E", "output": "minor" }, { "input": "D# H G#", "output": "minor" }, { "input": "C F A", "output": "major" }, { "input": "H E G#", "output": "major" }, { "input": "G D# B", "output": "major" }, { "input": "E C G", "output": "major" }, { "input": "G# C# F", "output": "major" }, { "input": "D# C G#", "output": "major" }, { "input": "C# F B", "output": "minor" }, { "input": "D# C G", "output": "minor" }, { "input": "A D F", "output": "minor" }, { "input": "F# H D", "output": "minor" }, { "input": "D A F", "output": "minor" }, { "input": "D A F#", "output": "major" }, { "input": "C# B F", "output": "minor" }, { "input": "A C F", "output": "major" }, { "input": "D F# H", "output": "minor" }, { "input": "H G# D#", "output": "minor" }, { "input": "A D F#", "output": "major" }, { "input": "H E G#", "output": "major" }, { "input": "D# B F#", "output": "minor" }, { "input": "D# H F#", "output": "major" }, { "input": "A D F#", "output": "major" }, { "input": "B G D#", "output": "major" }, { "input": "E A C#", "output": "major" }, { "input": "D H G", "output": "major" }, { "input": "H D F#", "output": "minor" }, { "input": "G D# C", "output": "minor" }, { "input": "H D G", "output": "major" }, { "input": "E C G", "output": "major" }, { "input": "D# A E", "output": "strange" }, { "input": "A F E", "output": "strange" }, { "input": "C E F", "output": "strange" }, { "input": "A B C", "output": "strange" }, { "input": "E F D#", "output": "strange" }, { "input": "C G# G#", "output": "strange" }, { "input": "F D# G#", "output": "strange" }, { "input": "B G D#", "output": "major" }, { "input": "E E G#", "output": "strange" }, { "input": "A G H", "output": "strange" }, { "input": "F E A", "output": "strange" }, { "input": "D B E", "output": "strange" }, { "input": "G# C# D", "output": "strange" }, { "input": "D# D# F#", "output": "strange" }, { "input": "H B G", "output": "strange" }, { "input": "D C B", "output": "strange" }, { "input": "D B B", "output": "strange" }, { "input": "C A H", "output": "strange" }, { "input": "F# H F#", "output": "strange" }, { "input": "A F F#", "output": "strange" }, { "input": "C D C", "output": "strange" }, { "input": "G F# A", "output": "strange" }, { "input": "C C# D", "output": "strange" }, { "input": "C A E", "output": "minor" }, { "input": "A H B", "output": "strange" }, { "input": "B E F#", "output": "strange" }, { "input": "G# G# A", "output": "strange" }, { "input": "B C# C#", "output": "strange" }, { "input": "C G D#", "output": "minor" }, { "input": "C B D#", "output": "strange" }, { "input": "F H F", "output": "strange" }, { "input": "E G# C", "output": "strange" }, { "input": "F# F# F#", "output": "strange" }, { "input": "F C F", "output": "strange" }, { "input": "A F D", "output": "minor" } ]
1,590,084,238
2,147,483,647
Python 3
OK
TESTS
75
218
307,200
s = input().split() chords = ['C', 'C#', 'D', 'D#', 'E', 'F', 'F#', 'G', 'G#', 'A', 'B', 'H'] c1 = chords.index(s[0]) c2 = chords.index(s[1]) c3 = chords.index(s[2]) #print(c1,c2,c3) printed = False possibilities = [[c1,c2,c3],[c2,c1,c3],[c1,c3,c2],[c2,c3,c1],[c3,c1,c2],[c3,c2,c1]] for t in possibilities: #print(t) if (t[0]+4)%12 == t[1] and (t[0] + 7)%12 == t[2] : print('major') printed = True break if (t[0]+3)%12 == t[1] and (t[0] + 7)%12 == t[2]: print('minor') printed = True break if not printed: print('strange')
Title: Chord Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Vasya studies music. He has learned lots of interesting stuff. For example, he knows that there are 12 notes: C, C#, D, D#, E, F, F#, G, G#, A, B, H. He also knows that the notes are repeated cyclically: after H goes C again, and before C stands H. We will consider the C note in the row's beginning and the C note after the H similar and we will identify them with each other. The distance between the notes along the musical scale is measured in tones: between two consecutive notes there's exactly one semitone, that is, 0.5 tone. The distance is taken from the lowest tone to the uppest one, that is, the distance between C and E is 4 semitones and between E and C is 8 semitones Vasya also knows what a chord is. A chord is an unordered set of no less than three notes. However, for now Vasya only works with triads, that is with the chords that consist of exactly three notes. He can already distinguish between two types of triads — major and minor. Let's define a major triad. Let the triad consist of notes *X*, *Y* and *Z*. If we can order the notes so as the distance along the musical scale between *X* and *Y* equals 4 semitones and the distance between *Y* and *Z* is 3 semitones, then the triad is major. The distance between *X* and *Z*, accordingly, equals 7 semitones. A minor triad is different in that the distance between *X* and *Y* should be 3 semitones and between *Y* and *Z* — 4 semitones. For example, the triad "C E G" is major: between C and E are 4 semitones, and between E and G are 3 semitones. And the triplet "C# B F" is minor, because if we order the notes as "B C# F", than between B and C# will be 3 semitones, and between C# and F — 4 semitones. Help Vasya classify the triad the teacher has given to him. Input Specification: The only line contains 3 space-separated notes in the above-given notation. Output Specification: Print "major" if the chord is major, "minor" if it is minor, and "strange" if the teacher gave Vasya some weird chord which is neither major nor minor. Vasya promises you that the answer will always be unambiguous. That is, there are no chords that are both major and minor simultaneously. Demo Input: ['C E G\n', 'C# B F\n', 'A B H\n'] Demo Output: ['major\n', 'minor\n', 'strange\n'] Note: none
```python s = input().split() chords = ['C', 'C#', 'D', 'D#', 'E', 'F', 'F#', 'G', 'G#', 'A', 'B', 'H'] c1 = chords.index(s[0]) c2 = chords.index(s[1]) c3 = chords.index(s[2]) #print(c1,c2,c3) printed = False possibilities = [[c1,c2,c3],[c2,c1,c3],[c1,c3,c2],[c2,c3,c1],[c3,c1,c2],[c3,c2,c1]] for t in possibilities: #print(t) if (t[0]+4)%12 == t[1] and (t[0] + 7)%12 == t[2] : print('major') printed = True break if (t[0]+3)%12 == t[1] and (t[0] + 7)%12 == t[2]: print('minor') printed = True break if not printed: print('strange') ```
3.944928
4
A
Watermelon
PROGRAMMING
800
[ "brute force", "math" ]
A. Watermelon
1
64
One hot summer day Pete and his friend Billy decided to buy a watermelon. They chose the biggest and the ripest one, in their opinion. After that the watermelon was weighed, and the scales showed *w* kilos. They rushed home, dying of thirst, and decided to divide the berry, however they faced a hard problem. Pete and Billy are great fans of even numbers, that's why they want to divide the watermelon in such a way that each of the two parts weighs even number of kilos, at the same time it is not obligatory that the parts are equal. The boys are extremely tired and want to start their meal as soon as possible, that's why you should help them and find out, if they can divide the watermelon in the way they want. For sure, each of them should get a part of positive weight.
The first (and the only) input line contains integer number *w* (1<=≤<=*w*<=≤<=100) — the weight of the watermelon bought by the boys.
Print YES, if the boys can divide the watermelon into two parts, each of them weighing even number of kilos; and NO in the opposite case.
[ "8\n" ]
[ "YES\n" ]
For example, the boys can divide the watermelon into two parts of 2 and 6 kilos respectively (another variant — two parts of 4 and 4 kilos).
0
[ { "input": "8", "output": "YES" }, { "input": "5", "output": "NO" }, { "input": "4", "output": "YES" }, { "input": "3", "output": "NO" }, { "input": "2", "output": "NO" }, { "input": "1", "output": "NO" }, { "input": "7", "output": "NO" }, { "input": "6", "output": "YES" }, { "input": "10", "output": "YES" }, { "input": "9", "output": "NO" }, { "input": "53", "output": "NO" }, { "input": "77", "output": "NO" }, { "input": "32", "output": "YES" }, { "input": "44", "output": "YES" }, { "input": "98", "output": "YES" }, { "input": "99", "output": "NO" }, { "input": "90", "output": "YES" }, { "input": "67", "output": "NO" }, { "input": "100", "output": "YES" }, { "input": "88", "output": "YES" } ]
1,690,580,387
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
4
62
0
weight = int(input()) if weight%2 == 0: print("YES") else: print("NO")
Title: Watermelon Time Limit: 1 seconds Memory Limit: 64 megabytes Problem Description: One hot summer day Pete and his friend Billy decided to buy a watermelon. They chose the biggest and the ripest one, in their opinion. After that the watermelon was weighed, and the scales showed *w* kilos. They rushed home, dying of thirst, and decided to divide the berry, however they faced a hard problem. Pete and Billy are great fans of even numbers, that's why they want to divide the watermelon in such a way that each of the two parts weighs even number of kilos, at the same time it is not obligatory that the parts are equal. The boys are extremely tired and want to start their meal as soon as possible, that's why you should help them and find out, if they can divide the watermelon in the way they want. For sure, each of them should get a part of positive weight. Input Specification: The first (and the only) input line contains integer number *w* (1<=≤<=*w*<=≤<=100) — the weight of the watermelon bought by the boys. Output Specification: Print YES, if the boys can divide the watermelon into two parts, each of them weighing even number of kilos; and NO in the opposite case. Demo Input: ['8\n'] Demo Output: ['YES\n'] Note: For example, the boys can divide the watermelon into two parts of 2 and 6 kilos respectively (another variant — two parts of 4 and 4 kilos).
```python weight = int(input()) if weight%2 == 0: print("YES") else: print("NO") ```
0
2
A
Winner
PROGRAMMING
1,500
[ "hashing", "implementation" ]
A. Winner
1
64
The winner of the card game popular in Berland "Berlogging" is determined according to the following rules. If at the end of the game there is only one player with the maximum number of points, he is the winner. The situation becomes more difficult if the number of such players is more than one. During each round a player gains or loses a particular number of points. In the course of the game the number of points is registered in the line "name score", where name is a player's name, and score is the number of points gained in this round, which is an integer number. If score is negative, this means that the player has lost in the round. So, if two or more players have the maximum number of points (say, it equals to *m*) at the end of the game, than wins the one of them who scored at least *m* points first. Initially each player has 0 points. It's guaranteed that at the end of the game at least one player has a positive number of points.
The first line contains an integer number *n* (1<=<=≤<=<=*n*<=<=≤<=<=1000), *n* is the number of rounds played. Then follow *n* lines, containing the information about the rounds in "name score" format in chronological order, where name is a string of lower-case Latin letters with the length from 1 to 32, and score is an integer number between -1000 and 1000, inclusive.
Print the name of the winner.
[ "3\nmike 3\nandrew 5\nmike 2\n", "3\nandrew 3\nandrew 2\nmike 5\n" ]
[ "andrew\n", "andrew\n" ]
none
0
[ { "input": "3\nmike 3\nandrew 5\nmike 2", "output": "andrew" }, { "input": "3\nandrew 3\nandrew 2\nmike 5", "output": "andrew" }, { "input": "5\nkaxqybeultn -352\nmgochgrmeyieyskhuourfg -910\nkaxqybeultn 691\nmgochgrmeyieyskhuourfg -76\nkaxqybeultn -303", "output": "kaxqybeultn" }, { "input": "7\nksjuuerbnlklcfdjeyq 312\ndthjlkrvvbyahttifpdewvyslsh -983\nksjuuerbnlklcfdjeyq 268\ndthjlkrvvbyahttifpdewvyslsh 788\nksjuuerbnlklcfdjeyq -79\nksjuuerbnlklcfdjeyq -593\nksjuuerbnlklcfdjeyq 734", "output": "ksjuuerbnlklcfdjeyq" }, { "input": "12\natrtthfpcvishmqbakprquvnejr 185\natrtthfpcvishmqbakprquvnejr -699\natrtthfpcvishmqbakprquvnejr -911\natrtthfpcvishmqbakprquvnejr -220\nfcgslzkicjrpbqaifgweyzreajjfdo 132\nfcgslzkicjrpbqaifgweyzreajjfdo -242\nm 177\nm -549\natrtthfpcvishmqbakprquvnejr -242\nm 38\natrtthfpcvishmqbakprquvnejr -761\nfcgslzkicjrpbqaifgweyzreajjfdo 879", "output": "fcgslzkicjrpbqaifgweyzreajjfdo" }, { "input": "15\naawtvezfntstrcpgbzjbf 681\nzhahpvqiptvksnbjkdvmknb -74\naawtvezfntstrcpgbzjbf 661\njpdwmyke 474\naawtvezfntstrcpgbzjbf -547\naawtvezfntstrcpgbzjbf 600\nzhahpvqiptvksnbjkdvmknb -11\njpdwmyke 711\nbjmj 652\naawtvezfntstrcpgbzjbf -1000\naawtvezfntstrcpgbzjbf -171\nbjmj -302\naawtvezfntstrcpgbzjbf 961\nzhahpvqiptvksnbjkdvmknb 848\nbjmj -735", "output": "aawtvezfntstrcpgbzjbf" }, { "input": "17\nqdplghhx -649\nivhgbxiv 424\nivhgbxiv -72\nivhgbxiv 479\nugowfouebkelargxylsxn 83\nugowfouebkelargxylsxn -840\nivhgbxiv 382\nqdplghhx -904\nivhgbxiv -710\nugowfouebkelargxylsxn -838\nqdplghhx -974\nqdplghhx 571\nivhgbxiv -145\nugowfouebkelargxylsxn -788\nivhgbxiv 209\nivhgbxiv 420\nqdplghhx -158", "output": "ivhgbxiv" } ]
1,690,619,494
2,147,483,647
PyPy 3-64
WRONG_ANSWER
TESTS
0
60
0
input_v = [] for _ in range(int(input())): (name, score) = input().split(' ') input_v.append((name, int(score))) max = 0 score = dict() for i, (n,s) in enumerate(input_v): score[n] = score.get(n, 0) + s if score[n] > max: max = score[n] score = dict() for n,s in input_v: score[n] = score.get(n, 0) + s if score[n] >= max: print(n + " wins") break
Title: Winner Time Limit: 1 seconds Memory Limit: 64 megabytes Problem Description: The winner of the card game popular in Berland "Berlogging" is determined according to the following rules. If at the end of the game there is only one player with the maximum number of points, he is the winner. The situation becomes more difficult if the number of such players is more than one. During each round a player gains or loses a particular number of points. In the course of the game the number of points is registered in the line "name score", where name is a player's name, and score is the number of points gained in this round, which is an integer number. If score is negative, this means that the player has lost in the round. So, if two or more players have the maximum number of points (say, it equals to *m*) at the end of the game, than wins the one of them who scored at least *m* points first. Initially each player has 0 points. It's guaranteed that at the end of the game at least one player has a positive number of points. Input Specification: The first line contains an integer number *n* (1<=<=≤<=<=*n*<=<=≤<=<=1000), *n* is the number of rounds played. Then follow *n* lines, containing the information about the rounds in "name score" format in chronological order, where name is a string of lower-case Latin letters with the length from 1 to 32, and score is an integer number between -1000 and 1000, inclusive. Output Specification: Print the name of the winner. Demo Input: ['3\nmike 3\nandrew 5\nmike 2\n', '3\nandrew 3\nandrew 2\nmike 5\n'] Demo Output: ['andrew\n', 'andrew\n'] Note: none
```python input_v = [] for _ in range(int(input())): (name, score) = input().split(' ') input_v.append((name, int(score))) max = 0 score = dict() for i, (n,s) in enumerate(input_v): score[n] = score.get(n, 0) + s if score[n] > max: max = score[n] score = dict() for n,s in input_v: score[n] = score.get(n, 0) + s if score[n] >= max: print(n + " wins") break ```
0
129
A
Cookies
PROGRAMMING
900
[ "implementation" ]
null
null
Olga came to visit the twins Anna and Maria and saw that they have many cookies. The cookies are distributed into bags. As there are many cookies, Olga decided that it's no big deal if she steals a bag. However, she doesn't want the sisters to quarrel because of nothing when they divide the cookies. That's why Olga wants to steal a bag with cookies so that the number of cookies in the remaining bags was even, that is, so that Anna and Maria could evenly divide it into two (even 0 remaining cookies will do, just as any other even number). How many ways there are to steal exactly one cookie bag so that the total number of cookies in the remaining bags was even?
The first line contains the only integer *n* (1<=≤<=*n*<=≤<=100) — the number of cookie bags Anna and Maria have. The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=100) — the number of cookies in the *i*-th bag.
Print in the only line the only number — the sought number of ways. If there are no such ways print 0.
[ "1\n1\n", "10\n1 2 2 3 4 4 4 2 2 2\n", "11\n2 2 2 2 2 2 2 2 2 2 99\n" ]
[ "1\n", "8\n", "1\n" ]
In the first sample Olga should take the only bag so that the twins ended up with the even number of cookies. In the second sample Olga can take any of five bags with two cookies or any of three bags with four cookies — 5 + 3 = 8 ways in total. In the third sample, no matter which bag with two cookies Olga chooses, the twins are left with 2 * 9 + 99 = 117 cookies. Thus, Olga has only one option: to take the bag with 99 cookies.
500
[ { "input": "1\n1", "output": "1" }, { "input": "10\n1 2 2 3 4 4 4 2 2 2", "output": "8" }, { "input": "11\n2 2 2 2 2 2 2 2 2 2 99", "output": "1" }, { "input": "2\n1 1", "output": "0" }, { "input": "2\n2 2", "output": "2" }, { "input": "2\n1 2", "output": "1" }, { "input": "7\n7 7 7 7 7 7 7", "output": "7" }, { "input": "8\n1 2 3 4 5 6 7 8", "output": "4" }, { "input": "100\n1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2", "output": "50" }, { "input": "99\n99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99", "output": "49" }, { "input": "82\n43 44 96 33 23 42 33 66 53 87 8 90 43 91 40 88 51 18 48 62 59 10 22 20 54 6 13 63 2 56 31 52 98 42 54 32 26 77 9 24 33 91 16 30 39 34 78 82 73 90 12 15 67 76 30 18 44 86 84 98 65 54 100 79 28 34 40 56 11 43 72 35 86 59 89 40 30 33 7 19 44 15", "output": "50" }, { "input": "17\n50 14 17 77 74 74 38 76 41 27 45 29 66 98 38 73 38", "output": "7" }, { "input": "94\n81 19 90 99 26 11 86 44 78 36 80 59 99 90 78 72 71 20 94 56 42 40 71 84 10 85 10 70 52 27 39 55 90 16 48 25 7 79 99 100 38 10 99 56 3 4 78 9 16 57 14 40 52 54 57 70 30 86 56 84 97 60 59 69 49 66 23 92 90 46 86 73 53 47 1 83 14 20 24 66 13 45 41 14 86 75 55 88 48 95 82 24 47 87", "output": "39" }, { "input": "88\n64 95 12 90 40 65 98 45 52 54 79 7 81 25 98 19 68 82 41 53 35 50 5 22 32 21 8 39 8 6 72 27 81 30 12 79 21 42 60 2 66 87 46 93 62 78 52 71 76 32 78 94 86 85 55 15 34 76 41 20 32 26 94 81 89 45 74 49 11 40 40 39 49 46 80 85 90 23 80 40 86 58 70 26 48 93 23 53", "output": "37" }, { "input": "84\n95 9 43 43 13 84 60 90 1 8 97 99 54 34 59 83 33 15 51 26 40 12 66 65 19 30 29 78 92 60 25 13 19 84 71 73 12 24 54 49 16 41 11 40 57 59 34 40 39 9 71 83 1 77 79 53 94 47 78 55 77 85 29 52 80 90 53 77 97 97 27 79 28 23 83 25 26 22 49 86 63 56 3 32", "output": "51" }, { "input": "47\n61 97 76 94 91 22 2 68 62 73 90 47 16 79 44 71 98 68 43 6 53 52 40 27 68 67 43 96 14 91 60 61 96 24 97 13 32 65 85 96 81 77 34 18 23 14 80", "output": "21" }, { "input": "69\n71 1 78 74 58 89 30 6 100 90 22 61 11 59 14 74 27 25 78 61 45 19 25 33 37 4 52 43 53 38 9 100 56 67 69 38 76 91 63 60 93 52 28 61 9 98 8 14 57 63 89 64 98 51 36 66 36 86 13 82 50 91 52 64 86 78 78 83 81", "output": "37" }, { "input": "52\n38 78 36 75 19 3 56 1 39 97 24 79 84 16 93 55 96 64 12 24 1 86 80 29 12 32 36 36 73 39 76 65 53 98 30 20 28 8 86 43 70 22 75 69 62 65 81 25 53 40 71 59", "output": "28" }, { "input": "74\n81 31 67 97 26 75 69 81 11 13 13 74 77 88 52 20 52 64 66 75 72 28 41 54 26 75 41 91 75 15 18 36 13 83 63 61 14 48 53 63 19 67 35 48 23 65 73 100 44 55 92 88 99 17 73 25 83 7 31 89 12 80 98 39 42 75 14 29 81 35 77 87 33 94", "output": "47" }, { "input": "44\n46 56 31 31 37 71 94 2 14 100 45 72 36 72 80 3 38 54 42 98 50 32 31 42 62 31 45 50 95 100 18 17 64 22 18 25 52 56 70 57 43 40 81 28", "output": "15" }, { "input": "22\n28 57 40 74 51 4 45 84 99 12 95 14 92 60 47 81 84 51 31 91 59 42", "output": "11" }, { "input": "59\n73 45 94 76 41 49 65 13 74 66 36 25 47 75 40 23 92 72 11 32 32 8 81 26 68 56 41 8 76 47 96 55 70 11 84 14 83 18 70 22 30 39 28 100 48 11 92 45 78 69 86 1 54 90 98 91 13 17 35", "output": "33" }, { "input": "63\n20 18 44 94 68 57 16 43 74 55 68 24 21 95 76 84 50 50 47 86 86 12 58 55 28 72 86 18 34 45 81 88 3 72 41 9 60 90 81 93 12 6 9 6 2 41 1 7 9 29 81 14 64 80 20 36 67 54 7 5 35 81 22", "output": "37" }, { "input": "28\n49 84 48 19 44 91 11 82 96 95 88 90 71 82 87 25 31 23 18 13 98 45 26 65 35 12 31 14", "output": "15" }, { "input": "61\n34 18 28 64 28 45 9 77 77 20 63 92 79 16 16 100 86 2 91 91 57 15 31 95 10 88 84 5 82 83 53 98 59 17 97 80 76 80 81 3 91 81 87 93 61 46 10 49 6 22 21 75 63 89 21 81 30 19 67 38 77", "output": "35" }, { "input": "90\n41 90 43 1 28 75 90 50 3 70 76 64 81 63 25 69 83 82 29 91 59 66 21 61 7 55 72 49 38 69 72 20 64 58 30 81 61 29 96 14 39 5 100 20 29 98 75 29 44 78 97 45 26 77 73 59 22 99 41 6 3 96 71 20 9 18 96 18 90 62 34 78 54 5 41 6 73 33 2 54 26 21 18 6 45 57 43 73 95 75", "output": "42" }, { "input": "45\n93 69 4 27 20 14 71 48 79 3 32 26 49 30 57 88 13 56 49 61 37 32 47 41 41 70 45 68 82 18 8 6 25 20 15 13 71 99 28 6 52 34 19 59 26", "output": "23" }, { "input": "33\n29 95 48 49 91 10 83 71 47 25 66 36 51 12 34 10 54 74 41 96 89 26 89 1 42 33 1 62 9 32 49 65 78", "output": "15" }, { "input": "34\n98 24 42 36 41 82 28 58 89 34 77 70 76 44 74 54 66 100 13 79 4 88 21 1 11 45 91 29 87 100 29 54 82 78", "output": "13" }, { "input": "29\n91 84 26 84 9 63 52 9 65 56 90 2 36 7 67 33 91 14 65 38 53 36 81 83 85 14 33 95 51", "output": "17" }, { "input": "100\n2 88 92 82 87 100 78 28 84 43 78 32 43 33 97 19 15 52 29 84 57 72 54 13 99 28 82 79 40 70 34 92 91 53 9 88 27 43 14 92 72 37 26 37 20 95 19 34 49 64 33 37 34 27 80 79 9 54 99 68 25 4 68 73 46 66 24 78 3 87 26 52 50 84 4 95 23 83 39 58 86 36 33 16 98 2 84 19 53 12 69 60 10 11 78 17 79 92 77 59", "output": "45" }, { "input": "100\n2 95 45 73 9 54 20 97 57 82 88 26 18 71 25 27 75 54 31 11 58 85 69 75 72 91 76 5 25 80 45 49 4 73 8 81 81 38 5 12 53 77 7 96 90 35 28 80 73 94 19 69 96 17 94 49 69 9 32 19 5 12 46 29 26 40 59 59 6 95 82 50 72 2 45 69 12 5 72 29 39 72 23 96 81 28 28 56 68 58 37 41 30 1 90 84 15 24 96 43", "output": "53" }, { "input": "100\n27 72 35 91 13 10 35 45 24 55 83 84 63 96 29 79 34 67 63 92 48 83 18 77 28 27 49 66 29 88 55 15 6 58 14 67 94 36 77 7 7 64 61 52 71 18 36 99 76 6 50 67 16 13 41 7 89 73 61 51 78 22 78 32 76 100 3 31 89 71 63 53 15 85 77 54 89 33 68 74 3 23 57 5 43 89 75 35 9 86 90 11 31 46 48 37 74 17 77 8", "output": "40" }, { "input": "100\n69 98 69 88 11 49 55 8 25 91 17 81 47 26 15 73 96 71 18 42 42 61 48 14 92 78 35 72 4 27 62 75 83 79 17 16 46 80 96 90 82 54 37 69 85 21 67 70 96 10 46 63 21 59 56 92 54 88 77 30 75 45 44 29 86 100 51 11 65 69 66 56 82 63 27 1 51 51 13 10 3 55 26 85 34 16 87 72 13 100 81 71 90 95 86 50 83 55 55 54", "output": "53" }, { "input": "100\n34 35 99 64 2 66 78 93 20 48 12 79 19 10 87 7 42 92 60 79 5 2 24 89 57 48 63 92 74 4 16 51 7 12 90 48 87 17 18 73 51 58 97 97 25 38 15 97 96 73 67 91 6 75 14 13 87 79 75 3 15 55 35 95 71 45 10 13 20 37 82 26 2 22 13 83 97 84 39 79 43 100 54 59 98 8 61 34 7 65 75 44 24 77 73 88 34 95 44 77", "output": "55" }, { "input": "100\n15 86 3 1 51 26 74 85 37 87 64 58 10 6 57 26 30 47 85 65 24 72 50 40 12 35 91 47 91 60 47 87 95 34 80 91 26 3 36 39 14 86 28 70 51 44 28 21 72 79 57 61 16 71 100 94 57 67 36 74 24 21 89 85 25 2 97 67 76 53 76 80 97 64 35 13 8 32 21 52 62 61 67 14 74 73 66 44 55 76 24 3 43 42 99 61 36 80 38 66", "output": "52" }, { "input": "100\n45 16 54 54 80 94 74 93 75 85 58 95 79 30 81 2 84 4 57 23 92 64 78 1 50 36 13 27 56 54 10 77 87 1 5 38 85 74 94 82 30 45 72 83 82 30 81 82 82 3 69 82 7 92 39 60 94 42 41 5 3 17 67 21 79 44 79 96 28 3 53 68 79 89 63 83 1 44 4 31 84 15 73 77 19 66 54 6 73 1 67 24 91 11 86 45 96 82 20 89", "output": "51" }, { "input": "100\n84 23 50 32 90 71 92 43 58 70 6 82 7 55 85 19 70 89 12 26 29 56 74 30 2 27 4 39 63 67 91 81 11 33 75 10 82 88 39 43 43 80 68 35 55 67 53 62 73 65 86 74 43 51 14 48 42 92 83 57 22 33 24 99 5 27 78 96 7 28 11 15 8 38 85 67 5 92 24 96 57 59 14 95 91 4 9 18 45 33 74 83 64 85 14 51 51 94 29 2", "output": "53" }, { "input": "100\n77 56 56 45 73 55 32 37 39 50 30 95 79 21 44 34 51 43 86 91 39 30 85 15 35 93 100 14 57 31 80 79 38 40 88 4 91 54 7 95 76 26 62 84 17 33 67 47 6 82 69 51 17 2 59 24 11 12 31 90 12 11 55 38 72 49 30 50 42 46 5 97 9 9 30 45 86 23 19 82 40 42 5 40 35 98 35 32 60 60 5 28 84 35 21 49 68 53 68 23", "output": "48" }, { "input": "100\n78 38 79 61 45 86 83 83 86 90 74 69 2 84 73 39 2 5 20 71 24 80 54 89 58 34 77 40 39 62 2 47 28 53 97 75 88 98 94 96 33 71 44 90 47 36 19 89 87 98 90 87 5 85 34 79 82 3 42 88 89 63 35 7 89 30 40 48 12 41 56 76 83 60 80 80 39 56 77 4 72 96 30 55 57 51 7 19 11 1 66 1 91 87 11 62 95 85 79 25", "output": "48" }, { "input": "100\n5 34 23 20 76 75 19 51 17 82 60 13 83 6 65 16 20 43 66 54 87 10 87 73 50 24 16 98 33 28 80 52 54 82 26 92 14 13 84 92 94 29 61 21 60 20 48 94 24 20 75 70 58 27 68 45 86 89 29 8 67 38 83 48 18 100 11 22 46 84 52 97 70 19 50 75 3 7 52 53 72 41 18 31 1 38 49 53 11 64 99 76 9 87 48 12 100 32 44 71", "output": "58" }, { "input": "100\n76 89 68 78 24 72 73 95 98 72 58 15 2 5 56 32 9 65 50 70 94 31 29 54 89 52 31 93 43 56 26 35 72 95 51 55 78 70 11 92 17 5 54 94 81 31 78 95 73 91 95 37 59 9 53 48 65 55 84 8 45 97 64 37 96 34 36 53 66 17 72 48 99 23 27 18 92 84 44 73 60 78 53 29 68 99 19 39 61 40 69 6 77 12 47 29 15 4 8 45", "output": "53" }, { "input": "100\n82 40 31 53 8 50 85 93 3 84 54 17 96 59 51 42 18 19 35 84 79 31 17 46 54 82 72 49 35 73 26 89 61 73 3 50 12 29 25 77 88 21 58 24 22 89 96 54 82 29 96 56 77 16 1 68 90 93 20 23 57 22 31 18 92 90 51 14 50 72 31 54 12 50 66 62 2 34 17 45 68 50 87 97 23 71 1 72 17 82 42 15 20 78 4 49 66 59 10 17", "output": "54" }, { "input": "100\n32 82 82 24 39 53 48 5 29 24 9 37 91 37 91 95 1 97 84 52 12 56 93 47 22 20 14 17 40 22 79 34 24 2 69 30 69 29 3 89 21 46 60 92 39 29 18 24 49 18 40 22 60 13 77 50 39 64 50 70 99 8 66 31 90 38 20 54 7 21 5 56 41 68 69 20 54 89 69 62 9 53 43 89 81 97 15 2 52 78 89 65 16 61 59 42 56 25 32 52", "output": "49" }, { "input": "100\n72 54 23 24 97 14 99 87 15 25 7 23 17 87 72 31 71 87 34 82 51 77 74 85 62 38 24 7 84 48 98 21 29 71 70 84 25 58 67 92 18 44 32 9 81 15 53 29 63 18 86 16 7 31 38 99 70 32 89 16 23 11 66 96 69 82 97 59 6 9 49 80 85 19 6 9 52 51 85 74 53 46 73 55 31 63 78 61 34 80 77 65 87 77 92 52 89 8 52 31", "output": "44" }, { "input": "100\n56 88 8 19 7 15 11 54 35 50 19 57 63 72 51 43 50 19 57 90 40 100 8 92 11 96 30 32 59 65 93 47 62 3 50 41 30 50 72 83 61 46 83 60 20 46 33 1 5 18 83 22 34 16 41 95 63 63 7 59 55 95 91 29 64 60 64 81 45 45 10 9 88 37 69 85 21 82 41 76 42 34 47 78 51 83 65 100 13 22 59 76 63 1 26 86 36 94 99 74", "output": "46" }, { "input": "100\n27 89 67 60 62 80 43 50 28 88 72 5 94 11 63 91 18 78 99 3 71 26 12 97 74 62 23 24 22 3 100 72 98 7 94 32 12 75 61 88 42 48 10 14 45 9 48 56 73 76 70 70 79 90 35 39 96 37 81 11 19 65 99 39 23 79 34 61 35 74 90 37 73 23 46 21 94 84 73 58 11 89 13 9 10 85 42 78 73 32 53 39 49 90 43 5 28 31 97 75", "output": "53" }, { "input": "100\n33 24 97 96 1 14 99 51 13 65 67 20 46 88 42 44 20 49 5 89 98 83 15 40 74 83 58 3 10 79 34 2 69 28 37 100 55 52 14 8 44 94 97 89 6 42 11 28 30 33 55 56 20 57 52 25 75 1 87 42 62 41 37 12 54 85 95 80 42 36 94 96 28 76 54 36 4 17 26 24 62 15 17 79 84 36 92 78 74 91 96 77 54 92 81 91 62 98 37 37", "output": "43" }, { "input": "100\n86 24 61 15 11 85 1 31 47 36 23 36 59 34 3 27 16 29 82 28 58 52 52 66 71 61 98 39 60 20 67 41 67 90 73 29 92 17 70 95 58 98 58 32 21 73 46 56 87 72 80 75 40 27 94 31 59 92 93 37 14 99 96 21 97 23 81 91 52 52 96 94 92 28 38 29 52 16 57 27 17 24 91 21 79 55 96 98 95 94 23 78 79 12 77 35 32 75 100 82", "output": "51" }, { "input": "100\n88 85 41 37 69 21 7 69 36 5 92 26 64 75 22 46 67 20 70 22 62 66 38 24 47 49 68 30 90 31 67 86 86 82 9 51 43 45 48 42 73 44 31 94 45 60 54 66 20 87 11 94 34 32 87 66 56 28 75 39 37 90 72 93 55 72 31 42 30 71 87 61 4 12 12 81 23 61 56 98 71 32 30 33 96 63 92 16 8 78 47 91 47 54 49 3 81 82 41 5", "output": "47" }, { "input": "1\n2", "output": "1" }, { "input": "5\n1 1 3 2 2", "output": "3" } ]
1,602,573,524
2,147,483,647
Python 3
OK
TESTS
52
216
0
n,a=int(input()),list(map(int,input().split())) even=list(filter(lambda x:x%2==0,a)) odd=n-len(even) if odd%2!=0: print(odd) else: print(len(even))
Title: Cookies Time Limit: None seconds Memory Limit: None megabytes Problem Description: Olga came to visit the twins Anna and Maria and saw that they have many cookies. The cookies are distributed into bags. As there are many cookies, Olga decided that it's no big deal if she steals a bag. However, she doesn't want the sisters to quarrel because of nothing when they divide the cookies. That's why Olga wants to steal a bag with cookies so that the number of cookies in the remaining bags was even, that is, so that Anna and Maria could evenly divide it into two (even 0 remaining cookies will do, just as any other even number). How many ways there are to steal exactly one cookie bag so that the total number of cookies in the remaining bags was even? Input Specification: The first line contains the only integer *n* (1<=≤<=*n*<=≤<=100) — the number of cookie bags Anna and Maria have. The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=100) — the number of cookies in the *i*-th bag. Output Specification: Print in the only line the only number — the sought number of ways. If there are no such ways print 0. Demo Input: ['1\n1\n', '10\n1 2 2 3 4 4 4 2 2 2\n', '11\n2 2 2 2 2 2 2 2 2 2 99\n'] Demo Output: ['1\n', '8\n', '1\n'] Note: In the first sample Olga should take the only bag so that the twins ended up with the even number of cookies. In the second sample Olga can take any of five bags with two cookies or any of three bags with four cookies — 5 + 3 = 8 ways in total. In the third sample, no matter which bag with two cookies Olga chooses, the twins are left with 2 * 9 + 99 = 117 cookies. Thus, Olga has only one option: to take the bag with 99 cookies.
```python n,a=int(input()),list(map(int,input().split())) even=list(filter(lambda x:x%2==0,a)) odd=n-len(even) if odd%2!=0: print(odd) else: print(len(even)) ```
3
78
B
Easter Eggs
PROGRAMMING
1,200
[ "constructive algorithms", "implementation" ]
B. Easter Eggs
2
256
The Easter Rabbit laid *n* eggs in a circle and is about to paint them. Each egg should be painted one color out of 7: red, orange, yellow, green, blue, indigo or violet. Also, the following conditions should be satisfied: - Each of the seven colors should be used to paint at least one egg. - Any four eggs lying sequentially should be painted different colors. Help the Easter Rabbit paint the eggs in the required manner. We know that it is always possible.
The only line contains an integer *n* — the amount of eggs (7<=≤<=*n*<=≤<=100).
Print one line consisting of *n* characters. The *i*-th character should describe the color of the *i*-th egg in the order they lie in the circle. The colors should be represented as follows: "R" stands for red, "O" stands for orange, "Y" stands for yellow, "G" stands for green, "B" stands for blue, "I" stands for indigo, "V" stands for violet. If there are several answers, print any of them.
[ "8\n", "13\n" ]
[ "ROYGRBIV\n", "ROYGBIVGBIVYG\n" ]
The way the eggs will be painted in the first sample is shown on the picture:
1,000
[ { "input": "8", "output": "ROYGBIVG" }, { "input": "13", "output": "ROYGBIVOYGBIV" }, { "input": "7", "output": "ROYGBIV" }, { "input": "10", "output": "ROYGBIVYGB" }, { "input": "14", "output": "ROYGBIVROYGBIV" }, { "input": "50", "output": "ROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVG" }, { "input": "9", "output": "ROYGBIVGB" }, { "input": "11", "output": "ROYGBIVYGBI" }, { "input": "12", "output": "ROYGBIVOYGBI" }, { "input": "15", "output": "ROYGBIVROYGBIVG" }, { "input": "16", "output": "ROYGBIVROYGBIVGB" }, { "input": "17", "output": "ROYGBIVROYGBIVYGB" }, { "input": "18", "output": "ROYGBIVROYGBIVYGBI" }, { "input": "19", "output": "ROYGBIVROYGBIVOYGBI" }, { "input": "20", "output": "ROYGBIVROYGBIVOYGBIV" }, { "input": "21", "output": "ROYGBIVROYGBIVROYGBIV" }, { "input": "22", "output": "ROYGBIVROYGBIVROYGBIVG" }, { "input": "23", "output": "ROYGBIVROYGBIVROYGBIVGB" }, { "input": "24", "output": "ROYGBIVROYGBIVROYGBIVYGB" }, { "input": "25", "output": "ROYGBIVROYGBIVROYGBIVYGBI" }, { "input": "28", "output": "ROYGBIVROYGBIVROYGBIVROYGBIV" }, { "input": "29", "output": "ROYGBIVROYGBIVROYGBIVROYGBIVG" }, { "input": "34", "output": "ROYGBIVROYGBIVROYGBIVROYGBIVOYGBIV" }, { "input": "43", "output": "ROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVG" }, { "input": "61", "output": "ROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVOYGBI" }, { "input": "79", "output": "ROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVGB" }, { "input": "81", "output": "ROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVYGBI" }, { "input": "92", "output": "ROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVG" }, { "input": "95", "output": "ROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVYGBI" }, { "input": "96", "output": "ROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVOYGBI" }, { "input": "97", "output": "ROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVOYGBIV" }, { "input": "98", "output": "ROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIV" }, { "input": "99", "output": "ROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVG" }, { "input": "100", "output": "ROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVGB" } ]
1,595,657,883
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
0
186
6,656,000
import math n = int(input()) base = 'ROYGBIV' additional = ['R', 'O', 'Y', 'G', 'B', 'I', 'V'] ans = base for i in range(n - 7): for col in additional: if col not in ans[0] and col not in ans[1] and col not in ans[-1] and col not in ans[-2]: ans += col print(ans[:n])
Title: Easter Eggs Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: The Easter Rabbit laid *n* eggs in a circle and is about to paint them. Each egg should be painted one color out of 7: red, orange, yellow, green, blue, indigo or violet. Also, the following conditions should be satisfied: - Each of the seven colors should be used to paint at least one egg. - Any four eggs lying sequentially should be painted different colors. Help the Easter Rabbit paint the eggs in the required manner. We know that it is always possible. Input Specification: The only line contains an integer *n* — the amount of eggs (7<=≤<=*n*<=≤<=100). Output Specification: Print one line consisting of *n* characters. The *i*-th character should describe the color of the *i*-th egg in the order they lie in the circle. The colors should be represented as follows: "R" stands for red, "O" stands for orange, "Y" stands for yellow, "G" stands for green, "B" stands for blue, "I" stands for indigo, "V" stands for violet. If there are several answers, print any of them. Demo Input: ['8\n', '13\n'] Demo Output: ['ROYGRBIV\n', 'ROYGBIVGBIVYG\n'] Note: The way the eggs will be painted in the first sample is shown on the picture:
```python import math n = int(input()) base = 'ROYGBIV' additional = ['R', 'O', 'Y', 'G', 'B', 'I', 'V'] ans = base for i in range(n - 7): for col in additional: if col not in ans[0] and col not in ans[1] and col not in ans[-1] and col not in ans[-2]: ans += col print(ans[:n]) ```
0
337
A
Puzzles
PROGRAMMING
900
[ "greedy" ]
null
null
The end of the school year is near and Ms. Manana, the teacher, will soon have to say goodbye to a yet another class. She decided to prepare a goodbye present for her *n* students and give each of them a jigsaw puzzle (which, as wikipedia states, is a tiling puzzle that requires the assembly of numerous small, often oddly shaped, interlocking and tessellating pieces). The shop assistant told the teacher that there are *m* puzzles in the shop, but they might differ in difficulty and size. Specifically, the first jigsaw puzzle consists of *f*1 pieces, the second one consists of *f*2 pieces and so on. Ms. Manana doesn't want to upset the children, so she decided that the difference between the numbers of pieces in her presents must be as small as possible. Let *A* be the number of pieces in the largest puzzle that the teacher buys and *B* be the number of pieces in the smallest such puzzle. She wants to choose such *n* puzzles that *A*<=-<=*B* is minimum possible. Help the teacher and find the least possible value of *A*<=-<=*B*.
The first line contains space-separated integers *n* and *m* (2<=≤<=*n*<=≤<=*m*<=≤<=50). The second line contains *m* space-separated integers *f*1,<=*f*2,<=...,<=*f**m* (4<=≤<=*f**i*<=≤<=1000) — the quantities of pieces in the puzzles sold in the shop.
Print a single integer — the least possible difference the teacher can obtain.
[ "4 6\n10 12 10 7 5 22\n" ]
[ "5\n" ]
Sample 1. The class has 4 students. The shop sells 6 puzzles. If Ms. Manana buys the first four puzzles consisting of 10, 12, 10 and 7 pieces correspondingly, then the difference between the sizes of the largest and the smallest puzzle will be equal to 5. It is impossible to obtain a smaller difference. Note that the teacher can also buy puzzles 1, 3, 4 and 5 to obtain the difference 5.
500
[ { "input": "4 6\n10 12 10 7 5 22", "output": "5" }, { "input": "2 2\n4 4", "output": "0" }, { "input": "2 10\n4 5 6 7 8 9 10 11 12 12", "output": "0" }, { "input": "4 5\n818 136 713 59 946", "output": "759" }, { "input": "3 20\n446 852 783 313 549 965 40 88 86 617 479 118 768 34 47 826 366 957 463 903", "output": "13" }, { "input": "2 25\n782 633 152 416 432 825 115 97 386 357 836 310 530 413 354 373 847 882 913 682 729 582 671 674 94", "output": "3" }, { "input": "4 25\n226 790 628 528 114 64 239 279 619 39 894 763 763 847 525 93 882 697 999 643 650 244 159 884 190", "output": "31" }, { "input": "2 50\n971 889 628 39 253 157 925 694 129 516 660 272 738 319 611 816 142 717 514 392 41 105 132 676 958 118 306 768 600 685 103 857 704 346 857 309 23 718 618 161 176 379 846 834 640 468 952 878 164 997", "output": "0" }, { "input": "25 50\n582 146 750 905 313 509 402 21 488 512 32 898 282 64 579 869 37 996 377 929 975 697 666 837 311 205 116 992 533 298 648 268 54 479 792 595 152 69 267 417 184 433 894 603 988 712 24 414 301 176", "output": "412" }, { "input": "49 50\n58 820 826 960 271 294 473 102 925 318 729 672 244 914 796 646 868 6 893 882 726 203 528 498 271 195 355 459 721 680 547 147 631 116 169 804 145 996 133 559 110 257 771 476 576 251 607 314 427 886", "output": "938" }, { "input": "50 50\n374 573 323 744 190 806 485 247 628 336 491 606 702 321 991 678 337 579 86 240 993 208 668 686 855 205 363 177 719 249 896 919 782 434 59 647 787 996 286 216 636 212 546 903 958 559 544 126 608 993", "output": "937" }, { "input": "6 50\n6 8 7 8 5 4 4 5 7 8 6 5 7 4 7 7 7 8 6 4 6 6 8 8 7 7 8 7 5 8 5 4 4 7 8 4 4 6 6 6 8 7 4 7 6 6 5 8 4 7", "output": "0" }, { "input": "37 50\n14 5 11 17 8 20 19 16 20 11 17 20 16 9 14 14 13 18 11 20 8 8 8 5 19 17 6 18 10 20 9 7 12 6 14 17 4 4 10 13 7 4 11 6 20 19 12 12 15 19", "output": "12" }, { "input": "40 50\n4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4", "output": "0" }, { "input": "40 50\n17 20 43 26 41 37 14 8 30 35 30 24 43 8 42 9 41 50 41 35 27 32 35 43 28 36 31 16 5 7 23 16 14 29 8 39 12 16 36 18 49 39 33 37 38 6 6 27 23 17", "output": "31" }, { "input": "2 2\n1000 4", "output": "996" }, { "input": "2 3\n4 502 1000", "output": "498" }, { "input": "3 3\n4 1000 4", "output": "996" } ]
1,679,489,646
2,147,483,647
Python 3
COMPILATION_ERROR
TESTS
0
0
0
#include <bits/stdc++.h> using namespace std; int main() { #ifndef ONLINE_JUDGE freopen("input.txt", "r", stdin); freopen("output.txt", "w", stdout); #endif int n, m; cin >> n >> m; int arr[m]; for (int i = 0; i < n; ++i) { cin >> arr[i]; } sort(arr, arr + m); int least = arr[n - 1] - arr[0]; for (int i = 0; i < m - n; ++i) { if (arr[n + i - 1] - arr[i] < least) least = arr[n + i - 1] - arr[i]; } cout << least; }
Title: Puzzles Time Limit: None seconds Memory Limit: None megabytes Problem Description: The end of the school year is near and Ms. Manana, the teacher, will soon have to say goodbye to a yet another class. She decided to prepare a goodbye present for her *n* students and give each of them a jigsaw puzzle (which, as wikipedia states, is a tiling puzzle that requires the assembly of numerous small, often oddly shaped, interlocking and tessellating pieces). The shop assistant told the teacher that there are *m* puzzles in the shop, but they might differ in difficulty and size. Specifically, the first jigsaw puzzle consists of *f*1 pieces, the second one consists of *f*2 pieces and so on. Ms. Manana doesn't want to upset the children, so she decided that the difference between the numbers of pieces in her presents must be as small as possible. Let *A* be the number of pieces in the largest puzzle that the teacher buys and *B* be the number of pieces in the smallest such puzzle. She wants to choose such *n* puzzles that *A*<=-<=*B* is minimum possible. Help the teacher and find the least possible value of *A*<=-<=*B*. Input Specification: The first line contains space-separated integers *n* and *m* (2<=≤<=*n*<=≤<=*m*<=≤<=50). The second line contains *m* space-separated integers *f*1,<=*f*2,<=...,<=*f**m* (4<=≤<=*f**i*<=≤<=1000) — the quantities of pieces in the puzzles sold in the shop. Output Specification: Print a single integer — the least possible difference the teacher can obtain. Demo Input: ['4 6\n10 12 10 7 5 22\n'] Demo Output: ['5\n'] Note: Sample 1. The class has 4 students. The shop sells 6 puzzles. If Ms. Manana buys the first four puzzles consisting of 10, 12, 10 and 7 pieces correspondingly, then the difference between the sizes of the largest and the smallest puzzle will be equal to 5. It is impossible to obtain a smaller difference. Note that the teacher can also buy puzzles 1, 3, 4 and 5 to obtain the difference 5.
```python #include <bits/stdc++.h> using namespace std; int main() { #ifndef ONLINE_JUDGE freopen("input.txt", "r", stdin); freopen("output.txt", "w", stdout); #endif int n, m; cin >> n >> m; int arr[m]; for (int i = 0; i < n; ++i) { cin >> arr[i]; } sort(arr, arr + m); int least = arr[n - 1] - arr[0]; for (int i = 0; i < m - n; ++i) { if (arr[n + i - 1] - arr[i] < least) least = arr[n + i - 1] - arr[i]; } cout << least; } ```
-1
868
C
Qualification Rounds
PROGRAMMING
1,500
[ "bitmasks", "brute force", "constructive algorithms", "dp" ]
null
null
Snark and Philip are preparing the problemset for the upcoming pre-qualification round for semi-quarter-finals. They have a bank of *n* problems, and they want to select any non-empty subset of it as a problemset. *k* experienced teams are participating in the contest. Some of these teams already know some of the problems. To make the contest interesting for them, each of the teams should know at most half of the selected problems. Determine if Snark and Philip can make an interesting problemset!
The first line contains two integers *n*, *k* (1<=≤<=*n*<=≤<=105, 1<=≤<=*k*<=≤<=4) — the number of problems and the number of experienced teams. Each of the next *n* lines contains *k* integers, each equal to 0 or 1. The *j*-th number in the *i*-th line is 1 if *j*-th team knows *i*-th problem and 0 otherwise.
Print "YES" (quotes for clarity), if it is possible to make an interesting problemset, and "NO" otherwise. You can print each character either upper- or lowercase ("YeS" and "yes" are valid when the answer is "YES").
[ "5 3\n1 0 1\n1 1 0\n1 0 0\n1 0 0\n1 0 0\n", "3 2\n1 0\n1 1\n0 1\n" ]
[ "NO\n", "YES\n" ]
In the first example you can't make any interesting problemset, because the first team knows all problems. In the second example you can choose the first and the third problems.
1,000
[ { "input": "5 3\n1 0 1\n1 1 0\n1 0 0\n1 0 0\n1 0 0", "output": "NO" }, { "input": "3 2\n1 0\n1 1\n0 1", "output": "YES" }, { "input": "10 2\n1 0\n1 0\n0 0\n1 1\n0 0\n1 1\n0 0\n1 1\n0 1\n0 1", "output": "YES" }, { "input": "10 3\n1 0 0\n0 1 1\n1 0 0\n0 1 0\n0 0 1\n1 0 1\n0 1 1\n1 0 0\n1 1 0\n0 0 0", "output": "YES" }, { "input": "10 4\n1 0 1 0\n1 0 0 1\n1 1 0 1\n1 0 1 1\n1 1 0 1\n1 0 1 0\n0 0 0 0\n0 0 1 0\n1 0 1 0\n0 0 1 1", "output": "YES" }, { "input": "2 2\n0 0\n1 0", "output": "YES" }, { "input": "3 3\n1 0 1\n1 0 0\n1 1 1", "output": "NO" }, { "input": "4 4\n0 0 0 0\n1 1 0 0\n1 1 1 1\n1 0 1 1", "output": "YES" }, { "input": "4 1\n1\n1\n0\n0", "output": "YES" }, { "input": "1 4\n0 0 0 0", "output": "YES" }, { "input": "3 3\n0 0 1\n0 1 1\n1 0 0", "output": "YES" }, { "input": "2 3\n0 0 1\n1 0 0", "output": "YES" }, { "input": "1 1\n0", "output": "YES" }, { "input": "2 4\n0 1 1 1\n1 0 0 0", "output": "YES" }, { "input": "2 4\n1 0 1 0\n0 1 0 1", "output": "YES" }, { "input": "2 4\n1 0 0 0\n0 0 0 1", "output": "YES" }, { "input": "2 3\n0 1 0\n0 0 1", "output": "YES" }, { "input": "3 4\n1 0 1 0\n0 1 0 1\n1 1 1 1", "output": "YES" }, { "input": "3 4\n0 0 1 1\n1 1 1 0\n1 1 0 1", "output": "NO" }, { "input": "4 4\n0 0 0 1\n0 0 0 1\n0 0 1 0\n0 0 1 0", "output": "YES" }, { "input": "2 4\n1 1 0 0\n0 0 1 1", "output": "YES" }, { "input": "2 4\n1 0 0 0\n0 1 0 0", "output": "YES" }, { "input": "2 3\n1 0 0\n0 0 1", "output": "YES" }, { "input": "3 4\n1 0 1 0\n0 1 1 1\n1 0 0 0", "output": "YES" }, { "input": "1 2\n0 0", "output": "YES" }, { "input": "6 3\n0 1 1\n1 0 1\n1 1 1\n0 1 0\n1 0 1\n1 1 0", "output": "YES" }, { "input": "1 4\n0 0 1 1", "output": "NO" }, { "input": "3 3\n1 0 0\n0 1 0\n0 0 1", "output": "YES" }, { "input": "3 4\n1 0 0 0\n1 1 0 0\n0 1 1 1", "output": "YES" }, { "input": "3 2\n0 0\n0 0\n0 0", "output": "YES" }, { "input": "2 4\n1 0 0 0\n1 0 1 1", "output": "NO" }, { "input": "2 4\n0 0 0 1\n1 0 0 0", "output": "YES" }, { "input": "2 4\n1 0 0 0\n0 1 1 1", "output": "YES" }, { "input": "4 4\n1 1 1 1\n0 0 0 1\n0 0 1 1\n1 0 1 1", "output": "NO" }, { "input": "6 3\n1 0 0\n1 1 1\n1 1 1\n0 1 0\n0 1 0\n1 0 0", "output": "YES" }, { "input": "4 4\n0 1 0 0\n1 1 1 1\n1 1 1 1\n1 0 1 1", "output": "YES" }, { "input": "1 3\n0 0 0", "output": "YES" }, { "input": "3 3\n1 0 0\n0 1 0\n0 0 0", "output": "YES" }, { "input": "2 4\n0 1 1 0\n0 0 0 0", "output": "YES" }, { "input": "1 4\n0 0 0 1", "output": "NO" }, { "input": "4 4\n0 0 0 1\n0 0 0 1\n0 0 1 1\n1 1 1 0", "output": "YES" }, { "input": "2 3\n1 0 0\n0 1 1", "output": "YES" }, { "input": "3 2\n0 1\n0 1\n1 0", "output": "YES" }, { "input": "4 3\n1 1 0\n1 1 1\n0 0 1\n0 0 1", "output": "YES" }, { "input": "2 1\n0\n0", "output": "YES" }, { "input": "2 4\n1 1 1 0\n0 0 0 1", "output": "YES" }, { "input": "5 4\n1 1 1 0\n1 1 0 1\n1 0 1 1\n0 1 1 1\n1 1 0 0", "output": "NO" }, { "input": "3 4\n0 1 1 0\n0 1 0 1\n0 0 1 1", "output": "NO" }, { "input": "1 1\n1", "output": "NO" }, { "input": "3 4\n1 0 0 0\n1 0 0 0\n0 1 1 1", "output": "YES" }, { "input": "2 3\n1 1 0\n0 0 1", "output": "YES" }, { "input": "3 3\n0 0 1\n1 1 1\n1 1 0", "output": "YES" }, { "input": "4 4\n0 1 1 1\n1 0 1 0\n1 1 0 1\n1 0 1 0", "output": "NO" }, { "input": "3 3\n1 0 0\n0 0 0\n1 0 0", "output": "YES" }, { "input": "3 4\n1 1 0 0\n1 1 0 0\n0 0 1 1", "output": "YES" }, { "input": "2 4\n1 0 0 1\n0 0 1 0", "output": "YES" }, { "input": "2 4\n0 0 1 1\n1 1 0 0", "output": "YES" }, { "input": "2 3\n0 0 1\n0 1 0", "output": "YES" }, { "input": "2 3\n1 0 0\n0 1 0", "output": "YES" }, { "input": "3 2\n1 0\n0 1\n0 1", "output": "YES" }, { "input": "3 4\n1 1 0 1\n0 0 1 1\n1 0 1 0", "output": "NO" }, { "input": "3 4\n0 0 1 1\n0 1 1 0\n1 1 0 0", "output": "YES" }, { "input": "3 4\n0 0 0 1\n0 0 0 1\n1 1 1 0", "output": "YES" }, { "input": "3 4\n1 1 1 0\n1 1 0 1\n0 0 1 0", "output": "YES" }, { "input": "8 4\n0 0 0 1\n0 0 1 1\n0 0 1 1\n0 0 1 1\n0 0 1 1\n0 0 1 1\n0 0 1 1\n1 1 1 0", "output": "YES" }, { "input": "3 4\n1 0 1 1\n1 1 1 0\n0 1 0 1", "output": "NO" }, { "input": "2 4\n1 1 0 0\n0 0 0 1", "output": "YES" }, { "input": "10 4\n1 0 1 0\n1 0 1 0\n0 1 1 1\n1 0 1 1\n1 1 0 1\n1 0 0 1\n0 1 1 1\n0 0 0 1\n1 1 1 1\n1 0 1 0", "output": "YES" }, { "input": "2 4\n0 1 0 0\n0 0 1 1", "output": "YES" }, { "input": "3 3\n1 1 0\n1 0 1\n0 1 1", "output": "NO" }, { "input": "3 3\n1 1 0\n0 0 1\n1 1 1", "output": "YES" }, { "input": "4 4\n1 1 0 0\n1 0 1 0\n0 1 1 0\n0 0 1 1", "output": "YES" }, { "input": "4 4\n1 0 0 0\n1 0 0 1\n1 0 0 1\n0 1 1 1", "output": "YES" }, { "input": "4 3\n1 0 0\n1 0 0\n1 0 0\n0 1 1", "output": "YES" }, { "input": "2 4\n0 0 1 0\n0 1 0 0", "output": "YES" }, { "input": "1 2\n0 1", "output": "NO" }, { "input": "3 4\n1 1 1 0\n0 0 1 1\n1 1 0 0", "output": "YES" }, { "input": "3 4\n0 0 1 1\n0 1 0 1\n1 0 1 0", "output": "YES" }, { "input": "2 3\n1 0 1\n0 1 0", "output": "YES" }, { "input": "2 4\n0 0 0 0\n0 0 0 0", "output": "YES" }, { "input": "3 4\n0 1 0 1\n0 1 1 0\n1 0 0 1", "output": "YES" }, { "input": "3 4\n0 1 1 0\n1 1 0 0\n0 0 1 1", "output": "YES" }, { "input": "2 4\n1 0 1 0\n0 0 0 1", "output": "YES" }, { "input": "2 3\n0 1 0\n1 0 0", "output": "YES" }, { "input": "4 4\n0 0 1 1\n0 1 1 0\n1 1 0 0\n1 0 0 1", "output": "YES" }, { "input": "10 4\n0 0 1 1\n0 0 1 1\n0 0 1 1\n0 0 1 1\n0 0 1 1\n0 0 1 1\n0 0 1 1\n0 0 1 1\n0 0 1 1\n1 1 0 0", "output": "YES" }, { "input": "3 3\n1 1 0\n0 1 1\n1 0 1", "output": "NO" }, { "input": "2 3\n0 0 1\n1 1 0", "output": "YES" }, { "input": "4 4\n0 0 0 1\n0 0 1 1\n1 1 0 1\n1 1 1 0", "output": "YES" }, { "input": "3 4\n0 0 1 1\n1 0 1 0\n0 1 0 1", "output": "YES" }, { "input": "5 4\n1 1 1 0\n1 1 0 1\n1 0 1 1\n0 1 1 1\n0 0 1 1", "output": "NO" }, { "input": "3 4\n1 0 0 0\n1 1 0 0\n0 1 1 0", "output": "YES" } ]
1,638,953,264
2,147,483,647
PyPy 3-64
OK
TESTS
143
732
4,812,800
n, k = map(int, input().split()) k21 = (1 << k) - 1 inn = [0] * (1 << k) for i in range(n): inn[int("".join(input().split()), 2)] = 1 if inn[0]: print("YES") exit() a = [] for i, x in enumerate(inn): if x: a.append(i) for i in range(1, len(a)): for j in range(i): if a[i] & a[j] == 0: print("YES") exit() print("NO")
Title: Qualification Rounds Time Limit: None seconds Memory Limit: None megabytes Problem Description: Snark and Philip are preparing the problemset for the upcoming pre-qualification round for semi-quarter-finals. They have a bank of *n* problems, and they want to select any non-empty subset of it as a problemset. *k* experienced teams are participating in the contest. Some of these teams already know some of the problems. To make the contest interesting for them, each of the teams should know at most half of the selected problems. Determine if Snark and Philip can make an interesting problemset! Input Specification: The first line contains two integers *n*, *k* (1<=≤<=*n*<=≤<=105, 1<=≤<=*k*<=≤<=4) — the number of problems and the number of experienced teams. Each of the next *n* lines contains *k* integers, each equal to 0 or 1. The *j*-th number in the *i*-th line is 1 if *j*-th team knows *i*-th problem and 0 otherwise. Output Specification: Print "YES" (quotes for clarity), if it is possible to make an interesting problemset, and "NO" otherwise. You can print each character either upper- or lowercase ("YeS" and "yes" are valid when the answer is "YES"). Demo Input: ['5 3\n1 0 1\n1 1 0\n1 0 0\n1 0 0\n1 0 0\n', '3 2\n1 0\n1 1\n0 1\n'] Demo Output: ['NO\n', 'YES\n'] Note: In the first example you can't make any interesting problemset, because the first team knows all problems. In the second example you can choose the first and the third problems.
```python n, k = map(int, input().split()) k21 = (1 << k) - 1 inn = [0] * (1 << k) for i in range(n): inn[int("".join(input().split()), 2)] = 1 if inn[0]: print("YES") exit() a = [] for i, x in enumerate(inn): if x: a.append(i) for i in range(1, len(a)): for j in range(i): if a[i] & a[j] == 0: print("YES") exit() print("NO") ```
3
864
B
Polycarp and Letters
PROGRAMMING
1,000
[ "brute force", "implementation", "strings" ]
null
null
Polycarp loves lowercase letters and dislikes uppercase ones. Once he got a string *s* consisting only of lowercase and uppercase Latin letters. Let *A* be a set of positions in the string. Let's call it pretty if following conditions are met: - letters on positions from *A* in the string are all distinct and lowercase; - there are no uppercase letters in the string which are situated between positions from *A* (i.e. there is no such *j* that *s*[*j*] is an uppercase letter, and *a*1<=&lt;<=*j*<=&lt;<=*a*2 for some *a*1 and *a*2 from *A*). Write a program that will determine the maximum number of elements in a pretty set of positions.
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=200) — length of string *s*. The second line contains a string *s* consisting of lowercase and uppercase Latin letters.
Print maximum number of elements in pretty set of positions for string *s*.
[ "11\naaaaBaabAbA\n", "12\nzACaAbbaazzC\n", "3\nABC\n" ]
[ "2\n", "3\n", "0\n" ]
In the first example the desired positions might be 6 and 8 or 7 and 8. Positions 6 and 7 contain letters 'a', position 8 contains letter 'b'. The pair of positions 1 and 8 is not suitable because there is an uppercase letter 'B' between these position. In the second example desired positions can be 7, 8 and 11. There are other ways to choose pretty set consisting of three elements. In the third example the given string *s* does not contain any lowercase letters, so the answer is 0.
1,000
[ { "input": "11\naaaaBaabAbA", "output": "2" }, { "input": "12\nzACaAbbaazzC", "output": "3" }, { "input": "3\nABC", "output": "0" }, { "input": "1\na", "output": "1" }, { "input": "2\naz", "output": "2" }, { "input": "200\nXbTJZqcbpYuZQEoUrbxlPXAPCtVLrRExpQzxzqzcqsqzsiisswqitswzCtJQxOavicSdBIodideVRKHPojCNHmbnrLgwJlwOpyrJJIhrUePszxSjJGeUgTtOfewPQnPVWhZAtogRPrJLwyShNQaeNsvrJwjuuBOMPCeSckBMISQzGngfOmeyfDObncyeNsihYVtQbSEh", "output": "8" }, { "input": "2\nAZ", "output": "0" }, { "input": "28\nAabcBabcCBNMaaaaabbbbbcccccc", "output": "3" }, { "input": "200\nrsgraosldglhdoorwhkrsehjpuxrjuwgeanjgezhekprzarelduuaxdnspzjuooguuwnzkowkuhzduakdrzpnslauejhrrkalwpurpuuswdgeadlhjwzjgegwpknepazwwleulppwrlgrgedlwdzuodzropsrrkxusjnuzshdkjrxxpgzanzdrpnggdwxarpwohxdepJ", "output": "17" }, { "input": "1\nk", "output": "1" }, { "input": "1\nH", "output": "0" }, { "input": "2\nzG", "output": "1" }, { "input": "2\ngg", "output": "1" }, { "input": "2\nai", "output": "2" }, { "input": "20\npEjVrKWLIFCZjIHgggVU", "output": "1" }, { "input": "20\niFSiiigiYFSKmDnMGcgM", "output": "2" }, { "input": "20\nedxedxxxCQiIVmYEUtLi", "output": "3" }, { "input": "20\nprnchweyabjvzkoqiltm", "output": "20" }, { "input": "35\nQLDZNKFXKVSVLUVHRTDPQYMSTDXBELXBOTS", "output": "0" }, { "input": "35\nbvZWiitgxodztelnYUyljYGnCoWluXTvBLp", "output": "10" }, { "input": "35\nBTexnaeplecllxwlanarpcollawHLVMHIIF", "output": "10" }, { "input": "35\nhhwxqysolegsthsvfcqiryenbujbrrScobu", "output": "20" }, { "input": "26\npbgfqosklxjuzmdheyvawrictn", "output": "26" }, { "input": "100\nchMRWwymTDuZDZuSTvUmmuxvSscnTasyjlwwodhzcoifeahnbmcifyeobbydwparebduoLDCgHlOsPtVRbYGGQXfnkdvrWKIwCRl", "output": "20" }, { "input": "100\nhXYLXKUMBrGkjqQJTGbGWAfmztqqapdbjbhcualhypgnaieKXmhzGMnqXVlcPesskfaEVgvWQTTShRRnEtFahWDyuBzySMpugxCM", "output": "19" }, { "input": "100\nucOgELrgjMrFOgtHzqgvUgtHngKJxdMFKBjfcCppciqmGZXXoiSZibgpadshyljqrwxbomzeutvnhTLGVckZUmyiFPLlwuLBFito", "output": "23" }, { "input": "200\nWTCKAKLVGXSYFVMVJDUYERXNMVNTGWXUGRFCGMYXJQGLODYZTUIDENHYEGFKXFIEUILAMESAXAWZXVCZPJPEYUXBITHMTZOTMKWITGRSFHODKVJHPAHVVWTCTHIVAWAREQXWMPUWQSTPPJFHKGKELBTPUYDAVIUMGASPUEDIODRYXIWCORHOSLIBLOZUNJPHHMXEXOAY", "output": "0" }, { "input": "200\neLCCuYMPPwQoNlCpPOtKWJaQJmWfHeZCKiMSpILHSKjFOYGpRMzMCfMXdDuQdBGNsCNrHIVJzEFfBZcNMwNcFjOFVJvEtUQmLbFNKVHgNDyFkFVQhUTUQDgXhMjJZgFSSiHhMKuTgZQYJqAqKBpHoHddddddddddddddddXSSYNKNnRrKuOjAVKZlRLzCjExPdHaDHBT", "output": "1" }, { "input": "200\nitSYxgOLlwOoAkkkkkzzzzzzzzkzkzkzkkkkkzkzzkzUDJSKybRPBvaIDsNuWImPJvrHkKiMeYukWmtHtgZSyQsgYanZvXNbKXBlFLSUcqRnGWSriAvKxsTkDJfROqaKdzXhvJsPEDATueCraWOGEvRDWjPwXuiNpWsEnCuhDcKWOQxjBkdBqmFatWFkgKsbZuLtRGtY", "output": "2" }, { "input": "200\noggqoqqogoqoggggoggqgooqggogogooogqqgggoqgggqoqogogggogggqgooqgqggqqqoqgqgoooqgqogqoggoqqgqoqgoooqoogooqoogqoqoqqgoqgoqgggogqqqoqoggoqoqqoqggqoggooqqqoqggoggqqqqqqqqqgogqgggggooogogqgggqogqgoqoqogoooq", "output": "3" }, { "input": "200\nCtclUtUnmqFniaLqGRmMoUMeLyFfAgWxIZxdrBarcRQprSOGcdUYsmDbooSuOvBLgrYlgaIjJtFgcxJKHGkCXpYfVKmUbouuIqGstFrrwJzYQqjjqqppqqqqqpqqqjpjjpjqjXRYkfPhGAatOigFuItkKxkjCBLdiNMVGjmdWNMgOOvmaJEdGsWNoaERrINNKqKeQajv", "output": "3" }, { "input": "200\nmeZNrhqtSTSmktGQnnNOTcnyAMTKSixxKQKiagrMqRYBqgbRlsbJhvtNeHVUuMCyZLCnsIixRYrYEAkfQOxSVqXkrPqeCZQksInzRsRKBgvIqlGVPxPQnypknSXjgMjsjElcqGsaJRbegJVAKtWcHoOnzHqzhoKReqBBsOhZYLaYJhmqOMQsizdCsQfjUDHcTtHoeYwu", "output": "4" }, { "input": "200\nvFAYTHJLZaivWzSYmiuDBDUFACDSVbkImnVaXBpCgrbgmTfXKJfoglIkZxWPSeVSFPnHZDNUAqLyhjLXSuAqGLskBlDxjxGPJyGdwzlPfIekwsblIrkxzfhJeNoHywdfAGlJzqXOfQaKceSqViVFTRJEGfACnsFeSFpOYisIHJciqTMNAmgeXeublTvfWoPnddtvKIyF", "output": "6" }, { "input": "200\ngnDdkqJjYvduVYDSsswZDvoCouyaYZTfhmpSakERWLhufZtthWsfbQdTGwhKYjEcrqWBOyxBbiFhdLlIjChLOPiOpYmcrJgDtXsJfmHtLrabyGKOfHQRukEtTzwoqBHfmyVXPebfcpGQacLkGWFwerszjdHpTBXGssYXmGHlcCBgBXyGJqxbVhvDffLyCrZnxonABEXV", "output": "7" }, { "input": "200\nBmggKNRZBXPtJqlJaXLdKKQLDJvXpDuQGupiRQfDwCJCJvAlDDGpPZNOvXkrdKOFOEFBVfrsZjWyHPoKGzXmTAyPJGEmxCyCXpeAdTwbrMtWLmlmGNqxvuxmqpmtpuhrmxxtrquSLFYVlnSYgRJDYHWgHBbziBLZRwCIJNvbtsEdLLxmTbnjkoqSPAuzEeTYLlmejOUH", "output": "9" }, { "input": "200\nMkuxcDWdcnqsrlTsejehQKrTwoOBRCUAywqSnZkDLRmVBDVoOqdZHbrInQQyeRFAjiYYmHGrBbWgWstCPfLPRdNVDXBdqFJsGQfSXbufsiogybEhKDlWfPazIuhpONwGzZWaQNwVnmhTqWdewaklgjwaumXYDGwjSeEcYXjkVtLiYSWULEnTFukIlWQGWsXwWRMJGTcI", "output": "10" }, { "input": "200\nOgMBgYeuMJdjPtLybvwmGDrQEOhliaabEtwulzNEjsfnaznXUMoBbbxkLEwSQzcLrlJdjJCLGVNBxorghPxTYCoqniySJMcilpsqpBAbqdzqRUDVaYOgqGhGrxlIJkyYgkOdTUgRZwpgIkeZFXojLXpDilzirHVVadiHaMrxhzodzpdvhvrzdzxbhmhdpxqqpoDegfFQ", "output": "11" }, { "input": "200\nOLaJOtwultZLiZPSYAVGIbYvbIuZkqFZXwfsqpsavCDmBMStAuUFLBVknWDXNzmiuUYIsUMGxtoadWlPYPqvqSvpYdOiJRxFzGGnnmstniltvitnrmyrblnqyruylummmlsqtqitlbulvtuitiqimuintbimqyurviuntqnnvslynlNYMpYVKYwKVTbIUVdlNGrcFZON", "output": "12" }, { "input": "200\nGAcmlaqfjSAQLvXlkhxujXgSbxdFAwnoxDuldDvYmpUhTWJdcEQSdARLrozJzIgFVCkzPUztWIpaGfiKeqzoXinEjVuoKqyBHmtFjBWcRdBmyjviNlGAIkpikjAimmBgayfphrstfbjexjbttzfzfzaysxfyrjazfhtpghnbbeffjhxrjxpttesgzrnrfbgzzsRsCgmz", "output": "15" }, { "input": "200\nYRvIopNqSTYDhViTqCLMwEbTTIdHkoeuBmAJWhgtOgVxlcHSsavDNzMfpwTghkBvYEtCYQxicLUxdgAcaCzOOgbQYsfnaTXFlFxbeEiGwdNvxwHzkTdKtWlqzalwniDDBDipkxfflpaqkfkgfezbkxdvzemlfohwtgytzzywmwhvzUgPlPdeAVqTPAUZbogQheRXetvT", "output": "20" }, { "input": "200\nNcYVomemswLCUqVRSDKHCknlBmqeSWhVyRzQrnZaOANnTGqsRFMjpczllcEVebqpxdavzppvztxsnfmtcharzqlginndyjkawzurqkxJLXiXKNZTIIxhSQghDpjwzatEqnLMTLxwoEKpHytvWkKFDUcZjLShCiVdocxRvvJtbXHCDGpJvMwRKWLhcTFtswdLUHkbhfau", "output": "25" }, { "input": "200\nDxNZuvkTkQEqdWIkLzcKAwfqvZQiptnTazydSCTIfGjDhLMrlPZiKEsqIdDhgKPAlEvXyzNwWtYorotgkcwydpabjqnzubaksdchucxtkmjzfretdmvlxgklyvicrtftvztsbiUaQorfNIYUOdwQDRsKpxLUiLknbLbinilpPXPTTwLAnXVpMHBaAcKWgDBeOFabPtXU", "output": "26" }, { "input": "4\nabbc", "output": "3" }, { "input": "3\naaa", "output": "1" }, { "input": "3\naba", "output": "2" }, { "input": "3\nabb", "output": "2" }, { "input": "3\nbba", "output": "2" }, { "input": "3\nAaa", "output": "1" }, { "input": "3\nAba", "output": "2" }, { "input": "3\naBa", "output": "1" }, { "input": "3\naAa", "output": "1" }, { "input": "3\naAb", "output": "1" }, { "input": "3\nAaA", "output": "1" }, { "input": "5\naBacd", "output": "3" }, { "input": "5\naAabc", "output": "3" } ]
1,624,728,271
2,147,483,647
Python 3
OK
TESTS
55
77
0
n=int(input()) s=input() x=-1 for i in range(n): if ord(s[i])>=65 and ord(s[i])<=90: x=i break if x==-1: print(len(set(s))) elif x+1==n: print(len(set(s[:n-1]))) else: h=len(set(s[:x])) l,r=x+1,x for i in range(x,n): if ord(s[i])>=65 and ord(s[i])<=90: t=len(set(s[l:i])) if t>h: h=t l=i+1 t=len(set(s[l:])) if t>h: h=t print(h)
Title: Polycarp and Letters Time Limit: None seconds Memory Limit: None megabytes Problem Description: Polycarp loves lowercase letters and dislikes uppercase ones. Once he got a string *s* consisting only of lowercase and uppercase Latin letters. Let *A* be a set of positions in the string. Let's call it pretty if following conditions are met: - letters on positions from *A* in the string are all distinct and lowercase; - there are no uppercase letters in the string which are situated between positions from *A* (i.e. there is no such *j* that *s*[*j*] is an uppercase letter, and *a*1<=&lt;<=*j*<=&lt;<=*a*2 for some *a*1 and *a*2 from *A*). Write a program that will determine the maximum number of elements in a pretty set of positions. Input Specification: The first line contains a single integer *n* (1<=≤<=*n*<=≤<=200) — length of string *s*. The second line contains a string *s* consisting of lowercase and uppercase Latin letters. Output Specification: Print maximum number of elements in pretty set of positions for string *s*. Demo Input: ['11\naaaaBaabAbA\n', '12\nzACaAbbaazzC\n', '3\nABC\n'] Demo Output: ['2\n', '3\n', '0\n'] Note: In the first example the desired positions might be 6 and 8 or 7 and 8. Positions 6 and 7 contain letters 'a', position 8 contains letter 'b'. The pair of positions 1 and 8 is not suitable because there is an uppercase letter 'B' between these position. In the second example desired positions can be 7, 8 and 11. There are other ways to choose pretty set consisting of three elements. In the third example the given string *s* does not contain any lowercase letters, so the answer is 0.
```python n=int(input()) s=input() x=-1 for i in range(n): if ord(s[i])>=65 and ord(s[i])<=90: x=i break if x==-1: print(len(set(s))) elif x+1==n: print(len(set(s[:n-1]))) else: h=len(set(s[:x])) l,r=x+1,x for i in range(x,n): if ord(s[i])>=65 and ord(s[i])<=90: t=len(set(s[l:i])) if t>h: h=t l=i+1 t=len(set(s[l:])) if t>h: h=t print(h) ```
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<=≤<=*n*<=≤<=1000) — 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 — 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" } ]
1,696,326,911
2,147,483,647
PyPy 3-64
OK
TESTS
21
154
1,638,400
rounds = int(input()) sum = 0 for index in range(rounds): petya, vasya, tonya = [int(oneorzero) for oneorzero in input().split()] if petya + vasya + tonya >= 2: sum += 1 print(sum)
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<=≤<=*n*<=≤<=1000) — 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 — 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 rounds = int(input()) sum = 0 for index in range(rounds): petya, vasya, tonya = [int(oneorzero) for oneorzero in input().split()] if petya + vasya + tonya >= 2: sum += 1 print(sum) ```
3
83
A
Magical Array
PROGRAMMING
1,300
[ "math" ]
A. Magical Array
2
256
Valery is very interested in magic. Magic attracts him so much that he sees it everywhere. He explains any strange and weird phenomenon through intervention of supernatural forces. But who would have thought that even in a regular array of numbers Valera manages to see something beautiful and magical. Valera absolutely accidentally got a piece of ancient parchment on which an array of numbers was written. He immediately thought that the numbers in this array were not random. As a result of extensive research Valera worked out a wonderful property that a magical array should have: an array is defined as magic if its minimum and maximum coincide. He decided to share this outstanding discovery with you, but he asks you for help in return. Despite the tremendous intelligence and wit, Valera counts very badly and so you will have to complete his work. All you have to do is count the number of magical subarrays of the original array of numbers, written on the parchment. Subarray is defined as non-empty sequence of consecutive elements.
The first line of the input data contains an integer *n* (1<=≤<=*n*<=≤<=105). The second line contains an array of original integers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=109<=≤<=*a**i*<=≤<=109).
Print on the single line the answer to the problem: the amount of subarrays, which are magical. Please do not use the %lld specificator to read or write 64-bit numbers in C++. It is recommended to use cin, cout streams (you can also use the %I64d specificator).
[ "4\n2 1 1 4\n", "5\n-2 -2 -2 0 1\n" ]
[ "5\n", "8\n" ]
Notes to sample tests: Magical subarrays are shown with pairs of indices [a;b] of the beginning and the end. In the first sample: [1;1], [2;2], [3;3], [4;4], [2;3]. In the second sample: [1;1], [2;2], [3;3], [4;4], [5;5], [1;2], [2;3], [1;3].
500
[ { "input": "4\n2 1 1 4", "output": "5" }, { "input": "5\n-2 -2 -2 0 1", "output": "8" }, { "input": "1\n10", "output": "1" }, { "input": "2\n5 6", "output": "2" }, { "input": "5\n5 5 4 5 5", "output": "7" }, { "input": "8\n1 2 0 0 0 0 3 3", "output": "15" }, { "input": "12\n-4 3 3 2 3 3 3 -4 2 -4 -4 -4", "output": "19" }, { "input": "10\n7 1 0 10 0 -5 -3 -2 0 0", "output": "11" }, { "input": "20\n6 0 0 -3 1 -3 0 -8 1 3 5 2 -1 -5 -1 9 0 6 -2 4", "output": "21" }, { "input": "100\n0 -18 -9 -15 3 16 -28 0 -28 0 28 -20 -9 9 -11 0 18 -15 -18 -26 0 -27 -25 -22 6 -5 8 14 -17 24 20 3 -6 24 -27 1 -23 0 4 12 -20 0 -10 30 22 -6 13 16 0 15 17 -8 -2 0 -5 13 11 23 -17 -29 10 15 -28 0 -23 4 20 17 -7 -5 -16 -17 16 2 20 19 -8 0 8 -5 12 0 0 -14 -15 -28 -10 20 0 8 -1 10 14 9 0 4 -16 15 13 -10", "output": "101" }, { "input": "50\n2 0 2 0 0 0 0 -1 -2 -2 -2 1 1 2 2 0 2 0 2 -3 0 0 0 0 3 1 -2 0 -1 0 -2 3 -1 2 0 2 0 0 0 0 2 0 1 0 0 3 0 0 -2 0", "output": "75" }, { "input": "2\n-510468670 0", "output": "2" }, { "input": "150\n0 -2 1 -2 0 0 0 0 -2 0 -2 -1 0 0 2 0 1 -2 1 -1 0 0 0 2 -2 2 -1 0 0 0 -2 0 2 0 1 0 -2 0 -2 -1 -1 -2 -2 2 0 0 1 -2 -2 -1 -2 0 2 1 1 -1 1 0 -2 2 0 0 0 1 -1 0 -2 -1 0 -2 2 1 1 0 0 2 0 0 2 -1 0 0 2 0 2 0 -2 -1 1 -2 1 0 0 -2 -1 -1 0 0 2 -1 -1 -1 -1 -2 0 0 2 -1 -1 1 0 0 1 -1 0 0 -1 2 2 0 0 0 -1 -2 0 1 0 -1 0 -1 -1 0 2 0 1 2 0 0 2 0 0 1 2 0 2 -2 2 0 2 2", "output": "196" } ]
1,626,783,750
2,147,483,647
PyPy 3
WRONG_ANSWER
TESTS
7
218
20,172,800
n=int(input()) a=[int(i) for i in input().split()] a.append(0) cnt=0 ans=0 for i in range(len(a)-1): cnt+=1 if a[i]!=a[i+1]: ans+=((cnt*(cnt+1))//2) cnt=0 print(ans)
Title: Magical Array Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: Valery is very interested in magic. Magic attracts him so much that he sees it everywhere. He explains any strange and weird phenomenon through intervention of supernatural forces. But who would have thought that even in a regular array of numbers Valera manages to see something beautiful and magical. Valera absolutely accidentally got a piece of ancient parchment on which an array of numbers was written. He immediately thought that the numbers in this array were not random. As a result of extensive research Valera worked out a wonderful property that a magical array should have: an array is defined as magic if its minimum and maximum coincide. He decided to share this outstanding discovery with you, but he asks you for help in return. Despite the tremendous intelligence and wit, Valera counts very badly and so you will have to complete his work. All you have to do is count the number of magical subarrays of the original array of numbers, written on the parchment. Subarray is defined as non-empty sequence of consecutive elements. Input Specification: The first line of the input data contains an integer *n* (1<=≤<=*n*<=≤<=105). The second line contains an array of original integers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=109<=≤<=*a**i*<=≤<=109). Output Specification: Print on the single line the answer to the problem: the amount of subarrays, which are magical. Please do not use the %lld specificator to read or write 64-bit numbers in C++. It is recommended to use cin, cout streams (you can also use the %I64d specificator). Demo Input: ['4\n2 1 1 4\n', '5\n-2 -2 -2 0 1\n'] Demo Output: ['5\n', '8\n'] Note: Notes to sample tests: Magical subarrays are shown with pairs of indices [a;b] of the beginning and the end. In the first sample: [1;1], [2;2], [3;3], [4;4], [2;3]. In the second sample: [1;1], [2;2], [3;3], [4;4], [5;5], [1;2], [2;3], [1;3].
```python n=int(input()) a=[int(i) for i in input().split()] a.append(0) cnt=0 ans=0 for i in range(len(a)-1): cnt+=1 if a[i]!=a[i+1]: ans+=((cnt*(cnt+1))//2) cnt=0 print(ans) ```
0
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 2·*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 2·*m* from left to right, then the *j*-th flat of the *i*-th floor has windows 2·*j*<=-<=1 and 2·*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<=≤<=*n*,<=*m*<=≤<=100) — 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 2·*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 integer — 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" } ]
1,586,445,534
2,147,483,647
Python 3
OK
TESTS
36
124
204,800
def Flats(): n, m = list(map(int, input().split())) apartment = [] for i in range(n): apartment.extend(list(map(int, input().split()))) times = 0 for i in range(0, len(apartment), 2): if 1 in apartment[i:i+2]: times += 1 print(times) Flats()
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 2·*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 2·*m* from left to right, then the *j*-th flat of the *i*-th floor has windows 2·*j*<=-<=1 and 2·*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<=≤<=*n*,<=*m*<=≤<=100) — 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 2·*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 integer — 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 def Flats(): n, m = list(map(int, input().split())) apartment = [] for i in range(n): apartment.extend(list(map(int, input().split()))) times = 0 for i in range(0, len(apartment), 2): if 1 in apartment[i:i+2]: times += 1 print(times) Flats() ```
3
41
A
Translation
PROGRAMMING
800
[ "implementation", "strings" ]
A. Translation
2
256
The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly.
The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols.
If the word *t* is a word *s*, written reversely, print YES, otherwise print NO.
[ "code\nedoc\n", "abb\naba\n", "code\ncode\n" ]
[ "YES\n", "NO\n", "NO\n" ]
none
500
[ { "input": "code\nedoc", "output": "YES" }, { "input": "abb\naba", "output": "NO" }, { "input": "code\ncode", "output": "NO" }, { "input": "abacaba\nabacaba", "output": "YES" }, { "input": "q\nq", "output": "YES" }, { "input": "asrgdfngfnmfgnhweratgjkk\nasrgdfngfnmfgnhweratgjkk", "output": "NO" }, { "input": "z\na", "output": "NO" }, { "input": "asd\ndsa", "output": "YES" }, { "input": "abcdef\nfecdba", "output": "NO" }, { "input": "ywjjbirapvskozubvxoemscfwl\ngnduubaogtfaiowjizlvjcu", "output": "NO" }, { "input": "mfrmqxtzvgaeuleubcmcxcfqyruwzenguhgrmkuhdgnhgtgkdszwqyd\nmfxufheiperjnhyczclkmzyhcxntdfskzkzdwzzujdinf", "output": "NO" }, { "input": "bnbnemvybqizywlnghlykniaxxxlkhftppbdeqpesrtgkcpoeqowjwhrylpsziiwcldodcoonpimudvrxejjo\ntiynnekmlalogyvrgptbinkoqdwzuiyjlrldxhzjmmp", "output": "NO" }, { "input": "pwlpubwyhzqvcitemnhvvwkmwcaawjvdiwtoxyhbhbxerlypelevasmelpfqwjk\nstruuzebbcenziscuoecywugxncdwzyfozhljjyizpqcgkyonyetarcpwkqhuugsqjuixsxptmbnlfupdcfigacdhhrzb", "output": "NO" }, { "input": "gdvqjoyxnkypfvdxssgrihnwxkeojmnpdeobpecytkbdwujqfjtxsqspxvxpqioyfagzjxupqqzpgnpnpxcuipweunqch\nkkqkiwwasbhezqcfeceyngcyuogrkhqecwsyerdniqiocjehrpkljiljophqhyaiefjpavoom", "output": "NO" }, { "input": "umeszdawsvgkjhlqwzents\nhxqhdungbylhnikwviuh", "output": "NO" }, { "input": "juotpscvyfmgntshcealgbsrwwksgrwnrrbyaqqsxdlzhkbugdyx\nibqvffmfktyipgiopznsqtrtxiijntdbgyy", "output": "NO" }, { "input": "zbwueheveouatecaglziqmudxemhrsozmaujrwlqmppzoumxhamwugedikvkblvmxwuofmpafdprbcftew\nulczwrqhctbtbxrhhodwbcxwimncnexosksujlisgclllxokrsbnozthajnnlilyffmsyko", "output": "NO" }, { "input": "nkgwuugukzcv\nqktnpxedwxpxkrxdvgmfgoxkdfpbzvwsduyiybynbkouonhvmzakeiruhfmvrktghadbfkmwxduoqv", "output": "NO" }, { "input": "incenvizhqpcenhjhehvjvgbsnfixbatrrjstxjzhlmdmxijztphxbrldlqwdfimweepkggzcxsrwelodpnryntepioqpvk\ndhjbjjftlvnxibkklxquwmzhjfvnmwpapdrslioxisbyhhfymyiaqhlgecpxamqnocizwxniubrmpyubvpenoukhcobkdojlybxd", "output": "NO" }, { "input": "w\nw", "output": "YES" }, { "input": "vz\nzv", "output": "YES" }, { "input": "ry\nyr", "output": "YES" }, { "input": "xou\nuox", "output": "YES" }, { "input": "axg\ngax", "output": "NO" }, { "input": "zdsl\nlsdz", "output": "YES" }, { "input": "kudl\nldku", "output": "NO" }, { "input": "zzlzwnqlcl\nlclqnwzlzz", "output": "YES" }, { "input": "vzzgicnzqooejpjzads\nsdazjpjeooqzncigzzv", "output": "YES" }, { "input": "raqhmvmzuwaykjpyxsykr\nxkysrypjkyawuzmvmhqar", "output": "NO" }, { "input": "ngedczubzdcqbxksnxuavdjaqtmdwncjnoaicvmodcqvhfezew\nwezefhvqcdomvciaonjcnwdmtqajdvauxnskxbqcdzbuzcdegn", "output": "YES" }, { "input": "muooqttvrrljcxbroizkymuidvfmhhsjtumksdkcbwwpfqdyvxtrlymofendqvznzlmim\nmimlznzvqdnefomylrtxvydqfpwwbckdskmutjshhmfvdiumykziorbxcjlrrvttqooum", "output": "YES" }, { "input": "vxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaivg\ngviayyikkitmuomcpiakhbxszgbnhvwyzkftwoagzixaearxpjacrnvpvbuzenvovehkmmxvblqyxvctroddksdsgebcmlluqpxv", "output": "YES" }, { "input": "mnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfdc\ncdfmkdgrdptkpewbsqvszipgxvgvuiuzbkkwuowbafkikgvnqdkxnayzdjygvezmtsgywnupocdntipiyiorblqkrzjpzatxahnm", "output": "NO" }, { "input": "dgxmzbqofstzcdgthbaewbwocowvhqpinehpjatnnbrijcolvsatbblsrxabzrpszoiecpwhfjmwuhqrapvtcgvikuxtzbftydkw\nwkdytfbztxukivgctvparqhuwmjfhwpceiozsprzbaxrslbbqasvlocjirbnntajphenipthvwocowbweabhtgdcztsfoqbzmxgd", "output": "NO" }, { "input": "gxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwgeh\nhegwxvocotmzstqfbmpjvijgkcyodlxyjawrpkczpmdspsuhoiruavnnnuwvtwohglkdxjetshkboalvzqbgjgthoteceixioxg", "output": "YES" }, { "input": "sihxuwvmaambplxvjfoskinghzicyfqebjtkysotattkahssumfcgrkheotdxwjckpvapbkaepqrxseyfrwtyaycmrzsrsngkh\nhkgnsrszrmcyaytwrfyesxrqpeakbpavpkcjwxdtoehkrgcfmusshakttatosyktjbeqfycizhgniksofjvxlpbmaamvwuxhis", "output": "YES" }, { "input": "ycnahksbughnonldzrhkysujmylcgcfuludjvjiahtkyzqvkopzqcnwhltbzfugzojqkjjlggmvnultascmygelkiktmfieok\nkoeifmtkiklegkmcsatlunvmggkjjlqjozgufzbtlhwncqzpokvqzykthaijvjdulufcgclymjusyyhrzdlnonhgubskhancy", "output": "NO" }, { "input": "wbqasaehtkfojruzyhrlgwmtyiovmzyfifslvlemhqheyaelzwnthrenjsbmntwaoryzwfbxmscmypvxlfmzpnkkjlvwvmtz\nztmvwvljkknpzmflxvpymcsmxbfwzyroawtnmbsjnerhtnwzleayehqhmelvlsfifyzmvoiytmwglrhyzurjofktheasaqbw", "output": "YES" }, { "input": "imippqurprbhfugngtgifelytadegwrgaefnfhbjjnmzikvjaccotqzemufqieqldgnbmviisgkynzeldlhqxuqphjfmyij\njiymfjhpquxqhldleznykgsiivmbngdlqeiqfumezqtoccajvkizmnjjbhfnfeagrwgedatylefigtgngufhbrpruqppimi", "output": "YES" }, { "input": "bikydffiuisckpvzqlteqfhegsagimodb\nbdomigasgehfqetlqzvpkcsiuiffdykib", "output": "YES" } ]
1,594,348,190
2,147,483,647
Python 3
OK
TESTS
40
218
6,656,000
s = input() t = input() t_arr = list(t) t_arr.reverse() t_final = ''.join(t_arr) if s == t_final: print('YES') else: print('NO')
Title: Translation Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly. Input Specification: The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols. Output Specification: If the word *t* is a word *s*, written reversely, print YES, otherwise print NO. Demo Input: ['code\nedoc\n', 'abb\naba\n', 'code\ncode\n'] Demo Output: ['YES\n', 'NO\n', 'NO\n'] Note: none
```python s = input() t = input() t_arr = list(t) t_arr.reverse() t_final = ''.join(t_arr) if s == t_final: print('YES') else: print('NO') ```
3.933102
84
A
Toy Army
PROGRAMMING
900
[ "math", "number theory" ]
A. Toy Army
2
256
The hero of our story, Valera, and his best friend Arcady are still in school, and therefore they spend all the free time playing turn-based strategy "GAGA: Go And Go Again". The gameplay is as follows. There are two armies on the playing field each of which consists of *n* men (*n* is always even). The current player specifies for each of her soldiers an enemy's soldier he will shoot (a target) and then all the player's soldiers shot simultaneously. This is a game world, and so each soldier shoots perfectly, that is he absolutely always hits the specified target. If an enemy soldier is hit, he will surely die. It may happen that several soldiers had been indicated the same target. Killed soldiers do not participate in the game anymore. The game "GAGA" consists of three steps: first Valera makes a move, then Arcady, then Valera again and the game ends. You are asked to calculate the maximum total number of soldiers that may be killed during the game.
The input data consist of a single integer *n* (2<=≤<=*n*<=≤<=108, *n* is even). Please note that before the game starts there are 2*n* soldiers on the fields.
Print a single number — a maximum total number of soldiers that could be killed in the course of the game in three turns.
[ "2\n", "4\n" ]
[ "3\n", "6\n" ]
The first sample test: 1) Valera's soldiers 1 and 2 shoot at Arcady's soldier 1. 2) Arcady's soldier 2 shoots at Valera's soldier 1. 3) Valera's soldier 1 shoots at Arcady's soldier 2. There are 3 soldiers killed in total: Valera's soldier 1 and Arcady's soldiers 1 and 2.
500
[ { "input": "2", "output": "3" }, { "input": "4", "output": "6" }, { "input": "6", "output": "9" }, { "input": "8", "output": "12" }, { "input": "10", "output": "15" }, { "input": "140", "output": "210" }, { "input": "500", "output": "750" }, { "input": "1000", "output": "1500" }, { "input": "2000", "output": "3000" }, { "input": "50000", "output": "75000" }, { "input": "10000", "output": "15000" }, { "input": "25460", "output": "38190" }, { "input": "54646", "output": "81969" }, { "input": "59790", "output": "89685" }, { "input": "578456", "output": "867684" }, { "input": "56798056", "output": "85197084" }, { "input": "8457980", "output": "12686970" }, { "input": "5687986", "output": "8531979" }, { "input": "10984932", "output": "16477398" }, { "input": "99999994", "output": "149999991" }, { "input": "99999996", "output": "149999994" }, { "input": "99999998", "output": "149999997" }, { "input": "100000000", "output": "150000000" } ]
1,679,122,018
2,147,483,647
Python 3
OK
TESTS
23
92
0
n=eval(input()) if n%2==0: print(int(n/2*3)) else: print(4*((n-1)%2)-2)
Title: Toy Army Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: The hero of our story, Valera, and his best friend Arcady are still in school, and therefore they spend all the free time playing turn-based strategy "GAGA: Go And Go Again". The gameplay is as follows. There are two armies on the playing field each of which consists of *n* men (*n* is always even). The current player specifies for each of her soldiers an enemy's soldier he will shoot (a target) and then all the player's soldiers shot simultaneously. This is a game world, and so each soldier shoots perfectly, that is he absolutely always hits the specified target. If an enemy soldier is hit, he will surely die. It may happen that several soldiers had been indicated the same target. Killed soldiers do not participate in the game anymore. The game "GAGA" consists of three steps: first Valera makes a move, then Arcady, then Valera again and the game ends. You are asked to calculate the maximum total number of soldiers that may be killed during the game. Input Specification: The input data consist of a single integer *n* (2<=≤<=*n*<=≤<=108, *n* is even). Please note that before the game starts there are 2*n* soldiers on the fields. Output Specification: Print a single number — a maximum total number of soldiers that could be killed in the course of the game in three turns. Demo Input: ['2\n', '4\n'] Demo Output: ['3\n', '6\n'] Note: The first sample test: 1) Valera's soldiers 1 and 2 shoot at Arcady's soldier 1. 2) Arcady's soldier 2 shoots at Valera's soldier 1. 3) Valera's soldier 1 shoots at Arcady's soldier 2. There are 3 soldiers killed in total: Valera's soldier 1 and Arcady's soldiers 1 and 2.
```python n=eval(input()) if n%2==0: print(int(n/2*3)) else: print(4*((n-1)%2)-2) ```
3.977
855
A
Tom Riddle's Diary
PROGRAMMING
800
[ "brute force", "implementation", "strings" ]
null
null
Harry Potter is on a mission to destroy You-Know-Who's Horcruxes. The first Horcrux that he encountered in the Chamber of Secrets is Tom Riddle's diary. The diary was with Ginny and it forced her to open the Chamber of Secrets. Harry wants to know the different people who had ever possessed the diary to make sure they are not under its influence. He has names of *n* people who possessed the diary in order. You need to tell, for each person, if he/she possessed the diary at some point before or not. Formally, for a name *s**i* in the *i*-th line, output "YES" (without quotes) if there exists an index *j* such that *s**i*<==<=*s**j* and *j*<=&lt;<=*i*, otherwise, output "NO" (without quotes).
First line of input contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of names in the list. Next *n* lines each contain a string *s**i*, consisting of lowercase English letters. The length of each string is between 1 and 100.
Output *n* lines each containing either "YES" or "NO" (without quotes), depending on whether this string was already present in the stream or not. You can print each letter in any case (upper or lower).
[ "6\ntom\nlucius\nginny\nharry\nginny\nharry\n", "3\na\na\na\n" ]
[ "NO\nNO\nNO\nNO\nYES\nYES\n", "NO\nYES\nYES\n" ]
In test case 1, for *i* = 5 there exists *j* = 3 such that *s*<sub class="lower-index">*i*</sub> = *s*<sub class="lower-index">*j*</sub> and *j* &lt; *i*, which means that answer for *i* = 5 is "YES".
500
[ { "input": "6\ntom\nlucius\nginny\nharry\nginny\nharry", "output": "NO\nNO\nNO\nNO\nYES\nYES" }, { "input": "3\na\na\na", "output": "NO\nYES\nYES" }, { "input": "1\nzn", "output": "NO" }, { "input": "9\nliyzmbjwnzryjokufuxcqtzwworjeoxkbaqrujrhdidqdvwdfzilwszgnzglnnbogaclckfnbqovtziuhwvyrqwmskx\nliyzmbjwnzryjokufuxcqtzwworjeoxkbaqrujrhdidqdvwdfzilwszgnzglnnbogaclckfnbqovtziuhwvyrqwmskx\nliyzmbjwnzryjokufuxcqtzwworjeoxkbaqrujrhdidqdvwdfzilwszgnzglnnbogaclckfnbqovtziuhwvyrqwmskx\nhrtm\nssjqvixduertmotgagizamvfucfwtxqnhuowbqbzctgznivehelpcyigwrbbdsxnewfqvcf\nhyrtxvozpbveexfkgalmguozzakitjiwsduqxonb\nwcyxteiwtcyuztaguilqpbiwcwjaiq\nwcyxteiwtcyuztaguilqpbiwcwjaiq\nbdbivqzvhggth", "output": "NO\nYES\nYES\nNO\nNO\nNO\nNO\nYES\nNO" }, { "input": "10\nkkiubdktydpdcbbttwpfdplhhjhrpqmpg\nkkiubdktydpdcbbttwpfdplhhjhrpqmpg\nmvutw\nqooeqoxzxwetlpecqiwgdbogiqqulttysyohwhzxzphvsfmnplizxoebzcvvfyppqbhxjksuzepuezqqzxlfmdanoeaoqmor\nmvutw\nvchawxjoreboqzuklifv\nvchawxjoreboqzuklifv\nnivijte\nrflybruq\nvchawxjoreboqzuklifv", "output": "NO\nYES\nNO\nNO\nYES\nNO\nYES\nNO\nNO\nYES" }, { "input": "1\nz", "output": "NO" }, { "input": "9\nl\ny\nm\nj\nn\nr\nj\nk\nf", "output": "NO\nNO\nNO\nNO\nNO\nNO\nYES\nNO\nNO" }, { "input": "14\nw\na\nh\np\nk\nw\ny\nv\ns\nf\nx\nd\nk\nr", "output": "NO\nNO\nNO\nNO\nNO\nYES\nNO\nNO\nNO\nNO\nNO\nNO\nYES\nNO" }, { "input": "25\np\nk\nu\nl\nf\nt\nc\ns\nq\nd\nb\nq\no\ni\ni\nd\ni\nw\nn\ng\nw\nt\na\ne\ni", "output": "NO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nYES\nNO\nNO\nYES\nYES\nYES\nNO\nNO\nNO\nYES\nYES\nNO\nNO\nYES" }, { "input": "20\nd\nh\ng\no\np\ne\nt\nj\nv\ni\nt\nh\ns\ni\nw\nf\nx\na\nl\ni", "output": "NO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nYES\nYES\nNO\nYES\nNO\nNO\nNO\nNO\nNO\nYES" }, { "input": "3\nbbbbbbb\nbbbbbbbbb\nbbbbbbbbbbbbbbbbbbbbbbbbbbbb", "output": "NO\nNO\nNO" }, { "input": "2\nab\nba", "output": "NO\nNO" }, { "input": "6\ntom\nlucius\nginnys\nharpy\nginny\nharry", "output": "NO\nNO\nNO\nNO\nNO\nNO" }, { "input": "2\nabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcde\nabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcdeabcde", "output": "NO\nYES" }, { "input": "42\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na", "output": "NO\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES" } ]
1,609,135,472
2,147,483,647
Python 3
OK
TESTS
55
109
0
n=int(input()) alist=[] for i in range(n): name=str(input()) alist.append(name) blist=[] blist.append("NO") for i in range(1,n): if alist[i] in alist[0:i]: blist.append("YES") else: blist.append("NO") for i in blist: print(i)
Title: Tom Riddle's Diary Time Limit: None seconds Memory Limit: None megabytes Problem Description: Harry Potter is on a mission to destroy You-Know-Who's Horcruxes. The first Horcrux that he encountered in the Chamber of Secrets is Tom Riddle's diary. The diary was with Ginny and it forced her to open the Chamber of Secrets. Harry wants to know the different people who had ever possessed the diary to make sure they are not under its influence. He has names of *n* people who possessed the diary in order. You need to tell, for each person, if he/she possessed the diary at some point before or not. Formally, for a name *s**i* in the *i*-th line, output "YES" (without quotes) if there exists an index *j* such that *s**i*<==<=*s**j* and *j*<=&lt;<=*i*, otherwise, output "NO" (without quotes). Input Specification: First line of input contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of names in the list. Next *n* lines each contain a string *s**i*, consisting of lowercase English letters. The length of each string is between 1 and 100. Output Specification: Output *n* lines each containing either "YES" or "NO" (without quotes), depending on whether this string was already present in the stream or not. You can print each letter in any case (upper or lower). Demo Input: ['6\ntom\nlucius\nginny\nharry\nginny\nharry\n', '3\na\na\na\n'] Demo Output: ['NO\nNO\nNO\nNO\nYES\nYES\n', 'NO\nYES\nYES\n'] Note: In test case 1, for *i* = 5 there exists *j* = 3 such that *s*<sub class="lower-index">*i*</sub> = *s*<sub class="lower-index">*j*</sub> and *j* &lt; *i*, which means that answer for *i* = 5 is "YES".
```python n=int(input()) alist=[] for i in range(n): name=str(input()) alist.append(name) blist=[] blist.append("NO") for i in range(1,n): if alist[i] in alist[0:i]: blist.append("YES") else: blist.append("NO") for i in blist: print(i) ```
3
699
A
Launch of Collider
PROGRAMMING
1,000
[ "implementation" ]
null
null
There will be a launch of a new, powerful and unusual collider very soon, which located along a straight line. *n* particles will be launched inside it. All of them are located in a straight line and there can not be two or more particles located in the same point. The coordinates of the particles coincide with the distance in meters from the center of the collider, *x**i* is the coordinate of the *i*-th particle and its position in the collider at the same time. All coordinates of particle positions are even integers. You know the direction of each particle movement — it will move to the right or to the left after the collider's launch start. All particles begin to move simultaneously at the time of the collider's launch start. Each particle will move straight to the left or straight to the right with the constant speed of 1 meter per microsecond. The collider is big enough so particles can not leave it in the foreseeable time. Write the program which finds the moment of the first collision of any two particles of the collider. In other words, find the number of microseconds before the first moment when any two particles are at the same point.
The first line contains the positive integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of particles. The second line contains *n* symbols "L" and "R". If the *i*-th symbol equals "L", then the *i*-th particle will move to the left, otherwise the *i*-th symbol equals "R" and the *i*-th particle will move to the right. The third line contains the sequence of pairwise distinct even integers *x*1,<=*x*2,<=...,<=*x**n* (0<=≤<=*x**i*<=≤<=109) — the coordinates of particles in the order from the left to the right. It is guaranteed that the coordinates of particles are given in the increasing order.
In the first line print the only integer — the first moment (in microseconds) when two particles are at the same point and there will be an explosion. Print the only integer -1, if the collision of particles doesn't happen.
[ "4\nRLRL\n2 4 6 10\n", "3\nLLR\n40 50 60\n" ]
[ "1\n", "-1\n" ]
In the first sample case the first explosion will happen in 1 microsecond because the particles number 1 and 2 will simultaneously be at the same point with the coordinate 3. In the second sample case there will be no explosion because there are no particles which will simultaneously be at the same point.
500
[ { "input": "4\nRLRL\n2 4 6 10", "output": "1" }, { "input": "3\nLLR\n40 50 60", "output": "-1" }, { "input": "4\nRLLR\n46 230 264 470", "output": "92" }, { "input": "6\nLLRLLL\n446 492 650 844 930 970", "output": "97" }, { "input": "8\nRRLLLLLL\n338 478 512 574 594 622 834 922", "output": "17" }, { "input": "10\nLRLRLLRRLR\n82 268 430 598 604 658 670 788 838 1000", "output": "3" }, { "input": "2\nRL\n0 1000000000", "output": "500000000" }, { "input": "12\nLRLLRRRRLRLL\n254 1260 1476 1768 2924 4126 4150 4602 5578 7142 8134 9082", "output": "108" }, { "input": "14\nRLLRRLRLLRLLLR\n698 2900 3476 3724 3772 3948 4320 4798 5680 6578 7754 8034 8300 8418", "output": "88" }, { "input": "16\nRRLLLRLRLLLLRLLR\n222 306 968 1060 1636 1782 2314 2710 3728 4608 5088 6790 6910 7156 7418 7668", "output": "123" }, { "input": "18\nRLRLLRRRLLLRLRRLRL\n1692 2028 2966 3008 3632 4890 5124 5838 6596 6598 6890 8294 8314 8752 8868 9396 9616 9808", "output": "10" }, { "input": "20\nRLLLLLLLRRRRLRRLRRLR\n380 902 1400 1834 2180 2366 2562 2596 2702 2816 3222 3238 3742 5434 6480 7220 7410 8752 9708 9970", "output": "252" }, { "input": "22\nLRRRRRRRRRRRLLRRRRRLRL\n1790 2150 2178 2456 2736 3282 3622 4114 4490 4772 5204 5240 5720 5840 5910 5912 6586 7920 8584 9404 9734 9830", "output": "48" }, { "input": "24\nLLRLRRLLRLRRRRLLRRLRLRRL\n100 360 864 1078 1360 1384 1438 2320 2618 3074 3874 3916 3964 5178 5578 6278 6630 6992 8648 8738 8922 8930 9276 9720", "output": "27" }, { "input": "26\nRLLLLLLLRLRRLRLRLRLRLLLRRR\n908 1826 2472 2474 2728 3654 3716 3718 3810 3928 4058 4418 4700 5024 5768 6006 6128 6386 6968 7040 7452 7774 7822 8726 9338 9402", "output": "59" }, { "input": "28\nRRLRLRRRRRRLLLRRLRRLLLRRLLLR\n156 172 1120 1362 2512 3326 3718 4804 4990 5810 6242 6756 6812 6890 6974 7014 7088 7724 8136 8596 8770 8840 9244 9250 9270 9372 9400 9626", "output": "10" }, { "input": "30\nRLLRLRLLRRRLRRRLLLLLLRRRLRRLRL\n128 610 1680 2436 2896 2994 3008 3358 3392 4020 4298 4582 4712 4728 5136 5900 6088 6232 6282 6858 6934 7186 7224 7256 7614 8802 8872 9170 9384 9794", "output": "7" }, { "input": "10\nLLLLRRRRRR\n0 2 4 6 8 10 12 14 16 18", "output": "-1" }, { "input": "5\nLLLLL\n0 10 20 30 40", "output": "-1" }, { "input": "6\nRRRRRR\n40 50 60 70 80 100", "output": "-1" }, { "input": "1\nR\n0", "output": "-1" }, { "input": "2\nRL\n2 1000000000", "output": "499999999" }, { "input": "2\nRL\n0 400000", "output": "200000" }, { "input": "2\nRL\n0 200002", "output": "100001" }, { "input": "2\nRL\n2 20000000", "output": "9999999" }, { "input": "4\nLLRL\n2 4 10 100", "output": "45" }, { "input": "4\nRLRL\n2 10 12 14", "output": "1" }, { "input": "2\nRL\n0 100000000", "output": "50000000" }, { "input": "2\nRL\n2 600002", "output": "300000" }, { "input": "1\nL\n0", "output": "-1" }, { "input": "2\nRL\n0 600000", "output": "300000" }, { "input": "5\nRRRRR\n0 2 4 6 8", "output": "-1" }, { "input": "2\nRL\n2 200000000", "output": "99999999" }, { "input": "2\nRL\n0 267382766", "output": "133691383" }, { "input": "3\nRRL\n4 8 999999998", "output": "499999995" }, { "input": "2\nRL\n0 2", "output": "1" }, { "input": "2\nRL\n2 400002", "output": "200000" }, { "input": "2\nLL\n2 4", "output": "-1" }, { "input": "2\nLL\n0 2", "output": "-1" }, { "input": "2\nRL\n0 100000", "output": "50000" }, { "input": "2\nRL\n2 200000020", "output": "100000009" }, { "input": "2\nRL\n2000000 4000000", "output": "1000000" }, { "input": "2\nRL\n0 199998", "output": "99999" }, { "input": "3\nLRR\n40 50 60", "output": "-1" }, { "input": "2\nRL\n200 400400", "output": "200100" }, { "input": "2\nRL\n2 400004", "output": "200001" }, { "input": "2\nRL\n0 200000000", "output": "100000000" } ]
1,622,744,896
2,147,483,647
Python 3
OK
TESTS
85
639
16,076,800
import re def calcMoments (r , l): return (l - r) / 2 moment = 0 many = int(input()) # 7 dirline = input() # RLLRRLR, R -> ++ L-> -- coor = [int(x) for x in input().split()] # 2 4 6 10 , even, increasing order 2 4 6 16 closest = -1 Rpos = [m.start() for m in re.finditer('R', dirline)] #[0, 3, 4, 6] index = 0 once = True expect = False lst = list() for r in Rpos: if r == len(dirline)-1: break if index+1 < len(Rpos) and r == Rpos[index+1]-1: index = index +1 continue index = index +1 expect = True # lst.append((r, r+1)) if once: closest = calcMoments(coor[r], coor[r+1]) once = False if calcMoments(coor[r], coor[r+1]) < closest: closest = calcMoments(coor[r], coor[r+1]) if not expect: print(-1) else: print(int(closest))
Title: Launch of Collider Time Limit: None seconds Memory Limit: None megabytes Problem Description: There will be a launch of a new, powerful and unusual collider very soon, which located along a straight line. *n* particles will be launched inside it. All of them are located in a straight line and there can not be two or more particles located in the same point. The coordinates of the particles coincide with the distance in meters from the center of the collider, *x**i* is the coordinate of the *i*-th particle and its position in the collider at the same time. All coordinates of particle positions are even integers. You know the direction of each particle movement — it will move to the right or to the left after the collider's launch start. All particles begin to move simultaneously at the time of the collider's launch start. Each particle will move straight to the left or straight to the right with the constant speed of 1 meter per microsecond. The collider is big enough so particles can not leave it in the foreseeable time. Write the program which finds the moment of the first collision of any two particles of the collider. In other words, find the number of microseconds before the first moment when any two particles are at the same point. Input Specification: The first line contains the positive integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of particles. The second line contains *n* symbols "L" and "R". If the *i*-th symbol equals "L", then the *i*-th particle will move to the left, otherwise the *i*-th symbol equals "R" and the *i*-th particle will move to the right. The third line contains the sequence of pairwise distinct even integers *x*1,<=*x*2,<=...,<=*x**n* (0<=≤<=*x**i*<=≤<=109) — the coordinates of particles in the order from the left to the right. It is guaranteed that the coordinates of particles are given in the increasing order. Output Specification: In the first line print the only integer — the first moment (in microseconds) when two particles are at the same point and there will be an explosion. Print the only integer -1, if the collision of particles doesn't happen. Demo Input: ['4\nRLRL\n2 4 6 10\n', '3\nLLR\n40 50 60\n'] Demo Output: ['1\n', '-1\n'] Note: In the first sample case the first explosion will happen in 1 microsecond because the particles number 1 and 2 will simultaneously be at the same point with the coordinate 3. In the second sample case there will be no explosion because there are no particles which will simultaneously be at the same point.
```python import re def calcMoments (r , l): return (l - r) / 2 moment = 0 many = int(input()) # 7 dirline = input() # RLLRRLR, R -> ++ L-> -- coor = [int(x) for x in input().split()] # 2 4 6 10 , even, increasing order 2 4 6 16 closest = -1 Rpos = [m.start() for m in re.finditer('R', dirline)] #[0, 3, 4, 6] index = 0 once = True expect = False lst = list() for r in Rpos: if r == len(dirline)-1: break if index+1 < len(Rpos) and r == Rpos[index+1]-1: index = index +1 continue index = index +1 expect = True # lst.append((r, r+1)) if once: closest = calcMoments(coor[r], coor[r+1]) once = False if calcMoments(coor[r], coor[r+1]) < closest: closest = calcMoments(coor[r], coor[r+1]) if not expect: print(-1) else: print(int(closest)) ```
3
452
A
Eevee
PROGRAMMING
1,000
[ "brute force", "implementation", "strings" ]
null
null
You are solving the crossword problem K from IPSC 2014. You solved all the clues except for one: who does Eevee evolve into? You are not very into pokemons, but quick googling helped you find out, that Eevee can evolve into eight different pokemons: Vaporeon, Jolteon, Flareon, Espeon, Umbreon, Leafeon, Glaceon, and Sylveon. You know the length of the word in the crossword, and you already know some letters. Designers of the crossword made sure that the answer is unambiguous, so you can assume that exactly one pokemon out of the 8 that Eevee evolves into fits the length and the letters given. Your task is to find it.
First line contains an integer *n* (6<=≤<=*n*<=≤<=8) – the length of the string. Next line contains a string consisting of *n* characters, each of which is either a lower case english letter (indicating a known letter) or a dot character (indicating an empty cell in the crossword).
Print a name of the pokemon that Eevee can evolve into that matches the pattern in the input. Use lower case letters only to print the name (in particular, do not capitalize the first letter).
[ "7\nj......\n", "7\n...feon\n", "7\n.l.r.o.\n" ]
[ "jolteon\n", "leafeon\n", "flareon\n" ]
Here's a set of names in a form you can paste into your solution: ["vaporeon", "jolteon", "flareon", "espeon", "umbreon", "leafeon", "glaceon", "sylveon"] {"vaporeon", "jolteon", "flareon", "espeon", "umbreon", "leafeon", "glaceon", "sylveon"}
500
[ { "input": "7\n...feon", "output": "leafeon" }, { "input": "7\n.l.r.o.", "output": "flareon" }, { "input": "6\n.s..o.", "output": "espeon" }, { "input": "7\nglaceon", "output": "glaceon" }, { "input": "8\n.a.o.e.n", "output": "vaporeon" }, { "input": "7\n.laceon", "output": "glaceon" }, { "input": "7\n..lveon", "output": "sylveon" }, { "input": "7\n.l.ceon", "output": "glaceon" }, { "input": "7\n..areon", "output": "flareon" } ]
1,557,837,811
2,147,483,647
Python 3
OK
TESTS
20
124
0
b=["vaporeon", "jolteon", "flareon" , "espeon", "umbreon", "leafeon", "glaceon", "sylveon"] n=int(input()) a=input() c=0 # for x in a: # for y in b: # for z in y: # if b=[x for x in b if len(x)==n] for word in b: for x in range(n): if a[x]=='.': continue elif a[x]!=word[x]: break if x+1==n: print(word)
Title: Eevee Time Limit: None seconds Memory Limit: None megabytes Problem Description: You are solving the crossword problem K from IPSC 2014. You solved all the clues except for one: who does Eevee evolve into? You are not very into pokemons, but quick googling helped you find out, that Eevee can evolve into eight different pokemons: Vaporeon, Jolteon, Flareon, Espeon, Umbreon, Leafeon, Glaceon, and Sylveon. You know the length of the word in the crossword, and you already know some letters. Designers of the crossword made sure that the answer is unambiguous, so you can assume that exactly one pokemon out of the 8 that Eevee evolves into fits the length and the letters given. Your task is to find it. Input Specification: First line contains an integer *n* (6<=≤<=*n*<=≤<=8) – the length of the string. Next line contains a string consisting of *n* characters, each of which is either a lower case english letter (indicating a known letter) or a dot character (indicating an empty cell in the crossword). Output Specification: Print a name of the pokemon that Eevee can evolve into that matches the pattern in the input. Use lower case letters only to print the name (in particular, do not capitalize the first letter). Demo Input: ['7\nj......\n', '7\n...feon\n', '7\n.l.r.o.\n'] Demo Output: ['jolteon\n', 'leafeon\n', 'flareon\n'] Note: Here's a set of names in a form you can paste into your solution: ["vaporeon", "jolteon", "flareon", "espeon", "umbreon", "leafeon", "glaceon", "sylveon"] {"vaporeon", "jolteon", "flareon", "espeon", "umbreon", "leafeon", "glaceon", "sylveon"}
```python b=["vaporeon", "jolteon", "flareon" , "espeon", "umbreon", "leafeon", "glaceon", "sylveon"] n=int(input()) a=input() c=0 # for x in a: # for y in b: # for z in y: # if b=[x for x in b if len(x)==n] for word in b: for x in range(n): if a[x]=='.': continue elif a[x]!=word[x]: break if x+1==n: print(word) ```
3
540
A
Combination Lock
PROGRAMMING
800
[ "implementation" ]
null
null
Scrooge McDuck keeps his most treasured savings in a home safe with a combination lock. Each time he wants to put there the treasures that he's earned fair and square, he has to open the lock. The combination lock is represented by *n* rotating disks with digits from 0 to 9 written on them. Scrooge McDuck has to turn some disks so that the combination of digits on the disks forms a secret combination. In one move, he can rotate one disk one digit forwards or backwards. In particular, in one move he can go from digit 0 to digit 9 and vice versa. What minimum number of actions does he need for that?
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of disks on the combination lock. The second line contains a string of *n* digits — the original state of the disks. The third line contains a string of *n* digits — Scrooge McDuck's combination that opens the lock.
Print a single integer — the minimum number of moves Scrooge McDuck needs to open the lock.
[ "5\n82195\n64723\n" ]
[ "13\n" ]
In the sample he needs 13 moves: - 1 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/b8967f65a723782358b93eff9ce69f336817cf70.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 2 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/07fa58573ece0d32c4d555e498d2b24d2f70f36a.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 3 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/cc2275d9252aae35a6867c6a5b4ba7596e9a7626.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 4 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/b100aea470fcaaab4e9529b234ba0d7875943c10.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 5 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/eb2cbe4324cebca65b85816262a85e473cd65967.png" style="max-width: 100.0%;max-height: 100.0%;"/>
500
[ { "input": "5\n82195\n64723", "output": "13" }, { "input": "12\n102021090898\n010212908089", "output": "16" }, { "input": "1\n8\n1", "output": "3" }, { "input": "2\n83\n57", "output": "7" }, { "input": "10\n0728592530\n1362615763", "output": "27" }, { "input": "100\n4176196363694273682807653052945037727131821799902563705176501742060696655282954944720643131654235909\n3459912084922154505910287499879975659298239371519889866585472674423008837878123067103005344986554746", "output": "245" }, { "input": "1\n8\n1", "output": "3" }, { "input": "2\n83\n57", "output": "7" }, { "input": "3\n607\n684", "output": "5" }, { "input": "4\n0809\n0636", "output": "8" }, { "input": "5\n84284\n08941", "output": "16" }, { "input": "25\n8037856825987124762280548\n9519431339078678836940020", "output": "72" }, { "input": "125\n23269567683904664184142384849516523616863461607751021071772615078579713054027902974007001544768640273491193035874486891541257\n47635110303703399505805044019026243695451609639556649012447370081552870340011971572363458960190590266459684717415349529509024", "output": "305" }, { "input": "5\n84284\n08941", "output": "16" }, { "input": "25\n8037856825987124762285484\n9519431339078678836940202", "output": "74" }, { "input": "125\n23269567689466418414238845152168634610771021717726157879713054270294007001544768647391193035874486891412573389247025830678706\n47635110307339950580504010224954516093956649124473708152870340117152363458960190596659684717415349529090241694059599629136831", "output": "357" }, { "input": "5\n84284\n08941", "output": "16" }, { "input": "25\n8378525987476228048406972\n9194339078883694020217816", "output": "55" }, { "input": "125\n23269576839046618414238484916523616863461607750210717761078579713054027902974007015447686027349193035874486891541257338624472\n47635103037033950580504401926243695451609639556490124437081552870340011971572363489601905026645984717415349529509024169604599", "output": "274" }, { "input": "1\n0\n0", "output": "0" }, { "input": "1\n7\n7", "output": "0" }, { "input": "1\n0\n5", "output": "5" }, { "input": "1\n2\n7", "output": "5" }, { "input": "1\n7\n9", "output": "2" }, { "input": "1\n9\n7", "output": "2" }, { "input": "1\n2\n9", "output": "3" }, { "input": "1\n9\n2", "output": "3" }, { "input": "25\n3164978461316464614169874\n9413979197249127496597357", "output": "66" }, { "input": "4\n9999\n9999", "output": "0" }, { "input": "2\n11\n11", "output": "0" } ]
1,684,420,639
2,147,483,647
Python 3
OK
TESTS
43
31
0
n=int(input()) a=input() b=input() count=0 for i in range(n): if((int(a[i])-int(b[i]))<0): if(((9-int(b[i])+int(a[i])-0)+1)<(int(b[i])-int(a[i]))): count+=9-int(b[i])+int(a[i])-0+1 else: count+=int(b[i])-int(a[i]) else: if(((9-int(a[i])+int(b[i])+1)-0)<(int(a[i])-int(b[i]))): count+=9-int(a[i])+int(b[i])-0+1 else: count+=int(a[i])-int(b[i]) print(count)
Title: Combination Lock Time Limit: None seconds Memory Limit: None megabytes Problem Description: Scrooge McDuck keeps his most treasured savings in a home safe with a combination lock. Each time he wants to put there the treasures that he's earned fair and square, he has to open the lock. The combination lock is represented by *n* rotating disks with digits from 0 to 9 written on them. Scrooge McDuck has to turn some disks so that the combination of digits on the disks forms a secret combination. In one move, he can rotate one disk one digit forwards or backwards. In particular, in one move he can go from digit 0 to digit 9 and vice versa. What minimum number of actions does he need for that? Input Specification: The first line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of disks on the combination lock. The second line contains a string of *n* digits — the original state of the disks. The third line contains a string of *n* digits — Scrooge McDuck's combination that opens the lock. Output Specification: Print a single integer — the minimum number of moves Scrooge McDuck needs to open the lock. Demo Input: ['5\n82195\n64723\n'] Demo Output: ['13\n'] Note: In the sample he needs 13 moves: - 1 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/b8967f65a723782358b93eff9ce69f336817cf70.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 2 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/07fa58573ece0d32c4d555e498d2b24d2f70f36a.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 3 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/cc2275d9252aae35a6867c6a5b4ba7596e9a7626.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 4 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/b100aea470fcaaab4e9529b234ba0d7875943c10.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 5 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/eb2cbe4324cebca65b85816262a85e473cd65967.png" style="max-width: 100.0%;max-height: 100.0%;"/>
```python n=int(input()) a=input() b=input() count=0 for i in range(n): if((int(a[i])-int(b[i]))<0): if(((9-int(b[i])+int(a[i])-0)+1)<(int(b[i])-int(a[i]))): count+=9-int(b[i])+int(a[i])-0+1 else: count+=int(b[i])-int(a[i]) else: if(((9-int(a[i])+int(b[i])+1)-0)<(int(a[i])-int(b[i]))): count+=9-int(a[i])+int(b[i])-0+1 else: count+=int(a[i])-int(b[i]) print(count) ```
3
493
B
Vasya and Wrestling
PROGRAMMING
1,400
[ "implementation" ]
null
null
Vasya has become interested in wrestling. In wrestling wrestlers use techniques for which they are awarded points by judges. The wrestler who gets the most points wins. When the numbers of points of both wrestlers are equal, the wrestler whose sequence of points is lexicographically greater, wins. If the sequences of the awarded points coincide, the wrestler who performed the last technique wins. Your task is to determine which wrestler won.
The first line contains number *n* — the number of techniques that the wrestlers have used (1<=≤<=*n*<=≤<=2·105). The following *n* lines contain integer numbers *a**i* (|*a**i*|<=≤<=109, *a**i*<=≠<=0). If *a**i* is positive, that means that the first wrestler performed the technique that was awarded with *a**i* points. And if *a**i* is negative, that means that the second wrestler performed the technique that was awarded with (<=-<=*a**i*) points. The techniques are given in chronological order.
If the first wrestler wins, print string "first", otherwise print "second"
[ "5\n1\n2\n-3\n-4\n3\n", "3\n-1\n-2\n3\n", "2\n4\n-4\n" ]
[ "second\n", "first\n", "second\n" ]
Sequence *x*  =  *x*<sub class="lower-index">1</sub>*x*<sub class="lower-index">2</sub>... *x*<sub class="lower-index">|*x*|</sub> is lexicographically larger than sequence *y*  =  *y*<sub class="lower-index">1</sub>*y*<sub class="lower-index">2</sub>... *y*<sub class="lower-index">|*y*|</sub>, if either |*x*|  &gt;  |*y*| and *x*<sub class="lower-index">1</sub>  =  *y*<sub class="lower-index">1</sub>,  *x*<sub class="lower-index">2</sub>  =  *y*<sub class="lower-index">2</sub>, ... ,  *x*<sub class="lower-index">|*y*|</sub>  =  *y*<sub class="lower-index">|*y*|</sub>, or there is such number *r* (*r*  &lt;  |*x*|, *r*  &lt;  |*y*|), that *x*<sub class="lower-index">1</sub>  =  *y*<sub class="lower-index">1</sub>,  *x*<sub class="lower-index">2</sub>  =  *y*<sub class="lower-index">2</sub>,  ... ,  *x*<sub class="lower-index">*r*</sub>  =  *y*<sub class="lower-index">*r*</sub> and *x*<sub class="lower-index">*r*  +  1</sub>  &gt;  *y*<sub class="lower-index">*r*  +  1</sub>. We use notation |*a*| to denote length of sequence *a*.
1,000
[ { "input": "5\n1\n2\n-3\n-4\n3", "output": "second" }, { "input": "3\n-1\n-2\n3", "output": "first" }, { "input": "2\n4\n-4", "output": "second" }, { "input": "7\n1\n2\n-3\n4\n5\n-6\n7", "output": "first" }, { "input": "14\n1\n2\n3\n4\n5\n6\n7\n-8\n-9\n-10\n-11\n-12\n-13\n-14", "output": "second" }, { "input": "4\n16\n12\n19\n-98", "output": "second" }, { "input": "5\n-6\n-1\n-1\n5\n3", "output": "second" }, { "input": "11\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1", "output": "first" }, { "input": "1\n-534365", "output": "second" }, { "input": "1\n10253033", "output": "first" }, { "input": "3\n-1\n-2\n3", "output": "first" }, { "input": "8\n1\n-2\n-3\n4\n5\n-6\n-7\n8", "output": "second" }, { "input": "2\n1\n-1", "output": "second" }, { "input": "5\n1\n2\n3\n4\n5", "output": "first" }, { "input": "5\n-1\n-2\n-3\n-4\n-5", "output": "second" }, { "input": "10\n-1\n-2\n-3\n-4\n-5\n5\n4\n3\n2\n1", "output": "first" }, { "input": "131\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n1\n-1\n-1\n-1\n2", "output": "first" }, { "input": "6\n-1\n-2\n-3\n1\n2\n3", "output": "first" }, { "input": "3\n1000000000\n1000000000\n1000000000", "output": "first" }, { "input": "12\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000", "output": "first" }, { "input": "4\n1000000000\n1000000000\n1000000000\n-1000000000", "output": "first" }, { "input": "20\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000", "output": "first" }, { "input": "5\n1000000000\n1000000000\n-1000000000\n-1000000000\n-1000000000", "output": "second" }, { "input": "4\n1\n-1000000000\n-1000000000\n-1000000000", "output": "second" }, { "input": "5\n1000000000\n1000000000\n1000000000\n-1000000000\n-1000000000", "output": "first" }, { "input": "4\n-1\n1000000000\n1000000000\n1000000000", "output": "first" }, { "input": "11\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000", "output": "first" }, { "input": "2\n-4\n4", "output": "first" }, { "input": "3\n-12\n3\n9", "output": "second" }, { "input": "3\n9\n1\n-10", "output": "second" }, { "input": "3\n1\n2\n-3", "output": "second" }, { "input": "4\n55\n5\n-5\n-55", "output": "first" }, { "input": "4\n5\n-1\n1\n-5", "output": "first" }, { "input": "2\n-5\n6", "output": "first" }, { "input": "4\n5\n-4\n3\n-40", "output": "second" }, { "input": "4\n1000000000\n1000000000\n1000000000\n-5", "output": "first" }, { "input": "6\n3\n2\n1\n-3\n-1\n-2", "output": "first" }, { "input": "5\n4\n1\n1\n-3\n-3", "output": "first" }, { "input": "5\n208\n-52\n-52\n-52\n-52", "output": "first" }, { "input": "3\n-100\n-200\n300", "output": "first" }, { "input": "3\n400\n-200\n-200", "output": "first" }, { "input": "3\n208\n-207\n-1", "output": "first" }, { "input": "3\n98888887\n98888888\n-197777775", "output": "second" } ]
1,608,620,997
2,147,483,647
PyPy 3
OK
TESTS
57
1,466
10,649,600
arr = [] brr = [] for _ in range(int(input())): n = int(input()) #print(n) if n > 0: arr.append(n) else: brr.append(-1*n) #print(arr,brr,n) if sum(arr) > sum(brr): print("first") elif sum(arr) < sum(brr): print("second") else: if arr > brr: print("first") elif arr < brr: print("second") else: if n > 0: print("first") else: print("second")
Title: Vasya and Wrestling Time Limit: None seconds Memory Limit: None megabytes Problem Description: Vasya has become interested in wrestling. In wrestling wrestlers use techniques for which they are awarded points by judges. The wrestler who gets the most points wins. When the numbers of points of both wrestlers are equal, the wrestler whose sequence of points is lexicographically greater, wins. If the sequences of the awarded points coincide, the wrestler who performed the last technique wins. Your task is to determine which wrestler won. Input Specification: The first line contains number *n* — the number of techniques that the wrestlers have used (1<=≤<=*n*<=≤<=2·105). The following *n* lines contain integer numbers *a**i* (|*a**i*|<=≤<=109, *a**i*<=≠<=0). If *a**i* is positive, that means that the first wrestler performed the technique that was awarded with *a**i* points. And if *a**i* is negative, that means that the second wrestler performed the technique that was awarded with (<=-<=*a**i*) points. The techniques are given in chronological order. Output Specification: If the first wrestler wins, print string "first", otherwise print "second" Demo Input: ['5\n1\n2\n-3\n-4\n3\n', '3\n-1\n-2\n3\n', '2\n4\n-4\n'] Demo Output: ['second\n', 'first\n', 'second\n'] Note: Sequence *x*  =  *x*<sub class="lower-index">1</sub>*x*<sub class="lower-index">2</sub>... *x*<sub class="lower-index">|*x*|</sub> is lexicographically larger than sequence *y*  =  *y*<sub class="lower-index">1</sub>*y*<sub class="lower-index">2</sub>... *y*<sub class="lower-index">|*y*|</sub>, if either |*x*|  &gt;  |*y*| and *x*<sub class="lower-index">1</sub>  =  *y*<sub class="lower-index">1</sub>,  *x*<sub class="lower-index">2</sub>  =  *y*<sub class="lower-index">2</sub>, ... ,  *x*<sub class="lower-index">|*y*|</sub>  =  *y*<sub class="lower-index">|*y*|</sub>, or there is such number *r* (*r*  &lt;  |*x*|, *r*  &lt;  |*y*|), that *x*<sub class="lower-index">1</sub>  =  *y*<sub class="lower-index">1</sub>,  *x*<sub class="lower-index">2</sub>  =  *y*<sub class="lower-index">2</sub>,  ... ,  *x*<sub class="lower-index">*r*</sub>  =  *y*<sub class="lower-index">*r*</sub> and *x*<sub class="lower-index">*r*  +  1</sub>  &gt;  *y*<sub class="lower-index">*r*  +  1</sub>. We use notation |*a*| to denote length of sequence *a*.
```python arr = [] brr = [] for _ in range(int(input())): n = int(input()) #print(n) if n > 0: arr.append(n) else: brr.append(-1*n) #print(arr,brr,n) if sum(arr) > sum(brr): print("first") elif sum(arr) < sum(brr): print("second") else: if arr > brr: print("first") elif arr < brr: print("second") else: if n > 0: print("first") else: print("second") ```
3
350
A
TL
PROGRAMMING
1,200
[ "brute force", "greedy", "implementation" ]
null
null
Valera wanted to prepare a Codesecrof round. He's already got one problem and he wants to set a time limit (TL) on it. Valera has written *n* correct solutions. For each correct solution, he knows its running time (in seconds). Valera has also wrote *m* wrong solutions and for each wrong solution he knows its running time (in seconds). Let's suppose that Valera will set *v* seconds TL in the problem. Then we can say that a solution passes the system testing if its running time is at most *v* seconds. We can also say that a solution passes the system testing with some "extra" time if for its running time, *a* seconds, an inequality 2*a*<=≤<=*v* holds. As a result, Valera decided to set *v* seconds TL, that the following conditions are met: 1. *v* is a positive integer; 1. all correct solutions pass the system testing; 1. at least one correct solution passes the system testing with some "extra" time; 1. all wrong solutions do not pass the system testing; 1. value *v* is minimum among all TLs, for which points 1, 2, 3, 4 hold. Help Valera and find the most suitable TL or else state that such TL doesn't exist.
The first line contains two integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=100). The second line contains *n* space-separated positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100) — the running time of each of the *n* correct solutions in seconds. The third line contains *m* space-separated positive integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=100) — the running time of each of *m* wrong solutions in seconds.
If there is a valid TL value, print it. Otherwise, print -1.
[ "3 6\n4 5 2\n8 9 6 10 7 11\n", "3 1\n3 4 5\n6\n" ]
[ "5", "-1\n" ]
none
500
[ { "input": "3 6\n4 5 2\n8 9 6 10 7 11", "output": "5" }, { "input": "3 1\n3 4 5\n6", "output": "-1" }, { "input": "2 5\n45 99\n49 41 77 83 45", "output": "-1" }, { "input": "50 50\n18 13 5 34 10 36 36 12 15 11 16 17 14 36 23 45 32 24 31 18 24 32 7 1 31 3 49 8 16 23 3 39 47 43 42 38 40 22 41 1 49 47 9 8 19 15 29 30 16 18\n91 58 86 51 94 94 73 84 98 69 74 56 52 80 88 61 53 99 88 50 55 95 65 84 87 79 51 52 69 60 74 73 93 61 73 59 64 56 95 78 86 72 79 70 93 78 54 61 71 50", "output": "49" }, { "input": "55 44\n93 17 74 15 34 16 41 80 26 54 94 94 86 93 20 44 63 72 39 43 67 4 37 49 76 94 5 51 64 74 11 47 77 97 57 30 42 72 71 26 8 14 67 64 49 57 30 23 40 4 76 78 87 78 79\n38 55 17 65 26 7 36 65 48 28 49 93 18 98 31 90 26 57 1 26 88 56 48 56 23 13 8 67 80 2 51 3 21 33 20 54 2 45 21 36 3 98 62 2", "output": "-1" }, { "input": "32 100\n30 8 4 35 18 41 18 12 33 39 39 18 39 19 33 46 45 33 34 27 14 39 40 21 38 9 42 35 27 10 14 14\n65 49 89 64 47 78 59 52 73 51 84 82 88 63 91 99 67 87 53 99 75 47 85 82 58 47 80 50 65 91 83 90 77 52 100 88 97 74 98 99 50 93 65 61 65 65 65 96 61 51 84 67 79 90 92 83 100 100 100 95 80 54 77 51 98 64 74 62 60 96 73 74 94 55 89 60 92 65 74 79 66 81 53 47 71 51 54 85 74 97 68 72 88 94 100 85 65 63 65 90", "output": "46" }, { "input": "1 50\n7\n65 52 99 78 71 19 96 72 80 15 50 94 20 35 79 95 44 41 45 53 77 50 74 66 59 96 26 84 27 48 56 84 36 78 89 81 67 34 79 74 99 47 93 92 90 96 72 28 78 66", "output": "14" }, { "input": "1 1\n4\n9", "output": "8" }, { "input": "1 1\n2\n4", "output": "-1" }, { "input": "22 56\n49 20 42 68 15 46 98 78 82 8 7 33 50 30 75 96 36 88 35 99 19 87\n15 18 81 24 35 89 25 32 23 3 48 24 52 69 18 32 23 61 48 98 50 38 5 17 70 20 38 32 49 54 68 11 51 81 46 22 19 59 29 38 45 83 18 13 91 17 84 62 25 60 97 32 23 13 83 58", "output": "-1" }, { "input": "1 1\n50\n100", "output": "-1" }, { "input": "1 1\n49\n100", "output": "98" }, { "input": "1 1\n100\n100", "output": "-1" }, { "input": "1 1\n99\n100", "output": "-1" }, { "input": "8 4\n1 2 49 99 99 95 78 98\n100 100 100 100", "output": "99" }, { "input": "68 85\n43 55 2 4 72 45 19 56 53 81 18 90 11 87 47 8 94 88 24 4 67 9 21 70 25 66 65 27 46 13 8 51 65 99 37 43 71 59 71 79 32 56 49 43 57 85 95 81 40 28 60 36 72 81 60 40 16 78 61 37 29 26 15 95 70 27 50 97\n6 6 48 72 54 31 1 50 29 64 93 9 29 93 66 63 25 90 52 1 66 13 70 30 24 87 32 90 84 72 44 13 25 45 31 16 92 60 87 40 62 7 20 63 86 78 73 88 5 36 74 100 64 34 9 5 62 29 58 48 81 46 84 56 27 1 60 14 54 88 31 93 62 7 9 69 27 48 10 5 33 10 53 66 2", "output": "-1" }, { "input": "5 100\n1 1 1 1 1\n77 53 38 29 97 33 64 17 78 100 27 12 42 44 20 24 44 68 58 57 65 90 8 24 4 6 74 68 61 43 25 69 8 62 36 85 67 48 69 30 35 41 42 12 87 66 50 92 53 76 38 67 85 7 80 78 53 76 94 8 37 50 4 100 4 71 10 48 34 47 83 42 25 81 64 72 25 51 53 75 43 98 53 77 94 38 81 15 89 91 72 76 7 36 27 41 88 18 19 75", "output": "2" }, { "input": "3 3\n2 3 4\n8 9 10", "output": "4" }, { "input": "2 1\n2 3\n15", "output": "4" }, { "input": "2 1\n2 4\n4", "output": "-1" }, { "input": "2 3\n4 5\n10 11 12", "output": "8" }, { "input": "3 1\n2 3 3\n5", "output": "4" }, { "input": "2 1\n9 10\n100", "output": "18" }, { "input": "3 3\n3 12 15\n7 8 9", "output": "-1" }, { "input": "2 2\n3 5\n7 8", "output": "6" }, { "input": "3 3\n4 5 6\n10 11 12", "output": "8" }, { "input": "3 5\n2 3 3\n6 6 6 6 2", "output": "-1" }, { "input": "3 6\n4 5 3\n8 9 7 10 7 11", "output": "6" }, { "input": "3 6\n4 5 2\n8 9 6 10 7 4", "output": "-1" }, { "input": "2 1\n4 6\n10", "output": "8" }, { "input": "1 2\n1\n3 1", "output": "-1" }, { "input": "2 1\n5 6\n20", "output": "10" }, { "input": "2 1\n1 5\n5", "output": "-1" }, { "input": "3 2\n10 20 30\n30 40", "output": "-1" }, { "input": "2 2\n5 6\n7 100", "output": "-1" }, { "input": "2 1\n2 5\n7", "output": "5" }, { "input": "1 1\n5\n20", "output": "10" }, { "input": "2 1\n10 11\n100", "output": "20" }, { "input": "1 1\n1\n10", "output": "2" }, { "input": "1 1\n10\n100", "output": "20" } ]
1,380,643,594
2,194
Python 3
WRONG_ANSWER
TESTS
7
92
0
n, m = map(int, input().split()) A = list(map(int, input().split())) B = list(map(int, input().split())) res = max(A) if min(B) > res and min(A) * 2 <= res: print(res) else: print(-1)
Title: TL Time Limit: None seconds Memory Limit: None megabytes Problem Description: Valera wanted to prepare a Codesecrof round. He's already got one problem and he wants to set a time limit (TL) on it. Valera has written *n* correct solutions. For each correct solution, he knows its running time (in seconds). Valera has also wrote *m* wrong solutions and for each wrong solution he knows its running time (in seconds). Let's suppose that Valera will set *v* seconds TL in the problem. Then we can say that a solution passes the system testing if its running time is at most *v* seconds. We can also say that a solution passes the system testing with some "extra" time if for its running time, *a* seconds, an inequality 2*a*<=≤<=*v* holds. As a result, Valera decided to set *v* seconds TL, that the following conditions are met: 1. *v* is a positive integer; 1. all correct solutions pass the system testing; 1. at least one correct solution passes the system testing with some "extra" time; 1. all wrong solutions do not pass the system testing; 1. value *v* is minimum among all TLs, for which points 1, 2, 3, 4 hold. Help Valera and find the most suitable TL or else state that such TL doesn't exist. Input Specification: The first line contains two integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=100). The second line contains *n* space-separated positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100) — the running time of each of the *n* correct solutions in seconds. The third line contains *m* space-separated positive integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=100) — the running time of each of *m* wrong solutions in seconds. Output Specification: If there is a valid TL value, print it. Otherwise, print -1. Demo Input: ['3 6\n4 5 2\n8 9 6 10 7 11\n', '3 1\n3 4 5\n6\n'] Demo Output: ['5', '-1\n'] Note: none
```python n, m = map(int, input().split()) A = list(map(int, input().split())) B = list(map(int, input().split())) res = max(A) if min(B) > res and min(A) * 2 <= res: print(res) else: print(-1) ```
0
242
B
Big Segment
PROGRAMMING
1,100
[ "implementation", "sortings" ]
null
null
A coordinate line has *n* segments, the *i*-th segment starts at the position *l**i* and ends at the position *r**i*. We will denote such a segment as [*l**i*,<=*r**i*]. You have suggested that one of the defined segments covers all others. In other words, there is such segment in the given set, which contains all other ones. Now you want to test your assumption. Find in the given set the segment which covers all other segments, and print its number. If such a segment doesn't exist, print -1. Formally we will assume that segment [*a*,<=*b*] covers segment [*c*,<=*d*], if they meet this condition *a*<=≤<=*c*<=≤<=*d*<=≤<=*b*.
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of segments. Next *n* lines contain the descriptions of the segments. The *i*-th line contains two space-separated integers *l**i*,<=*r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=109) — the borders of the *i*-th segment. It is guaranteed that no two segments coincide.
Print a single integer — the number of the segment that covers all other segments in the set. If there's no solution, print -1. The segments are numbered starting from 1 in the order in which they appear in the input.
[ "3\n1 1\n2 2\n3 3\n", "6\n1 5\n2 3\n1 10\n7 10\n7 7\n10 10\n" ]
[ "-1\n", "3\n" ]
none
1,000
[ { "input": "3\n1 1\n2 2\n3 3", "output": "-1" }, { "input": "6\n1 5\n2 3\n1 10\n7 10\n7 7\n10 10", "output": "3" }, { "input": "4\n1 5\n2 2\n2 4\n2 5", "output": "1" }, { "input": "5\n3 3\n1 3\n2 2\n2 3\n1 2", "output": "2" }, { "input": "7\n7 7\n8 8\n3 7\n1 6\n1 7\n4 7\n2 8", "output": "-1" }, { "input": "3\n2 5\n3 4\n2 3", "output": "1" }, { "input": "16\n15 15\n8 12\n6 9\n15 16\n8 14\n3 12\n7 19\n9 13\n5 16\n9 17\n10 15\n9 14\n9 9\n18 19\n5 15\n6 19", "output": "-1" }, { "input": "9\n1 10\n7 8\n6 7\n1 4\n5 9\n2 8\n3 10\n1 1\n2 3", "output": "1" }, { "input": "1\n1 100000", "output": "1" }, { "input": "6\n2 2\n3 3\n3 5\n4 5\n1 1\n1 5", "output": "6" }, { "input": "33\n2 18\n4 14\n2 16\n10 12\n4 6\n9 17\n2 8\n4 12\n8 20\n1 10\n11 14\n11 17\n8 15\n3 16\n3 4\n6 9\n6 19\n4 17\n17 19\n6 16\n3 12\n1 7\n6 20\n8 16\n12 19\n1 3\n12 18\n6 11\n7 20\n16 18\n4 15\n3 15\n15 19", "output": "-1" }, { "input": "34\n3 8\n5 9\n2 9\n1 4\n3 7\n3 3\n8 9\n6 10\n4 7\n6 7\n5 8\n5 10\n1 5\n8 8\n2 5\n3 5\n7 7\n2 8\n4 5\n1 1\n7 9\n5 6\n2 3\n1 2\n2 4\n8 10\n7 8\n1 3\n4 8\n9 10\n1 7\n10 10\n2 2\n1 8", "output": "-1" }, { "input": "55\n3 4\n6 8\n9 10\n3 9\n9 9\n2 5\n4 8\n3 8\n8 10\n1 1\n4 9\n10 10\n6 6\n8 8\n1 8\n5 5\n4 5\n5 9\n2 2\n3 10\n4 6\n3 6\n1 6\n1 7\n6 10\n2 6\n3 7\n2 4\n4 4\n5 10\n1 4\n2 9\n1 3\n7 9\n7 8\n1 9\n1 10\n2 8\n8 9\n6 7\n1 2\n6 9\n7 7\n4 7\n3 3\n2 7\n4 10\n7 10\n2 3\n2 10\n5 7\n3 5\n5 8\n1 5\n5 6", "output": "37" }, { "input": "1\n999999999 1000000000", "output": "1" }, { "input": "3\n1 20\n2 22\n3 18", "output": "-1" }, { "input": "1\n1000000000 1000000000", "output": "1" }, { "input": "2\n100001 100008\n100005 100006", "output": "1" }, { "input": "1\n1000000 10000000", "output": "1" }, { "input": "3\n3 6\n2 4\n1 5", "output": "-1" }, { "input": "2\n3 5\n1 2", "output": "-1" } ]
1,623,737,016
2,147,483,647
PyPy 3
OK
TESTS
45
1,808
11,571,200
n = int(input()) data = [] min_ , max_ = list(map(int,input().split())) data.append([min_,max_]) num = 0 for i in range(1,n): a,b = list(map(int,input().split())) data.append([a,b]) if a <= min_ and b >= max_ : min_ = a max_ = b num = i flag = 0 for i in range(n): a,b = data[i] if (a >= min_ and b <= max_) or i == num-1: pass else: flag = 1 break if flag == 0: print(num+1) else: print(-1)
Title: Big Segment Time Limit: None seconds Memory Limit: None megabytes Problem Description: A coordinate line has *n* segments, the *i*-th segment starts at the position *l**i* and ends at the position *r**i*. We will denote such a segment as [*l**i*,<=*r**i*]. You have suggested that one of the defined segments covers all others. In other words, there is such segment in the given set, which contains all other ones. Now you want to test your assumption. Find in the given set the segment which covers all other segments, and print its number. If such a segment doesn't exist, print -1. Formally we will assume that segment [*a*,<=*b*] covers segment [*c*,<=*d*], if they meet this condition *a*<=≤<=*c*<=≤<=*d*<=≤<=*b*. Input Specification: The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of segments. Next *n* lines contain the descriptions of the segments. The *i*-th line contains two space-separated integers *l**i*,<=*r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=109) — the borders of the *i*-th segment. It is guaranteed that no two segments coincide. Output Specification: Print a single integer — the number of the segment that covers all other segments in the set. If there's no solution, print -1. The segments are numbered starting from 1 in the order in which they appear in the input. Demo Input: ['3\n1 1\n2 2\n3 3\n', '6\n1 5\n2 3\n1 10\n7 10\n7 7\n10 10\n'] Demo Output: ['-1\n', '3\n'] Note: none
```python n = int(input()) data = [] min_ , max_ = list(map(int,input().split())) data.append([min_,max_]) num = 0 for i in range(1,n): a,b = list(map(int,input().split())) data.append([a,b]) if a <= min_ and b >= max_ : min_ = a max_ = b num = i flag = 0 for i in range(n): a,b = data[i] if (a >= min_ and b <= max_) or i == num-1: pass else: flag = 1 break if flag == 0: print(num+1) else: print(-1) ```
3
388
A
Fox and Box Accumulation
PROGRAMMING
1,400
[ "greedy", "sortings" ]
null
null
Fox Ciel has *n* boxes in her room. They have the same size and weight, but they might have different strength. The *i*-th box can hold at most *x**i* boxes on its top (we'll call *x**i* the strength of the box). Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For example, imagine Ciel has three boxes: the first has strength 2, the second has strength 1 and the third has strength 1. She cannot put the second and the third box simultaneously directly on the top of the first one. But she can put the second box directly on the top of the first one, and then the third box directly on the top of the second one. We will call such a construction of boxes a pile. Fox Ciel wants to construct piles from all the boxes. Each pile will contain some boxes from top to bottom, and there cannot be more than *x**i* boxes on the top of *i*-th box. What is the minimal number of piles she needs to construct?
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). The next line contains *n* integers *x*1,<=*x*2,<=...,<=*x**n* (0<=≤<=*x**i*<=≤<=100).
Output a single integer — the minimal possible number of piles.
[ "3\n0 0 10\n", "5\n0 1 2 3 4\n", "4\n0 0 0 0\n", "9\n0 1 0 2 0 1 1 2 10\n" ]
[ "2\n", "1\n", "4\n", "3\n" ]
In example 1, one optimal way is to build 2 piles: the first pile contains boxes 1 and 3 (from top to bottom), the second pile contains only box 2. In example 2, we can build only 1 pile that contains boxes 1, 2, 3, 4, 5 (from top to bottom).
500
[ { "input": "3\n0 0 10", "output": "2" }, { "input": "5\n0 1 2 3 4", "output": "1" }, { "input": "4\n0 0 0 0", "output": "4" }, { "input": "9\n0 1 0 2 0 1 1 2 10", "output": "3" }, { "input": "1\n0", "output": "1" }, { "input": "2\n0 0", "output": "2" }, { "input": "2\n0 1", "output": "1" }, { "input": "2\n100 99", "output": "1" }, { "input": "9\n0 1 1 0 2 0 3 45 4", "output": "3" }, { "input": "10\n1 1 1 1 2 2 2 2 2 2", "output": "4" }, { "input": "100\n50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50", "output": "2" }, { "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\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100", "output": "1" }, { "input": "11\n71 34 31 71 42 38 64 60 36 76 67", "output": "1" }, { "input": "39\n54 54 54 54 54 54 54 54 54 54 54 54 54 54 54 54 54 54 54 54 54 54 54 54 54 54 54 54 54 54 54 54 54 54 54 54 54 54 54", "output": "1" }, { "input": "59\n61 33 84 76 56 47 70 94 46 77 95 85 35 90 83 62 48 74 36 74 83 97 62 92 95 75 70 82 94 67 82 42 78 70 50 73 80 76 94 83 96 80 80 88 91 79 83 54 38 90 33 93 53 33 86 95 48 34 46", "output": "1" }, { "input": "87\n52 63 93 90 50 35 67 66 46 89 43 64 33 88 34 80 69 59 75 55 55 68 66 83 46 33 72 36 73 34 54 85 52 87 67 68 47 95 52 78 92 58 71 66 84 61 36 77 69 44 84 70 71 55 43 91 33 65 77 34 43 59 83 70 95 38 92 92 74 53 66 65 81 45 55 89 49 52 43 69 78 41 37 79 63 70 67", "output": "1" }, { "input": "15\n20 69 36 63 40 40 52 42 20 43 59 68 64 49 47", "output": "1" }, { "input": "39\n40 20 49 35 80 18 20 75 39 62 43 59 46 37 58 52 67 16 34 65 32 75 59 42 59 41 68 21 41 61 66 19 34 63 19 63 78 62 24", "output": "1" }, { "input": "18\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0", "output": "18" }, { "input": "46\n14 13 13 10 13 15 8 8 12 9 11 15 8 10 13 8 12 13 11 8 12 15 12 15 11 13 12 9 13 12 10 8 13 15 9 15 8 13 11 8 9 9 9 8 11 8", "output": "3" }, { "input": "70\n6 1 4 1 1 6 5 2 5 1 1 5 2 1 2 4 1 1 1 2 4 5 2 1 6 6 5 2 1 4 3 1 4 3 6 5 2 1 3 4 4 1 4 5 6 2 1 2 4 4 5 3 6 1 1 2 2 1 5 6 1 6 3 1 4 4 2 3 1 4", "output": "11" }, { "input": "94\n11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11", "output": "8" }, { "input": "18\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1", "output": "9" }, { "input": "46\n14 8 7 4 8 7 8 8 12 9 9 12 9 12 14 8 10 14 14 6 9 11 7 14 14 13 11 4 13 13 11 13 9 10 10 12 10 8 12 10 13 10 7 13 14 6", "output": "4" }, { "input": "74\n4 4 5 5 5 5 5 5 6 6 5 4 4 4 3 3 5 4 5 3 4 4 5 6 3 3 5 4 4 5 4 3 5 5 4 4 3 5 6 4 3 6 6 3 4 5 4 4 3 3 3 6 3 5 6 5 5 5 5 3 6 4 5 4 4 6 6 3 4 5 6 6 6 6", "output": "11" }, { "input": "100\n48 35 44 37 35 42 42 39 49 53 35 55 41 42 42 39 43 49 46 54 48 39 42 53 55 39 56 43 43 38 48 40 54 36 48 55 46 40 41 39 45 56 38 40 47 46 45 46 53 51 38 41 54 35 35 47 42 43 54 54 39 44 49 41 37 49 36 37 37 49 53 44 47 37 55 49 45 40 35 51 44 40 42 35 46 48 53 48 35 38 42 36 54 46 44 47 41 40 41 42", "output": "2" }, { "input": "100\n34 3 37 35 40 44 38 46 13 31 12 23 26 40 26 18 28 36 5 21 2 4 10 29 3 46 38 41 37 28 44 14 39 10 35 17 24 28 38 16 29 6 2 42 47 34 43 2 43 46 7 16 16 43 33 32 20 47 8 48 32 4 45 38 15 7 25 25 19 41 20 35 16 2 31 5 31 25 27 3 45 29 32 36 9 47 39 35 9 21 32 17 21 41 29 48 11 40 5 25", "output": "3" }, { "input": "100\n2 4 5 5 0 5 3 0 3 0 5 3 4 1 0 3 0 5 5 0 4 3 3 3 0 2 1 2 2 4 4 2 4 0 1 3 4 1 4 2 5 3 5 2 3 0 1 2 5 5 2 0 4 2 5 1 0 0 4 0 1 2 0 1 2 4 1 4 5 3 4 5 5 1 0 0 3 1 4 0 4 5 1 3 3 0 4 2 0 4 5 2 3 0 5 1 4 4 1 0", "output": "21" }, { "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": "17" }, { "input": "100\n1 1 1 2 2 2 2 2 2 1 1 1 2 0 2 2 0 0 0 0 0 2 0 0 2 2 1 0 2 0 2 1 1 2 2 1 2 2 1 2 1 2 2 1 2 0 1 2 2 0 2 2 2 2 1 0 1 0 0 0 2 0 2 0 1 1 0 2 2 2 2 1 1 1 2 1 1 2 1 1 1 2 1 0 2 1 0 1 2 0 1 1 2 0 0 1 1 0 1 1", "output": "34" }, { "input": "100\n0 3 1 0 3 2 1 2 2 1 2 1 3 2 1 2 1 3 2 0 0 2 3 0 0 2 1 2 2 3 1 2 2 2 0 3 3 2 0 0 1 0 1 2 3 1 0 3 3 3 0 2 1 3 0 1 3 2 2 2 2 3 3 2 0 2 0 1 0 1 3 0 1 2 0 1 3 2 0 3 1 1 2 3 1 3 1 0 3 0 3 0 2 1 1 1 2 2 0 1", "output": "26" }, { "input": "100\n1 0 2 2 2 2 1 0 1 2 2 2 0 1 0 1 2 1 2 1 0 1 2 2 2 1 0 1 0 2 1 2 0 2 1 1 2 1 1 0 1 2 1 1 2 1 1 0 2 2 0 0 1 2 0 2 0 0 1 1 0 0 2 1 2 1 0 2 2 2 2 2 2 1 2 0 1 2 1 2 1 0 1 0 1 0 1 1 0 2 1 0 0 1 2 2 1 0 0 1", "output": "34" }, { "input": "100\n3 4 4 4 3 3 3 3 3 4 4 4 3 3 3 4 3 4 4 4 3 4 3 4 3 4 3 3 4 4 3 4 4 3 4 4 4 4 4 3 4 3 3 3 4 3 3 4 3 4 3 4 3 3 4 4 4 3 3 3 3 3 4 4 3 4 4 3 4 3 3 3 4 4 3 3 3 3 3 4 3 4 4 3 3 4 3 4 3 4 4 4 3 3 3 4 4 4 4 3", "output": "20" }, { "input": "100\n8 7 9 10 2 7 8 11 11 4 7 10 2 5 8 9 10 3 9 4 10 5 5 6 3 8 8 9 6 9 5 5 4 11 4 2 11 8 3 5 6 6 11 9 8 11 9 8 3 3 8 9 8 9 4 8 6 11 4 4 4 9 7 5 3 4 11 3 9 11 8 10 3 5 5 7 6 9 4 5 2 11 3 6 2 10 9 4 6 10 5 11 8 10 10 8 9 8 5 3", "output": "9" }, { "input": "5\n4 1 1 1 1", "output": "2" } ]
1,548,450,740
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
2
109
0
#code n = int(input()) l = [int(i) for i in input().split()] l.sort() l.reverse() cnt = int(0) for i in range(1,n): if(l[i] > l[i - 1] or l[i] == 0): cnt += 1 print(cnt)
Title: Fox and Box Accumulation Time Limit: None seconds Memory Limit: None megabytes Problem Description: Fox Ciel has *n* boxes in her room. They have the same size and weight, but they might have different strength. The *i*-th box can hold at most *x**i* boxes on its top (we'll call *x**i* the strength of the box). Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For example, imagine Ciel has three boxes: the first has strength 2, the second has strength 1 and the third has strength 1. She cannot put the second and the third box simultaneously directly on the top of the first one. But she can put the second box directly on the top of the first one, and then the third box directly on the top of the second one. We will call such a construction of boxes a pile. Fox Ciel wants to construct piles from all the boxes. Each pile will contain some boxes from top to bottom, and there cannot be more than *x**i* boxes on the top of *i*-th box. What is the minimal number of piles she needs to construct? Input Specification: The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). The next line contains *n* integers *x*1,<=*x*2,<=...,<=*x**n* (0<=≤<=*x**i*<=≤<=100). Output Specification: Output a single integer — the minimal possible number of piles. Demo Input: ['3\n0 0 10\n', '5\n0 1 2 3 4\n', '4\n0 0 0 0\n', '9\n0 1 0 2 0 1 1 2 10\n'] Demo Output: ['2\n', '1\n', '4\n', '3\n'] Note: In example 1, one optimal way is to build 2 piles: the first pile contains boxes 1 and 3 (from top to bottom), the second pile contains only box 2. In example 2, we can build only 1 pile that contains boxes 1, 2, 3, 4, 5 (from top to bottom).
```python #code n = int(input()) l = [int(i) for i in input().split()] l.sort() l.reverse() cnt = int(0) for i in range(1,n): if(l[i] > l[i - 1] or l[i] == 0): cnt += 1 print(cnt) ```
0
476
B
Dreamoon and WiFi
PROGRAMMING
1,300
[ "bitmasks", "brute force", "combinatorics", "dp", "math", "probabilities" ]
null
null
Dreamoon is standing at the position 0 on a number line. Drazil is sending a list of commands through Wi-Fi to Dreamoon's smartphone and Dreamoon follows them. Each command is one of the following two types: 1. Go 1 unit towards the positive direction, denoted as '+' 1. Go 1 unit towards the negative direction, denoted as '-' But the Wi-Fi condition is so poor that Dreamoon's smartphone reports some of the commands can't be recognized and Dreamoon knows that some of them might even be wrong though successfully recognized. Dreamoon decides to follow every recognized command and toss a fair coin to decide those unrecognized ones (that means, he moves to the 1 unit to the negative or positive direction with the same probability 0.5). You are given an original list of commands sent by Drazil and list received by Dreamoon. What is the probability that Dreamoon ends in the position originally supposed to be final by Drazil's commands?
The first line contains a string *s*1 — the commands Drazil sends to Dreamoon, this string consists of only the characters in the set {'+', '-'}. The second line contains a string *s*2 — the commands Dreamoon's smartphone recognizes, this string consists of only the characters in the set {'+', '-', '?'}. '?' denotes an unrecognized command. Lengths of two strings are equal and do not exceed 10.
Output a single real number corresponding to the probability. The answer will be considered correct if its relative or absolute error doesn't exceed 10<=-<=9.
[ "++-+-\n+-+-+\n", "+-+-\n+-??\n", "+++\n??-\n" ]
[ "1.000000000000\n", "0.500000000000\n", "0.000000000000\n" ]
For the first sample, both *s*<sub class="lower-index">1</sub> and *s*<sub class="lower-index">2</sub> will lead Dreamoon to finish at the same position  + 1. For the second sample, *s*<sub class="lower-index">1</sub> will lead Dreamoon to finish at position 0, while there are four possibilites for *s*<sub class="lower-index">2</sub>: {"+-++", "+-+-", "+--+", "+---"} with ending position {+2, 0, 0, -2} respectively. So there are 2 correct cases out of 4, so the probability of finishing at the correct position is 0.5. For the third sample, *s*<sub class="lower-index">2</sub> could only lead us to finish at positions {+1, -1, -3}, so the probability to finish at the correct position  + 3 is 0.
1,500
[ { "input": "++-+-\n+-+-+", "output": "1.000000000000" }, { "input": "+-+-\n+-??", "output": "0.500000000000" }, { "input": "+++\n??-", "output": "0.000000000000" }, { "input": "++++++++++\n+++??++?++", "output": "0.125000000000" }, { "input": "--+++---+-\n??????????", "output": "0.205078125000" }, { "input": "+--+++--+-\n??????????", "output": "0.246093750000" }, { "input": "+\n+", "output": "1.000000000000" }, { "input": "-\n?", "output": "0.500000000000" }, { "input": "+\n-", "output": "0.000000000000" }, { "input": "-\n-", "output": "1.000000000000" }, { "input": "-\n+", "output": "0.000000000000" }, { "input": "+\n?", "output": "0.500000000000" }, { "input": "++++++++++\n++++++++++", "output": "1.000000000000" }, { "input": "++++++++++\n++++-+++++", "output": "0.000000000000" }, { "input": "----------\n++++++++++", "output": "0.000000000000" }, { "input": "++++++++++\n++++??++++", "output": "0.250000000000" }, { "input": "----------\n+++?++++-+", "output": "0.000000000000" }, { "input": "++++++++++\n++++++++?+", "output": "0.500000000000" }, { "input": "--++++--+\n?-+?-??+-", "output": "0.250000000000" }, { "input": "----------\n??????????", "output": "0.000976562500" }, { "input": "+--++\n+--+-", "output": "0.000000000000" }, { "input": "-----++---\n????????+?", "output": "0.017578125000" }, { "input": "------+--+\n??????????", "output": "0.043945312500" }, { "input": "---++--\n???????", "output": "0.164062500000" }, { "input": "-----++\n???????", "output": "0.164062500000" }, { "input": "+---+--\n???????", "output": "0.164062500000" }, { "input": "---+-+\n??????", "output": "0.234375000000" }, { "input": "+++-+\n---++", "output": "0.000000000000" }, { "input": "++-+--+\n?-?+??+", "output": "0.375000000000" }, { "input": "----+++--\n-+?+++?--", "output": "0.000000000000" }, { "input": "+-----\n+?----", "output": "0.500000000000" } ]
1,691,173,575
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
3
31
0
import math def main(): orig_cmd = input() guess_cmd = input() orig_score = orig_cmd.count("+") - orig_cmd.count("-") guess_score = guess_cmd.count("+") - guess_cmd.count("-") num_guesses = guess_cmd.count("?") if abs(orig_score - guess_score) > num_guesses: chance = 0.0 else: chance = 1 / (2 ** (math.comb(num_guesses, math.floor(num_guesses/2)) - 1)) print("{:.9f}".format(chance)) if __name__ == "__main__": main()
Title: Dreamoon and WiFi Time Limit: None seconds Memory Limit: None megabytes Problem Description: Dreamoon is standing at the position 0 on a number line. Drazil is sending a list of commands through Wi-Fi to Dreamoon's smartphone and Dreamoon follows them. Each command is one of the following two types: 1. Go 1 unit towards the positive direction, denoted as '+' 1. Go 1 unit towards the negative direction, denoted as '-' But the Wi-Fi condition is so poor that Dreamoon's smartphone reports some of the commands can't be recognized and Dreamoon knows that some of them might even be wrong though successfully recognized. Dreamoon decides to follow every recognized command and toss a fair coin to decide those unrecognized ones (that means, he moves to the 1 unit to the negative or positive direction with the same probability 0.5). You are given an original list of commands sent by Drazil and list received by Dreamoon. What is the probability that Dreamoon ends in the position originally supposed to be final by Drazil's commands? Input Specification: The first line contains a string *s*1 — the commands Drazil sends to Dreamoon, this string consists of only the characters in the set {'+', '-'}. The second line contains a string *s*2 — the commands Dreamoon's smartphone recognizes, this string consists of only the characters in the set {'+', '-', '?'}. '?' denotes an unrecognized command. Lengths of two strings are equal and do not exceed 10. Output Specification: Output a single real number corresponding to the probability. The answer will be considered correct if its relative or absolute error doesn't exceed 10<=-<=9. Demo Input: ['++-+-\n+-+-+\n', '+-+-\n+-??\n', '+++\n??-\n'] Demo Output: ['1.000000000000\n', '0.500000000000\n', '0.000000000000\n'] Note: For the first sample, both *s*<sub class="lower-index">1</sub> and *s*<sub class="lower-index">2</sub> will lead Dreamoon to finish at the same position  + 1. For the second sample, *s*<sub class="lower-index">1</sub> will lead Dreamoon to finish at position 0, while there are four possibilites for *s*<sub class="lower-index">2</sub>: {"+-++", "+-+-", "+--+", "+---"} with ending position {+2, 0, 0, -2} respectively. So there are 2 correct cases out of 4, so the probability of finishing at the correct position is 0.5. For the third sample, *s*<sub class="lower-index">2</sub> could only lead us to finish at positions {+1, -1, -3}, so the probability to finish at the correct position  + 3 is 0.
```python import math def main(): orig_cmd = input() guess_cmd = input() orig_score = orig_cmd.count("+") - orig_cmd.count("-") guess_score = guess_cmd.count("+") - guess_cmd.count("-") num_guesses = guess_cmd.count("?") if abs(orig_score - guess_score) > num_guesses: chance = 0.0 else: chance = 1 / (2 ** (math.comb(num_guesses, math.floor(num_guesses/2)) - 1)) print("{:.9f}".format(chance)) if __name__ == "__main__": main() ```
0
34
B
Sale
PROGRAMMING
900
[ "greedy", "sortings" ]
B. Sale
2
256
Once Bob got to a sale of old TV sets. There were *n* TV sets at that sale. TV set with index *i* costs *a**i* bellars. Some TV sets have a negative price — their owners are ready to pay Bob if he buys their useless apparatus. Bob can «buy» any TV sets he wants. Though he's very strong, Bob can carry at most *m* TV sets, and he has no desire to go to the sale for the second time. Please, help Bob find out the maximum sum of money that he can earn.
The first line contains two space-separated integers *n* and *m* (1<=≤<=*m*<=≤<=*n*<=≤<=100) — amount of TV sets at the sale, and amount of TV sets that Bob can carry. The following line contains *n* space-separated integers *a**i* (<=-<=1000<=≤<=*a**i*<=≤<=1000) — prices of the TV sets.
Output the only number — the maximum sum of money that Bob can earn, given that he can carry at most *m* TV sets.
[ "5 3\n-6 0 35 -2 4\n", "4 2\n7 0 0 -7\n" ]
[ "8\n", "7\n" ]
none
1,000
[ { "input": "5 3\n-6 0 35 -2 4", "output": "8" }, { "input": "4 2\n7 0 0 -7", "output": "7" }, { "input": "6 6\n756 -611 251 -66 572 -818", "output": "1495" }, { "input": "5 5\n976 437 937 788 518", "output": "0" }, { "input": "5 3\n-2 -2 -2 -2 -2", "output": "6" }, { "input": "5 1\n998 997 985 937 998", "output": "0" }, { "input": "2 2\n-742 -187", "output": "929" }, { "input": "3 3\n522 597 384", "output": "0" }, { "input": "4 2\n-215 -620 192 647", "output": "835" }, { "input": "10 6\n557 605 685 231 910 633 130 838 -564 -85", "output": "649" }, { "input": "20 14\n932 442 960 943 624 624 955 998 631 910 850 517 715 123 1000 155 -10 961 966 59", "output": "10" }, { "input": "30 5\n991 997 996 967 977 999 991 986 1000 965 984 997 998 1000 958 983 974 1000 991 999 1000 978 961 992 990 998 998 978 998 1000", "output": "0" }, { "input": "50 20\n-815 -947 -946 -993 -992 -846 -884 -954 -963 -733 -940 -746 -766 -930 -821 -937 -937 -999 -914 -938 -936 -975 -939 -981 -977 -952 -925 -901 -952 -978 -994 -957 -946 -896 -905 -836 -994 -951 -887 -939 -859 -953 -985 -988 -946 -829 -956 -842 -799 -886", "output": "19441" }, { "input": "88 64\n999 999 1000 1000 999 996 995 1000 1000 999 1000 997 998 1000 999 1000 997 1000 993 998 994 999 998 996 1000 997 1000 1000 1000 997 1000 998 997 1000 1000 998 1000 998 999 1000 996 999 999 999 996 995 999 1000 998 999 1000 999 999 1000 1000 1000 996 1000 1000 1000 997 1000 1000 997 999 1000 1000 1000 1000 1000 999 999 1000 1000 996 999 1000 1000 995 999 1000 996 1000 998 999 999 1000 999", "output": "0" }, { "input": "99 17\n-993 -994 -959 -989 -991 -995 -976 -997 -990 -1000 -996 -994 -999 -995 -1000 -983 -979 -1000 -989 -968 -994 -992 -962 -993 -999 -983 -991 -979 -995 -993 -973 -999 -995 -995 -999 -993 -995 -992 -947 -1000 -999 -998 -982 -988 -979 -993 -963 -988 -980 -990 -979 -976 -995 -999 -981 -988 -998 -999 -970 -1000 -983 -994 -943 -975 -998 -977 -973 -997 -959 -999 -983 -985 -950 -977 -977 -991 -998 -973 -987 -985 -985 -986 -984 -994 -978 -998 -989 -989 -988 -970 -985 -974 -997 -981 -962 -972 -995 -988 -993", "output": "16984" }, { "input": "100 37\n205 19 -501 404 912 -435 -322 -469 -655 880 -804 -470 793 312 -108 586 -642 -928 906 605 -353 -800 745 -440 -207 752 -50 -28 498 -800 -62 -195 602 -833 489 352 536 404 -775 23 145 -512 524 759 651 -461 -427 -557 684 -366 62 592 -563 -811 64 418 -881 -308 591 -318 -145 -261 -321 -216 -18 595 -202 960 -4 219 226 -238 -882 -963 425 970 -434 -160 243 -672 -4 873 8 -633 904 -298 -151 -377 -61 -72 -677 -66 197 -716 3 -870 -30 152 -469 981", "output": "21743" }, { "input": "100 99\n-931 -806 -830 -828 -916 -962 -660 -867 -952 -966 -820 -906 -724 -982 -680 -717 -488 -741 -897 -613 -986 -797 -964 -939 -808 -932 -810 -860 -641 -916 -858 -628 -821 -929 -917 -976 -664 -985 -778 -665 -624 -928 -940 -958 -884 -757 -878 -896 -634 -526 -514 -873 -990 -919 -988 -878 -650 -973 -774 -783 -733 -648 -756 -895 -833 -974 -832 -725 -841 -748 -806 -613 -924 -867 -881 -943 -864 -991 -809 -926 -777 -817 -998 -682 -910 -996 -241 -722 -964 -904 -821 -920 -835 -699 -805 -632 -779 -317 -915 -654", "output": "81283" }, { "input": "100 14\n995 994 745 684 510 737 984 690 979 977 542 933 871 603 758 653 962 997 747 974 773 766 975 770 527 960 841 989 963 865 974 967 950 984 757 685 986 809 982 959 931 880 978 867 805 562 970 900 834 782 616 885 910 608 974 918 576 700 871 980 656 941 978 759 767 840 573 859 841 928 693 853 716 927 976 851 962 962 627 797 707 873 869 988 993 533 665 887 962 880 929 980 877 887 572 790 721 883 848 782", "output": "0" }, { "input": "100 84\n768 946 998 752 931 912 826 1000 991 910 875 962 901 952 958 733 959 908 872 840 923 826 952 980 974 980 947 955 959 822 997 963 966 933 829 923 971 999 926 932 865 984 974 858 994 855 949 941 992 861 951 949 991 711 763 728 935 485 716 907 869 952 960 859 909 963 978 942 968 933 923 909 997 962 687 764 924 774 875 1000 961 951 987 974 848 921 966 859 995 997 974 931 886 941 974 986 906 978 998 823", "output": "0" }, { "input": "100 80\n-795 -994 -833 -930 -974 -980 -950 -940 -788 -927 -583 -956 -945 -949 -809 -974 -957 -736 -967 -908 -975 -961 -986 -983 -963 -771 -952 -847 -751 -741 -982 -959 -925 -931 -839 -937 -880 -914 -858 -998 -812 -911 -862 -965 -943 -984 -738 -920 -950 -998 -909 -998 -781 -901 -677 -940 -985 -951 -675 -952 -967 -949 -882 -641 -969 -937 -975 -993 -913 -941 -807 -851 -832 -960 -939 -943 -895 -929 -528 -880 -823 -930 -888 -862 -948 -966 -962 -857 -799 -969 -833 -998 -952 -878 -946 -971 -976 -974 -723 -992", "output": "75068" }, { "input": "1 1\n0", "output": "0" }, { "input": "1 1\n1", "output": "0" }, { "input": "1 1\n555", "output": "0" }, { "input": "1 1\n-1", "output": "1" }, { "input": "1 1\n-24", "output": "24" } ]
1,663,556,088
2,147,483,647
Python 3
OK
TESTS
25
92
0
n,m=map(int,input().split()) price=sorted(list(map(int,input().split()))) sum_=0 for i in range(m): if price[i]<0: sum_+=abs(price[i]) else: break print(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 — their owners are ready to pay Bob if he buys their useless apparatus. Bob can «buy» any TV sets he wants. Though he's very strong, Bob can carry at most *m* TV sets, and he has no desire to go to the sale for the second time. Please, help Bob find out the maximum sum of money that he can earn. Input Specification: The first line contains two space-separated integers *n* and *m* (1<=≤<=*m*<=≤<=*n*<=≤<=100) — amount of TV sets at the sale, and amount of TV sets that Bob can carry. The following line contains *n* space-separated integers *a**i* (<=-<=1000<=≤<=*a**i*<=≤<=1000) — prices of the TV sets. Output Specification: Output the only number — the maximum sum of money that Bob can earn, given that he can carry at most *m* TV sets. Demo Input: ['5 3\n-6 0 35 -2 4\n', '4 2\n7 0 0 -7\n'] Demo Output: ['8\n', '7\n'] Note: none
```python n,m=map(int,input().split()) price=sorted(list(map(int,input().split()))) sum_=0 for i in range(m): if price[i]<0: sum_+=abs(price[i]) else: break print(sum_) ```
3.977
339
B
Xenia and Ringroad
PROGRAMMING
1,000
[ "implementation" ]
null
null
Xenia lives in a city that has *n* houses built along the main ringroad. The ringroad houses are numbered 1 through *n* in the clockwise order. The ringroad traffic is one way and also is clockwise. Xenia has recently moved into the ringroad house number 1. As a result, she's got *m* things to do. In order to complete the *i*-th task, she needs to be in the house number *a**i* and complete all tasks with numbers less than *i*. Initially, Xenia is in the house number 1, find the minimum time she needs to complete all her tasks if moving from a house to a neighboring one along the ringroad takes one unit of time.
The first line contains two integers *n* and *m* (2<=≤<=*n*<=≤<=105,<=1<=≤<=*m*<=≤<=105). The second line contains *m* integers *a*1,<=*a*2,<=...,<=*a**m* (1<=≤<=*a**i*<=≤<=*n*). Note that Xenia can have multiple consecutive tasks in one house.
Print a single integer — the time Xenia needs to complete all tasks. Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
[ "4 3\n3 2 3\n", "4 3\n2 3 3\n" ]
[ "6\n", "2\n" ]
In the first test example the sequence of Xenia's moves along the ringroad looks as follows: 1 → 2 → 3 → 4 → 1 → 2 → 3. This is optimal sequence. So, she needs 6 time units.
1,000
[ { "input": "4 3\n3 2 3", "output": "6" }, { "input": "4 3\n2 3 3", "output": "2" }, { "input": "2 2\n1 1", "output": "0" }, { "input": "2 2\n1 2", "output": "1" }, { "input": "2 2\n1 2", "output": "1" }, { "input": "100 100\n56 46 1 47 5 86 45 35 81 1 31 70 67 70 62 99 100 47 44 33 78 35 32 37 92 12 95 18 3 22 54 24 22 90 25 22 78 88 51 92 46 84 15 29 28 40 8 5 93 68 77 47 45 76 85 39 84 94 52 69 93 64 31 60 99 17 51 59 62 37 46 47 86 60 88 14 68 22 47 93 50 10 55 87 46 50 43 63 44 43 61 65 91 43 33 97 67 57 66 70", "output": "4869" }, { "input": "78 58\n23 14 73 45 47 14 27 59 65 39 15 23 5 1 50 37 3 51 46 69 75 65 45 68 48 59 77 39 53 21 72 33 46 32 34 5 69 55 56 53 47 31 32 5 42 23 76 15 2 77 65 24 16 68 61 28 55 10", "output": "2505" }, { "input": "14 54\n9 13 14 9 5 12 4 7 3 14 5 12 13 1 1 11 10 2 7 9 5 2 2 8 10 7 3 9 5 11 2 2 6 12 11 5 4 11 11 6 2 11 14 13 8 7 13 9 4 9 11 3 7 13", "output": "362" }, { "input": "100 100\n48 73 63 16 49 88 36 17 66 6 87 13 94 52 58 70 71 52 7 70 25 42 24 36 57 9 79 26 75 39 13 14 38 26 33 66 88 28 75 98 53 48 67 54 63 25 69 87 88 32 72 17 36 35 29 67 74 89 70 47 20 90 78 13 94 57 32 73 29 74 45 78 85 64 81 56 12 65 19 67 34 86 55 71 41 33 76 13 100 47 44 76 86 78 37 15 26 98 83 98", "output": "4997" }, { "input": "99 100\n88 65 10 91 18 35 58 49 42 2 22 57 74 31 53 24 27 93 45 4 71 2 69 39 21 90 97 89 45 73 20 45 82 98 35 90 37 76 68 26 21 65 95 63 24 74 50 59 3 93 65 6 30 37 62 71 18 88 40 12 56 40 89 56 38 71 90 41 97 43 44 23 19 22 10 80 3 24 32 85 26 65 70 60 76 85 66 68 74 11 64 88 12 63 16 15 79 57 93 58", "output": "4809" }, { "input": "65 100\n53 14 5 10 32 60 31 52 52 56 38 6 8 17 52 23 59 3 18 28 15 2 46 26 8 2 40 6 58 30 28 46 49 23 47 24 9 53 3 47 55 12 36 49 12 24 54 55 58 7 50 42 15 4 58 49 34 40 19 4 59 19 31 17 35 65 36 50 45 5 33 11 29 52 55 40 48 11 32 41 31 7 46 55 32 41 56 51 39 13 5 59 58 34 38 50 55 10 43 30", "output": "3149" }, { "input": "10 100\n7 6 2 10 7 2 3 8 10 4 6 1 4 5 7 10 1 2 3 5 4 10 8 2 3 3 6 8 3 9 4 1 9 10 1 2 5 1 8 8 5 9 2 8 1 2 3 2 1 10 10 7 1 3 2 2 7 1 6 6 6 9 2 3 1 7 2 2 9 7 3 3 2 10 7 4 7 3 3 3 2 4 4 2 2 8 4 1 10 10 5 10 6 10 6 10 3 10 8 9", "output": "428" }, { "input": "2 100\n1 1 2 2 2 2 1 2 1 2 2 2 1 1 2 2 2 2 1 1 2 1 2 2 1 1 2 2 2 1 2 1 1 1 2 1 2 2 2 1 2 2 2 2 1 2 1 1 1 2 1 1 2 1 1 2 2 1 2 1 2 2 2 1 1 1 1 1 2 2 2 1 1 2 2 1 1 2 2 1 1 2 1 1 1 1 2 2 1 1 1 2 1 1 1 1 1 1 1 2", "output": "47" }, { "input": "67 100\n49 5 25 48 37 55 5 33 14 30 59 28 57 46 45 32 47 22 40 28 58 34 27 29 4 52 63 44 31 65 42 61 11 17 32 17 18 1 12 33 38 11 59 46 43 55 23 30 23 2 42 21 45 51 30 19 35 15 10 30 13 21 32 34 33 3 5 59 23 63 6 9 20 43 64 3 42 41 40 4 14 20 40 33 25 44 1 2 50 46 13 10 3 20 22 64 28 42 58 30", "output": "3245" }, { "input": "100 100\n81 62 26 90 7 87 60 35 75 81 54 94 53 71 64 80 58 83 53 70 40 96 87 50 37 63 5 77 83 75 53 28 22 98 52 45 72 70 62 21 1 99 37 93 16 98 68 27 96 37 21 83 92 23 62 96 85 31 85 66 38 77 45 16 90 86 6 21 75 42 18 60 2 17 51 61 94 68 97 79 17 59 66 14 87 78 70 40 63 78 58 22 87 7 90 17 20 97 9 31", "output": "5530" }, { "input": "2 1\n1", "output": "0" }, { "input": "2 10\n1 1 1 1 1 1 1 1 1 1", "output": "0" }, { "input": "2 10\n1 1 1 2 2 2 1 1 1 2", "output": "3" }, { "input": "3 6\n1 2 3 3 2 1", "output": "6" } ]
1,699,196,740
2,147,483,647
PyPy 3-64
WRONG_ANSWER
TESTS
5
124
0
n, m = map(int, input().split()) a = [int(i) for i in input().split()] ans = 0 for i in range(1, m): if a[i] < a[i - 1]: ans += n - 1 + a[i] - 1 if a[-1] >= a[-2]: ans += a[-1] - 1 print(ans)
Title: Xenia and Ringroad Time Limit: None seconds Memory Limit: None megabytes Problem Description: Xenia lives in a city that has *n* houses built along the main ringroad. The ringroad houses are numbered 1 through *n* in the clockwise order. The ringroad traffic is one way and also is clockwise. Xenia has recently moved into the ringroad house number 1. As a result, she's got *m* things to do. In order to complete the *i*-th task, she needs to be in the house number *a**i* and complete all tasks with numbers less than *i*. Initially, Xenia is in the house number 1, find the minimum time she needs to complete all her tasks if moving from a house to a neighboring one along the ringroad takes one unit of time. Input Specification: The first line contains two integers *n* and *m* (2<=≤<=*n*<=≤<=105,<=1<=≤<=*m*<=≤<=105). The second line contains *m* integers *a*1,<=*a*2,<=...,<=*a**m* (1<=≤<=*a**i*<=≤<=*n*). Note that Xenia can have multiple consecutive tasks in one house. Output Specification: Print a single integer — the time Xenia needs to complete all tasks. Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier. Demo Input: ['4 3\n3 2 3\n', '4 3\n2 3 3\n'] Demo Output: ['6\n', '2\n'] Note: In the first test example the sequence of Xenia's moves along the ringroad looks as follows: 1 → 2 → 3 → 4 → 1 → 2 → 3. This is optimal sequence. So, she needs 6 time units.
```python n, m = map(int, input().split()) a = [int(i) for i in input().split()] ans = 0 for i in range(1, m): if a[i] < a[i - 1]: ans += n - 1 + a[i] - 1 if a[-1] >= a[-2]: ans += a[-1] - 1 print(ans) ```
0
602
B
Approximating a Constant Range
PROGRAMMING
1,400
[ "dp", "implementation", "two pointers" ]
null
null
When Xellos was doing a practice course in university, he once had to measure the intensity of an effect that slowly approached equilibrium. A good way to determine the equilibrium intensity would be choosing a sufficiently large number of consecutive data points that seems as constant as possible and taking their average. Of course, with the usual sizes of data, it's nothing challenging — but why not make a similar programming contest problem while we're at it? You're given a sequence of *n* data points *a*1,<=...,<=*a**n*. There aren't any big jumps between consecutive data points — for each 1<=≤<=*i*<=&lt;<=*n*, it's guaranteed that |*a**i*<=+<=1<=-<=*a**i*|<=≤<=1. A range [*l*,<=*r*] of data points is said to be almost constant if the difference between the largest and the smallest value in that range is at most 1. Formally, let *M* be the maximum and *m* the minimum value of *a**i* for *l*<=≤<=*i*<=≤<=*r*; the range [*l*,<=*r*] is almost constant if *M*<=-<=*m*<=≤<=1. Find the length of the longest almost constant range.
The first line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=100<=000) — the number of data points. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100<=000).
Print a single number — the maximum length of an almost constant range of the given sequence.
[ "5\n1 2 3 3 2\n", "11\n5 4 5 5 6 7 8 8 8 7 6\n" ]
[ "4\n", "5\n" ]
In the first sample, the longest almost constant range is [2, 5]; its length (the number of data points in it) is 4. In the second sample, there are three almost constant ranges of length 4: [1, 4], [6, 9] and [7, 10]; the only almost constant range of the maximum length 5 is [6, 10].
1,000
[ { "input": "5\n1 2 3 3 2", "output": "4" }, { "input": "11\n5 4 5 5 6 7 8 8 8 7 6", "output": "5" }, { "input": "2\n3 2", "output": "2" }, { "input": "4\n1001 1000 1000 1001", "output": "4" }, { "input": "4\n1 1 2 3", "output": "3" }, { "input": "3\n1 2 1", "output": "3" }, { "input": "3\n1 2 3", "output": "2" }, { "input": "18\n10 9 8 7 6 5 4 3 2 1 2 3 4 5 6 7 8 9", "output": "3" }, { "input": "3\n1 2 2", "output": "3" }, { "input": "4\n10 9 10 9", "output": "4" }, { "input": "4\n4 3 2 3", "output": "3" }, { "input": "4\n8 8 7 7", "output": "4" }, { "input": "3\n99998 99999 100000", "output": "2" }, { "input": "3\n100000 99999 99998", "output": "2" }, { "input": "3\n1 1 1", "output": "3" }, { "input": "2\n99999 100000", "output": "2" }, { "input": "2\n100000 100000", "output": "2" }, { "input": "2\n1 1", "output": "2" }, { "input": "15\n1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000", "output": "15" } ]
1,691,913,883
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
0
30
0
def max_almost_constant_range(n, data_points): max_length = 0 current_length = 1 for i in range(1, n): if abs(data_points[i] - data_points[i - 1]) <= 1: current_length += 1 else: max_length = max(max_length, current_length) current_length = 1 max_length = max(max_length, current_length) return max_length # Read input n = int(input()) data_points = list(map(int, input().split())) # Calculate and print the result result = max_almost_constant_range(n, data_points) print(result)
Title: Approximating a Constant Range Time Limit: None seconds Memory Limit: None megabytes Problem Description: When Xellos was doing a practice course in university, he once had to measure the intensity of an effect that slowly approached equilibrium. A good way to determine the equilibrium intensity would be choosing a sufficiently large number of consecutive data points that seems as constant as possible and taking their average. Of course, with the usual sizes of data, it's nothing challenging — but why not make a similar programming contest problem while we're at it? You're given a sequence of *n* data points *a*1,<=...,<=*a**n*. There aren't any big jumps between consecutive data points — for each 1<=≤<=*i*<=&lt;<=*n*, it's guaranteed that |*a**i*<=+<=1<=-<=*a**i*|<=≤<=1. A range [*l*,<=*r*] of data points is said to be almost constant if the difference between the largest and the smallest value in that range is at most 1. Formally, let *M* be the maximum and *m* the minimum value of *a**i* for *l*<=≤<=*i*<=≤<=*r*; the range [*l*,<=*r*] is almost constant if *M*<=-<=*m*<=≤<=1. Find the length of the longest almost constant range. Input Specification: The first line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=100<=000) — the number of data points. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100<=000). Output Specification: Print a single number — the maximum length of an almost constant range of the given sequence. Demo Input: ['5\n1 2 3 3 2\n', '11\n5 4 5 5 6 7 8 8 8 7 6\n'] Demo Output: ['4\n', '5\n'] Note: In the first sample, the longest almost constant range is [2, 5]; its length (the number of data points in it) is 4. In the second sample, there are three almost constant ranges of length 4: [1, 4], [6, 9] and [7, 10]; the only almost constant range of the maximum length 5 is [6, 10].
```python def max_almost_constant_range(n, data_points): max_length = 0 current_length = 1 for i in range(1, n): if abs(data_points[i] - data_points[i - 1]) <= 1: current_length += 1 else: max_length = max(max_length, current_length) current_length = 1 max_length = max(max_length, current_length) return max_length # Read input n = int(input()) data_points = list(map(int, input().split())) # Calculate and print the result result = max_almost_constant_range(n, data_points) print(result) ```
0
462
A
Appleman and Easy Task
PROGRAMMING
1,000
[ "brute force", "implementation" ]
null
null
Toastman came up with a very easy task. He gives it to Appleman, but Appleman doesn't know how to solve it. Can you help him? Given a *n*<=×<=*n* checkerboard. Each cell of the board has either character 'x', or character 'o'. Is it true that each cell of the board has even number of adjacent cells with 'o'? Two cells of the board are adjacent if they share a side.
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Then *n* lines follow containing the description of the checkerboard. Each of them contains *n* characters (either 'x' or 'o') without spaces.
Print "YES" or "NO" (without the quotes) depending on the answer to the problem.
[ "3\nxxo\nxox\noxx\n", "4\nxxxo\nxoxo\noxox\nxxxx\n" ]
[ "YES\n", "NO\n" ]
none
500
[ { "input": "3\nxxo\nxox\noxx", "output": "YES" }, { "input": "4\nxxxo\nxoxo\noxox\nxxxx", "output": "NO" }, { "input": "1\no", "output": "YES" }, { "input": "2\nox\nxo", "output": "YES" }, { "input": "2\nxx\nxo", "output": "NO" }, { "input": "3\nooo\noxo\nxoo", "output": "NO" }, { "input": "3\nxxx\nxxo\nxxo", "output": "NO" }, { "input": "4\nxooo\nooxo\noxoo\nooox", "output": "YES" }, { "input": "4\noooo\noxxo\nxoxo\noooo", "output": "NO" }, { "input": "5\noxoxo\nxxxxx\noxoxo\nxxxxx\noxoxo", "output": "YES" }, { "input": "5\nxxxox\nxxxxo\nxoxox\noxoxx\nxoxxx", "output": "NO" }, { "input": "10\nxoxooooooo\noxxoxxxxxo\nxxooxoooxo\noooxxoxoxo\noxxxooooxo\noxooooxxxo\noxoxoxxooo\noxoooxooxx\noxxxxxoxxo\noooooooxox", "output": "YES" }, { "input": "10\nxxxxxxxoox\nxooxxooooo\noxoooxxooo\nxoxxxxxxxx\nxxoxooxxox\nooxoxxooox\nooxxxxxooo\nxxxxoxooox\nxoxxooxxxx\noooooxxoxo", "output": "NO" }, { "input": "19\noxoxoxoxooxoooxxoox\nxxxxxxxxoxxoxoooooo\noxoxoxooxxxooxxxooo\nxxoxxxooxooxxxoxxox\noxoxooxxxooooxxoxox\nxxxoooxoxxoxxoxxxoo\noxooxxxoooooxxoooxo\nxxooxooxoxxoxxoxxoo\noxxxxooooxxxooooxxx\nooxooxoxxoxxoxooxoo\nxxxooooxxxooooxoxox\noooxoxooxxoxooxooxx\nxxoooxxxooooxxoooxo\nooxxxooxoxooxooxxxx\nxoxoxxooxoxxxooxoxo\nxoxxoxoxooxooxxxxxx\noooxxxooxxxooxoxoxo\nxoooooxoxooxxxxxxxo\nxooxxoooxxoxoxoxoxx", "output": "NO" }, { "input": "12\nxxooxxoxxxoo\nxxoooxoxoxoo\nooxoxoxxooxx\nooxxooooxoxo\nxxxxxxxxoxxx\noxooooxxxooo\noxxoxoxoooxx\nxxxxxxxooxox\noxoooooxoxxx\nxxooxxoxxoxx\noxxxxxxxooxx\nooxoxooxxooo", "output": "NO" }, { "input": "2\noo\nxx", "output": "NO" } ]
1,680,599,439
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
5
30
0
def filtring_outside_the_grid(r , c , row , col ) : return 0<= r <row and 0<= c < col def matrix_generator () : row = int(input()) matrix = ["0"] * row for rowindx in range(row) : matrix[rowindx] = [i for i in input()] return matrix , row def all_postions (i , j , row , col , cnt = 4) : mov_i = [1 , 0 , -1 , 0 ] mov_j = [0 , 1 , 0 ,-1] return [(r,c) for d in range(cnt) if filtring_outside_the_grid( r:= i + mov_i[d] , c:= j +mov_j[d],row,col )] def main() : flag = 'YES' matrix , row = matrix_generator() col = row for r in range(row) : counter = 0 for c in range(col) : all_pos =all_postions(r,c,row,col) for i , j in all_pos: if matrix[i][j] =="o" and matrix[r][c] =="x": counter += 1 if counter % 2 != 0 : flag = "NO" break print(flag) main()
Title: Appleman and Easy Task Time Limit: None seconds Memory Limit: None megabytes Problem Description: Toastman came up with a very easy task. He gives it to Appleman, but Appleman doesn't know how to solve it. Can you help him? Given a *n*<=×<=*n* checkerboard. Each cell of the board has either character 'x', or character 'o'. Is it true that each cell of the board has even number of adjacent cells with 'o'? Two cells of the board are adjacent if they share a side. Input Specification: The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Then *n* lines follow containing the description of the checkerboard. Each of them contains *n* characters (either 'x' or 'o') without spaces. Output Specification: Print "YES" or "NO" (without the quotes) depending on the answer to the problem. Demo Input: ['3\nxxo\nxox\noxx\n', '4\nxxxo\nxoxo\noxox\nxxxx\n'] Demo Output: ['YES\n', 'NO\n'] Note: none
```python def filtring_outside_the_grid(r , c , row , col ) : return 0<= r <row and 0<= c < col def matrix_generator () : row = int(input()) matrix = ["0"] * row for rowindx in range(row) : matrix[rowindx] = [i for i in input()] return matrix , row def all_postions (i , j , row , col , cnt = 4) : mov_i = [1 , 0 , -1 , 0 ] mov_j = [0 , 1 , 0 ,-1] return [(r,c) for d in range(cnt) if filtring_outside_the_grid( r:= i + mov_i[d] , c:= j +mov_j[d],row,col )] def main() : flag = 'YES' matrix , row = matrix_generator() col = row for r in range(row) : counter = 0 for c in range(col) : all_pos =all_postions(r,c,row,col) for i , j in all_pos: if matrix[i][j] =="o" and matrix[r][c] =="x": counter += 1 if counter % 2 != 0 : flag = "NO" break print(flag) main() ```
0
5
C
Longest Regular Bracket Sequence
PROGRAMMING
1,900
[ "constructive algorithms", "data structures", "dp", "greedy", "sortings", "strings" ]
C. Longest Regular Bracket Sequence
2
256
This is yet another problem dealing with regular bracket sequences. We should remind you that a bracket sequence is called regular, if by inserting «+» and «1» into it we can get a correct mathematical expression. For example, sequences «(())()», «()» and «(()(()))» are regular, while «)(», «(()» and «(()))(» are not. You are given a string of «(» and «)» characters. You are to find its longest substring that is a regular bracket sequence. You are to find the number of such substrings as well.
The first line of the input file contains a non-empty string, consisting of «(» and «)» characters. Its length does not exceed 106.
Print the length of the longest substring that is a regular bracket sequence, and the number of such substrings. If there are no such substrings, write the only line containing "0 1".
[ ")((())))(()())\n", "))(\n" ]
[ "6 2\n", "0 1\n" ]
none
0
[ { "input": ")((())))(()())", "output": "6 2" }, { "input": "))(", "output": "0 1" }, { "input": "()(())()", "output": "8 1" }, { "input": "((((()(((", "output": "2 1" }, { "input": "))))()())))", "output": "4 1" }, { "input": "(()())()(())()()())())()((()(()(())()()())((()(())()(()()()()))()(())()(((()())()(()((())()(())(()))", "output": "28 1" }, { "input": ")(()(((((())))))(((()))))))((()(((((((())())()))))((((((((())))((((((())))))(()((((()))))(()))))))))", "output": "68 1" }, { "input": "())(((((())())((((()))(())))())())(((()(()()()())(())()))((()(())())()()))()(()())))))(()))((())((((", "output": "80 1" }, { "input": "))(((((((()((((())(((((((((()()))))(()())()((())())))((((()))(()()()))))(()()))())))))((())())()))((", "output": "92 1" } ]
1,553,851,903
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
5
218
409,600
from collections import Counter as cr st=input() c=0 s=[] ans=[] for i in st: if i=='(': if len(s)==0: ans.append(c) c=0 s.append(i) else: if len(s)!=0: c+=2 s.pop() else: ans.append(c) c=0 ans.append(')') ans.append(c) t=[0] for i in ans: if i==')': t.append(0) else: t.append(i+t[-1]) m=cr(t) c=max(t) if c==0: print("0 1") else: print(c,m[c])
Title: Longest Regular Bracket Sequence Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: This is yet another problem dealing with regular bracket sequences. We should remind you that a bracket sequence is called regular, if by inserting «+» and «1» into it we can get a correct mathematical expression. For example, sequences «(())()», «()» and «(()(()))» are regular, while «)(», «(()» and «(()))(» are not. You are given a string of «(» and «)» characters. You are to find its longest substring that is a regular bracket sequence. You are to find the number of such substrings as well. Input Specification: The first line of the input file contains a non-empty string, consisting of «(» and «)» characters. Its length does not exceed 106. Output Specification: Print the length of the longest substring that is a regular bracket sequence, and the number of such substrings. If there are no such substrings, write the only line containing "0 1". Demo Input: [')((())))(()())\n', '))(\n'] Demo Output: ['6 2\n', '0 1\n'] Note: none
```python from collections import Counter as cr st=input() c=0 s=[] ans=[] for i in st: if i=='(': if len(s)==0: ans.append(c) c=0 s.append(i) else: if len(s)!=0: c+=2 s.pop() else: ans.append(c) c=0 ans.append(')') ans.append(c) t=[0] for i in ans: if i==')': t.append(0) else: t.append(i+t[-1]) m=cr(t) c=max(t) if c==0: print("0 1") else: print(c,m[c]) ```
0
263
A
Beautiful Matrix
PROGRAMMING
800
[ "implementation" ]
null
null
You've got a 5<=×<=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<=≤<=*i*<=&lt;<=5). 1. Swap two neighboring matrix columns, that is, columns with indexes *j* and *j*<=+<=1 for some integer *j* (1<=≤<=*j*<=&lt;<=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 — 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" } ]
1,694,964,760
2,147,483,647
Python 3
RUNTIME_ERROR
TESTS
0
30
0
marrix=[] for i in range(5): row=list(map(int,input().split())) matrix.append(row) row_index,col_index=None,None for i in range(5): for j in range(5): if matrix[i][j] == 1: row_index,col_index=i,j moves=abs(2-row_index)+abs(2-col_index) print(moves)
Title: Beautiful Matrix Time Limit: None seconds Memory Limit: None megabytes Problem Description: You've got a 5<=×<=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<=≤<=*i*<=&lt;<=5). 1. Swap two neighboring matrix columns, that is, columns with indexes *j* and *j*<=+<=1 for some integer *j* (1<=≤<=*j*<=&lt;<=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 — 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 marrix=[] for i in range(5): row=list(map(int,input().split())) matrix.append(row) row_index,col_index=None,None for i in range(5): for j in range(5): if matrix[i][j] == 1: row_index,col_index=i,j moves=abs(2-row_index)+abs(2-col_index) print(moves) ```
-1
989
B
A Tide of Riverscape
PROGRAMMING
1,200
[ "constructive algorithms", "strings" ]
null
null
"Time," Mino thinks aloud. "What?" "Time and tide wait for no man," explains Mino. "My name, taken from the river, always reminds me of this." "And what are you recording?" "You see it, tide. Everything has its own period, and I think I've figured out this one," says Mino with confidence. Doubtfully, Kanno peeks at Mino's records. The records are expressed as a string $s$ of characters '0', '1' and '.', where '0' denotes a low tide, '1' denotes a high tide, and '.' denotes an unknown one (either high or low). You are to help Mino determine whether it's possible that after replacing each '.' independently with '0' or '1', a given integer $p$ is not a period of the resulting string. In case the answer is yes, please also show such a replacement to Mino. In this problem, a positive integer $p$ is considered a period of string $s$, if for all $1 \leq i \leq \lvert s \rvert - p$, the $i$-th and $(i + p)$-th characters of $s$ are the same. Here $\lvert s \rvert$ is the length of $s$.
The first line contains two space-separated integers $n$ and $p$ ($1 \leq p \leq n \leq 2000$) — the length of the given string and the supposed period, respectively. The second line contains a string $s$ of $n$ characters — Mino's records. $s$ only contains characters '0', '1' and '.', and contains at least one '.' character.
Output one line — if it's possible that $p$ is not a period of the resulting string, output any one of such strings; otherwise output "No" (without quotes, you can print letters in any case (upper or lower)).
[ "10 7\n1.0.1.0.1.\n", "10 6\n1.0.1.1000\n", "10 9\n1........1\n" ]
[ "1000100010\n", "1001101000\n", "No\n" ]
In the first example, $7$ is not a period of the resulting string because the $1$-st and $8$-th characters of it are different. In the second example, $6$ is not a period of the resulting string because the $4$-th and $10$-th characters of it are different. In the third example, $9$ is always a period because the only constraint that the first and last characters are the same is already satisfied. Note that there are multiple acceptable answers for the first two examples, you can print any of them.
1,000
[ { "input": "10 7\n1.0.1.0.1.", "output": "1000100010" }, { "input": "10 6\n1.0.1.1000", "output": "1001101000" }, { "input": "10 9\n1........1", "output": "No" }, { "input": "1 1\n.", "output": "No" }, { "input": "5 1\n0...1", "output": "00001" }, { "input": "17 10\n..1.100..1..0.100", "output": "00101000010000100" }, { "input": "2 1\n0.", "output": "01" }, { "input": "2 1\n..", "output": "01" }, { "input": "3 1\n.0.", "output": "001" }, { "input": "3 1\n00.", "output": "001" }, { "input": "3 2\n0..", "output": "001" }, { "input": "3 2\n0.0", "output": "No" }, { "input": "3 2\n1..", "output": "100" }, { "input": "3 2\n.1.", "output": "011" }, { "input": "3 2\n1.0", "output": "100" }, { "input": "3 3\n1..", "output": "No" }, { "input": "3 3\n.00", "output": "No" }, { "input": "5 3\n0.000", "output": "01000" }, { "input": "10 6\n10010.1001", "output": "No" }, { "input": "75 38\n00.0.1.0.0110.1.00010..100.1110..110..00.0.1.0.0110.1.00010..100.1110..110.", "output": "000001000011001000010001000111000110000000010000110010000100010001110001101" }, { "input": "128 108\n01100.110...000.0001.1.11.11.010010.01100.0.1.01.0.0011.11001.000101...1.0.0..100.0110.0110.0.0101.0.0.0001.01100.110...100.0001", "output": "01100011000000000001010110110010010001100000100100000110110010000101000100000010000110001100000101000000001001100011000010000001" }, { "input": "5 4\n.101.", "output": "01011" }, { "input": "4 2\n101.", "output": "1011" }, { "input": "5 4\n.1011", "output": "01011" }, { "input": "2 1\n..", "output": "01" }, { "input": "5 3\n00.11", "output": "00011" }, { "input": "10 8\n1111.00000", "output": "1111000000" }, { "input": "10 3\n11111111.1", "output": "1111111101" }, { "input": "3 2\n1.0", "output": "100" }, { "input": "6 4\n11..10", "output": "110010" }, { "input": "4 2\n.111", "output": "0111" }, { "input": "3 2\n01.", "output": "011" }, { "input": "5 4\n10.00", "output": "10000" }, { "input": "10 9\n1........0", "output": "1000000000" }, { "input": "2 1\n0.", "output": "01" }, { "input": "8 4\n111111..", "output": "11111100" }, { "input": "3 2\n0.1", "output": "001" }, { "input": "4 1\n111.", "output": "1110" }, { "input": "3 1\n01.", "output": "010" }, { "input": "10 7\n000....111", "output": "0000000111" } ]
1,530,039,640
2,147,483,647
PyPy 3
WRONG_ANSWER
TESTS
6
109
204,800
p,n = map(int,input().split()) s = input() flag = 0 pos = 0 for j in range(p-n): a = s[j] for i in range(n,len(s),n+1): if(s[i] != a ): flag = 1 pos = i break if(flag == 1): t = "" for i in range(p): if(s[i] == '.'): if(i == pos): if(a == '0'): t = t + '1' else: t = t + '0' else: t = t + '1' else: t = t + s[i] print(t) break if(flag == 0): print("No")
Title: A Tide of Riverscape Time Limit: None seconds Memory Limit: None megabytes Problem Description: "Time," Mino thinks aloud. "What?" "Time and tide wait for no man," explains Mino. "My name, taken from the river, always reminds me of this." "And what are you recording?" "You see it, tide. Everything has its own period, and I think I've figured out this one," says Mino with confidence. Doubtfully, Kanno peeks at Mino's records. The records are expressed as a string $s$ of characters '0', '1' and '.', where '0' denotes a low tide, '1' denotes a high tide, and '.' denotes an unknown one (either high or low). You are to help Mino determine whether it's possible that after replacing each '.' independently with '0' or '1', a given integer $p$ is not a period of the resulting string. In case the answer is yes, please also show such a replacement to Mino. In this problem, a positive integer $p$ is considered a period of string $s$, if for all $1 \leq i \leq \lvert s \rvert - p$, the $i$-th and $(i + p)$-th characters of $s$ are the same. Here $\lvert s \rvert$ is the length of $s$. Input Specification: The first line contains two space-separated integers $n$ and $p$ ($1 \leq p \leq n \leq 2000$) — the length of the given string and the supposed period, respectively. The second line contains a string $s$ of $n$ characters — Mino's records. $s$ only contains characters '0', '1' and '.', and contains at least one '.' character. Output Specification: Output one line — if it's possible that $p$ is not a period of the resulting string, output any one of such strings; otherwise output "No" (without quotes, you can print letters in any case (upper or lower)). Demo Input: ['10 7\n1.0.1.0.1.\n', '10 6\n1.0.1.1000\n', '10 9\n1........1\n'] Demo Output: ['1000100010\n', '1001101000\n', 'No\n'] Note: In the first example, $7$ is not a period of the resulting string because the $1$-st and $8$-th characters of it are different. In the second example, $6$ is not a period of the resulting string because the $4$-th and $10$-th characters of it are different. In the third example, $9$ is always a period because the only constraint that the first and last characters are the same is already satisfied. Note that there are multiple acceptable answers for the first two examples, you can print any of them.
```python p,n = map(int,input().split()) s = input() flag = 0 pos = 0 for j in range(p-n): a = s[j] for i in range(n,len(s),n+1): if(s[i] != a ): flag = 1 pos = i break if(flag == 1): t = "" for i in range(p): if(s[i] == '.'): if(i == pos): if(a == '0'): t = t + '1' else: t = t + '0' else: t = t + '1' else: t = t + s[i] print(t) break if(flag == 0): print("No") ```
0
69
A
Young Physicist
PROGRAMMING
1,000
[ "implementation", "math" ]
A. Young Physicist
2
256
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100).
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
[ "3\n4 1 7\n-2 4 -1\n1 -5 -3\n", "3\n3 -1 7\n-5 2 -4\n2 -1 -3\n" ]
[ "NO", "YES" ]
none
500
[ { "input": "3\n4 1 7\n-2 4 -1\n1 -5 -3", "output": "NO" }, { "input": "3\n3 -1 7\n-5 2 -4\n2 -1 -3", "output": "YES" }, { "input": "10\n21 32 -46\n43 -35 21\n42 2 -50\n22 40 20\n-27 -9 38\n-4 1 1\n-40 6 -31\n-13 -2 34\n-21 34 -12\n-32 -29 41", "output": "NO" }, { "input": "10\n25 -33 43\n-27 -42 28\n-35 -20 19\n41 -42 -1\n49 -39 -4\n-49 -22 7\n-19 29 41\n8 -27 -43\n8 34 9\n-11 -3 33", "output": "NO" }, { "input": "10\n-6 21 18\n20 -11 -8\n37 -11 41\n-5 8 33\n29 23 32\n30 -33 -11\n39 -49 -36\n28 34 -49\n22 29 -34\n-18 -6 7", "output": "NO" }, { "input": "10\n47 -2 -27\n0 26 -14\n5 -12 33\n2 18 3\n45 -30 -49\n4 -18 8\n-46 -44 -41\n-22 -10 -40\n-35 -21 26\n33 20 38", "output": "NO" }, { "input": "13\n-3 -36 -46\n-11 -50 37\n42 -11 -15\n9 42 44\n-29 -12 24\n3 9 -40\n-35 13 50\n14 43 18\n-13 8 24\n-48 -15 10\n50 9 -50\n21 0 -50\n0 0 -6", "output": "YES" }, { "input": "14\n43 23 17\n4 17 44\n5 -5 -16\n-43 -7 -6\n47 -48 12\n50 47 -45\n2 14 43\n37 -30 15\n4 -17 -11\n17 9 -45\n-50 -3 -8\n-50 0 0\n-50 0 0\n-16 0 0", "output": "YES" }, { "input": "13\n29 49 -11\n38 -11 -20\n25 1 -40\n-11 28 11\n23 -19 1\n45 -41 -17\n-3 0 -19\n-13 -33 49\n-30 0 28\n34 17 45\n-50 9 -27\n-50 0 0\n-37 0 0", "output": "YES" }, { "input": "12\n3 28 -35\n-32 -44 -17\n9 -25 -6\n-42 -22 20\n-19 15 38\n-21 38 48\n-1 -37 -28\n-10 -13 -50\n-5 21 29\n34 28 50\n50 11 -49\n34 0 0", "output": "YES" }, { "input": "37\n-64 -79 26\n-22 59 93\n-5 39 -12\n77 -9 76\n55 -86 57\n83 100 -97\n-70 94 84\n-14 46 -94\n26 72 35\n14 78 -62\n17 82 92\n-57 11 91\n23 15 92\n-80 -1 1\n12 39 18\n-23 -99 -75\n-34 50 19\n-39 84 -7\n45 -30 -39\n-60 49 37\n45 -16 -72\n33 -51 -56\n-48 28 5\n97 91 88\n45 -82 -11\n-21 -15 -90\n-53 73 -26\n-74 85 -90\n-40 23 38\n100 -13 49\n32 -100 -100\n0 -100 -70\n0 -100 0\n0 -100 0\n0 -100 0\n0 -100 0\n0 -37 0", "output": "YES" }, { "input": "4\n68 3 100\n68 21 -100\n-100 -24 0\n-36 0 0", "output": "YES" }, { "input": "33\n-1 -46 -12\n45 -16 -21\n-11 45 -21\n-60 -42 -93\n-22 -45 93\n37 96 85\n-76 26 83\n-4 9 55\n7 -52 -9\n66 8 -85\n-100 -54 11\n-29 59 74\n-24 12 2\n-56 81 85\n-92 69 -52\n-26 -97 91\n54 59 -51\n58 21 -57\n7 68 56\n-47 -20 -51\n-59 77 -13\n-85 27 91\n79 60 -56\n66 -80 5\n21 -99 42\n-31 -29 98\n66 93 76\n-49 45 61\n100 -100 -100\n100 -100 -100\n66 -75 -100\n0 0 -100\n0 0 -87", "output": "YES" }, { "input": "3\n1 2 3\n3 2 1\n0 0 0", "output": "NO" }, { "input": "2\n5 -23 12\n0 0 0", "output": "NO" }, { "input": "1\n0 0 0", "output": "YES" }, { "input": "1\n1 -2 0", "output": "NO" }, { "input": "2\n-23 77 -86\n23 -77 86", "output": "YES" }, { "input": "26\n86 7 20\n-57 -64 39\n-45 6 -93\n-44 -21 100\n-11 -49 21\n73 -71 -80\n-2 -89 56\n-65 -2 7\n5 14 84\n57 41 13\n-12 69 54\n40 -25 27\n-17 -59 0\n64 -91 -30\n-53 9 42\n-54 -8 14\n-35 82 27\n-48 -59 -80\n88 70 79\n94 57 97\n44 63 25\n84 -90 -40\n-100 100 -100\n-92 100 -100\n0 10 -100\n0 0 -82", "output": "YES" }, { "input": "42\n11 27 92\n-18 -56 -57\n1 71 81\n33 -92 30\n82 83 49\n-87 -61 -1\n-49 45 49\n73 26 15\n-22 22 -77\n29 -93 87\n-68 44 -90\n-4 -84 20\n85 67 -6\n-39 26 77\n-28 -64 20\n65 -97 24\n-72 -39 51\n35 -75 -91\n39 -44 -8\n-25 -27 -57\n91 8 -46\n-98 -94 56\n94 -60 59\n-9 -95 18\n-53 -37 98\n-8 -94 -84\n-52 55 60\n15 -14 37\n65 -43 -25\n94 12 66\n-8 -19 -83\n29 81 -78\n-58 57 33\n24 86 -84\n-53 32 -88\n-14 7 3\n89 97 -53\n-5 -28 -91\n-100 100 -6\n-84 100 0\n0 100 0\n0 70 0", "output": "YES" }, { "input": "3\n96 49 -12\n2 -66 28\n-98 17 -16", "output": "YES" }, { "input": "5\n70 -46 86\n-100 94 24\n-27 63 -63\n57 -100 -47\n0 -11 0", "output": "YES" }, { "input": "18\n-86 -28 70\n-31 -89 42\n31 -48 -55\n95 -17 -43\n24 -95 -85\n-21 -14 31\n68 -18 81\n13 31 60\n-15 28 99\n-42 15 9\n28 -61 -62\n-16 71 29\n-28 75 -48\n-77 -67 36\n-100 83 89\n100 100 -100\n57 34 -100\n0 0 -53", "output": "YES" }, { "input": "44\n52 -54 -29\n-82 -5 -94\n-54 43 43\n91 16 71\n7 80 -91\n3 15 29\n-99 -6 -77\n-3 -77 -64\n73 67 34\n25 -10 -18\n-29 91 63\n-72 86 -16\n-68 85 -81\n-3 36 44\n-74 -14 -80\n34 -96 -97\n-76 -78 -33\n-24 44 -58\n98 12 77\n95 -63 -6\n-51 3 -90\n-92 -10 72\n7 3 -68\n57 -53 71\n29 57 -48\n35 -60 10\n79 -70 -61\n-20 77 55\n-86 -15 -35\n84 -88 -18\n100 -42 77\n-20 46 8\n-41 -43 -65\n38 -98 -23\n-100 65 45\n-7 -91 -63\n46 88 -85\n48 59 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 1", "output": "YES" }, { "input": "18\n-14 -64 -91\n-8 -66 -86\n-23 92 -40\n6 -3 -53\n57 41 78\n-79 42 -22\n-88 -17 45\n4 -45 44\n83 -18 -25\n34 86 -92\n75 -30 12\n44 99 11\n-67 -13 72\n22 83 -56\n-37 71 72\n-9 -100 100\n0 -100 31\n0 -58 0", "output": "YES" }, { "input": "23\n-70 37 78\n42 84 6\n28 -94 -24\n-49 76 95\n-67 18 84\n-53 78 -5\n65 -63 -64\n-66 -64 -69\n81 -93 95\n10 52 -79\n-89 -61 -64\n-64 -47 43\n-81 -35 55\n80 82 73\n-60 -81 -18\n49 -10 -19\n-58 70 12\n-24 -15 -93\n98 -93 -54\n-28 -75 11\n100 100 -63\n100 100 0\n56 34 0", "output": "YES" }, { "input": "3\n0 2 -2\n1 -1 3\n-3 0 0", "output": "NO" } ]
1,666,765,513
2,147,483,647
PyPy 3-64
OK
TESTS
81
124
0
n=int(input()) x0,y0,z0=0,0,0 for i in range(n): x,y,z=[int(s) for s in input().split()] x0+=x y0+=y z0+=z if(x0==y0==z0==0): print("YES") else: print("NO")
Title: Young Physicist Time Limit: 2 seconds Memory Limit: 256 megabytes Problem Description: A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces. Input Specification: The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100). Output Specification: Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not. Demo Input: ['3\n4 1 7\n-2 4 -1\n1 -5 -3\n', '3\n3 -1 7\n-5 2 -4\n2 -1 -3\n'] Demo Output: ['NO', 'YES'] Note: none
```python n=int(input()) x0,y0,z0=0,0,0 for i in range(n): x,y,z=[int(s) for s in input().split()] x0+=x y0+=y z0+=z if(x0==y0==z0==0): print("YES") else: print("NO") ```
3.969
264
A
Escape from Stones
PROGRAMMING
1,200
[ "constructive algorithms", "data structures", "implementation", "two pointers" ]
null
null
Squirrel Liss lived in a forest peacefully, but unexpected trouble happens. Stones fall from a mountain. Initially Squirrel Liss occupies an interval [0,<=1]. Next, *n* stones will fall and Liss will escape from the stones. The stones are numbered from 1 to *n* in order. The stones always fall to the center of Liss's interval. When Liss occupies the interval [*k*<=-<=*d*,<=*k*<=+<=*d*] and a stone falls to *k*, she will escape to the left or to the right. If she escapes to the left, her new interval will be [*k*<=-<=*d*,<=*k*]. If she escapes to the right, her new interval will be [*k*,<=*k*<=+<=*d*]. You are given a string *s* of length *n*. If the *i*-th character of *s* is "l" or "r", when the *i*-th stone falls Liss will escape to the left or to the right, respectively. Find the sequence of stones' numbers from left to right after all the *n* stones falls.
The input consists of only one line. The only line contains the string *s* (1<=≤<=|*s*|<=≤<=106). Each character in *s* will be either "l" or "r".
Output *n* lines — on the *i*-th line you should print the *i*-th stone's number from the left.
[ "llrlr\n", "rrlll\n", "lrlrr\n" ]
[ "3\n5\n4\n2\n1\n", "1\n2\n5\n4\n3\n", "2\n4\n5\n3\n1\n" ]
In the first example, the positions of stones 1, 2, 3, 4, 5 will be <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/58fdb5684df807bfcb705a9da9ce175613362b7d.png" style="max-width: 100.0%;max-height: 100.0%;"/>, respectively. So you should print the sequence: 3, 5, 4, 2, 1.
500
[ { "input": "llrlr", "output": "3\n5\n4\n2\n1" }, { "input": "rrlll", "output": "1\n2\n5\n4\n3" }, { "input": "lrlrr", "output": "2\n4\n5\n3\n1" }, { "input": "lllrlrllrl", "output": "4\n6\n9\n10\n8\n7\n5\n3\n2\n1" }, { "input": "llrlrrrlrr", "output": "3\n5\n6\n7\n9\n10\n8\n4\n2\n1" }, { "input": "rlrrrllrrr", "output": "1\n3\n4\n5\n8\n9\n10\n7\n6\n2" }, { "input": "lrrlrrllrrrrllllllrr", "output": "2\n3\n5\n6\n9\n10\n11\n12\n19\n20\n18\n17\n16\n15\n14\n13\n8\n7\n4\n1" }, { "input": "rlrrrlrrrllrrllrlrll", "output": "1\n3\n4\n5\n7\n8\n9\n12\n13\n16\n18\n20\n19\n17\n15\n14\n11\n10\n6\n2" }, { "input": "lllrrlrlrllrrrrrllrl", "output": "4\n5\n7\n9\n12\n13\n14\n15\n16\n19\n20\n18\n17\n11\n10\n8\n6\n3\n2\n1" }, { "input": "rrrllrrrlllrlllrlrrr", "output": "1\n2\n3\n6\n7\n8\n12\n16\n18\n19\n20\n17\n15\n14\n13\n11\n10\n9\n5\n4" }, { "input": "rrlllrrrlrrlrrrlllrlrlrrrlllrllrrllrllrrlrlrrllllrlrrrrlrlllrlrrrlrlrllrlrlrrlrrllrrrlrlrlllrrllllrl", "output": "1\n2\n6\n7\n8\n10\n11\n13\n14\n15\n19\n21\n23\n24\n25\n29\n32\n33\n36\n39\n40\n42\n44\n45\n50\n52\n53\n54\n55\n57\n61\n63\n64\n65\n67\n69\n72\n74\n76\n77\n79\n80\n83\n84\n85\n87\n89\n93\n94\n99\n100\n98\n97\n96\n95\n92\n91\n90\n88\n86\n82\n81\n78\n75\n73\n71\n70\n68\n66\n62\n60\n59\n58\n56\n51\n49\n48\n47\n46\n43\n41\n38\n37\n35\n34\n31\n30\n28\n27\n26\n22\n20\n18\n17\n16\n12\n9\n5\n4\n3" }, { "input": "llrlrlllrrllrllllrlrrlrlrrllrlrlrrlrrrrrrlllrrlrrrrrlrrrlrlrlrrlllllrrrrllrrlrlrrrllllrlrrlrrlrlrrll", "output": "3\n5\n9\n10\n13\n18\n20\n21\n23\n25\n26\n29\n31\n33\n34\n36\n37\n38\n39\n40\n41\n45\n46\n48\n49\n50\n51\n52\n54\n55\n56\n58\n60\n62\n63\n69\n70\n71\n72\n75\n76\n78\n80\n81\n82\n87\n89\n90\n92\n93\n95\n97\n98\n100\n99\n96\n94\n91\n88\n86\n85\n84\n83\n79\n77\n74\n73\n68\n67\n66\n65\n64\n61\n59\n57\n53\n47\n44\n43\n42\n35\n32\n30\n28\n27\n24\n22\n19\n17\n16\n15\n14\n12\n11\n8\n7\n6\n4\n2\n1" }, { "input": "llrrrrllrrlllrlrllrlrllllllrrrrrrrrllrrrrrrllrlrrrlllrrrrrrlllllllrrlrrllrrrllllrrlllrrrlrlrrlrlrllr", "output": "3\n4\n5\n6\n9\n10\n14\n16\n19\n21\n28\n29\n30\n31\n32\n33\n34\n35\n38\n39\n40\n41\n42\n43\n46\n48\n49\n50\n54\n55\n56\n57\n58\n59\n67\n68\n70\n71\n74\n75\n76\n81\n82\n86\n87\n88\n90\n92\n93\n95\n97\n100\n99\n98\n96\n94\n91\n89\n85\n84\n83\n80\n79\n78\n77\n73\n72\n69\n66\n65\n64\n63\n62\n61\n60\n53\n52\n51\n47\n45\n44\n37\n36\n27\n26\n25\n24\n23\n22\n20\n18\n17\n15\n13\n12\n11\n8\n7\n2\n1" }, { "input": "lllllrllrrlllrrrllrrrrlrrlrllllrrrrrllrlrllllllrrlrllrlrllrlrrlrlrrlrrrlrrrrllrlrrrrrrrllrllrrlrllrl", "output": "6\n9\n10\n14\n15\n16\n19\n20\n21\n22\n24\n25\n27\n32\n33\n34\n35\n36\n39\n41\n48\n49\n51\n54\n56\n59\n61\n62\n64\n66\n67\n69\n70\n71\n73\n74\n75\n76\n79\n81\n82\n83\n84\n85\n86\n87\n90\n93\n94\n96\n99\n100\n98\n97\n95\n92\n91\n89\n88\n80\n78\n77\n72\n68\n65\n63\n60\n58\n57\n55\n53\n52\n50\n47\n46\n45\n44\n43\n42\n40\n38\n37\n31\n30\n29\n28\n26\n23\n18\n17\n13\n12\n11\n8\n7\n5\n4\n3\n2\n1" }, { "input": "llrlrlrlrlrlrrlllllllrllllrllrrrlllrrllrllrrlllrrlllrlrrllllrrlllrrllrrllllrrlllrlllrrrllrrrrrrllrrl", "output": "3\n5\n7\n9\n11\n13\n14\n22\n27\n30\n31\n32\n36\n37\n40\n43\n44\n48\n49\n53\n55\n56\n61\n62\n66\n67\n70\n71\n76\n77\n81\n85\n86\n87\n90\n91\n92\n93\n94\n95\n98\n99\n100\n97\n96\n89\n88\n84\n83\n82\n80\n79\n78\n75\n74\n73\n72\n69\n68\n65\n64\n63\n60\n59\n58\n57\n54\n52\n51\n50\n47\n46\n45\n42\n41\n39\n38\n35\n34\n33\n29\n28\n26\n25\n24\n23\n21\n20\n19\n18\n17\n16\n15\n12\n10\n8\n6\n4\n2\n1" }, { "input": "l", "output": "1" }, { "input": "r", "output": "1" } ]
1,557,118,860
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
10
93
0
s=input() n=len(s) prevl=0 prevr=1 pos=[] pos.append([0,1/2]) curr=1/2 for i in range(0,n-1): if s[i]=='l': pos.append([i+1,(prevl+curr)/2]) prevr=curr curr=(prevl+curr)/2 else: pos.append([i+1,(curr+prevr)/2]) prevl=curr curr=(curr+prevr)/2 pos.sort(key=lambda x:x[1]) for i in range(n): print(pos[i][0]+1)
Title: Escape from Stones Time Limit: None seconds Memory Limit: None megabytes Problem Description: Squirrel Liss lived in a forest peacefully, but unexpected trouble happens. Stones fall from a mountain. Initially Squirrel Liss occupies an interval [0,<=1]. Next, *n* stones will fall and Liss will escape from the stones. The stones are numbered from 1 to *n* in order. The stones always fall to the center of Liss's interval. When Liss occupies the interval [*k*<=-<=*d*,<=*k*<=+<=*d*] and a stone falls to *k*, she will escape to the left or to the right. If she escapes to the left, her new interval will be [*k*<=-<=*d*,<=*k*]. If she escapes to the right, her new interval will be [*k*,<=*k*<=+<=*d*]. You are given a string *s* of length *n*. If the *i*-th character of *s* is "l" or "r", when the *i*-th stone falls Liss will escape to the left or to the right, respectively. Find the sequence of stones' numbers from left to right after all the *n* stones falls. Input Specification: The input consists of only one line. The only line contains the string *s* (1<=≤<=|*s*|<=≤<=106). Each character in *s* will be either "l" or "r". Output Specification: Output *n* lines — on the *i*-th line you should print the *i*-th stone's number from the left. Demo Input: ['llrlr\n', 'rrlll\n', 'lrlrr\n'] Demo Output: ['3\n5\n4\n2\n1\n', '1\n2\n5\n4\n3\n', '2\n4\n5\n3\n1\n'] Note: In the first example, the positions of stones 1, 2, 3, 4, 5 will be <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/58fdb5684df807bfcb705a9da9ce175613362b7d.png" style="max-width: 100.0%;max-height: 100.0%;"/>, respectively. So you should print the sequence: 3, 5, 4, 2, 1.
```python s=input() n=len(s) prevl=0 prevr=1 pos=[] pos.append([0,1/2]) curr=1/2 for i in range(0,n-1): if s[i]=='l': pos.append([i+1,(prevl+curr)/2]) prevr=curr curr=(prevl+curr)/2 else: pos.append([i+1,(curr+prevr)/2]) prevl=curr curr=(curr+prevr)/2 pos.sort(key=lambda x:x[1]) for i in range(n): print(pos[i][0]+1) ```
0
4
A
Watermelon
PROGRAMMING
800
[ "brute force", "math" ]
A. Watermelon
1
64
One hot summer day Pete and his friend Billy decided to buy a watermelon. They chose the biggest and the ripest one, in their opinion. After that the watermelon was weighed, and the scales showed *w* kilos. They rushed home, dying of thirst, and decided to divide the berry, however they faced a hard problem. Pete and Billy are great fans of even numbers, that's why they want to divide the watermelon in such a way that each of the two parts weighs even number of kilos, at the same time it is not obligatory that the parts are equal. The boys are extremely tired and want to start their meal as soon as possible, that's why you should help them and find out, if they can divide the watermelon in the way they want. For sure, each of them should get a part of positive weight.
The first (and the only) input line contains integer number *w* (1<=≤<=*w*<=≤<=100) — the weight of the watermelon bought by the boys.
Print YES, if the boys can divide the watermelon into two parts, each of them weighing even number of kilos; and NO in the opposite case.
[ "8\n" ]
[ "YES\n" ]
For example, the boys can divide the watermelon into two parts of 2 and 6 kilos respectively (another variant — two parts of 4 and 4 kilos).
0
[ { "input": "8", "output": "YES" }, { "input": "5", "output": "NO" }, { "input": "4", "output": "YES" }, { "input": "3", "output": "NO" }, { "input": "2", "output": "NO" }, { "input": "1", "output": "NO" }, { "input": "7", "output": "NO" }, { "input": "6", "output": "YES" }, { "input": "10", "output": "YES" }, { "input": "9", "output": "NO" }, { "input": "53", "output": "NO" }, { "input": "77", "output": "NO" }, { "input": "32", "output": "YES" }, { "input": "44", "output": "YES" }, { "input": "98", "output": "YES" }, { "input": "99", "output": "NO" }, { "input": "90", "output": "YES" }, { "input": "67", "output": "NO" }, { "input": "100", "output": "YES" }, { "input": "88", "output": "YES" } ]
1,687,884,743
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
4
62
0
n = int(input()) x = n%2 y= n/2 if x == 0: if y%2 == 0: print("Yes") else: print("No")
Title: Watermelon Time Limit: 1 seconds Memory Limit: 64 megabytes Problem Description: One hot summer day Pete and his friend Billy decided to buy a watermelon. They chose the biggest and the ripest one, in their opinion. After that the watermelon was weighed, and the scales showed *w* kilos. They rushed home, dying of thirst, and decided to divide the berry, however they faced a hard problem. Pete and Billy are great fans of even numbers, that's why they want to divide the watermelon in such a way that each of the two parts weighs even number of kilos, at the same time it is not obligatory that the parts are equal. The boys are extremely tired and want to start their meal as soon as possible, that's why you should help them and find out, if they can divide the watermelon in the way they want. For sure, each of them should get a part of positive weight. Input Specification: The first (and the only) input line contains integer number *w* (1<=≤<=*w*<=≤<=100) — the weight of the watermelon bought by the boys. Output Specification: Print YES, if the boys can divide the watermelon into two parts, each of them weighing even number of kilos; and NO in the opposite case. Demo Input: ['8\n'] Demo Output: ['YES\n'] Note: For example, the boys can divide the watermelon into two parts of 2 and 6 kilos respectively (another variant — two parts of 4 and 4 kilos).
```python n = int(input()) x = n%2 y= n/2 if x == 0: if y%2 == 0: print("Yes") else: print("No") ```
0
1
A
Theatre Square
PROGRAMMING
1,000
[ "math" ]
A. Theatre Square
1
256
Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=×<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=×<=*a*. What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square.
The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109).
Write the needed number of flagstones.
[ "6 6 4\n" ]
[ "4\n" ]
none
0
[ { "input": "6 6 4", "output": "4" }, { "input": "1 1 1", "output": "1" }, { "input": "2 1 1", "output": "2" }, { "input": "1 2 1", "output": "2" }, { "input": "2 2 1", "output": "4" }, { "input": "2 1 2", "output": "1" }, { "input": "1 1 3", "output": "1" }, { "input": "2 3 4", "output": "1" }, { "input": "1000000000 1000000000 1", "output": "1000000000000000000" }, { "input": "12 13 4", "output": "12" }, { "input": "222 332 5", "output": "3015" }, { "input": "1000 1000 10", "output": "10000" }, { "input": "1001 1000 10", "output": "10100" }, { "input": "100 10001 1000000000", "output": "1" }, { "input": "1000000000 1000000000 1000000000", "output": "1" }, { "input": "1000000000 1000000000 999999999", "output": "4" }, { "input": "1000000000 1000000000 192", "output": "27126743055556" }, { "input": "1000000000 987654321 1", "output": "987654321000000000" }, { "input": "456784567 1000000000 51", "output": "175618850864484" }, { "input": "39916800 134217728 40320", "output": "3295710" } ]
1,664,376,509
2,147,483,647
Python 3
RUNTIME_ERROR
TESTS
0
15
0
m,n,a=map(int,input().split()) t=m+0.9*a s=n+0.9*b x=int(s/a) y=int(t/a) print(x*y)
Title: Theatre Square Time Limit: 1 seconds Memory Limit: 256 megabytes Problem Description: Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=×<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=×<=*a*. What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square. Input Specification: The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109). Output Specification: Write the needed number of flagstones. Demo Input: ['6 6 4\n'] Demo Output: ['4\n'] Note: none
```python m,n,a=map(int,input().split()) t=m+0.9*a s=n+0.9*b x=int(s/a) y=int(t/a) print(x*y) ```
-1
523
A
Rotate, Flip and Zoom
PROGRAMMING
1,200
[ "*special", "implementation" ]
null
null
Polycarp is writing the prototype of a graphic editor. He has already made up his mind that the basic image transformations in his editor will be: rotate the image 90 degrees clockwise, flip the image horizontally (symmetry relative to the vertical line, that is, the right part of the image moves to the left, and vice versa) and zooming on the image. He is sure that that there is a large number of transformations that can be expressed through these three. He has recently stopped implementing all three transformations for monochrome images. To test this feature, he asked you to write a code that will consecutively perform three actions with a monochrome image: first it will rotate the image 90 degrees clockwise, then it will flip the image horizontally and finally, it will zoom in twice on the image (that is, it will double all the linear sizes). Implement this feature to help Polycarp test his editor.
The first line contains two integers, *w* and *h* (1<=≤<=*w*,<=*h*<=≤<=100) — the width and height of an image in pixels. The picture is given in *h* lines, each line contains *w* characters — each character encodes the color of the corresponding pixel of the image. The line consists only of characters "." and "*", as the image is monochrome.
Print 2*w* lines, each containing 2*h* characters — the result of consecutive implementing of the three transformations, described above.
[ "3 2\n.*.\n.*.\n", "9 20\n**.......\n****.....\n******...\n*******..\n..******.\n....****.\n......***\n*.....***\n*********\n*********\n*********\n*********\n....**...\n...****..\n..******.\n.********\n****..***\n***...***\n**.....**\n*.......*\n" ]
[ "....\n....\n****\n****\n....\n....\n", "********......**********........********\n********......**********........********\n********........********......********..\n********........********......********..\n..********......********....********....\n..********......********....********....\n..********......********..********......\n..********......********..********......\n....********....****************........\n....********....****************........\n....********....****************........\n....********....****************........\n......******************..**********....\n......******************..**********....\n........****************....**********..\n........****************....**********..\n............************......**********\n............************......**********\n" ]
none
500
[ { "input": "3 2\n.*.\n.*.", "output": "....\n....\n****\n****\n....\n...." }, { "input": "9 20\n**.......\n****.....\n******...\n*******..\n..******.\n....****.\n......***\n*.....***\n*********\n*********\n*********\n*********\n....**...\n...****..\n..******.\n.********\n****..***\n***...***\n**.....**\n*.......*", "output": "********......**********........********\n********......**********........********\n********........********......********..\n********........********......********..\n..********......********....********....\n..********......********....********....\n..********......********..********......\n..********......********..********......\n....********....****************........\n....********....****************........\n....********....****************........\n....********....****************........\n......*..." }, { "input": "1 100\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.", "output": "........................................................................................................................................................................................................\n........................................................................................................................................................................................................" }, { "input": "1 100\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*", "output": "********************************************************************************************************************************************************************************************************\n********************************************************************************************************************************************************************************************************" }, { "input": "1 100\n.\n*\n.\n.\n.\n*\n.\n.\n.\n*\n*\n*\n.\n.\n.\n.\n.\n.\n*\n.\n.\n.\n*\n.\n*\n.\n.\n*\n*\n.\n*\n.\n.\n*\n.\n.\n*\n*\n.\n.\n.\n.\n.\n*\n.\n*\n.\n*\n.\n.\n.\n.\n*\n*\n*\n.\n.\n.\n.\n*\n.\n.\n*\n*\n*\n*\n.\n*\n*\n*\n*\n*\n.\n*\n*\n*\n*\n*\n*\n*\n*\n*\n*\n.\n.\n*\n*\n*\n*\n*\n*\n*\n.\n.\n*\n.\n.\n*\n*\n.", "output": "..**......**......******............**......**..**....****..**....**....****..........**..**..**........******........**....********..**********..********************....**************....**....****..\n..**......**......******............**......**..**....****..**....**....****..........**..**..**........******........**....********..**********..********************....**************....**....****.." }, { "input": "100 1\n****************************************************************************************************", "output": "**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n..." }, { "input": "100 1\n*...***.....**.*...*.*.**.************.**..**.*..**..**.*.**...***.*...*.*..*.*.*......**..*..*...**", "output": "**\n**\n..\n..\n..\n..\n..\n..\n**\n**\n**\n**\n**\n**\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n**\n**\n**\n**\n..\n..\n**\n**\n..\n..\n..\n..\n..\n..\n**\n**\n..\n..\n**\n**\n..\n..\n**\n**\n**\n**\n..\n..\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n**\n..\n..\n**\n**\n**\n**\n..\n..\n..\n..\n**\n**\n**\n**\n..\n..\n**\n**\n..\n..\n..\n..\n**\n**\n**\n**\n..\n..\n..\n..\n**\n**\n**\n**\n..\n..\n**\n**\n..\n..\n**\n**\n**\n**\n..\n..\n..\n..\n..\n..\n**\n**\n..." }, { "input": "1 1\n.", "output": "..\n.." }, { "input": "1 1\n*", "output": "**\n**" }, { "input": "2 2\n.*\n*.", "output": "..**\n..**\n**..\n**.." }, { "input": "1 2\n*\n.", "output": "**..\n**.." }, { "input": "2 1\n*.", "output": "**\n**\n..\n.." } ]
1,455,737,761
2,147,483,647
Python 3
OK
TESTS
24
93
307,200
def rot(x): matrix = [[0]*h for i in range(w)] for i in range(w): for j in range(h): matrix[i][j] = x[h-1-j][i] #for i in matrix:(print(i)) #print() return matrix def mir(x): matrix = [[0]*h for i in range(w)] for i in range(w): for j in range(h): matrix[i][j] = x[i][h-j-1] #for i in matrix:(print(i)) #print() return matrix def scale(x): matrix = [[0]*2*h for i in range(w*2)] for i in range(2*w): for j in range(2*h): matrix[i][j] = x[i//2][j//2] #for i in matrix:(print(i)) #print() return matrix w, h = map(int,input().split()) x = [] for i in range(h): x.append(list(input())) x = scale(mir(rot(x))) for i in x: print(''.join(i))
Title: Rotate, Flip and Zoom Time Limit: None seconds Memory Limit: None megabytes Problem Description: Polycarp is writing the prototype of a graphic editor. He has already made up his mind that the basic image transformations in his editor will be: rotate the image 90 degrees clockwise, flip the image horizontally (symmetry relative to the vertical line, that is, the right part of the image moves to the left, and vice versa) and zooming on the image. He is sure that that there is a large number of transformations that can be expressed through these three. He has recently stopped implementing all three transformations for monochrome images. To test this feature, he asked you to write a code that will consecutively perform three actions with a monochrome image: first it will rotate the image 90 degrees clockwise, then it will flip the image horizontally and finally, it will zoom in twice on the image (that is, it will double all the linear sizes). Implement this feature to help Polycarp test his editor. Input Specification: The first line contains two integers, *w* and *h* (1<=≤<=*w*,<=*h*<=≤<=100) — the width and height of an image in pixels. The picture is given in *h* lines, each line contains *w* characters — each character encodes the color of the corresponding pixel of the image. The line consists only of characters "." and "*", as the image is monochrome. Output Specification: Print 2*w* lines, each containing 2*h* characters — the result of consecutive implementing of the three transformations, described above. Demo Input: ['3 2\n.*.\n.*.\n', '9 20\n**.......\n****.....\n******...\n*******..\n..******.\n....****.\n......***\n*.....***\n*********\n*********\n*********\n*********\n....**...\n...****..\n..******.\n.********\n****..***\n***...***\n**.....**\n*.......*\n'] Demo Output: ['....\n....\n****\n****\n....\n....\n', '********......**********........********\n********......**********........********\n********........********......********..\n********........********......********..\n..********......********....********....\n..********......********....********....\n..********......********..********......\n..********......********..********......\n....********....****************........\n....********....****************........\n....********....****************........\n....********....****************........\n......******************..**********....\n......******************..**********....\n........****************....**********..\n........****************....**********..\n............************......**********\n............************......**********\n'] Note: none
```python def rot(x): matrix = [[0]*h for i in range(w)] for i in range(w): for j in range(h): matrix[i][j] = x[h-1-j][i] #for i in matrix:(print(i)) #print() return matrix def mir(x): matrix = [[0]*h for i in range(w)] for i in range(w): for j in range(h): matrix[i][j] = x[i][h-j-1] #for i in matrix:(print(i)) #print() return matrix def scale(x): matrix = [[0]*2*h for i in range(w*2)] for i in range(2*w): for j in range(2*h): matrix[i][j] = x[i//2][j//2] #for i in matrix:(print(i)) #print() return matrix w, h = map(int,input().split()) x = [] for i in range(h): x.append(list(input())) x = scale(mir(rot(x))) for i in x: print(''.join(i)) ```
3
218
A
Mountain Scenery
PROGRAMMING
1,100
[ "brute force", "constructive algorithms", "implementation" ]
null
null
Little Bolek has found a picture with *n* mountain peaks painted on it. The *n* painted peaks are represented by a non-closed polyline, consisting of 2*n* segments. The segments go through 2*n*<=+<=1 points with coordinates (1,<=*y*1), (2,<=*y*2), ..., (2*n*<=+<=1,<=*y*2*n*<=+<=1), with the *i*-th segment connecting the point (*i*,<=*y**i*) and the point (*i*<=+<=1,<=*y**i*<=+<=1). For any even *i* (2<=≤<=*i*<=≤<=2*n*) the following condition holds: *y**i*<=-<=1<=&lt;<=*y**i* and *y**i*<=&gt;<=*y**i*<=+<=1. We shall call a vertex of a polyline with an even *x* coordinate a mountain peak. Bolek fancied a little mischief. He chose exactly *k* mountain peaks, rubbed out the segments that went through those peaks and increased each peak's height by one (that is, he increased the *y* coordinate of the corresponding points). Then he painted the missing segments to get a new picture of mountain peaks. Let us denote the points through which the new polyline passes on Bolek's new picture as (1,<=*r*1), (2,<=*r*2), ..., (2*n*<=+<=1,<=*r*2*n*<=+<=1). Given Bolek's final picture, restore the initial one.
The first line contains two space-separated integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=100). The next line contains 2*n*<=+<=1 space-separated integers *r*1,<=*r*2,<=...,<=*r*2*n*<=+<=1 (0<=≤<=*r**i*<=≤<=100) — the *y* coordinates of the polyline vertices on Bolek's picture. It is guaranteed that we can obtain the given picture after performing the described actions on some picture of mountain peaks.
Print 2*n*<=+<=1 integers *y*1,<=*y*2,<=...,<=*y*2*n*<=+<=1 — the *y* coordinates of the vertices of the polyline on the initial picture. If there are multiple answers, output any one of them.
[ "3 2\n0 5 3 5 1 5 2\n", "1 1\n0 2 0\n" ]
[ "0 5 3 4 1 4 2 \n", "0 1 0 \n" ]
none
500
[ { "input": "3 2\n0 5 3 5 1 5 2", "output": "0 5 3 4 1 4 2 " }, { "input": "1 1\n0 2 0", "output": "0 1 0 " }, { "input": "1 1\n1 100 0", "output": "1 99 0 " }, { "input": "3 1\n0 1 0 1 0 2 0", "output": "0 1 0 1 0 1 0 " }, { "input": "3 1\n0 1 0 2 0 1 0", "output": "0 1 0 1 0 1 0 " }, { "input": "3 3\n0 100 35 67 40 60 3", "output": "0 99 35 66 40 59 3 " }, { "input": "7 3\n1 2 1 3 1 2 1 2 1 3 1 3 1 2 1", "output": "1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 " }, { "input": "100 100\n1 3 1 3 1 3 0 2 0 3 1 3 1 3 1 3 0 3 1 3 0 2 0 2 0 3 0 2 0 2 0 3 1 3 1 3 1 3 1 3 0 2 0 3 1 3 0 2 0 2 0 2 0 2 0 2 0 3 0 3 0 3 0 3 0 2 0 3 1 3 1 3 1 3 0 3 0 2 0 2 0 2 0 2 0 3 0 3 1 3 0 3 1 3 1 3 0 3 1 3 0 3 1 3 1 3 0 3 1 3 0 3 1 3 0 2 0 3 1 3 0 3 1 3 0 2 0 3 1 3 0 3 0 2 0 3 1 3 0 3 0 3 0 2 0 2 0 2 0 3 0 3 1 3 1 3 0 3 1 3 1 3 1 3 0 2 0 3 0 2 0 3 1 3 0 3 0 3 1 3 0 2 0 3 0 2 0 2 0 2 0 2 0 3 1 3 0 3 1 3 1", "output": "1 2 1 2 1 2 0 1 0 2 1 2 1 2 1 2 0 2 1 2 0 1 0 1 0 2 0 1 0 1 0 2 1 2 1 2 1 2 1 2 0 1 0 2 1 2 0 1 0 1 0 1 0 1 0 1 0 2 0 2 0 2 0 2 0 1 0 2 1 2 1 2 1 2 0 2 0 1 0 1 0 1 0 1 0 2 0 2 1 2 0 2 1 2 1 2 0 2 1 2 0 2 1 2 1 2 0 2 1 2 0 2 1 2 0 1 0 2 1 2 0 2 1 2 0 1 0 2 1 2 0 2 0 1 0 2 1 2 0 2 0 2 0 1 0 1 0 1 0 2 0 2 1 2 1 2 0 2 1 2 1 2 1 2 0 1 0 2 0 1 0 2 1 2 0 2 0 2 1 2 0 1 0 2 0 1 0 1 0 1 0 1 0 2 1 2 0 2 1 2 1 " }, { "input": "30 20\n1 3 1 3 0 2 0 4 1 3 0 3 1 3 1 4 2 3 1 2 0 4 2 4 0 4 1 3 0 4 1 4 2 4 2 4 0 3 1 2 1 4 0 3 0 4 1 3 1 4 1 3 0 1 0 4 0 3 2 3 1", "output": "1 3 1 3 0 2 0 4 1 2 0 2 1 2 1 3 2 3 1 2 0 3 2 3 0 3 1 2 0 3 1 3 2 3 2 3 0 2 1 2 1 3 0 2 0 3 1 2 1 3 1 2 0 1 0 3 0 3 2 3 1 " }, { "input": "10 6\n0 5 2 4 1 5 2 5 2 4 2 5 3 5 0 2 0 1 0 1 0", "output": "0 5 2 4 1 4 2 4 2 3 2 4 3 4 0 1 0 1 0 1 0 " }, { "input": "11 6\n3 5 1 4 3 5 0 2 0 2 0 4 0 3 0 4 1 5 2 4 0 4 0", "output": "3 5 1 4 3 5 0 2 0 2 0 3 0 2 0 3 1 4 2 3 0 3 0 " }, { "input": "12 6\n1 2 1 5 0 2 0 4 1 3 1 4 2 4 0 4 0 4 2 4 0 4 0 5 3", "output": "1 2 1 5 0 2 0 4 1 3 1 4 2 3 0 3 0 3 2 3 0 3 0 4 3 " }, { "input": "13 6\n3 5 2 5 0 3 0 1 0 2 0 1 0 1 0 2 1 4 3 5 1 3 1 3 2 3 1", "output": "3 4 2 4 0 2 0 1 0 1 0 1 0 1 0 2 1 4 3 4 1 2 1 3 2 3 1 " }, { "input": "24 7\n3 4 2 4 1 4 3 4 3 5 1 3 1 3 0 3 0 3 1 4 0 3 0 1 0 1 0 3 2 3 2 3 1 2 1 3 2 5 1 3 0 1 0 2 0 3 1 3 1", "output": "3 4 2 4 1 4 3 4 3 5 1 3 1 3 0 3 0 3 1 3 0 2 0 1 0 1 0 3 2 3 2 3 1 2 1 3 2 4 1 2 0 1 0 1 0 2 1 2 1 " }, { "input": "25 8\n3 5 2 4 2 4 0 1 0 1 0 1 0 2 1 5 2 4 2 4 2 3 1 2 0 1 0 2 0 3 2 5 3 5 0 4 2 3 2 4 1 4 0 4 1 4 0 1 0 4 2", "output": "3 5 2 4 2 4 0 1 0 1 0 1 0 2 1 5 2 4 2 4 2 3 1 2 0 1 0 2 0 3 2 4 3 4 0 3 2 3 2 3 1 3 0 3 1 3 0 1 0 3 2 " }, { "input": "26 9\n3 4 2 3 1 3 1 3 2 4 0 1 0 2 1 3 1 3 0 5 1 4 3 5 0 5 2 3 0 3 1 4 1 3 1 4 2 3 1 4 3 4 1 3 2 4 1 3 2 5 1 2 0", "output": "3 4 2 3 1 3 1 3 2 4 0 1 0 2 1 3 1 3 0 4 1 4 3 4 0 4 2 3 0 2 1 3 1 2 1 3 2 3 1 4 3 4 1 3 2 3 1 3 2 4 1 2 0 " }, { "input": "27 10\n3 5 3 5 3 4 1 3 1 3 1 3 2 3 2 3 2 4 2 3 0 4 2 5 3 4 3 4 1 5 3 4 1 2 1 5 0 3 0 5 0 5 3 4 0 1 0 2 0 2 1 4 0 2 1", "output": "3 5 3 5 3 4 1 3 1 3 1 3 2 3 2 3 2 3 2 3 0 3 2 4 3 4 3 4 1 4 3 4 1 2 1 4 0 2 0 4 0 4 3 4 0 1 0 1 0 2 1 3 0 2 1 " }, { "input": "40 1\n0 2 1 2 0 2 1 2 1 2 1 2 1 2 1 3 0 1 0 1 0 1 0 2 0 2 1 2 0 2 1 2 1 2 1 2 1 2 0 2 1 2 1 2 0 1 0 2 0 2 0 1 0 1 0 1 0 1 0 1 0 2 0 2 0 2 0 1 0 2 0 1 0 2 0 1 0 2 1 2 0", "output": "0 2 1 2 0 2 1 2 1 2 1 2 1 2 1 3 0 1 0 1 0 1 0 2 0 2 1 2 0 2 1 2 1 2 1 2 1 2 0 2 1 2 1 2 0 1 0 2 0 2 0 1 0 1 0 1 0 1 0 1 0 2 0 2 0 2 0 1 0 2 0 1 0 1 0 1 0 2 1 2 0 " }, { "input": "40 2\n0 3 1 2 1 2 0 1 0 2 1 3 0 2 0 3 0 3 0 1 0 2 0 3 1 2 0 2 1 2 0 2 0 1 0 1 0 2 0 2 1 3 0 2 0 1 0 1 0 1 0 3 1 3 1 2 1 2 0 3 0 1 0 3 0 2 1 2 0 1 0 2 0 3 1 2 1 3 1 3 0", "output": "0 3 1 2 1 2 0 1 0 2 1 3 0 2 0 3 0 3 0 1 0 2 0 3 1 2 0 2 1 2 0 2 0 1 0 1 0 2 0 2 1 3 0 2 0 1 0 1 0 1 0 3 1 3 1 2 1 2 0 3 0 1 0 3 0 2 1 2 0 1 0 2 0 3 1 2 1 2 1 2 0 " }, { "input": "40 3\n1 3 1 2 0 4 1 2 0 1 0 1 0 3 0 3 2 3 0 3 1 3 0 4 1 3 2 3 0 2 1 3 0 2 0 1 0 3 1 3 2 3 2 3 0 1 0 2 0 1 0 1 0 3 1 3 0 3 1 3 1 2 0 1 0 3 0 2 0 3 0 1 0 2 0 3 1 2 0 3 0", "output": "1 3 1 2 0 4 1 2 0 1 0 1 0 3 0 3 2 3 0 3 1 3 0 4 1 3 2 3 0 2 1 3 0 2 0 1 0 3 1 3 2 3 2 3 0 1 0 2 0 1 0 1 0 3 1 3 0 3 1 3 1 2 0 1 0 3 0 2 0 3 0 1 0 1 0 2 1 2 0 2 0 " }, { "input": "50 40\n1 4 2 4 1 2 1 4 1 4 2 3 1 2 1 4 1 3 0 2 1 4 0 1 0 3 1 3 1 3 0 4 2 4 2 4 2 4 2 4 2 4 2 4 0 4 1 3 1 3 0 4 1 4 2 3 2 3 0 3 0 3 0 4 1 4 1 3 1 4 1 3 0 4 0 3 0 2 0 2 0 4 1 4 0 2 0 4 1 4 0 3 0 2 1 3 0 2 0 4 0", "output": "1 4 2 4 1 2 1 3 1 3 2 3 1 2 1 3 1 2 0 2 1 3 0 1 0 2 1 2 1 2 0 3 2 3 2 3 2 3 2 3 2 3 2 3 0 3 1 2 1 2 0 3 1 3 2 3 2 3 0 2 0 2 0 3 1 3 1 2 1 3 1 2 0 3 0 2 0 1 0 1 0 3 1 3 0 1 0 3 1 3 0 2 0 2 1 2 0 1 0 3 0 " }, { "input": "100 2\n1 3 1 2 1 3 2 3 1 3 1 3 1 3 1 2 0 3 0 2 0 3 2 3 0 3 1 2 1 2 0 3 0 1 0 1 0 3 2 3 1 2 0 1 0 2 0 1 0 2 1 3 1 2 1 3 2 3 1 3 1 2 0 3 2 3 0 2 1 3 1 2 0 3 2 3 1 3 2 3 0 4 0 3 0 1 0 3 0 1 0 1 0 2 0 2 1 3 1 2 1 2 0 2 0 1 0 2 0 2 1 3 1 3 2 3 0 2 1 2 0 3 0 1 0 2 0 3 2 3 1 3 0 3 1 2 0 1 0 3 0 1 0 1 0 1 0 2 0 1 0 2 1 2 1 2 1 3 0 1 0 2 1 3 0 2 1 3 0 2 1 2 0 3 1 3 1 3 0 2 1 2 1 3 0 2 1 3 2 3 1 2 0 3 1 2 0 3 1 2 0", "output": "1 3 1 2 1 3 2 3 1 3 1 3 1 3 1 2 0 3 0 2 0 3 2 3 0 3 1 2 1 2 0 3 0 1 0 1 0 3 2 3 1 2 0 1 0 2 0 1 0 2 1 3 1 2 1 3 2 3 1 3 1 2 0 3 2 3 0 2 1 3 1 2 0 3 2 3 1 3 2 3 0 4 0 3 0 1 0 3 0 1 0 1 0 2 0 2 1 3 1 2 1 2 0 2 0 1 0 2 0 2 1 3 1 3 2 3 0 2 1 2 0 3 0 1 0 2 0 3 2 3 1 3 0 3 1 2 0 1 0 3 0 1 0 1 0 1 0 2 0 1 0 2 1 2 1 2 1 3 0 1 0 2 1 3 0 2 1 3 0 2 1 2 0 3 1 3 1 3 0 2 1 2 1 3 0 2 1 3 2 3 1 2 0 2 1 2 0 2 1 2 0 " }, { "input": "100 3\n0 2 1 2 0 1 0 1 0 3 0 2 1 3 1 3 2 3 0 2 0 1 0 2 0 1 0 3 2 3 2 3 1 2 1 3 1 2 1 3 2 3 2 3 0 3 2 3 2 3 2 3 0 2 0 3 0 3 2 3 2 3 2 3 2 3 0 3 0 1 0 2 1 3 0 2 1 2 0 3 2 3 2 3 1 3 0 3 1 3 0 3 0 1 0 1 0 2 0 2 1 2 0 3 1 3 0 3 2 3 2 3 2 3 2 3 0 1 0 1 0 1 0 2 1 2 0 2 1 3 2 3 0 1 0 1 0 1 0 1 0 2 0 1 0 3 1 2 1 2 1 3 1 2 0 3 0 2 1 2 1 3 2 3 1 3 2 3 0 1 0 1 0 1 0 1 0 3 0 1 0 2 1 2 0 3 1 3 2 3 0 3 1 2 1 3 1 3 1 3 0", "output": "0 2 1 2 0 1 0 1 0 3 0 2 1 3 1 3 2 3 0 2 0 1 0 2 0 1 0 3 2 3 2 3 1 2 1 3 1 2 1 3 2 3 2 3 0 3 2 3 2 3 2 3 0 2 0 3 0 3 2 3 2 3 2 3 2 3 0 3 0 1 0 2 1 3 0 2 1 2 0 3 2 3 2 3 1 3 0 3 1 3 0 3 0 1 0 1 0 2 0 2 1 2 0 3 1 3 0 3 2 3 2 3 2 3 2 3 0 1 0 1 0 1 0 2 1 2 0 2 1 3 2 3 0 1 0 1 0 1 0 1 0 2 0 1 0 3 1 2 1 2 1 3 1 2 0 3 0 2 1 2 1 3 2 3 1 3 2 3 0 1 0 1 0 1 0 1 0 3 0 1 0 2 1 2 0 3 1 3 2 3 0 3 1 2 1 2 1 2 1 2 0 " }, { "input": "100 20\n0 1 0 3 0 3 2 3 2 4 0 2 0 3 1 3 0 2 0 2 0 3 0 1 0 3 2 4 0 1 0 2 0 2 1 2 1 4 2 4 1 2 0 1 0 2 1 3 0 2 1 3 2 3 1 2 0 2 1 4 0 3 0 2 0 1 0 1 0 1 0 2 1 3 2 3 2 3 2 3 0 1 0 1 0 4 2 3 2 3 0 3 1 2 0 2 0 2 1 3 2 3 1 4 0 1 0 2 1 2 0 2 0 3 2 3 0 2 0 2 1 4 2 3 1 3 0 3 0 2 0 2 1 2 1 3 0 3 1 2 1 3 1 3 1 2 1 2 0 2 1 3 0 2 0 3 0 1 0 3 0 3 0 1 0 4 1 3 0 1 0 1 0 2 1 2 0 2 1 4 1 3 0 2 1 3 1 3 1 3 0 3 0 2 0 1 0 2 1 2 1", "output": "0 1 0 3 0 3 2 3 2 4 0 2 0 3 1 3 0 2 0 2 0 3 0 1 0 3 2 4 0 1 0 2 0 2 1 2 1 4 2 4 1 2 0 1 0 2 1 3 0 2 1 3 2 3 1 2 0 2 1 4 0 3 0 2 0 1 0 1 0 1 0 2 1 3 2 3 2 3 2 3 0 1 0 1 0 4 2 3 2 3 0 3 1 2 0 2 0 2 1 3 2 3 1 4 0 1 0 2 1 2 0 2 0 3 2 3 0 2 0 2 1 4 2 3 1 3 0 2 0 1 0 2 1 2 1 2 0 2 1 2 1 2 1 2 1 2 1 2 0 2 1 2 0 1 0 2 0 1 0 2 0 2 0 1 0 3 1 2 0 1 0 1 0 2 1 2 0 2 1 3 1 2 0 2 1 2 1 2 1 2 0 2 0 1 0 1 0 2 1 2 1 " }, { "input": "100 20\n2 3 0 4 0 1 0 6 3 4 3 6 4 6 0 9 0 6 2 7 3 8 7 10 2 9 3 9 5 6 5 10 3 7 1 5 2 8 3 7 2 3 1 6 0 8 3 8 0 4 1 8 3 7 1 9 5 9 5 8 7 8 5 6 5 8 1 9 8 9 8 10 7 10 5 8 6 10 2 6 3 9 2 6 3 10 5 9 3 10 1 3 2 11 8 9 8 10 1 8 7 11 0 9 5 8 4 5 0 7 3 7 5 9 5 10 1 7 1 9 1 6 3 8 2 4 1 4 2 6 0 4 2 4 2 7 6 9 0 1 0 4 0 4 0 9 2 7 6 7 2 8 0 8 2 7 5 10 1 2 0 2 0 4 3 5 4 7 0 10 2 10 3 6 3 7 1 4 0 9 1 4 3 8 1 10 1 10 0 3 2 5 3 9 0 7 4 5 0 1 0", "output": "2 3 0 4 0 1 0 6 3 4 3 6 4 6 0 9 0 6 2 7 3 8 7 10 2 9 3 9 5 6 5 10 3 7 1 5 2 8 3 7 2 3 1 6 0 8 3 8 0 4 1 8 3 7 1 9 5 9 5 8 7 8 5 6 5 8 1 9 8 9 8 10 7 10 5 8 6 10 2 6 3 9 2 6 3 10 5 9 3 10 1 3 2 11 8 9 8 10 1 8 7 11 0 9 5 8 4 5 0 7 3 7 5 9 5 10 1 7 1 9 1 6 3 8 2 4 1 4 2 6 0 4 2 4 2 7 6 9 0 1 0 4 0 3 0 8 2 7 6 7 2 7 0 7 2 6 5 9 1 2 0 1 0 4 3 5 4 6 0 9 2 9 3 5 3 6 1 3 0 8 1 4 3 7 1 9 1 9 0 3 2 4 3 8 0 6 4 5 0 1 0 " }, { "input": "98 3\n1 2 1 2 0 2 0 2 1 2 0 1 0 2 1 2 0 2 1 2 1 2 0 1 0 2 1 2 1 2 0 2 1 2 0 2 0 2 0 1 0 1 0 1 0 2 1 3 1 2 1 2 1 2 1 2 1 2 1 2 0 2 0 2 1 2 1 2 0 2 1 2 0 1 0 1 0 1 0 1 0 2 0 1 0 2 0 2 1 2 1 2 1 2 0 1 0 1 0 1 0 2 1 2 0 2 1 2 0 2 0 1 0 2 1 2 0 1 0 2 1 2 1 2 1 2 0 2 1 2 1 2 1 2 0 2 1 2 1 2 0 1 0 2 0 2 0 1 0 2 0 2 0 1 0 1 0 1 0 2 0 2 1 2 0 1 0 2 0 2 0 1 0 2 1 2 1 2 1 2 0 2 1 2 1 2 1 2 0 1 0 1 0 2 0 2 0", "output": "1 2 1 2 0 2 0 2 1 2 0 1 0 2 1 2 0 2 1 2 1 2 0 1 0 2 1 2 1 2 0 2 1 2 0 2 0 2 0 1 0 1 0 1 0 2 1 3 1 2 1 2 1 2 1 2 1 2 1 2 0 2 0 2 1 2 1 2 0 2 1 2 0 1 0 1 0 1 0 1 0 2 0 1 0 2 0 2 1 2 1 2 1 2 0 1 0 1 0 1 0 2 1 2 0 2 1 2 0 2 0 1 0 2 1 2 0 1 0 2 1 2 1 2 1 2 0 2 1 2 1 2 1 2 0 2 1 2 1 2 0 1 0 2 0 2 0 1 0 2 0 2 0 1 0 1 0 1 0 2 0 2 1 2 0 1 0 2 0 1 0 1 0 2 1 2 1 2 1 2 0 2 1 2 1 2 1 2 0 1 0 1 0 1 0 1 0 " }, { "input": "2 1\n0 2 1 4 1", "output": "0 2 1 3 1 " }, { "input": "2 1\n0 2 1 5 1", "output": "0 2 1 4 1 " }, { "input": "3 3\n1 12 9 11 6 8 1", "output": "1 11 9 10 6 7 1 " }, { "input": "3 2\n0 7 4 7 1 3 2", "output": "0 6 4 6 1 3 2 " }, { "input": "2 1\n1 3 2 4 1", "output": "1 3 2 3 1 " }, { "input": "4 1\n5 6 5 6 5 6 1 3 1", "output": "5 6 5 6 5 6 1 2 1 " }, { "input": "2 1\n0 2 1 3 0", "output": "0 2 1 2 0 " }, { "input": "2 2\n98 100 1 7 2", "output": "98 99 1 6 2 " }, { "input": "3 1\n8 10 9 10 3 5 1", "output": "8 10 9 10 3 4 1 " }, { "input": "3 2\n0 4 3 5 2 5 2", "output": "0 4 3 4 2 4 2 " }, { "input": "2 1\n4 5 2 4 2", "output": "4 5 2 3 2 " }, { "input": "3 1\n0 2 1 2 0 2 0", "output": "0 2 1 2 0 1 0 " }, { "input": "1 1\n5 7 2", "output": "5 6 2 " }, { "input": "2 1\n3 4 1 3 1", "output": "3 4 1 2 1 " }, { "input": "3 1\n0 4 3 5 0 5 0", "output": "0 4 3 5 0 4 0 " }, { "input": "3 1\n1 3 2 3 1 3 1", "output": "1 3 2 3 1 2 1 " }, { "input": "2 1\n0 8 7 100 0", "output": "0 8 7 99 0 " }, { "input": "2 1\n1 3 2 5 1", "output": "1 3 2 4 1 " } ]
1,690,633,001
2,147,483,647
Python 3
WRONG_ANSWER
TESTS
0
62
0
def mountain_scenery(n, k, r): y = [0] * (2 * n + 1) for i in range(2 * n + 1): y[i] = r[i] - 1 if i % 2 == 0 else r[i] + 1 for i in range(1, 2 * n + 1, 2): if y[i] == y[i - 1] or y[i] == y[i + 1]: y[i] -= 1 return y if __name__ == "__main__": n, k = map(int, input().split()) r = list(map(int, input().split())) y = mountain_scenery(n, k, r) print(*y)
Title: Mountain Scenery Time Limit: None seconds Memory Limit: None megabytes Problem Description: Little Bolek has found a picture with *n* mountain peaks painted on it. The *n* painted peaks are represented by a non-closed polyline, consisting of 2*n* segments. The segments go through 2*n*<=+<=1 points with coordinates (1,<=*y*1), (2,<=*y*2), ..., (2*n*<=+<=1,<=*y*2*n*<=+<=1), with the *i*-th segment connecting the point (*i*,<=*y**i*) and the point (*i*<=+<=1,<=*y**i*<=+<=1). For any even *i* (2<=≤<=*i*<=≤<=2*n*) the following condition holds: *y**i*<=-<=1<=&lt;<=*y**i* and *y**i*<=&gt;<=*y**i*<=+<=1. We shall call a vertex of a polyline with an even *x* coordinate a mountain peak. Bolek fancied a little mischief. He chose exactly *k* mountain peaks, rubbed out the segments that went through those peaks and increased each peak's height by one (that is, he increased the *y* coordinate of the corresponding points). Then he painted the missing segments to get a new picture of mountain peaks. Let us denote the points through which the new polyline passes on Bolek's new picture as (1,<=*r*1), (2,<=*r*2), ..., (2*n*<=+<=1,<=*r*2*n*<=+<=1). Given Bolek's final picture, restore the initial one. Input Specification: The first line contains two space-separated integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=100). The next line contains 2*n*<=+<=1 space-separated integers *r*1,<=*r*2,<=...,<=*r*2*n*<=+<=1 (0<=≤<=*r**i*<=≤<=100) — the *y* coordinates of the polyline vertices on Bolek's picture. It is guaranteed that we can obtain the given picture after performing the described actions on some picture of mountain peaks. Output Specification: Print 2*n*<=+<=1 integers *y*1,<=*y*2,<=...,<=*y*2*n*<=+<=1 — the *y* coordinates of the vertices of the polyline on the initial picture. If there are multiple answers, output any one of them. Demo Input: ['3 2\n0 5 3 5 1 5 2\n', '1 1\n0 2 0\n'] Demo Output: ['0 5 3 4 1 4 2 \n', '0 1 0 \n'] Note: none
```python def mountain_scenery(n, k, r): y = [0] * (2 * n + 1) for i in range(2 * n + 1): y[i] = r[i] - 1 if i % 2 == 0 else r[i] + 1 for i in range(1, 2 * n + 1, 2): if y[i] == y[i - 1] or y[i] == y[i + 1]: y[i] -= 1 return y if __name__ == "__main__": n, k = map(int, input().split()) r = list(map(int, input().split())) y = mountain_scenery(n, k, r) print(*y) ```
0