message stringlengths 2 67k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 463 109k | cluster float64 19 19 | __index_level_0__ int64 926 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a legend in the IT City college. A student that failed to answer all questions on the game theory exam is given one more chance by his professor. The student has to play a game with the professor.
The game is played on a square field consisting of n Γ n cells. Initially all cells are empty. On each turn a player chooses and paint an empty cell that has no common sides with previously painted cells. Adjacent corner of painted cells is allowed. On the next turn another player does the same, then the first one and so on. The player with no cells to paint on his turn loses.
The professor have chosen the field size n and allowed the student to choose to be the first or the second player in the game. What should the student choose to win the game? Both players play optimally.
Input
The only line of the input contains one integer n (1 β€ n β€ 1018) β the size of the field.
Output
Output number 1, if the player making the first turn wins when both players play optimally, otherwise print number 2.
Examples
Input
1
Output
1
Input
2
Output
2
Submitted Solution:
```
n = int(input())
if n&1:
print(1)
else:
print(2)
``` | instruction | 0 | 61,912 | 19 | 123,824 |
Yes | output | 1 | 61,912 | 19 | 123,825 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a legend in the IT City college. A student that failed to answer all questions on the game theory exam is given one more chance by his professor. The student has to play a game with the professor.
The game is played on a square field consisting of n Γ n cells. Initially all cells are empty. On each turn a player chooses and paint an empty cell that has no common sides with previously painted cells. Adjacent corner of painted cells is allowed. On the next turn another player does the same, then the first one and so on. The player with no cells to paint on his turn loses.
The professor have chosen the field size n and allowed the student to choose to be the first or the second player in the game. What should the student choose to win the game? Both players play optimally.
Input
The only line of the input contains one integer n (1 β€ n β€ 1018) β the size of the field.
Output
Output number 1, if the player making the first turn wins when both players play optimally, otherwise print number 2.
Examples
Input
1
Output
1
Input
2
Output
2
Submitted Solution:
```
import sys
import math
# sys.stdin = open('input.txt')
# sys.stdout = open('output.txt', 'w')
def main():
n = int(input())
if n % 2 == 0:
print(2)
else:
print(1)
main()
``` | instruction | 0 | 61,913 | 19 | 123,826 |
Yes | output | 1 | 61,913 | 19 | 123,827 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a legend in the IT City college. A student that failed to answer all questions on the game theory exam is given one more chance by his professor. The student has to play a game with the professor.
The game is played on a square field consisting of n Γ n cells. Initially all cells are empty. On each turn a player chooses and paint an empty cell that has no common sides with previously painted cells. Adjacent corner of painted cells is allowed. On the next turn another player does the same, then the first one and so on. The player with no cells to paint on his turn loses.
The professor have chosen the field size n and allowed the student to choose to be the first or the second player in the game. What should the student choose to win the game? Both players play optimally.
Input
The only line of the input contains one integer n (1 β€ n β€ 1018) β the size of the field.
Output
Output number 1, if the player making the first turn wins when both players play optimally, otherwise print number 2.
Examples
Input
1
Output
1
Input
2
Output
2
Submitted Solution:
```
print(2 if int(input()) % 2 == 0 else 1)
``` | instruction | 0 | 61,914 | 19 | 123,828 |
Yes | output | 1 | 61,914 | 19 | 123,829 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a legend in the IT City college. A student that failed to answer all questions on the game theory exam is given one more chance by his professor. The student has to play a game with the professor.
The game is played on a square field consisting of n Γ n cells. Initially all cells are empty. On each turn a player chooses and paint an empty cell that has no common sides with previously painted cells. Adjacent corner of painted cells is allowed. On the next turn another player does the same, then the first one and so on. The player with no cells to paint on his turn loses.
The professor have chosen the field size n and allowed the student to choose to be the first or the second player in the game. What should the student choose to win the game? Both players play optimally.
Input
The only line of the input contains one integer n (1 β€ n β€ 1018) β the size of the field.
Output
Output number 1, if the player making the first turn wins when both players play optimally, otherwise print number 2.
Examples
Input
1
Output
1
Input
2
Output
2
Submitted Solution:
```
print((1 if int(input()) % 2 == 1 else 2))
``` | instruction | 0 | 61,915 | 19 | 123,830 |
Yes | output | 1 | 61,915 | 19 | 123,831 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a legend in the IT City college. A student that failed to answer all questions on the game theory exam is given one more chance by his professor. The student has to play a game with the professor.
The game is played on a square field consisting of n Γ n cells. Initially all cells are empty. On each turn a player chooses and paint an empty cell that has no common sides with previously painted cells. Adjacent corner of painted cells is allowed. On the next turn another player does the same, then the first one and so on. The player with no cells to paint on his turn loses.
The professor have chosen the field size n and allowed the student to choose to be the first or the second player in the game. What should the student choose to win the game? Both players play optimally.
Input
The only line of the input contains one integer n (1 β€ n β€ 1018) β the size of the field.
Output
Output number 1, if the player making the first turn wins when both players play optimally, otherwise print number 2.
Examples
Input
1
Output
1
Input
2
Output
2
Submitted Solution:
```
n=int(input())
if n%2==0:
print(1)
else:
print(2)
``` | instruction | 0 | 61,916 | 19 | 123,832 |
No | output | 1 | 61,916 | 19 | 123,833 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a legend in the IT City college. A student that failed to answer all questions on the game theory exam is given one more chance by his professor. The student has to play a game with the professor.
The game is played on a square field consisting of n Γ n cells. Initially all cells are empty. On each turn a player chooses and paint an empty cell that has no common sides with previously painted cells. Adjacent corner of painted cells is allowed. On the next turn another player does the same, then the first one and so on. The player with no cells to paint on his turn loses.
The professor have chosen the field size n and allowed the student to choose to be the first or the second player in the game. What should the student choose to win the game? Both players play optimally.
Input
The only line of the input contains one integer n (1 β€ n β€ 1018) β the size of the field.
Output
Output number 1, if the player making the first turn wins when both players play optimally, otherwise print number 2.
Examples
Input
1
Output
1
Input
2
Output
2
Submitted Solution:
```
x=int(input())
print(2-(((x%3)>0)&(x&1)))
``` | instruction | 0 | 61,917 | 19 | 123,834 |
No | output | 1 | 61,917 | 19 | 123,835 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a legend in the IT City college. A student that failed to answer all questions on the game theory exam is given one more chance by his professor. The student has to play a game with the professor.
The game is played on a square field consisting of n Γ n cells. Initially all cells are empty. On each turn a player chooses and paint an empty cell that has no common sides with previously painted cells. Adjacent corner of painted cells is allowed. On the next turn another player does the same, then the first one and so on. The player with no cells to paint on his turn loses.
The professor have chosen the field size n and allowed the student to choose to be the first or the second player in the game. What should the student choose to win the game? Both players play optimally.
Input
The only line of the input contains one integer n (1 β€ n β€ 1018) β the size of the field.
Output
Output number 1, if the player making the first turn wins when both players play optimally, otherwise print number 2.
Examples
Input
1
Output
1
Input
2
Output
2
Submitted Solution:
```
n = int(input())
if (n == 1):
print(1)
else:
print(2)
``` | instruction | 0 | 61,918 | 19 | 123,836 |
No | output | 1 | 61,918 | 19 | 123,837 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a legend in the IT City college. A student that failed to answer all questions on the game theory exam is given one more chance by his professor. The student has to play a game with the professor.
The game is played on a square field consisting of n Γ n cells. Initially all cells are empty. On each turn a player chooses and paint an empty cell that has no common sides with previously painted cells. Adjacent corner of painted cells is allowed. On the next turn another player does the same, then the first one and so on. The player with no cells to paint on his turn loses.
The professor have chosen the field size n and allowed the student to choose to be the first or the second player in the game. What should the student choose to win the game? Both players play optimally.
Input
The only line of the input contains one integer n (1 β€ n β€ 1018) β the size of the field.
Output
Output number 1, if the player making the first turn wins when both players play optimally, otherwise print number 2.
Examples
Input
1
Output
1
Input
2
Output
2
Submitted Solution:
```
a = int(input())
if a%2 == 0:
print(1)
else:
print(2)
``` | instruction | 0 | 61,919 | 19 | 123,838 |
No | output | 1 | 61,919 | 19 | 123,839 |
Provide a correct Python 3 solution for this coding contest problem.
A group of people played a game. All players had distinct scores, which are positive integers.
Takahashi knows N facts on the players' scores. The i-th fact is as follows: the A_i-th highest score among the players is B_i.
Find the maximum possible number of players in the game.
Constraints
* 1 \leq N \leq 10^5
* 1 \leq A_i \leq 10^9(1\leq i\leq N)
* 0 \leq B_i \leq 10^9(1\leq i\leq N)
* If i β j, A_i β A_j.
* There exists a possible outcome of the game that are consistent with the facts.
* All input values are integers.
Inputs
Input is given from Standard Input in the following format:
N
A_1 B_1
:
A_N B_N
Outputs
Print the maximum possible number of players in the game.
Examples
Input
3
4 7
2 9
6 2
Output
8
Input
5
1 10
3 6
5 2
4 4
2 8
Output
7
Input
2
1 1000000000
1000000000 1
Output
1000000001 | instruction | 0 | 62,142 | 19 | 124,284 |
"Correct Solution:
```
n=int(input())
arr=[list(map(int,input().split())) for _ in range(n)]
arr=sorted(arr,key=lambda x:x[0])
print(arr[-1][0]+arr[-1][1])
``` | output | 1 | 62,142 | 19 | 124,285 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A group of people played a game. All players had distinct scores, which are positive integers.
Takahashi knows N facts on the players' scores. The i-th fact is as follows: the A_i-th highest score among the players is B_i.
Find the maximum possible number of players in the game.
Constraints
* 1 \leq N \leq 10^5
* 1 \leq A_i \leq 10^9(1\leq i\leq N)
* 0 \leq B_i \leq 10^9(1\leq i\leq N)
* If i β j, A_i β A_j.
* There exists a possible outcome of the game that are consistent with the facts.
* All input values are integers.
Inputs
Input is given from Standard Input in the following format:
N
A_1 B_1
:
A_N B_N
Outputs
Print the maximum possible number of players in the game.
Examples
Input
3
4 7
2 9
6 2
Output
8
Input
5
1 10
3 6
5 2
4 4
2 8
Output
7
Input
2
1 1000000000
1000000000 1
Output
1000000001
Submitted Solution:
```
import numpy as np
N=int(input())
A=np.empty((N,2),int)
for i in range(N):
A=np.append(A,np.array([list(map(int,input().split(' ')))]),axis=0)
Aind=np.argmax(A[:,0])
print(Aind+A[Aind,1])
``` | instruction | 0 | 62,155 | 19 | 124,310 |
No | output | 1 | 62,155 | 19 | 124,311 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After all the events in Orlando we all know, Sasha and Roma decided to find out who is still the team's biggest loser. Thankfully, Masha found somewhere a revolver with a rotating cylinder of n bullet slots able to contain exactly k bullets, now the boys have a chance to resolve the problem once and for all.
Sasha selects any k out of n slots he wishes and puts bullets there. Roma spins the cylinder so that every of n possible cylinder's shifts is equiprobable. Then the game starts, the players take turns, Sasha starts: he puts the gun to his head and shoots. If there was no bullet in front of the trigger, the cylinder shifts by one position and the weapon is given to Roma for make the same move. The game continues until someone is shot, the survivor is the winner.
Sasha does not want to lose, so he must choose slots for bullets in such a way as to minimize the probability of its own loss. Of all the possible variant he wants to select the lexicographically minimal one, where an empty slot is lexicographically less than a charged one.
More formally, the cylinder of n bullet slots able to contain k bullets can be represented as a string of n characters. Exactly k of them are "X" (charged slots) and the others are "." (uncharged slots).
Let us describe the process of a shot. Suppose that the trigger is in front of the first character of the string (the first slot). If a shot doesn't kill anyone and the cylinder shifts, then the string shifts left. So the first character becomes the last one, the second character becomes the first one, and so on. But the trigger doesn't move. It will be in front of the first character of the resulting string.
Among all the strings that give the minimal probability of loss, Sasha choose the lexicographically minimal one. According to this very string, he charges the gun. You have to help Sasha to charge the gun. For that, each xi query must be answered: is there a bullet in the positions xi?
Input
The first line contains three integers n, k and p (1 β€ n β€ 1018, 0 β€ k β€ n, 1 β€ p β€ 1000) β the number of slots in the cylinder, the number of bullets and the number of queries. Then follow p lines; they are the queries. Each line contains one integer xi (1 β€ xi β€ n) the number of slot to describe.
Please do not use the %lld specificator to read or write 64-bit numbers in Π‘++. It is preferred to use cin, cout streams or the %I64d specificator.
Output
For each query print "." if the slot should be empty and "X" if the slot should be charged.
Examples
Input
3 1 3
1
2
3
Output
..X
Input
6 3 6
1
2
3
4
5
6
Output
.X.X.X
Input
5 2 5
1
2
3
4
5
Output
...XX
Note
The lexicographical comparison of is performed by the < operator in modern programming languages. The a string is lexicographically less that the b string, if there exists such i (1 β€ i β€ n), that ai < bi, and for any j (1 β€ j < i) aj = bj. | instruction | 0 | 62,286 | 19 | 124,572 |
Tags: constructive algorithms, greedy
Correct Solution:
```
#!/usr/bin/env python3
n, k, p = map(int, input().strip().split())
if k == 0:
ak = 0
an = n
else:
ak = k - 1 if n % 2 == 1 else k
an = n - (n % 2)
ans = ''
for i in range(p):
v = int(input().rstrip())
if k == 0:
print('.', end='')
else:
if v == n:
print('X', end='')
else:
idx = (an - v) / 2
idx += (v % 2) * (an / 2)
if idx >= ak:
print('.', end='')
else:
print('X', end='')
``` | output | 1 | 62,286 | 19 | 124,573 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After all the events in Orlando we all know, Sasha and Roma decided to find out who is still the team's biggest loser. Thankfully, Masha found somewhere a revolver with a rotating cylinder of n bullet slots able to contain exactly k bullets, now the boys have a chance to resolve the problem once and for all.
Sasha selects any k out of n slots he wishes and puts bullets there. Roma spins the cylinder so that every of n possible cylinder's shifts is equiprobable. Then the game starts, the players take turns, Sasha starts: he puts the gun to his head and shoots. If there was no bullet in front of the trigger, the cylinder shifts by one position and the weapon is given to Roma for make the same move. The game continues until someone is shot, the survivor is the winner.
Sasha does not want to lose, so he must choose slots for bullets in such a way as to minimize the probability of its own loss. Of all the possible variant he wants to select the lexicographically minimal one, where an empty slot is lexicographically less than a charged one.
More formally, the cylinder of n bullet slots able to contain k bullets can be represented as a string of n characters. Exactly k of them are "X" (charged slots) and the others are "." (uncharged slots).
Let us describe the process of a shot. Suppose that the trigger is in front of the first character of the string (the first slot). If a shot doesn't kill anyone and the cylinder shifts, then the string shifts left. So the first character becomes the last one, the second character becomes the first one, and so on. But the trigger doesn't move. It will be in front of the first character of the resulting string.
Among all the strings that give the minimal probability of loss, Sasha choose the lexicographically minimal one. According to this very string, he charges the gun. You have to help Sasha to charge the gun. For that, each xi query must be answered: is there a bullet in the positions xi?
Input
The first line contains three integers n, k and p (1 β€ n β€ 1018, 0 β€ k β€ n, 1 β€ p β€ 1000) β the number of slots in the cylinder, the number of bullets and the number of queries. Then follow p lines; they are the queries. Each line contains one integer xi (1 β€ xi β€ n) the number of slot to describe.
Please do not use the %lld specificator to read or write 64-bit numbers in Π‘++. It is preferred to use cin, cout streams or the %I64d specificator.
Output
For each query print "." if the slot should be empty and "X" if the slot should be charged.
Examples
Input
3 1 3
1
2
3
Output
..X
Input
6 3 6
1
2
3
4
5
6
Output
.X.X.X
Input
5 2 5
1
2
3
4
5
Output
...XX
Note
The lexicographical comparison of is performed by the < operator in modern programming languages. The a string is lexicographically less that the b string, if there exists such i (1 β€ i β€ n), that ai < bi, and for any j (1 β€ j < i) aj = bj. | instruction | 0 | 62,287 | 19 | 124,574 |
Tags: constructive algorithms, greedy
Correct Solution:
```
n,k,p=map(int,input().split())
if n%2:n,k=n-1,k-1
for i in range(p):
x=int(input())
z=((k<=n//2 and x%2==0 and x>n-2*k) or (k>n//2 and(x%2==0 or (x>n-2*(k-n//2)))) or (x>n and k>=0))
print(['.','X'][z],end='')
``` | output | 1 | 62,287 | 19 | 124,575 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After all the events in Orlando we all know, Sasha and Roma decided to find out who is still the team's biggest loser. Thankfully, Masha found somewhere a revolver with a rotating cylinder of n bullet slots able to contain exactly k bullets, now the boys have a chance to resolve the problem once and for all.
Sasha selects any k out of n slots he wishes and puts bullets there. Roma spins the cylinder so that every of n possible cylinder's shifts is equiprobable. Then the game starts, the players take turns, Sasha starts: he puts the gun to his head and shoots. If there was no bullet in front of the trigger, the cylinder shifts by one position and the weapon is given to Roma for make the same move. The game continues until someone is shot, the survivor is the winner.
Sasha does not want to lose, so he must choose slots for bullets in such a way as to minimize the probability of its own loss. Of all the possible variant he wants to select the lexicographically minimal one, where an empty slot is lexicographically less than a charged one.
More formally, the cylinder of n bullet slots able to contain k bullets can be represented as a string of n characters. Exactly k of them are "X" (charged slots) and the others are "." (uncharged slots).
Let us describe the process of a shot. Suppose that the trigger is in front of the first character of the string (the first slot). If a shot doesn't kill anyone and the cylinder shifts, then the string shifts left. So the first character becomes the last one, the second character becomes the first one, and so on. But the trigger doesn't move. It will be in front of the first character of the resulting string.
Among all the strings that give the minimal probability of loss, Sasha choose the lexicographically minimal one. According to this very string, he charges the gun. You have to help Sasha to charge the gun. For that, each xi query must be answered: is there a bullet in the positions xi?
Input
The first line contains three integers n, k and p (1 β€ n β€ 1018, 0 β€ k β€ n, 1 β€ p β€ 1000) β the number of slots in the cylinder, the number of bullets and the number of queries. Then follow p lines; they are the queries. Each line contains one integer xi (1 β€ xi β€ n) the number of slot to describe.
Please do not use the %lld specificator to read or write 64-bit numbers in Π‘++. It is preferred to use cin, cout streams or the %I64d specificator.
Output
For each query print "." if the slot should be empty and "X" if the slot should be charged.
Examples
Input
3 1 3
1
2
3
Output
..X
Input
6 3 6
1
2
3
4
5
6
Output
.X.X.X
Input
5 2 5
1
2
3
4
5
Output
...XX
Note
The lexicographical comparison of is performed by the < operator in modern programming languages. The a string is lexicographically less that the b string, if there exists such i (1 β€ i β€ n), that ai < bi, and for any j (1 β€ j < i) aj = bj. | instruction | 0 | 62,288 | 19 | 124,576 |
Tags: constructive algorithms, greedy
Correct Solution:
```
n, k, p = map(int, input().split())
for i in range(p):
pos = int(input())
if n % 2:
if pos == n:
print('X' if k > 0 else '.', end='')
else:
if k * 2 > n + 1:
print('X' if (pos & 1) == 0 or (n - pos) // 2 \
+ n // 2 + 1<= k else '.', end='')
else:
print('X' if (pos & 1) == 0 and \
(n + 1 - pos) // 2 < k else '.', end='')
else:
if k * 2 > n:
print('X' if (pos & 1) == 0 or (n - pos + 1) // 2 + \
n // 2 <= k else '.', end='')
else:
print('X' if (pos & 1) == 0 and (n - pos + 2) // 2 <= k \
else '.', end='')
# Made By Mostafa_Khaled
``` | output | 1 | 62,288 | 19 | 124,577 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After all the events in Orlando we all know, Sasha and Roma decided to find out who is still the team's biggest loser. Thankfully, Masha found somewhere a revolver with a rotating cylinder of n bullet slots able to contain exactly k bullets, now the boys have a chance to resolve the problem once and for all.
Sasha selects any k out of n slots he wishes and puts bullets there. Roma spins the cylinder so that every of n possible cylinder's shifts is equiprobable. Then the game starts, the players take turns, Sasha starts: he puts the gun to his head and shoots. If there was no bullet in front of the trigger, the cylinder shifts by one position and the weapon is given to Roma for make the same move. The game continues until someone is shot, the survivor is the winner.
Sasha does not want to lose, so he must choose slots for bullets in such a way as to minimize the probability of its own loss. Of all the possible variant he wants to select the lexicographically minimal one, where an empty slot is lexicographically less than a charged one.
More formally, the cylinder of n bullet slots able to contain k bullets can be represented as a string of n characters. Exactly k of them are "X" (charged slots) and the others are "." (uncharged slots).
Let us describe the process of a shot. Suppose that the trigger is in front of the first character of the string (the first slot). If a shot doesn't kill anyone and the cylinder shifts, then the string shifts left. So the first character becomes the last one, the second character becomes the first one, and so on. But the trigger doesn't move. It will be in front of the first character of the resulting string.
Among all the strings that give the minimal probability of loss, Sasha choose the lexicographically minimal one. According to this very string, he charges the gun. You have to help Sasha to charge the gun. For that, each xi query must be answered: is there a bullet in the positions xi?
Input
The first line contains three integers n, k and p (1 β€ n β€ 1018, 0 β€ k β€ n, 1 β€ p β€ 1000) β the number of slots in the cylinder, the number of bullets and the number of queries. Then follow p lines; they are the queries. Each line contains one integer xi (1 β€ xi β€ n) the number of slot to describe.
Please do not use the %lld specificator to read or write 64-bit numbers in Π‘++. It is preferred to use cin, cout streams or the %I64d specificator.
Output
For each query print "." if the slot should be empty and "X" if the slot should be charged.
Examples
Input
3 1 3
1
2
3
Output
..X
Input
6 3 6
1
2
3
4
5
6
Output
.X.X.X
Input
5 2 5
1
2
3
4
5
Output
...XX
Note
The lexicographical comparison of is performed by the < operator in modern programming languages. The a string is lexicographically less that the b string, if there exists such i (1 β€ i β€ n), that ai < bi, and for any j (1 β€ j < i) aj = bj. | instruction | 0 | 62,289 | 19 | 124,578 |
Tags: constructive algorithms, greedy
Correct Solution:
```
n, k, p = map(int, input().split())
for i in range(p):
pos = int(input())
if n % 2:
if pos == n:
print('X' if k > 0 else '.', end='')
else:
if k * 2 > n + 1:
print('X' if (pos & 1) == 0 or (n - pos) // 2 \
+ n // 2 + 1<= k else '.', end='')
else:
print('X' if (pos & 1) == 0 and \
(n + 1 - pos) // 2 < k else '.', end='')
else:
if k * 2 > n:
print('X' if (pos & 1) == 0 or (n - pos + 1) // 2 + \
n // 2 <= k else '.', end='')
else:
print('X' if (pos & 1) == 0 and (n - pos + 2) // 2 <= k \
else '.', end='')
``` | output | 1 | 62,289 | 19 | 124,579 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After all the events in Orlando we all know, Sasha and Roma decided to find out who is still the team's biggest loser. Thankfully, Masha found somewhere a revolver with a rotating cylinder of n bullet slots able to contain exactly k bullets, now the boys have a chance to resolve the problem once and for all.
Sasha selects any k out of n slots he wishes and puts bullets there. Roma spins the cylinder so that every of n possible cylinder's shifts is equiprobable. Then the game starts, the players take turns, Sasha starts: he puts the gun to his head and shoots. If there was no bullet in front of the trigger, the cylinder shifts by one position and the weapon is given to Roma for make the same move. The game continues until someone is shot, the survivor is the winner.
Sasha does not want to lose, so he must choose slots for bullets in such a way as to minimize the probability of its own loss. Of all the possible variant he wants to select the lexicographically minimal one, where an empty slot is lexicographically less than a charged one.
More formally, the cylinder of n bullet slots able to contain k bullets can be represented as a string of n characters. Exactly k of them are "X" (charged slots) and the others are "." (uncharged slots).
Let us describe the process of a shot. Suppose that the trigger is in front of the first character of the string (the first slot). If a shot doesn't kill anyone and the cylinder shifts, then the string shifts left. So the first character becomes the last one, the second character becomes the first one, and so on. But the trigger doesn't move. It will be in front of the first character of the resulting string.
Among all the strings that give the minimal probability of loss, Sasha choose the lexicographically minimal one. According to this very string, he charges the gun. You have to help Sasha to charge the gun. For that, each xi query must be answered: is there a bullet in the positions xi?
Input
The first line contains three integers n, k and p (1 β€ n β€ 1018, 0 β€ k β€ n, 1 β€ p β€ 1000) β the number of slots in the cylinder, the number of bullets and the number of queries. Then follow p lines; they are the queries. Each line contains one integer xi (1 β€ xi β€ n) the number of slot to describe.
Please do not use the %lld specificator to read or write 64-bit numbers in Π‘++. It is preferred to use cin, cout streams or the %I64d specificator.
Output
For each query print "." if the slot should be empty and "X" if the slot should be charged.
Examples
Input
3 1 3
1
2
3
Output
..X
Input
6 3 6
1
2
3
4
5
6
Output
.X.X.X
Input
5 2 5
1
2
3
4
5
Output
...XX
Note
The lexicographical comparison of is performed by the < operator in modern programming languages. The a string is lexicographically less that the b string, if there exists such i (1 β€ i β€ n), that ai < bi, and for any j (1 β€ j < i) aj = bj.
Submitted Solution:
```
#!/usr/bin/env python3
n, k, p = map(int, input().strip().split())
if k == 0:
ak = 0
an = n
else:
ak = k - 1 if n % 2 == 1 else k
an = n - (n % 2)
ans = ''
for i in range(p):
v = int(input().rstrip())
if k == 0:
print(0)
else:
if v == n:
print('X', end='')
else:
idx = (an - v) / 2
idx += (v % 2) * (an / 2)
if idx >= ak:
print('.', end='')
else:
print('X', end='')
``` | instruction | 0 | 62,290 | 19 | 124,580 |
No | output | 1 | 62,290 | 19 | 124,581 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After all the events in Orlando we all know, Sasha and Roma decided to find out who is still the team's biggest loser. Thankfully, Masha found somewhere a revolver with a rotating cylinder of n bullet slots able to contain exactly k bullets, now the boys have a chance to resolve the problem once and for all.
Sasha selects any k out of n slots he wishes and puts bullets there. Roma spins the cylinder so that every of n possible cylinder's shifts is equiprobable. Then the game starts, the players take turns, Sasha starts: he puts the gun to his head and shoots. If there was no bullet in front of the trigger, the cylinder shifts by one position and the weapon is given to Roma for make the same move. The game continues until someone is shot, the survivor is the winner.
Sasha does not want to lose, so he must choose slots for bullets in such a way as to minimize the probability of its own loss. Of all the possible variant he wants to select the lexicographically minimal one, where an empty slot is lexicographically less than a charged one.
More formally, the cylinder of n bullet slots able to contain k bullets can be represented as a string of n characters. Exactly k of them are "X" (charged slots) and the others are "." (uncharged slots).
Let us describe the process of a shot. Suppose that the trigger is in front of the first character of the string (the first slot). If a shot doesn't kill anyone and the cylinder shifts, then the string shifts left. So the first character becomes the last one, the second character becomes the first one, and so on. But the trigger doesn't move. It will be in front of the first character of the resulting string.
Among all the strings that give the minimal probability of loss, Sasha choose the lexicographically minimal one. According to this very string, he charges the gun. You have to help Sasha to charge the gun. For that, each xi query must be answered: is there a bullet in the positions xi?
Input
The first line contains three integers n, k and p (1 β€ n β€ 1018, 0 β€ k β€ n, 1 β€ p β€ 1000) β the number of slots in the cylinder, the number of bullets and the number of queries. Then follow p lines; they are the queries. Each line contains one integer xi (1 β€ xi β€ n) the number of slot to describe.
Please do not use the %lld specificator to read or write 64-bit numbers in Π‘++. It is preferred to use cin, cout streams or the %I64d specificator.
Output
For each query print "." if the slot should be empty and "X" if the slot should be charged.
Examples
Input
3 1 3
1
2
3
Output
..X
Input
6 3 6
1
2
3
4
5
6
Output
.X.X.X
Input
5 2 5
1
2
3
4
5
Output
...XX
Note
The lexicographical comparison of is performed by the < operator in modern programming languages. The a string is lexicographically less that the b string, if there exists such i (1 β€ i β€ n), that ai < bi, and for any j (1 β€ j < i) aj = bj.
Submitted Solution:
```
#!/usr/bin/env python3
n, k, p = map(int, input().strip().split())
if k == 0:
ak = 0
an = n
else:
ak = k - 1 if n % 2 == 1 else k
an = n - (n % 2)
for i in range(p):
v = int(input().rstrip())
if k == 0:
print(0)
else:
if v == n:
print('X')
else:
idx = (an - v) / 2
idx += (v % 2) * (an / 2)
if idx >= ak:
print('.')
else:
print('X')
``` | instruction | 0 | 62,291 | 19 | 124,582 |
No | output | 1 | 62,291 | 19 | 124,583 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After all the events in Orlando we all know, Sasha and Roma decided to find out who is still the team's biggest loser. Thankfully, Masha found somewhere a revolver with a rotating cylinder of n bullet slots able to contain exactly k bullets, now the boys have a chance to resolve the problem once and for all.
Sasha selects any k out of n slots he wishes and puts bullets there. Roma spins the cylinder so that every of n possible cylinder's shifts is equiprobable. Then the game starts, the players take turns, Sasha starts: he puts the gun to his head and shoots. If there was no bullet in front of the trigger, the cylinder shifts by one position and the weapon is given to Roma for make the same move. The game continues until someone is shot, the survivor is the winner.
Sasha does not want to lose, so he must choose slots for bullets in such a way as to minimize the probability of its own loss. Of all the possible variant he wants to select the lexicographically minimal one, where an empty slot is lexicographically less than a charged one.
More formally, the cylinder of n bullet slots able to contain k bullets can be represented as a string of n characters. Exactly k of them are "X" (charged slots) and the others are "." (uncharged slots).
Let us describe the process of a shot. Suppose that the trigger is in front of the first character of the string (the first slot). If a shot doesn't kill anyone and the cylinder shifts, then the string shifts left. So the first character becomes the last one, the second character becomes the first one, and so on. But the trigger doesn't move. It will be in front of the first character of the resulting string.
Among all the strings that give the minimal probability of loss, Sasha choose the lexicographically minimal one. According to this very string, he charges the gun. You have to help Sasha to charge the gun. For that, each xi query must be answered: is there a bullet in the positions xi?
Input
The first line contains three integers n, k and p (1 β€ n β€ 1018, 0 β€ k β€ n, 1 β€ p β€ 1000) β the number of slots in the cylinder, the number of bullets and the number of queries. Then follow p lines; they are the queries. Each line contains one integer xi (1 β€ xi β€ n) the number of slot to describe.
Please do not use the %lld specificator to read or write 64-bit numbers in Π‘++. It is preferred to use cin, cout streams or the %I64d specificator.
Output
For each query print "." if the slot should be empty and "X" if the slot should be charged.
Examples
Input
3 1 3
1
2
3
Output
..X
Input
6 3 6
1
2
3
4
5
6
Output
.X.X.X
Input
5 2 5
1
2
3
4
5
Output
...XX
Note
The lexicographical comparison of is performed by the < operator in modern programming languages. The a string is lexicographically less that the b string, if there exists such i (1 β€ i β€ n), that ai < bi, and for any j (1 β€ j < i) aj = bj.
Submitted Solution:
```
n, k, p = map(int, input().split())
for i in range(p):
pos = int(input())
if n % 2:
print('.' if n - pos >= k else 'X', end='')
else:
if k * 2 > n:
print('X' if (pos & 1) == 0 or (n - pos + 1) // 2 + \
n // 2 <= k else '.', end='')
else:
print('X' if (pos & 1) == 0 and (n - pos + 2) // 2 <= k \
else '.', end='')
``` | instruction | 0 | 62,292 | 19 | 124,584 |
No | output | 1 | 62,292 | 19 | 124,585 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently, Tokitsukaze found an interesting game. Tokitsukaze had n items at the beginning of this game. However, she thought there were too many items, so now she wants to discard m (1 β€ m β€ n) special items of them.
These n items are marked with indices from 1 to n. In the beginning, the item with index i is placed on the i-th position. Items are divided into several pages orderly, such that each page contains exactly k positions and the last positions on the last page may be left empty.
Tokitsukaze would do the following operation: focus on the first special page that contains at least one special item, and at one time, Tokitsukaze would discard all special items on this page. After an item is discarded or moved, its old position would be empty, and then the item below it, if exists, would move up to this empty position. The movement may bring many items forward and even into previous pages, so Tokitsukaze would keep waiting until all the items stop moving, and then do the operation (i.e. check the special page and discard the special items) repeatedly until there is no item need to be discarded.
<image> Consider the first example from the statement: n=10, m=4, k=5, p=[3, 5, 7, 10]. The are two pages. Initially, the first page is special (since it is the first page containing a special item). So Tokitsukaze discards the special items with indices 3 and 5. After, the first page remains to be special. It contains [1, 2, 4, 6, 7], Tokitsukaze discards the special item with index 7. After, the second page is special (since it is the first page containing a special item). It contains [9, 10], Tokitsukaze discards the special item with index 10.
Tokitsukaze wants to know the number of operations she would do in total.
Input
The first line contains three integers n, m and k (1 β€ n β€ 10^{18}, 1 β€ m β€ 10^5, 1 β€ m, k β€ n) β the number of items, the number of special items to be discarded and the number of positions in each page.
The second line contains m distinct integers p_1, p_2, β¦, p_m (1 β€ p_1 < p_2 < β¦ < p_m β€ n) β the indices of special items which should be discarded.
Output
Print a single integer β the number of operations that Tokitsukaze would do in total.
Examples
Input
10 4 5
3 5 7 10
Output
3
Input
13 4 5
7 8 9 10
Output
1
Note
For the first example:
* In the first operation, Tokitsukaze would focus on the first page [1, 2, 3, 4, 5] and discard items with indices 3 and 5;
* In the second operation, Tokitsukaze would focus on the first page [1, 2, 4, 6, 7] and discard item with index 7;
* In the third operation, Tokitsukaze would focus on the second page [9, 10] and discard item with index 10.
For the second example, Tokitsukaze would focus on the second page [6, 7, 8, 9, 10] and discard all special items at once.
Submitted Solution:
```
n, m, k = map(int, input().split())
p = list(map(int, input().split()))
page_num = 1
shifted = 0
removed_without_shift = 0
num_of_operation = 1
for e in p:
relative_pos = e - shifted
if relative_pos <= page_num * k: # this special pos removed without any shift
removed_without_shift += 1
else:
if removed_without_shift > 0:
num_of_operation += 1
shifted += removed_without_shift
relative_pos = e - shifted
page_num = (relative_pos + k - 1) // k
removed_without_shift = 1
print(num_of_operation)
``` | instruction | 0 | 62,363 | 19 | 124,726 |
Yes | output | 1 | 62,363 | 19 | 124,727 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently, Tokitsukaze found an interesting game. Tokitsukaze had n items at the beginning of this game. However, she thought there were too many items, so now she wants to discard m (1 β€ m β€ n) special items of them.
These n items are marked with indices from 1 to n. In the beginning, the item with index i is placed on the i-th position. Items are divided into several pages orderly, such that each page contains exactly k positions and the last positions on the last page may be left empty.
Tokitsukaze would do the following operation: focus on the first special page that contains at least one special item, and at one time, Tokitsukaze would discard all special items on this page. After an item is discarded or moved, its old position would be empty, and then the item below it, if exists, would move up to this empty position. The movement may bring many items forward and even into previous pages, so Tokitsukaze would keep waiting until all the items stop moving, and then do the operation (i.e. check the special page and discard the special items) repeatedly until there is no item need to be discarded.
<image> Consider the first example from the statement: n=10, m=4, k=5, p=[3, 5, 7, 10]. The are two pages. Initially, the first page is special (since it is the first page containing a special item). So Tokitsukaze discards the special items with indices 3 and 5. After, the first page remains to be special. It contains [1, 2, 4, 6, 7], Tokitsukaze discards the special item with index 7. After, the second page is special (since it is the first page containing a special item). It contains [9, 10], Tokitsukaze discards the special item with index 10.
Tokitsukaze wants to know the number of operations she would do in total.
Input
The first line contains three integers n, m and k (1 β€ n β€ 10^{18}, 1 β€ m β€ 10^5, 1 β€ m, k β€ n) β the number of items, the number of special items to be discarded and the number of positions in each page.
The second line contains m distinct integers p_1, p_2, β¦, p_m (1 β€ p_1 < p_2 < β¦ < p_m β€ n) β the indices of special items which should be discarded.
Output
Print a single integer β the number of operations that Tokitsukaze would do in total.
Examples
Input
10 4 5
3 5 7 10
Output
3
Input
13 4 5
7 8 9 10
Output
1
Note
For the first example:
* In the first operation, Tokitsukaze would focus on the first page [1, 2, 3, 4, 5] and discard items with indices 3 and 5;
* In the second operation, Tokitsukaze would focus on the first page [1, 2, 4, 6, 7] and discard item with index 7;
* In the third operation, Tokitsukaze would focus on the second page [9, 10] and discard item with index 10.
For the second example, Tokitsukaze would focus on the second page [6, 7, 8, 9, 10] and discard all special items at once.
Submitted Solution:
```
n, m, k = map(int, input().split())
arr = [int(x) for x in input().split()]
modulo = 0
tmp = 0
op = 1
cur = (arr[0] - 1) // k
for i in range(m):
if (arr[i] - 1 - modulo) // k != cur:
modulo += tmp
cur = (arr[i] - 1 - modulo) // k
tmp = 0
op += 1
tmp += 1
print(op)
``` | instruction | 0 | 62,364 | 19 | 124,728 |
Yes | output | 1 | 62,364 | 19 | 124,729 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently, Tokitsukaze found an interesting game. Tokitsukaze had n items at the beginning of this game. However, she thought there were too many items, so now she wants to discard m (1 β€ m β€ n) special items of them.
These n items are marked with indices from 1 to n. In the beginning, the item with index i is placed on the i-th position. Items are divided into several pages orderly, such that each page contains exactly k positions and the last positions on the last page may be left empty.
Tokitsukaze would do the following operation: focus on the first special page that contains at least one special item, and at one time, Tokitsukaze would discard all special items on this page. After an item is discarded or moved, its old position would be empty, and then the item below it, if exists, would move up to this empty position. The movement may bring many items forward and even into previous pages, so Tokitsukaze would keep waiting until all the items stop moving, and then do the operation (i.e. check the special page and discard the special items) repeatedly until there is no item need to be discarded.
<image> Consider the first example from the statement: n=10, m=4, k=5, p=[3, 5, 7, 10]. The are two pages. Initially, the first page is special (since it is the first page containing a special item). So Tokitsukaze discards the special items with indices 3 and 5. After, the first page remains to be special. It contains [1, 2, 4, 6, 7], Tokitsukaze discards the special item with index 7. After, the second page is special (since it is the first page containing a special item). It contains [9, 10], Tokitsukaze discards the special item with index 10.
Tokitsukaze wants to know the number of operations she would do in total.
Input
The first line contains three integers n, m and k (1 β€ n β€ 10^{18}, 1 β€ m β€ 10^5, 1 β€ m, k β€ n) β the number of items, the number of special items to be discarded and the number of positions in each page.
The second line contains m distinct integers p_1, p_2, β¦, p_m (1 β€ p_1 < p_2 < β¦ < p_m β€ n) β the indices of special items which should be discarded.
Output
Print a single integer β the number of operations that Tokitsukaze would do in total.
Examples
Input
10 4 5
3 5 7 10
Output
3
Input
13 4 5
7 8 9 10
Output
1
Note
For the first example:
* In the first operation, Tokitsukaze would focus on the first page [1, 2, 3, 4, 5] and discard items with indices 3 and 5;
* In the second operation, Tokitsukaze would focus on the first page [1, 2, 4, 6, 7] and discard item with index 7;
* In the third operation, Tokitsukaze would focus on the second page [9, 10] and discard item with index 10.
For the second example, Tokitsukaze would focus on the second page [6, 7, 8, 9, 10] and discard all special items at once.
Submitted Solution:
```
import math
n , m , k = map(int,input().split())
lis=list(map(int,input().split()))
c=0
ans=1
a=1
page = (lis[0]-a)//k
for i in range(m):
if (lis[i]-a)//k==page:
c+=1
else:
a+=c
ans+=1
c=1
page = (lis[i]-a)//k
print(ans)
``` | instruction | 0 | 62,365 | 19 | 124,730 |
Yes | output | 1 | 62,365 | 19 | 124,731 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently, Tokitsukaze found an interesting game. Tokitsukaze had n items at the beginning of this game. However, she thought there were too many items, so now she wants to discard m (1 β€ m β€ n) special items of them.
These n items are marked with indices from 1 to n. In the beginning, the item with index i is placed on the i-th position. Items are divided into several pages orderly, such that each page contains exactly k positions and the last positions on the last page may be left empty.
Tokitsukaze would do the following operation: focus on the first special page that contains at least one special item, and at one time, Tokitsukaze would discard all special items on this page. After an item is discarded or moved, its old position would be empty, and then the item below it, if exists, would move up to this empty position. The movement may bring many items forward and even into previous pages, so Tokitsukaze would keep waiting until all the items stop moving, and then do the operation (i.e. check the special page and discard the special items) repeatedly until there is no item need to be discarded.
<image> Consider the first example from the statement: n=10, m=4, k=5, p=[3, 5, 7, 10]. The are two pages. Initially, the first page is special (since it is the first page containing a special item). So Tokitsukaze discards the special items with indices 3 and 5. After, the first page remains to be special. It contains [1, 2, 4, 6, 7], Tokitsukaze discards the special item with index 7. After, the second page is special (since it is the first page containing a special item). It contains [9, 10], Tokitsukaze discards the special item with index 10.
Tokitsukaze wants to know the number of operations she would do in total.
Input
The first line contains three integers n, m and k (1 β€ n β€ 10^{18}, 1 β€ m β€ 10^5, 1 β€ m, k β€ n) β the number of items, the number of special items to be discarded and the number of positions in each page.
The second line contains m distinct integers p_1, p_2, β¦, p_m (1 β€ p_1 < p_2 < β¦ < p_m β€ n) β the indices of special items which should be discarded.
Output
Print a single integer β the number of operations that Tokitsukaze would do in total.
Examples
Input
10 4 5
3 5 7 10
Output
3
Input
13 4 5
7 8 9 10
Output
1
Note
For the first example:
* In the first operation, Tokitsukaze would focus on the first page [1, 2, 3, 4, 5] and discard items with indices 3 and 5;
* In the second operation, Tokitsukaze would focus on the first page [1, 2, 4, 6, 7] and discard item with index 7;
* In the third operation, Tokitsukaze would focus on the second page [9, 10] and discard item with index 10.
For the second example, Tokitsukaze would focus on the second page [6, 7, 8, 9, 10] and discard all special items at once.
Submitted Solution:
```
n,m,k = map(int,input().split())
p = list(map(int,input().split()))
ind =0
done =0
q = 0
while ind <m:
st = (p[ind]+k-1-done)//k
sr = ind
while ind<m and (p[ind]+k-1-done)//k==st:
ind+=1
done +=ind -sr
q+=1
print(q)
``` | instruction | 0 | 62,366 | 19 | 124,732 |
Yes | output | 1 | 62,366 | 19 | 124,733 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently, Tokitsukaze found an interesting game. Tokitsukaze had n items at the beginning of this game. However, she thought there were too many items, so now she wants to discard m (1 β€ m β€ n) special items of them.
These n items are marked with indices from 1 to n. In the beginning, the item with index i is placed on the i-th position. Items are divided into several pages orderly, such that each page contains exactly k positions and the last positions on the last page may be left empty.
Tokitsukaze would do the following operation: focus on the first special page that contains at least one special item, and at one time, Tokitsukaze would discard all special items on this page. After an item is discarded or moved, its old position would be empty, and then the item below it, if exists, would move up to this empty position. The movement may bring many items forward and even into previous pages, so Tokitsukaze would keep waiting until all the items stop moving, and then do the operation (i.e. check the special page and discard the special items) repeatedly until there is no item need to be discarded.
<image> Consider the first example from the statement: n=10, m=4, k=5, p=[3, 5, 7, 10]. The are two pages. Initially, the first page is special (since it is the first page containing a special item). So Tokitsukaze discards the special items with indices 3 and 5. After, the first page remains to be special. It contains [1, 2, 4, 6, 7], Tokitsukaze discards the special item with index 7. After, the second page is special (since it is the first page containing a special item). It contains [9, 10], Tokitsukaze discards the special item with index 10.
Tokitsukaze wants to know the number of operations she would do in total.
Input
The first line contains three integers n, m and k (1 β€ n β€ 10^{18}, 1 β€ m β€ 10^5, 1 β€ m, k β€ n) β the number of items, the number of special items to be discarded and the number of positions in each page.
The second line contains m distinct integers p_1, p_2, β¦, p_m (1 β€ p_1 < p_2 < β¦ < p_m β€ n) β the indices of special items which should be discarded.
Output
Print a single integer β the number of operations that Tokitsukaze would do in total.
Examples
Input
10 4 5
3 5 7 10
Output
3
Input
13 4 5
7 8 9 10
Output
1
Note
For the first example:
* In the first operation, Tokitsukaze would focus on the first page [1, 2, 3, 4, 5] and discard items with indices 3 and 5;
* In the second operation, Tokitsukaze would focus on the first page [1, 2, 4, 6, 7] and discard item with index 7;
* In the third operation, Tokitsukaze would focus on the second page [9, 10] and discard item with index 10.
For the second example, Tokitsukaze would focus on the second page [6, 7, 8, 9, 10] and discard all special items at once.
Submitted Solution:
```
# from random import randint
# a = list(map(int, input().split()))
# a, b, c, d = map(int, input().split())
# q = int(input())
def gcd(a, b):
if b == 0:
return a
else:
return gcd(b, a % b)
n, m, k = map(int, input().split())
p = list(map(int, input().split()))
res = 1
s = 1
q = 0
a = (p[0] - s) // k
for i in range(m):
if (p[i] - s) // k == a:
q += 1
else:
res += 1
s += q
q = 0
a = (p[0] - s) // k
print(res)
``` | instruction | 0 | 62,367 | 19 | 124,734 |
No | output | 1 | 62,367 | 19 | 124,735 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently, Tokitsukaze found an interesting game. Tokitsukaze had n items at the beginning of this game. However, she thought there were too many items, so now she wants to discard m (1 β€ m β€ n) special items of them.
These n items are marked with indices from 1 to n. In the beginning, the item with index i is placed on the i-th position. Items are divided into several pages orderly, such that each page contains exactly k positions and the last positions on the last page may be left empty.
Tokitsukaze would do the following operation: focus on the first special page that contains at least one special item, and at one time, Tokitsukaze would discard all special items on this page. After an item is discarded or moved, its old position would be empty, and then the item below it, if exists, would move up to this empty position. The movement may bring many items forward and even into previous pages, so Tokitsukaze would keep waiting until all the items stop moving, and then do the operation (i.e. check the special page and discard the special items) repeatedly until there is no item need to be discarded.
<image> Consider the first example from the statement: n=10, m=4, k=5, p=[3, 5, 7, 10]. The are two pages. Initially, the first page is special (since it is the first page containing a special item). So Tokitsukaze discards the special items with indices 3 and 5. After, the first page remains to be special. It contains [1, 2, 4, 6, 7], Tokitsukaze discards the special item with index 7. After, the second page is special (since it is the first page containing a special item). It contains [9, 10], Tokitsukaze discards the special item with index 10.
Tokitsukaze wants to know the number of operations she would do in total.
Input
The first line contains three integers n, m and k (1 β€ n β€ 10^{18}, 1 β€ m β€ 10^5, 1 β€ m, k β€ n) β the number of items, the number of special items to be discarded and the number of positions in each page.
The second line contains m distinct integers p_1, p_2, β¦, p_m (1 β€ p_1 < p_2 < β¦ < p_m β€ n) β the indices of special items which should be discarded.
Output
Print a single integer β the number of operations that Tokitsukaze would do in total.
Examples
Input
10 4 5
3 5 7 10
Output
3
Input
13 4 5
7 8 9 10
Output
1
Note
For the first example:
* In the first operation, Tokitsukaze would focus on the first page [1, 2, 3, 4, 5] and discard items with indices 3 and 5;
* In the second operation, Tokitsukaze would focus on the first page [1, 2, 4, 6, 7] and discard item with index 7;
* In the third operation, Tokitsukaze would focus on the second page [9, 10] and discard item with index 10.
For the second example, Tokitsukaze would focus on the second page [6, 7, 8, 9, 10] and discard all special items at once.
Submitted Solution:
```
n, m, k = map(int, input().split())
a = list(map(int, input().split()))
for i in range(m):
a[i] -= 1
l = 0
ct = 0
moves = 0
while l < m:
cur = (a[l]-ct)%m
r = l+1
while r < m and (a[r]-ct)%m > (a[l]-ct)%m:
r += 1
ct += 1
moves += 1
l = r
print(moves)
``` | instruction | 0 | 62,368 | 19 | 124,736 |
No | output | 1 | 62,368 | 19 | 124,737 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently, Tokitsukaze found an interesting game. Tokitsukaze had n items at the beginning of this game. However, she thought there were too many items, so now she wants to discard m (1 β€ m β€ n) special items of them.
These n items are marked with indices from 1 to n. In the beginning, the item with index i is placed on the i-th position. Items are divided into several pages orderly, such that each page contains exactly k positions and the last positions on the last page may be left empty.
Tokitsukaze would do the following operation: focus on the first special page that contains at least one special item, and at one time, Tokitsukaze would discard all special items on this page. After an item is discarded or moved, its old position would be empty, and then the item below it, if exists, would move up to this empty position. The movement may bring many items forward and even into previous pages, so Tokitsukaze would keep waiting until all the items stop moving, and then do the operation (i.e. check the special page and discard the special items) repeatedly until there is no item need to be discarded.
<image> Consider the first example from the statement: n=10, m=4, k=5, p=[3, 5, 7, 10]. The are two pages. Initially, the first page is special (since it is the first page containing a special item). So Tokitsukaze discards the special items with indices 3 and 5. After, the first page remains to be special. It contains [1, 2, 4, 6, 7], Tokitsukaze discards the special item with index 7. After, the second page is special (since it is the first page containing a special item). It contains [9, 10], Tokitsukaze discards the special item with index 10.
Tokitsukaze wants to know the number of operations she would do in total.
Input
The first line contains three integers n, m and k (1 β€ n β€ 10^{18}, 1 β€ m β€ 10^5, 1 β€ m, k β€ n) β the number of items, the number of special items to be discarded and the number of positions in each page.
The second line contains m distinct integers p_1, p_2, β¦, p_m (1 β€ p_1 < p_2 < β¦ < p_m β€ n) β the indices of special items which should be discarded.
Output
Print a single integer β the number of operations that Tokitsukaze would do in total.
Examples
Input
10 4 5
3 5 7 10
Output
3
Input
13 4 5
7 8 9 10
Output
1
Note
For the first example:
* In the first operation, Tokitsukaze would focus on the first page [1, 2, 3, 4, 5] and discard items with indices 3 and 5;
* In the second operation, Tokitsukaze would focus on the first page [1, 2, 4, 6, 7] and discard item with index 7;
* In the third operation, Tokitsukaze would focus on the second page [9, 10] and discard item with index 10.
For the second example, Tokitsukaze would focus on the second page [6, 7, 8, 9, 10] and discard all special items at once.
Submitted Solution:
```
n, m, k = [int(i) for i in input().split()]
lst = [int(j) for j in input().split()]
result = 0
for i in range(m):
count = i
while i + 1 < m and (lst[i + 1] - count) // k == (lst[i] - count) // k:
i += 1
result += 1
print(result)
``` | instruction | 0 | 62,369 | 19 | 124,738 |
No | output | 1 | 62,369 | 19 | 124,739 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently, Tokitsukaze found an interesting game. Tokitsukaze had n items at the beginning of this game. However, she thought there were too many items, so now she wants to discard m (1 β€ m β€ n) special items of them.
These n items are marked with indices from 1 to n. In the beginning, the item with index i is placed on the i-th position. Items are divided into several pages orderly, such that each page contains exactly k positions and the last positions on the last page may be left empty.
Tokitsukaze would do the following operation: focus on the first special page that contains at least one special item, and at one time, Tokitsukaze would discard all special items on this page. After an item is discarded or moved, its old position would be empty, and then the item below it, if exists, would move up to this empty position. The movement may bring many items forward and even into previous pages, so Tokitsukaze would keep waiting until all the items stop moving, and then do the operation (i.e. check the special page and discard the special items) repeatedly until there is no item need to be discarded.
<image> Consider the first example from the statement: n=10, m=4, k=5, p=[3, 5, 7, 10]. The are two pages. Initially, the first page is special (since it is the first page containing a special item). So Tokitsukaze discards the special items with indices 3 and 5. After, the first page remains to be special. It contains [1, 2, 4, 6, 7], Tokitsukaze discards the special item with index 7. After, the second page is special (since it is the first page containing a special item). It contains [9, 10], Tokitsukaze discards the special item with index 10.
Tokitsukaze wants to know the number of operations she would do in total.
Input
The first line contains three integers n, m and k (1 β€ n β€ 10^{18}, 1 β€ m β€ 10^5, 1 β€ m, k β€ n) β the number of items, the number of special items to be discarded and the number of positions in each page.
The second line contains m distinct integers p_1, p_2, β¦, p_m (1 β€ p_1 < p_2 < β¦ < p_m β€ n) β the indices of special items which should be discarded.
Output
Print a single integer β the number of operations that Tokitsukaze would do in total.
Examples
Input
10 4 5
3 5 7 10
Output
3
Input
13 4 5
7 8 9 10
Output
1
Note
For the first example:
* In the first operation, Tokitsukaze would focus on the first page [1, 2, 3, 4, 5] and discard items with indices 3 and 5;
* In the second operation, Tokitsukaze would focus on the first page [1, 2, 4, 6, 7] and discard item with index 7;
* In the third operation, Tokitsukaze would focus on the second page [9, 10] and discard item with index 10.
For the second example, Tokitsukaze would focus on the second page [6, 7, 8, 9, 10] and discard all special items at once.
Submitted Solution:
```
n,m,k=list(map(int,input().split()))
ar=list(map(int,input().split()))
i=0
r=0
size=k
while(i<len(ar)):
c=0
if(ar[i]<=k):
while(i<len(ar) and ar[i]<=k):
c+=1
i+=1
r+=1
else:
a=ar[i]-k
k+=size*(a//size+1)
k+=c
print(r)
``` | instruction | 0 | 62,370 | 19 | 124,740 |
No | output | 1 | 62,370 | 19 | 124,741 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice and Bob are playing yet another card game. This time the rules are the following. There are n cards lying in a row in front of them. The i-th card has value a_i.
First, Alice chooses a non-empty consecutive segment of cards [l; r] (l β€ r). After that Bob removes a single card j from that segment (l β€ j β€ r). The score of the game is the total value of the remaining cards on the segment (a_l + a_{l + 1} + ... + a_{j - 1} + a_{j + 1} + ... + a_{r - 1} + a_r). In particular, if Alice chooses a segment with just one element, then the score after Bob removes the only card is 0.
Alice wants to make the score as big as possible. Bob takes such a card that the score is as small as possible.
What segment should Alice choose so that the score is maximum possible? Output the maximum score.
Input
The first line contains a single integer n (1 β€ n β€ 10^5) β the number of cards.
The second line contains n integers a_1, a_2, ..., a_n (-30 β€ a_i β€ 30) β the values on the cards.
Output
Print a single integer β the final score of the game.
Examples
Input
5
5 -2 10 -1 4
Output
6
Input
8
5 2 5 3 -30 -30 6 9
Output
10
Input
3
-10 6 -15
Output
0
Note
In the first example Alice chooses a segment [1;5] β the entire row of cards. Bob removes card 3 with the value 10 from the segment. Thus, the final score is 5 + (-2) + (-1) + 4 = 6.
In the second example Alice chooses a segment [1;4], so that Bob removes either card 1 or 3 with the value 5, making the answer 5 + 2 + 3 = 10.
In the third example Alice can choose any of the segments of length 1: [1;1], [2;2] or [3;3]. Bob removes the only card, so the score is 0. If Alice chooses some other segment then the answer will be less than 0. | instruction | 0 | 62,456 | 19 | 124,912 |
Tags: data structures, dp, implementation, two pointers
Correct Solution:
```
# import sys
# sys = input.stdin.readline
n=int(input())
arr=list(map(int,input().split()))
ans=0
for i in range(1,31):
res=0
for j in arr:
if(j>i):
res = 0
else:
res += j
res=max(res,0)
ans=max(ans,res-i)
print(ans)
``` | output | 1 | 62,456 | 19 | 124,913 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice and Bob are playing yet another card game. This time the rules are the following. There are n cards lying in a row in front of them. The i-th card has value a_i.
First, Alice chooses a non-empty consecutive segment of cards [l; r] (l β€ r). After that Bob removes a single card j from that segment (l β€ j β€ r). The score of the game is the total value of the remaining cards on the segment (a_l + a_{l + 1} + ... + a_{j - 1} + a_{j + 1} + ... + a_{r - 1} + a_r). In particular, if Alice chooses a segment with just one element, then the score after Bob removes the only card is 0.
Alice wants to make the score as big as possible. Bob takes such a card that the score is as small as possible.
What segment should Alice choose so that the score is maximum possible? Output the maximum score.
Input
The first line contains a single integer n (1 β€ n β€ 10^5) β the number of cards.
The second line contains n integers a_1, a_2, ..., a_n (-30 β€ a_i β€ 30) β the values on the cards.
Output
Print a single integer β the final score of the game.
Examples
Input
5
5 -2 10 -1 4
Output
6
Input
8
5 2 5 3 -30 -30 6 9
Output
10
Input
3
-10 6 -15
Output
0
Note
In the first example Alice chooses a segment [1;5] β the entire row of cards. Bob removes card 3 with the value 10 from the segment. Thus, the final score is 5 + (-2) + (-1) + 4 = 6.
In the second example Alice chooses a segment [1;4], so that Bob removes either card 1 or 3 with the value 5, making the answer 5 + 2 + 3 = 10.
In the third example Alice can choose any of the segments of length 1: [1;1], [2;2] or [3;3]. Bob removes the only card, so the score is 0. If Alice chooses some other segment then the answer will be less than 0. | instruction | 0 | 62,457 | 19 | 124,914 |
Tags: data structures, dp, implementation, two pointers
Correct Solution:
```
# Enter your code here. Read input from STDIN. Print output to STDOUT# ===============================================================================================
# importing some useful libraries.
from __future__ import division, print_function
from fractions import Fraction
import sys
import os
from io import BytesIO, IOBase
from itertools import *
import bisect
from heapq import *
from math import ceil, floor
from copy import *
from collections import deque, defaultdict
from collections import Counter as counter # Counter(list) return a dict with {key: count}
from itertools import combinations # if a = [1,2,3] then print(list(comb(a,2))) -----> [(1, 2), (1, 3), (2, 3)]
from itertools import permutations as permutate
from bisect import bisect_left as bl
from operator import *
# If the element is already present in the list,
# the left most position where element has to be inserted is returned.
from bisect import bisect_right as br
from bisect import bisect
# If the element is already present in the list,
# the right most position where element has to be inserted is returned
# ==============================================================================================
# fast I/O region
BUFSIZE = 8192
from sys import stderr
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
# inp = lambda: sys.stdin.readline().rstrip("\r\n")
# ===============================================================================================
### START ITERATE RECURSION ###
from types import GeneratorType
def iterative(f, stack=[]):
def wrapped_func(*args, **kwargs):
if stack: return f(*args, **kwargs)
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
continue
stack.pop()
if not stack: break
to = stack[-1].send(to)
return to
return wrapped_func
#### END ITERATE RECURSION ####
###########################
# Sorted list
class SortedList:
def __init__(self, iterable=[], _load=200):
"""Initialize sorted list instance."""
values = sorted(iterable)
self._len = _len = len(values)
self._load = _load
self._lists = _lists = [values[i:i + _load] for i in range(0, _len, _load)]
self._list_lens = [len(_list) for _list in _lists]
self._mins = [_list[0] for _list in _lists]
self._fen_tree = []
self._rebuild = True
def _fen_build(self):
"""Build a fenwick tree instance."""
self._fen_tree[:] = self._list_lens
_fen_tree = self._fen_tree
for i in range(len(_fen_tree)):
if i | i + 1 < len(_fen_tree):
_fen_tree[i | i + 1] += _fen_tree[i]
self._rebuild = False
def _fen_update(self, index, value):
"""Update `fen_tree[index] += value`."""
if not self._rebuild:
_fen_tree = self._fen_tree
while index < len(_fen_tree):
_fen_tree[index] += value
index |= index + 1
def _fen_query(self, end):
"""Return `sum(_fen_tree[:end])`."""
if self._rebuild:
self._fen_build()
_fen_tree = self._fen_tree
x = 0
while end:
x += _fen_tree[end - 1]
end &= end - 1
return x
def _fen_findkth(self, k):
"""Return a pair of (the largest `idx` such that `sum(_fen_tree[:idx]) <= k`, `k - sum(_fen_tree[:idx])`)."""
_list_lens = self._list_lens
if k < _list_lens[0]:
return 0, k
if k >= self._len - _list_lens[-1]:
return len(_list_lens) - 1, k + _list_lens[-1] - self._len
if self._rebuild:
self._fen_build()
_fen_tree = self._fen_tree
idx = -1
for d in reversed(range(len(_fen_tree).bit_length())):
right_idx = idx + (1 << d)
if right_idx < len(_fen_tree) and k >= _fen_tree[right_idx]:
idx = right_idx
k -= _fen_tree[idx]
return idx + 1, k
def _delete(self, pos, idx):
"""Delete value at the given `(pos, idx)`."""
_lists = self._lists
_mins = self._mins
_list_lens = self._list_lens
self._len -= 1
self._fen_update(pos, -1)
del _lists[pos][idx]
_list_lens[pos] -= 1
if _list_lens[pos]:
_mins[pos] = _lists[pos][0]
else:
del _lists[pos]
del _list_lens[pos]
del _mins[pos]
self._rebuild = True
def _loc_left(self, value):
"""Return an index pair that corresponds to the first position of `value` in the sorted list."""
if not self._len:
return 0, 0
_lists = self._lists
_mins = self._mins
lo, pos = -1, len(_lists) - 1
while lo + 1 < pos:
mi = (lo + pos) >> 1
if value <= _mins[mi]:
pos = mi
else:
lo = mi
if pos and value <= _lists[pos - 1][-1]:
pos -= 1
_list = _lists[pos]
lo, idx = -1, len(_list)
while lo + 1 < idx:
mi = (lo + idx) >> 1
if value <= _list[mi]:
idx = mi
else:
lo = mi
return pos, idx
def _loc_right(self, value):
"""Return an index pair that corresponds to the last position of `value` in the sorted list."""
if not self._len:
return 0, 0
_lists = self._lists
_mins = self._mins
pos, hi = 0, len(_lists)
while pos + 1 < hi:
mi = (pos + hi) >> 1
if value < _mins[mi]:
hi = mi
else:
pos = mi
_list = _lists[pos]
lo, idx = -1, len(_list)
while lo + 1 < idx:
mi = (lo + idx) >> 1
if value < _list[mi]:
idx = mi
else:
lo = mi
return pos, idx
def add(self, value):
"""Add `value` to sorted list."""
_load = self._load
_lists = self._lists
_mins = self._mins
_list_lens = self._list_lens
self._len += 1
if _lists:
pos, idx = self._loc_right(value)
self._fen_update(pos, 1)
_list = _lists[pos]
_list.insert(idx, value)
_list_lens[pos] += 1
_mins[pos] = _list[0]
if _load + _load < len(_list):
_lists.insert(pos + 1, _list[_load:])
_list_lens.insert(pos + 1, len(_list) - _load)
_mins.insert(pos + 1, _list[_load])
_list_lens[pos] = _load
del _list[_load:]
self._rebuild = True
else:
_lists.append([value])
_mins.append(value)
_list_lens.append(1)
self._rebuild = True
def discard(self, value):
"""Remove `value` from sorted list if it is a member."""
_lists = self._lists
if _lists:
pos, idx = self._loc_right(value)
if idx and _lists[pos][idx - 1] == value:
self._delete(pos, idx - 1)
def remove(self, value):
"""Remove `value` from sorted list; `value` must be a member."""
_len = self._len
self.discard(value)
if _len == self._len:
raise ValueError('{0!r} not in list'.format(value))
def pop(self, index=-1):
"""Remove and return value at `index` in sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
value = self._lists[pos][idx]
self._delete(pos, idx)
return value
def bisect_left(self, value):
"""Return the first index to insert `value` in the sorted list."""
pos, idx = self._loc_left(value)
return self._fen_query(pos) + idx
def bisect_right(self, value):
"""Return the last index to insert `value` in the sorted list."""
pos, idx = self._loc_right(value)
return self._fen_query(pos) + idx
def count(self, value):
"""Return number of occurrences of `value` in the sorted list."""
return self.bisect_right(value) - self.bisect_left(value)
def __len__(self):
"""Return the size of the sorted list."""
return self._len
def __getitem__(self, index):
"""Lookup value at `index` in sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
return self._lists[pos][idx]
def __delitem__(self, index):
"""Remove value at `index` from sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
self._delete(pos, idx)
def __contains__(self, value):
"""Return true if `value` is an element of the sorted list."""
_lists = self._lists
if _lists:
pos, idx = self._loc_left(value)
return idx < len(_lists[pos]) and _lists[pos][idx] == value
return False
def __iter__(self):
"""Return an iterator over the sorted list."""
return (value for _list in self._lists for value in _list)
def __reversed__(self):
"""Return a reverse iterator over the sorted list."""
return (value for _list in reversed(self._lists) for value in reversed(_list))
def __repr__(self):
"""Return string representation of sorted list."""
return 'SortedList({0})'.format(list(self))
# ===============================================================================================
# some shortcuts
mod = 1000000007
def inp(): return sys.stdin.readline().rstrip("\r\n") # for fast input
def out(var): sys.stdout.write(str(var)) # for fast output, always take string
def lis(): return list(map(int, inp().split()))
def stringlis(): return list(map(str, inp().split()))
def sep(): return map(int, inp().split())
def strsep(): return map(str, inp().split())
def fsep(): return map(float, inp().split())
def nextline(): out("\n") # as stdout.write always print sring.
def testcase(t):
for p in range(t):
solve()
def pow(x, y, p):
res = 1 # Initialize result
x = x % p # Update x if it is more , than or equal to p
if (x == 0):
return 0
while (y > 0):
if ((y & 1) == 1): # If y is odd, multiply, x with result
res = (res * x) % p
y = y >> 1 # y = y/2
x = (x * x) % p
return res
from functools import reduce
def factors(n):
return set(reduce(list.__add__,
([i, n // i] for i in range(1, int(n ** 0.5) + 1) if n % i == 0)))
def gcd(a, b):
if a == b: return a
while b > 0: a, b = b, a % b
return a
# discrete binary search
# minimise:
# def search():
# l = 0
# r = 10 ** 15
#
# for i in range(200):
# if isvalid(l):
# return l
# if l == r:
# return l
# m = (l + r) // 2
# if isvalid(m) and not isvalid(m - 1):
# return m
# if isvalid(m):
# r = m + 1
# else:
# l = m
# return m
# maximise:
# def search():
# l = 0
# r = 10 ** 15
#
# for i in range(200):
# # print(l,r)
# if isvalid(r):
# return r
# if l == r:
# return l
# m = (l + r) // 2
# if isvalid(m) and not isvalid(m + 1):
# return m
# if isvalid(m):
# l = m
# else:
# r = m - 1
# return m
##############Find sum of product of subsets of size k in a array
# ar=[0,1,2,3]
# k=3
# n=len(ar)-1
# dp=[0]*(n+1)
# dp[0]=1
# for pos in range(1,n+1):
# dp[pos]=0
# l=max(1,k+pos-n-1)
# for j in range(min(pos,k),l-1,-1):
# dp[j]=dp[j]+ar[pos]*dp[j-1]
# print(dp[k])
def prefix_sum(ar): # [1,2,3,4]->[1,3,6,10]
return list(accumulate(ar))
def suffix_sum(ar): # [1,2,3,4]->[10,9,7,4]
return list(accumulate(ar[::-1]))[::-1]
def N():
return int(inp())
dx = [0, 0, 1, -1]
dy = [1, -1, 0, 0]
def YES():
print("YES")
def NO():
print("NO")
def Yes():
print("Yes")
def No():
print("No")
# =========================================================================================
from collections import defaultdict
def numberOfSetBits(i):
i = i - ((i >> 1) & 0x55555555)
i = (i & 0x33333333) + ((i >> 2) & 0x33333333)
return (((i + (i >> 4) & 0xF0F0F0F) * 0x1010101) & 0xffffffff) >> 24
class MergeFind:
def __init__(self, n):
self.parent = list(range(n))
self.size = [1] * n
self.num_sets = n
# self.lista = [[_] for _ in range(n)]
def find(self, a):
to_update = []
while a != self.parent[a]:
to_update.append(a)
a = self.parent[a]
for b in to_update:
self.parent[b] = a
return self.parent[a]
def merge(self, a, b):
a = self.find(a)
b = self.find(b)
if a == b:
return
self.num_sets -= 1
self.parent[a] = b
self.size[b] += self.size[a]
# self.lista[a] += self.lista[b]
# self.lista[b] = []
def set_size(self, a):
return self.size[self.find(a)]
def __len__(self):
return self.num_sets
def lcm(a, b):
return abs((a // gcd(a, b)) * b)
#
# # to find factorial and ncr
# tot = 400005
# mod = 10 ** 9 + 7
# fac = [1, 1]
# finv = [1, 1]
# inv = [0, 1]
#
# for i in range(2, tot + 1):
# fac.append((fac[-1] * i) % mod)
# inv.append(mod - (inv[mod % i] * (mod // i) % mod))
# finv.append(finv[-1] * inv[-1] % mod)
#
#
# def comb(n, r):
# if n < r:
# return 0
# else:
# return fac[n] * (finv[r] * finv[n - r] % mod) % mod
def solve():
n=N()
ar=lis()
inf=float("inf")
def find(ar,n):
dp=[ar[0]]
for i in range(1,n):
dp.append(max(dp[-1]+ar[i],ar[i]))
return max(dp)
m=-inf
for i in range(-30,31):
new=[]
for j in range(n):
if ar[j]>i:
new.append(-inf)
else:
new.append(ar[j])
v=find(new,n)
m=max(v-i,m)
print(m)
solve()
#testcase(int(inp()))
``` | output | 1 | 62,457 | 19 | 124,915 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice and Bob are playing yet another card game. This time the rules are the following. There are n cards lying in a row in front of them. The i-th card has value a_i.
First, Alice chooses a non-empty consecutive segment of cards [l; r] (l β€ r). After that Bob removes a single card j from that segment (l β€ j β€ r). The score of the game is the total value of the remaining cards on the segment (a_l + a_{l + 1} + ... + a_{j - 1} + a_{j + 1} + ... + a_{r - 1} + a_r). In particular, if Alice chooses a segment with just one element, then the score after Bob removes the only card is 0.
Alice wants to make the score as big as possible. Bob takes such a card that the score is as small as possible.
What segment should Alice choose so that the score is maximum possible? Output the maximum score.
Input
The first line contains a single integer n (1 β€ n β€ 10^5) β the number of cards.
The second line contains n integers a_1, a_2, ..., a_n (-30 β€ a_i β€ 30) β the values on the cards.
Output
Print a single integer β the final score of the game.
Examples
Input
5
5 -2 10 -1 4
Output
6
Input
8
5 2 5 3 -30 -30 6 9
Output
10
Input
3
-10 6 -15
Output
0
Note
In the first example Alice chooses a segment [1;5] β the entire row of cards. Bob removes card 3 with the value 10 from the segment. Thus, the final score is 5 + (-2) + (-1) + 4 = 6.
In the second example Alice chooses a segment [1;4], so that Bob removes either card 1 or 3 with the value 5, making the answer 5 + 2 + 3 = 10.
In the third example Alice can choose any of the segments of length 1: [1;1], [2;2] or [3;3]. Bob removes the only card, so the score is 0. If Alice chooses some other segment then the answer will be less than 0. | instruction | 0 | 62,458 | 19 | 124,916 |
Tags: data structures, dp, implementation, two pointers
Correct Solution:
```
def calc(x,mx):
if x > mx:
return float("-inf")
else:
return x
def calcmax(arr):
max_so_far=float("-inf")
currmax=0
for i in arr:
currmax+=i
if currmax > max_so_far:
max_so_far =currmax
if currmax < 0:
currmax= 0
return max_so_far
n=int(input())
arr=[int(c) for c in input().split()]
ans=float("-inf")
for i in range(-30,31):
a=[calc(x,i) for x in arr]
ans=max(ans,calcmax(a)-i)
print(ans)
``` | output | 1 | 62,458 | 19 | 124,917 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice and Bob are playing yet another card game. This time the rules are the following. There are n cards lying in a row in front of them. The i-th card has value a_i.
First, Alice chooses a non-empty consecutive segment of cards [l; r] (l β€ r). After that Bob removes a single card j from that segment (l β€ j β€ r). The score of the game is the total value of the remaining cards on the segment (a_l + a_{l + 1} + ... + a_{j - 1} + a_{j + 1} + ... + a_{r - 1} + a_r). In particular, if Alice chooses a segment with just one element, then the score after Bob removes the only card is 0.
Alice wants to make the score as big as possible. Bob takes such a card that the score is as small as possible.
What segment should Alice choose so that the score is maximum possible? Output the maximum score.
Input
The first line contains a single integer n (1 β€ n β€ 10^5) β the number of cards.
The second line contains n integers a_1, a_2, ..., a_n (-30 β€ a_i β€ 30) β the values on the cards.
Output
Print a single integer β the final score of the game.
Examples
Input
5
5 -2 10 -1 4
Output
6
Input
8
5 2 5 3 -30 -30 6 9
Output
10
Input
3
-10 6 -15
Output
0
Note
In the first example Alice chooses a segment [1;5] β the entire row of cards. Bob removes card 3 with the value 10 from the segment. Thus, the final score is 5 + (-2) + (-1) + 4 = 6.
In the second example Alice chooses a segment [1;4], so that Bob removes either card 1 or 3 with the value 5, making the answer 5 + 2 + 3 = 10.
In the third example Alice can choose any of the segments of length 1: [1;1], [2;2] or [3;3]. Bob removes the only card, so the score is 0. If Alice chooses some other segment then the answer will be less than 0. | instruction | 0 | 62,459 | 19 | 124,918 |
Tags: data structures, dp, implementation, two pointers
Correct Solution:
```
INF=-1000000000
n=int(input())
a=list(map(int,input().split()))
ans=0
for mx in range(31):
cur=0
best=0
for i in range(n):
if (a[i]>mx):
val=INF
else:
val=a[i]
cur+=val
best=min(best,cur)
ans=max(ans,(cur-best)-mx)
print(ans)
``` | output | 1 | 62,459 | 19 | 124,919 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice and Bob are playing yet another card game. This time the rules are the following. There are n cards lying in a row in front of them. The i-th card has value a_i.
First, Alice chooses a non-empty consecutive segment of cards [l; r] (l β€ r). After that Bob removes a single card j from that segment (l β€ j β€ r). The score of the game is the total value of the remaining cards on the segment (a_l + a_{l + 1} + ... + a_{j - 1} + a_{j + 1} + ... + a_{r - 1} + a_r). In particular, if Alice chooses a segment with just one element, then the score after Bob removes the only card is 0.
Alice wants to make the score as big as possible. Bob takes such a card that the score is as small as possible.
What segment should Alice choose so that the score is maximum possible? Output the maximum score.
Input
The first line contains a single integer n (1 β€ n β€ 10^5) β the number of cards.
The second line contains n integers a_1, a_2, ..., a_n (-30 β€ a_i β€ 30) β the values on the cards.
Output
Print a single integer β the final score of the game.
Examples
Input
5
5 -2 10 -1 4
Output
6
Input
8
5 2 5 3 -30 -30 6 9
Output
10
Input
3
-10 6 -15
Output
0
Note
In the first example Alice chooses a segment [1;5] β the entire row of cards. Bob removes card 3 with the value 10 from the segment. Thus, the final score is 5 + (-2) + (-1) + 4 = 6.
In the second example Alice chooses a segment [1;4], so that Bob removes either card 1 or 3 with the value 5, making the answer 5 + 2 + 3 = 10.
In the third example Alice can choose any of the segments of length 1: [1;1], [2;2] or [3;3]. Bob removes the only card, so the score is 0. If Alice chooses some other segment then the answer will be less than 0. | instruction | 0 | 62,460 | 19 | 124,920 |
Tags: data structures, dp, implementation, two pointers
Correct Solution:
```
def inp():
return(int(input()))
import math
def inlt():
return(list(map(int,input().split())))
def insr():
s = input()
return(list(s[:len(s) - 1]))
def invr():
return(map(int,input().split()))
import bisect
#python3 codeforces_1359D.py
N = inp()
big = 0
li = inlt()
for j in range(31):
val = 0
for item in li:
if item <= j:
val = max(0, val + item)
else:
val = 0
big = max(big, val - j)
if big < 0:
print(0)
else:
print(big)
#1 -26 -6 26 7 10 -2 -4 -6 14 -9 25 9 -12 -28
``` | output | 1 | 62,460 | 19 | 124,921 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice and Bob are playing yet another card game. This time the rules are the following. There are n cards lying in a row in front of them. The i-th card has value a_i.
First, Alice chooses a non-empty consecutive segment of cards [l; r] (l β€ r). After that Bob removes a single card j from that segment (l β€ j β€ r). The score of the game is the total value of the remaining cards on the segment (a_l + a_{l + 1} + ... + a_{j - 1} + a_{j + 1} + ... + a_{r - 1} + a_r). In particular, if Alice chooses a segment with just one element, then the score after Bob removes the only card is 0.
Alice wants to make the score as big as possible. Bob takes such a card that the score is as small as possible.
What segment should Alice choose so that the score is maximum possible? Output the maximum score.
Input
The first line contains a single integer n (1 β€ n β€ 10^5) β the number of cards.
The second line contains n integers a_1, a_2, ..., a_n (-30 β€ a_i β€ 30) β the values on the cards.
Output
Print a single integer β the final score of the game.
Examples
Input
5
5 -2 10 -1 4
Output
6
Input
8
5 2 5 3 -30 -30 6 9
Output
10
Input
3
-10 6 -15
Output
0
Note
In the first example Alice chooses a segment [1;5] β the entire row of cards. Bob removes card 3 with the value 10 from the segment. Thus, the final score is 5 + (-2) + (-1) + 4 = 6.
In the second example Alice chooses a segment [1;4], so that Bob removes either card 1 or 3 with the value 5, making the answer 5 + 2 + 3 = 10.
In the third example Alice can choose any of the segments of length 1: [1;1], [2;2] or [3;3]. Bob removes the only card, so the score is 0. If Alice chooses some other segment then the answer will be less than 0. | instruction | 0 | 62,461 | 19 | 124,922 |
Tags: data structures, dp, implementation, two pointers
Correct Solution:
```
from bisect import bisect_left as lower_bound, bisect_right as upper_bound
from sys import stdin, stdout
from collections import defaultdict
N = 10**5 + 7
minn = [0 for _ in range(2*N)]
maxx = [0 for _ in range(2*N)]
def build(p, n):
for i in range(n):
maxx[n+i] = p[i]
minn[n+i] = p[i]
for i in range(n-1, 0, -1):
maxx[i] = max(maxx[i<<1], maxx[i<<1|1])
minn[i] = min(minn[i<<1], minn[i<<1|1])
def query(l, r, n):
l += n
r += n
retminn, retmaxx = float('inf'), -float('inf')
while l < r:
if l&1:
retminn = min(retminn, minn[l])
retmaxx = max(retmaxx, maxx[l])
l += 1
if r&1:
r -= 1
retminn = min(retminn, minn[r])
retmaxx = max(retmaxx, maxx[r])
l >>= 1
r >>= 1
return retminn, retmaxx
def main():
n = int(input())
a = list(map(int, input().strip().split()))
left = [-1]*n
right = [n]*n
p = [a[i] for i in range(n)]
for i in range(1, n): p[i] += p[i-1]
build(p, n)
st = [0]
for i in range(1, n):
while st and a[st[-1]] <= a[i]: st.pop()
if st: left[i] = st[-1]
st.append(i)
st = [n-1]
for i in range(n-2, -1, -1):
while st and a[st[-1]] < a[i]: st.pop()
if st: right[i] = st[-1]
st.append(i)
#print(left, right, p)
#
if max(a) <= 0:
print(0)
return None
ret = -float('inf')
for i in range(n):
l, r = left[i], right[i]
_, rmaxx = query(i, r, n)
lminn, _ = query(max(0, l), i, n)
if l < 0: lminn = min(lminn, 0)
s = rmaxx - lminn
#print(i, s, a[i], l, r)
ret = max(ret, s-a[i])
print(ret)
if __name__ == "__main__": main()
``` | output | 1 | 62,461 | 19 | 124,923 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice and Bob are playing yet another card game. This time the rules are the following. There are n cards lying in a row in front of them. The i-th card has value a_i.
First, Alice chooses a non-empty consecutive segment of cards [l; r] (l β€ r). After that Bob removes a single card j from that segment (l β€ j β€ r). The score of the game is the total value of the remaining cards on the segment (a_l + a_{l + 1} + ... + a_{j - 1} + a_{j + 1} + ... + a_{r - 1} + a_r). In particular, if Alice chooses a segment with just one element, then the score after Bob removes the only card is 0.
Alice wants to make the score as big as possible. Bob takes such a card that the score is as small as possible.
What segment should Alice choose so that the score is maximum possible? Output the maximum score.
Input
The first line contains a single integer n (1 β€ n β€ 10^5) β the number of cards.
The second line contains n integers a_1, a_2, ..., a_n (-30 β€ a_i β€ 30) β the values on the cards.
Output
Print a single integer β the final score of the game.
Examples
Input
5
5 -2 10 -1 4
Output
6
Input
8
5 2 5 3 -30 -30 6 9
Output
10
Input
3
-10 6 -15
Output
0
Note
In the first example Alice chooses a segment [1;5] β the entire row of cards. Bob removes card 3 with the value 10 from the segment. Thus, the final score is 5 + (-2) + (-1) + 4 = 6.
In the second example Alice chooses a segment [1;4], so that Bob removes either card 1 or 3 with the value 5, making the answer 5 + 2 + 3 = 10.
In the third example Alice can choose any of the segments of length 1: [1;1], [2;2] or [3;3]. Bob removes the only card, so the score is 0. If Alice chooses some other segment then the answer will be less than 0. | instruction | 0 | 62,462 | 19 | 124,924 |
Tags: data structures, dp, implementation, two pointers
Correct Solution:
```
n = int(input())
l = list(map(int,input().split()))
ans = 0
s = set()
for i in l:
if i > 0:
s.add(i)
for i in s:
now = 0
for j in l:
if j > i:
now -= float("INF")
else:
now += j
if now <= 0:
now = 0
else:
ans = max(ans,now-i)
print(ans)
``` | output | 1 | 62,462 | 19 | 124,925 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice and Bob are playing yet another card game. This time the rules are the following. There are n cards lying in a row in front of them. The i-th card has value a_i.
First, Alice chooses a non-empty consecutive segment of cards [l; r] (l β€ r). After that Bob removes a single card j from that segment (l β€ j β€ r). The score of the game is the total value of the remaining cards on the segment (a_l + a_{l + 1} + ... + a_{j - 1} + a_{j + 1} + ... + a_{r - 1} + a_r). In particular, if Alice chooses a segment with just one element, then the score after Bob removes the only card is 0.
Alice wants to make the score as big as possible. Bob takes such a card that the score is as small as possible.
What segment should Alice choose so that the score is maximum possible? Output the maximum score.
Input
The first line contains a single integer n (1 β€ n β€ 10^5) β the number of cards.
The second line contains n integers a_1, a_2, ..., a_n (-30 β€ a_i β€ 30) β the values on the cards.
Output
Print a single integer β the final score of the game.
Examples
Input
5
5 -2 10 -1 4
Output
6
Input
8
5 2 5 3 -30 -30 6 9
Output
10
Input
3
-10 6 -15
Output
0
Note
In the first example Alice chooses a segment [1;5] β the entire row of cards. Bob removes card 3 with the value 10 from the segment. Thus, the final score is 5 + (-2) + (-1) + 4 = 6.
In the second example Alice chooses a segment [1;4], so that Bob removes either card 1 or 3 with the value 5, making the answer 5 + 2 + 3 = 10.
In the third example Alice can choose any of the segments of length 1: [1;1], [2;2] or [3;3]. Bob removes the only card, so the score is 0. If Alice chooses some other segment then the answer will be less than 0. | instruction | 0 | 62,463 | 19 | 124,926 |
Tags: data structures, dp, implementation, two pointers
Correct Solution:
```
import sys
def I():
return sys.stdin.readline().rstrip()
def f(l, st, en):
m = -1
for i in range(st, en):
m = max(m, l[i])
if m <= 0:
return 0
s = 0
ms = 0
for i in range(st, en):
x = l[i]
s += x
ms = max(ms, s)
s = max(s, 0)
ans = max(ms - m, 0)
le = st
ri = st - 1
while le < en:
le = ri + 1
ri = le
while ri < en and l[ri] < m:
ri += 1
if le < ri - 1:
ans = max( ans, f( l, le, ri ) )
return ans
n = int(I())
print(f(list(map(int, I().split())), 0, n))
``` | output | 1 | 62,463 | 19 | 124,927 |
Provide tags and a correct Python 2 solution for this coding contest problem.
Alice and Bob are playing yet another card game. This time the rules are the following. There are n cards lying in a row in front of them. The i-th card has value a_i.
First, Alice chooses a non-empty consecutive segment of cards [l; r] (l β€ r). After that Bob removes a single card j from that segment (l β€ j β€ r). The score of the game is the total value of the remaining cards on the segment (a_l + a_{l + 1} + ... + a_{j - 1} + a_{j + 1} + ... + a_{r - 1} + a_r). In particular, if Alice chooses a segment with just one element, then the score after Bob removes the only card is 0.
Alice wants to make the score as big as possible. Bob takes such a card that the score is as small as possible.
What segment should Alice choose so that the score is maximum possible? Output the maximum score.
Input
The first line contains a single integer n (1 β€ n β€ 10^5) β the number of cards.
The second line contains n integers a_1, a_2, ..., a_n (-30 β€ a_i β€ 30) β the values on the cards.
Output
Print a single integer β the final score of the game.
Examples
Input
5
5 -2 10 -1 4
Output
6
Input
8
5 2 5 3 -30 -30 6 9
Output
10
Input
3
-10 6 -15
Output
0
Note
In the first example Alice chooses a segment [1;5] β the entire row of cards. Bob removes card 3 with the value 10 from the segment. Thus, the final score is 5 + (-2) + (-1) + 4 = 6.
In the second example Alice chooses a segment [1;4], so that Bob removes either card 1 or 3 with the value 5, making the answer 5 + 2 + 3 = 10.
In the third example Alice can choose any of the segments of length 1: [1;1], [2;2] or [3;3]. Bob removes the only card, so the score is 0. If Alice chooses some other segment then the answer will be less than 0. | instruction | 0 | 62,464 | 19 | 124,928 |
Tags: data structures, dp, implementation, two pointers
Correct Solution:
```
from sys import stdin, stdout
from collections import Counter, defaultdict
from itertools import permutations, combinations
from fractions import Fraction
raw_input = stdin.readline
pr = stdout.write
mod=10**9+7
def ni():
return int(raw_input())
def li():
return map(int,raw_input().split())
def pn(n):
stdout.write(str(n)+'\n')
def pa(arr):
pr(' '.join(map(str,arr))+'\n')
# fast read function for total integer input
def inp():
# this function returns whole input of
# space/line seperated integers
# Use Ctrl+D to flush stdin.
return map(int,stdin.read().split())
range = xrange # not for python 3.0+
# main code
n=ni()
l=li()
ans=0
for j in range(1,31):
sm=0
for i in range(n):
if l[i]>j:
sm=0
elif sm<0:
sm=l[i]
else:
sm+=l[i]
ans=max(ans,sm-j)
pn(ans)
``` | output | 1 | 62,464 | 19 | 124,929 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice and Bob are playing yet another card game. This time the rules are the following. There are n cards lying in a row in front of them. The i-th card has value a_i.
First, Alice chooses a non-empty consecutive segment of cards [l; r] (l β€ r). After that Bob removes a single card j from that segment (l β€ j β€ r). The score of the game is the total value of the remaining cards on the segment (a_l + a_{l + 1} + ... + a_{j - 1} + a_{j + 1} + ... + a_{r - 1} + a_r). In particular, if Alice chooses a segment with just one element, then the score after Bob removes the only card is 0.
Alice wants to make the score as big as possible. Bob takes such a card that the score is as small as possible.
What segment should Alice choose so that the score is maximum possible? Output the maximum score.
Input
The first line contains a single integer n (1 β€ n β€ 10^5) β the number of cards.
The second line contains n integers a_1, a_2, ..., a_n (-30 β€ a_i β€ 30) β the values on the cards.
Output
Print a single integer β the final score of the game.
Examples
Input
5
5 -2 10 -1 4
Output
6
Input
8
5 2 5 3 -30 -30 6 9
Output
10
Input
3
-10 6 -15
Output
0
Note
In the first example Alice chooses a segment [1;5] β the entire row of cards. Bob removes card 3 with the value 10 from the segment. Thus, the final score is 5 + (-2) + (-1) + 4 = 6.
In the second example Alice chooses a segment [1;4], so that Bob removes either card 1 or 3 with the value 5, making the answer 5 + 2 + 3 = 10.
In the third example Alice can choose any of the segments of length 1: [1;1], [2;2] or [3;3]. Bob removes the only card, so the score is 0. If Alice chooses some other segment then the answer will be less than 0.
Submitted Solution:
```
# from math import factorial as fac
from collections import defaultdict
# from copy import deepcopy
import sys, math
f = None
try:
f = open('q1.input', 'r')
except IOError:
f = sys.stdin
if 'xrange' in dir(__builtins__):
range = xrange
# print(f.readline())
# sys.setrecursionlimit(10**6)
def print_case_iterable(case_num, iterable):
print("Case #{}: {}".format(case_num," ".join(map(str,iterable))))
def print_case_number(case_num, iterable):
print("Case #{}: {}".format(case_num,iterable))
def print_iterable(A):
print (' '.join(A))
def read_int():
return int(f.readline().strip())
def read_int_array():
return [int(x) for x in f.readline().strip().split(" ")]
def rns():
a = [x for x in f.readline().split(" ")]
return int(a[0]), a[1].strip()
def read_string():
return list(f.readline().strip())
def bi(x):
return bin(x)[2:]
from collections import deque
import math
def regular_bs(L,x,l=None,r=None):
l = 0 if l == None else l
r = len(L) if r == None else r
while (r-l) > 1:
m = (r+l)//2
l = m if L[m] <= x else l
r = m if L[m] > x else r
cnt = 0
return l
def solution(A,n):
total_max = 0
d = defaultdict(list)
elems = set(A)
for x in elems:
if x > 0:
d[x].append(-1)
for i in range(n):
if A[i] <= 0:
continue
for x in range(1,A[i]):
if x in elems:
d[x].append(i)
for x in elems:
if x > 0:
d[x].append(n)
# print(d)
for m in elems:
for j in range(1,len(d[m])):
start = d[m][j-1]+1
end = d[m][j]
max_so_far = 0
max_end_here = 0
saw_max = False
for i in range(start,end):
x = A[i]
if x == m and not saw_max:
saw_max = True
max_so_far = max(max_so_far,max_end_here)
continue
max_end_here = max_end_here + x
if saw_max:
max_so_far = max(max_so_far,max_end_here)
if saw_max and max_end_here <= -m or ((not saw_max) and max_end_here<=0):
saw_max = False
max_end_here = 0
total_max = max(total_max,max_so_far)
return total_max
def main():
T = 1
for i in range(T):
n = read_int()
A = read_int_array()
x = solution(A,n)
if 'xrange' not in dir(__builtins__):
print(x)
else:
print >>output,str(x)# "Case #"+str(i+1)+':',
if 'xrange' in dir(__builtins__):
print(output.getvalue())
output.close()
if 'xrange' in dir(__builtins__):
import cStringIO
output = cStringIO.StringIO()
#example usage:
# for l in res:
# print >>output, str(len(l)) + ' ' + ' '.join(l)
if __name__ == '__main__':
main()
``` | instruction | 0 | 62,465 | 19 | 124,930 |
Yes | output | 1 | 62,465 | 19 | 124,931 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice and Bob are playing yet another card game. This time the rules are the following. There are n cards lying in a row in front of them. The i-th card has value a_i.
First, Alice chooses a non-empty consecutive segment of cards [l; r] (l β€ r). After that Bob removes a single card j from that segment (l β€ j β€ r). The score of the game is the total value of the remaining cards on the segment (a_l + a_{l + 1} + ... + a_{j - 1} + a_{j + 1} + ... + a_{r - 1} + a_r). In particular, if Alice chooses a segment with just one element, then the score after Bob removes the only card is 0.
Alice wants to make the score as big as possible. Bob takes such a card that the score is as small as possible.
What segment should Alice choose so that the score is maximum possible? Output the maximum score.
Input
The first line contains a single integer n (1 β€ n β€ 10^5) β the number of cards.
The second line contains n integers a_1, a_2, ..., a_n (-30 β€ a_i β€ 30) β the values on the cards.
Output
Print a single integer β the final score of the game.
Examples
Input
5
5 -2 10 -1 4
Output
6
Input
8
5 2 5 3 -30 -30 6 9
Output
10
Input
3
-10 6 -15
Output
0
Note
In the first example Alice chooses a segment [1;5] β the entire row of cards. Bob removes card 3 with the value 10 from the segment. Thus, the final score is 5 + (-2) + (-1) + 4 = 6.
In the second example Alice chooses a segment [1;4], so that Bob removes either card 1 or 3 with the value 5, making the answer 5 + 2 + 3 = 10.
In the third example Alice can choose any of the segments of length 1: [1;1], [2;2] or [3;3]. Bob removes the only card, so the score is 0. If Alice chooses some other segment then the answer will be less than 0.
Submitted Solution:
```
import sys
input = sys.stdin.readline
from itertools import accumulate
n = int(input())
a = [0]+list(map(int,input().split()))
acc = list(accumulate(a))
mxls = [0]
ans = 0
l = 0
acmn = 1000000000
acmnls = [acmn]*(n+1)
for i in range(1,n+1):
if mxls[i-1] < a[i]:
t = i
mxls.append(a[i])
while t >= 2 and mxls[t-1] < a[i]:
mxls[t-1] = a[i]
t -= 1
acmn = acmnls[t-1]
for j in range(t,i+1):
acmn = min(acmn,acc[j-1]+mxls[j])
acmnls[j] = acmn
else:
mxls.append(a[i])
acmn = min(acmn,acc[i-1]+mxls[i])
acmnls[i] = acmn
ans = max(ans,acc[i]-acmn)
print(ans)
``` | instruction | 0 | 62,466 | 19 | 124,932 |
Yes | output | 1 | 62,466 | 19 | 124,933 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice and Bob are playing yet another card game. This time the rules are the following. There are n cards lying in a row in front of them. The i-th card has value a_i.
First, Alice chooses a non-empty consecutive segment of cards [l; r] (l β€ r). After that Bob removes a single card j from that segment (l β€ j β€ r). The score of the game is the total value of the remaining cards on the segment (a_l + a_{l + 1} + ... + a_{j - 1} + a_{j + 1} + ... + a_{r - 1} + a_r). In particular, if Alice chooses a segment with just one element, then the score after Bob removes the only card is 0.
Alice wants to make the score as big as possible. Bob takes such a card that the score is as small as possible.
What segment should Alice choose so that the score is maximum possible? Output the maximum score.
Input
The first line contains a single integer n (1 β€ n β€ 10^5) β the number of cards.
The second line contains n integers a_1, a_2, ..., a_n (-30 β€ a_i β€ 30) β the values on the cards.
Output
Print a single integer β the final score of the game.
Examples
Input
5
5 -2 10 -1 4
Output
6
Input
8
5 2 5 3 -30 -30 6 9
Output
10
Input
3
-10 6 -15
Output
0
Note
In the first example Alice chooses a segment [1;5] β the entire row of cards. Bob removes card 3 with the value 10 from the segment. Thus, the final score is 5 + (-2) + (-1) + 4 = 6.
In the second example Alice chooses a segment [1;4], so that Bob removes either card 1 or 3 with the value 5, making the answer 5 + 2 + 3 = 10.
In the third example Alice can choose any of the segments of length 1: [1;1], [2;2] or [3;3]. Bob removes the only card, so the score is 0. If Alice chooses some other segment then the answer will be less than 0.
Submitted Solution:
```
import sys
input=sys.stdin.readline
from collections import defaultdict
def kadane(hi,lo,sub_max):
max_so_far = 0
max_ending_here = 0
for i in range(hi,lo):
max_ending_here = max_ending_here + l[i]
if max_ending_here < 0:
max_ending_here = 0
elif (max_so_far < max_ending_here):
max_so_far = max_ending_here
return(max_so_far-sub_max)
n = int(input())
l = [int(i) for i in input().split()]
maxl=max(l)
ind=defaultdict(list)
stack=[]
ind_arr=[0]*(31)
for i in range(len(l)):
if l[i]>=0:
ind_arr[l[i]]=i
while len(stack)>0 and stack[-1]<l[i]:
temp=stack.pop()
ind[temp][-1].append(i)
if len(stack)==0:
ind[l[i]].append([0])
elif stack[-1]!=l[i]:
ind[l[i]].append([ind_arr[stack[-1]]+1])
stack.append(l[i])
#print(stack)
ans=[0]
#print(ind)
for i in ind:
for j in ind[i]:
if len(j)==1:
ans.append(kadane(j[0],n,i))
#print(j[0],n,i,ans)
else:
ans.append(kadane(j[0],j[1],i))
sys.stdout.write(str(max(ans))+'\n')
``` | instruction | 0 | 62,467 | 19 | 124,934 |
Yes | output | 1 | 62,467 | 19 | 124,935 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice and Bob are playing yet another card game. This time the rules are the following. There are n cards lying in a row in front of them. The i-th card has value a_i.
First, Alice chooses a non-empty consecutive segment of cards [l; r] (l β€ r). After that Bob removes a single card j from that segment (l β€ j β€ r). The score of the game is the total value of the remaining cards on the segment (a_l + a_{l + 1} + ... + a_{j - 1} + a_{j + 1} + ... + a_{r - 1} + a_r). In particular, if Alice chooses a segment with just one element, then the score after Bob removes the only card is 0.
Alice wants to make the score as big as possible. Bob takes such a card that the score is as small as possible.
What segment should Alice choose so that the score is maximum possible? Output the maximum score.
Input
The first line contains a single integer n (1 β€ n β€ 10^5) β the number of cards.
The second line contains n integers a_1, a_2, ..., a_n (-30 β€ a_i β€ 30) β the values on the cards.
Output
Print a single integer β the final score of the game.
Examples
Input
5
5 -2 10 -1 4
Output
6
Input
8
5 2 5 3 -30 -30 6 9
Output
10
Input
3
-10 6 -15
Output
0
Note
In the first example Alice chooses a segment [1;5] β the entire row of cards. Bob removes card 3 with the value 10 from the segment. Thus, the final score is 5 + (-2) + (-1) + 4 = 6.
In the second example Alice chooses a segment [1;4], so that Bob removes either card 1 or 3 with the value 5, making the answer 5 + 2 + 3 = 10.
In the third example Alice can choose any of the segments of length 1: [1;1], [2;2] or [3;3]. Bob removes the only card, so the score is 0. If Alice chooses some other segment then the answer will be less than 0.
Submitted Solution:
```
from sys import stdin
input = stdin.buffer.readline
n = int(input())
*a, = map(int, input().split())
ans = 0
for x in range(0, 31):
if not x in a:
continue
mx = a[0] if a[0] <= x else 0
sm, mn = 0, 0
for i in range(n):
if a[i] > x:
continue
sm += a[i]
mx = max(mx,sm - mn)
mn = min(mn,sm)
ans = max(ans,mx - x)
print(ans)
``` | instruction | 0 | 62,468 | 19 | 124,936 |
Yes | output | 1 | 62,468 | 19 | 124,937 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice and Bob are playing yet another card game. This time the rules are the following. There are n cards lying in a row in front of them. The i-th card has value a_i.
First, Alice chooses a non-empty consecutive segment of cards [l; r] (l β€ r). After that Bob removes a single card j from that segment (l β€ j β€ r). The score of the game is the total value of the remaining cards on the segment (a_l + a_{l + 1} + ... + a_{j - 1} + a_{j + 1} + ... + a_{r - 1} + a_r). In particular, if Alice chooses a segment with just one element, then the score after Bob removes the only card is 0.
Alice wants to make the score as big as possible. Bob takes such a card that the score is as small as possible.
What segment should Alice choose so that the score is maximum possible? Output the maximum score.
Input
The first line contains a single integer n (1 β€ n β€ 10^5) β the number of cards.
The second line contains n integers a_1, a_2, ..., a_n (-30 β€ a_i β€ 30) β the values on the cards.
Output
Print a single integer β the final score of the game.
Examples
Input
5
5 -2 10 -1 4
Output
6
Input
8
5 2 5 3 -30 -30 6 9
Output
10
Input
3
-10 6 -15
Output
0
Note
In the first example Alice chooses a segment [1;5] β the entire row of cards. Bob removes card 3 with the value 10 from the segment. Thus, the final score is 5 + (-2) + (-1) + 4 = 6.
In the second example Alice chooses a segment [1;4], so that Bob removes either card 1 or 3 with the value 5, making the answer 5 + 2 + 3 = 10.
In the third example Alice can choose any of the segments of length 1: [1;1], [2;2] or [3;3]. Bob removes the only card, so the score is 0. If Alice chooses some other segment then the answer will be less than 0.
Submitted Solution:
```
col = int(input())
arr = list(map(int, input().split()))
if len(arr) > col:
print("error")
else:
end = 0
obr = 0
a = 0
count = len(arr)
for i in range(len(arr)):
for j in range (count):
obr = sum(arr[j:end+count]) - max(arr[j:end+count])
if obr > a:
a = obr
else:
continue
count -= 1
print(arr)
print(a)
``` | instruction | 0 | 62,469 | 19 | 124,938 |
No | output | 1 | 62,469 | 19 | 124,939 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice and Bob are playing yet another card game. This time the rules are the following. There are n cards lying in a row in front of them. The i-th card has value a_i.
First, Alice chooses a non-empty consecutive segment of cards [l; r] (l β€ r). After that Bob removes a single card j from that segment (l β€ j β€ r). The score of the game is the total value of the remaining cards on the segment (a_l + a_{l + 1} + ... + a_{j - 1} + a_{j + 1} + ... + a_{r - 1} + a_r). In particular, if Alice chooses a segment with just one element, then the score after Bob removes the only card is 0.
Alice wants to make the score as big as possible. Bob takes such a card that the score is as small as possible.
What segment should Alice choose so that the score is maximum possible? Output the maximum score.
Input
The first line contains a single integer n (1 β€ n β€ 10^5) β the number of cards.
The second line contains n integers a_1, a_2, ..., a_n (-30 β€ a_i β€ 30) β the values on the cards.
Output
Print a single integer β the final score of the game.
Examples
Input
5
5 -2 10 -1 4
Output
6
Input
8
5 2 5 3 -30 -30 6 9
Output
10
Input
3
-10 6 -15
Output
0
Note
In the first example Alice chooses a segment [1;5] β the entire row of cards. Bob removes card 3 with the value 10 from the segment. Thus, the final score is 5 + (-2) + (-1) + 4 = 6.
In the second example Alice chooses a segment [1;4], so that Bob removes either card 1 or 3 with the value 5, making the answer 5 + 2 + 3 = 10.
In the third example Alice can choose any of the segments of length 1: [1;1], [2;2] or [3;3]. Bob removes the only card, so the score is 0. If Alice chooses some other segment then the answer will be less than 0.
Submitted Solution:
```
n = input()
s = input()
a = []
[a.append(int(i)) for i in s.split(' ')]
m = 0
for j in range(int(n)):
i = j
isDone = False
while not isDone:
if i == len(a) - 1 or a[i] < 0 and sum(a[i:]) < 0:
break
i += 1
if a[i] > 0 or i == 0:
i += 1
if sum(a[:i]) - max(a[:i]) > m and j != i:
m = sum(a[j:i]) - max(a[j:i])
print(m)
``` | instruction | 0 | 62,470 | 19 | 124,940 |
No | output | 1 | 62,470 | 19 | 124,941 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice and Bob are playing yet another card game. This time the rules are the following. There are n cards lying in a row in front of them. The i-th card has value a_i.
First, Alice chooses a non-empty consecutive segment of cards [l; r] (l β€ r). After that Bob removes a single card j from that segment (l β€ j β€ r). The score of the game is the total value of the remaining cards on the segment (a_l + a_{l + 1} + ... + a_{j - 1} + a_{j + 1} + ... + a_{r - 1} + a_r). In particular, if Alice chooses a segment with just one element, then the score after Bob removes the only card is 0.
Alice wants to make the score as big as possible. Bob takes such a card that the score is as small as possible.
What segment should Alice choose so that the score is maximum possible? Output the maximum score.
Input
The first line contains a single integer n (1 β€ n β€ 10^5) β the number of cards.
The second line contains n integers a_1, a_2, ..., a_n (-30 β€ a_i β€ 30) β the values on the cards.
Output
Print a single integer β the final score of the game.
Examples
Input
5
5 -2 10 -1 4
Output
6
Input
8
5 2 5 3 -30 -30 6 9
Output
10
Input
3
-10 6 -15
Output
0
Note
In the first example Alice chooses a segment [1;5] β the entire row of cards. Bob removes card 3 with the value 10 from the segment. Thus, the final score is 5 + (-2) + (-1) + 4 = 6.
In the second example Alice chooses a segment [1;4], so that Bob removes either card 1 or 3 with the value 5, making the answer 5 + 2 + 3 = 10.
In the third example Alice can choose any of the segments of length 1: [1;1], [2;2] or [3;3]. Bob removes the only card, so the score is 0. If Alice chooses some other segment then the answer will be less than 0.
Submitted Solution:
```
import sys
def input():
return sys.stdin.readline().rstrip()
n = int(input())
arr = [int(i) for i in input().split()]
prefix_sum = [0]
freqs = []
freqs.append([0 for i in range(-30, 31)])
# maxes = [-float('inf')]
for a in arr:
prefix_sum.append(a + prefix_sum[-1])
temp = freqs[-1].copy()
temp[a] += 1
freqs.append(temp)
# maxes.append()
# maxes_with_freqs = []
# maxes_with_freqs.append((-60, 0))
# for i in range(n):
# prev_max, freq = maxes_with_freqs[-1]
# if arr[i] == prev_max:
# maxes_with_freqs.append((prev_max,freq+1))
# elif:
# arr[i] < prev_max:
# maxes_with_freqs.append((arr[i],1))
# else:
# maxes_with_freqs.append((prev_max,freq))
def fetch_sum(left, right):
global prefix_sum
return prefix_sum[right + 1] - prefix_sum[left]
def fetch_max(left, right):
global freqs
freqs1 = freqs[left]
freqs2 = freqs[right+1]
for i in range(-30, 31):
if freqs2[i] > freqs1[i]:
best = i
return best
# max1 , f1 = maxes_with_freqs[right + 1]
# max2, f2 = maxes_with_freqs[left]
# if max1 > max2:
# return max1
# elif max1 == max2:
# if f1 == f2:
# return
left = 0
right = -1
current_sum = 0
# current_score = 0
best = 0
while(True):
if right + 1 >= len(arr):
break
#keep on inc left
right = right + 1
prospect = arr[right]
if prospect > 0 or (prospect + current_sum) > 0:
current_sum += prospect
# current_max = fetch_max(left, right)
# if prospect > current_max:
# current_score += current_max
# else:
# current_score += prospect
current_score = fetch_sum(left, right) - fetch_max(left, right)
# best = max(current_score, best)
if current_score > best:
best = current_score
best_left, best_right = left, right
else:
#prospect is highly negative
left = right + 1
right = right
current_sum = 0
# current_score = 0
#continue
print(best)
``` | instruction | 0 | 62,471 | 19 | 124,942 |
No | output | 1 | 62,471 | 19 | 124,943 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice and Bob are playing yet another card game. This time the rules are the following. There are n cards lying in a row in front of them. The i-th card has value a_i.
First, Alice chooses a non-empty consecutive segment of cards [l; r] (l β€ r). After that Bob removes a single card j from that segment (l β€ j β€ r). The score of the game is the total value of the remaining cards on the segment (a_l + a_{l + 1} + ... + a_{j - 1} + a_{j + 1} + ... + a_{r - 1} + a_r). In particular, if Alice chooses a segment with just one element, then the score after Bob removes the only card is 0.
Alice wants to make the score as big as possible. Bob takes such a card that the score is as small as possible.
What segment should Alice choose so that the score is maximum possible? Output the maximum score.
Input
The first line contains a single integer n (1 β€ n β€ 10^5) β the number of cards.
The second line contains n integers a_1, a_2, ..., a_n (-30 β€ a_i β€ 30) β the values on the cards.
Output
Print a single integer β the final score of the game.
Examples
Input
5
5 -2 10 -1 4
Output
6
Input
8
5 2 5 3 -30 -30 6 9
Output
10
Input
3
-10 6 -15
Output
0
Note
In the first example Alice chooses a segment [1;5] β the entire row of cards. Bob removes card 3 with the value 10 from the segment. Thus, the final score is 5 + (-2) + (-1) + 4 = 6.
In the second example Alice chooses a segment [1;4], so that Bob removes either card 1 or 3 with the value 5, making the answer 5 + 2 + 3 = 10.
In the third example Alice can choose any of the segments of length 1: [1;1], [2;2] or [3;3]. Bob removes the only card, so the score is 0. If Alice chooses some other segment then the answer will be less than 0.
Submitted Solution:
```
## necessary imports
import sys
input = sys.stdin.readline
from math import ceil, floor, factorial;
# swap_array function
def swaparr(arr, a,b):
temp = arr[a];
arr[a] = arr[b];
arr[b] = temp
## gcd function
def gcd(a,b):
if a == 0:
return b
return gcd(b%a, a)
## nCr function efficient using Binomial Cofficient
def nCr(n, k):
if(k > n - k):
k = n - k
res = 1
for i in range(k):
res = res * (n - i)
res = res / (i + 1)
return int(res)
## upper bound function code -- such that e in a[:i] e < x;
def upper_bound(a, x, lo=0):
hi = len(a)
while lo < hi:
mid = (lo+hi)//2
if a[mid] < x:
lo = mid+1
else:
hi = mid
return lo
## prime factorization
def primefs(n):
## if n == 1 ## calculating primes
primes = {}
while(n%2 == 0):
primes[2] = primes.get(2, 0) + 1
n = n//2
for i in range(3, int(n**0.5)+2, 2):
while(n%i == 0):
primes[i] = primes.get(i, 0) + 1
n = n//i
if n > 2:
primes[n] = primes.get(n, 0) + 1
## prime factoriazation of n is stored in dictionary
## primes and can be accesed. O(sqrt n)
return primes
## MODULAR EXPONENTIATION FUNCTION
def power(x, y, p):
res = 1
x = x % p
if (x == 0) :
return 0
while (y > 0) :
if ((y & 1) == 1) :
res = (res * x) % p
y = y >> 1
x = (x * x) % p
return res
## DISJOINT SET UNINON FUNCTIONS
def swap(a,b):
temp = a
a = b
b = temp
return a,b
# find function with path compression included (recursive)
# def find(x, link):
# if link[x] == x:
# return x
# link[x] = find(link[x], link);
# return link[x];
# find function with path compression (ITERATIVE)
def find(x, link):
p = x;
while( p != link[p]):
p = link[p];
while( x != p):
nex = link[x];
link[x] = p;
x = nex;
return p;
# the union function which makes union(x,y)
# of two nodes x and y
def union(x, y, link, size):
x = find(x, link)
y = find(y, link)
if size[x] < size[y]:
x,y = swap(x,y)
if x != y:
size[x] += size[y]
link[y] = x
## returns an array of boolean if primes or not USING SIEVE OF ERATOSTHANES
def sieve(n):
prime = [True for i in range(n+1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
return prime
#### PRIME FACTORIZATION IN O(log n) using Sieve ####
MAXN = int(1e6 + 5)
def spf_sieve():
spf[1] = 1;
for i in range(2, MAXN):
spf[i] = i;
for i in range(4, MAXN, 2):
spf[i] = 2;
for i in range(3, ceil(MAXN ** 0.5), 2):
if spf[i] == i:
for j in range(i*i, MAXN, i):
if spf[j] == j:
spf[j] = i;
## function for storing smallest prime factors (spf) in the array
################## un-comment below 2 lines when using factorization #################
# spf = [0 for i in range(MAXN)]
# spf_sieve()
def factoriazation(x):
ret = {};
while x != 1:
ret[spf[x]] = ret.get(spf[x], 0) + 1;
x = x//spf[x]
return ret
## this function is useful for multiple queries only, o/w use
## primefs function above. complexity O(log n)
## taking integer array input
def int_array():
return list(map(int, input().strip().split()))
## taking string array input
def str_array():
return input().strip().split();
#defining a couple constants
MOD = int(1e9)+7;
CMOD = 998244353;
INF = float('inf'); NINF = -float('inf');
################### ---------------- TEMPLATE ENDS HERE ---------------- ###################
n = int(input()); a = int_array();
ans = 0;
for i in range(1, 31):
s = 0;
for j in a:
if j <= i and s >= 0:
s += j;
else:
s = 0;
ans = max(ans, s-i);
print(ans);
``` | instruction | 0 | 62,472 | 19 | 124,944 |
No | output | 1 | 62,472 | 19 | 124,945 |
Provide tags and a correct Python 3 solution for this coding contest problem.
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
The first line contains integer n (1 β€ n β€ 1000) β the number of cards on the table. The second line contains space-separated numbers on the cards from left to right. The numbers on the cards are distinct integers from 1 to 1000.
Output
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.
Examples
Input
4
4 1 2 10
Output
12 5
Input
7
1 2 3 4 5 6 7
Output
16 12
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. | instruction | 0 | 62,653 | 19 | 125,306 |
Tags: greedy, implementation, two pointers
Correct Solution:
```
n=int(input())
a=list(map(int,input().split()))
b=True
l=0
r=n-1
ans1=0
ans2=0
while l<=r:
if a[l]>a[r]:
if b:
ans1+=a[l]
b=False
else:
ans2+=a[l]
b=True
l+=1
else:
if b:
ans1+=a[r]
b=False
else:
ans2+=a[r]
b=True
r-=1
print(ans1,ans2)
``` | output | 1 | 62,653 | 19 | 125,307 |
Provide tags and a correct Python 3 solution for this coding contest problem.
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
The first line contains integer n (1 β€ n β€ 1000) β the number of cards on the table. The second line contains space-separated numbers on the cards from left to right. The numbers on the cards are distinct integers from 1 to 1000.
Output
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.
Examples
Input
4
4 1 2 10
Output
12 5
Input
7
1 2 3 4 5 6 7
Output
16 12
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. | instruction | 0 | 62,654 | 19 | 125,308 |
Tags: greedy, implementation, two pointers
Correct Solution:
```
# import sys
# sys.stdin=open("input.in",'r')
# sys.stdout=open("outp.out",'w')
n=int(input())
a=list(map(int,input().split()))
i=0
s,d=0,0
while len(a):
x=max(a[0],a[len(a)-1])
s+=x
a.remove(x)
if len(a):
y=max(a[0],a[len(a)-1])
d+=y
a.remove(y)
else:
break
i+=1
print(s,d)
``` | output | 1 | 62,654 | 19 | 125,309 |
Provide tags and a correct Python 3 solution for this coding contest problem.
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
The first line contains integer n (1 β€ n β€ 1000) β the number of cards on the table. The second line contains space-separated numbers on the cards from left to right. The numbers on the cards are distinct integers from 1 to 1000.
Output
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.
Examples
Input
4
4 1 2 10
Output
12 5
Input
7
1 2 3 4 5 6 7
Output
16 12
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. | instruction | 0 | 62,655 | 19 | 125,310 |
Tags: greedy, implementation, two pointers
Correct Solution:
```
num = int(input())
cartas = list(map(int, input().split()))
sereja, dima = 0, 0
while len(cartas)>0:
if cartas[len(cartas)-1]>cartas[0]:
sereja+=cartas.pop()
if len(cartas)>0 and cartas[len(cartas)-1]>cartas[0]:
dima+=cartas.pop()
elif len(cartas)>0:
dima+=cartas.pop(0)
else:
sereja+=cartas.pop(0)
if len(cartas)>0 and cartas[len(cartas)-1]>cartas[0]:
dima+=cartas.pop()
elif len(cartas)>0:
dima+=cartas.pop(0)
'''
if len(cartas)%2==1:
sereja+=cartas.pop(cartas.index(max(cartas)))
while len(cartas)>0:
dima+=cartas.pop(cartas.index(max(cartas)))
sereja+=cartas.pop(cartas.index(max(cartas)))
else:
while len(cartas)>0:
sereja+=cartas.pop(cartas.index(max(cartas)))
dima+=cartas.pop(cartas.index(max(cartas)))
'''
print(sereja, dima)
``` | output | 1 | 62,655 | 19 | 125,311 |
Provide tags and a correct Python 3 solution for this coding contest problem.
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
The first line contains integer n (1 β€ n β€ 1000) β the number of cards on the table. The second line contains space-separated numbers on the cards from left to right. The numbers on the cards are distinct integers from 1 to 1000.
Output
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.
Examples
Input
4
4 1 2 10
Output
12 5
Input
7
1 2 3 4 5 6 7
Output
16 12
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. | instruction | 0 | 62,656 | 19 | 125,312 |
Tags: greedy, implementation, two pointers
Correct Solution:
```
n=int(input())
sereja=0
dima=0
a=list(map(int,input().split()))[:n]
for i in range(n):
b=max(a[0],a[-1])
if i%2==0:
sereja+=b
else:
dima+=b
a.remove(b)
print(sereja, dima)
``` | output | 1 | 62,656 | 19 | 125,313 |
Provide tags and a correct Python 3 solution for this coding contest problem.
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
The first line contains integer n (1 β€ n β€ 1000) β the number of cards on the table. The second line contains space-separated numbers on the cards from left to right. The numbers on the cards are distinct integers from 1 to 1000.
Output
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.
Examples
Input
4
4 1 2 10
Output
12 5
Input
7
1 2 3 4 5 6 7
Output
16 12
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. | instruction | 0 | 62,657 | 19 | 125,314 |
Tags: greedy, implementation, two pointers
Correct Solution:
```
t = int(input())
arr = list(map(int,input().split(' ')))
s,d = 0,0
i,j = 0,t-1
chance = 's'
while s+d!=sum(arr):
if chance == 's':
if max(arr[i],arr[j]) == arr[i]:
s = s + arr[i]
i += 1
elif max(arr[i],arr[j]) == arr[j]:
s = s + arr[j]
j -= 1
chance = 'd'
elif chance == 'd':
if max(arr[i],arr[j]) == arr[i]:
d = d + arr[i]
i += 1
elif max(arr[i],arr[j]) == arr[j]:
d = d + arr[j]
j -= 1
chance = 's'
print(s,d)
``` | output | 1 | 62,657 | 19 | 125,315 |
Provide tags and a correct Python 3 solution for this coding contest problem.
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
The first line contains integer n (1 β€ n β€ 1000) β the number of cards on the table. The second line contains space-separated numbers on the cards from left to right. The numbers on the cards are distinct integers from 1 to 1000.
Output
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.
Examples
Input
4
4 1 2 10
Output
12 5
Input
7
1 2 3 4 5 6 7
Output
16 12
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. | instruction | 0 | 62,658 | 19 | 125,316 |
Tags: greedy, implementation, two pointers
Correct Solution:
```
n=int(input())
a=list(map(int,input().split()))
s=0
d=0
while(len(a)>0):
m=max(a[0],a[len(a)-1])
s+=m
a.remove(m)
if len(a)>0:
l=max(a[0],a[len(a)-1])
d+=l
a.remove(l)
else:
break
print(s,d)
``` | output | 1 | 62,658 | 19 | 125,317 |
Provide tags and a correct Python 3 solution for this coding contest problem.
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
The first line contains integer n (1 β€ n β€ 1000) β the number of cards on the table. The second line contains space-separated numbers on the cards from left to right. The numbers on the cards are distinct integers from 1 to 1000.
Output
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.
Examples
Input
4
4 1 2 10
Output
12 5
Input
7
1 2 3 4 5 6 7
Output
16 12
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. | instruction | 0 | 62,659 | 19 | 125,318 |
Tags: greedy, implementation, two pointers
Correct Solution:
```
a=0
b=0
n=int(input())
crd=input().split(" ")
crd = [int(l) for l in crd]
x=0
while x<n:
x=x+1
if crd[0]>crd[-1]:
if x%2==0:
b=b+crd[0]
else:
a=a+crd[0]
del crd[0]
else:
if x%2==0:
b=b+crd[-1]
else:
a=a+crd[-1]
del crd[-1]
print(a,b)
``` | output | 1 | 62,659 | 19 | 125,319 |
Provide tags and a correct Python 3 solution for this coding contest problem.
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
The first line contains integer n (1 β€ n β€ 1000) β the number of cards on the table. The second line contains space-separated numbers on the cards from left to right. The numbers on the cards are distinct integers from 1 to 1000.
Output
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.
Examples
Input
4
4 1 2 10
Output
12 5
Input
7
1 2 3 4 5 6 7
Output
16 12
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. | instruction | 0 | 62,660 | 19 | 125,320 |
Tags: greedy, implementation, two pointers
Correct Solution:
```
def mod():
n=int(input())
x=list(map(int,input().split()))
i=0
j=n-1
f=[]
s=[]
d=[]
while i<=j:
if x[i]>x[j]:
f.append(x[i])
i=i+1
else:
f.append(x[j])
j = j - 1
for k in range(len(f)):
if k%2==0:
s.append(f[k])
else:
d.append(f[k])
print(sum(s),sum(d))
mod()
``` | output | 1 | 62,660 | 19 | 125,321 |
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