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.
Omkar is playing his favorite pixelated video game, Bed Wars! In Bed Wars, there are n players arranged in a circle, so that for all j such that 2 ≤ j ≤ n, player j - 1 is to the left of the player j, and player j is to the right of player j - 1. Additionally, player n is to the left of player 1, and player 1 is to the right of player n.
Currently, each player is attacking either the player to their left or the player to their right. This means that each player is currently being attacked by either 0, 1, or 2 other players. A key element of Bed Wars strategy is that if a player is being attacked by exactly 1 other player, then they should logically attack that player in response. If instead a player is being attacked by 0 or 2 other players, then Bed Wars strategy says that the player can logically attack either of the adjacent players.
Unfortunately, it might be that some players in this game are not following Bed Wars strategy correctly. Omkar is aware of whom each player is currently attacking, and he can talk to any amount of the n players in the game to make them instead attack another player — i. e. if they are currently attacking the player to their left, Omkar can convince them to instead attack the player to their right; if they are currently attacking the player to their right, Omkar can convince them to instead attack the player to their left.
Omkar would like all players to be acting logically. Calculate the minimum amount of players that Omkar needs to talk to so that after all players he talked to (if any) have changed which player they are attacking, all players are acting logically according to Bed Wars strategy.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). The descriptions of the test cases follows.
The first line of each test case contains one integer n (3 ≤ n ≤ 2 ⋅ 10^5) — the amount of players (and therefore beds) in this game of Bed Wars.
The second line of each test case contains a string s of length n. The j-th character of s is equal to L if the j-th player is attacking the player to their left, and R if the j-th player is attacking the player to their right.
It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case, output one integer: the minimum number of players Omkar needs to talk to to make it so that all players are acting logically according to Bed Wars strategy.
It can be proven that it is always possible for Omkar to achieve this under the given constraints.
Example
Input
5
4
RLRL
6
LRRRRL
8
RLLRRRLL
12
LLLLRRLRRRLL
5
RRRRR
Output
0
1
1
3
2
Note
In the first test case, players 1 and 2 are attacking each other, and players 3 and 4 are attacking each other. Each player is being attacked by exactly 1 other player, and each player is attacking the player that is attacking them, so all players are already being logical according to Bed Wars strategy and Omkar does not need to talk to any of them, making the answer 0.
In the second test case, not every player acts logically: for example, player 3 is attacked only by player 2, but doesn't attack him in response. Omkar can talk to player 3 to convert the attack arrangement to LRLRRL, in which you can see that all players are being logical according to Bed Wars strategy, making the answer 1.
Submitted Solution:
```
from sys import stdin
def inp():
return stdin.buffer.readline().rstrip().decode('utf8')
def itg():
return int(stdin.buffer.readline())
def mpint():
return map(int, stdin.buffer.readline().split())
# ############################## import
# ############################## main
# def solve():
def check_idx(x):
if x == -1:
return n - 1
if x == n:
return 0
return x
def is_good(x):
left = s[check_idx(x-1)]
right = s[check_idx(x+1)]
if s[x] == 'L':
return left == 'R' or right == 'R'
return left == 'L' or right == 'L'
for case in range(1, itg() + 1):
n = itg()
s = list(inp())
ans = 0
is_bad = set(i for i in range(n) if not is_good(i))
if not is_bad:
print(0)
continue
while is_bad:
start = temp = is_bad.pop()
count = 1
while is_bad: # right
temp = check_idx(temp + 1)
if temp in is_bad:
is_bad.remove(temp)
count += 1
else:
break
temp = start
while is_bad: # left
temp = check_idx(temp - 1)
if temp in is_bad:
is_bad.remove(temp)
count += 1
else:
break
count = 1 if count == 1 else count >> 1
ans += count
print(ans)
# Please check!
``` | instruction | 0 | 79,331 | 19 | 158,662 |
No | output | 1 | 79,331 | 19 | 158,663 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Omkar is playing his favorite pixelated video game, Bed Wars! In Bed Wars, there are n players arranged in a circle, so that for all j such that 2 ≤ j ≤ n, player j - 1 is to the left of the player j, and player j is to the right of player j - 1. Additionally, player n is to the left of player 1, and player 1 is to the right of player n.
Currently, each player is attacking either the player to their left or the player to their right. This means that each player is currently being attacked by either 0, 1, or 2 other players. A key element of Bed Wars strategy is that if a player is being attacked by exactly 1 other player, then they should logically attack that player in response. If instead a player is being attacked by 0 or 2 other players, then Bed Wars strategy says that the player can logically attack either of the adjacent players.
Unfortunately, it might be that some players in this game are not following Bed Wars strategy correctly. Omkar is aware of whom each player is currently attacking, and he can talk to any amount of the n players in the game to make them instead attack another player — i. e. if they are currently attacking the player to their left, Omkar can convince them to instead attack the player to their right; if they are currently attacking the player to their right, Omkar can convince them to instead attack the player to their left.
Omkar would like all players to be acting logically. Calculate the minimum amount of players that Omkar needs to talk to so that after all players he talked to (if any) have changed which player they are attacking, all players are acting logically according to Bed Wars strategy.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). The descriptions of the test cases follows.
The first line of each test case contains one integer n (3 ≤ n ≤ 2 ⋅ 10^5) — the amount of players (and therefore beds) in this game of Bed Wars.
The second line of each test case contains a string s of length n. The j-th character of s is equal to L if the j-th player is attacking the player to their left, and R if the j-th player is attacking the player to their right.
It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case, output one integer: the minimum number of players Omkar needs to talk to to make it so that all players are acting logically according to Bed Wars strategy.
It can be proven that it is always possible for Omkar to achieve this under the given constraints.
Example
Input
5
4
RLRL
6
LRRRRL
8
RLLRRRLL
12
LLLLRRLRRRLL
5
RRRRR
Output
0
1
1
3
2
Note
In the first test case, players 1 and 2 are attacking each other, and players 3 and 4 are attacking each other. Each player is being attacked by exactly 1 other player, and each player is attacking the player that is attacking them, so all players are already being logical according to Bed Wars strategy and Omkar does not need to talk to any of them, making the answer 0.
In the second test case, not every player acts logically: for example, player 3 is attacked only by player 2, but doesn't attack him in response. Omkar can talk to player 3 to convert the attack arrangement to LRLRRL, in which you can see that all players are being logical according to Bed Wars strategy, making the answer 1.
Submitted Solution:
```
for test in range(int(input())):
n = int(input())
s1 = input()
#n = len(s1)
s = []
tt = 0
opi = 0
cou = 0
for i in range(len(s1)):
s.append(s1[i])
if s[i] == s[i-1]:
cou+=1
else:
if cou > 5:
opi += 1
cou = 1
if cou > 5:
opi += 1
ans = 0
for i in range(n):
if s[i-1] == s[i] and s[i] == s[(i+1)%n]:
if s[i] == "R":
s[i] = "L"
else:
s[i] = "R"
ans += 1
#print(''.join(s))
#print(ans)
print(ans-opi)
``` | instruction | 0 | 79,332 | 19 | 158,664 |
No | output | 1 | 79,332 | 19 | 158,665 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Omkar is playing his favorite pixelated video game, Bed Wars! In Bed Wars, there are n players arranged in a circle, so that for all j such that 2 ≤ j ≤ n, player j - 1 is to the left of the player j, and player j is to the right of player j - 1. Additionally, player n is to the left of player 1, and player 1 is to the right of player n.
Currently, each player is attacking either the player to their left or the player to their right. This means that each player is currently being attacked by either 0, 1, or 2 other players. A key element of Bed Wars strategy is that if a player is being attacked by exactly 1 other player, then they should logically attack that player in response. If instead a player is being attacked by 0 or 2 other players, then Bed Wars strategy says that the player can logically attack either of the adjacent players.
Unfortunately, it might be that some players in this game are not following Bed Wars strategy correctly. Omkar is aware of whom each player is currently attacking, and he can talk to any amount of the n players in the game to make them instead attack another player — i. e. if they are currently attacking the player to their left, Omkar can convince them to instead attack the player to their right; if they are currently attacking the player to their right, Omkar can convince them to instead attack the player to their left.
Omkar would like all players to be acting logically. Calculate the minimum amount of players that Omkar needs to talk to so that after all players he talked to (if any) have changed which player they are attacking, all players are acting logically according to Bed Wars strategy.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). The descriptions of the test cases follows.
The first line of each test case contains one integer n (3 ≤ n ≤ 2 ⋅ 10^5) — the amount of players (and therefore beds) in this game of Bed Wars.
The second line of each test case contains a string s of length n. The j-th character of s is equal to L if the j-th player is attacking the player to their left, and R if the j-th player is attacking the player to their right.
It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case, output one integer: the minimum number of players Omkar needs to talk to to make it so that all players are acting logically according to Bed Wars strategy.
It can be proven that it is always possible for Omkar to achieve this under the given constraints.
Example
Input
5
4
RLRL
6
LRRRRL
8
RLLRRRLL
12
LLLLRRLRRRLL
5
RRRRR
Output
0
1
1
3
2
Note
In the first test case, players 1 and 2 are attacking each other, and players 3 and 4 are attacking each other. Each player is being attacked by exactly 1 other player, and each player is attacking the player that is attacking them, so all players are already being logical according to Bed Wars strategy and Omkar does not need to talk to any of them, making the answer 0.
In the second test case, not every player acts logically: for example, player 3 is attacked only by player 2, but doesn't attack him in response. Omkar can talk to player 3 to convert the attack arrangement to LRLRRL, in which you can see that all players are being logical according to Bed Wars strategy, making the answer 1.
Submitted Solution:
```
for t in range(int(input())):
n=int(input())
s=input()
if len(list(set(list(s))))==1:
print((n+2)//3)
else:
a=list(s)
c=0
ans=[]
temp=[]
i=1
temp.append(a[0])
while i<len(a):
if a[i]==a[i-1]:
temp.append(a[i])
else:
if a[i]=='L':
ans.append(temp)
else:
ans.append(temp)
temp=[]
temp.append(a[i])
i+=1
for i in range(len(ans)):
if len(ans[i])>=3:
c+=(len(ans[i])//3)
if len(temp)>=3:
c+=len(temp)//3
if set(ans[0])==set(temp):
if len(temp)>1:
if len(ans[0])%3!=0 and len(temp)%3!=0 and len(ans[0])<3:
c+=1
if len(ans[0])==1 and len(temp)>3 and len(temp)%3!=2:
c-=1
else:
if len(ans[0])>=2 and len(ans[0])%3==2:
c+=1
print(c)
``` | instruction | 0 | 79,333 | 19 | 158,666 |
No | output | 1 | 79,333 | 19 | 158,667 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Iahub got bored, so he invented a game to be played on paper.
He writes n integers a1, a2, ..., an. Each of those integers can be either 0 or 1. He's allowed to do exactly one move: he chooses two indices i and j (1 ≤ i ≤ j ≤ n) and flips all values ak for which their positions are in range [i, j] (that is i ≤ k ≤ j). Flip the value of x means to apply operation x = 1 - x.
The goal of the game is that after exactly one move to obtain the maximum number of ones. Write a program to solve the little game of Iahub.
Input
The first line of the input contains an integer n (1 ≤ n ≤ 100). In the second line of the input there are n integers: a1, a2, ..., an. It is guaranteed that each of those n values is either 0 or 1.
Output
Print an integer — the maximal number of 1s that can be obtained after exactly one move.
Examples
Input
5
1 0 0 1 0
Output
4
Input
4
1 0 0 1
Output
4
Note
In the first case, flip the segment from 2 to 5 (i = 2, j = 5). That flip changes the sequence, it becomes: [1 1 1 0 1]. So, it contains four ones. There is no way to make the whole sequence equal to [1 1 1 1 1].
In the second case, flipping only the second and the third element (i = 2, j = 3) will turn all numbers into 1. | instruction | 0 | 79,434 | 19 | 158,868 |
Tags: brute force, dp, implementation
Correct Solution:
```
n=int(input())
s=input().split()
M=[[0]*n for _ in range(n)]
Max=-1000
num=0
for i in range(n):
if s[i]=='0':
M[i][i]=1
else:
M[i][i]=-1
num+=1
Max=max(M[i][i],Max)
for j in range(n-1,-1,-1):
for i in range(j-1,-1,-1):
if s[i]=='1':
M[i][j]=M[i+1][j]-1
else:
M[i][j]=M[i+1][j]+1
Max=max(M[i][j],Max)
print(num+Max)
``` | output | 1 | 79,434 | 19 | 158,869 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Iahub got bored, so he invented a game to be played on paper.
He writes n integers a1, a2, ..., an. Each of those integers can be either 0 or 1. He's allowed to do exactly one move: he chooses two indices i and j (1 ≤ i ≤ j ≤ n) and flips all values ak for which their positions are in range [i, j] (that is i ≤ k ≤ j). Flip the value of x means to apply operation x = 1 - x.
The goal of the game is that after exactly one move to obtain the maximum number of ones. Write a program to solve the little game of Iahub.
Input
The first line of the input contains an integer n (1 ≤ n ≤ 100). In the second line of the input there are n integers: a1, a2, ..., an. It is guaranteed that each of those n values is either 0 or 1.
Output
Print an integer — the maximal number of 1s that can be obtained after exactly one move.
Examples
Input
5
1 0 0 1 0
Output
4
Input
4
1 0 0 1
Output
4
Note
In the first case, flip the segment from 2 to 5 (i = 2, j = 5). That flip changes the sequence, it becomes: [1 1 1 0 1]. So, it contains four ones. There is no way to make the whole sequence equal to [1 1 1 1 1].
In the second case, flipping only the second and the third element (i = 2, j = 3) will turn all numbers into 1. | instruction | 0 | 79,437 | 19 | 158,874 |
Tags: brute force, dp, implementation
Correct Solution:
```
class CodeforcesTask327ASolution:
def __init__(self):
self.result = ''
self.n = 0
self.a = []
def read_input(self):
self.n = int(input())
self.a = [int(x) for x in input().split(" ")]
def process_task(self):
mx = 0
base = sum(self.a)
for x in range(self.n):
for y in range(x + 1, self.n + 1):
n = sum(self.a[x:y])
nb = base - n + (y - x - n)
#print(self.a[x:y], x, y, n, nb, mx)
mx = max(mx, nb)
self.result = str(mx)
def get_result(self):
return self.result
if __name__ == "__main__":
Solution = CodeforcesTask327ASolution()
Solution.read_input()
Solution.process_task()
print(Solution.get_result())
``` | output | 1 | 79,437 | 19 | 158,875 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Iahub got bored, so he invented a game to be played on paper.
He writes n integers a1, a2, ..., an. Each of those integers can be either 0 or 1. He's allowed to do exactly one move: he chooses two indices i and j (1 ≤ i ≤ j ≤ n) and flips all values ak for which their positions are in range [i, j] (that is i ≤ k ≤ j). Flip the value of x means to apply operation x = 1 - x.
The goal of the game is that after exactly one move to obtain the maximum number of ones. Write a program to solve the little game of Iahub.
Input
The first line of the input contains an integer n (1 ≤ n ≤ 100). In the second line of the input there are n integers: a1, a2, ..., an. It is guaranteed that each of those n values is either 0 or 1.
Output
Print an integer — the maximal number of 1s that can be obtained after exactly one move.
Examples
Input
5
1 0 0 1 0
Output
4
Input
4
1 0 0 1
Output
4
Note
In the first case, flip the segment from 2 to 5 (i = 2, j = 5). That flip changes the sequence, it becomes: [1 1 1 0 1]. So, it contains four ones. There is no way to make the whole sequence equal to [1 1 1 1 1].
In the second case, flipping only the second and the third element (i = 2, j = 3) will turn all numbers into 1. | instruction | 0 | 79,438 | 19 | 158,876 |
Tags: brute force, dp, implementation
Correct Solution:
```
n = int(input())
a = [int(i) for i in input().split()][:n]
print(max(sum(a)-2*sum(a[i:j])+j-i for i in range(n) for j in range(i+1,n+1)))
``` | output | 1 | 79,438 | 19 | 158,877 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Iahub got bored, so he invented a game to be played on paper.
He writes n integers a1, a2, ..., an. Each of those integers can be either 0 or 1. He's allowed to do exactly one move: he chooses two indices i and j (1 ≤ i ≤ j ≤ n) and flips all values ak for which their positions are in range [i, j] (that is i ≤ k ≤ j). Flip the value of x means to apply operation x = 1 - x.
The goal of the game is that after exactly one move to obtain the maximum number of ones. Write a program to solve the little game of Iahub.
Input
The first line of the input contains an integer n (1 ≤ n ≤ 100). In the second line of the input there are n integers: a1, a2, ..., an. It is guaranteed that each of those n values is either 0 or 1.
Output
Print an integer — the maximal number of 1s that can be obtained after exactly one move.
Examples
Input
5
1 0 0 1 0
Output
4
Input
4
1 0 0 1
Output
4
Note
In the first case, flip the segment from 2 to 5 (i = 2, j = 5). That flip changes the sequence, it becomes: [1 1 1 0 1]. So, it contains four ones. There is no way to make the whole sequence equal to [1 1 1 1 1].
In the second case, flipping only the second and the third element (i = 2, j = 3) will turn all numbers into 1. | instruction | 0 | 79,439 | 19 | 158,878 |
Tags: brute force, dp, implementation
Correct Solution:
```
n=int(input())
l=[int(j) for j in input().split()]
t=sum(l)
m=0
for i in range(0,n):
for j in range(i,n):
d=sum(l[i:j+1])
f=t-d
r=f+j-i+1 - d
m=max(m,r)
print(m)
``` | output | 1 | 79,439 | 19 | 158,879 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Iahub got bored, so he invented a game to be played on paper.
He writes n integers a1, a2, ..., an. Each of those integers can be either 0 or 1. He's allowed to do exactly one move: he chooses two indices i and j (1 ≤ i ≤ j ≤ n) and flips all values ak for which their positions are in range [i, j] (that is i ≤ k ≤ j). Flip the value of x means to apply operation x = 1 - x.
The goal of the game is that after exactly one move to obtain the maximum number of ones. Write a program to solve the little game of Iahub.
Input
The first line of the input contains an integer n (1 ≤ n ≤ 100). In the second line of the input there are n integers: a1, a2, ..., an. It is guaranteed that each of those n values is either 0 or 1.
Output
Print an integer — the maximal number of 1s that can be obtained after exactly one move.
Examples
Input
5
1 0 0 1 0
Output
4
Input
4
1 0 0 1
Output
4
Note
In the first case, flip the segment from 2 to 5 (i = 2, j = 5). That flip changes the sequence, it becomes: [1 1 1 0 1]. So, it contains four ones. There is no way to make the whole sequence equal to [1 1 1 1 1].
In the second case, flipping only the second and the third element (i = 2, j = 3) will turn all numbers into 1. | instruction | 0 | 79,440 | 19 | 158,880 |
Tags: brute force, dp, implementation
Correct Solution:
```
n=int(input())
l=[int(i) for i in input().split()]
total=sum(l)
w=0
z=0
for i in range(n):
c=0
k=total
z=0
for j in range(i,n):
if l[j]==0:
c+=1
else:
k-=1
z=max(z,c+k)
w=max(w,z)
print(w)
``` | output | 1 | 79,440 | 19 | 158,881 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Iahub got bored, so he invented a game to be played on paper.
He writes n integers a1, a2, ..., an. Each of those integers can be either 0 or 1. He's allowed to do exactly one move: he chooses two indices i and j (1 ≤ i ≤ j ≤ n) and flips all values ak for which their positions are in range [i, j] (that is i ≤ k ≤ j). Flip the value of x means to apply operation x = 1 - x.
The goal of the game is that after exactly one move to obtain the maximum number of ones. Write a program to solve the little game of Iahub.
Input
The first line of the input contains an integer n (1 ≤ n ≤ 100). In the second line of the input there are n integers: a1, a2, ..., an. It is guaranteed that each of those n values is either 0 or 1.
Output
Print an integer — the maximal number of 1s that can be obtained after exactly one move.
Examples
Input
5
1 0 0 1 0
Output
4
Input
4
1 0 0 1
Output
4
Note
In the first case, flip the segment from 2 to 5 (i = 2, j = 5). That flip changes the sequence, it becomes: [1 1 1 0 1]. So, it contains four ones. There is no way to make the whole sequence equal to [1 1 1 1 1].
In the second case, flipping only the second and the third element (i = 2, j = 3) will turn all numbers into 1. | instruction | 0 | 79,441 | 19 | 158,882 |
Tags: brute force, dp, implementation
Correct Solution:
```
n=int(input())
N = [int(num) for num in input().split(" ", n-1)]
A=[0]*n
i=p=0
while(i<n):
j=i
while j<n :
if(N[j]==0):
p+=1
elif(N[j]==1 and p>0):
p-=1
A[i]=max(A[i],p)
j+=1
p=0
i+=1
x=max(A)
y=sum(N)
if(x>0):
print(x+y)
else:
print(y-1)
``` | output | 1 | 79,441 | 19 | 158,883 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Iahub got bored, so he invented a game to be played on paper.
He writes n integers a1, a2, ..., an. Each of those integers can be either 0 or 1. He's allowed to do exactly one move: he chooses two indices i and j (1 ≤ i ≤ j ≤ n) and flips all values ak for which their positions are in range [i, j] (that is i ≤ k ≤ j). Flip the value of x means to apply operation x = 1 - x.
The goal of the game is that after exactly one move to obtain the maximum number of ones. Write a program to solve the little game of Iahub.
Input
The first line of the input contains an integer n (1 ≤ n ≤ 100). In the second line of the input there are n integers: a1, a2, ..., an. It is guaranteed that each of those n values is either 0 or 1.
Output
Print an integer — the maximal number of 1s that can be obtained after exactly one move.
Examples
Input
5
1 0 0 1 0
Output
4
Input
4
1 0 0 1
Output
4
Note
In the first case, flip the segment from 2 to 5 (i = 2, j = 5). That flip changes the sequence, it becomes: [1 1 1 0 1]. So, it contains four ones. There is no way to make the whole sequence equal to [1 1 1 1 1].
In the second case, flipping only the second and the third element (i = 2, j = 3) will turn all numbers into 1.
Submitted Solution:
```
n = int(input())
a = list(map(int,input().split()))
total = 0
maxv = 0
for i in range(n):
if a[i] == 1:
total += 1
temp = 0
for i in range(n):
for j in range(i+1,n):
temp = 0
for k in range(i,j+1):
if a[k] == 0:
temp += 1
else:
temp -= 1
if maxv < temp:
maxv = temp
if n == 1:
print(1-a[0])
elif maxv == 0:
print(total-1)
else:
print(total+maxv)
``` | instruction | 0 | 79,442 | 19 | 158,884 |
Yes | output | 1 | 79,442 | 19 | 158,885 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Iahub got bored, so he invented a game to be played on paper.
He writes n integers a1, a2, ..., an. Each of those integers can be either 0 or 1. He's allowed to do exactly one move: he chooses two indices i and j (1 ≤ i ≤ j ≤ n) and flips all values ak for which their positions are in range [i, j] (that is i ≤ k ≤ j). Flip the value of x means to apply operation x = 1 - x.
The goal of the game is that after exactly one move to obtain the maximum number of ones. Write a program to solve the little game of Iahub.
Input
The first line of the input contains an integer n (1 ≤ n ≤ 100). In the second line of the input there are n integers: a1, a2, ..., an. It is guaranteed that each of those n values is either 0 or 1.
Output
Print an integer — the maximal number of 1s that can be obtained after exactly one move.
Examples
Input
5
1 0 0 1 0
Output
4
Input
4
1 0 0 1
Output
4
Note
In the first case, flip the segment from 2 to 5 (i = 2, j = 5). That flip changes the sequence, it becomes: [1 1 1 0 1]. So, it contains four ones. There is no way to make the whole sequence equal to [1 1 1 1 1].
In the second case, flipping only the second and the third element (i = 2, j = 3) will turn all numbers into 1.
Submitted Solution:
```
n=int(input())
line=input().split()
m=[int(i) for i in line]
one=m.count(1)
if 0 not in m:
print(n-1)
else:
two=0
a=0
for i in range(n):
if m[i]==1:
a-=1
elif m[i]==0:
a+=1
if a<0:
a=0
if a>two:
two=a
print(one+two)
``` | instruction | 0 | 79,443 | 19 | 158,886 |
Yes | output | 1 | 79,443 | 19 | 158,887 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Iahub got bored, so he invented a game to be played on paper.
He writes n integers a1, a2, ..., an. Each of those integers can be either 0 or 1. He's allowed to do exactly one move: he chooses two indices i and j (1 ≤ i ≤ j ≤ n) and flips all values ak for which their positions are in range [i, j] (that is i ≤ k ≤ j). Flip the value of x means to apply operation x = 1 - x.
The goal of the game is that after exactly one move to obtain the maximum number of ones. Write a program to solve the little game of Iahub.
Input
The first line of the input contains an integer n (1 ≤ n ≤ 100). In the second line of the input there are n integers: a1, a2, ..., an. It is guaranteed that each of those n values is either 0 or 1.
Output
Print an integer — the maximal number of 1s that can be obtained after exactly one move.
Examples
Input
5
1 0 0 1 0
Output
4
Input
4
1 0 0 1
Output
4
Note
In the first case, flip the segment from 2 to 5 (i = 2, j = 5). That flip changes the sequence, it becomes: [1 1 1 0 1]. So, it contains four ones. There is no way to make the whole sequence equal to [1 1 1 1 1].
In the second case, flipping only the second and the third element (i = 2, j = 3) will turn all numbers into 1.
Submitted Solution:
```
n = int(input())
count0 = 0
count1 = 0
maxRun = -1
for i in input().split():
t = int(i)
if(t == 1):
count1 += 1
if(count0 > 0):
count0 -= 1
else:
count0 += 1
if(count0 > maxRun):
maxRun = count0
ans = count1 + maxRun
print(ans)
``` | instruction | 0 | 79,444 | 19 | 158,888 |
Yes | output | 1 | 79,444 | 19 | 158,889 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Iahub got bored, so he invented a game to be played on paper.
He writes n integers a1, a2, ..., an. Each of those integers can be either 0 or 1. He's allowed to do exactly one move: he chooses two indices i and j (1 ≤ i ≤ j ≤ n) and flips all values ak for which their positions are in range [i, j] (that is i ≤ k ≤ j). Flip the value of x means to apply operation x = 1 - x.
The goal of the game is that after exactly one move to obtain the maximum number of ones. Write a program to solve the little game of Iahub.
Input
The first line of the input contains an integer n (1 ≤ n ≤ 100). In the second line of the input there are n integers: a1, a2, ..., an. It is guaranteed that each of those n values is either 0 or 1.
Output
Print an integer — the maximal number of 1s that can be obtained after exactly one move.
Examples
Input
5
1 0 0 1 0
Output
4
Input
4
1 0 0 1
Output
4
Note
In the first case, flip the segment from 2 to 5 (i = 2, j = 5). That flip changes the sequence, it becomes: [1 1 1 0 1]. So, it contains four ones. There is no way to make the whole sequence equal to [1 1 1 1 1].
In the second case, flipping only the second and the third element (i = 2, j = 3) will turn all numbers into 1.
Submitted Solution:
```
n = int(input())
a = list(map(int, input().split()))
s = 0
v = []
for i in a:
s += i
if i:
v.append(-1)
else:
v.append(1)
x = v[0]
ans = x
mn = x
for i in range(len(v)):
x += v[i]
ans = max(ans, x - mn)
mn = min(mn, x)
print(s + ans)
``` | instruction | 0 | 79,445 | 19 | 158,890 |
Yes | output | 1 | 79,445 | 19 | 158,891 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Iahub got bored, so he invented a game to be played on paper.
He writes n integers a1, a2, ..., an. Each of those integers can be either 0 or 1. He's allowed to do exactly one move: he chooses two indices i and j (1 ≤ i ≤ j ≤ n) and flips all values ak for which their positions are in range [i, j] (that is i ≤ k ≤ j). Flip the value of x means to apply operation x = 1 - x.
The goal of the game is that after exactly one move to obtain the maximum number of ones. Write a program to solve the little game of Iahub.
Input
The first line of the input contains an integer n (1 ≤ n ≤ 100). In the second line of the input there are n integers: a1, a2, ..., an. It is guaranteed that each of those n values is either 0 or 1.
Output
Print an integer — the maximal number of 1s that can be obtained after exactly one move.
Examples
Input
5
1 0 0 1 0
Output
4
Input
4
1 0 0 1
Output
4
Note
In the first case, flip the segment from 2 to 5 (i = 2, j = 5). That flip changes the sequence, it becomes: [1 1 1 0 1]. So, it contains four ones. There is no way to make the whole sequence equal to [1 1 1 1 1].
In the second case, flipping only the second and the third element (i = 2, j = 3) will turn all numbers into 1.
Submitted Solution:
```
if __name__ == "__main__":
n = int(input())
a = list(map(int, input().split()))
maxSoFar = sum(1 for i in a if i == 1)
S = maxSoFar
maxHere = maxSoFar
inverse_dp = [1 if i == 0 else -1 for i in a]
for i in range(n):
maxHere += a[i]
if maxSoFar < maxHere:
maxSoFar = maxHere
if maxHere < S:
maxHere = S
maxSoFar = max(maxSoFar,maxHere)
print(maxSoFar)
``` | instruction | 0 | 79,446 | 19 | 158,892 |
No | output | 1 | 79,446 | 19 | 158,893 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Iahub got bored, so he invented a game to be played on paper.
He writes n integers a1, a2, ..., an. Each of those integers can be either 0 or 1. He's allowed to do exactly one move: he chooses two indices i and j (1 ≤ i ≤ j ≤ n) and flips all values ak for which their positions are in range [i, j] (that is i ≤ k ≤ j). Flip the value of x means to apply operation x = 1 - x.
The goal of the game is that after exactly one move to obtain the maximum number of ones. Write a program to solve the little game of Iahub.
Input
The first line of the input contains an integer n (1 ≤ n ≤ 100). In the second line of the input there are n integers: a1, a2, ..., an. It is guaranteed that each of those n values is either 0 or 1.
Output
Print an integer — the maximal number of 1s that can be obtained after exactly one move.
Examples
Input
5
1 0 0 1 0
Output
4
Input
4
1 0 0 1
Output
4
Note
In the first case, flip the segment from 2 to 5 (i = 2, j = 5). That flip changes the sequence, it becomes: [1 1 1 0 1]. So, it contains four ones. There is no way to make the whole sequence equal to [1 1 1 1 1].
In the second case, flipping only the second and the third element (i = 2, j = 3) will turn all numbers into 1.
Submitted Solution:
```
n = int(input())
a = input().split()
l = [0]
zbr = 0
for i in range (n):
if (a[i] == "1"):
zbr += 1
l.append (zbr)
for i in range (n):
r = []
for j in range (i, n):
p = l[i - 1]
s = l[j - i]
k = l[-1] - l[j]
r.append(p + s + k)
l.append(max(r))
print (max(l))
``` | instruction | 0 | 79,447 | 19 | 158,894 |
No | output | 1 | 79,447 | 19 | 158,895 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Iahub got bored, so he invented a game to be played on paper.
He writes n integers a1, a2, ..., an. Each of those integers can be either 0 or 1. He's allowed to do exactly one move: he chooses two indices i and j (1 ≤ i ≤ j ≤ n) and flips all values ak for which their positions are in range [i, j] (that is i ≤ k ≤ j). Flip the value of x means to apply operation x = 1 - x.
The goal of the game is that after exactly one move to obtain the maximum number of ones. Write a program to solve the little game of Iahub.
Input
The first line of the input contains an integer n (1 ≤ n ≤ 100). In the second line of the input there are n integers: a1, a2, ..., an. It is guaranteed that each of those n values is either 0 or 1.
Output
Print an integer — the maximal number of 1s that can be obtained after exactly one move.
Examples
Input
5
1 0 0 1 0
Output
4
Input
4
1 0 0 1
Output
4
Note
In the first case, flip the segment from 2 to 5 (i = 2, j = 5). That flip changes the sequence, it becomes: [1 1 1 0 1]. So, it contains four ones. There is no way to make the whole sequence equal to [1 1 1 1 1].
In the second case, flipping only the second and the third element (i = 2, j = 3) will turn all numbers into 1.
Submitted Solution:
```
n=int(input())
l=input().split()
a=0
b=0
m=[]
for i in range(n):
if int(l[i])==0:
a+=1
else:
m.append(a)
b+=1
a=0
f=max(m)+b
print(f)
``` | instruction | 0 | 79,448 | 19 | 158,896 |
No | output | 1 | 79,448 | 19 | 158,897 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Iahub got bored, so he invented a game to be played on paper.
He writes n integers a1, a2, ..., an. Each of those integers can be either 0 or 1. He's allowed to do exactly one move: he chooses two indices i and j (1 ≤ i ≤ j ≤ n) and flips all values ak for which their positions are in range [i, j] (that is i ≤ k ≤ j). Flip the value of x means to apply operation x = 1 - x.
The goal of the game is that after exactly one move to obtain the maximum number of ones. Write a program to solve the little game of Iahub.
Input
The first line of the input contains an integer n (1 ≤ n ≤ 100). In the second line of the input there are n integers: a1, a2, ..., an. It is guaranteed that each of those n values is either 0 or 1.
Output
Print an integer — the maximal number of 1s that can be obtained after exactly one move.
Examples
Input
5
1 0 0 1 0
Output
4
Input
4
1 0 0 1
Output
4
Note
In the first case, flip the segment from 2 to 5 (i = 2, j = 5). That flip changes the sequence, it becomes: [1 1 1 0 1]. So, it contains four ones. There is no way to make the whole sequence equal to [1 1 1 1 1].
In the second case, flipping only the second and the third element (i = 2, j = 3) will turn all numbers into 1.
Submitted Solution:
```
n=int(input())
p=[int(x) for x in input().split()]
if n==1:
if p[0]==1:
print(0)
else:
print(1)
else:
q=[0]*n
for i in range (1,n):
if p[i]==1:
q[i]=q[i-1]+1
else:
q[i]=q[i-1]-1
print(p.count(1)+max(q)-min(q))
``` | instruction | 0 | 79,449 | 19 | 158,898 |
No | output | 1 | 79,449 | 19 | 158,899 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya and Vasya decided to play a little. They found n red cubes and m blue cubes. The game goes like that: the players take turns to choose a cube of some color (red or blue) and put it in a line from left to right (overall the line will have n + m cubes). Petya moves first. Petya's task is to get as many pairs of neighbouring cubes of the same color as possible. Vasya's task is to get as many pairs of neighbouring cubes of different colors as possible.
The number of Petya's points in the game is the number of pairs of neighboring cubes of the same color in the line, the number of Vasya's points in the game is the number of neighbouring cubes of the different color in the line. Your task is to calculate the score at the end of the game (Petya's and Vasya's points, correspondingly), if both boys are playing optimally well. To "play optimally well" first of all means to maximize the number of one's points, and second — to minimize the number of the opponent's points.
Input
The only line contains two space-separated integers n and m (1 ≤ n, m ≤ 105) — the number of red and blue cubes, correspondingly.
Output
On a single line print two space-separated integers — the number of Petya's and Vasya's points correspondingly provided that both players play optimally well.
Examples
Input
3 1
Output
2 1
Input
2 4
Output
3 2
Note
In the first test sample the optimal strategy for Petya is to put the blue cube in the line. After that there will be only red cubes left, so by the end of the game the line of cubes from left to right will look as [blue, red, red, red]. So, Petya gets 2 points and Vasya gets 1 point.
If Petya would choose the red cube during his first move, then, provided that both boys play optimally well, Petya would get 1 point and Vasya would get 2 points. | instruction | 0 | 80,233 | 19 | 160,466 |
Tags: games, greedy, implementation
Correct Solution:
```
n,m=map(int,input().split())
res=n+m-1
diff=min(n,m)
z=res-diff
print(z,diff)
``` | output | 1 | 80,233 | 19 | 160,467 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya and Vasya decided to play a little. They found n red cubes and m blue cubes. The game goes like that: the players take turns to choose a cube of some color (red or blue) and put it in a line from left to right (overall the line will have n + m cubes). Petya moves first. Petya's task is to get as many pairs of neighbouring cubes of the same color as possible. Vasya's task is to get as many pairs of neighbouring cubes of different colors as possible.
The number of Petya's points in the game is the number of pairs of neighboring cubes of the same color in the line, the number of Vasya's points in the game is the number of neighbouring cubes of the different color in the line. Your task is to calculate the score at the end of the game (Petya's and Vasya's points, correspondingly), if both boys are playing optimally well. To "play optimally well" first of all means to maximize the number of one's points, and second — to minimize the number of the opponent's points.
Input
The only line contains two space-separated integers n and m (1 ≤ n, m ≤ 105) — the number of red and blue cubes, correspondingly.
Output
On a single line print two space-separated integers — the number of Petya's and Vasya's points correspondingly provided that both players play optimally well.
Examples
Input
3 1
Output
2 1
Input
2 4
Output
3 2
Note
In the first test sample the optimal strategy for Petya is to put the blue cube in the line. After that there will be only red cubes left, so by the end of the game the line of cubes from left to right will look as [blue, red, red, red]. So, Petya gets 2 points and Vasya gets 1 point.
If Petya would choose the red cube during his first move, then, provided that both boys play optimally well, Petya would get 1 point and Vasya would get 2 points. | instruction | 0 | 80,234 | 19 | 160,468 |
Tags: games, greedy, implementation
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
from math import ceil, floor, pow, sqrt, gcd
from collections import Counter, defaultdict
from itertools import permutations, combinations
from time import time, sleep
BUFSIZE = 8192
MOD = 1000000007
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")
stdin, stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
from os import path
if path.exists('tc.txt'):
stdin = open('tc.txt', 'r')
def gmi(): return map(int, stdin.readline().strip().split())
def gms(): return map(str, stdin.readline().strip().split())
def gari(): return list(map(int, stdin.readline().strip().split()))
def gart(): return tuple(map(int, stdin.readline().strip().split()))
def gars(): return list(map(str, stdin.readline().strip().split()))
def gs(): return stdin.readline().strip()
def gls(): return list(stdin.readline().strip())
def gi(): return int(stdin.readline())
def pri(i, end="\n"): stdout.write(f"{i}" + end)
def priar(ar, end="\n"): stdout.write(" ".join(map(str, ar)) + end)
def solve():
r, b = gmi()
return [max(b, r)-1, min(b, r)]
if __name__ == "__main__":
tc = 1
for i in range(tc):
priar(solve())
``` | output | 1 | 80,234 | 19 | 160,469 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya and Vasya decided to play a little. They found n red cubes and m blue cubes. The game goes like that: the players take turns to choose a cube of some color (red or blue) and put it in a line from left to right (overall the line will have n + m cubes). Petya moves first. Petya's task is to get as many pairs of neighbouring cubes of the same color as possible. Vasya's task is to get as many pairs of neighbouring cubes of different colors as possible.
The number of Petya's points in the game is the number of pairs of neighboring cubes of the same color in the line, the number of Vasya's points in the game is the number of neighbouring cubes of the different color in the line. Your task is to calculate the score at the end of the game (Petya's and Vasya's points, correspondingly), if both boys are playing optimally well. To "play optimally well" first of all means to maximize the number of one's points, and second — to minimize the number of the opponent's points.
Input
The only line contains two space-separated integers n and m (1 ≤ n, m ≤ 105) — the number of red and blue cubes, correspondingly.
Output
On a single line print two space-separated integers — the number of Petya's and Vasya's points correspondingly provided that both players play optimally well.
Examples
Input
3 1
Output
2 1
Input
2 4
Output
3 2
Note
In the first test sample the optimal strategy for Petya is to put the blue cube in the line. After that there will be only red cubes left, so by the end of the game the line of cubes from left to right will look as [blue, red, red, red]. So, Petya gets 2 points and Vasya gets 1 point.
If Petya would choose the red cube during his first move, then, provided that both boys play optimally well, Petya would get 1 point and Vasya would get 2 points. | instruction | 0 | 80,235 | 19 | 160,470 |
Tags: games, greedy, implementation
Correct Solution:
```
# your code goes here
arr = input().split()
a = int(arr[0])
b = int(arr[1])
#print('a = ' + str(a) + 'b = ' + str(b))
if a%2 == 1 and b%2 == 1:
if a < b:
# print('a')
a -= 1
flag = 1
else:
# print('b')
b -= 1
flag = 2
elif a%2 == 1:
a -= 1
flag = 1
elif b%2 == 1:
b -= 1
flag = 2
else:
if a > b:
a -= 1
flag = 1
else:
b -= 1
flag = 2
if flag == 1:
s = 'a'
else:
s = 'b'
play = 2
while a > 0 and b > 0:
if flag == 1:
if play == 2:
s += 'b'
b -= 1
play = 1
flag = 2
else:
s += 'a'
a -= 1
play = 2
else:
if play == 2:
s += 'a'
a -= 1
play = 1
flag = 1
else:
s += 'b'
b -= 1
play = 2
while a > 0:
s += 'a'
a -= 1
while b > 0:
s += 'b'
b -= 1
s = list(s)
prev = s[0]
n = len(s)
a = 0
b = 0
for i in range(1,n):
if s[i] == prev:
a += 1
else:
b += 1
prev = s[i]
print(str(a) + ' ' + str(b))
``` | output | 1 | 80,235 | 19 | 160,471 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya and Vasya decided to play a little. They found n red cubes and m blue cubes. The game goes like that: the players take turns to choose a cube of some color (red or blue) and put it in a line from left to right (overall the line will have n + m cubes). Petya moves first. Petya's task is to get as many pairs of neighbouring cubes of the same color as possible. Vasya's task is to get as many pairs of neighbouring cubes of different colors as possible.
The number of Petya's points in the game is the number of pairs of neighboring cubes of the same color in the line, the number of Vasya's points in the game is the number of neighbouring cubes of the different color in the line. Your task is to calculate the score at the end of the game (Petya's and Vasya's points, correspondingly), if both boys are playing optimally well. To "play optimally well" first of all means to maximize the number of one's points, and second — to minimize the number of the opponent's points.
Input
The only line contains two space-separated integers n and m (1 ≤ n, m ≤ 105) — the number of red and blue cubes, correspondingly.
Output
On a single line print two space-separated integers — the number of Petya's and Vasya's points correspondingly provided that both players play optimally well.
Examples
Input
3 1
Output
2 1
Input
2 4
Output
3 2
Note
In the first test sample the optimal strategy for Petya is to put the blue cube in the line. After that there will be only red cubes left, so by the end of the game the line of cubes from left to right will look as [blue, red, red, red]. So, Petya gets 2 points and Vasya gets 1 point.
If Petya would choose the red cube during his first move, then, provided that both boys play optimally well, Petya would get 1 point and Vasya would get 2 points. | instruction | 0 | 80,236 | 19 | 160,472 |
Tags: games, greedy, implementation
Correct Solution:
```
n, m = map(int, input().split())
seq = []
if n%2 and m%2:
print(n+m-1-min(n, m), min(n, m))
elif not n%2 and not m%2:
print(n+m-1-min(n, m), min(n, m))
else:
print(n+m-1-min(n, m), min(n, m))
``` | output | 1 | 80,236 | 19 | 160,473 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya and Vasya decided to play a little. They found n red cubes and m blue cubes. The game goes like that: the players take turns to choose a cube of some color (red or blue) and put it in a line from left to right (overall the line will have n + m cubes). Petya moves first. Petya's task is to get as many pairs of neighbouring cubes of the same color as possible. Vasya's task is to get as many pairs of neighbouring cubes of different colors as possible.
The number of Petya's points in the game is the number of pairs of neighboring cubes of the same color in the line, the number of Vasya's points in the game is the number of neighbouring cubes of the different color in the line. Your task is to calculate the score at the end of the game (Petya's and Vasya's points, correspondingly), if both boys are playing optimally well. To "play optimally well" first of all means to maximize the number of one's points, and second — to minimize the number of the opponent's points.
Input
The only line contains two space-separated integers n and m (1 ≤ n, m ≤ 105) — the number of red and blue cubes, correspondingly.
Output
On a single line print two space-separated integers — the number of Petya's and Vasya's points correspondingly provided that both players play optimally well.
Examples
Input
3 1
Output
2 1
Input
2 4
Output
3 2
Note
In the first test sample the optimal strategy for Petya is to put the blue cube in the line. After that there will be only red cubes left, so by the end of the game the line of cubes from left to right will look as [blue, red, red, red]. So, Petya gets 2 points and Vasya gets 1 point.
If Petya would choose the red cube during his first move, then, provided that both boys play optimally well, Petya would get 1 point and Vasya would get 2 points. | instruction | 0 | 80,237 | 19 | 160,474 |
Tags: games, greedy, implementation
Correct Solution:
```
"""Things to do if you are stuck:-
1.Read the problem statement again, maybe you've read something wrong.
2.See the explanation for the sample input .
3.If the solution is getting too complex in cases where no. of submissions
are high ,then drop that idea because there is something simple which you are
missing.
4.Check for runtime errors if unexpected o/p is seen.
5.Check on edge cases before submitting.
6.Ensure that you have read all the inputs before returning for a test case.
7.Try to think of brute force first if nothing is striking.
8.Take more examples
9.Don't give up , maybe you're just one statement away! """
#pyrival orz
import os
import sys
import math
from io import BytesIO, IOBase
input = sys.stdin.readline
############ ---- Input Functions ---- ############
def inp():
return(int(input()))
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()))
############ ---- Dijkstra with path ---- ############
def dijkstra(start, distance, path, n):
# requires n == number of vertices in graph,
# adj == adjacency list with weight of graph
visited = [False for _ in range(n)] # To keep track of vertices that are visited
distance[start] = 0 # distance of start node from itself is 0
for i in range(n):
v = -1 # Initialize v == vertex from which its neighboring vertices' distance will be calculated
for j in range(n):
# If it has not been visited and has the lowest distance from start
if not visited[v] and (v == -1 or distance[j] < distance[v]):
v = j
if distance[v] == math.inf:
break
visited[v] = True # Mark as visited
for edge in adj[v]:
destination = edge[0] # Neighbor of the vertex
weight = edge[1] # Its corresponding weight
if distance[v] + weight < distance[destination]: # If its distance is less than the stored distance
distance[destination] = distance[v] + weight # Update the distance
path[destination] = v # Update the path
def gcd(a, b):
if b == 0:
return a
else:
return gcd(b, a%b)
def lcm(a, b):
return (a*b)//gcd(a, b)
def ncr(n, r):
return math.factorial(n)//(math.factorial(n-r)*math.factorial(r))
def npr(n, r):
return math.factorial(n)//math.factorial(n-r)
def seive(n):
primes = [True]*(n+1)
ans = []
for i in range(2, n):
if not primes[i]:
continue
j = 2*i
while j <= n:
primes[j] = False
j += i
for p in range(2, n+1):
if primes[p]:
ans += [p]
return ans
def factors(n):
factors = []
x = 1
while x*x <= n:
if n % x == 0:
if n // x == x:
factors.append(x)
else:
factors.append(x)
factors.append(n//x)
x += 1
return factors
# Functions: list of factors, seive of primes, gcd of two numbers,
# lcm of two numbers, npr, ncr
def main():
try:
n, m = invr()
print(max(n, m) - 1, min(n, m))
except Exception as e:
print(e)
# region fastio
BUFSIZE = 8192
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")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
main()
``` | output | 1 | 80,237 | 19 | 160,475 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya and Vasya decided to play a little. They found n red cubes and m blue cubes. The game goes like that: the players take turns to choose a cube of some color (red or blue) and put it in a line from left to right (overall the line will have n + m cubes). Petya moves first. Petya's task is to get as many pairs of neighbouring cubes of the same color as possible. Vasya's task is to get as many pairs of neighbouring cubes of different colors as possible.
The number of Petya's points in the game is the number of pairs of neighboring cubes of the same color in the line, the number of Vasya's points in the game is the number of neighbouring cubes of the different color in the line. Your task is to calculate the score at the end of the game (Petya's and Vasya's points, correspondingly), if both boys are playing optimally well. To "play optimally well" first of all means to maximize the number of one's points, and second — to minimize the number of the opponent's points.
Input
The only line contains two space-separated integers n and m (1 ≤ n, m ≤ 105) — the number of red and blue cubes, correspondingly.
Output
On a single line print two space-separated integers — the number of Petya's and Vasya's points correspondingly provided that both players play optimally well.
Examples
Input
3 1
Output
2 1
Input
2 4
Output
3 2
Note
In the first test sample the optimal strategy for Petya is to put the blue cube in the line. After that there will be only red cubes left, so by the end of the game the line of cubes from left to right will look as [blue, red, red, red]. So, Petya gets 2 points and Vasya gets 1 point.
If Petya would choose the red cube during his first move, then, provided that both boys play optimally well, Petya would get 1 point and Vasya would get 2 points. | instruction | 0 | 80,238 | 19 | 160,476 |
Tags: games, greedy, implementation
Correct Solution:
```
from sys import stdin, stdout
if not __debug__:
stdin = open("input.txt", "r")
tcs = int(stdin.readline()) if not __debug__ else 1
t = 1
while t<=tcs:
n, m = map(int, stdin.readline().split())
big = max(n, m)
small = min(n, m)
print(big-1, small)
t += 1
``` | output | 1 | 80,238 | 19 | 160,477 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya and Vasya decided to play a little. They found n red cubes and m blue cubes. The game goes like that: the players take turns to choose a cube of some color (red or blue) and put it in a line from left to right (overall the line will have n + m cubes). Petya moves first. Petya's task is to get as many pairs of neighbouring cubes of the same color as possible. Vasya's task is to get as many pairs of neighbouring cubes of different colors as possible.
The number of Petya's points in the game is the number of pairs of neighboring cubes of the same color in the line, the number of Vasya's points in the game is the number of neighbouring cubes of the different color in the line. Your task is to calculate the score at the end of the game (Petya's and Vasya's points, correspondingly), if both boys are playing optimally well. To "play optimally well" first of all means to maximize the number of one's points, and second — to minimize the number of the opponent's points.
Input
The only line contains two space-separated integers n and m (1 ≤ n, m ≤ 105) — the number of red and blue cubes, correspondingly.
Output
On a single line print two space-separated integers — the number of Petya's and Vasya's points correspondingly provided that both players play optimally well.
Examples
Input
3 1
Output
2 1
Input
2 4
Output
3 2
Note
In the first test sample the optimal strategy for Petya is to put the blue cube in the line. After that there will be only red cubes left, so by the end of the game the line of cubes from left to right will look as [blue, red, red, red]. So, Petya gets 2 points and Vasya gets 1 point.
If Petya would choose the red cube during his first move, then, provided that both boys play optimally well, Petya would get 1 point and Vasya would get 2 points. | instruction | 0 | 80,239 | 19 | 160,478 |
Tags: games, greedy, implementation
Correct Solution:
```
# ---------------------------iye ha aam zindegi---------------------------------------------
import math
import random
import heapq, bisect
import sys
from collections import deque, defaultdict
from fractions import Fraction
import sys
import threading
from collections import defaultdict
threading.stack_size(10**8)
mod = 10 ** 9 + 7
mod1 = 998244353
# ------------------------------warmup----------------------------
import os
import sys
from io import BytesIO, IOBase
sys.setrecursionlimit(300000)
BUFSIZE = 8192
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")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# -------------------game starts now----------------------------------------------------import math
class TreeNode:
def __init__(self, k, v):
self.key = k
self.value = v
self.left = None
self.right = None
self.parent = None
self.height = 1
self.num_left = 1
self.num_total = 1
class AvlTree:
def __init__(self):
self._tree = None
def add(self, k, v):
if not self._tree:
self._tree = TreeNode(k, v)
return
node = self._add(k, v)
if node:
self._rebalance(node)
def _add(self, k, v):
node = self._tree
while node:
if k < node.key:
if node.left:
node = node.left
else:
node.left = TreeNode(k, v)
node.left.parent = node
return node.left
elif node.key < k:
if node.right:
node = node.right
else:
node.right = TreeNode(k, v)
node.right.parent = node
return node.right
else:
node.value = v
return
@staticmethod
def get_height(x):
return x.height if x else 0
@staticmethod
def get_num_total(x):
return x.num_total if x else 0
def _rebalance(self, node):
n = node
while n:
lh = self.get_height(n.left)
rh = self.get_height(n.right)
n.height = max(lh, rh) + 1
balance_factor = lh - rh
n.num_total = 1 + self.get_num_total(n.left) + self.get_num_total(n.right)
n.num_left = 1 + self.get_num_total(n.left)
if balance_factor > 1:
if self.get_height(n.left.left) < self.get_height(n.left.right):
self._rotate_left(n.left)
self._rotate_right(n)
elif balance_factor < -1:
if self.get_height(n.right.right) < self.get_height(n.right.left):
self._rotate_right(n.right)
self._rotate_left(n)
else:
n = n.parent
def _remove_one(self, node):
"""
Side effect!!! Changes node. Node should have exactly one child
"""
replacement = node.left or node.right
if node.parent:
if AvlTree._is_left(node):
node.parent.left = replacement
else:
node.parent.right = replacement
replacement.parent = node.parent
node.parent = None
else:
self._tree = replacement
replacement.parent = None
node.left = None
node.right = None
node.parent = None
self._rebalance(replacement)
def _remove_leaf(self, node):
if node.parent:
if AvlTree._is_left(node):
node.parent.left = None
else:
node.parent.right = None
self._rebalance(node.parent)
else:
self._tree = None
node.parent = None
node.left = None
node.right = None
def remove(self, k):
node = self._get_node(k)
if not node:
return
if AvlTree._is_leaf(node):
self._remove_leaf(node)
return
if node.left and node.right:
nxt = AvlTree._get_next(node)
node.key = nxt.key
node.value = nxt.value
if self._is_leaf(nxt):
self._remove_leaf(nxt)
else:
self._remove_one(nxt)
self._rebalance(node)
else:
self._remove_one(node)
def get(self, k):
node = self._get_node(k)
return node.value if node else -1
def _get_node(self, k):
if not self._tree:
return None
node = self._tree
while node:
if k < node.key:
node = node.left
elif node.key < k:
node = node.right
else:
return node
return None
def get_at(self, pos):
x = pos + 1
node = self._tree
while node:
if x < node.num_left:
node = node.left
elif node.num_left < x:
x -= node.num_left
node = node.right
else:
return (node.key, node.value)
raise IndexError("Out of ranges")
@staticmethod
def _is_left(node):
return node.parent.left and node.parent.left == node
@staticmethod
def _is_leaf(node):
return node.left is None and node.right is None
def _rotate_right(self, node):
if not node.parent:
self._tree = node.left
node.left.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.left
node.left.parent = node.parent
else:
node.parent.right = node.left
node.left.parent = node.parent
bk = node.left.right
node.left.right = node
node.parent = node.left
node.left = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
def _rotate_left(self, node):
if not node.parent:
self._tree = node.right
node.right.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.right
node.right.parent = node.parent
else:
node.parent.right = node.right
node.right.parent = node.parent
bk = node.right.left
node.right.left = node
node.parent = node.right
node.right = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
@staticmethod
def _get_next(node):
if not node.right:
return node.parent
n = node.right
while n.left:
n = n.left
return n
# -----------------------------------------------binary seacrh tree---------------------------------------
class SegmentTree1:
def __init__(self, data, default=2**51, func=lambda a, b: a & b):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------game starts now----------------------------------------------------import math
class SegmentTree:
def __init__(self, data, default=0, func=lambda a, b: a + b):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------------------iye ha chutiya zindegi-------------------------------------
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
# --------------------------------------iye ha combinations ka zindegi---------------------------------
def powm(a, n, m):
if a == 1 or n == 0:
return 1
if n % 2 == 0:
s = powm(a, n // 2, m)
return s * s % m
else:
return a * powm(a, n - 1, m) % m
# --------------------------------------iye ha power ka zindegi---------------------------------
def sort_list(list1, list2):
zipped_pairs = zip(list2, list1)
z = [x for _, x in sorted(zipped_pairs)]
return z
# --------------------------------------------------product----------------------------------------
def product(l):
por = 1
for i in range(len(l)):
por *= l[i]
return por
# --------------------------------------------------binary----------------------------------------
def binarySearchCount(arr, n, key):
left = 0
right = n - 1
count = 0
while (left <= right):
mid = int((right + left) / 2)
# Check if middle element is
# less than or equal to key
if (arr[mid] <=key):
count = mid + 1
left = mid + 1
# If key is smaller, ignore right half
else:
right = mid - 1
return count
# --------------------------------------------------binary----------------------------------------
def countdig(n):
c = 0
while (n > 0):
n //= 10
c += 1
return c
def binary(x, length):
y = bin(x)[2:]
return y if len(y) >= length else "0" * (length - len(y)) + y
def countGreater(arr, n, k):
l = 0
r = n - 1
# Stores the index of the left most element
# from the array which is greater than k
leftGreater = n
# Finds number of elements greater than k
while (l <= r):
m = int(l + (r - l) / 2)
if (arr[m] >= k):
leftGreater = m
r = m - 1
# If mid element is less than
# or equal to k update l
else:
l = m + 1
# Return the count of elements
# greater than k
return (n - leftGreater)
# --------------------------------------------------binary------------------------------------
n,m=map(int,input().split())
t=1
last=0
w=[n-1,m]
an=0
ans=0
an1=0
ans1=0
for i in range(1,n+m):
if t==1:
if w[1-last]>0:
w[1-last]-=1
an+=1
last=1-last
else:
w[last]-=1
ans+=1
else:
if w[last]>0:
w[last]-=1
ans+=1
else:
w[1-last]-=1
an+=1
t=1-t
w=[m-1,n]
last=0
t=1
for i in range(1,n+m):
if t==1:
if w[1-last]>0:
w[1-last]-=1
an1+=1
last=1-last
else:
w[last]-=1
ans1+=1
else:
if w[last]>0:
w[last]-=1
ans1+=1
else:
w[1-last]-=1
an1+=1
t=1-t
if ans>ans1:
print(ans,an)
elif ans<ans1:
print(ans1,an1)
else:
if an>an1:
print(ans,an)
else:
print(ans1,an1)
``` | output | 1 | 80,239 | 19 | 160,479 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya and Vasya decided to play a little. They found n red cubes and m blue cubes. The game goes like that: the players take turns to choose a cube of some color (red or blue) and put it in a line from left to right (overall the line will have n + m cubes). Petya moves first. Petya's task is to get as many pairs of neighbouring cubes of the same color as possible. Vasya's task is to get as many pairs of neighbouring cubes of different colors as possible.
The number of Petya's points in the game is the number of pairs of neighboring cubes of the same color in the line, the number of Vasya's points in the game is the number of neighbouring cubes of the different color in the line. Your task is to calculate the score at the end of the game (Petya's and Vasya's points, correspondingly), if both boys are playing optimally well. To "play optimally well" first of all means to maximize the number of one's points, and second — to minimize the number of the opponent's points.
Input
The only line contains two space-separated integers n and m (1 ≤ n, m ≤ 105) — the number of red and blue cubes, correspondingly.
Output
On a single line print two space-separated integers — the number of Petya's and Vasya's points correspondingly provided that both players play optimally well.
Examples
Input
3 1
Output
2 1
Input
2 4
Output
3 2
Note
In the first test sample the optimal strategy for Petya is to put the blue cube in the line. After that there will be only red cubes left, so by the end of the game the line of cubes from left to right will look as [blue, red, red, red]. So, Petya gets 2 points and Vasya gets 1 point.
If Petya would choose the red cube during his first move, then, provided that both boys play optimally well, Petya would get 1 point and Vasya would get 2 points. | instruction | 0 | 80,240 | 19 | 160,480 |
Tags: games, greedy, implementation
Correct Solution:
```
n,m=map(int,input().split())
if min(n,m)==1:
print (n+m-2,1)
else:
x=min(n,m)
a=x//2+x//2-1
b=2*(x//2)
x=x%2
y=max(n,m)-(2*(x//2))
a=a+1
y=y-1
if x==1:
b=b+1
a=a+(n+m-1-a-b)
print (a,b)
``` | output | 1 | 80,240 | 19 | 160,481 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya and Vasya decided to play a little. They found n red cubes and m blue cubes. The game goes like that: the players take turns to choose a cube of some color (red or blue) and put it in a line from left to right (overall the line will have n + m cubes). Petya moves first. Petya's task is to get as many pairs of neighbouring cubes of the same color as possible. Vasya's task is to get as many pairs of neighbouring cubes of different colors as possible.
The number of Petya's points in the game is the number of pairs of neighboring cubes of the same color in the line, the number of Vasya's points in the game is the number of neighbouring cubes of the different color in the line. Your task is to calculate the score at the end of the game (Petya's and Vasya's points, correspondingly), if both boys are playing optimally well. To "play optimally well" first of all means to maximize the number of one's points, and second — to minimize the number of the opponent's points.
Input
The only line contains two space-separated integers n and m (1 ≤ n, m ≤ 105) — the number of red and blue cubes, correspondingly.
Output
On a single line print two space-separated integers — the number of Petya's and Vasya's points correspondingly provided that both players play optimally well.
Examples
Input
3 1
Output
2 1
Input
2 4
Output
3 2
Note
In the first test sample the optimal strategy for Petya is to put the blue cube in the line. After that there will be only red cubes left, so by the end of the game the line of cubes from left to right will look as [blue, red, red, red]. So, Petya gets 2 points and Vasya gets 1 point.
If Petya would choose the red cube during his first move, then, provided that both boys play optimally well, Petya would get 1 point and Vasya would get 2 points.
Submitted Solution:
```
n, m = map(int, input().split())
ans, n1, m1 = [], n, m
for j in ['r', 'b']:
ch, p1, p2 = j, 0, 0
if ch == 'r':
n -= 1
else:
m -= 1
for i in range(1, n1 + m1):
if i % 2:
if ch == 'r':
if m:
ch = 'b'
m -= 1
p2 += 1
else:
n -= 1
p1 += 1
else:
if n:
ch = 'r'
n -= 1
p2 += 1
else:
m -= 1
p1 += 1
else:
if ch == 'r':
if n:
n -= 1
p1 += 1
else:
ch = 'b'
m -= 1
p2 += 1
else:
if m:
m -= 1
p1 += 1
else:
ch = 'r'
n -= 1
p2 += 1
ans.append([p1, p2])
n, m = n1, m1
ans.sort(key=lambda x: x[0], reverse=True)
print(*ans[0])
``` | instruction | 0 | 80,241 | 19 | 160,482 |
Yes | output | 1 | 80,241 | 19 | 160,483 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya and Vasya decided to play a little. They found n red cubes and m blue cubes. The game goes like that: the players take turns to choose a cube of some color (red or blue) and put it in a line from left to right (overall the line will have n + m cubes). Petya moves first. Petya's task is to get as many pairs of neighbouring cubes of the same color as possible. Vasya's task is to get as many pairs of neighbouring cubes of different colors as possible.
The number of Petya's points in the game is the number of pairs of neighboring cubes of the same color in the line, the number of Vasya's points in the game is the number of neighbouring cubes of the different color in the line. Your task is to calculate the score at the end of the game (Petya's and Vasya's points, correspondingly), if both boys are playing optimally well. To "play optimally well" first of all means to maximize the number of one's points, and second — to minimize the number of the opponent's points.
Input
The only line contains two space-separated integers n and m (1 ≤ n, m ≤ 105) — the number of red and blue cubes, correspondingly.
Output
On a single line print two space-separated integers — the number of Petya's and Vasya's points correspondingly provided that both players play optimally well.
Examples
Input
3 1
Output
2 1
Input
2 4
Output
3 2
Note
In the first test sample the optimal strategy for Petya is to put the blue cube in the line. After that there will be only red cubes left, so by the end of the game the line of cubes from left to right will look as [blue, red, red, red]. So, Petya gets 2 points and Vasya gets 1 point.
If Petya would choose the red cube during his first move, then, provided that both boys play optimally well, Petya would get 1 point and Vasya would get 2 points.
Submitted Solution:
```
def fun(r,b,col):
P = 0
R = 0
lst = [col]
chance = "R"
if(col == "red"):
d1,d2 = r-1,b
else:
d1,d2 = r,b-1
i = 0
while(d1>0 and d2>0):
if(chance == "P"):
if(lst[i] == "blue"):
lst.append("blue")
d2 -= 1
else:
lst.append("red")
d1 -= 1
P += 1
chance = "R"
if(chance == "R"):
if(lst[i] == "red"):
lst.append("blue")
d2 -= 1
else:
lst.append("red")
d1 -= 1
R += 1
chance = "P"
i += 1
if(d1 > 0):
if(lst[i] == "red"):
P += 1
else:
R += 1
P += d1 - 1
if(d2 > 0):
if(lst[i] == "blue"):
P += 1
else:
R += 1
P += d2 - 1
return str(P) + " " + str(R)
pass
r,b = [int(i) for i in input().split()]
P1,R1 = [int(i) for i in fun(r,b,"red").split()]
P2,R2 = [int(i) for i in fun(r,b,"blue").split()]
if(P1>P2):
print(P1,R1)
else:
print(P2,R2)
``` | instruction | 0 | 80,242 | 19 | 160,484 |
Yes | output | 1 | 80,242 | 19 | 160,485 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya and Vasya decided to play a little. They found n red cubes and m blue cubes. The game goes like that: the players take turns to choose a cube of some color (red or blue) and put it in a line from left to right (overall the line will have n + m cubes). Petya moves first. Petya's task is to get as many pairs of neighbouring cubes of the same color as possible. Vasya's task is to get as many pairs of neighbouring cubes of different colors as possible.
The number of Petya's points in the game is the number of pairs of neighboring cubes of the same color in the line, the number of Vasya's points in the game is the number of neighbouring cubes of the different color in the line. Your task is to calculate the score at the end of the game (Petya's and Vasya's points, correspondingly), if both boys are playing optimally well. To "play optimally well" first of all means to maximize the number of one's points, and second — to minimize the number of the opponent's points.
Input
The only line contains two space-separated integers n and m (1 ≤ n, m ≤ 105) — the number of red and blue cubes, correspondingly.
Output
On a single line print two space-separated integers — the number of Petya's and Vasya's points correspondingly provided that both players play optimally well.
Examples
Input
3 1
Output
2 1
Input
2 4
Output
3 2
Note
In the first test sample the optimal strategy for Petya is to put the blue cube in the line. After that there will be only red cubes left, so by the end of the game the line of cubes from left to right will look as [blue, red, red, red]. So, Petya gets 2 points and Vasya gets 1 point.
If Petya would choose the red cube during his first move, then, provided that both boys play optimally well, Petya would get 1 point and Vasya would get 2 points.
Submitted Solution:
```
n,m=map(int,input().split())
n,m=min(n,m),max(n,m)
score1=0
score2=0
score3=0
score4=0
l1=[]
l2=[]
x=n//2
n-=x*2
m-=x*2
score2=x*2
score1=score2-1
if n%2==0:
score1+=(m)
else :
score2+=1
score1+=m
print(score1,score2)
``` | instruction | 0 | 80,243 | 19 | 160,486 |
Yes | output | 1 | 80,243 | 19 | 160,487 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya and Vasya decided to play a little. They found n red cubes and m blue cubes. The game goes like that: the players take turns to choose a cube of some color (red or blue) and put it in a line from left to right (overall the line will have n + m cubes). Petya moves first. Petya's task is to get as many pairs of neighbouring cubes of the same color as possible. Vasya's task is to get as many pairs of neighbouring cubes of different colors as possible.
The number of Petya's points in the game is the number of pairs of neighboring cubes of the same color in the line, the number of Vasya's points in the game is the number of neighbouring cubes of the different color in the line. Your task is to calculate the score at the end of the game (Petya's and Vasya's points, correspondingly), if both boys are playing optimally well. To "play optimally well" first of all means to maximize the number of one's points, and second — to minimize the number of the opponent's points.
Input
The only line contains two space-separated integers n and m (1 ≤ n, m ≤ 105) — the number of red and blue cubes, correspondingly.
Output
On a single line print two space-separated integers — the number of Petya's and Vasya's points correspondingly provided that both players play optimally well.
Examples
Input
3 1
Output
2 1
Input
2 4
Output
3 2
Note
In the first test sample the optimal strategy for Petya is to put the blue cube in the line. After that there will be only red cubes left, so by the end of the game the line of cubes from left to right will look as [blue, red, red, red]. So, Petya gets 2 points and Vasya gets 1 point.
If Petya would choose the red cube during his first move, then, provided that both boys play optimally well, Petya would get 1 point and Vasya would get 2 points.
Submitted Solution:
```
n,m=map(int,input().split())
print(n+m-min(n,m)-1,min(n,m))
``` | instruction | 0 | 80,244 | 19 | 160,488 |
Yes | output | 1 | 80,244 | 19 | 160,489 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya and Vasya decided to play a little. They found n red cubes and m blue cubes. The game goes like that: the players take turns to choose a cube of some color (red or blue) and put it in a line from left to right (overall the line will have n + m cubes). Petya moves first. Petya's task is to get as many pairs of neighbouring cubes of the same color as possible. Vasya's task is to get as many pairs of neighbouring cubes of different colors as possible.
The number of Petya's points in the game is the number of pairs of neighboring cubes of the same color in the line, the number of Vasya's points in the game is the number of neighbouring cubes of the different color in the line. Your task is to calculate the score at the end of the game (Petya's and Vasya's points, correspondingly), if both boys are playing optimally well. To "play optimally well" first of all means to maximize the number of one's points, and second — to minimize the number of the opponent's points.
Input
The only line contains two space-separated integers n and m (1 ≤ n, m ≤ 105) — the number of red and blue cubes, correspondingly.
Output
On a single line print two space-separated integers — the number of Petya's and Vasya's points correspondingly provided that both players play optimally well.
Examples
Input
3 1
Output
2 1
Input
2 4
Output
3 2
Note
In the first test sample the optimal strategy for Petya is to put the blue cube in the line. After that there will be only red cubes left, so by the end of the game the line of cubes from left to right will look as [blue, red, red, red]. So, Petya gets 2 points and Vasya gets 1 point.
If Petya would choose the red cube during his first move, then, provided that both boys play optimally well, Petya would get 1 point and Vasya would get 2 points.
Submitted Solution:
```
a, b = map(int, input().split())
if a >= b:
print(a + b - 1 - (b | 1), (b | 1))
else:
print((a | 1), a + b - 1 - (a | 1))
``` | instruction | 0 | 80,245 | 19 | 160,490 |
No | output | 1 | 80,245 | 19 | 160,491 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya and Vasya decided to play a little. They found n red cubes and m blue cubes. The game goes like that: the players take turns to choose a cube of some color (red or blue) and put it in a line from left to right (overall the line will have n + m cubes). Petya moves first. Petya's task is to get as many pairs of neighbouring cubes of the same color as possible. Vasya's task is to get as many pairs of neighbouring cubes of different colors as possible.
The number of Petya's points in the game is the number of pairs of neighboring cubes of the same color in the line, the number of Vasya's points in the game is the number of neighbouring cubes of the different color in the line. Your task is to calculate the score at the end of the game (Petya's and Vasya's points, correspondingly), if both boys are playing optimally well. To "play optimally well" first of all means to maximize the number of one's points, and second — to minimize the number of the opponent's points.
Input
The only line contains two space-separated integers n and m (1 ≤ n, m ≤ 105) — the number of red and blue cubes, correspondingly.
Output
On a single line print two space-separated integers — the number of Petya's and Vasya's points correspondingly provided that both players play optimally well.
Examples
Input
3 1
Output
2 1
Input
2 4
Output
3 2
Note
In the first test sample the optimal strategy for Petya is to put the blue cube in the line. After that there will be only red cubes left, so by the end of the game the line of cubes from left to right will look as [blue, red, red, red]. So, Petya gets 2 points and Vasya gets 1 point.
If Petya would choose the red cube during his first move, then, provided that both boys play optimally well, Petya would get 1 point and Vasya would get 2 points.
Submitted Solution:
```
def pairs(s):
petya = 0
vasya = 0
for i in range(len(s)-1):
if s[i] == s[i+1]:
petya += 1
else:
vasya += 1
print(petya,vasya)
def main():
n,m = map(int,input().split())
s = ''
if n == min(n,m):
s = 'r'
n -= 1
else:
s = 'b'
m -= 1
while n > 0 and m > 0:
if s[-1] == 'r':
s += 'b'
m -= 1
else:
s += 'r'
n -= 1
while n > 0:
s += 'r'
n -= 1
while m > 0:
s += 'b'
m -= 1
pairs(s)
main()
``` | instruction | 0 | 80,246 | 19 | 160,492 |
No | output | 1 | 80,246 | 19 | 160,493 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya and Vasya decided to play a little. They found n red cubes and m blue cubes. The game goes like that: the players take turns to choose a cube of some color (red or blue) and put it in a line from left to right (overall the line will have n + m cubes). Petya moves first. Petya's task is to get as many pairs of neighbouring cubes of the same color as possible. Vasya's task is to get as many pairs of neighbouring cubes of different colors as possible.
The number of Petya's points in the game is the number of pairs of neighboring cubes of the same color in the line, the number of Vasya's points in the game is the number of neighbouring cubes of the different color in the line. Your task is to calculate the score at the end of the game (Petya's and Vasya's points, correspondingly), if both boys are playing optimally well. To "play optimally well" first of all means to maximize the number of one's points, and second — to minimize the number of the opponent's points.
Input
The only line contains two space-separated integers n and m (1 ≤ n, m ≤ 105) — the number of red and blue cubes, correspondingly.
Output
On a single line print two space-separated integers — the number of Petya's and Vasya's points correspondingly provided that both players play optimally well.
Examples
Input
3 1
Output
2 1
Input
2 4
Output
3 2
Note
In the first test sample the optimal strategy for Petya is to put the blue cube in the line. After that there will be only red cubes left, so by the end of the game the line of cubes from left to right will look as [blue, red, red, red]. So, Petya gets 2 points and Vasya gets 1 point.
If Petya would choose the red cube during his first move, then, provided that both boys play optimally well, Petya would get 1 point and Vasya would get 2 points.
Submitted Solution:
```
ar = []
for i in input().split(' '):
ar.append(int(i))
red = ar[0]
blue = ar[1]
res = []
while(red!=0 or blue!=0):
if red == 0:
res.append(0)
blue = blue - 1
elif blue == 0:
res.append(1)
red = red - 1
elif len(res) == 0:
res.append(1)
red = red - 1
elif res[-1] == 1:
res.append(0)
blue = blue - 1
else:
res.append(1)
red = red - 1
p = 0
v = 0
for i in range(1, len(res)):
if res[i] == res[i-1]:
p = p + 1
else:
v = v + 1
print(v, p)
``` | instruction | 0 | 80,247 | 19 | 160,494 |
No | output | 1 | 80,247 | 19 | 160,495 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya and Vasya decided to play a little. They found n red cubes and m blue cubes. The game goes like that: the players take turns to choose a cube of some color (red or blue) and put it in a line from left to right (overall the line will have n + m cubes). Petya moves first. Petya's task is to get as many pairs of neighbouring cubes of the same color as possible. Vasya's task is to get as many pairs of neighbouring cubes of different colors as possible.
The number of Petya's points in the game is the number of pairs of neighboring cubes of the same color in the line, the number of Vasya's points in the game is the number of neighbouring cubes of the different color in the line. Your task is to calculate the score at the end of the game (Petya's and Vasya's points, correspondingly), if both boys are playing optimally well. To "play optimally well" first of all means to maximize the number of one's points, and second — to minimize the number of the opponent's points.
Input
The only line contains two space-separated integers n and m (1 ≤ n, m ≤ 105) — the number of red and blue cubes, correspondingly.
Output
On a single line print two space-separated integers — the number of Petya's and Vasya's points correspondingly provided that both players play optimally well.
Examples
Input
3 1
Output
2 1
Input
2 4
Output
3 2
Note
In the first test sample the optimal strategy for Petya is to put the blue cube in the line. After that there will be only red cubes left, so by the end of the game the line of cubes from left to right will look as [blue, red, red, red]. So, Petya gets 2 points and Vasya gets 1 point.
If Petya would choose the red cube during his first move, then, provided that both boys play optimally well, Petya would get 1 point and Vasya would get 2 points.
Submitted Solution:
```
def main():
def score(l):
(a, b) = (0, 0)
for i in range(len(l) - 1):
if l[i] == l[i + 1]:
a += 1
else:
b += 1
return "{} {}".format(a, b)
# red, blue
(n, m) = map(int, input().split(' '))
l = []
i = 0
if n > m:
l.append('red')
n -= 1
else:
l.append('blue')
m -= 1
while n > 0 and m > 0:
moved = False
if i == 0:
if l[-1] == 'red' and n > 0:
l.append('red')
n -= 1
elif l[-1] == 'blue' and m > 0:
l.append('blue')
m -= 1
elif l[0] == 'red':
l = ['red'] + l
n -= 1
elif l[0] == 'blue':
l = ['blue'] + l
m -= 1
else:
assert(False) # oops
else:
if l[-1] == 'red' and m > 0:
l.append('blue')
m -= 1
elif l[-1] == 'blue' and n > 0:
l.append('red')
n -= 1
elif l[0] == 'red':
l = ['blue'] + l
m -= 1
elif l[0] == 'blue':
l = ['red'] + l
n -= 1
else:
assert(False) # oops
i ^= 1
while n > 0:
if (i == 0 and l[-1] == 'red') or (i == 1 and l[0] == 'blue'):
l.append('red')
elif i == 0 and l[0] == 'red' or (i == 1 and l[-1] == 'blue'):
l = ['red'] + l
n -= 1
i ^= 1
while m > 0:
if (i == 0 and l[-1] == 'blue') or (i == 1 and l[-1] == 'red'):
l.append('blue')
elif (i == 0 and l[0] == 'blue') or (i == 1 and l[0] == 'red'):
l = ['blue'] + l
m -= 1
i ^= 1
return score(l)
print(main())
``` | instruction | 0 | 80,248 | 19 | 160,496 |
No | output | 1 | 80,248 | 19 | 160,497 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As we know, DZY loves playing games. One day DZY decided to play with a n × m matrix. To be more precise, he decided to modify the matrix with exactly k operations.
Each modification is one of the following:
1. Pick some row of the matrix and decrease each element of the row by p. This operation brings to DZY the value of pleasure equal to the sum of elements of the row before the decreasing.
2. Pick some column of the matrix and decrease each element of the column by p. This operation brings to DZY the value of pleasure equal to the sum of elements of the column before the decreasing.
DZY wants to know: what is the largest total value of pleasure he could get after performing exactly k modifications? Please, help him to calculate this value.
Input
The first line contains four space-separated integers n, m, k and p (1 ≤ n, m ≤ 103; 1 ≤ k ≤ 106; 1 ≤ p ≤ 100).
Then n lines follow. Each of them contains m integers representing aij (1 ≤ aij ≤ 103) — the elements of the current row of the matrix.
Output
Output a single integer — the maximum possible total pleasure value DZY could get.
Examples
Input
2 2 2 2
1 3
2 4
Output
11
Input
2 2 5 2
1 3
2 4
Output
11
Note
For the first sample test, we can modify: column 2, row 2. After that the matrix becomes:
1 1
0 0
For the second sample test, we can modify: column 2, row 2, row 1, column 1, column 2. After that the matrix becomes:
-3 -3
-2 -2
Submitted Solution:
```
import sys
n, m, k, p = map(int, input().split())
d = [0, 0]
a = [[0]*(n+1), [0]*(m+1)]
f = [0]*(k+1)
g = [0]*(k+1)
for i in range(n):
r = list(map(int, input().split()))
for j in range(len(r)):
a[0][i+1] += r[j]
a[1][j+1] += r[j]
def popup(l, x):
while x>1 and l[x]>l[x>>1]:
l[x], l[x>>1] = l[x>>1], l[x]
x>>=1
for i in range(n,0,-1):
popup(a[0], i)
for i in range(m,0,-1):
popup(a[1], i)
def bigger(b, x, y, n):
if y>=n: return True
if b[x]>=b[y]: return True
return False
def get(b, n, m):
res = b[1]
b[1]=b[n]
x = 1
while (x*2<n):
if bigger(b,x,x*2,n) and bigger(b,x,x*2+1,n): break
if bigger(b, x*2, x*2+1, n):
b[x*2], b[x] = b[x], b[x*2]
x*=2
else:
b[x*2+1], b[x] = b[x], b[x*2+1]
x=x*2+1
b[n]=res-m*p
popup(b, n)
return res
ans = 0
for fuck in range(1, k+1):
f[fuck] = f[fuck-1]+get(a[0],n,m)
g[fuck] = g[fuck-1]+get(a[1],m,n)
for fuck in range(k+1):
ans=max(ans, f[fuck]-fuck*(k-fuck)*p+g[k-fuck])
print(ans)
``` | instruction | 0 | 80,350 | 19 | 160,700 |
No | output | 1 | 80,350 | 19 | 160,701 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As we know, DZY loves playing games. One day DZY decided to play with a n × m matrix. To be more precise, he decided to modify the matrix with exactly k operations.
Each modification is one of the following:
1. Pick some row of the matrix and decrease each element of the row by p. This operation brings to DZY the value of pleasure equal to the sum of elements of the row before the decreasing.
2. Pick some column of the matrix and decrease each element of the column by p. This operation brings to DZY the value of pleasure equal to the sum of elements of the column before the decreasing.
DZY wants to know: what is the largest total value of pleasure he could get after performing exactly k modifications? Please, help him to calculate this value.
Input
The first line contains four space-separated integers n, m, k and p (1 ≤ n, m ≤ 103; 1 ≤ k ≤ 106; 1 ≤ p ≤ 100).
Then n lines follow. Each of them contains m integers representing aij (1 ≤ aij ≤ 103) — the elements of the current row of the matrix.
Output
Output a single integer — the maximum possible total pleasure value DZY could get.
Examples
Input
2 2 2 2
1 3
2 4
Output
11
Input
2 2 5 2
1 3
2 4
Output
11
Note
For the first sample test, we can modify: column 2, row 2. After that the matrix becomes:
1 1
0 0
For the second sample test, we can modify: column 2, row 2, row 1, column 1, column 2. After that the matrix becomes:
-3 -3
-2 -2
Submitted Solution:
```
import sys
n, m, k, p = map(int, input().split())
d = [0, 0]
a = [[0]*(n+1), [0]*(m+1)]
for i in range(n):
r = list(map(int, input().split()))
for j in range(len(r)):
a[0][i+1] += r[j]
a[1][j+1] += r[j]
def popup(l, x):
while x>1 and l[x]>l[x>>1]:
l[x], l[x>>1] = l[x>>1], l[x]
x>>=1
for i in range(n,0,-1):
popup(a[0], i)
for i in range(m,0,-1):
popup(a[1], i)
def smaller(b, x, y, n):
if y>=n: return False
if b[x]<b[y]: return True
return False
def get(b, n, m):
res = b[1]
b[1]=b[n]
x = 1
while (x*2<n):
if not smaller(b,x,x*2,n) and not smaller(b,x,x*2+1,n): break
if x*2+1 >= n or b[x*2]<b[x*2+1]:
b[x*2], b[x] = b[x], b[x*2]
x*=2
else:
b[x*2+1], b[x] = b[x], b[x*2+1]
x=x*2+1
b[n]=res-m*p
popup(b, n)
return res
ans = 0
for fuck in range(k):
if a[0][1]-d[0]>a[1][1]-d[1]:
ans+=get(a[0], n, m)-d[0]
d[1]+=p
else:
ans+=get(a[1], m, n)-d[1]
d[0]+=p
print(ans)
``` | instruction | 0 | 80,352 | 19 | 160,704 |
No | output | 1 | 80,352 | 19 | 160,705 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As we know, DZY loves playing games. One day DZY decided to play with a n × m matrix. To be more precise, he decided to modify the matrix with exactly k operations.
Each modification is one of the following:
1. Pick some row of the matrix and decrease each element of the row by p. This operation brings to DZY the value of pleasure equal to the sum of elements of the row before the decreasing.
2. Pick some column of the matrix and decrease each element of the column by p. This operation brings to DZY the value of pleasure equal to the sum of elements of the column before the decreasing.
DZY wants to know: what is the largest total value of pleasure he could get after performing exactly k modifications? Please, help him to calculate this value.
Input
The first line contains four space-separated integers n, m, k and p (1 ≤ n, m ≤ 103; 1 ≤ k ≤ 106; 1 ≤ p ≤ 100).
Then n lines follow. Each of them contains m integers representing aij (1 ≤ aij ≤ 103) — the elements of the current row of the matrix.
Output
Output a single integer — the maximum possible total pleasure value DZY could get.
Examples
Input
2 2 2 2
1 3
2 4
Output
11
Input
2 2 5 2
1 3
2 4
Output
11
Note
For the first sample test, we can modify: column 2, row 2. After that the matrix becomes:
1 1
0 0
For the second sample test, we can modify: column 2, row 2, row 1, column 1, column 2. After that the matrix becomes:
-3 -3
-2 -2
Submitted Solution:
```
# ---------------------------iye ha aam zindegi---------------------------------------------
import math
import heapq, bisect
import sys
from collections import deque, defaultdict
from fractions import Fraction
import sys
mod = 10 ** 9 + 7
mod1 = 998244353
sys.setrecursionlimit(300000)
# ------------------------------warmup----------------------------
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
sys.setrecursionlimit(300000)
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")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# -------------------game starts now----------------------------------------------------import math
class TreeNode:
def __init__(self, k, v):
self.key = k
self.value = v
self.left = None
self.right = None
self.parent = None
self.height = 1
self.num_left = 1
self.num_total = 1
class AvlTree:
def __init__(self):
self._tree = None
def add(self, k, v):
if not self._tree:
self._tree = TreeNode(k, v)
return
node = self._add(k, v)
if node:
self._rebalance(node)
def _add(self, k, v):
node = self._tree
while node:
if k < node.key:
if node.left:
node = node.left
else:
node.left = TreeNode(k, v)
node.left.parent = node
return node.left
elif node.key < k:
if node.right:
node = node.right
else:
node.right = TreeNode(k, v)
node.right.parent = node
return node.right
else:
node.value = v
return
@staticmethod
def get_height(x):
return x.height if x else 0
@staticmethod
def get_num_total(x):
return x.num_total if x else 0
def _rebalance(self, node):
n = node
while n:
lh = self.get_height(n.left)
rh = self.get_height(n.right)
n.height = max(lh, rh) + 1
balance_factor = lh - rh
n.num_total = 1 + self.get_num_total(n.left) + self.get_num_total(n.right)
n.num_left = 1 + self.get_num_total(n.left)
if balance_factor > 1:
if self.get_height(n.left.left) < self.get_height(n.left.right):
self._rotate_left(n.left)
self._rotate_right(n)
elif balance_factor < -1:
if self.get_height(n.right.right) < self.get_height(n.right.left):
self._rotate_right(n.right)
self._rotate_left(n)
else:
n = n.parent
def _remove_one(self, node):
"""
Side effect!!! Changes node. Node should have exactly one child
"""
replacement = node.left or node.right
if node.parent:
if AvlTree._is_left(node):
node.parent.left = replacement
else:
node.parent.right = replacement
replacement.parent = node.parent
node.parent = None
else:
self._tree = replacement
replacement.parent = None
node.left = None
node.right = None
node.parent = None
self._rebalance(replacement)
def _remove_leaf(self, node):
if node.parent:
if AvlTree._is_left(node):
node.parent.left = None
else:
node.parent.right = None
self._rebalance(node.parent)
else:
self._tree = None
node.parent = None
node.left = None
node.right = None
def remove(self, k):
node = self._get_node(k)
if not node:
return
if AvlTree._is_leaf(node):
self._remove_leaf(node)
return
if node.left and node.right:
nxt = AvlTree._get_next(node)
node.key = nxt.key
node.value = nxt.value
if self._is_leaf(nxt):
self._remove_leaf(nxt)
else:
self._remove_one(nxt)
self._rebalance(node)
else:
self._remove_one(node)
def get(self, k):
node = self._get_node(k)
return node.value if node else -1
def _get_node(self, k):
if not self._tree:
return None
node = self._tree
while node:
if k < node.key:
node = node.left
elif node.key < k:
node = node.right
else:
return node
return None
def get_at(self, pos):
x = pos + 1
node = self._tree
while node:
if x < node.num_left:
node = node.left
elif node.num_left < x:
x -= node.num_left
node = node.right
else:
return (node.key, node.value)
raise IndexError("Out of ranges")
@staticmethod
def _is_left(node):
return node.parent.left and node.parent.left == node
@staticmethod
def _is_leaf(node):
return node.left is None and node.right is None
def _rotate_right(self, node):
if not node.parent:
self._tree = node.left
node.left.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.left
node.left.parent = node.parent
else:
node.parent.right = node.left
node.left.parent = node.parent
bk = node.left.right
node.left.right = node
node.parent = node.left
node.left = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
def _rotate_left(self, node):
if not node.parent:
self._tree = node.right
node.right.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.right
node.right.parent = node.parent
else:
node.parent.right = node.right
node.right.parent = node.parent
bk = node.right.left
node.right.left = node
node.parent = node.right
node.right = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
@staticmethod
def _get_next(node):
if not node.right:
return node.parent
n = node.right
while n.left:
n = n.left
return n
# -----------------------------------------------binary seacrh tree---------------------------------------
class SegmentTree1:
def __init__(self, data, default=0, func=lambda a, b: max(a, b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------game starts now----------------------------------------------------import math
class SegmentTree:
def __init__(self, data, default=0, func=lambda a, b: a + b):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------------------iye ha chutiya zindegi-------------------------------------
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
# --------------------------------------iye ha combinations ka zindegi---------------------------------
def powm(a, n, m):
if a == 1 or n == 0:
return 1
if n % 2 == 0:
s = powm(a, n // 2, m)
return s * s % m
else:
return a * powm(a, n - 1, m) % m
# --------------------------------------iye ha power ka zindegi---------------------------------
def sort_list(list1, list2):
zipped_pairs = zip(list2, list1)
z = [x for _, x in sorted(zipped_pairs)]
return z
# --------------------------------------------------product----------------------------------------
def product(l):
por = 1
for i in range(len(l)):
por *= l[i]
return por
# --------------------------------------------------binary----------------------------------------
def binarySearchCount(arr, n, key):
left = 0
right = n - 1
count = 0
while (left <= right):
mid = int((right + left) / 2)
# Check if middle element is
# less than or equal to key
if (arr[mid] < key):
count = mid + 1
left = mid + 1
# If key is smaller, ignore right half
else:
right = mid - 1
return count
# --------------------------------------------------binary----------------------------------------
def countdig(n):
c = 0
while (n > 0):
n //= 10
c += 1
return c
def binary(x, length):
y = bin(x)[2:]
return y if len(y) >= length else "0" * (length - len(y)) + y
def countGreater(arr, n, k):
l = 0
r = n - 1
# Stores the index of the left most element
# from the array which is greater than k
leftGreater = n
# Finds number of elements greater than k
while (l <= r):
m = int(l + (r - l) / 2)
if (arr[m] >= k):
leftGreater = m
r = m - 1
# If mid element is less than
# or equal to k update l
else:
l = m + 1
# Return the count of elements
# greater than k
return (n - leftGreater)
# --------------------------------------------------binary------------------------------------
n,m,k,p=map(int,input().split())
l=[]
ros=[]
cos=[]
for i in range(n):
l.append(list(map(int,input().split())))
for i in range(n):
row=0
for j in range(m):
row+=l[i][j]
ros.append(-row)
for i in range(m):
col=0
for j in range(n):
col+=l[j][i]
cos.append(-col)
heapq.heapify(cos)
heapq.heapify(ros)
ans=0
rs=0
cs=0
while(k>0):
k-=1
e=-heapq.heappop(cos)-cs
e1=-heapq.heappop(ros)-rs
ans+=max(e,e1)
if e1>=e:
heapq.heappush(cos,-e-cs)
cs+=p
heapq.heappush(ros,-e1+m*p-rs)
else:
heapq.heappush(ros, -e1 - rs)
rs += p
heapq.heappush(cos, -e +n*p-cs)
print(ans)
``` | instruction | 0 | 80,353 | 19 | 160,706 |
No | output | 1 | 80,353 | 19 | 160,707 |
Provide a correct Python 3 solution for this coding contest problem.
Sugoroku
problem
JOI is playing sugoroku alone. There are N squares in a straight line in this sugoroku, and each has a movement instruction written on it. The starting point is the 1st square and the goal is the Nth square. JOI repeats the following until he reaches the goal.
Roll the dice and proceed from the current square by the number of rolls, and follow the instructions of that square. Do not follow the instructions of the square to which you moved according to the instructions.
The goal is not only when you stop at the Nth square, but also when the destination exceeds the Nth square.
Create a program that outputs how many times you roll the dice to reach the goal when you are given a sugoroku board and M dice rolls.
input
The input consists of multiple datasets. Each dataset is given in the following format.
Each dataset consists of 1 + N + M rows.
Two integers N, M (2 ≤ N ≤ 1000, 1 ≤ M ≤ 1000) are written on the first line of the input, separated by blanks. N represents the number of sugoroku squares, and M represents the number of dice given.
In the following N lines, integers between -999 and 999 are written one by one. The integer on the 1 + i line (1 ≤ i ≤ N) represents the indication of the sugoroku i-th cell. Let X be the written integer. When X = 0, it indicates "do nothing", when X> 0, it indicates "advance X mass", and when X <0, it indicates "| X | mass return". However, | X | represents the absolute value of X.
In the following M line, integers from 1 to 6 are written one by one, and the number on the 1 + N + j line (1 ≤ j ≤ M) represents the dice roll that appears on the jth time.
However, the number of lines 2 and 1 + N is always 0. There is no cell with instructions to move to the cell before the first cell. In addition, the number of times the dice are rolled is M or less in any scoring input data.
When both N and M are 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each data set, an integer indicating how many times the dice are rolled to reach the goal is output on one line.
Input / output example
Input example
10 5
0
0
Five
6
-3
8
1
8
-Four
0
1
3
Five
1
Five
10 10
0
-1
-1
Four
Four
-Five
0
1
-6
0
1
Five
2
Four
6
Five
Five
Four
1
6
0 0
Output example
Five
6
The following figure shows the first input example.
<image>
The following figure shows the second input example.
<image>
The above question sentences and the data used for the automatic referee are the question sentences created and published by the Japan Committee for Information Olympics and the test data for scoring.
Example
Input
10 5
0
0
5
6
-3
8
1
8
-4
0
1
3
5
1
5
10 10
0
-1
-1
4
4
-5
0
1
-6
0
1
5
2
4
6
5
5
4
1
6
0 0
Output
5
6 | instruction | 0 | 80,743 | 19 | 161,486 |
"Correct Solution:
```
for e in iter(input,'0 0'):
N,M=map(int,e.split())
k,p=1,0
S=[int(input()) for _ in[0]*N]
for d in[int(input())for _ in[0]*M]:
p+=d if N<=p+d else d+S[p+d]
if N<=p+1:break
k+=1
print(k)
``` | output | 1 | 80,743 | 19 | 161,487 |
Provide a correct Python 3 solution for this coding contest problem.
Sugoroku
problem
JOI is playing sugoroku alone. There are N squares in a straight line in this sugoroku, and each has a movement instruction written on it. The starting point is the 1st square and the goal is the Nth square. JOI repeats the following until he reaches the goal.
Roll the dice and proceed from the current square by the number of rolls, and follow the instructions of that square. Do not follow the instructions of the square to which you moved according to the instructions.
The goal is not only when you stop at the Nth square, but also when the destination exceeds the Nth square.
Create a program that outputs how many times you roll the dice to reach the goal when you are given a sugoroku board and M dice rolls.
input
The input consists of multiple datasets. Each dataset is given in the following format.
Each dataset consists of 1 + N + M rows.
Two integers N, M (2 ≤ N ≤ 1000, 1 ≤ M ≤ 1000) are written on the first line of the input, separated by blanks. N represents the number of sugoroku squares, and M represents the number of dice given.
In the following N lines, integers between -999 and 999 are written one by one. The integer on the 1 + i line (1 ≤ i ≤ N) represents the indication of the sugoroku i-th cell. Let X be the written integer. When X = 0, it indicates "do nothing", when X> 0, it indicates "advance X mass", and when X <0, it indicates "| X | mass return". However, | X | represents the absolute value of X.
In the following M line, integers from 1 to 6 are written one by one, and the number on the 1 + N + j line (1 ≤ j ≤ M) represents the dice roll that appears on the jth time.
However, the number of lines 2 and 1 + N is always 0. There is no cell with instructions to move to the cell before the first cell. In addition, the number of times the dice are rolled is M or less in any scoring input data.
When both N and M are 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each data set, an integer indicating how many times the dice are rolled to reach the goal is output on one line.
Input / output example
Input example
10 5
0
0
Five
6
-3
8
1
8
-Four
0
1
3
Five
1
Five
10 10
0
-1
-1
Four
Four
-Five
0
1
-6
0
1
Five
2
Four
6
Five
Five
Four
1
6
0 0
Output example
Five
6
The following figure shows the first input example.
<image>
The following figure shows the second input example.
<image>
The above question sentences and the data used for the automatic referee are the question sentences created and published by the Japan Committee for Information Olympics and the test data for scoring.
Example
Input
10 5
0
0
5
6
-3
8
1
8
-4
0
1
3
5
1
5
10 10
0
-1
-1
4
4
-5
0
1
-6
0
1
5
2
4
6
5
5
4
1
6
0 0
Output
5
6 | instruction | 0 | 80,744 | 19 | 161,488 |
"Correct Solution:
```
# -*- coding: utf-8 -*-
"""
http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=0544
"""
import sys
from sys import stdin
input = stdin.readline
def move(pos, result):
pos = pos + result
try:
if squares[pos] != 0:
pos += squares[pos]
return pos
except:
return float('inf')
squares = []
def main(args):
global squares
while True:
N, M = map(int, input().split())
if N == 0 and M == 0:
break
squares = [int(input()) for _ in range(N)]
dice = [int(input()) for _ in range(M)]
turn = 0
pos = 0
for d in dice:
turn += 1
pos = move(pos, d)
if pos >= (N - 1):
break
print(turn)
if __name__ == '__main__':
main(sys.argv[1:])
``` | output | 1 | 80,744 | 19 | 161,489 |
Provide a correct Python 3 solution for this coding contest problem.
Sugoroku
problem
JOI is playing sugoroku alone. There are N squares in a straight line in this sugoroku, and each has a movement instruction written on it. The starting point is the 1st square and the goal is the Nth square. JOI repeats the following until he reaches the goal.
Roll the dice and proceed from the current square by the number of rolls, and follow the instructions of that square. Do not follow the instructions of the square to which you moved according to the instructions.
The goal is not only when you stop at the Nth square, but also when the destination exceeds the Nth square.
Create a program that outputs how many times you roll the dice to reach the goal when you are given a sugoroku board and M dice rolls.
input
The input consists of multiple datasets. Each dataset is given in the following format.
Each dataset consists of 1 + N + M rows.
Two integers N, M (2 ≤ N ≤ 1000, 1 ≤ M ≤ 1000) are written on the first line of the input, separated by blanks. N represents the number of sugoroku squares, and M represents the number of dice given.
In the following N lines, integers between -999 and 999 are written one by one. The integer on the 1 + i line (1 ≤ i ≤ N) represents the indication of the sugoroku i-th cell. Let X be the written integer. When X = 0, it indicates "do nothing", when X> 0, it indicates "advance X mass", and when X <0, it indicates "| X | mass return". However, | X | represents the absolute value of X.
In the following M line, integers from 1 to 6 are written one by one, and the number on the 1 + N + j line (1 ≤ j ≤ M) represents the dice roll that appears on the jth time.
However, the number of lines 2 and 1 + N is always 0. There is no cell with instructions to move to the cell before the first cell. In addition, the number of times the dice are rolled is M or less in any scoring input data.
When both N and M are 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each data set, an integer indicating how many times the dice are rolled to reach the goal is output on one line.
Input / output example
Input example
10 5
0
0
Five
6
-3
8
1
8
-Four
0
1
3
Five
1
Five
10 10
0
-1
-1
Four
Four
-Five
0
1
-6
0
1
Five
2
Four
6
Five
Five
Four
1
6
0 0
Output example
Five
6
The following figure shows the first input example.
<image>
The following figure shows the second input example.
<image>
The above question sentences and the data used for the automatic referee are the question sentences created and published by the Japan Committee for Information Olympics and the test data for scoring.
Example
Input
10 5
0
0
5
6
-3
8
1
8
-4
0
1
3
5
1
5
10 10
0
-1
-1
4
4
-5
0
1
-6
0
1
5
2
4
6
5
5
4
1
6
0 0
Output
5
6 | instruction | 0 | 80,745 | 19 | 161,490 |
"Correct Solution:
```
while 1:
n, m = map(int, input().split())
if n == 0:
break
masu = []
for _ in range(n):
x = int(input())
masu.append(x)
dice = []
for _ in range(m):
x = int(input())
dice.append(x)
p = 0
for i, d in enumerate(dice):
p += d
if p >= n:
break
p += masu[p]
if p >= n:
break
print(i+1)
``` | output | 1 | 80,745 | 19 | 161,491 |
Provide a correct Python 3 solution for this coding contest problem.
Sugoroku
problem
JOI is playing sugoroku alone. There are N squares in a straight line in this sugoroku, and each has a movement instruction written on it. The starting point is the 1st square and the goal is the Nth square. JOI repeats the following until he reaches the goal.
Roll the dice and proceed from the current square by the number of rolls, and follow the instructions of that square. Do not follow the instructions of the square to which you moved according to the instructions.
The goal is not only when you stop at the Nth square, but also when the destination exceeds the Nth square.
Create a program that outputs how many times you roll the dice to reach the goal when you are given a sugoroku board and M dice rolls.
input
The input consists of multiple datasets. Each dataset is given in the following format.
Each dataset consists of 1 + N + M rows.
Two integers N, M (2 ≤ N ≤ 1000, 1 ≤ M ≤ 1000) are written on the first line of the input, separated by blanks. N represents the number of sugoroku squares, and M represents the number of dice given.
In the following N lines, integers between -999 and 999 are written one by one. The integer on the 1 + i line (1 ≤ i ≤ N) represents the indication of the sugoroku i-th cell. Let X be the written integer. When X = 0, it indicates "do nothing", when X> 0, it indicates "advance X mass", and when X <0, it indicates "| X | mass return". However, | X | represents the absolute value of X.
In the following M line, integers from 1 to 6 are written one by one, and the number on the 1 + N + j line (1 ≤ j ≤ M) represents the dice roll that appears on the jth time.
However, the number of lines 2 and 1 + N is always 0. There is no cell with instructions to move to the cell before the first cell. In addition, the number of times the dice are rolled is M or less in any scoring input data.
When both N and M are 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each data set, an integer indicating how many times the dice are rolled to reach the goal is output on one line.
Input / output example
Input example
10 5
0
0
Five
6
-3
8
1
8
-Four
0
1
3
Five
1
Five
10 10
0
-1
-1
Four
Four
-Five
0
1
-6
0
1
Five
2
Four
6
Five
Five
Four
1
6
0 0
Output example
Five
6
The following figure shows the first input example.
<image>
The following figure shows the second input example.
<image>
The above question sentences and the data used for the automatic referee are the question sentences created and published by the Japan Committee for Information Olympics and the test data for scoring.
Example
Input
10 5
0
0
5
6
-3
8
1
8
-4
0
1
3
5
1
5
10 10
0
-1
-1
4
4
-5
0
1
-6
0
1
5
2
4
6
5
5
4
1
6
0 0
Output
5
6 | instruction | 0 | 80,746 | 19 | 161,492 |
"Correct Solution:
```
while True:
n, m = map(int, input().split())
if not n:
break
stops = [int(input()) for _ in range(n)] + [0] * 5
dices = [int(input()) for _ in range(m)]
i, p = 1, 0
g = n - 1
for d in dices:
p += d + stops[p + d]
if p >= g:
break
i += 1
print(i)
``` | output | 1 | 80,746 | 19 | 161,493 |
Provide a correct Python 3 solution for this coding contest problem.
Sugoroku
problem
JOI is playing sugoroku alone. There are N squares in a straight line in this sugoroku, and each has a movement instruction written on it. The starting point is the 1st square and the goal is the Nth square. JOI repeats the following until he reaches the goal.
Roll the dice and proceed from the current square by the number of rolls, and follow the instructions of that square. Do not follow the instructions of the square to which you moved according to the instructions.
The goal is not only when you stop at the Nth square, but also when the destination exceeds the Nth square.
Create a program that outputs how many times you roll the dice to reach the goal when you are given a sugoroku board and M dice rolls.
input
The input consists of multiple datasets. Each dataset is given in the following format.
Each dataset consists of 1 + N + M rows.
Two integers N, M (2 ≤ N ≤ 1000, 1 ≤ M ≤ 1000) are written on the first line of the input, separated by blanks. N represents the number of sugoroku squares, and M represents the number of dice given.
In the following N lines, integers between -999 and 999 are written one by one. The integer on the 1 + i line (1 ≤ i ≤ N) represents the indication of the sugoroku i-th cell. Let X be the written integer. When X = 0, it indicates "do nothing", when X> 0, it indicates "advance X mass", and when X <0, it indicates "| X | mass return". However, | X | represents the absolute value of X.
In the following M line, integers from 1 to 6 are written one by one, and the number on the 1 + N + j line (1 ≤ j ≤ M) represents the dice roll that appears on the jth time.
However, the number of lines 2 and 1 + N is always 0. There is no cell with instructions to move to the cell before the first cell. In addition, the number of times the dice are rolled is M or less in any scoring input data.
When both N and M are 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each data set, an integer indicating how many times the dice are rolled to reach the goal is output on one line.
Input / output example
Input example
10 5
0
0
Five
6
-3
8
1
8
-Four
0
1
3
Five
1
Five
10 10
0
-1
-1
Four
Four
-Five
0
1
-6
0
1
Five
2
Four
6
Five
Five
Four
1
6
0 0
Output example
Five
6
The following figure shows the first input example.
<image>
The following figure shows the second input example.
<image>
The above question sentences and the data used for the automatic referee are the question sentences created and published by the Japan Committee for Information Olympics and the test data for scoring.
Example
Input
10 5
0
0
5
6
-3
8
1
8
-4
0
1
3
5
1
5
10 10
0
-1
-1
4
4
-5
0
1
-6
0
1
5
2
4
6
5
5
4
1
6
0 0
Output
5
6 | instruction | 0 | 80,747 | 19 | 161,494 |
"Correct Solution:
```
for e in iter(input,'0 0'):
N,M=map(int,e.split())
S=[int(input())for _ in[0]*N]
p=b=1
for i in range(M):
d=int(input())
p+=d
if N<=p:
if b:print(-~i);b=0
continue
p+=S[~-p]
if(N<=p)*b:print(-~i);b=0
``` | output | 1 | 80,747 | 19 | 161,495 |
Provide a correct Python 3 solution for this coding contest problem.
Sugoroku
problem
JOI is playing sugoroku alone. There are N squares in a straight line in this sugoroku, and each has a movement instruction written on it. The starting point is the 1st square and the goal is the Nth square. JOI repeats the following until he reaches the goal.
Roll the dice and proceed from the current square by the number of rolls, and follow the instructions of that square. Do not follow the instructions of the square to which you moved according to the instructions.
The goal is not only when you stop at the Nth square, but also when the destination exceeds the Nth square.
Create a program that outputs how many times you roll the dice to reach the goal when you are given a sugoroku board and M dice rolls.
input
The input consists of multiple datasets. Each dataset is given in the following format.
Each dataset consists of 1 + N + M rows.
Two integers N, M (2 ≤ N ≤ 1000, 1 ≤ M ≤ 1000) are written on the first line of the input, separated by blanks. N represents the number of sugoroku squares, and M represents the number of dice given.
In the following N lines, integers between -999 and 999 are written one by one. The integer on the 1 + i line (1 ≤ i ≤ N) represents the indication of the sugoroku i-th cell. Let X be the written integer. When X = 0, it indicates "do nothing", when X> 0, it indicates "advance X mass", and when X <0, it indicates "| X | mass return". However, | X | represents the absolute value of X.
In the following M line, integers from 1 to 6 are written one by one, and the number on the 1 + N + j line (1 ≤ j ≤ M) represents the dice roll that appears on the jth time.
However, the number of lines 2 and 1 + N is always 0. There is no cell with instructions to move to the cell before the first cell. In addition, the number of times the dice are rolled is M or less in any scoring input data.
When both N and M are 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each data set, an integer indicating how many times the dice are rolled to reach the goal is output on one line.
Input / output example
Input example
10 5
0
0
Five
6
-3
8
1
8
-Four
0
1
3
Five
1
Five
10 10
0
-1
-1
Four
Four
-Five
0
1
-6
0
1
Five
2
Four
6
Five
Five
Four
1
6
0 0
Output example
Five
6
The following figure shows the first input example.
<image>
The following figure shows the second input example.
<image>
The above question sentences and the data used for the automatic referee are the question sentences created and published by the Japan Committee for Information Olympics and the test data for scoring.
Example
Input
10 5
0
0
5
6
-3
8
1
8
-4
0
1
3
5
1
5
10 10
0
-1
-1
4
4
-5
0
1
-6
0
1
5
2
4
6
5
5
4
1
6
0 0
Output
5
6 | instruction | 0 | 80,748 | 19 | 161,496 |
"Correct Solution:
```
def solve():
while True:
n, m = map(int, input().split())
if not n:
break
smap = [int(input()) for _ in range(n)]
ind = 0
cnt = 0
flag = True
for _ in range(m):
ind += int(input())
if flag:
cnt += 1
if ind >= n - 1:
print(cnt)
flag = False
else:
ind += smap[ind]
if ind >= n:
print(cnt)
flag = False
solve()
``` | output | 1 | 80,748 | 19 | 161,497 |
Provide a correct Python 3 solution for this coding contest problem.
Sugoroku
problem
JOI is playing sugoroku alone. There are N squares in a straight line in this sugoroku, and each has a movement instruction written on it. The starting point is the 1st square and the goal is the Nth square. JOI repeats the following until he reaches the goal.
Roll the dice and proceed from the current square by the number of rolls, and follow the instructions of that square. Do not follow the instructions of the square to which you moved according to the instructions.
The goal is not only when you stop at the Nth square, but also when the destination exceeds the Nth square.
Create a program that outputs how many times you roll the dice to reach the goal when you are given a sugoroku board and M dice rolls.
input
The input consists of multiple datasets. Each dataset is given in the following format.
Each dataset consists of 1 + N + M rows.
Two integers N, M (2 ≤ N ≤ 1000, 1 ≤ M ≤ 1000) are written on the first line of the input, separated by blanks. N represents the number of sugoroku squares, and M represents the number of dice given.
In the following N lines, integers between -999 and 999 are written one by one. The integer on the 1 + i line (1 ≤ i ≤ N) represents the indication of the sugoroku i-th cell. Let X be the written integer. When X = 0, it indicates "do nothing", when X> 0, it indicates "advance X mass", and when X <0, it indicates "| X | mass return". However, | X | represents the absolute value of X.
In the following M line, integers from 1 to 6 are written one by one, and the number on the 1 + N + j line (1 ≤ j ≤ M) represents the dice roll that appears on the jth time.
However, the number of lines 2 and 1 + N is always 0. There is no cell with instructions to move to the cell before the first cell. In addition, the number of times the dice are rolled is M or less in any scoring input data.
When both N and M are 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each data set, an integer indicating how many times the dice are rolled to reach the goal is output on one line.
Input / output example
Input example
10 5
0
0
Five
6
-3
8
1
8
-Four
0
1
3
Five
1
Five
10 10
0
-1
-1
Four
Four
-Five
0
1
-6
0
1
Five
2
Four
6
Five
Five
Four
1
6
0 0
Output example
Five
6
The following figure shows the first input example.
<image>
The following figure shows the second input example.
<image>
The above question sentences and the data used for the automatic referee are the question sentences created and published by the Japan Committee for Information Olympics and the test data for scoring.
Example
Input
10 5
0
0
5
6
-3
8
1
8
-4
0
1
3
5
1
5
10 10
0
-1
-1
4
4
-5
0
1
-6
0
1
5
2
4
6
5
5
4
1
6
0 0
Output
5
6 | instruction | 0 | 80,749 | 19 | 161,498 |
"Correct Solution:
```
while True:
N, M = map(int, input().split())
if N==0 and M==0: break
n = [int(input()) for _ in range(N)]
m = [int(input()) for _ in range(M)]
j = 0 #現在のマス目
for i in range(M):
j += m[i]
if j>=N: break
j += n[j]
if j>=N: break
print(i+1)
``` | output | 1 | 80,749 | 19 | 161,499 |
Provide a correct Python 3 solution for this coding contest problem.
Sugoroku
problem
JOI is playing sugoroku alone. There are N squares in a straight line in this sugoroku, and each has a movement instruction written on it. The starting point is the 1st square and the goal is the Nth square. JOI repeats the following until he reaches the goal.
Roll the dice and proceed from the current square by the number of rolls, and follow the instructions of that square. Do not follow the instructions of the square to which you moved according to the instructions.
The goal is not only when you stop at the Nth square, but also when the destination exceeds the Nth square.
Create a program that outputs how many times you roll the dice to reach the goal when you are given a sugoroku board and M dice rolls.
input
The input consists of multiple datasets. Each dataset is given in the following format.
Each dataset consists of 1 + N + M rows.
Two integers N, M (2 ≤ N ≤ 1000, 1 ≤ M ≤ 1000) are written on the first line of the input, separated by blanks. N represents the number of sugoroku squares, and M represents the number of dice given.
In the following N lines, integers between -999 and 999 are written one by one. The integer on the 1 + i line (1 ≤ i ≤ N) represents the indication of the sugoroku i-th cell. Let X be the written integer. When X = 0, it indicates "do nothing", when X> 0, it indicates "advance X mass", and when X <0, it indicates "| X | mass return". However, | X | represents the absolute value of X.
In the following M line, integers from 1 to 6 are written one by one, and the number on the 1 + N + j line (1 ≤ j ≤ M) represents the dice roll that appears on the jth time.
However, the number of lines 2 and 1 + N is always 0. There is no cell with instructions to move to the cell before the first cell. In addition, the number of times the dice are rolled is M or less in any scoring input data.
When both N and M are 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each data set, an integer indicating how many times the dice are rolled to reach the goal is output on one line.
Input / output example
Input example
10 5
0
0
Five
6
-3
8
1
8
-Four
0
1
3
Five
1
Five
10 10
0
-1
-1
Four
Four
-Five
0
1
-6
0
1
Five
2
Four
6
Five
Five
Four
1
6
0 0
Output example
Five
6
The following figure shows the first input example.
<image>
The following figure shows the second input example.
<image>
The above question sentences and the data used for the automatic referee are the question sentences created and published by the Japan Committee for Information Olympics and the test data for scoring.
Example
Input
10 5
0
0
5
6
-3
8
1
8
-4
0
1
3
5
1
5
10 10
0
-1
-1
4
4
-5
0
1
-6
0
1
5
2
4
6
5
5
4
1
6
0 0
Output
5
6 | instruction | 0 | 80,750 | 19 | 161,500 |
"Correct Solution:
```
def main():
while True:
N, M = map(int, input().split())
if not N:
return
order = [int(input()) for _ in range(N)]
draw = [int(input()) for _ in range(M)]
p = 0
ans = 0
for i in range(M):
p += draw[i]
if p >= N - 1:
ans = i + 1
break
p += order[p]
if p >= N - 1:
ans = i + 1
break
print(ans)
if __name__ == '__main__':
main()
``` | output | 1 | 80,750 | 19 | 161,501 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sugoroku
problem
JOI is playing sugoroku alone. There are N squares in a straight line in this sugoroku, and each has a movement instruction written on it. The starting point is the 1st square and the goal is the Nth square. JOI repeats the following until he reaches the goal.
Roll the dice and proceed from the current square by the number of rolls, and follow the instructions of that square. Do not follow the instructions of the square to which you moved according to the instructions.
The goal is not only when you stop at the Nth square, but also when the destination exceeds the Nth square.
Create a program that outputs how many times you roll the dice to reach the goal when you are given a sugoroku board and M dice rolls.
input
The input consists of multiple datasets. Each dataset is given in the following format.
Each dataset consists of 1 + N + M rows.
Two integers N, M (2 ≤ N ≤ 1000, 1 ≤ M ≤ 1000) are written on the first line of the input, separated by blanks. N represents the number of sugoroku squares, and M represents the number of dice given.
In the following N lines, integers between -999 and 999 are written one by one. The integer on the 1 + i line (1 ≤ i ≤ N) represents the indication of the sugoroku i-th cell. Let X be the written integer. When X = 0, it indicates "do nothing", when X> 0, it indicates "advance X mass", and when X <0, it indicates "| X | mass return". However, | X | represents the absolute value of X.
In the following M line, integers from 1 to 6 are written one by one, and the number on the 1 + N + j line (1 ≤ j ≤ M) represents the dice roll that appears on the jth time.
However, the number of lines 2 and 1 + N is always 0. There is no cell with instructions to move to the cell before the first cell. In addition, the number of times the dice are rolled is M or less in any scoring input data.
When both N and M are 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each data set, an integer indicating how many times the dice are rolled to reach the goal is output on one line.
Input / output example
Input example
10 5
0
0
Five
6
-3
8
1
8
-Four
0
1
3
Five
1
Five
10 10
0
-1
-1
Four
Four
-Five
0
1
-6
0
1
Five
2
Four
6
Five
Five
Four
1
6
0 0
Output example
Five
6
The following figure shows the first input example.
<image>
The following figure shows the second input example.
<image>
The above question sentences and the data used for the automatic referee are the question sentences created and published by the Japan Committee for Information Olympics and the test data for scoring.
Example
Input
10 5
0
0
5
6
-3
8
1
8
-4
0
1
3
5
1
5
10 10
0
-1
-1
4
4
-5
0
1
-6
0
1
5
2
4
6
5
5
4
1
6
0 0
Output
5
6
Submitted Solution:
```
while True:
n,m = map(int,input().split())
if n==0: break
mapp = []
for _ in range(n):
mapp.append(int(input()))
mas = 0
ans = 0
for dice in range(m):
mas += int(input())
if mas >= n-1 and ans == 0: ans = dice+1
if mas < n-1 : mas += mapp[mas]
if mas >= n-1 and ans == 0: ans = dice+1
print(ans)
``` | instruction | 0 | 80,751 | 19 | 161,502 |
Yes | output | 1 | 80,751 | 19 | 161,503 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sugoroku
problem
JOI is playing sugoroku alone. There are N squares in a straight line in this sugoroku, and each has a movement instruction written on it. The starting point is the 1st square and the goal is the Nth square. JOI repeats the following until he reaches the goal.
Roll the dice and proceed from the current square by the number of rolls, and follow the instructions of that square. Do not follow the instructions of the square to which you moved according to the instructions.
The goal is not only when you stop at the Nth square, but also when the destination exceeds the Nth square.
Create a program that outputs how many times you roll the dice to reach the goal when you are given a sugoroku board and M dice rolls.
input
The input consists of multiple datasets. Each dataset is given in the following format.
Each dataset consists of 1 + N + M rows.
Two integers N, M (2 ≤ N ≤ 1000, 1 ≤ M ≤ 1000) are written on the first line of the input, separated by blanks. N represents the number of sugoroku squares, and M represents the number of dice given.
In the following N lines, integers between -999 and 999 are written one by one. The integer on the 1 + i line (1 ≤ i ≤ N) represents the indication of the sugoroku i-th cell. Let X be the written integer. When X = 0, it indicates "do nothing", when X> 0, it indicates "advance X mass", and when X <0, it indicates "| X | mass return". However, | X | represents the absolute value of X.
In the following M line, integers from 1 to 6 are written one by one, and the number on the 1 + N + j line (1 ≤ j ≤ M) represents the dice roll that appears on the jth time.
However, the number of lines 2 and 1 + N is always 0. There is no cell with instructions to move to the cell before the first cell. In addition, the number of times the dice are rolled is M or less in any scoring input data.
When both N and M are 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each data set, an integer indicating how many times the dice are rolled to reach the goal is output on one line.
Input / output example
Input example
10 5
0
0
Five
6
-3
8
1
8
-Four
0
1
3
Five
1
Five
10 10
0
-1
-1
Four
Four
-Five
0
1
-6
0
1
Five
2
Four
6
Five
Five
Four
1
6
0 0
Output example
Five
6
The following figure shows the first input example.
<image>
The following figure shows the second input example.
<image>
The above question sentences and the data used for the automatic referee are the question sentences created and published by the Japan Committee for Information Olympics and the test data for scoring.
Example
Input
10 5
0
0
5
6
-3
8
1
8
-4
0
1
3
5
1
5
10 10
0
-1
-1
4
4
-5
0
1
-6
0
1
5
2
4
6
5
5
4
1
6
0 0
Output
5
6
Submitted Solution:
```
import sys
def line():return sys.stdin.readline().strip()
def cin():return sys.stdin.readline().strip().split()
while True:
N,M = [int(i) for i in cin()]
if N == 0:break
X = [int(line()) for _ in range(N)]
Y = [int(line()) for _ in range(M)]
c = 1
for i in range(M):
c+=int(Y[i])
if c < N and X[c - 1] != 0:
c+=X[c - 1]
if c >= N:
print(i + 1)
break
``` | instruction | 0 | 80,752 | 19 | 161,504 |
Yes | output | 1 | 80,752 | 19 | 161,505 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sugoroku
problem
JOI is playing sugoroku alone. There are N squares in a straight line in this sugoroku, and each has a movement instruction written on it. The starting point is the 1st square and the goal is the Nth square. JOI repeats the following until he reaches the goal.
Roll the dice and proceed from the current square by the number of rolls, and follow the instructions of that square. Do not follow the instructions of the square to which you moved according to the instructions.
The goal is not only when you stop at the Nth square, but also when the destination exceeds the Nth square.
Create a program that outputs how many times you roll the dice to reach the goal when you are given a sugoroku board and M dice rolls.
input
The input consists of multiple datasets. Each dataset is given in the following format.
Each dataset consists of 1 + N + M rows.
Two integers N, M (2 ≤ N ≤ 1000, 1 ≤ M ≤ 1000) are written on the first line of the input, separated by blanks. N represents the number of sugoroku squares, and M represents the number of dice given.
In the following N lines, integers between -999 and 999 are written one by one. The integer on the 1 + i line (1 ≤ i ≤ N) represents the indication of the sugoroku i-th cell. Let X be the written integer. When X = 0, it indicates "do nothing", when X> 0, it indicates "advance X mass", and when X <0, it indicates "| X | mass return". However, | X | represents the absolute value of X.
In the following M line, integers from 1 to 6 are written one by one, and the number on the 1 + N + j line (1 ≤ j ≤ M) represents the dice roll that appears on the jth time.
However, the number of lines 2 and 1 + N is always 0. There is no cell with instructions to move to the cell before the first cell. In addition, the number of times the dice are rolled is M or less in any scoring input data.
When both N and M are 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each data set, an integer indicating how many times the dice are rolled to reach the goal is output on one line.
Input / output example
Input example
10 5
0
0
Five
6
-3
8
1
8
-Four
0
1
3
Five
1
Five
10 10
0
-1
-1
Four
Four
-Five
0
1
-6
0
1
Five
2
Four
6
Five
Five
Four
1
6
0 0
Output example
Five
6
The following figure shows the first input example.
<image>
The following figure shows the second input example.
<image>
The above question sentences and the data used for the automatic referee are the question sentences created and published by the Japan Committee for Information Olympics and the test data for scoring.
Example
Input
10 5
0
0
5
6
-3
8
1
8
-4
0
1
3
5
1
5
10 10
0
-1
-1
4
4
-5
0
1
-6
0
1
5
2
4
6
5
5
4
1
6
0 0
Output
5
6
Submitted Solution:
```
while True:
n, m = map(int, input().split())
if n == m == 0:
break
a = []
b = []
for i in range(n):
a.append(int(input()))
masu = 1
cnt = 0
for i in range(m):
b.append(int(input()))
for i in range(m):
cnt += 1
masu += b[i]
if masu >= n:
print(cnt)
break
masu += a[masu-1]
if masu >= n:
print(cnt)
break
``` | instruction | 0 | 80,753 | 19 | 161,506 |
Yes | output | 1 | 80,753 | 19 | 161,507 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sugoroku
problem
JOI is playing sugoroku alone. There are N squares in a straight line in this sugoroku, and each has a movement instruction written on it. The starting point is the 1st square and the goal is the Nth square. JOI repeats the following until he reaches the goal.
Roll the dice and proceed from the current square by the number of rolls, and follow the instructions of that square. Do not follow the instructions of the square to which you moved according to the instructions.
The goal is not only when you stop at the Nth square, but also when the destination exceeds the Nth square.
Create a program that outputs how many times you roll the dice to reach the goal when you are given a sugoroku board and M dice rolls.
input
The input consists of multiple datasets. Each dataset is given in the following format.
Each dataset consists of 1 + N + M rows.
Two integers N, M (2 ≤ N ≤ 1000, 1 ≤ M ≤ 1000) are written on the first line of the input, separated by blanks. N represents the number of sugoroku squares, and M represents the number of dice given.
In the following N lines, integers between -999 and 999 are written one by one. The integer on the 1 + i line (1 ≤ i ≤ N) represents the indication of the sugoroku i-th cell. Let X be the written integer. When X = 0, it indicates "do nothing", when X> 0, it indicates "advance X mass", and when X <0, it indicates "| X | mass return". However, | X | represents the absolute value of X.
In the following M line, integers from 1 to 6 are written one by one, and the number on the 1 + N + j line (1 ≤ j ≤ M) represents the dice roll that appears on the jth time.
However, the number of lines 2 and 1 + N is always 0. There is no cell with instructions to move to the cell before the first cell. In addition, the number of times the dice are rolled is M or less in any scoring input data.
When both N and M are 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each data set, an integer indicating how many times the dice are rolled to reach the goal is output on one line.
Input / output example
Input example
10 5
0
0
Five
6
-3
8
1
8
-Four
0
1
3
Five
1
Five
10 10
0
-1
-1
Four
Four
-Five
0
1
-6
0
1
Five
2
Four
6
Five
Five
Four
1
6
0 0
Output example
Five
6
The following figure shows the first input example.
<image>
The following figure shows the second input example.
<image>
The above question sentences and the data used for the automatic referee are the question sentences created and published by the Japan Committee for Information Olympics and the test data for scoring.
Example
Input
10 5
0
0
5
6
-3
8
1
8
-4
0
1
3
5
1
5
10 10
0
-1
-1
4
4
-5
0
1
-6
0
1
5
2
4
6
5
5
4
1
6
0 0
Output
5
6
Submitted Solution:
```
while True:
n,m = map(int,input().split())
if n == m == 0:
break
N = [int(input()) for i in range(n)]
M = [int(input()) for i in range(m)]
x = 1
for i in range(m):
x += M[i]
if n <= x:
print(i+1)
break
else:
x += N[x-1]
if n <= x:
print(i+1)
break
``` | instruction | 0 | 80,754 | 19 | 161,508 |
Yes | output | 1 | 80,754 | 19 | 161,509 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sugoroku
problem
JOI is playing sugoroku alone. There are N squares in a straight line in this sugoroku, and each has a movement instruction written on it. The starting point is the 1st square and the goal is the Nth square. JOI repeats the following until he reaches the goal.
Roll the dice and proceed from the current square by the number of rolls, and follow the instructions of that square. Do not follow the instructions of the square to which you moved according to the instructions.
The goal is not only when you stop at the Nth square, but also when the destination exceeds the Nth square.
Create a program that outputs how many times you roll the dice to reach the goal when you are given a sugoroku board and M dice rolls.
input
The input consists of multiple datasets. Each dataset is given in the following format.
Each dataset consists of 1 + N + M rows.
Two integers N, M (2 ≤ N ≤ 1000, 1 ≤ M ≤ 1000) are written on the first line of the input, separated by blanks. N represents the number of sugoroku squares, and M represents the number of dice given.
In the following N lines, integers between -999 and 999 are written one by one. The integer on the 1 + i line (1 ≤ i ≤ N) represents the indication of the sugoroku i-th cell. Let X be the written integer. When X = 0, it indicates "do nothing", when X> 0, it indicates "advance X mass", and when X <0, it indicates "| X | mass return". However, | X | represents the absolute value of X.
In the following M line, integers from 1 to 6 are written one by one, and the number on the 1 + N + j line (1 ≤ j ≤ M) represents the dice roll that appears on the jth time.
However, the number of lines 2 and 1 + N is always 0. There is no cell with instructions to move to the cell before the first cell. In addition, the number of times the dice are rolled is M or less in any scoring input data.
When both N and M are 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each data set, an integer indicating how many times the dice are rolled to reach the goal is output on one line.
Input / output example
Input example
10 5
0
0
Five
6
-3
8
1
8
-Four
0
1
3
Five
1
Five
10 10
0
-1
-1
Four
Four
-Five
0
1
-6
0
1
Five
2
Four
6
Five
Five
Four
1
6
0 0
Output example
Five
6
The following figure shows the first input example.
<image>
The following figure shows the second input example.
<image>
The above question sentences and the data used for the automatic referee are the question sentences created and published by the Japan Committee for Information Olympics and the test data for scoring.
Example
Input
10 5
0
0
5
6
-3
8
1
8
-4
0
1
3
5
1
5
10 10
0
-1
-1
4
4
-5
0
1
-6
0
1
5
2
4
6
5
5
4
1
6
0 0
Output
5
6
Submitted Solution:
```
field=[]
x=0
y=0
i=0
n,m=map(int,input().split(" "))
for i in range(0,n):
field.append(int(input()))
for i in range(0,m):
y=int(input())
if((n-1)<=x+y):
break
x+=field[x+y]+y
y=i+1
for n in range(i+1,m):
x=input()
print(y)
``` | instruction | 0 | 80,755 | 19 | 161,510 |
No | output | 1 | 80,755 | 19 | 161,511 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sugoroku
problem
JOI is playing sugoroku alone. There are N squares in a straight line in this sugoroku, and each has a movement instruction written on it. The starting point is the 1st square and the goal is the Nth square. JOI repeats the following until he reaches the goal.
Roll the dice and proceed from the current square by the number of rolls, and follow the instructions of that square. Do not follow the instructions of the square to which you moved according to the instructions.
The goal is not only when you stop at the Nth square, but also when the destination exceeds the Nth square.
Create a program that outputs how many times you roll the dice to reach the goal when you are given a sugoroku board and M dice rolls.
input
The input consists of multiple datasets. Each dataset is given in the following format.
Each dataset consists of 1 + N + M rows.
Two integers N, M (2 ≤ N ≤ 1000, 1 ≤ M ≤ 1000) are written on the first line of the input, separated by blanks. N represents the number of sugoroku squares, and M represents the number of dice given.
In the following N lines, integers between -999 and 999 are written one by one. The integer on the 1 + i line (1 ≤ i ≤ N) represents the indication of the sugoroku i-th cell. Let X be the written integer. When X = 0, it indicates "do nothing", when X> 0, it indicates "advance X mass", and when X <0, it indicates "| X | mass return". However, | X | represents the absolute value of X.
In the following M line, integers from 1 to 6 are written one by one, and the number on the 1 + N + j line (1 ≤ j ≤ M) represents the dice roll that appears on the jth time.
However, the number of lines 2 and 1 + N is always 0. There is no cell with instructions to move to the cell before the first cell. In addition, the number of times the dice are rolled is M or less in any scoring input data.
When both N and M are 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each data set, an integer indicating how many times the dice are rolled to reach the goal is output on one line.
Input / output example
Input example
10 5
0
0
Five
6
-3
8
1
8
-Four
0
1
3
Five
1
Five
10 10
0
-1
-1
Four
Four
-Five
0
1
-6
0
1
Five
2
Four
6
Five
Five
Four
1
6
0 0
Output example
Five
6
The following figure shows the first input example.
<image>
The following figure shows the second input example.
<image>
The above question sentences and the data used for the automatic referee are the question sentences created and published by the Japan Committee for Information Olympics and the test data for scoring.
Example
Input
10 5
0
0
5
6
-3
8
1
8
-4
0
1
3
5
1
5
10 10
0
-1
-1
4
4
-5
0
1
-6
0
1
5
2
4
6
5
5
4
1
6
0 0
Output
5
6
Submitted Solution:
```
while True:
N, M = map(int, input().split())
if N == 0:
break
maps = [int(input()) for _ in range(N)]
mas = 0
ans = 0
for dice in range(M):
mas += int(input())
if mas >= n - 1 and ans == 0:
ans = dice + 1
if mas < n - 1:
mas += maps[ans]
if mas >= n - 1 and ans == 0:
ans = dice + 1
print(ans)
``` | instruction | 0 | 80,756 | 19 | 161,512 |
No | output | 1 | 80,756 | 19 | 161,513 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.