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 |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
As we all know, Max is the best video game player among her friends. Her friends were so jealous of hers, that they created an actual game just to prove that she's not the best at games. The game is played on a directed acyclic graph (a DAG) with n vertices and m edges. There's a character written on each edge, a lowercase English letter.
<image>
Max and Lucas are playing the game. Max goes first, then Lucas, then Max again and so on. Each player has a marble, initially located at some vertex. Each player in his/her turn should move his/her marble along some edge (a player can move the marble from vertex v to vertex u if there's an outgoing edge from v to u). If the player moves his/her marble from vertex v to vertex u, the "character" of that round is the character written on the edge from v to u. There's one additional rule; the ASCII code of character of round i should be greater than or equal to the ASCII code of character of round i - 1 (for i > 1). The rounds are numbered for both players together, i. e. Max goes in odd numbers, Lucas goes in even numbers. The player that can't make a move loses the game. The marbles may be at the same vertex at the same time.
Since the game could take a while and Lucas and Max have to focus on finding Dart, they don't have time to play. So they asked you, if they both play optimally, who wins the game?
You have to determine the winner of the game for all initial positions of the marbles.
Input
The first line of input contains two integers n and m (2 ≤ n ≤ 100, <image>).
The next m lines contain the edges. Each line contains two integers v, u and a lowercase English letter c, meaning there's an edge from v to u written c on it (1 ≤ v, u ≤ n, v ≠ u). There's at most one edge between any pair of vertices. It is guaranteed that the graph is acyclic.
Output
Print n lines, a string of length n in each one. The j-th character in i-th line should be 'A' if Max will win the game in case her marble is initially at vertex i and Lucas's marble is initially at vertex j, and 'B' otherwise.
Examples
Input
4 4
1 2 b
1 3 a
2 4 c
3 4 b
Output
BAAA
ABAA
BBBA
BBBB
Input
5 8
5 3 h
1 2 c
3 1 c
3 2 r
5 1 r
4 3 z
5 4 r
5 2 h
Output
BABBB
BBBBB
AABBB
AAABA
AAAAB
Note
Here's the graph in the first sample test case:
<image>
Here's the graph in the second sample test case:
<image> | instruction | 0 | 55,362 | 19 | 110,724 |
Tags: dfs and similar, dp, games, graphs
Correct Solution:
```
from functools import lru_cache
def readline(): return list(map(int, input().split()))
def main():
n, m = readline()
edges = [list() for __ in range(n)]
for __ in range(m):
tokens = input().split()
begin, end = map(int, tokens[:2])
weight = ord(tokens[2])
edges[begin - 1].append((end - 1, weight))
@lru_cache(maxsize=None)
def first_wins(first, second, lower=0):
for (nxt, w) in edges[first]:
if w >= lower:
if not first_wins(second, nxt, w):
return True
return False
for i in range(n):
print("".join("BA"[first_wins(i, j)] for j in range(n)))
main()
``` | output | 1 | 55,362 | 19 | 110,725 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As we all know, Max is the best video game player among her friends. Her friends were so jealous of hers, that they created an actual game just to prove that she's not the best at games. The game is played on a directed acyclic graph (a DAG) with n vertices and m edges. There's a character written on each edge, a lowercase English letter.
<image>
Max and Lucas are playing the game. Max goes first, then Lucas, then Max again and so on. Each player has a marble, initially located at some vertex. Each player in his/her turn should move his/her marble along some edge (a player can move the marble from vertex v to vertex u if there's an outgoing edge from v to u). If the player moves his/her marble from vertex v to vertex u, the "character" of that round is the character written on the edge from v to u. There's one additional rule; the ASCII code of character of round i should be greater than or equal to the ASCII code of character of round i - 1 (for i > 1). The rounds are numbered for both players together, i. e. Max goes in odd numbers, Lucas goes in even numbers. The player that can't make a move loses the game. The marbles may be at the same vertex at the same time.
Since the game could take a while and Lucas and Max have to focus on finding Dart, they don't have time to play. So they asked you, if they both play optimally, who wins the game?
You have to determine the winner of the game for all initial positions of the marbles.
Input
The first line of input contains two integers n and m (2 ≤ n ≤ 100, <image>).
The next m lines contain the edges. Each line contains two integers v, u and a lowercase English letter c, meaning there's an edge from v to u written c on it (1 ≤ v, u ≤ n, v ≠ u). There's at most one edge between any pair of vertices. It is guaranteed that the graph is acyclic.
Output
Print n lines, a string of length n in each one. The j-th character in i-th line should be 'A' if Max will win the game in case her marble is initially at vertex i and Lucas's marble is initially at vertex j, and 'B' otherwise.
Examples
Input
4 4
1 2 b
1 3 a
2 4 c
3 4 b
Output
BAAA
ABAA
BBBA
BBBB
Input
5 8
5 3 h
1 2 c
3 1 c
3 2 r
5 1 r
4 3 z
5 4 r
5 2 h
Output
BABBB
BBBBB
AABBB
AAABA
AAAAB
Note
Here's the graph in the first sample test case:
<image>
Here's the graph in the second sample test case:
<image> | instruction | 0 | 55,363 | 19 | 110,726 |
Tags: dfs and similar, dp, games, graphs
Correct Solution:
```
def mat(shape, inital_val=None):
if len(shape) > 1:
return [mat(shape[1:], inital_val) for _ in range(shape[0])]
else:
return [inital_val] * shape[0]
def main():
n, m = [int(x) for x in input().split()]
graph = [{} for _ in range(n)]
for _ in range(m):
v, u, c = input().split()
graph[int(v) - 1][int(u) - 1] = c
winner_table = mat([n, n, 26])
def get_winner(u, v, char_to_beat):
"""
Args:
u: The position of current turn's player.
v: The position of next turn's player.
char_to_beat: The character played in the previous round.
Returns:
'A' if current turn's player wins, 'B' otherwise.
"""
char_idx = ord(char_to_beat) - ord('a')
if not winner_table[u][v][char_idx]:
winner = 'B'
for w, c in graph[u].items():
if c >= char_to_beat and get_winner(v, w, c) == 'B':
winner = 'A'
break
winner_table[u][v][char_idx] = winner
return winner_table[u][v][char_idx]
for i in range(n):
print(''.join(get_winner(i, j, 'a') for j in range(n)))
if __name__ == '__main__':
main()
``` | output | 1 | 55,363 | 19 | 110,727 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As we all know, Max is the best video game player among her friends. Her friends were so jealous of hers, that they created an actual game just to prove that she's not the best at games. The game is played on a directed acyclic graph (a DAG) with n vertices and m edges. There's a character written on each edge, a lowercase English letter.
<image>
Max and Lucas are playing the game. Max goes first, then Lucas, then Max again and so on. Each player has a marble, initially located at some vertex. Each player in his/her turn should move his/her marble along some edge (a player can move the marble from vertex v to vertex u if there's an outgoing edge from v to u). If the player moves his/her marble from vertex v to vertex u, the "character" of that round is the character written on the edge from v to u. There's one additional rule; the ASCII code of character of round i should be greater than or equal to the ASCII code of character of round i - 1 (for i > 1). The rounds are numbered for both players together, i. e. Max goes in odd numbers, Lucas goes in even numbers. The player that can't make a move loses the game. The marbles may be at the same vertex at the same time.
Since the game could take a while and Lucas and Max have to focus on finding Dart, they don't have time to play. So they asked you, if they both play optimally, who wins the game?
You have to determine the winner of the game for all initial positions of the marbles.
Input
The first line of input contains two integers n and m (2 ≤ n ≤ 100, <image>).
The next m lines contain the edges. Each line contains two integers v, u and a lowercase English letter c, meaning there's an edge from v to u written c on it (1 ≤ v, u ≤ n, v ≠ u). There's at most one edge between any pair of vertices. It is guaranteed that the graph is acyclic.
Output
Print n lines, a string of length n in each one. The j-th character in i-th line should be 'A' if Max will win the game in case her marble is initially at vertex i and Lucas's marble is initially at vertex j, and 'B' otherwise.
Examples
Input
4 4
1 2 b
1 3 a
2 4 c
3 4 b
Output
BAAA
ABAA
BBBA
BBBB
Input
5 8
5 3 h
1 2 c
3 1 c
3 2 r
5 1 r
4 3 z
5 4 r
5 2 h
Output
BABBB
BBBBB
AABBB
AAABA
AAAAB
Note
Here's the graph in the first sample test case:
<image>
Here's the graph in the second sample test case:
<image>
Submitted Solution:
```
s = input()
n, ans = len(s), 0
for i in range(n):
open, open_q, close_q = 0, 0, 0
for j in range(i, n):
if (s[j] == '('):
open+=1
elif (s[j] == '?'):
if (open):
open-=1
close_q+=1
else:
open+=1
open_q+=1
else:
if (open):
open-=1
else:
if (close_q):
close_q-=1
open_q+=1
open+=1
else:
break
if (open == 0):
ans+=1
print(ans)
``` | instruction | 0 | 55,364 | 19 | 110,728 |
No | output | 1 | 55,364 | 19 | 110,729 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As we all know, Max is the best video game player among her friends. Her friends were so jealous of hers, that they created an actual game just to prove that she's not the best at games. The game is played on a directed acyclic graph (a DAG) with n vertices and m edges. There's a character written on each edge, a lowercase English letter.
<image>
Max and Lucas are playing the game. Max goes first, then Lucas, then Max again and so on. Each player has a marble, initially located at some vertex. Each player in his/her turn should move his/her marble along some edge (a player can move the marble from vertex v to vertex u if there's an outgoing edge from v to u). If the player moves his/her marble from vertex v to vertex u, the "character" of that round is the character written on the edge from v to u. There's one additional rule; the ASCII code of character of round i should be greater than or equal to the ASCII code of character of round i - 1 (for i > 1). The rounds are numbered for both players together, i. e. Max goes in odd numbers, Lucas goes in even numbers. The player that can't make a move loses the game. The marbles may be at the same vertex at the same time.
Since the game could take a while and Lucas and Max have to focus on finding Dart, they don't have time to play. So they asked you, if they both play optimally, who wins the game?
You have to determine the winner of the game for all initial positions of the marbles.
Input
The first line of input contains two integers n and m (2 ≤ n ≤ 100, <image>).
The next m lines contain the edges. Each line contains two integers v, u and a lowercase English letter c, meaning there's an edge from v to u written c on it (1 ≤ v, u ≤ n, v ≠ u). There's at most one edge between any pair of vertices. It is guaranteed that the graph is acyclic.
Output
Print n lines, a string of length n in each one. The j-th character in i-th line should be 'A' if Max will win the game in case her marble is initially at vertex i and Lucas's marble is initially at vertex j, and 'B' otherwise.
Examples
Input
4 4
1 2 b
1 3 a
2 4 c
3 4 b
Output
BAAA
ABAA
BBBA
BBBB
Input
5 8
5 3 h
1 2 c
3 1 c
3 2 r
5 1 r
4 3 z
5 4 r
5 2 h
Output
BABBB
BBBBB
AABBB
AAABA
AAAAB
Note
Here's the graph in the first sample test case:
<image>
Here's the graph in the second sample test case:
<image>
Submitted Solution:
```
def lessee(ref,one, two):
from collections import deque
queue = deque()
maxa1, maxa2 = 0,0
queue.append( (one, two,'A',0) )
while len(queue)!= 0:
a = queue.popleft()
if a[0] in ref.keys() and a[1] not in ref.keys():
#print('in')
for i in ref[a[0]]:
if a[2] <= i[1]:
maxa1 = max(maxa1, a[3]+1)
maxa2 = max(maxa2, a[3])
elif a[0] in ref.keys() and a[1] in ref.keys():
for i in ref[a[0]]:
for j in ref[a[1]]:
if a[2] <= i[1] and j[1] >= i[1] :
queue.append((i[0], j[0], j[1], a[3]+1) )
maxa1 = max(maxa1, a[3]+1)
maxa2 = max(maxa2, a[3]+1)
elif a[2] <= i[1] and j[1] < i[1] :
maxa1 = max(maxa1, a[3]+1)
return maxa1, maxa2
def final(arr,n):
ref = {}
for i in arr:
if i[0]-1 not in ref.keys():
ref[i[0]-1] = [(i[1]-1, i[2])]
else:
ref[i[0]-1].append((i[1]-1, i[2]))
finalAns = [[0 for i in range(n)] for j in range(n)]
#print(lessee(ref,0,0))
for i in range(n):
for j in range(n):
p,q = lessee(ref, i,j)
#print(i+1, j+1,'adfa', p,q)
if p > q:
finalAns[i][j] = 'A'
else:
finalAns[i][j] = 'B'
return finalAns
import sys
c = -1
for line in sys.stdin:
if c == -1:
t = line.split()
n= int(t[0])
k = int(t[1])
arr = []
c = 0
else:
c += 1
temp = line.split()
arr.append([int(temp[0]), int(temp[1]), temp[2]])
if c == k:
ans = final(arr, n)
for i in ans:
tem = ''.join(g for g in i)
print(tem)
``` | instruction | 0 | 55,365 | 19 | 110,730 |
No | output | 1 | 55,365 | 19 | 110,731 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As we all know, Max is the best video game player among her friends. Her friends were so jealous of hers, that they created an actual game just to prove that she's not the best at games. The game is played on a directed acyclic graph (a DAG) with n vertices and m edges. There's a character written on each edge, a lowercase English letter.
<image>
Max and Lucas are playing the game. Max goes first, then Lucas, then Max again and so on. Each player has a marble, initially located at some vertex. Each player in his/her turn should move his/her marble along some edge (a player can move the marble from vertex v to vertex u if there's an outgoing edge from v to u). If the player moves his/her marble from vertex v to vertex u, the "character" of that round is the character written on the edge from v to u. There's one additional rule; the ASCII code of character of round i should be greater than or equal to the ASCII code of character of round i - 1 (for i > 1). The rounds are numbered for both players together, i. e. Max goes in odd numbers, Lucas goes in even numbers. The player that can't make a move loses the game. The marbles may be at the same vertex at the same time.
Since the game could take a while and Lucas and Max have to focus on finding Dart, they don't have time to play. So they asked you, if they both play optimally, who wins the game?
You have to determine the winner of the game for all initial positions of the marbles.
Input
The first line of input contains two integers n and m (2 ≤ n ≤ 100, <image>).
The next m lines contain the edges. Each line contains two integers v, u and a lowercase English letter c, meaning there's an edge from v to u written c on it (1 ≤ v, u ≤ n, v ≠ u). There's at most one edge between any pair of vertices. It is guaranteed that the graph is acyclic.
Output
Print n lines, a string of length n in each one. The j-th character in i-th line should be 'A' if Max will win the game in case her marble is initially at vertex i and Lucas's marble is initially at vertex j, and 'B' otherwise.
Examples
Input
4 4
1 2 b
1 3 a
2 4 c
3 4 b
Output
BAAA
ABAA
BBBA
BBBB
Input
5 8
5 3 h
1 2 c
3 1 c
3 2 r
5 1 r
4 3 z
5 4 r
5 2 h
Output
BABBB
BBBBB
AABBB
AAABA
AAAAB
Note
Here's the graph in the first sample test case:
<image>
Here's the graph in the second sample test case:
<image>
Submitted Solution:
```
def recur(node, height, ref2):
#print(node)
if node not in ref2.keys():
if height%2 ==0:
return 1
return 2
for i in ref2[node]:
if i[1] > height:
a = recur(i[0], i[1], ref2)
if height%2==1 and a ==1:
return 1
if height%2==0 and a==2:
return 2
if height%2==1:
return 2
else:
return 1
def final(ref, one, two):
prev = 'A'
arr = []
temp = []
temp2 = []
height = 0
for i in ref[one]:
arr.append([one, i[0],i[1],height+1])
temp.append(len(arr)-1)
for i in temp:
c = arr[i]
for j in ref[two]:
if j[1] >= c[2]:
arr.append([c[1], j[0], j[1],height+2])
temp2.append(len(arr)-1)
#print(arr)
while True:
height+=2
temp = []
for i in temp2:
c = arr[i]
for j in ref[c[0]]:
if j[1] >= c[2]:
arr.append([c[1], j[0], j[1],height+1])
temp.append(len(arr)-1)
temp2 = []
for i in temp:
c = arr[i]
for j in ref[c[0]]:
if j[1] >= c[2]:
arr.append([c[1], j[0], j[1],height+2])
temp2.append(len(arr)-1)
if len(temp2) == 0:
break
ref2 = {}
for i in arr:
if i[0] not in ref2.keys():
ref2[i[0]] = [(i[1],i[3])]
else:
ref2[i[0]].append((i[1],i[3]))
#print('ref2',one, two,ref2)
#return 1
a= recur(one,0,ref2)
return a
def finalfinal(ref,n):
finalAns = [[0 for i in range(n)] for j in range(n)]
#print(lessee(ref,0,0))
for i in range(1,n+1):
for j in range(1,n+1):
curr = final(ref, i,j)
#print('OVER', curr)
#print(i+1, j+1,'adfa', p,q)
if curr == 2:
finalAns[i-1][j-1] = 'A'
else:
finalAns[i-1][j-1] = 'B'
return finalAns
import sys
sys.setrecursionlimit(150000)
#with open("/content/madmax2.txt", "r") as file:
c = -1
for line in sys.stdin:
if c == -1:
t = line.split()
n= int(t[0])
k = int(t[1])
#arr = []
c = 0
ref = {}
for i in range(1,n+1):
ref[i] = []
else:
c += 1
temp = line.split()
#arr.append([int(temp[0]), int(temp[1]), temp[2]])
if int(temp[0]) not in ref.keys():
ref[int(temp[0])] = [(int(temp[1]), temp[2])]
else:
ref[int(temp[0])].append((int(temp[1]), temp[2]))
if c == k:
#print('ref',ref)
#print(final(ref,3,2))
ans = finalfinal(ref, n)
for i in ans:
tem = ''.join(g for g in i)
print(tem)
``` | instruction | 0 | 55,366 | 19 | 110,732 |
No | output | 1 | 55,366 | 19 | 110,733 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As we all know, Max is the best video game player among her friends. Her friends were so jealous of hers, that they created an actual game just to prove that she's not the best at games. The game is played on a directed acyclic graph (a DAG) with n vertices and m edges. There's a character written on each edge, a lowercase English letter.
<image>
Max and Lucas are playing the game. Max goes first, then Lucas, then Max again and so on. Each player has a marble, initially located at some vertex. Each player in his/her turn should move his/her marble along some edge (a player can move the marble from vertex v to vertex u if there's an outgoing edge from v to u). If the player moves his/her marble from vertex v to vertex u, the "character" of that round is the character written on the edge from v to u. There's one additional rule; the ASCII code of character of round i should be greater than or equal to the ASCII code of character of round i - 1 (for i > 1). The rounds are numbered for both players together, i. e. Max goes in odd numbers, Lucas goes in even numbers. The player that can't make a move loses the game. The marbles may be at the same vertex at the same time.
Since the game could take a while and Lucas and Max have to focus on finding Dart, they don't have time to play. So they asked you, if they both play optimally, who wins the game?
You have to determine the winner of the game for all initial positions of the marbles.
Input
The first line of input contains two integers n and m (2 ≤ n ≤ 100, <image>).
The next m lines contain the edges. Each line contains two integers v, u and a lowercase English letter c, meaning there's an edge from v to u written c on it (1 ≤ v, u ≤ n, v ≠ u). There's at most one edge between any pair of vertices. It is guaranteed that the graph is acyclic.
Output
Print n lines, a string of length n in each one. The j-th character in i-th line should be 'A' if Max will win the game in case her marble is initially at vertex i and Lucas's marble is initially at vertex j, and 'B' otherwise.
Examples
Input
4 4
1 2 b
1 3 a
2 4 c
3 4 b
Output
BAAA
ABAA
BBBA
BBBB
Input
5 8
5 3 h
1 2 c
3 1 c
3 2 r
5 1 r
4 3 z
5 4 r
5 2 h
Output
BABBB
BBBBB
AABBB
AAABA
AAAAB
Note
Here's the graph in the first sample test case:
<image>
Here's the graph in the second sample test case:
<image>
Submitted Solution:
```
import sys
import os
import math
import re
n,m = map(int,input().split())
edges = []
for _ in range(m):
a,b,c = input().split()
edges.append((a,b,c))
#print (edges)
def bestMove(curRound,pos):
#print('Determining best move')
global edges
potentialMoves = []
for edge in edges:
if edge[0] == pos:
#print('got here')
if edge[2] >= curRound:
potentialMoves.append((edge[1],edge[2]))
if not potentialMoves:
return []
potentialMoves = sorted(potentialMoves,key = lambda x: x[1],reverse = True)
tempMove = potentialMoves[0]
tempy = 'a'
for move in potentialMoves:
for edge in edges:
if edge[0] == move[0]:
if edge[2] > tempy:
tempy = edge[2]
tempMove = move
return tempMove
def determineWinner(i,j):
global edges
moves = 0
maxy = str(i)
curRound = 'a'
lucas = str(j)
while(True):
move = bestMove(curRound,maxy)
if not move:
break;
moves+=1;
#moveL = bestMove(move[len(move)-1][0],lucas)
#moveL2 = bestMove(move[0][0],lucas)
maxy = move[0]
curRound = move[1]
move = bestMove(curRound,lucas)
if not move:
break;
moves += 1
lucas = move[0]
curRound = move[1]
#print('ROUND ',curRound)
if moves % 2 ==0:
return 'B'
else:
return 'A'
for i in range(1,n+1):
for j in range(1,n+1):
print(determineWinner(i,j),end="")
print('\n',end="")
``` | instruction | 0 | 55,367 | 19 | 110,734 |
No | output | 1 | 55,367 | 19 | 110,735 |
Provide a correct Python 3 solution for this coding contest problem.
There are N piles of stones. The i-th pile has A_i stones.
Aoki and Takahashi are about to use them to play the following game:
* Starting with Aoki, the two players alternately do the following operation:
* Operation: Choose one pile of stones, and remove one or more stones from it.
* When a player is unable to do the operation, he loses, and the other player wins.
When the two players play optimally, there are two possibilities in this game: the player who moves first always wins, or the player who moves second always wins, only depending on the initial number of stones in each pile.
In such a situation, Takahashi, the second player to act, is trying to guarantee his win by moving at least zero and at most (A_1 - 1) stones from the 1-st pile to the 2-nd pile before the game begins.
If this is possible, print the minimum number of stones to move to guarantee his victory; otherwise, print `-1` instead.
Constraints
* 2 \leq N \leq 300
* 1 \leq A_i \leq 10^{12}
Input
Input is given from Standard Input in the following format:
N
A_1 \ldots A_N
Output
Print the minimum number of stones to move to guarantee Takahashi's win; otherwise, print `-1` instead.
Examples
Input
2
5 3
Output
1
Input
2
3 5
Output
-1
Input
3
1 1 2
Output
-1
Input
8
10 9 8 7 6 5 4 3
Output
3
Input
3
4294967297 8589934593 12884901890
Output
1 | instruction | 0 | 55,374 | 19 | 110,748 |
"Correct Solution:
```
def main():
import sys
def input(): return sys.stdin.readline().rstrip()
n = int(input())
a = list(map(int, input().split()))
x = 0
for i in range(2, n):
x ^= a[i]
d = a[0]+a[1]-x
if d%2 == 1 or d < 0:
print(-1)
return
d >>= 1
if d&x != 0 or d > a[0]:
print(-1)
return
k = x.bit_length()
tmp = d
# d^tmp はd&x=0からd|tmpと一緒
for i in range(k, -1, -1):
if (x >> i) & 1:
if tmp|1<<i <= a[0]:
tmp |= 1<<i
if 0 < tmp <= a[0]:
print(a[0]-tmp)
else:
print(-1)
if __name__ == '__main__':
main()
``` | output | 1 | 55,374 | 19 | 110,749 |
Provide a correct Python 3 solution for this coding contest problem.
There are N piles of stones. The i-th pile has A_i stones.
Aoki and Takahashi are about to use them to play the following game:
* Starting with Aoki, the two players alternately do the following operation:
* Operation: Choose one pile of stones, and remove one or more stones from it.
* When a player is unable to do the operation, he loses, and the other player wins.
When the two players play optimally, there are two possibilities in this game: the player who moves first always wins, or the player who moves second always wins, only depending on the initial number of stones in each pile.
In such a situation, Takahashi, the second player to act, is trying to guarantee his win by moving at least zero and at most (A_1 - 1) stones from the 1-st pile to the 2-nd pile before the game begins.
If this is possible, print the minimum number of stones to move to guarantee his victory; otherwise, print `-1` instead.
Constraints
* 2 \leq N \leq 300
* 1 \leq A_i \leq 10^{12}
Input
Input is given from Standard Input in the following format:
N
A_1 \ldots A_N
Output
Print the minimum number of stones to move to guarantee Takahashi's win; otherwise, print `-1` instead.
Examples
Input
2
5 3
Output
1
Input
2
3 5
Output
-1
Input
3
1 1 2
Output
-1
Input
8
10 9 8 7 6 5 4 3
Output
3
Input
3
4294967297 8589934593 12884901890
Output
1 | instruction | 0 | 55,375 | 19 | 110,750 |
"Correct Solution:
```
def compute(S, X):
if (S-X)%2==1:
print(-1)
return
A = (S - X)//2
a = 0
ambiguous=[0]*64
for i in range(63,-1,-1):
Xi = (X & (1 << i))
Ai = (A & (1 << i))
if (Xi == 0 and Ai == 0):
pass
elif (Xi == 0 and Ai > 0):
a = ((1 << i) | a)
elif (Xi > 0 and Ai == 0):
ambiguous[i]=1
else:
print(-1)
return
for i in range(63,-1,-1):
if ambiguous[i]==1:
R=((1 << i) | a)
if R<=a1[0]:
a=R
else:
a = ((0 << i) | a)
if a>a1[0]:
print(-1)
else:
if a==0 and a1[0]!=0:
print(-1)
else:
print(a1[0]-a)
import sys
n=int(input())
a1=list(map(int,input().split()))
nimsum=0
for i in range(2,n):
nimsum=a1[i]^nimsum
compute(a1[0]+a1[1],nimsum)
``` | output | 1 | 55,375 | 19 | 110,751 |
Provide a correct Python 3 solution for this coding contest problem.
There are N piles of stones. The i-th pile has A_i stones.
Aoki and Takahashi are about to use them to play the following game:
* Starting with Aoki, the two players alternately do the following operation:
* Operation: Choose one pile of stones, and remove one or more stones from it.
* When a player is unable to do the operation, he loses, and the other player wins.
When the two players play optimally, there are two possibilities in this game: the player who moves first always wins, or the player who moves second always wins, only depending on the initial number of stones in each pile.
In such a situation, Takahashi, the second player to act, is trying to guarantee his win by moving at least zero and at most (A_1 - 1) stones from the 1-st pile to the 2-nd pile before the game begins.
If this is possible, print the minimum number of stones to move to guarantee his victory; otherwise, print `-1` instead.
Constraints
* 2 \leq N \leq 300
* 1 \leq A_i \leq 10^{12}
Input
Input is given from Standard Input in the following format:
N
A_1 \ldots A_N
Output
Print the minimum number of stones to move to guarantee Takahashi's win; otherwise, print `-1` instead.
Examples
Input
2
5 3
Output
1
Input
2
3 5
Output
-1
Input
3
1 1 2
Output
-1
Input
8
10 9 8 7 6 5 4 3
Output
3
Input
3
4294967297 8589934593 12884901890
Output
1 | instruction | 0 | 55,376 | 19 | 110,752 |
"Correct Solution:
```
from functools import lru_cache
N, = map(int, input().split())
X = list(map(int, input().split()))
t = 0
for x in X:
t = t^x
inf = float('inf')
@lru_cache(None)
def it(x, y, z):
xm, ym, zm = x%2, y%2, z%2
if xm ^ ym != zm:
return inf
if x^y == z:
return 0
Rz = 2*it(x>>1, y>>1, z>>1)
Ro = inf
if x:
Ro = 2*it((x-1)>>1, (y+1)>>1, z>>1)+1
Rz = min(Rz, Ro)
return Rz
r = it(X[0], X[1], t^X[0]^X[1])
if 0 <= r < X[0]:
print(r)
else:
print(-1)
``` | output | 1 | 55,376 | 19 | 110,753 |
Provide a correct Python 3 solution for this coding contest problem.
There are N piles of stones. The i-th pile has A_i stones.
Aoki and Takahashi are about to use them to play the following game:
* Starting with Aoki, the two players alternately do the following operation:
* Operation: Choose one pile of stones, and remove one or more stones from it.
* When a player is unable to do the operation, he loses, and the other player wins.
When the two players play optimally, there are two possibilities in this game: the player who moves first always wins, or the player who moves second always wins, only depending on the initial number of stones in each pile.
In such a situation, Takahashi, the second player to act, is trying to guarantee his win by moving at least zero and at most (A_1 - 1) stones from the 1-st pile to the 2-nd pile before the game begins.
If this is possible, print the minimum number of stones to move to guarantee his victory; otherwise, print `-1` instead.
Constraints
* 2 \leq N \leq 300
* 1 \leq A_i \leq 10^{12}
Input
Input is given from Standard Input in the following format:
N
A_1 \ldots A_N
Output
Print the minimum number of stones to move to guarantee Takahashi's win; otherwise, print `-1` instead.
Examples
Input
2
5 3
Output
1
Input
2
3 5
Output
-1
Input
3
1 1 2
Output
-1
Input
8
10 9 8 7 6 5 4 3
Output
3
Input
3
4294967297 8589934593 12884901890
Output
1 | instruction | 0 | 55,377 | 19 | 110,754 |
"Correct Solution:
```
import bisect
N=int(input())
A=list(map(int,input().split()))
x=0
for i in range(2,N):
x^=A[i]
#print(x)
if (A[0]+A[1]-x)%2!=0:
print(-1)
else:
AND=(A[0]+A[1]-x)//2
u=AND
v=AND
if AND&x!=0:
print(-1)
exit()
t=[i for i in range(40) if x>>i &1==1]
#print(t)
k=len(t)//2
data1=[]
data2=[]
for i in range(2**k):
temp=0
for j in range(k):
if i>>j&1==1:
temp+=2**t[j]
data1.append(temp)
for i in range(2**(len(t)-k)):
temp=0
for j in range(len(t)-k):
if i>>j&1==1:
temp+=2**t[k+j]
data2.append(temp)
#print(data1,data2)
ans=-1
val=float("inf")
for test in data1:
temp=test+u
#print(temp)
id=bisect.bisect_right(data2,A[0]-temp)
if id!=0:
temp+=data2[id-1]
if val>A[0]-temp and A[0]>=temp and temp!=0:
val=A[0]-temp
ans=temp
else:
if val>A[0]-temp and A[0]>=temp and temp!=0:
val=A[0]-temp
ans=temp
#print(val,"jfkdjfd")
#print(data1,data2,temp)
#print(data1,data2)
if val!=float("inf"):
print(val)
else:
print(-1)
``` | output | 1 | 55,377 | 19 | 110,755 |
Provide a correct Python 3 solution for this coding contest problem.
There are N piles of stones. The i-th pile has A_i stones.
Aoki and Takahashi are about to use them to play the following game:
* Starting with Aoki, the two players alternately do the following operation:
* Operation: Choose one pile of stones, and remove one or more stones from it.
* When a player is unable to do the operation, he loses, and the other player wins.
When the two players play optimally, there are two possibilities in this game: the player who moves first always wins, or the player who moves second always wins, only depending on the initial number of stones in each pile.
In such a situation, Takahashi, the second player to act, is trying to guarantee his win by moving at least zero and at most (A_1 - 1) stones from the 1-st pile to the 2-nd pile before the game begins.
If this is possible, print the minimum number of stones to move to guarantee his victory; otherwise, print `-1` instead.
Constraints
* 2 \leq N \leq 300
* 1 \leq A_i \leq 10^{12}
Input
Input is given from Standard Input in the following format:
N
A_1 \ldots A_N
Output
Print the minimum number of stones to move to guarantee Takahashi's win; otherwise, print `-1` instead.
Examples
Input
2
5 3
Output
1
Input
2
3 5
Output
-1
Input
3
1 1 2
Output
-1
Input
8
10 9 8 7 6 5 4 3
Output
3
Input
3
4294967297 8589934593 12884901890
Output
1 | instruction | 0 | 55,378 | 19 | 110,756 |
"Correct Solution:
```
import sys
from functools import lru_cache
def solve(n, aaa):
if n == 2:
h, m = divmod(sum(aaa), 2)
if m == 1 or h > aaa[0]:
return -1
return aaa[0] - h
x = 0
for a in aaa[2:]:
x ^= a
a0, a1 = aaa[:2]
s = a0 + a1
if s & 1 != x & 1:
return -1
b = 1 << 40
MASK = (b << 1) - 1
@lru_cache(maxsize=None)
def dfs(p, b):
if b == 0:
return p
m = (MASK ^ (b - 1)) << 1
q = p ^ x & m
if x & b == 0:
if s >= p + q + 2 * b:
if p | b <= a0:
return dfs(p | b, b >> 1)
else:
return -1
return dfs(p, b >> 1)
else:
if s < p + q + b:
return -1
if p | b <= a0:
ret = dfs(p | b, b >> 1)
if ret != -1:
return ret
return dfs(p, b >> 1)
ret = dfs(0, b)
if ret <= 0:
return -1
return a0 - ret
n, *aaa = map(int, sys.stdin.buffer.read().split())
print(solve(n, aaa))
``` | output | 1 | 55,378 | 19 | 110,757 |
Provide a correct Python 3 solution for this coding contest problem.
There are N piles of stones. The i-th pile has A_i stones.
Aoki and Takahashi are about to use them to play the following game:
* Starting with Aoki, the two players alternately do the following operation:
* Operation: Choose one pile of stones, and remove one or more stones from it.
* When a player is unable to do the operation, he loses, and the other player wins.
When the two players play optimally, there are two possibilities in this game: the player who moves first always wins, or the player who moves second always wins, only depending on the initial number of stones in each pile.
In such a situation, Takahashi, the second player to act, is trying to guarantee his win by moving at least zero and at most (A_1 - 1) stones from the 1-st pile to the 2-nd pile before the game begins.
If this is possible, print the minimum number of stones to move to guarantee his victory; otherwise, print `-1` instead.
Constraints
* 2 \leq N \leq 300
* 1 \leq A_i \leq 10^{12}
Input
Input is given from Standard Input in the following format:
N
A_1 \ldots A_N
Output
Print the minimum number of stones to move to guarantee Takahashi's win; otherwise, print `-1` instead.
Examples
Input
2
5 3
Output
1
Input
2
3 5
Output
-1
Input
3
1 1 2
Output
-1
Input
8
10 9 8 7 6 5 4 3
Output
3
Input
3
4294967297 8589934593 12884901890
Output
1 | instruction | 0 | 55,379 | 19 | 110,758 |
"Correct Solution:
```
N = map(int, input().split())
As = list(map(int, input().split()))
added = As[0] + As[1]
xored = 0
for A in As[2:]:
xored ^= A
anded = added - xored
if anded % 2 == 1:
print(-1)
exit()
else:
anded >>= 1
left_stones_l_A0 = '' # <= A[0] for sure.
left_stones_e_A0 = '' #not sure if this is <= A[0].
for i, (a, x) in enumerate(zip(format(anded, '045b'), format(xored, '045b'))):
if x == '0':
if a == '0':
left_stones_l_A0 += '0'
left_stones_e_A0 += '0'
else:
left_stones_l_A0 += '1'
left_stones_e_A0 += '1'
else:
if a == '0':
if int(left_stones_l_A0 + '1' + '1' * (44 - i), base = 2) <= As[0]:
left_stones_l_A0 += '1'
else:
left_stones_l_A0 += '0'
if int(left_stones_e_A0 + '1' + '1' * (44 - i), base = 2) <= As[0]:
left_stones_e_A0 += '1'
left_stones_l_A0 = left_stones_e_A0
elif int(left_stones_e_A0 + '1' + '0' * (44 - i), base = 2) <= As[0]:
left_stones_l_A0 = left_stones_e_A0 + '0'
left_stones_e_A0 += '1'
elif int(left_stones_e_A0 + '0' + '0' * (44 - i), base = 2) <= As[0]:
left_stones_e_A0 += '0'
else:
left_stones_e_A0 = left_stones_l_A0
else:
print(-1)
exit()
if int(left_stones_e_A0, base = 2) <= As[0]:
ans = As[0] - int(left_stones_e_A0, base = 2)
else:
ans = As[0] - int(left_stones_l_A0, base = 2)
if 0 <= ans < As[0]:
print(ans)
else:
print(-1)
``` | output | 1 | 55,379 | 19 | 110,759 |
Provide a correct Python 3 solution for this coding contest problem.
There are N piles of stones. The i-th pile has A_i stones.
Aoki and Takahashi are about to use them to play the following game:
* Starting with Aoki, the two players alternately do the following operation:
* Operation: Choose one pile of stones, and remove one or more stones from it.
* When a player is unable to do the operation, he loses, and the other player wins.
When the two players play optimally, there are two possibilities in this game: the player who moves first always wins, or the player who moves second always wins, only depending on the initial number of stones in each pile.
In such a situation, Takahashi, the second player to act, is trying to guarantee his win by moving at least zero and at most (A_1 - 1) stones from the 1-st pile to the 2-nd pile before the game begins.
If this is possible, print the minimum number of stones to move to guarantee his victory; otherwise, print `-1` instead.
Constraints
* 2 \leq N \leq 300
* 1 \leq A_i \leq 10^{12}
Input
Input is given from Standard Input in the following format:
N
A_1 \ldots A_N
Output
Print the minimum number of stones to move to guarantee Takahashi's win; otherwise, print `-1` instead.
Examples
Input
2
5 3
Output
1
Input
2
3 5
Output
-1
Input
3
1 1 2
Output
-1
Input
8
10 9 8 7 6 5 4 3
Output
3
Input
3
4294967297 8589934593 12884901890
Output
1 | instruction | 0 | 55,380 | 19 | 110,760 |
"Correct Solution:
```
n = int(input())
a = list(map(int,input().split()))
c = 0
for i in range(2,n):
c = c^a[i]
start = a[0]
a,b = a[0],a[1]
if((a^b)==c):
print(0)
exit()
dif = a+b-c
if(dif%2==1):
print(-1)
exit()
dif = dif//2
for i in range(40):
if((c >> i)&1)&((dif >> i)&1):
print(-1)
exit()
if(dif > start):
print(-1)
exit()
goal = dif
for i in range(40,-1,-1):
if(c >> i)&1:
if(goal + 2**i <= start):
goal += 2**i
if(goal==0):
print(-1)
exit()
print(start-goal)
``` | output | 1 | 55,380 | 19 | 110,761 |
Provide a correct Python 3 solution for this coding contest problem.
There are N piles of stones. The i-th pile has A_i stones.
Aoki and Takahashi are about to use them to play the following game:
* Starting with Aoki, the two players alternately do the following operation:
* Operation: Choose one pile of stones, and remove one or more stones from it.
* When a player is unable to do the operation, he loses, and the other player wins.
When the two players play optimally, there are two possibilities in this game: the player who moves first always wins, or the player who moves second always wins, only depending on the initial number of stones in each pile.
In such a situation, Takahashi, the second player to act, is trying to guarantee his win by moving at least zero and at most (A_1 - 1) stones from the 1-st pile to the 2-nd pile before the game begins.
If this is possible, print the minimum number of stones to move to guarantee his victory; otherwise, print `-1` instead.
Constraints
* 2 \leq N \leq 300
* 1 \leq A_i \leq 10^{12}
Input
Input is given from Standard Input in the following format:
N
A_1 \ldots A_N
Output
Print the minimum number of stones to move to guarantee Takahashi's win; otherwise, print `-1` instead.
Examples
Input
2
5 3
Output
1
Input
2
3 5
Output
-1
Input
3
1 1 2
Output
-1
Input
8
10 9 8 7 6 5 4 3
Output
3
Input
3
4294967297 8589934593 12884901890
Output
1 | instruction | 0 | 55,381 | 19 | 110,762 |
"Correct Solution:
```
#写経
#https://atcoder.jp/contests/abc172/submissions/14777611
def resolve():
N = int(input())
A = [int(a) for a in input().split()]
x = 0
for a in A[2:]:
x ^= a # ^はXOR計算
s = sum(A[:2])
if s < x or (s ^ x) & 1:
print(-1)
return
d = (s-x) // 2
a = A[0]
n = 0
for i in range(40)[::-1]:
b,c = (x >> i) & 1, (d >> i) & 1 # >> i は右方向iビットシフト
if b == c == 1:
print(-1)
return
if b == 0 and c == 1:
n += 1 << i
for i in range(40)[::-1]:
b,c = (x >> i) & 1, (d >> i) & 1
if b == 1 and c == 0:
if n + (1 << i) <= a:
n += 1 << i
if n > a or n == 0:
print(-1)
else:
print(a-n)
resolve()
``` | output | 1 | 55,381 | 19 | 110,763 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N piles of stones. The i-th pile has A_i stones.
Aoki and Takahashi are about to use them to play the following game:
* Starting with Aoki, the two players alternately do the following operation:
* Operation: Choose one pile of stones, and remove one or more stones from it.
* When a player is unable to do the operation, he loses, and the other player wins.
When the two players play optimally, there are two possibilities in this game: the player who moves first always wins, or the player who moves second always wins, only depending on the initial number of stones in each pile.
In such a situation, Takahashi, the second player to act, is trying to guarantee his win by moving at least zero and at most (A_1 - 1) stones from the 1-st pile to the 2-nd pile before the game begins.
If this is possible, print the minimum number of stones to move to guarantee his victory; otherwise, print `-1` instead.
Constraints
* 2 \leq N \leq 300
* 1 \leq A_i \leq 10^{12}
Input
Input is given from Standard Input in the following format:
N
A_1 \ldots A_N
Output
Print the minimum number of stones to move to guarantee Takahashi's win; otherwise, print `-1` instead.
Examples
Input
2
5 3
Output
1
Input
2
3 5
Output
-1
Input
3
1 1 2
Output
-1
Input
8
10 9 8 7 6 5 4 3
Output
3
Input
3
4294967297 8589934593 12884901890
Output
1
Submitted Solution:
```
def main():
n = int(input())
a = list(map(int, input().split()))
x, y = a[0], a[1]
nim = 0
for i in range(2, n):
nim ^= a[i]
ans = 0
while((x ^ y) != nim):
for i in range(50, -1, -1):
j = (nim >> i) & 1
xy = ((x >> i) ^ (y >> i)) & 1
if j != xy:
m = min(x % (2**i)+1, 2**i-(y % (2**i)))
if x % (2**i)+1 == 2**i-(y % (2**i)):
print(-1)
return
x -= m
y += m
ans += m
if x <= 0:
print(-1)
return
break
else:
if (x ^ y) != nim:
print(-1)
return
print(ans)
main()
``` | instruction | 0 | 55,382 | 19 | 110,764 |
Yes | output | 1 | 55,382 | 19 | 110,765 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N piles of stones. The i-th pile has A_i stones.
Aoki and Takahashi are about to use them to play the following game:
* Starting with Aoki, the two players alternately do the following operation:
* Operation: Choose one pile of stones, and remove one or more stones from it.
* When a player is unable to do the operation, he loses, and the other player wins.
When the two players play optimally, there are two possibilities in this game: the player who moves first always wins, or the player who moves second always wins, only depending on the initial number of stones in each pile.
In such a situation, Takahashi, the second player to act, is trying to guarantee his win by moving at least zero and at most (A_1 - 1) stones from the 1-st pile to the 2-nd pile before the game begins.
If this is possible, print the minimum number of stones to move to guarantee his victory; otherwise, print `-1` instead.
Constraints
* 2 \leq N \leq 300
* 1 \leq A_i \leq 10^{12}
Input
Input is given from Standard Input in the following format:
N
A_1 \ldots A_N
Output
Print the minimum number of stones to move to guarantee Takahashi's win; otherwise, print `-1` instead.
Examples
Input
2
5 3
Output
1
Input
2
3 5
Output
-1
Input
3
1 1 2
Output
-1
Input
8
10 9 8 7 6 5 4 3
Output
3
Input
3
4294967297 8589934593 12884901890
Output
1
Submitted Solution:
```
def main():
n=int(input())
A=list(map(int,input().split()))
x=A[0]^A[1]
for i in A:
x^=i
s=A[0]+A[1]
dp=[[[-1]*2 for _ in range(2)] for _ in range(43)]
dp[0][0][0]=0
v=1
a=A[0]
for i in range(42):
cx,ca,cs=x&1,a&1,s&1
for j in range(2):
for k in range(2):
if dp[i][j][k]==-1:
continue
for na in range(2):
for nb in range(2):
if na^nb!=cx:
continue
ni,nj,nk,ns=i+1,0,k,na+nb+j
if ns%2!=cs:
continue
if ns>=2:
nj=1
if ca<na:
nk=1
elif ca==na:
nk=k
else:
nk=0
dp[ni][nj][nk]=max(dp[ni][nj][nk],dp[i][j][k]|(v*na))
x>>=1
s>>=1
a>>=1
v<<=1
a=dp[42][0][0]
if a==0 or a==-1:
ans=-1
else:
ans=A[0]-a
print(ans)
if __name__ == '__main__':
main()
``` | instruction | 0 | 55,383 | 19 | 110,766 |
Yes | output | 1 | 55,383 | 19 | 110,767 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N piles of stones. The i-th pile has A_i stones.
Aoki and Takahashi are about to use them to play the following game:
* Starting with Aoki, the two players alternately do the following operation:
* Operation: Choose one pile of stones, and remove one or more stones from it.
* When a player is unable to do the operation, he loses, and the other player wins.
When the two players play optimally, there are two possibilities in this game: the player who moves first always wins, or the player who moves second always wins, only depending on the initial number of stones in each pile.
In such a situation, Takahashi, the second player to act, is trying to guarantee his win by moving at least zero and at most (A_1 - 1) stones from the 1-st pile to the 2-nd pile before the game begins.
If this is possible, print the minimum number of stones to move to guarantee his victory; otherwise, print `-1` instead.
Constraints
* 2 \leq N \leq 300
* 1 \leq A_i \leq 10^{12}
Input
Input is given from Standard Input in the following format:
N
A_1 \ldots A_N
Output
Print the minimum number of stones to move to guarantee Takahashi's win; otherwise, print `-1` instead.
Examples
Input
2
5 3
Output
1
Input
2
3 5
Output
-1
Input
3
1 1 2
Output
-1
Input
8
10 9 8 7 6 5 4 3
Output
3
Input
3
4294967297 8589934593 12884901890
Output
1
Submitted Solution:
```
N = int(input())
A = list(map(int, input().split()))
P = 41
X = 0
for i in range(2, N):
X ^= A[i]
S = A[0] + A[1]
A1 = A[0]
D, M = divmod(S - X, 2)
ans = -1
if M % 2 == 0 and D & X == 0 and D <= A1:
Y = 0
for p in range(P, -1, -1):
if (X >> p) & 1 == 1:
new_Y = Y ^ (1 << p)
if D ^ new_Y <= A1:
Y = new_Y
ans = A1 - (D ^ Y)
if ans == A1:
ans = -1
print(ans)
``` | instruction | 0 | 55,384 | 19 | 110,768 |
Yes | output | 1 | 55,384 | 19 | 110,769 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N piles of stones. The i-th pile has A_i stones.
Aoki and Takahashi are about to use them to play the following game:
* Starting with Aoki, the two players alternately do the following operation:
* Operation: Choose one pile of stones, and remove one or more stones from it.
* When a player is unable to do the operation, he loses, and the other player wins.
When the two players play optimally, there are two possibilities in this game: the player who moves first always wins, or the player who moves second always wins, only depending on the initial number of stones in each pile.
In such a situation, Takahashi, the second player to act, is trying to guarantee his win by moving at least zero and at most (A_1 - 1) stones from the 1-st pile to the 2-nd pile before the game begins.
If this is possible, print the minimum number of stones to move to guarantee his victory; otherwise, print `-1` instead.
Constraints
* 2 \leq N \leq 300
* 1 \leq A_i \leq 10^{12}
Input
Input is given from Standard Input in the following format:
N
A_1 \ldots A_N
Output
Print the minimum number of stones to move to guarantee Takahashi's win; otherwise, print `-1` instead.
Examples
Input
2
5 3
Output
1
Input
2
3 5
Output
-1
Input
3
1 1 2
Output
-1
Input
8
10 9 8 7 6 5 4 3
Output
3
Input
3
4294967297 8589934593 12884901890
Output
1
Submitted Solution:
```
import sys
from functools import lru_cache
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
N, *A = map(int, read().split())
a, b = A[:2]
xor = 0
for x in A[2:]:
xor ^= x
INF = 10**13
@lru_cache(None)
def F(a, b, xor):
if (a & 1) ^ (b & 1) != (xor & 1):
return INF
if xor == 0:
return INF if a < b else (a - b) // 2
# 1の位を放置
x = 2 * F(a // 2, b // 2, xor // 2)
# 1の位を移動
y = 2 * F((a - 1) // 2, (b + 1) // 2, xor // 2) + 1
x = min(x, y)
if x >= INF:
x = INF
return x
x = F(a, b, xor)
if x >= a:
x = -1
print(x)
``` | instruction | 0 | 55,385 | 19 | 110,770 |
Yes | output | 1 | 55,385 | 19 | 110,771 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N piles of stones. The i-th pile has A_i stones.
Aoki and Takahashi are about to use them to play the following game:
* Starting with Aoki, the two players alternately do the following operation:
* Operation: Choose one pile of stones, and remove one or more stones from it.
* When a player is unable to do the operation, he loses, and the other player wins.
When the two players play optimally, there are two possibilities in this game: the player who moves first always wins, or the player who moves second always wins, only depending on the initial number of stones in each pile.
In such a situation, Takahashi, the second player to act, is trying to guarantee his win by moving at least zero and at most (A_1 - 1) stones from the 1-st pile to the 2-nd pile before the game begins.
If this is possible, print the minimum number of stones to move to guarantee his victory; otherwise, print `-1` instead.
Constraints
* 2 \leq N \leq 300
* 1 \leq A_i \leq 10^{12}
Input
Input is given from Standard Input in the following format:
N
A_1 \ldots A_N
Output
Print the minimum number of stones to move to guarantee Takahashi's win; otherwise, print `-1` instead.
Examples
Input
2
5 3
Output
1
Input
2
3 5
Output
-1
Input
3
1 1 2
Output
-1
Input
8
10 9 8 7 6 5 4 3
Output
3
Input
3
4294967297 8589934593 12884901890
Output
1
Submitted Solution:
```
import sys
sys.setrecursionlimit(10**7) #再帰関数の上限,10**5以上の場合python
import math
from copy import copy, deepcopy
from copy import deepcopy as dcp
from operator import itemgetter
from bisect import bisect_left, bisect, bisect_right#2分探索
#bisect_left(l,x), bisect(l,x)#aはソート済みである必要あり。aの中からx未満の要素数を返す。rightだと以下
from collections import deque
#deque(l), pop(), append(x), popleft(), appendleft(x)
#q.rotate(n)で → にn回ローテート
from collections import Counter#文字列を個数カウント辞書に、
#S=Counter(l),S.most_common(x),S.keys(),S.values(),S.items()
from itertools import accumulate,combinations,permutations#累積和
#list(accumulate(l))
from heapq import heapify,heappop,heappush
#heapify(q),heappush(q,a),heappop(q) #q=heapify(q)としないこと、返り値はNone
#import fractions#古いatcoderコンテストの場合GCDなどはここからimportする
from functools import lru_cache#pypyでもうごく
#@lru_cache(maxsize = None)#maxsizeは保存するデータ数の最大値、2**nが最も高効率
from decimal import Decimal
def input():
x=sys.stdin.readline()
return x[:-1] if x[-1]=="\n" else x
def printl(li): _=print(*li, sep="\n") if li else None
def argsort(s, return_sorted=False):
inds=sorted(range(len(s)), key=lambda k: s[k])
if return_sorted: return inds, [s[i] for i in inds]
return inds
def alp2num(c,cap=False): return ord(c)-97 if not cap else ord(c)-65
def num2alp(i,cap=False): return chr(i+97) if not cap else chr(i+65)
def matmat(A,B):
K,N,M=len(B),len(A),len(B[0])
return [[sum([(A[i][k]*B[k][j]) for k in range(K)]) for j in range(M)] for i in range(N)]
def matvec(M,v):
N,size=len(v),len(M)
return [sum([M[i][j]*v[j] for j in range(N)]) for i in range(size)]
def T(M):
n,m=len(M),len(M[0])
return [[M[j][i] for j in range(n)] for i in range(m)]
def main():
mod = 1000000007
#w.sort(key=itemgetter(1),reversed=True) #二個目の要素で降順並び替え
N = int(input())
#N, K = map(int, input().split())
A = tuple(map(int, input().split())) #1行ベクトル
#L = tuple(int(input()) for i in range(N)) #改行ベクトル
#S = tuple(tuple(map(int, input().split())) for i in range(N)) #改行行列
ref=0
for i in range(2,N):
ref^=A[i]
a0=A[0]
a1=A[1]
ans=0
#print(bin(a0),bin(a1),bin(ref))
def dfs(a0,a1,ref):
q=[(0,a0,a1,ref,0)]
visited=set()
while q:
tot,a0,a1,ref,j=heappop(q)#ここをpopleftにすると幅優先探索BFSになる
if a0 in visited:
continue
if a0^a1==ref:
return tot
visited.add(a0)
if ((a0>>j)&1)^((a1>>j)&1)!=(ref>>j)&1:
heappush(q,(tot,a0,a1,ref,j+1))
a0-=1<<j
a1+=1<<j
if a0>0:
heappush(q,(tot+(1<<j),a0,a1,ref,j+1))
return -1
print(dfs(a0,a1,ref))
# for j in range(0,41):
# a0j=(a0>>j)&1
# a1j=(a1>>j)&1
# refj=(ref>>j)&1
# #print(a0j,a1j,refj)
# if a0j^a1j==refj:
# nref=(ref>>(j+1))&1
# na0=((a0>>(j+1))&1)
# na1=((a1>>(j+1))&1)
# if ((a0>>(j+1))&1)^((a1>>(j+1))&1)!=nref:
# #print("a",na0,na1,nref)
# shif=1<<j
# if a0>0 and (((a0-shif))^(((a1+shif))==ref:
# ans+=shif
# a0-=shif
# a1+=shif
# continue
# else:
# shif=1<<j
# ans+=shif
# a0-=shif
# a1+=shif
# #print(shif,a0,a1)
# if a0<0:
# print(-1)
# return
if __name__ == "__main__":
main()
``` | instruction | 0 | 55,386 | 19 | 110,772 |
No | output | 1 | 55,386 | 19 | 110,773 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N piles of stones. The i-th pile has A_i stones.
Aoki and Takahashi are about to use them to play the following game:
* Starting with Aoki, the two players alternately do the following operation:
* Operation: Choose one pile of stones, and remove one or more stones from it.
* When a player is unable to do the operation, he loses, and the other player wins.
When the two players play optimally, there are two possibilities in this game: the player who moves first always wins, or the player who moves second always wins, only depending on the initial number of stones in each pile.
In such a situation, Takahashi, the second player to act, is trying to guarantee his win by moving at least zero and at most (A_1 - 1) stones from the 1-st pile to the 2-nd pile before the game begins.
If this is possible, print the minimum number of stones to move to guarantee his victory; otherwise, print `-1` instead.
Constraints
* 2 \leq N \leq 300
* 1 \leq A_i \leq 10^{12}
Input
Input is given from Standard Input in the following format:
N
A_1 \ldots A_N
Output
Print the minimum number of stones to move to guarantee Takahashi's win; otherwise, print `-1` instead.
Examples
Input
2
5 3
Output
1
Input
2
3 5
Output
-1
Input
3
1 1 2
Output
-1
Input
8
10 9 8 7 6 5 4 3
Output
3
Input
3
4294967297 8589934593 12884901890
Output
1
Submitted Solution:
```
from functools import lru_cache
def main():
N = int(input())
A = list(map(int, input().split()))
x = 0
for a in A[2:]:
x ^= a
a, b = A[0], A[1]
if (a ^ b ^ x) & 1 != 0:
return -1
@lru_cache(None)
def helper(a, b, x):
if a == 0 or b == 0:
return 0 if a ^ b == x else None
if (a ^ b) & 1 != x & 1:
return None
t1 = helper(a // 2, b // 2, x // 2)
t2 = helper((a - 1) // 2, (b + 1) // 2, x // 2)
if t1 is None and t2 is None:
return None
if t1 is None:
return t2 * 2 + 1
if t2 is None:
return t1 * 2
t1 = t1 * 2
t2 = t2 * 2 + 1
return min(t1, t2)
r = helper(a, b, x)
if r is None or r == a:
return -1
return r
print(main())
``` | instruction | 0 | 55,387 | 19 | 110,774 |
No | output | 1 | 55,387 | 19 | 110,775 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N piles of stones. The i-th pile has A_i stones.
Aoki and Takahashi are about to use them to play the following game:
* Starting with Aoki, the two players alternately do the following operation:
* Operation: Choose one pile of stones, and remove one or more stones from it.
* When a player is unable to do the operation, he loses, and the other player wins.
When the two players play optimally, there are two possibilities in this game: the player who moves first always wins, or the player who moves second always wins, only depending on the initial number of stones in each pile.
In such a situation, Takahashi, the second player to act, is trying to guarantee his win by moving at least zero and at most (A_1 - 1) stones from the 1-st pile to the 2-nd pile before the game begins.
If this is possible, print the minimum number of stones to move to guarantee his victory; otherwise, print `-1` instead.
Constraints
* 2 \leq N \leq 300
* 1 \leq A_i \leq 10^{12}
Input
Input is given from Standard Input in the following format:
N
A_1 \ldots A_N
Output
Print the minimum number of stones to move to guarantee Takahashi's win; otherwise, print `-1` instead.
Examples
Input
2
5 3
Output
1
Input
2
3 5
Output
-1
Input
3
1 1 2
Output
-1
Input
8
10 9 8 7 6 5 4 3
Output
3
Input
3
4294967297 8589934593 12884901890
Output
1
Submitted Solution:
```
import random
N = int(input())
A = list(map(int, input().split()))
X = A[0]
Y = A[1]
Z = 0
for a in A[2:]:
Z ^= a
if ((X^Y)&1)!=(Z&1):
print(-1)
exit(1)
def solve(X2, Y2):
if ((X2^Y2)&1)!=(Z&1):
return -1
d = 0
for i in range(1, 100):
if (((X2-d)^(Y2+d))>>i&1)!=(Z>>i&1):
d += 1<<(i-1)
if d<X2:
return d
else:
return -1
ans = solve(X, Y)
for i in range(10000):
if ans<=2:
break
r = random.randint(1, ans-1)
a = solve(X-r, Y+r)
if a!=-1 and r+a<ans:
ans = r+a
print(ans)
``` | instruction | 0 | 55,388 | 19 | 110,776 |
No | output | 1 | 55,388 | 19 | 110,777 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N piles of stones. The i-th pile has A_i stones.
Aoki and Takahashi are about to use them to play the following game:
* Starting with Aoki, the two players alternately do the following operation:
* Operation: Choose one pile of stones, and remove one or more stones from it.
* When a player is unable to do the operation, he loses, and the other player wins.
When the two players play optimally, there are two possibilities in this game: the player who moves first always wins, or the player who moves second always wins, only depending on the initial number of stones in each pile.
In such a situation, Takahashi, the second player to act, is trying to guarantee his win by moving at least zero and at most (A_1 - 1) stones from the 1-st pile to the 2-nd pile before the game begins.
If this is possible, print the minimum number of stones to move to guarantee his victory; otherwise, print `-1` instead.
Constraints
* 2 \leq N \leq 300
* 1 \leq A_i \leq 10^{12}
Input
Input is given from Standard Input in the following format:
N
A_1 \ldots A_N
Output
Print the minimum number of stones to move to guarantee Takahashi's win; otherwise, print `-1` instead.
Examples
Input
2
5 3
Output
1
Input
2
3 5
Output
-1
Input
3
1 1 2
Output
-1
Input
8
10 9 8 7 6 5 4 3
Output
3
Input
3
4294967297 8589934593 12884901890
Output
1
Submitted Solution:
```
n = int(input())
lis = list(map(int,input().split()))
num = 0
for i in range(2,n):
num ^= lis[i]
if lis[0] + lis[1] < num:
print(-1)
exit()
a = lis[0]
b = lis[1]
nu = a + b-num
if nu % 2 != 0:
print(-1)
exit()
a = format(a,"b")
b = format(b,"b")
num = format(num,"b")
l = max(len(a),len(b),len(num))
a = int(a,2)
b = int(b,2)
num = int(num,2)
a = format(a,"b").zfill(l)
b = format(b,"b").zfill(l)
num = format(num,"b").zfill(l)
ans = 0
for i in range(len(a)):
if b[-(i+1)] == "0" and num[-(i+1)] == "1":
if a[-(i+1)] != "1":
print(-1)
exit()
else:
ans += 2 ** (i+1)-1
a = int(a,2)
a -= 2 ** i
a = format(a,"b").zfill(l)
elif b[-(i+1)] == "1" and num[-(i+1)] == "0":
ans += 2 ** i
b = int(b,2)
b += 2 ** i
b = format(b,"b").zfill(l)
a = int(a,2)
a -= 2 ** i
if a < 0:
print(-1)
exit()
a = format(a,"b").zfill(l)
print(ans)
``` | instruction | 0 | 55,389 | 19 | 110,778 |
No | output | 1 | 55,389 | 19 | 110,779 |
Provide a correct Python 3 solution for this coding contest problem.
Let N be a positive odd number.
There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.
Taro has tossed all the N coins. Find the probability of having more heads than tails.
Constraints
* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1
Input
Input is given from Standard Input in the following format:
N
p_1 p_2 \ldots p_N
Output
Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.
Examples
Input
3
0.30 0.60 0.80
Output
0.612
Input
1
0.50
Output
0.5
Input
5
0.42 0.01 0.42 0.99 0.42
Output
0.3821815872 | instruction | 0 | 55,438 | 19 | 110,876 |
"Correct Solution:
```
N=int(input())
p=[float(x) for x in input().split()]
dp=[[0 for j in range(N+5)] for i in range(N+5)]
dp[0][0]=1
for i in range(N):
for j in range(i+1):
dp[i+1][j+1] += dp[i][j]*p[i]
dp[i+1][j] += dp[i][j]*(1-p[i])
M =(N+1)//2
result = 0
for i in range(M,N+1):
result += dp[N][i]
print(result)
``` | output | 1 | 55,438 | 19 | 110,877 |
Provide a correct Python 3 solution for this coding contest problem.
Let N be a positive odd number.
There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.
Taro has tossed all the N coins. Find the probability of having more heads than tails.
Constraints
* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1
Input
Input is given from Standard Input in the following format:
N
p_1 p_2 \ldots p_N
Output
Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.
Examples
Input
3
0.30 0.60 0.80
Output
0.612
Input
1
0.50
Output
0.5
Input
5
0.42 0.01 0.42 0.99 0.42
Output
0.3821815872 | instruction | 0 | 55,439 | 19 | 110,878 |
"Correct Solution:
```
input()
a=[1]
for i in input().split():
a=[x+(y-x)*float(i) for x,y in zip(a+[0],[0]+a)]
print(sum(a[len(a)//2:]))
``` | output | 1 | 55,439 | 19 | 110,879 |
Provide a correct Python 3 solution for this coding contest problem.
Let N be a positive odd number.
There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.
Taro has tossed all the N coins. Find the probability of having more heads than tails.
Constraints
* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1
Input
Input is given from Standard Input in the following format:
N
p_1 p_2 \ldots p_N
Output
Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.
Examples
Input
3
0.30 0.60 0.80
Output
0.612
Input
1
0.50
Output
0.5
Input
5
0.42 0.01 0.42 0.99 0.42
Output
0.3821815872 | instruction | 0 | 55,440 | 19 | 110,880 |
"Correct Solution:
```
N = int(input())
p = list(map(float,input().split()))
dp = [[0]*(N+1) for i in range(N)]
dp[0][0] = 1-p[0]
dp[0][1] = p[0]
for i in range(1,N):
dp[i][0] = (1-p[i])*dp[i-1][0]
for j in range(1,i+2):
dp[i][j] = dp[i-1][j-1]*p[i] + dp[i-1][j]*(1-p[i])
print(sum(dp[-1][(N+1)//2:]))
``` | output | 1 | 55,440 | 19 | 110,881 |
Provide a correct Python 3 solution for this coding contest problem.
Let N be a positive odd number.
There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.
Taro has tossed all the N coins. Find the probability of having more heads than tails.
Constraints
* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1
Input
Input is given from Standard Input in the following format:
N
p_1 p_2 \ldots p_N
Output
Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.
Examples
Input
3
0.30 0.60 0.80
Output
0.612
Input
1
0.50
Output
0.5
Input
5
0.42 0.01 0.42 0.99 0.42
Output
0.3821815872 | instruction | 0 | 55,441 | 19 | 110,882 |
"Correct Solution:
```
N = int(input())
P = list(map(float, input().split()))
dp = [[0] * (N + 1) for _ in range(N + 1)]
dp[0][0] = 1
for i in range(N):
p = P[i]
q = 1 - p
dp[i + 1][0] = dp[i][0] * q
for j in range(1, i + 2):
dp[i + 1][j] = dp[i][j - 1] * p + dp[i][j] * q
print (sum(dp[N][(N + 1) // 2:]))
``` | output | 1 | 55,441 | 19 | 110,883 |
Provide a correct Python 3 solution for this coding contest problem.
Let N be a positive odd number.
There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.
Taro has tossed all the N coins. Find the probability of having more heads than tails.
Constraints
* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1
Input
Input is given from Standard Input in the following format:
N
p_1 p_2 \ldots p_N
Output
Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.
Examples
Input
3
0.30 0.60 0.80
Output
0.612
Input
1
0.50
Output
0.5
Input
5
0.42 0.01 0.42 0.99 0.42
Output
0.3821815872 | instruction | 0 | 55,442 | 19 | 110,884 |
"Correct Solution:
```
n=int(input())
p=list(map(float,input().split()))
dp=[p[0],1-p[0]]
if n==1:
print(p[0])
exit()
for i in range(n-1):
tmp=[]
tmp.append(dp[0]*p[i+1])
for j in range(i+1):
tmp.append(dp[j]*(1-p[i+1])+dp[j+1]*p[i+1])
tmp.append(dp[-1]*(1-p[i+1]))
dp=tmp
print(sum(dp[:n//2+1]))
``` | output | 1 | 55,442 | 19 | 110,885 |
Provide a correct Python 3 solution for this coding contest problem.
Let N be a positive odd number.
There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.
Taro has tossed all the N coins. Find the probability of having more heads than tails.
Constraints
* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1
Input
Input is given from Standard Input in the following format:
N
p_1 p_2 \ldots p_N
Output
Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.
Examples
Input
3
0.30 0.60 0.80
Output
0.612
Input
1
0.50
Output
0.5
Input
5
0.42 0.01 0.42 0.99 0.42
Output
0.3821815872 | instruction | 0 | 55,443 | 19 | 110,886 |
"Correct Solution:
```
# -*- coding: utf-8 -*-
N = int(input())
P = list(map(float, input().split()))
dp = [[1 if k == 0 else 0 for k in range(N+1)] for _ in range(N+1)]
for i in range(1, N+1):
for j in range(1, i+1):
dp[i][j] = dp[i-1][j-1] * P[i-1] + dp[i-1][j] * (1-P[i-1])
print(dp[N][(N+1)//2])
``` | output | 1 | 55,443 | 19 | 110,887 |
Provide a correct Python 3 solution for this coding contest problem.
Let N be a positive odd number.
There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.
Taro has tossed all the N coins. Find the probability of having more heads than tails.
Constraints
* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1
Input
Input is given from Standard Input in the following format:
N
p_1 p_2 \ldots p_N
Output
Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.
Examples
Input
3
0.30 0.60 0.80
Output
0.612
Input
1
0.50
Output
0.5
Input
5
0.42 0.01 0.42 0.99 0.42
Output
0.3821815872 | instruction | 0 | 55,444 | 19 | 110,888 |
"Correct Solution:
```
N = int(input())
p = list(map(float, input().split()))
dp = [[0.] * (N+1) for _ in range(N+1)]
dp[0][0] = 1.
for i in range(1, N+1):
for j in range(i+1):
prob = dp[i-1][j-1]*p[i-1]
if i > 0:
prob += dp[i-1][j]*(1.-p[i-1])
dp[i][j] = max(dp[i][j], prob)
print(sum(dp[-1][N//2+1:]))
``` | output | 1 | 55,444 | 19 | 110,889 |
Provide a correct Python 3 solution for this coding contest problem.
Let N be a positive odd number.
There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.
Taro has tossed all the N coins. Find the probability of having more heads than tails.
Constraints
* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1
Input
Input is given from Standard Input in the following format:
N
p_1 p_2 \ldots p_N
Output
Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.
Examples
Input
3
0.30 0.60 0.80
Output
0.612
Input
1
0.50
Output
0.5
Input
5
0.42 0.01 0.42 0.99 0.42
Output
0.3821815872 | instruction | 0 | 55,445 | 19 | 110,890 |
"Correct Solution:
```
n=int(input())
p=list(map(float,input().split()))
dp=[[0]*(n+1) for _ in range(n+1)]
dp[0][0]=1
for i in range(1,n+1):
for j in range(i+1):
if j>0:
dp[i][j]+=dp[i-1][j-1]*p[i-1]
dp[i][j]+=dp[i-1][j]*(1-p[i-1])
print(sum(dp[n][n//2+1:]))
``` | output | 1 | 55,445 | 19 | 110,891 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let N be a positive odd number.
There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.
Taro has tossed all the N coins. Find the probability of having more heads than tails.
Constraints
* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1
Input
Input is given from Standard Input in the following format:
N
p_1 p_2 \ldots p_N
Output
Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.
Examples
Input
3
0.30 0.60 0.80
Output
0.612
Input
1
0.50
Output
0.5
Input
5
0.42 0.01 0.42 0.99 0.42
Output
0.3821815872
Submitted Solution:
```
N=int(input())
p=list(map(float,input().split()))
dp=[[0 for i in range(N+1)] for j in range(N+1)]
dp[0][0]=1
for i in range(1,N+1):
for j in range(N+1):
if j>0:
dp[i][j]+=dp[i-1][j]*(1-p[i-1])+dp[i-1][j-1]*p[i-1]
else:
dp[i][j]+=dp[i-1][j]*(1-p[i-1])
print(sum(dp[N][N//2+1:]))
``` | instruction | 0 | 55,446 | 19 | 110,892 |
Yes | output | 1 | 55,446 | 19 | 110,893 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let N be a positive odd number.
There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.
Taro has tossed all the N coins. Find the probability of having more heads than tails.
Constraints
* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1
Input
Input is given from Standard Input in the following format:
N
p_1 p_2 \ldots p_N
Output
Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.
Examples
Input
3
0.30 0.60 0.80
Output
0.612
Input
1
0.50
Output
0.5
Input
5
0.42 0.01 0.42 0.99 0.42
Output
0.3821815872
Submitted Solution:
```
n=int(input())
dp=[0]*(n+1)
dp[0]=1
pbl=list(map(float,input().split()))
for i in range(n):
for j in range(i+1,-1,-1):
dp[j]=dp[j-1]*pbl[i]+dp[j]*(1-pbl[i])
# print(dp)
ans=0
for i in range((n+1)//2,n+1):
ans+=dp[i]
print(ans)
``` | instruction | 0 | 55,447 | 19 | 110,894 |
Yes | output | 1 | 55,447 | 19 | 110,895 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let N be a positive odd number.
There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.
Taro has tossed all the N coins. Find the probability of having more heads than tails.
Constraints
* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1
Input
Input is given from Standard Input in the following format:
N
p_1 p_2 \ldots p_N
Output
Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.
Examples
Input
3
0.30 0.60 0.80
Output
0.612
Input
1
0.50
Output
0.5
Input
5
0.42 0.01 0.42 0.99 0.42
Output
0.3821815872
Submitted Solution:
```
n = int(input())
plst = list(map(float, input().split()))
dp = [0] * (n + 1)
dp[0] = 1
for p in plst:
for i in range(n, -1, -1):
if i != 0:
dp[i] = dp[i] * (1 - p) + dp[i - 1] * p
if i == 0:
dp[i] = dp[i] * (1 - p)
print(sum(dp[n // 2 + 1:]))
``` | instruction | 0 | 55,448 | 19 | 110,896 |
Yes | output | 1 | 55,448 | 19 | 110,897 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let N be a positive odd number.
There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.
Taro has tossed all the N coins. Find the probability of having more heads than tails.
Constraints
* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1
Input
Input is given from Standard Input in the following format:
N
p_1 p_2 \ldots p_N
Output
Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.
Examples
Input
3
0.30 0.60 0.80
Output
0.612
Input
1
0.50
Output
0.5
Input
5
0.42 0.01 0.42 0.99 0.42
Output
0.3821815872
Submitted Solution:
```
N = int(input())
P = tuple(map(float, input().split()))
dp = [[0 for _ in range(N + 1)] for __ in range(N + 1)]
dp[0][0] = 1
for i in range(N):
for j in range(N):
dp[i + 1][j + 1] += dp[i][j] * P[i]
dp[i + 1][j] += dp[i][j] * (1 - P[i])
print(sum(dp[N][(N + 1) // 2:]))
``` | instruction | 0 | 55,449 | 19 | 110,898 |
Yes | output | 1 | 55,449 | 19 | 110,899 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let N be a positive odd number.
There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.
Taro has tossed all the N coins. Find the probability of having more heads than tails.
Constraints
* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1
Input
Input is given from Standard Input in the following format:
N
p_1 p_2 \ldots p_N
Output
Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.
Examples
Input
3
0.30 0.60 0.80
Output
0.612
Input
1
0.50
Output
0.5
Input
5
0.42 0.01 0.42 0.99 0.42
Output
0.3821815872
Submitted Solution:
```
n=int(input())
p=list(map(float,input().split()))
dp=[[0] *(n+1) for _ in range(n+1)]
#print(dp)
dp[0][0]=1.0
for i in range(n):
for j in range(n):
dp[i+1][j+1]+=dp[i][j]*p[i]
dp[i+1][j]+=dp[i][j]*(1-p[i])
print(dp)
ans=0
for i in range((n+1)//2,n+1):
ans+=dp[n][i]
print("dp[n][i]",dp[n][i],"ans",ans)
print(ans)
``` | instruction | 0 | 55,450 | 19 | 110,900 |
No | output | 1 | 55,450 | 19 | 110,901 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let N be a positive odd number.
There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.
Taro has tossed all the N coins. Find the probability of having more heads than tails.
Constraints
* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1
Input
Input is given from Standard Input in the following format:
N
p_1 p_2 \ldots p_N
Output
Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.
Examples
Input
3
0.30 0.60 0.80
Output
0.612
Input
1
0.50
Output
0.5
Input
5
0.42 0.01 0.42 0.99 0.42
Output
0.3821815872
Submitted Solution:
```
n = int(input())
p = [float(x) for x in input().split()]
dp = [[0]*(n+1) for _ in range(n)] #dp[i][j] Prob of j heads in i tries
dp[0][1] = p[0]
dp[0][0] = 1-p[0]
for i in range(1, n):
dp[i][0] = dp[i-1][0]*(1-p[i])
for j in range(1,n + 1):
if j <= i + 1:
dp[i][j] = dp[i - 1][j - 1] * p[i] + dp[i - 1][j] * (1 - p[i])
sum = 0
for i in range(n + 1):
if i > n - i:
sum += dp[n-1][i]
print(sum)
``` | instruction | 0 | 55,451 | 19 | 110,902 |
No | output | 1 | 55,451 | 19 | 110,903 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let N be a positive odd number.
There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.
Taro has tossed all the N coins. Find the probability of having more heads than tails.
Constraints
* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1
Input
Input is given from Standard Input in the following format:
N
p_1 p_2 \ldots p_N
Output
Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.
Examples
Input
3
0.30 0.60 0.80
Output
0.612
Input
1
0.50
Output
0.5
Input
5
0.42 0.01 0.42 0.99 0.42
Output
0.3821815872
Submitted Solution:
```
n = int(input())
a = list(map(float, input().split()))
dp = [[None] * 1501 for i in range(3000)]
def dfs(i=0, at=(n + 1) // 2):
if i == n:
return int(at <= 0)
if dp[i][at] is not None:
return dp[i][at]
r = dfs(i + 1, at - 1) * a[i] + dfs(i + 1, at) * (1 - a[i])
dp[i][at] = r
return r
print(dfs())
``` | instruction | 0 | 55,452 | 19 | 110,904 |
No | output | 1 | 55,452 | 19 | 110,905 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let N be a positive odd number.
There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.
Taro has tossed all the N coins. Find the probability of having more heads than tails.
Constraints
* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1
Input
Input is given from Standard Input in the following format:
N
p_1 p_2 \ldots p_N
Output
Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.
Examples
Input
3
0.30 0.60 0.80
Output
0.612
Input
1
0.50
Output
0.5
Input
5
0.42 0.01 0.42 0.99 0.42
Output
0.3821815872
Submitted Solution:
```
# cook your dish here
N = int(input())
p = list(map(float, input().split()))
#p = [0.42, 0.01, 0.42, 0.99, 0.42]
dp = [0]*(len(p)+1)
dp[0] = 1
answer = 0
for curr_toss in range(1, len(p)+1):
for hd_tn in range(curr_toss, 0,-1):
#print(curr_toss, hd_tn)
dp[hd_tn] = p[curr_toss-1]*dp[hd_tn-1] + (1-p[curr_toss-1])*dp[hd_tn]
dp[hd_tn-1] = (1-p[curr_toss-1])*dp[hd_tn-1]
#print(dp)
for heads in range(1, len(p)+1):
tails = len(p)-heads
if heads > tails:
answer += dp[heads]
print(answer)
``` | instruction | 0 | 55,453 | 19 | 110,906 |
No | output | 1 | 55,453 | 19 | 110,907 |
Provide a correct Python 3 solution for this coding contest problem.
Taro and Hanako are playing card games. They have n cards each, and they compete n turns. At each turn Taro and Hanako respectively puts out a card. The name of the animal consisting of alphabetical letters is written on each card, and the bigger one in lexicographical order becomes the winner of that turn. The winner obtains 3 points. In the case of a draw, they obtain 1 point each.
Write a program which reads a sequence of cards Taro and Hanako have and reports the final scores of the game.
Constraints
* n ≤ 1000
* The length of the string ≤ 100
Input
In the first line, the number of cards n is given. In the following n lines, the cards for n turns are given respectively. For each line, the first string represents the Taro's card and the second one represents Hanako's card.
Output
Print the final scores of Taro and Hanako respectively. Put a single space character between them.
Example
Input
3
cat dog
fish fish
lion tiger
Output
1 7 | instruction | 0 | 55,587 | 19 | 111,174 |
"Correct Solution:
```
n = int(input())
ts,hs = 0,0
for i in range(n):
t,h = input().split()
if t > h:
ts += 3
elif t < h:
hs += 3
else:
ts += 1
hs += 1
print(ts,hs)
``` | output | 1 | 55,587 | 19 | 111,175 |
Provide a correct Python 3 solution for this coding contest problem.
Taro and Hanako are playing card games. They have n cards each, and they compete n turns. At each turn Taro and Hanako respectively puts out a card. The name of the animal consisting of alphabetical letters is written on each card, and the bigger one in lexicographical order becomes the winner of that turn. The winner obtains 3 points. In the case of a draw, they obtain 1 point each.
Write a program which reads a sequence of cards Taro and Hanako have and reports the final scores of the game.
Constraints
* n ≤ 1000
* The length of the string ≤ 100
Input
In the first line, the number of cards n is given. In the following n lines, the cards for n turns are given respectively. For each line, the first string represents the Taro's card and the second one represents Hanako's card.
Output
Print the final scores of Taro and Hanako respectively. Put a single space character between them.
Example
Input
3
cat dog
fish fish
lion tiger
Output
1 7 | instruction | 0 | 55,588 | 19 | 111,176 |
"Correct Solution:
```
N=int(input())
t=int(0)
h=int(0)
for i in range(N):
T,H=input().split()
if T>H:
t=t+3
elif T<H:
h=h+3
elif H==T:
t=t+1
h=h+1
print (t,h)
``` | output | 1 | 55,588 | 19 | 111,177 |
Provide a correct Python 3 solution for this coding contest problem.
Taro and Hanako are playing card games. They have n cards each, and they compete n turns. At each turn Taro and Hanako respectively puts out a card. The name of the animal consisting of alphabetical letters is written on each card, and the bigger one in lexicographical order becomes the winner of that turn. The winner obtains 3 points. In the case of a draw, they obtain 1 point each.
Write a program which reads a sequence of cards Taro and Hanako have and reports the final scores of the game.
Constraints
* n ≤ 1000
* The length of the string ≤ 100
Input
In the first line, the number of cards n is given. In the following n lines, the cards for n turns are given respectively. For each line, the first string represents the Taro's card and the second one represents Hanako's card.
Output
Print the final scores of Taro and Hanako respectively. Put a single space character between them.
Example
Input
3
cat dog
fish fish
lion tiger
Output
1 7 | instruction | 0 | 55,589 | 19 | 111,178 |
"Correct Solution:
```
num=int(input())
H=0
T=0
for i in range(num):
a=input().split()
if a[0]<a[1]:
T+=3
elif a[1]<a[0]:
H+=3
else:
T+=1
H+=1
print(str(H)+' '+str(T))
``` | output | 1 | 55,589 | 19 | 111,179 |
Provide a correct Python 3 solution for this coding contest problem.
Taro and Hanako are playing card games. They have n cards each, and they compete n turns. At each turn Taro and Hanako respectively puts out a card. The name of the animal consisting of alphabetical letters is written on each card, and the bigger one in lexicographical order becomes the winner of that turn. The winner obtains 3 points. In the case of a draw, they obtain 1 point each.
Write a program which reads a sequence of cards Taro and Hanako have and reports the final scores of the game.
Constraints
* n ≤ 1000
* The length of the string ≤ 100
Input
In the first line, the number of cards n is given. In the following n lines, the cards for n turns are given respectively. For each line, the first string represents the Taro's card and the second one represents Hanako's card.
Output
Print the final scores of Taro and Hanako respectively. Put a single space character between them.
Example
Input
3
cat dog
fish fish
lion tiger
Output
1 7 | instruction | 0 | 55,590 | 19 | 111,180 |
"Correct Solution:
```
t = h = 0
for _ in range(int(input())):
a, b = input().split()
if a > b: t += 3
elif a < b: h += 3
else: t += 1; h += 1
print(t, h)
``` | output | 1 | 55,590 | 19 | 111,181 |
Provide a correct Python 3 solution for this coding contest problem.
Taro and Hanako are playing card games. They have n cards each, and they compete n turns. At each turn Taro and Hanako respectively puts out a card. The name of the animal consisting of alphabetical letters is written on each card, and the bigger one in lexicographical order becomes the winner of that turn. The winner obtains 3 points. In the case of a draw, they obtain 1 point each.
Write a program which reads a sequence of cards Taro and Hanako have and reports the final scores of the game.
Constraints
* n ≤ 1000
* The length of the string ≤ 100
Input
In the first line, the number of cards n is given. In the following n lines, the cards for n turns are given respectively. For each line, the first string represents the Taro's card and the second one represents Hanako's card.
Output
Print the final scores of Taro and Hanako respectively. Put a single space character between them.
Example
Input
3
cat dog
fish fish
lion tiger
Output
1 7 | instruction | 0 | 55,591 | 19 | 111,182 |
"Correct Solution:
```
n=int(input())
t=0
h=0
for i in range(n):
taro,hana=map(str,input().split())
if taro>hana:
t+=3
elif hana>taro:
h+=3
else:
h+=1
t+=1
print("%s %d"%(t,h))
``` | output | 1 | 55,591 | 19 | 111,183 |
Provide a correct Python 3 solution for this coding contest problem.
Taro and Hanako are playing card games. They have n cards each, and they compete n turns. At each turn Taro and Hanako respectively puts out a card. The name of the animal consisting of alphabetical letters is written on each card, and the bigger one in lexicographical order becomes the winner of that turn. The winner obtains 3 points. In the case of a draw, they obtain 1 point each.
Write a program which reads a sequence of cards Taro and Hanako have and reports the final scores of the game.
Constraints
* n ≤ 1000
* The length of the string ≤ 100
Input
In the first line, the number of cards n is given. In the following n lines, the cards for n turns are given respectively. For each line, the first string represents the Taro's card and the second one represents Hanako's card.
Output
Print the final scores of Taro and Hanako respectively. Put a single space character between them.
Example
Input
3
cat dog
fish fish
lion tiger
Output
1 7 | instruction | 0 | 55,592 | 19 | 111,184 |
"Correct Solution:
```
n = int(input())
tp, hp = 0, 0
for _ in range(n):
t, h = input().split()
if t == h:
tp += 1
hp += 1
elif t > h:
tp += 3
else:
hp += 3
print(tp, hp)
``` | output | 1 | 55,592 | 19 | 111,185 |
Provide a correct Python 3 solution for this coding contest problem.
Taro and Hanako are playing card games. They have n cards each, and they compete n turns. At each turn Taro and Hanako respectively puts out a card. The name of the animal consisting of alphabetical letters is written on each card, and the bigger one in lexicographical order becomes the winner of that turn. The winner obtains 3 points. In the case of a draw, they obtain 1 point each.
Write a program which reads a sequence of cards Taro and Hanako have and reports the final scores of the game.
Constraints
* n ≤ 1000
* The length of the string ≤ 100
Input
In the first line, the number of cards n is given. In the following n lines, the cards for n turns are given respectively. For each line, the first string represents the Taro's card and the second one represents Hanako's card.
Output
Print the final scores of Taro and Hanako respectively. Put a single space character between them.
Example
Input
3
cat dog
fish fish
lion tiger
Output
1 7 | instruction | 0 | 55,593 | 19 | 111,186 |
"Correct Solution:
```
n = int(input())
tp = hp = 0
for i in range(n):
t, h = input().split()
if t == h:
tp += 1
hp += 1
elif t > h:
tp += 3
else:
hp += 3
print(f"{tp} {hp}")
``` | output | 1 | 55,593 | 19 | 111,187 |
Provide a correct Python 3 solution for this coding contest problem.
Taro and Hanako are playing card games. They have n cards each, and they compete n turns. At each turn Taro and Hanako respectively puts out a card. The name of the animal consisting of alphabetical letters is written on each card, and the bigger one in lexicographical order becomes the winner of that turn. The winner obtains 3 points. In the case of a draw, they obtain 1 point each.
Write a program which reads a sequence of cards Taro and Hanako have and reports the final scores of the game.
Constraints
* n ≤ 1000
* The length of the string ≤ 100
Input
In the first line, the number of cards n is given. In the following n lines, the cards for n turns are given respectively. For each line, the first string represents the Taro's card and the second one represents Hanako's card.
Output
Print the final scores of Taro and Hanako respectively. Put a single space character between them.
Example
Input
3
cat dog
fish fish
lion tiger
Output
1 7 | instruction | 0 | 55,594 | 19 | 111,188 |
"Correct Solution:
```
n=int(input())
a=0
b=0
for i in range(n):
A,B=map(str,input().split())
if A>B:
a+=3
elif A==B:
a+=1
b+=1
else:
b+=3
print(a,b)
``` | output | 1 | 55,594 | 19 | 111,189 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Taro and Hanako are playing card games. They have n cards each, and they compete n turns. At each turn Taro and Hanako respectively puts out a card. The name of the animal consisting of alphabetical letters is written on each card, and the bigger one in lexicographical order becomes the winner of that turn. The winner obtains 3 points. In the case of a draw, they obtain 1 point each.
Write a program which reads a sequence of cards Taro and Hanako have and reports the final scores of the game.
Constraints
* n ≤ 1000
* The length of the string ≤ 100
Input
In the first line, the number of cards n is given. In the following n lines, the cards for n turns are given respectively. For each line, the first string represents the Taro's card and the second one represents Hanako's card.
Output
Print the final scores of Taro and Hanako respectively. Put a single space character between them.
Example
Input
3
cat dog
fish fish
lion tiger
Output
1 7
Submitted Solution:
```
n=int(input())
A=0
B=0
for i in range(n):
a,b=input().split()
if a>b:
A+=3
elif a==b:
A+=1
B+=1
else:
B+=3
print(A,B)
``` | instruction | 0 | 55,595 | 19 | 111,190 |
Yes | output | 1 | 55,595 | 19 | 111,191 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Taro and Hanako are playing card games. They have n cards each, and they compete n turns. At each turn Taro and Hanako respectively puts out a card. The name of the animal consisting of alphabetical letters is written on each card, and the bigger one in lexicographical order becomes the winner of that turn. The winner obtains 3 points. In the case of a draw, they obtain 1 point each.
Write a program which reads a sequence of cards Taro and Hanako have and reports the final scores of the game.
Constraints
* n ≤ 1000
* The length of the string ≤ 100
Input
In the first line, the number of cards n is given. In the following n lines, the cards for n turns are given respectively. For each line, the first string represents the Taro's card and the second one represents Hanako's card.
Output
Print the final scores of Taro and Hanako respectively. Put a single space character between them.
Example
Input
3
cat dog
fish fish
lion tiger
Output
1 7
Submitted Solution:
```
n = int(input())
xx = 0
yy = 0
for i in range(n):
x,y=map(str,input().split())
if x>y:
xx+=3
elif x<y:
yy+=3
else:
xx+=1
yy+=1
print(xx,yy)
``` | instruction | 0 | 55,596 | 19 | 111,192 |
Yes | output | 1 | 55,596 | 19 | 111,193 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Taro and Hanako are playing card games. They have n cards each, and they compete n turns. At each turn Taro and Hanako respectively puts out a card. The name of the animal consisting of alphabetical letters is written on each card, and the bigger one in lexicographical order becomes the winner of that turn. The winner obtains 3 points. In the case of a draw, they obtain 1 point each.
Write a program which reads a sequence of cards Taro and Hanako have and reports the final scores of the game.
Constraints
* n ≤ 1000
* The length of the string ≤ 100
Input
In the first line, the number of cards n is given. In the following n lines, the cards for n turns are given respectively. For each line, the first string represents the Taro's card and the second one represents Hanako's card.
Output
Print the final scores of Taro and Hanako respectively. Put a single space character between them.
Example
Input
3
cat dog
fish fish
lion tiger
Output
1 7
Submitted Solution:
```
N=int(input())
s=0
t=0
for i in range(N):
a,b=input().split()
if a<b:
s+=3
elif a==b:
s+=1
t+=1
else:
t+=3
print(t,s)
``` | instruction | 0 | 55,597 | 19 | 111,194 |
Yes | output | 1 | 55,597 | 19 | 111,195 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Taro and Hanako are playing card games. They have n cards each, and they compete n turns. At each turn Taro and Hanako respectively puts out a card. The name of the animal consisting of alphabetical letters is written on each card, and the bigger one in lexicographical order becomes the winner of that turn. The winner obtains 3 points. In the case of a draw, they obtain 1 point each.
Write a program which reads a sequence of cards Taro and Hanako have and reports the final scores of the game.
Constraints
* n ≤ 1000
* The length of the string ≤ 100
Input
In the first line, the number of cards n is given. In the following n lines, the cards for n turns are given respectively. For each line, the first string represents the Taro's card and the second one represents Hanako's card.
Output
Print the final scores of Taro and Hanako respectively. Put a single space character between them.
Example
Input
3
cat dog
fish fish
lion tiger
Output
1 7
Submitted Solution:
```
t = int(input())
q,w = 0,0
for i in range(t):
a,b = input().split()
if a == b:
q += 1
w += 1
elif a<b :
w+=3
elif a>b:
q+=3
print(q,w)
``` | instruction | 0 | 55,598 | 19 | 111,196 |
Yes | output | 1 | 55,598 | 19 | 111,197 |
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