message stringlengths 2 19.9k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 322 108k | cluster float64 15 15 | __index_level_0__ int64 644 217k |
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Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This morning Tolik has understood that while he was sleeping he had invented an incredible problem which will be a perfect fit for Codeforces! But, as a "Discuss tasks" project hasn't been born yet (in English, well), he decides to test a problem and asks his uncle.
After a long time thinking, Tolik's uncle hasn't any ideas on how to solve it. But, he doesn't want to tell Tolik about his inability to solve it, so he hasn't found anything better than asking you how to solve this task.
In this task you are given a cell field n ⋅ m, consisting of n rows and m columns, where point's coordinates (x, y) mean it is situated in the x-th row and y-th column, considering numeration from one (1 ≤ x ≤ n, 1 ≤ y ≤ m). Initially, you stand in the cell (1, 1). Every move you can jump from cell (x, y), which you stand in, by any non-zero vector (dx, dy), thus you will stand in the (x+dx, y+dy) cell. Obviously, you can't leave the field, but also there is one more important condition — you're not allowed to use one vector twice. Your task is to visit each cell of the field exactly once (the initial cell is considered as already visited).
Tolik's uncle is a very respectful person. Help him to solve this task!
Input
The first and only line contains two positive integers n, m (1 ≤ n ⋅ m ≤ 10^{6}) — the number of rows and columns of the field respectively.
Output
Print "-1" (without quotes) if it is impossible to visit every cell exactly once.
Else print n ⋅ m pairs of integers, i-th from them should contain two integers x_i, y_i (1 ≤ x_i ≤ n, 1 ≤ y_i ≤ m) — cells of the field in order of visiting, so that all of them are distinct and vectors of jumps between them are distinct too.
Notice that the first cell should have (1, 1) coordinates, according to the statement.
Examples
Input
2 3
Output
1 1
1 3
1 2
2 2
2 3
2 1
Input
1 1
Output
1 1
Note
The vectors from the first example in the order of making jumps are (0, 2), (0, -1), (1, 0), (0, 1), (0, -2).
Submitted Solution:
```
# [0,0,0]
# [0,0,0]
# [0,0,0]
# [1,0,0]
# [0,0,0]
# [0,0,0]
# [1,0,1]
# [0,0,0]
# [0,0,0]
# [1,1,1]
# [0,0,0]
# [0,0,0]
# [1,1,1]
# [0,0,1]
# [0,0,0]
# [0,0,0] 3,2
#
# 0,0 3,2
# 0,1 3,1
# 0,2 3,0
# 1,0 2,2
# 1,1 2,1
# 1,2 2,0
n, m = map(int, input().split())
if n % 2 == 0:
for i in range(int(n / 2)):
for j in range(m):
print(f'{i+1} {j+1}')
print(f'{n-i} {m-j}')
else:
for i in range(int(n / 2)):
for j in range(m):
print(f'{i+1} {j+1}')
print(f'{n-i} {m-j}')
mid = int(n/2)
for j in range(m//2):
print(f'{mid+1} {j+1}')
print(f'{mid+1} {m-j}')
if m % 2 != 0:
print(f'{n//2+1} {m//2+1}')
"""
0 0 0
0 0 0
1 0 0
0 0 0
1 0 0
0 0 1
1 1 0
0 0 1
1 1 0
0 1 1
1 1 1
0 1 1
1 1 1
1 1 1
"""
``` | instruction | 0 | 88,446 | 15 | 176,892 |
No | output | 1 | 88,446 | 15 | 176,893 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Two chess pieces, a rook and a knight, stand on a standard chessboard 8 × 8 in size. The positions in which they are situated are known. It is guaranteed that none of them beats the other one.
Your task is to find the number of ways to place another knight on the board so that none of the three pieces on the board beat another one. A new piece can only be placed on an empty square.
Input
The first input line contains the description of the rook's position on the board. This description is a line which is 2 in length. Its first symbol is a lower-case Latin letter from a to h, and its second symbol is a number from 1 to 8. The second line contains the description of the knight's position in a similar way. It is guaranteed that their positions do not coincide.
Output
Print a single number which is the required number of ways.
Examples
Input
a1
b2
Output
44
Input
a8
d4
Output
38 | instruction | 0 | 88,767 | 15 | 177,534 |
Tags: brute force, implementation, math
Correct Solution:
```
#!/usr/bin/env python3
def is_valid(r, k, new):
if r[0] == new[0]:
return False
if r[1] == new[1]:
return False
if abs(r[0] - new[0]) == 2 and abs(r[1] - new[1]) == 1:
return False
if abs(r[0] - new[0]) == 1 and abs(r[1] - new[1]) == 2:
return False
if abs(k[0] - new[0]) == 2 and abs(k[1] - new[1]) == 1:
return False
if abs(k[0] - new[0]) == 1 and abs(k[1] - new[1]) == 2:
return False
return True
rook = [ord(c) for c in input()]
knight = [ord(c) for c in input()]
ok = 0
for i in range(ord('a'), ord('i')):
for j in range(ord('1'), ord('9')):
if is_valid(rook, knight, [i, j]):
ok += 1
print(ok-1)
``` | output | 1 | 88,767 | 15 | 177,535 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Two chess pieces, a rook and a knight, stand on a standard chessboard 8 × 8 in size. The positions in which they are situated are known. It is guaranteed that none of them beats the other one.
Your task is to find the number of ways to place another knight on the board so that none of the three pieces on the board beat another one. A new piece can only be placed on an empty square.
Input
The first input line contains the description of the rook's position on the board. This description is a line which is 2 in length. Its first symbol is a lower-case Latin letter from a to h, and its second symbol is a number from 1 to 8. The second line contains the description of the knight's position in a similar way. It is guaranteed that their positions do not coincide.
Output
Print a single number which is the required number of ways.
Examples
Input
a1
b2
Output
44
Input
a8
d4
Output
38 | instruction | 0 | 88,768 | 15 | 177,536 |
Tags: brute force, implementation, math
Correct Solution:
```
def knightFields(i,j):
result, pos =[], [(i+2, j+1), (i+2, j-1), (i-2, j+1), (i-2, j-1), (i+1, j+2), (i-1, j+2), (i+1, j-2), (i-1, j-2)]
for k in range(8):
if pos[k][0] >= 0 and pos[k][1] >= 0 and pos[k][0] < 8 and pos[k][1] < 8: result.append(pos[k])
#print(result)
return result;
rook,knight,textToInt = input(),input(),{'a':0,'b':1,'c':2,'d':3,'e':4,'f':5,'g':6,'h':7}
rook = (textToInt[rook[0]], int(rook[1])-1)
knight = (textToInt[knight[0]], int(knight[1])-1)
fields = [[True for _ in range(8)] for __ in range(8)]
fields[knight[0]][knight[1]] =False
for i in range(8):
for j in range(8):
if i == rook[0]: fields[i][j] = False
elif j == rook[1]: fields[i][j] = False
for field in knightFields(knight[0],knight[1]):
fields[field[0]][field[1]] = False
for field in knightFields(rook[0],rook[1]):
fields[field[0]][field[1]] = False
count = 0
for field in fields:
# print(field)
for f in field:
if f: count +=1
print(count)
``` | output | 1 | 88,768 | 15 | 177,537 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Two chess pieces, a rook and a knight, stand on a standard chessboard 8 × 8 in size. The positions in which they are situated are known. It is guaranteed that none of them beats the other one.
Your task is to find the number of ways to place another knight on the board so that none of the three pieces on the board beat another one. A new piece can only be placed on an empty square.
Input
The first input line contains the description of the rook's position on the board. This description is a line which is 2 in length. Its first symbol is a lower-case Latin letter from a to h, and its second symbol is a number from 1 to 8. The second line contains the description of the knight's position in a similar way. It is guaranteed that their positions do not coincide.
Output
Print a single number which is the required number of ways.
Examples
Input
a1
b2
Output
44
Input
a8
d4
Output
38 | instruction | 0 | 88,769 | 15 | 177,538 |
Tags: brute force, implementation, math
Correct Solution:
```
rook, horse = input(), input()
beat = {i: {j: False for j in range(1, 9)} for i in range(1, 9)}
key = {'a': 1, 'b': 2, 'c': 3, 'd': 4, 'e': 5, 'f': 6, 'g': 7, 'h': 8}
r = (key[rook[0]], int(rook[1]))
h = (key[horse[0]], int(horse[1]))
dx = (2, 1, -1, -2, -2, -1, 1, 2)
dy = (1, 2, 2, 1, -1, -2, -2, -1)
for i in range(1, 9):
beat[i][r[1]] = True
beat[r[0]][i] = True
for i in range(8):
x, y = r[0] + dx[i], r[1] + dy[i]
if 1 <= x <= 8 and 1 <= y <= 8:
beat[x][y] = True
x, y = h[0] + dx[i], h[1] + dy[i]
if 1 <= x <= 8 and 1 <= y <= 8:
beat[x][y] = True
beat[h[0]][h[1]] = True
ans = 0
for i in range(1, 9):
for j in range(1, 9):
if beat[i][j] == False:
ans += 1
print(ans)
``` | output | 1 | 88,769 | 15 | 177,539 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Two chess pieces, a rook and a knight, stand on a standard chessboard 8 × 8 in size. The positions in which they are situated are known. It is guaranteed that none of them beats the other one.
Your task is to find the number of ways to place another knight on the board so that none of the three pieces on the board beat another one. A new piece can only be placed on an empty square.
Input
The first input line contains the description of the rook's position on the board. This description is a line which is 2 in length. Its first symbol is a lower-case Latin letter from a to h, and its second symbol is a number from 1 to 8. The second line contains the description of the knight's position in a similar way. It is guaranteed that their positions do not coincide.
Output
Print a single number which is the required number of ways.
Examples
Input
a1
b2
Output
44
Input
a8
d4
Output
38 | instruction | 0 | 88,770 | 15 | 177,540 |
Tags: brute force, implementation, math
Correct Solution:
```
def chk(x,y,x1,y1,x2,y2):
if x==x1 or y==y1:
return False
if x==x2 and y==y2:
return False
if abs(x-x1)*abs(y-y1)==2:
return False
if abs(x-x2)*abs(y-y2)==2:
return False
return True
s=input()
x1=ord(s[0])-ord('a')
y1=int(s[1])-1
s=input()
x2=ord(s[0])-ord('a')
y2=int(s[1])-1
ans=0
for i in range(8):
for j in range(8):
if chk(i,j,x1,y1,x2,y2):
ans+=1
print(ans)
``` | output | 1 | 88,770 | 15 | 177,541 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Two chess pieces, a rook and a knight, stand on a standard chessboard 8 × 8 in size. The positions in which they are situated are known. It is guaranteed that none of them beats the other one.
Your task is to find the number of ways to place another knight on the board so that none of the three pieces on the board beat another one. A new piece can only be placed on an empty square.
Input
The first input line contains the description of the rook's position on the board. This description is a line which is 2 in length. Its first symbol is a lower-case Latin letter from a to h, and its second symbol is a number from 1 to 8. The second line contains the description of the knight's position in a similar way. It is guaranteed that their positions do not coincide.
Output
Print a single number which is the required number of ways.
Examples
Input
a1
b2
Output
44
Input
a8
d4
Output
38 | instruction | 0 | 88,771 | 15 | 177,542 |
Tags: brute force, implementation, math
Correct Solution:
```
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""
Created on Fri Jan 17 15:11:07 2020
@author: thom
"""
pos_rook = input()
pos_knight = input()
#pos_rook = "a1"
#pos_knight = "b2"
import string
letter_rook = string.ascii_lowercase.index(pos_rook[0])
number_rook = int(pos_rook[1]) - 1
letter_knight = string.ascii_lowercase.index(pos_knight[0])
number_knight = int(pos_knight[1]) - 1
#rook = [[0,0,0,0,0,0,0,0],[0,0,0,0,0,0,0,0],[0,0,0,0,0,0,0,0],[0,0,0,0,0,0,0,0],
# [0,0,0,0,0,0,0,0],[0,0,0,0,0,0,0,0],[0,0,0,0,0,0,0,0],[0,0,0,0,0,0,0,0]]
rook = [[0 for i in range(8)] for j in range(8)]
knight = [[0 for i in range(8)] for j in range(8)]
def prohibit_knight (board, letter, number):
for i in range(8):
for j in range(8):
if ((i - letter)**2 + (j-number)**2) == 5:
board[i][j] = 1
if (i ==letter and j == number):
board[i][j] = 1
return(board)
def prohibit_rook (board, letter, number):
prohibit_knight(board, letter, number)
for i in range(8):
board[letter][i] = 1
for i in range(8):
board[i][number] = 1
return(board)
v1 = prohibit_knight(knight, letter_knight, number_knight)
v2 = prohibit_rook(rook, letter_rook, number_rook)
v3 = [[0 for i in range(8)] for j in range(8)]
count = 0
for i in range(8):
for j in range(8):
v3[i][j] = v1[i][j] or v2[i][j]
count = count + v3[i][j]
print(64 - count)
``` | output | 1 | 88,771 | 15 | 177,543 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Two chess pieces, a rook and a knight, stand on a standard chessboard 8 × 8 in size. The positions in which they are situated are known. It is guaranteed that none of them beats the other one.
Your task is to find the number of ways to place another knight on the board so that none of the three pieces on the board beat another one. A new piece can only be placed on an empty square.
Input
The first input line contains the description of the rook's position on the board. This description is a line which is 2 in length. Its first symbol is a lower-case Latin letter from a to h, and its second symbol is a number from 1 to 8. The second line contains the description of the knight's position in a similar way. It is guaranteed that their positions do not coincide.
Output
Print a single number which is the required number of ways.
Examples
Input
a1
b2
Output
44
Input
a8
d4
Output
38 | instruction | 0 | 88,772 | 15 | 177,544 |
Tags: brute force, implementation, math
Correct Solution:
```
import sys
import math
import bisect
def is_knight_attack(a, b):
if abs(a[0] - b[0]) == 1 and abs(a[1] - b[1]) == 2:
return True
if abs(a[0] - b[0]) == 2 and abs(a[1] - b[1]) == 1:
return True
return False
def solve(a, b):
ans = 0
for i in range(8):
for j in range(8):
c = (i, j)
if c != a and c != b:
ok = True
#print('a: %s, b: %s, c: %s' % (str(a), str(b), str(c)))
if c[0] == a[0]:
ok = False
if c[1] == a[1]:
ok = False
if is_knight_attack(a, c):
ok = False
if is_knight_attack(b, c):
ok = False
if ok:
ans += 1
return ans
def main():
s = input()
A = (ord(s[0]) - ord('a'), int(s[1]) - 1)
s = input()
B = (ord(s[0]) - ord('a'), int(s[1]) - 1)
print(solve(A, B))
if __name__ == "__main__":
main()
``` | output | 1 | 88,772 | 15 | 177,545 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Two chess pieces, a rook and a knight, stand on a standard chessboard 8 × 8 in size. The positions in which they are situated are known. It is guaranteed that none of them beats the other one.
Your task is to find the number of ways to place another knight on the board so that none of the three pieces on the board beat another one. A new piece can only be placed on an empty square.
Input
The first input line contains the description of the rook's position on the board. This description is a line which is 2 in length. Its first symbol is a lower-case Latin letter from a to h, and its second symbol is a number from 1 to 8. The second line contains the description of the knight's position in a similar way. It is guaranteed that their positions do not coincide.
Output
Print a single number which is the required number of ways.
Examples
Input
a1
b2
Output
44
Input
a8
d4
Output
38 | instruction | 0 | 88,773 | 15 | 177,546 |
Tags: brute force, implementation, math
Correct Solution:
```
import string
import re
pattern = re.compile('^[abcdefgh][12345678]$')
r = input()
k = input()
r_d = [string.ascii_lowercase.index(r[0]), int(r[1])]
k_d = [string.ascii_lowercase.index(k[0]), int(k[1])]
bad1 = [r[0]+str(i) for i in range(1,9)] # rook can take knight 2
bad2 = [j+r[1] for j in 'abcdefgh'] # rook can take knight 2
bad3a = [string.ascii_lowercase[k_d[0] + l] + str(k_d[1] + n) for l in [-2, 2] for n in [-1,1]] # knight 1 can take knight 2
bad3b = [string.ascii_lowercase[k_d[0] + l] + str(k_d[1] + n) for l in [-1, 1] for n in [-2,2]] # knight 1 can take knight 2
bad4a = [string.ascii_lowercase[r_d[0] + l] + str(r_d[1] + n) for l in [-2, 2] for n in [-1,1]] # knight 2 can take knight 1
bad4b = [string.ascii_lowercase[r_d[0] + l] + str(r_d[1] + n) for l in [-1, 1] for n in [-2,2]] # knight 2 can take knight 1
bad = set(bad1+bad2+bad3a+bad3b+bad4a+bad4b+[k])
print(64 - sum([bool(pattern.match(i)) for i in bad]))
``` | output | 1 | 88,773 | 15 | 177,547 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Two chess pieces, a rook and a knight, stand on a standard chessboard 8 × 8 in size. The positions in which they are situated are known. It is guaranteed that none of them beats the other one.
Your task is to find the number of ways to place another knight on the board so that none of the three pieces on the board beat another one. A new piece can only be placed on an empty square.
Input
The first input line contains the description of the rook's position on the board. This description is a line which is 2 in length. Its first symbol is a lower-case Latin letter from a to h, and its second symbol is a number from 1 to 8. The second line contains the description of the knight's position in a similar way. It is guaranteed that their positions do not coincide.
Output
Print a single number which is the required number of ways.
Examples
Input
a1
b2
Output
44
Input
a8
d4
Output
38 | instruction | 0 | 88,774 | 15 | 177,548 |
Tags: brute force, implementation, math
Correct Solution:
```
def is_valid(r, c): return (0 <= r <= 7) and (0 <= c <= 7)
rook = input()
knight = input()
r = 0
knight_moves = ((0, 0), (-2, -1), (-2, 1), (-1, -2), (-1, 2), (1, -2), (1, 2), (2, -1), (2, 1))
cols_dict = {'a': 0, 'b': 1, 'c': 2, 'd': 3, 'e': 4, 'f': 5, 'g': 6, 'h': 7}
rook = (int(rook[1])-1, cols_dict[rook[0]])
knight = (int(knight[1])-1, cols_dict[knight[0]])
for i in range(8):
for j in range(8):
curr_pos = (i, j)
f = True
for k in range(-8, 8):
if (curr_pos[0]+k, curr_pos[1]) == rook: f = False; break
if (curr_pos[0], curr_pos[1]+k) == rook: f = False; break
if not f: continue
for m in knight_moves:
curr_possible_pos = (curr_pos[0]+m[0], curr_pos[1]+m[1])
if curr_possible_pos == knight or curr_possible_pos == rook: f = False; break
if f: r+= 1
print(r)
``` | output | 1 | 88,774 | 15 | 177,549 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Two chess pieces, a rook and a knight, stand on a standard chessboard 8 × 8 in size. The positions in which they are situated are known. It is guaranteed that none of them beats the other one.
Your task is to find the number of ways to place another knight on the board so that none of the three pieces on the board beat another one. A new piece can only be placed on an empty square.
Input
The first input line contains the description of the rook's position on the board. This description is a line which is 2 in length. Its first symbol is a lower-case Latin letter from a to h, and its second symbol is a number from 1 to 8. The second line contains the description of the knight's position in a similar way. It is guaranteed that their positions do not coincide.
Output
Print a single number which is the required number of ways.
Examples
Input
a1
b2
Output
44
Input
a8
d4
Output
38
Submitted Solution:
```
import string
class Position:
def __init__(self, notation):
self.column = string.ascii_letters.index(notation[0])
self.row = int(notation[1]) - 1
def __str__(self):
return '%d %d' % (self.row, self.column)
rook = Position(input().strip())
knight = Position(input().strip())
moves = [(-2, -1), (-2, 1), (-1, -2), (-1, 2),
(2, -1), (2, 1), (1, -2), (1, 2)]
result = 0
for r in range(8):
for c in range(8):
if rook.row == r or rook.column == c:
continue
feasible = True
for piece in [rook, knight]:
if r == piece.row and c == piece.column:
feasible = False
break
for (dr, dc) in moves:
if r + dr == piece.row and c + dc == piece.column:
feasible = False
break
if feasible:
result += 1
print(result)
``` | instruction | 0 | 88,775 | 15 | 177,550 |
Yes | output | 1 | 88,775 | 15 | 177,551 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Two chess pieces, a rook and a knight, stand on a standard chessboard 8 × 8 in size. The positions in which they are situated are known. It is guaranteed that none of them beats the other one.
Your task is to find the number of ways to place another knight on the board so that none of the three pieces on the board beat another one. A new piece can only be placed on an empty square.
Input
The first input line contains the description of the rook's position on the board. This description is a line which is 2 in length. Its first symbol is a lower-case Latin letter from a to h, and its second symbol is a number from 1 to 8. The second line contains the description of the knight's position in a similar way. It is guaranteed that their positions do not coincide.
Output
Print a single number which is the required number of ways.
Examples
Input
a1
b2
Output
44
Input
a8
d4
Output
38
Submitted Solution:
```
ab = lambda x: -x if x<0 else x
def f(l):
s = "abcdefgh"
r,k = l
rx,kx = s.index(r[0]), s.index(k[0])
ry,ky = int(r[1])-1, int(k[1])-1
b = [1]*64 #1=valid; 0=invalid
for i in range(64):
ix,iy = i//8,i%8
dx,dy = ab(ix-kx),ab(iy-ky)
ex,ey = ab(ix-rx),ab(iy-ry)
if ix==rx or iy==ry or dx+dy==0 or (dx<3 and dy<3 and dx+dy==3) or (ex<3 and ey<3 and ex+ey==3):
b[i]=0
return sum(b)
l = [input() for _ in range(2)]
print(f(l))
``` | instruction | 0 | 88,776 | 15 | 177,552 |
Yes | output | 1 | 88,776 | 15 | 177,553 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Two chess pieces, a rook and a knight, stand on a standard chessboard 8 × 8 in size. The positions in which they are situated are known. It is guaranteed that none of them beats the other one.
Your task is to find the number of ways to place another knight on the board so that none of the three pieces on the board beat another one. A new piece can only be placed on an empty square.
Input
The first input line contains the description of the rook's position on the board. This description is a line which is 2 in length. Its first symbol is a lower-case Latin letter from a to h, and its second symbol is a number from 1 to 8. The second line contains the description of the knight's position in a similar way. It is guaranteed that their positions do not coincide.
Output
Print a single number which is the required number of ways.
Examples
Input
a1
b2
Output
44
Input
a8
d4
Output
38
Submitted Solution:
```
def notation(s):
if s == 'a':
return 1
if s == 'b':
return 2
if s == 'c':
return 3
if s == 'd':
return 4
if s == 'e':
return 5
if s == 'f':
return 6
if s == 'g':
return 7
if s == 'h':
return 8
def take(x1, y1, x2, y2):
if abs(x1 - x2) == 1 and abs(y1 - y2) == 2:
return True
elif abs(y1 - y2) == 1 and abs(x1 - x2) == 2:
return True
elif x1 == x2 and y1 == y2:
return True
else:
return False
rookList = list(input())
knightList = list(input())
rList = notation(rookList[0]), int(rookList[1])
kList = notation(knightList[0]), int(knightList[1])
sm = 0
for i in range(8):
if i + 1 == rList[0]:
continue
for j in range(8):
if j + 1 == rList[1]:
continue
if take(i + 1, j + 1, rList[0], rList[1]):
continue
if take(i + 1, j + 1, kList[0], kList[1]):
continue
sm += 1
print(sm)
``` | instruction | 0 | 88,777 | 15 | 177,554 |
Yes | output | 1 | 88,777 | 15 | 177,555 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Two chess pieces, a rook and a knight, stand on a standard chessboard 8 × 8 in size. The positions in which they are situated are known. It is guaranteed that none of them beats the other one.
Your task is to find the number of ways to place another knight on the board so that none of the three pieces on the board beat another one. A new piece can only be placed on an empty square.
Input
The first input line contains the description of the rook's position on the board. This description is a line which is 2 in length. Its first symbol is a lower-case Latin letter from a to h, and its second symbol is a number from 1 to 8. The second line contains the description of the knight's position in a similar way. It is guaranteed that their positions do not coincide.
Output
Print a single number which is the required number of ways.
Examples
Input
a1
b2
Output
44
Input
a8
d4
Output
38
Submitted Solution:
```
def c(M,x,y):
p=[-2,+2]
q=[-1,+1]
l=[M[x+p[i]][y+q[j]]!=0 for i in range(len(p)) for j in range(len(q)) if(0<=x+p[i]<8 and 0<=y+q[j]<8)]
l+=[M[x+q[i]][y+p[j]]!=0 for i in range(len(q)) for j in range(len(p)) if(0<=x+q[i]<8 and 0<=y+p[j]<8)]
return 1 in l
mp=['a','b','c','d','e','f','g','h']
i=input
n=i()
m=i()
M=[[0 for x in range(8)]for y in range(8)]
l=[[mp.index(n[0]),int(n[1])-1],[mp.index(m[0]),int(m[1])-1]]
M[l[0][0]][l[0][1]]=2
M[l[1][0]][l[1][1]]=1
r=0
for x in range(8):
for y in range(8):
if(0==M[x][y]):
if(x==l[0][0]or y==l[0][1]):continue
if(c(M,x,y)==False):r+=1
print(r)
``` | instruction | 0 | 88,778 | 15 | 177,556 |
Yes | output | 1 | 88,778 | 15 | 177,557 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Two chess pieces, a rook and a knight, stand on a standard chessboard 8 × 8 in size. The positions in which they are situated are known. It is guaranteed that none of them beats the other one.
Your task is to find the number of ways to place another knight on the board so that none of the three pieces on the board beat another one. A new piece can only be placed on an empty square.
Input
The first input line contains the description of the rook's position on the board. This description is a line which is 2 in length. Its first symbol is a lower-case Latin letter from a to h, and its second symbol is a number from 1 to 8. The second line contains the description of the knight's position in a similar way. It is guaranteed that their positions do not coincide.
Output
Print a single number which is the required number of ways.
Examples
Input
a1
b2
Output
44
Input
a8
d4
Output
38
Submitted Solution:
```
def rmvKn(mgc, x, y):
mgc.add((x,y))
if not x - 1 < 0 and not y + 2 > 7: mgc.add((x-1, y+2))
if not x - 2 < 0 and not y - 1 < 0: mgc.add((x-2, y-1))
if not x - 2 < 0 and not y + 1 > 7: mgc.add((x-2, y+1))
if not x - 1 < 0 and not y - 2 < 0: mgc.add((x-1, y-2))
if not x + 1 > 7 and not y - 2 < 0: mgc.add((x+1, y-2))
if not x + 2 > 7 and not y - 1 < 0: mgc.add((x+2, y-1))
if not x + 2 > 7 and not y + 1 > 7: mgc.add((x+2, y+1))
if not x + 1 > 7 and not y + 2 > 7: mgc.add((x+1, y+2))
def rmvR(mgc, x, y):
for i in range(8):
mgc.add((i, y))
mgc.add((x, i))
r=input()
kn=input()
R=[ord(r[0])-97, int(r[1])-1]
Kn=[ord(kn[0])-97, int(kn[1])-1]
mgc=set()
rmvR(mgc,R[0],R[1])
rmvKn(mgc,Kn[0],Kn[1])
count = 0
for x in mgc:
count += 1
print(64-count-2)
``` | instruction | 0 | 88,779 | 15 | 177,558 |
No | output | 1 | 88,779 | 15 | 177,559 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Two chess pieces, a rook and a knight, stand on a standard chessboard 8 × 8 in size. The positions in which they are situated are known. It is guaranteed that none of them beats the other one.
Your task is to find the number of ways to place another knight on the board so that none of the three pieces on the board beat another one. A new piece can only be placed on an empty square.
Input
The first input line contains the description of the rook's position on the board. This description is a line which is 2 in length. Its first symbol is a lower-case Latin letter from a to h, and its second symbol is a number from 1 to 8. The second line contains the description of the knight's position in a similar way. It is guaranteed that their positions do not coincide.
Output
Print a single number which is the required number of ways.
Examples
Input
a1
b2
Output
44
Input
a8
d4
Output
38
Submitted Solution:
```
f={'a':1,'b':2,'c':3,'d':4,'e':5,'f':6,'g':7,'h':8}
a = input()
b =input()
A = [f[a[0]],int(a[-1])]
B = [f[b[0]],int(b[-1]) ]
u=[]
if 1<=A[0]-1 <=8 and 1<=A[1]-2<=8:
u.append([A[0]-1,A[1]-2])
if 1<=A[0]+1 <=8 and 1<=A[1]-2<=8:
u.append([A[0]+1,A[1]-2])
if 1<=A[0]+2 <=8 and 1<=A[1]-1<=8:
u.append([A[0]+2,A[1]-1])
if 1<=A[0]+2 <=8 and 1<=A[1]+1<=8:
u.append([A[0]+1,A[1]+1])
if 1<=A[0]+1 <=8 and 1<=A[1]+2<=8:
u.append([A[0]+1,A[1]+2])
if 1<=A[0]-1 <=8 and 1<=A[1]+2<=8:
u.append([A[0]-1,A[1]+2])
if 1<=A[0]-2 <=8 and 1<=A[1]+1<=8:
u.append([A[0]-2,A[1]+1])
if 1<=A[0]-2 <=8 and 1<=A[1]-1<=8:
u.append([A[0]-2,A[1]-1])
#hghj
if 1<=B[0]-1 <=8 and 1<=B[1]-2<=8:
u.append([B[0]-1,B[1]-2])
if 1<=B[0]+1 <=8 and 1<=B[1]-2<=8:
u.append([B[0]+1,B[1]-2])
if 1<=B[0]+2 <=8 and 1<=B[1]-1<=8:
u.append([B[0]+2,B[1]-1])
if 1<=B[0]+2 <=8 and 1<=B[1]+1<=8:
u.append([B[0]+1,B[1]+1])
if 1<=B[0]+1 <=8 and 1<=B[1]+2<=8:
u.append([B[0]+1,B[1]+2])
if 1<=B[0]-1 <=8 and 1<=B[1]+2<=8:
u.append([B[0]-1,B[1]+2])
if 1<=B[0]-2 <=8 and 1<=B[1]+1<=8:
u.append([B[0]-2,B[1]+1])
if 1<=B[0]-2 <=8 and 1<=B[1]-1<=8:
u.append([B[0]-2,B[1]-1])
for k in range(1,9):
if [k,B[1]] not in u:
u.append([k,B[1]])
for j in range(1,9):
if [B[0],j] not in u:
u.append([B[0],j])
f=[]
p=0
for j in u:
if j not in f:
p+=1
f.append(j)
if B not in f:
p+=1
if A not in f:
p+=1
print(64-(p))
``` | instruction | 0 | 88,780 | 15 | 177,560 |
No | output | 1 | 88,780 | 15 | 177,561 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Two chess pieces, a rook and a knight, stand on a standard chessboard 8 × 8 in size. The positions in which they are situated are known. It is guaranteed that none of them beats the other one.
Your task is to find the number of ways to place another knight on the board so that none of the three pieces on the board beat another one. A new piece can only be placed on an empty square.
Input
The first input line contains the description of the rook's position on the board. This description is a line which is 2 in length. Its first symbol is a lower-case Latin letter from a to h, and its second symbol is a number from 1 to 8. The second line contains the description of the knight's position in a similar way. It is guaranteed that their positions do not coincide.
Output
Print a single number which is the required number of ways.
Examples
Input
a1
b2
Output
44
Input
a8
d4
Output
38
Submitted Solution:
```
p = []
for i in range(12):
p.append([1] * 12)
for i in range(2, 10):
for t in range(2, 10):
p[i][t] = 0
dic = 'abcdefgh'
a = input()
b = input()
p1x = dic.index(a[0]) + 2
p1y = int(a[1]) + 1
p2x = dic.index(b[0]) + 2
p2y = int(b[1]) + 1
for i in range(12):
p[i][p1y] = 1
p[p1x][i] = 1
x = p2x
y = p2y
p[x + 1][y + 2] = 1
p[x + 2][y + 1] = 1
p[x + 2][y - 1] = 1
p[x + 1][y - 2] = 1
p[x - 1][y - 2] = 1
p[x - 2][y - 1] = 1
p[x - 2][y + 1] = 1
p[x - 1][y + 2] = 1
sum = 0
for i in range(12):
sum += p[i].count(0)
print(sum - 3)
``` | instruction | 0 | 88,781 | 15 | 177,562 |
No | output | 1 | 88,781 | 15 | 177,563 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Two chess pieces, a rook and a knight, stand on a standard chessboard 8 × 8 in size. The positions in which they are situated are known. It is guaranteed that none of them beats the other one.
Your task is to find the number of ways to place another knight on the board so that none of the three pieces on the board beat another one. A new piece can only be placed on an empty square.
Input
The first input line contains the description of the rook's position on the board. This description is a line which is 2 in length. Its first symbol is a lower-case Latin letter from a to h, and its second symbol is a number from 1 to 8. The second line contains the description of the knight's position in a similar way. It is guaranteed that their positions do not coincide.
Output
Print a single number which is the required number of ways.
Examples
Input
a1
b2
Output
44
Input
a8
d4
Output
38
Submitted Solution:
```
rook = input()
knight = input()
rook = list(rook)
knight = list(knight)
rook[0] = ord(rook[0])-96
knight[0] = ord(knight[0])-96
s = set()
s.add((int(rook[0]),int(rook[1])))
s.add((int(knight[0]),int(knight[1])))
# print(s)
for i in range(1,9):
s.add((int(rook[0]),i))
for i in range(1,9):
s.add((i,int(rook[1])))
knight[1] = int(knight[1])
rook[1] = int(rook[1])
# print(s)
if knight[0]-1 > 0 and knight[1]-2 > 0:
s.add((knight[0]-1,knight[1]-2))
if knight[0]-1 > 0 and knight[1]+2 <= 8:
s.add((knight[0]-1,knight[1]+2))
if knight[0]+1 <= 8 and knight[1]-2 > 0:
s.add((knight[0]+1,knight[1]-2))
if knight[0]+1 <= 8 and knight[1]+2 <= 8:
s.add((knight[0]+1,knight[1]+2))
if knight[0]+2 <= 8 and knight[1]-1 > 0:
s.add((knight[0]+2,knight[1]-1))
if knight[0]+2 <= 8 and knight[1]+1 <= 8:
s.add((knight[0]+2,knight[1]+1))
if knight[0]-2 > 0 and knight[1]+1 <= 8:
s.add((knight[0]-2,knight[1]+1))
if knight[0]-2 > 0 and knight[1]-1 > 0:
s.add((knight[0]-2,knight[1]-1))
if rook[0]-1 > 0 and rook[1]-2 > 0:
s.add((rook[0]-1,rook[1]-2))
if rook[0]-1 > 0 and rook[1]+2 <= 8:
s.add((rook[0]-1,rook[1]-2))
if rook[0]+1 <= 8 and rook[1]+2 <= 8:
s.add((rook[0]+1,rook[1]+2))
if rook[0]+1 <= 8 and rook[1]-2 > 0:
s.add((rook[0]+1,rook[1]+2))
if rook[0]-2 > 0 and rook[1]+1 <= 8:
s.add((rook[0]-2,rook[1]+1))
if rook[0]-2 > 0 and rook[1]-1 > 0:
s.add((rook[0]-2,rook[1]-1))
if rook[0]+2 <= 8 and rook[1]+1 <= 8:
s.add((rook[0]+2,rook[1]+1))
if rook[0]+2 <= 8 and rook[1]-1 > 0:
s.add((rook[0]+2,rook[1]-1))
print(64-len(s))
``` | instruction | 0 | 88,782 | 15 | 177,564 |
No | output | 1 | 88,782 | 15 | 177,565 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In one very old text file there was written Great Wisdom. This Wisdom was so Great that nobody could decipher it, even Phong — the oldest among the inhabitants of Mainframe. But still he managed to get some information from there. For example, he managed to learn that User launches games for pleasure — and then terrible Game Cubes fall down on the city, bringing death to those modules, who cannot win the game...
For sure, as guard Bob appeared in Mainframe many modules stopped fearing Game Cubes. Because Bob (as he is alive yet) has never been defeated by User, and he always meddles with Game Cubes, because he is programmed to this.
However, unpleasant situations can happen, when a Game Cube falls down on Lost Angles. Because there lives a nasty virus — Hexadecimal, who is... mmm... very strange. And she likes to play very much. So, willy-nilly, Bob has to play with her first, and then with User.
This time Hexadecimal invented the following entertainment: Bob has to leap over binary search trees with n nodes. We should remind you that a binary search tree is a binary tree, each node has a distinct key, for each node the following is true: the left sub-tree of a node contains only nodes with keys less than the node's key, the right sub-tree of a node contains only nodes with keys greater than the node's key. All the keys are different positive integer numbers from 1 to n. Each node of such a tree can have up to two children, or have no children at all (in the case when a node is a leaf).
In Hexadecimal's game all the trees are different, but the height of each is not lower than h. In this problem «height» stands for the maximum amount of nodes on the way from the root to the remotest leaf, the root node and the leaf itself included. When Bob leaps over a tree, it disappears. Bob gets the access to a Cube, when there are no trees left. He knows how many trees he will have to leap over in the worst case. And you?
Input
The input data contains two space-separated positive integer numbers n and h (n ≤ 35, h ≤ n).
Output
Output one number — the answer to the problem. It is guaranteed that it does not exceed 9·1018.
Examples
Input
3 2
Output
5
Input
3 3
Output
4 | instruction | 0 | 88,975 | 15 | 177,950 |
Tags: combinatorics, divide and conquer, dp
Correct Solution:
```
def t(n, h):
if dp[n][h] == -1:
dp[n][h] = sum(
t(m - 1, h - 1) * sum(t(n - m, i) for i in range(h)) +
t(n - m, h - 1) * sum(t(m - 1, i) for i in range(h - 1))
for m in range(1, n + 1)
)
return dp[n][h]
n, h = map(int, input().split())
def fill(i, j):
if i == 0 and j == 0:
return 1
if i == 0 or j == 0:
return 0
return -1
dp = [[fill(i, j) for j in range(n + 1)] for i in range(n + 1)]
for i in range(n + 1):
for j in range(n + 1):
t(i, j)
print(sum(dp[n][i] for i in range(h, n + 1)))
``` | output | 1 | 88,975 | 15 | 177,951 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In one very old text file there was written Great Wisdom. This Wisdom was so Great that nobody could decipher it, even Phong — the oldest among the inhabitants of Mainframe. But still he managed to get some information from there. For example, he managed to learn that User launches games for pleasure — and then terrible Game Cubes fall down on the city, bringing death to those modules, who cannot win the game...
For sure, as guard Bob appeared in Mainframe many modules stopped fearing Game Cubes. Because Bob (as he is alive yet) has never been defeated by User, and he always meddles with Game Cubes, because he is programmed to this.
However, unpleasant situations can happen, when a Game Cube falls down on Lost Angles. Because there lives a nasty virus — Hexadecimal, who is... mmm... very strange. And she likes to play very much. So, willy-nilly, Bob has to play with her first, and then with User.
This time Hexadecimal invented the following entertainment: Bob has to leap over binary search trees with n nodes. We should remind you that a binary search tree is a binary tree, each node has a distinct key, for each node the following is true: the left sub-tree of a node contains only nodes with keys less than the node's key, the right sub-tree of a node contains only nodes with keys greater than the node's key. All the keys are different positive integer numbers from 1 to n. Each node of such a tree can have up to two children, or have no children at all (in the case when a node is a leaf).
In Hexadecimal's game all the trees are different, but the height of each is not lower than h. In this problem «height» stands for the maximum amount of nodes on the way from the root to the remotest leaf, the root node and the leaf itself included. When Bob leaps over a tree, it disappears. Bob gets the access to a Cube, when there are no trees left. He knows how many trees he will have to leap over in the worst case. And you?
Input
The input data contains two space-separated positive integer numbers n and h (n ≤ 35, h ≤ n).
Output
Output one number — the answer to the problem. It is guaranteed that it does not exceed 9·1018.
Examples
Input
3 2
Output
5
Input
3 3
Output
4 | instruction | 0 | 88,976 | 15 | 177,952 |
Tags: combinatorics, divide and conquer, dp
Correct Solution:
```
n, h = list(map(int, input().split(" ")))
dp = [[0 for j in range(h + 1)] for i in range(n + 1)]
dp[0][0] = 1
for i in range(1, n + 1):
dp[i][0] = dp[i - 1][0] * 2 * (2 * i - 1) // (i + 1)
for i in range(1, n + 1):
for j in range(1, h + 1):
for k in range(1, i + 1):
dp[i][j] += dp[k - 1][0] * dp[i - k][j - 1] + dp[k - 1][j - 1] * dp[i - k][0] - dp[i - k][j - 1] * dp[k - 1][j - 1]
print(dp[n][h])
``` | output | 1 | 88,976 | 15 | 177,953 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In one very old text file there was written Great Wisdom. This Wisdom was so Great that nobody could decipher it, even Phong — the oldest among the inhabitants of Mainframe. But still he managed to get some information from there. For example, he managed to learn that User launches games for pleasure — and then terrible Game Cubes fall down on the city, bringing death to those modules, who cannot win the game...
For sure, as guard Bob appeared in Mainframe many modules stopped fearing Game Cubes. Because Bob (as he is alive yet) has never been defeated by User, and he always meddles with Game Cubes, because he is programmed to this.
However, unpleasant situations can happen, when a Game Cube falls down on Lost Angles. Because there lives a nasty virus — Hexadecimal, who is... mmm... very strange. And she likes to play very much. So, willy-nilly, Bob has to play with her first, and then with User.
This time Hexadecimal invented the following entertainment: Bob has to leap over binary search trees with n nodes. We should remind you that a binary search tree is a binary tree, each node has a distinct key, for each node the following is true: the left sub-tree of a node contains only nodes with keys less than the node's key, the right sub-tree of a node contains only nodes with keys greater than the node's key. All the keys are different positive integer numbers from 1 to n. Each node of such a tree can have up to two children, or have no children at all (in the case when a node is a leaf).
In Hexadecimal's game all the trees are different, but the height of each is not lower than h. In this problem «height» stands for the maximum amount of nodes on the way from the root to the remotest leaf, the root node and the leaf itself included. When Bob leaps over a tree, it disappears. Bob gets the access to a Cube, when there are no trees left. He knows how many trees he will have to leap over in the worst case. And you?
Input
The input data contains two space-separated positive integer numbers n and h (n ≤ 35, h ≤ n).
Output
Output one number — the answer to the problem. It is guaranteed that it does not exceed 9·1018.
Examples
Input
3 2
Output
5
Input
3 3
Output
4 | instruction | 0 | 88,977 | 15 | 177,954 |
Tags: combinatorics, divide and conquer, dp
Correct Solution:
```
MAX = 71
C=[[0]*MAX for i in range(MAX)]
dp=[[0]*MAX for i in range(MAX)]
C[0][0]=1
for i in range(1, MAX):
C[i][0]=1
for i in range(1, MAX):
for j in range(1, i+1):
C[i][j]=C[i-1][j]+C[i-1][j-1]
def catalan(n):
return C[2*n][n]//(n+1)
def solve(n, h):
if(n==0): return 0
if(n<h): return 0
if(h<=1): return catalan(n)
if(dp[n][h]): return dp[n][h];
ans=0
for k in range(0, n):
if(dp[n-k-1][h-1]==0):
dp[n-k-1][h-1]=solve(n-k-1, h-1)
if(dp[k][h-1]==0):
dp[k][h-1]=solve(k, h-1)
ans+=dp[n-k-1][h-1]*( catalan(k)-dp[k][h-1]) + catalan(n-k-1)*dp[k][h-1]
dp[n][h]=ans
return ans
n, h=map(int, input().split())
ans=solve(n, h)
print(ans)
``` | output | 1 | 88,977 | 15 | 177,955 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In one very old text file there was written Great Wisdom. This Wisdom was so Great that nobody could decipher it, even Phong — the oldest among the inhabitants of Mainframe. But still he managed to get some information from there. For example, he managed to learn that User launches games for pleasure — and then terrible Game Cubes fall down on the city, bringing death to those modules, who cannot win the game...
For sure, as guard Bob appeared in Mainframe many modules stopped fearing Game Cubes. Because Bob (as he is alive yet) has never been defeated by User, and he always meddles with Game Cubes, because he is programmed to this.
However, unpleasant situations can happen, when a Game Cube falls down on Lost Angles. Because there lives a nasty virus — Hexadecimal, who is... mmm... very strange. And she likes to play very much. So, willy-nilly, Bob has to play with her first, and then with User.
This time Hexadecimal invented the following entertainment: Bob has to leap over binary search trees with n nodes. We should remind you that a binary search tree is a binary tree, each node has a distinct key, for each node the following is true: the left sub-tree of a node contains only nodes with keys less than the node's key, the right sub-tree of a node contains only nodes with keys greater than the node's key. All the keys are different positive integer numbers from 1 to n. Each node of such a tree can have up to two children, or have no children at all (in the case when a node is a leaf).
In Hexadecimal's game all the trees are different, but the height of each is not lower than h. In this problem «height» stands for the maximum amount of nodes on the way from the root to the remotest leaf, the root node and the leaf itself included. When Bob leaps over a tree, it disappears. Bob gets the access to a Cube, when there are no trees left. He knows how many trees he will have to leap over in the worst case. And you?
Input
The input data contains two space-separated positive integer numbers n and h (n ≤ 35, h ≤ n).
Output
Output one number — the answer to the problem. It is guaranteed that it does not exceed 9·1018.
Examples
Input
3 2
Output
5
Input
3 3
Output
4 | instruction | 0 | 88,978 | 15 | 177,956 |
Tags: combinatorics, divide and conquer, dp
Correct Solution:
```
n, h = map(int,input().split())
dp = [[0 for j in range(n + 1)] for i in range(n + 1)]
dp[0] = [1 for j in range(n + 1)]
for i in range(n + 1):
for j in range(1, n + 1):
for k in range(i):
dp[i][j] += dp[k][j - 1] * dp[i - 1 - k][j - 1]
print(dp[n][n] - (h > 0 and [dp[n][h - 1]] or [0])[0])
``` | output | 1 | 88,978 | 15 | 177,957 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In one very old text file there was written Great Wisdom. This Wisdom was so Great that nobody could decipher it, even Phong — the oldest among the inhabitants of Mainframe. But still he managed to get some information from there. For example, he managed to learn that User launches games for pleasure — and then terrible Game Cubes fall down on the city, bringing death to those modules, who cannot win the game...
For sure, as guard Bob appeared in Mainframe many modules stopped fearing Game Cubes. Because Bob (as he is alive yet) has never been defeated by User, and he always meddles with Game Cubes, because he is programmed to this.
However, unpleasant situations can happen, when a Game Cube falls down on Lost Angles. Because there lives a nasty virus — Hexadecimal, who is... mmm... very strange. And she likes to play very much. So, willy-nilly, Bob has to play with her first, and then with User.
This time Hexadecimal invented the following entertainment: Bob has to leap over binary search trees with n nodes. We should remind you that a binary search tree is a binary tree, each node has a distinct key, for each node the following is true: the left sub-tree of a node contains only nodes with keys less than the node's key, the right sub-tree of a node contains only nodes with keys greater than the node's key. All the keys are different positive integer numbers from 1 to n. Each node of such a tree can have up to two children, or have no children at all (in the case when a node is a leaf).
In Hexadecimal's game all the trees are different, but the height of each is not lower than h. In this problem «height» stands for the maximum amount of nodes on the way from the root to the remotest leaf, the root node and the leaf itself included. When Bob leaps over a tree, it disappears. Bob gets the access to a Cube, when there are no trees left. He knows how many trees he will have to leap over in the worst case. And you?
Input
The input data contains two space-separated positive integer numbers n and h (n ≤ 35, h ≤ n).
Output
Output one number — the answer to the problem. It is guaranteed that it does not exceed 9·1018.
Examples
Input
3 2
Output
5
Input
3 3
Output
4 | instruction | 0 | 88,979 | 15 | 177,958 |
Tags: combinatorics, divide and conquer, dp
Correct Solution:
```
n, h = map(int, input().split())
ceq = [[0] * (n + 1) for i in range(n + 1)]
cle = [[0] * (n + 1) for i in range(n + 1)]
ceq[0][0] = 1
cle[0] = [1] * (n + 1)
for i in range(1, n + 1):
for j in range(n):
for left in range(i):
ceq[i][j + 1] += ceq[left][j] * cle[i - left - 1][j] + cle[left][j] * ceq[i - left - 1][j] - ceq[left][j] * ceq[i - left - 1][j]
cle[i][j + 1] = cle[i][j] + ceq[i][j + 1]
print(cle[n][n] - cle[n][h - 1])
``` | output | 1 | 88,979 | 15 | 177,959 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In one very old text file there was written Great Wisdom. This Wisdom was so Great that nobody could decipher it, even Phong — the oldest among the inhabitants of Mainframe. But still he managed to get some information from there. For example, he managed to learn that User launches games for pleasure — and then terrible Game Cubes fall down on the city, bringing death to those modules, who cannot win the game...
For sure, as guard Bob appeared in Mainframe many modules stopped fearing Game Cubes. Because Bob (as he is alive yet) has never been defeated by User, and he always meddles with Game Cubes, because he is programmed to this.
However, unpleasant situations can happen, when a Game Cube falls down on Lost Angles. Because there lives a nasty virus — Hexadecimal, who is... mmm... very strange. And she likes to play very much. So, willy-nilly, Bob has to play with her first, and then with User.
This time Hexadecimal invented the following entertainment: Bob has to leap over binary search trees with n nodes. We should remind you that a binary search tree is a binary tree, each node has a distinct key, for each node the following is true: the left sub-tree of a node contains only nodes with keys less than the node's key, the right sub-tree of a node contains only nodes with keys greater than the node's key. All the keys are different positive integer numbers from 1 to n. Each node of such a tree can have up to two children, or have no children at all (in the case when a node is a leaf).
In Hexadecimal's game all the trees are different, but the height of each is not lower than h. In this problem «height» stands for the maximum amount of nodes on the way from the root to the remotest leaf, the root node and the leaf itself included. When Bob leaps over a tree, it disappears. Bob gets the access to a Cube, when there are no trees left. He knows how many trees he will have to leap over in the worst case. And you?
Input
The input data contains two space-separated positive integer numbers n and h (n ≤ 35, h ≤ n).
Output
Output one number — the answer to the problem. It is guaranteed that it does not exceed 9·1018.
Examples
Input
3 2
Output
5
Input
3 3
Output
4 | instruction | 0 | 88,980 | 15 | 177,960 |
Tags: combinatorics, divide and conquer, dp
Correct Solution:
```
# Enter your code here. Read input from STDIN. Print output to STDOUT
MAX = 71
C=[[0]*MAX for i in range(MAX)]
dp=[[0]*MAX for i in range(MAX)]
C[0][0]=1
for i in range(1, MAX):
C[i][0]=1
for i in range(1, MAX):
for j in range(1, i+1):
C[i][j]=C[i-1][j]+C[i-1][j-1]
def catalan(n):
return C[2*n][n]//(n+1)
def solve(n, h):
if(n==0): return 0
if(n<h): return 0
if(h<=1): return catalan(n)
if(dp[n][h]): return dp[n][h];
ans=0
for k in range(0, n):
if(dp[n-k-1][h-1]==0):
dp[n-k-1][h-1]=solve(n-k-1, h-1)
if(dp[k][h-1]==0):
dp[k][h-1]=solve(k, h-1)
ans+=dp[n-k-1][h-1]*( catalan(k)-dp[k][h-1]) + catalan(n-k-1)*dp[k][h-1]
dp[n][h]=ans
return ans
n, h=map(int, input().split())
ans=solve(n, h)
print(ans)
``` | output | 1 | 88,980 | 15 | 177,961 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In one very old text file there was written Great Wisdom. This Wisdom was so Great that nobody could decipher it, even Phong — the oldest among the inhabitants of Mainframe. But still he managed to get some information from there. For example, he managed to learn that User launches games for pleasure — and then terrible Game Cubes fall down on the city, bringing death to those modules, who cannot win the game...
For sure, as guard Bob appeared in Mainframe many modules stopped fearing Game Cubes. Because Bob (as he is alive yet) has never been defeated by User, and he always meddles with Game Cubes, because he is programmed to this.
However, unpleasant situations can happen, when a Game Cube falls down on Lost Angles. Because there lives a nasty virus — Hexadecimal, who is... mmm... very strange. And she likes to play very much. So, willy-nilly, Bob has to play with her first, and then with User.
This time Hexadecimal invented the following entertainment: Bob has to leap over binary search trees with n nodes. We should remind you that a binary search tree is a binary tree, each node has a distinct key, for each node the following is true: the left sub-tree of a node contains only nodes with keys less than the node's key, the right sub-tree of a node contains only nodes with keys greater than the node's key. All the keys are different positive integer numbers from 1 to n. Each node of such a tree can have up to two children, or have no children at all (in the case when a node is a leaf).
In Hexadecimal's game all the trees are different, but the height of each is not lower than h. In this problem «height» stands for the maximum amount of nodes on the way from the root to the remotest leaf, the root node and the leaf itself included. When Bob leaps over a tree, it disappears. Bob gets the access to a Cube, when there are no trees left. He knows how many trees he will have to leap over in the worst case. And you?
Input
The input data contains two space-separated positive integer numbers n and h (n ≤ 35, h ≤ n).
Output
Output one number — the answer to the problem. It is guaranteed that it does not exceed 9·1018.
Examples
Input
3 2
Output
5
Input
3 3
Output
4 | instruction | 0 | 88,981 | 15 | 177,962 |
Tags: combinatorics, divide and conquer, dp
Correct Solution:
```
dic={}
def dfs(a,res):
if a<=1 and res<=a:
return 1
elif a*100+res in dic:
return dic[a*100+res]
else:
cnt=0
for i in range(a):
if i<res-1 and a-i-1<res-1:
continue
else:
cnt=cnt+dfs(i,0)*dfs(a-i-1,res-1)+dfs(i,res-1)*dfs(a-i-1,0)-dfs(i,res-1)*dfs(a-i-1,res-1)
dic[a*100+res]=cnt
return cnt
inp=input().split()
print(dfs(int(inp[0]),int(inp[1])))
``` | output | 1 | 88,981 | 15 | 177,963 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In one very old text file there was written Great Wisdom. This Wisdom was so Great that nobody could decipher it, even Phong — the oldest among the inhabitants of Mainframe. But still he managed to get some information from there. For example, he managed to learn that User launches games for pleasure — and then terrible Game Cubes fall down on the city, bringing death to those modules, who cannot win the game...
For sure, as guard Bob appeared in Mainframe many modules stopped fearing Game Cubes. Because Bob (as he is alive yet) has never been defeated by User, and he always meddles with Game Cubes, because he is programmed to this.
However, unpleasant situations can happen, when a Game Cube falls down on Lost Angles. Because there lives a nasty virus — Hexadecimal, who is... mmm... very strange. And she likes to play very much. So, willy-nilly, Bob has to play with her first, and then with User.
This time Hexadecimal invented the following entertainment: Bob has to leap over binary search trees with n nodes. We should remind you that a binary search tree is a binary tree, each node has a distinct key, for each node the following is true: the left sub-tree of a node contains only nodes with keys less than the node's key, the right sub-tree of a node contains only nodes with keys greater than the node's key. All the keys are different positive integer numbers from 1 to n. Each node of such a tree can have up to two children, or have no children at all (in the case when a node is a leaf).
In Hexadecimal's game all the trees are different, but the height of each is not lower than h. In this problem «height» stands for the maximum amount of nodes on the way from the root to the remotest leaf, the root node and the leaf itself included. When Bob leaps over a tree, it disappears. Bob gets the access to a Cube, when there are no trees left. He knows how many trees he will have to leap over in the worst case. And you?
Input
The input data contains two space-separated positive integer numbers n and h (n ≤ 35, h ≤ n).
Output
Output one number — the answer to the problem. It is guaranteed that it does not exceed 9·1018.
Examples
Input
3 2
Output
5
Input
3 3
Output
4 | instruction | 0 | 88,982 | 15 | 177,964 |
Tags: combinatorics, divide and conquer, dp
Correct Solution:
```
def main():
def rec(n, h):
if sht[n][h]:
return dp[n][h]
sht[n][h] = 1
if n == 0:
dp[n][h] = h == 0
return dp[n][h]
if h == 0:
return dp[n][h]
if n < h:
return dp[n][h]
for i in range(n):
for j in range(h):
dp[n][h] += rec(n - 1 - i, h - 1) * rec(i, j) + rec(n - 1 - i, j) * rec(i, h - 1)
dp[n][h] -= rec(n - 1 - i, h - 1) * rec(i, h - 1)
return dp[n][h]
dp = [[0] * 40 for _ in range(40)]
sht = [[0] * 40 for _ in range(40)]
n, h = map(int, input().split())
print(sum([rec(n, j) for j in range(h, n + 1)]))
return 0
main()
``` | output | 1 | 88,982 | 15 | 177,965 |
Provide a correct Python 3 solution for this coding contest problem.
We have a grid with N rows and M columns of squares. Each integer from 1 to NM is written in this grid once. The number written in the square at the i-th row from the top and the j-th column from the left is A_{ij}.
You need to rearrange these numbers as follows:
1. First, for each of the N rows, rearrange the numbers written in it as you like.
2. Second, for each of the M columns, rearrange the numbers written in it as you like.
3. Finally, for each of the N rows, rearrange the numbers written in it as you like.
After rearranging the numbers, you want the number written in the square at the i-th row from the top and the j-th column from the left to be M\times (i-1)+j. Construct one such way to rearrange the numbers. The constraints guarantee that it is always possible to achieve the objective.
Constraints
* 1 \leq N,M \leq 100
* 1 \leq A_{ij} \leq NM
* A_{ij} are distinct.
Input
Input is given from Standard Input in the following format:
N M
A_{11} A_{12} ... A_{1M}
:
A_{N1} A_{N2} ... A_{NM}
Output
Print one way to rearrange the numbers in the following format:
B_{11} B_{12} ... B_{1M}
:
B_{N1} B_{N2} ... B_{NM}
C_{11} C_{12} ... C_{1M}
:
C_{N1} C_{N2} ... C_{NM}
Here B_{ij} is the number written in the square at the i-th row from the top and the j-th column from the left after Step 1, and C_{ij} is the number written in that square after Step 2.
Examples
Input
3 2
2 6
4 3
1 5
Output
2 6
4 3
5 1
2 1
4 3
5 6
Input
3 4
1 4 7 10
2 5 8 11
3 6 9 12
Output
1 4 7 10
5 8 11 2
9 12 3 6
1 4 3 2
5 8 7 6
9 12 11 10 | instruction | 0 | 89,039 | 15 | 178,078 |
"Correct Solution:
```
from collections import defaultdict, deque
N, M = map(int, input().split())
A = [[0] * M for _ in range(N)]
B = [[0] * M for _ in range(N)]
C = [[0] * M for _ in range(N)]
G = defaultdict(lambda: defaultdict(int))
F = defaultdict(lambda: defaultdict(int))
rows = [0] * (N * M + 1)
for i in range(N):
A_list = list(map(int, input().split()))
A[i] = A_list
for a in A_list:
G[i + 1][N + (a + M - 1) // M] += 1
G[N + (a + M - 1) // M][i + 1] = 0
# dfs
for t1 in range(M):
for i in range(1, N + 1):
G[0][i] = 1
F[0][i] = 0
G[N + i][2 * N + 1] = 1
F[N + i][2 * N + 1] = 0
for t2 in range(N):
queue = deque([0])
searched = [0] * (2 * N + 2)
searched[0] = 1
parent = [0] * (2 * N + 2)
while len(queue) > 0:
p = queue.pop()
for q in G[p].keys():
if (G[p][q] > 0 or F[q][p] > 0) and searched[q] == 0:
parent[q] = p
searched[q] = 1
queue.append(q)
if searched[2 * N + 1] == 1:
break
path = [2 * N + 1]
while True:
path.append(parent[path[-1]])
if path[-1] == 0:
break
for i in range(len(path) - 1, 0, -1):
p, q = path[i], path[i-1]
if G[p][q] > 0:
G[p][q] -= 1
F[p][q] += 1
elif F[q][p] > 0:
F[q][p] -= 1
G[q][p] += 1
for i in range(1, N + 1):
ji = 0
for j in range(N + 1, 2 * N + 1):
if F[i][j] == 1:
ji = j
break
F[i][ji] -= 1
for a in A[i - 1]:
if N + (a + M - 1) // M == ji:
A[i - 1].remove(a)
B[i - 1][t1] = a
break
for j in range(M):
c_j = list(sorted([B[i][j] for i in range(N)]))
for i in range(N):
C[i][j] = c_j[i]
if __name__ == "__main__":
for i in range(N):
print(*B[i])
for i in range(N):
print(*C[i])
``` | output | 1 | 89,039 | 15 | 178,079 |
Provide a correct Python 3 solution for this coding contest problem.
We have a grid with N rows and M columns of squares. Each integer from 1 to NM is written in this grid once. The number written in the square at the i-th row from the top and the j-th column from the left is A_{ij}.
You need to rearrange these numbers as follows:
1. First, for each of the N rows, rearrange the numbers written in it as you like.
2. Second, for each of the M columns, rearrange the numbers written in it as you like.
3. Finally, for each of the N rows, rearrange the numbers written in it as you like.
After rearranging the numbers, you want the number written in the square at the i-th row from the top and the j-th column from the left to be M\times (i-1)+j. Construct one such way to rearrange the numbers. The constraints guarantee that it is always possible to achieve the objective.
Constraints
* 1 \leq N,M \leq 100
* 1 \leq A_{ij} \leq NM
* A_{ij} are distinct.
Input
Input is given from Standard Input in the following format:
N M
A_{11} A_{12} ... A_{1M}
:
A_{N1} A_{N2} ... A_{NM}
Output
Print one way to rearrange the numbers in the following format:
B_{11} B_{12} ... B_{1M}
:
B_{N1} B_{N2} ... B_{NM}
C_{11} C_{12} ... C_{1M}
:
C_{N1} C_{N2} ... C_{NM}
Here B_{ij} is the number written in the square at the i-th row from the top and the j-th column from the left after Step 1, and C_{ij} is the number written in that square after Step 2.
Examples
Input
3 2
2 6
4 3
1 5
Output
2 6
4 3
5 1
2 1
4 3
5 6
Input
3 4
1 4 7 10
2 5 8 11
3 6 9 12
Output
1 4 7 10
5 8 11 2
9 12 3 6
1 4 3 2
5 8 7 6
9 12 11 10 | instruction | 0 | 89,040 | 15 | 178,080 |
"Correct Solution:
```
import sys
readline = sys.stdin.readline
import collections
class Dinic:
def __init__(self, vnum):
self.edge = [[] for i in range(vnum)]
self.n = vnum
# infはint型の方が良いかもね
self.inf = float('inf')
def addedge(self, st, en, c):
self.edge[st].append([en, c, len(self.edge[en])])
self.edge[en].append([st, 0, len(self.edge[st])-1])
def bfs(self, vst):
dist = [-1]*self.n
dist[vst] = 0
Q = collections.deque([vst])
while Q:
nv = Q.popleft()
for vt, c, r in self.edge[nv]:
if dist[vt] == -1 and c > 0:
dist[vt] = dist[nv] + 1
Q.append(vt)
self.dist = dist
def dfs(self, nv, en, nf):
nextv = self.nextv
if nv == en:
return nf
dist = self.dist
ist = nextv[nv]
for i, (vt, c, r) in enumerate(self.edge[nv][ist:], ist):
if dist[nv] < dist[vt] and c > 0:
df = self.dfs(vt, en, min(nf, c))
if df > 0:
self.edge[nv][i][1] -= df
self.edge[vt][r][1] += df
return df
nextv[nv] += 1
return 0
def getmf(self, st, en):
mf = 0
while True:
self.bfs(st)
if self.dist[en] == -1:
break
self.nextv = [0]*self.n
while True:
fl = self.dfs(st, en, self.inf)
if fl > 0:
mf += fl
else:
break
return mf
N, M = map(int, readline().split())
A = [list(map(int, readline().split())) for _ in range(N)]
B = [[None]*M for _ in range(N)]
st = 2*N
en = 2*N+1
used = set()
for num in range(M):
T = Dinic(2*N+2)
for i in range(N):
T.addedge(st, i, 1)
T.addedge(N+i, en, 1)
for j in range(M):
aij = A[i][j]
if aij not in used:
T.addedge(i, N+(aij-1)//M, 1)
T.getmf(st, en)
for i in range(N):
candi = [e for e, cost, _ in T.edge[i+N] if cost == 1][0]
for j in range(M):
if A[candi][j] not in used and (A[candi][j]-1)//M == i:
used.add(A[candi][j])
B[candi][num] = A[candi][j]
break
C = list(map(list, zip(*B)))
C = [sorted(c) for c in C]
C = list(map(list, zip(*C)))
for b in B:
print(*b)
for c in C:
print(*c)
``` | output | 1 | 89,040 | 15 | 178,081 |
Provide a correct Python 3 solution for this coding contest problem.
We have a grid with N rows and M columns of squares. Each integer from 1 to NM is written in this grid once. The number written in the square at the i-th row from the top and the j-th column from the left is A_{ij}.
You need to rearrange these numbers as follows:
1. First, for each of the N rows, rearrange the numbers written in it as you like.
2. Second, for each of the M columns, rearrange the numbers written in it as you like.
3. Finally, for each of the N rows, rearrange the numbers written in it as you like.
After rearranging the numbers, you want the number written in the square at the i-th row from the top and the j-th column from the left to be M\times (i-1)+j. Construct one such way to rearrange the numbers. The constraints guarantee that it is always possible to achieve the objective.
Constraints
* 1 \leq N,M \leq 100
* 1 \leq A_{ij} \leq NM
* A_{ij} are distinct.
Input
Input is given from Standard Input in the following format:
N M
A_{11} A_{12} ... A_{1M}
:
A_{N1} A_{N2} ... A_{NM}
Output
Print one way to rearrange the numbers in the following format:
B_{11} B_{12} ... B_{1M}
:
B_{N1} B_{N2} ... B_{NM}
C_{11} C_{12} ... C_{1M}
:
C_{N1} C_{N2} ... C_{NM}
Here B_{ij} is the number written in the square at the i-th row from the top and the j-th column from the left after Step 1, and C_{ij} is the number written in that square after Step 2.
Examples
Input
3 2
2 6
4 3
1 5
Output
2 6
4 3
5 1
2 1
4 3
5 6
Input
3 4
1 4 7 10
2 5 8 11
3 6 9 12
Output
1 4 7 10
5 8 11 2
9 12 3 6
1 4 3 2
5 8 7 6
9 12 11 10 | instruction | 0 | 89,041 | 15 | 178,082 |
"Correct Solution:
```
"""
https://atcoder.jp/contests/agc037/tasks/agc037_d
まず分かりやすさのため全部-1してMで割る
2番目の操作後に上から0行目には全部0 , 1行目には全部1…となればおk
すなわち、1番目の操作では各列に1個づつ数字が入るようにする必要がある
→これは貪欲にやればok・・・じゃない
ある列をうまくはめれば必ず次も可能？
→最後の列を考えると、足りない要素と残りの要素のsetは等しいのでokそう
→2部マッチング！
=====解説=====
そもそも何をマッチングさせてた？
→各行の置けるaとa
→ 次数は不定 & 不定
→ Hallが成り立つと証明できない…
→ あれ???????正解解法もおかしくないか
"""
from collections import defaultdict
from collections import deque
route = []
def Ford_Fulkerson_Func(s,g,lines,cost):
global route
N = len(cost)
ans = 0
queue = deque([ [s,float("inf")] ])
ed = [True] * N
ed[s] = False
route = [0] * N
route[s] = -1
while queue:
now,flow = queue.pop()
for nex in lines[now]:
if ed[nex]:
flow = min(cost[now][nex],flow)
route[nex] = now
queue.append([nex,flow])
ed[nex] = False
if nex == g:
ans += flow
break
else:
continue
break
else:
return False,ans
t = g
s = route[t]
while s != -1:
cost[s][t] -= flow
if cost[s][t] == 0:
lines[s].remove(t)
if cost[t][s] == 0:
lines[t].add(s)
cost[t][s] += flow
t = s
s = route[t]
return True,ans
def Ford_Fulkerson(s,g,lines,cost):
ans = 0
while True:
fl,nans = Ford_Fulkerson_Func(s,g,lines,cost)
if fl:
ans += nans
continue
else:
break
return ans
from sys import stdin
import sys
sys.setrecursionlimit(10000)
N,M = map(int,stdin.readline().split())
B = [[None] * M for i in range(N)]
Bs = [[True] * M for i in range(N)]
A = []
for loop in range(N):
a = list(map(int,stdin.readline().split()))
A.append(a)
for loop in range(M):
lines = defaultdict(set)
cost = [ [0] * (2*N+2) for i in range(2*N+2) ]
for i in range(N):
lines[2*N].add(i)
cost[2*N][i] = 1
for i in range(N,2*N):
lines[i].add(2*N+1)
cost[i][2*N+1] = 1
for i in range(N):
for j in range(len(A[i])):
lines[i].add((A[i][j]-1)//M + N)
cost[i][(A[i][j]-1)//M + N] = 1
Ford_Fulkerson(2*N,2*N+1,lines,cost)
ans = []
for i in range(N):
for j in range(len(A[i])):
if cost[i][(A[i][j]-1)//M + N] == 0:
ans.append(A[i][j])
del A[i][j]
break
#print (ans)
for i in range(N):
B[i][loop] = ans[i]
for i in B:
print (*i)
C = [[None] * M for i in range(N)]
for j in range(M):
tmp = []
for i in range(N):
tmp.append(B[i][j])
tmp.sort()
for i in range(N):
C[i][j] = tmp[i]
for i in C:
print (*i)
``` | output | 1 | 89,041 | 15 | 178,083 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a grid with N rows and M columns of squares. Each integer from 1 to NM is written in this grid once. The number written in the square at the i-th row from the top and the j-th column from the left is A_{ij}.
You need to rearrange these numbers as follows:
1. First, for each of the N rows, rearrange the numbers written in it as you like.
2. Second, for each of the M columns, rearrange the numbers written in it as you like.
3. Finally, for each of the N rows, rearrange the numbers written in it as you like.
After rearranging the numbers, you want the number written in the square at the i-th row from the top and the j-th column from the left to be M\times (i-1)+j. Construct one such way to rearrange the numbers. The constraints guarantee that it is always possible to achieve the objective.
Constraints
* 1 \leq N,M \leq 100
* 1 \leq A_{ij} \leq NM
* A_{ij} are distinct.
Input
Input is given from Standard Input in the following format:
N M
A_{11} A_{12} ... A_{1M}
:
A_{N1} A_{N2} ... A_{NM}
Output
Print one way to rearrange the numbers in the following format:
B_{11} B_{12} ... B_{1M}
:
B_{N1} B_{N2} ... B_{NM}
C_{11} C_{12} ... C_{1M}
:
C_{N1} C_{N2} ... C_{NM}
Here B_{ij} is the number written in the square at the i-th row from the top and the j-th column from the left after Step 1, and C_{ij} is the number written in that square after Step 2.
Examples
Input
3 2
2 6
4 3
1 5
Output
2 6
4 3
5 1
2 1
4 3
5 6
Input
3 4
1 4 7 10
2 5 8 11
3 6 9 12
Output
1 4 7 10
5 8 11 2
9 12 3 6
1 4 3 2
5 8 7 6
9 12 11 10
Submitted Solution:
```
import sys
from collections import Counter
input = sys.stdin.readline
N, M = list(map(int,input().split())) # N:row M:column
A = [list(map(int,input().split())) for i in range(N)]
B = [list(map(lambda x:int((x-1)/M) ,A[i])) for i in range(N)]
rrow = [-1]*N
maxnum = [[i,0,-1] for i in range(N)]
order = 1
while min(rrow)==-1:
for i in range(N):
if rrow[i]==-1:
C = Counter(B[i])
value, num = C.most_common(order)[-1]
if maxnum[value][1]<num:
maxnum[value][1] = num
rrow[maxnum[value][2]] = -1
rrow[i] = value
maxnum[value][2] = i
order += 1
fix = [[0]*M for i in range(N)]
ss = [set(range(N)) for i in range(M)]
for iy in range(N):
t = rrow[iy]
col = 0
for ix in range(M):
if B[iy][ix]==t:
B[iy][col], B[iy][ix] = B[iy][ix], B[iy][col]
A[iy][col], A[iy][ix] = A[iy][ix], A[iy][col]
fix[iy][col] = 1
ss[col].remove(t)
col += 1
for ix in range(M):
while True:
ty = -1
minlen = 1000
for iy in range(N):
if fix[iy][ix]==0:
notfix = True
iset = ss[ix].intersection(B[iy][ix::])
ilen = len(iset)
if ilen<minlen:
ty = iy
tset = iset
minlen = ilen
if minlen==1: break
if ty==-1: break
jy, jx = ty, ix
while B[jy][jx] not in tset: jx+=1
ss[ix].remove(B[jy][jx])
B[jy][jx], B[jy][ix] = B[jy][ix], B[jy][jx]
A[jy][jx], A[jy][ix] = A[jy][ix], A[jy][jx]
fix[jy][ix] = 1
for i in A:print(*i)
for ix in range(M):
for iy in range(N):
row = B[iy][ix]
B[iy][ix], B[row][ix] = B[row][ix], B[iy][ix]
A[iy][ix], A[row][ix] = A[row][ix], A[iy][ix]
for i in A:print(*i)
``` | instruction | 0 | 89,042 | 15 | 178,084 |
No | output | 1 | 89,042 | 15 | 178,085 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a grid with N rows and M columns of squares. Each integer from 1 to NM is written in this grid once. The number written in the square at the i-th row from the top and the j-th column from the left is A_{ij}.
You need to rearrange these numbers as follows:
1. First, for each of the N rows, rearrange the numbers written in it as you like.
2. Second, for each of the M columns, rearrange the numbers written in it as you like.
3. Finally, for each of the N rows, rearrange the numbers written in it as you like.
After rearranging the numbers, you want the number written in the square at the i-th row from the top and the j-th column from the left to be M\times (i-1)+j. Construct one such way to rearrange the numbers. The constraints guarantee that it is always possible to achieve the objective.
Constraints
* 1 \leq N,M \leq 100
* 1 \leq A_{ij} \leq NM
* A_{ij} are distinct.
Input
Input is given from Standard Input in the following format:
N M
A_{11} A_{12} ... A_{1M}
:
A_{N1} A_{N2} ... A_{NM}
Output
Print one way to rearrange the numbers in the following format:
B_{11} B_{12} ... B_{1M}
:
B_{N1} B_{N2} ... B_{NM}
C_{11} C_{12} ... C_{1M}
:
C_{N1} C_{N2} ... C_{NM}
Here B_{ij} is the number written in the square at the i-th row from the top and the j-th column from the left after Step 1, and C_{ij} is the number written in that square after Step 2.
Examples
Input
3 2
2 6
4 3
1 5
Output
2 6
4 3
5 1
2 1
4 3
5 6
Input
3 4
1 4 7 10
2 5 8 11
3 6 9 12
Output
1 4 7 10
5 8 11 2
9 12 3 6
1 4 3 2
5 8 7 6
9 12 11 10
Submitted Solution:
```
import copy
n,m = list(map(int, input().split()))
A = [[int(i) for i in input().split()] for n in range(n)]
ngset = {}
for i in range(n):
tmp = []
for j in range(1,m+1):
tmp.append((i*m+j))
for k in tmp:
ngset[k] = tmp
ngs = {}
for i, aa in enumerate(A):
if i == 0:
for j, a in enumerate(aa):
ngs[j] = []
ngs[j].extend(ngset[a])
else:
for j, a in enumerate(aa):
if a in ngs[j]:
k = None
if j+1 < m:
for jtmp in range(j+1, m):
if A[i][j] not in ngs[jtmp] and A[i][jtmp] not in ngs[j]:
k = jtmp
break
if not k:
for jtmp in range(j, -1, -1):
if A[i][j] not in ngs[jtmp] and A[i][jtmp] not in ngs[j]:
k = jtmp
break
A[i][j] , A[i][k] = A[i][k], A[i][j]
for j, a in enumerate(aa):
print(ngs[j])
print(ngset[a])
``` | instruction | 0 | 89,043 | 15 | 178,086 |
No | output | 1 | 89,043 | 15 | 178,087 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a grid with N rows and M columns of squares. Each integer from 1 to NM is written in this grid once. The number written in the square at the i-th row from the top and the j-th column from the left is A_{ij}.
You need to rearrange these numbers as follows:
1. First, for each of the N rows, rearrange the numbers written in it as you like.
2. Second, for each of the M columns, rearrange the numbers written in it as you like.
3. Finally, for each of the N rows, rearrange the numbers written in it as you like.
After rearranging the numbers, you want the number written in the square at the i-th row from the top and the j-th column from the left to be M\times (i-1)+j. Construct one such way to rearrange the numbers. The constraints guarantee that it is always possible to achieve the objective.
Constraints
* 1 \leq N,M \leq 100
* 1 \leq A_{ij} \leq NM
* A_{ij} are distinct.
Input
Input is given from Standard Input in the following format:
N M
A_{11} A_{12} ... A_{1M}
:
A_{N1} A_{N2} ... A_{NM}
Output
Print one way to rearrange the numbers in the following format:
B_{11} B_{12} ... B_{1M}
:
B_{N1} B_{N2} ... B_{NM}
C_{11} C_{12} ... C_{1M}
:
C_{N1} C_{N2} ... C_{NM}
Here B_{ij} is the number written in the square at the i-th row from the top and the j-th column from the left after Step 1, and C_{ij} is the number written in that square after Step 2.
Examples
Input
3 2
2 6
4 3
1 5
Output
2 6
4 3
5 1
2 1
4 3
5 6
Input
3 4
1 4 7 10
2 5 8 11
3 6 9 12
Output
1 4 7 10
5 8 11 2
9 12 3 6
1 4 3 2
5 8 7 6
9 12 11 10
Submitted Solution:
```
import copy
n,m = list(map(int, input().split()))
A = [[int(i) for i in input().split()] for n in range(n)]
ngset = {}
for i in range(n):
tmp = []
for j in range(1,m+1):
tmp.append((i*m+j))
for k in tmp:
ngset[k] = tmp
ngs = {}
for i, aa in enumerate(A):
if i == 0:
for j, a in enumerate(aa):
ngs[j] = []
ngs[j].extend(ngset[a])
else:
for j, a in enumerate(aa):
if a in ngs[j]:
k = None
if j+1 < m:
for jtmp in range(j+1, m):
if A[i][j] not in ngs[jtmp] and A[i][jtmp] not in ngs[j]:
k = jtmp
break
if not k:
for jtmp in range(j, -1, -1):
if A[i][j] not in ngs[jtmp] and A[i][jtmp] not in ngs[j]:
k = jtmp
break
A[i][j] , A[i][k] = A[i][k], A[i][j]
for j, a in enumerate(aa):
ngs[j].extend(ngset[a])
B = copy.deepcopy(A)
import numpy as np
A = np.array(A)
for i in range(m):
A[:,i].sort()
for i in range(n):
text = ""
for j in range(m):
text += str(B[i][j]) + " "
print(text[:-1])
for i in range(n):
text = ""
for j in range(m):
text += str(A[i][j]) + " "
print(text[:-1])
``` | instruction | 0 | 89,044 | 15 | 178,088 |
No | output | 1 | 89,044 | 15 | 178,089 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a grid with N rows and M columns of squares. Each integer from 1 to NM is written in this grid once. The number written in the square at the i-th row from the top and the j-th column from the left is A_{ij}.
You need to rearrange these numbers as follows:
1. First, for each of the N rows, rearrange the numbers written in it as you like.
2. Second, for each of the M columns, rearrange the numbers written in it as you like.
3. Finally, for each of the N rows, rearrange the numbers written in it as you like.
After rearranging the numbers, you want the number written in the square at the i-th row from the top and the j-th column from the left to be M\times (i-1)+j. Construct one such way to rearrange the numbers. The constraints guarantee that it is always possible to achieve the objective.
Constraints
* 1 \leq N,M \leq 100
* 1 \leq A_{ij} \leq NM
* A_{ij} are distinct.
Input
Input is given from Standard Input in the following format:
N M
A_{11} A_{12} ... A_{1M}
:
A_{N1} A_{N2} ... A_{NM}
Output
Print one way to rearrange the numbers in the following format:
B_{11} B_{12} ... B_{1M}
:
B_{N1} B_{N2} ... B_{NM}
C_{11} C_{12} ... C_{1M}
:
C_{N1} C_{N2} ... C_{NM}
Here B_{ij} is the number written in the square at the i-th row from the top and the j-th column from the left after Step 1, and C_{ij} is the number written in that square after Step 2.
Examples
Input
3 2
2 6
4 3
1 5
Output
2 6
4 3
5 1
2 1
4 3
5 6
Input
3 4
1 4 7 10
2 5 8 11
3 6 9 12
Output
1 4 7 10
5 8 11 2
9 12 3 6
1 4 3 2
5 8 7 6
9 12 11 10
Submitted Solution:
```
import sys
input = sys.stdin.readline
sys.setrecursionlimit(10 ** 7)
"""
正則二部グラフの辺彩色
"""
import numpy as np
N,M = map(int,input().split())
grid = [[int(x) for x in input().split()] for _ in range(N)]
edge = []
for i,row in enumerate(grid):
for j,x in enumerate(row):
edge.append(((x-1)//M,N+i))
graph = [dict() for _ in range(N+N)] # 頂点ごろに、色->相手
rest_color = [set(range(M)) for _ in range(N+N)]
for u,v in edge:
c = rest_color[u].pop()
if c in rest_color[v]:
graph[u][c] = v
graph[v][c] = u
rest_color[v].remove(c)
continue
# 交互道を作って色c,dと当てていく
d = rest_color[v].pop()
cd = c + d
V = [u,v]
next_c = c
while next_c not in rest_color[v]:
v = graph[v][next_c]
V.append(v)
next_c = cd - next_c
rest_color[v].remove(next_c)
rest_color[v].add(cd - next_c)
for i,(u,v) in enumerate(zip(V,V[1:])):
if i%2 == 0:
graph[u][c] = v
graph[v][c] = u
else:
graph[u][d] = v
graph[v][d] = u
after = [[None] * M for _ in range(N)]
for i,row in enumerate(grid):
mod_to_x = [[] for _ in range(N)]
for x in row:
mod_to_x[(x-1)//M].append(x)
for color in range(M):
after[i][color] = mod_to_x[graph[i][color] - N].pop()
B = np.array(after)
C = B.copy()
C.sort(axis = 0)
print('\n'.join(' '.join(row) for row in B.astype(str)))
print('\n'.join(' '.join(row) for row in C.astype(str)))
``` | instruction | 0 | 89,045 | 15 | 178,090 |
No | output | 1 | 89,045 | 15 | 178,091 |
Provide a correct Python 3 solution for this coding contest problem.
We have a 2 \times N grid. We will denote the square at the i-th row and j-th column (1 \leq i \leq 2, 1 \leq j \leq N) as (i, j).
You are initially in the top-left square, (1, 1). You will travel to the bottom-right square, (2, N), by repeatedly moving right or down.
The square (i, j) contains A_{i, j} candies. You will collect all the candies you visit during the travel. The top-left and bottom-right squares also contain candies, and you will also collect them.
At most how many candies can you collect when you choose the best way to travel?
Constraints
* 1 \leq N \leq 100
* 1 \leq A_{i, j} \leq 100 (1 \leq i \leq 2, 1 \leq j \leq N)
Input
Input is given from Standard Input in the following format:
N
A_{1, 1} A_{1, 2} ... A_{1, N}
A_{2, 1} A_{2, 2} ... A_{2, N}
Output
Print the maximum number of candies that can be collected.
Examples
Input
5
3 2 2 4 1
1 2 2 2 1
Output
14
Input
4
1 1 1 1
1 1 1 1
Output
5
Input
7
3 3 4 5 4 5 3
5 3 4 4 2 3 2
Output
29
Input
1
2
3
Output
5 | instruction | 0 | 89,961 | 15 | 179,922 |
"Correct Solution:
```
N = int(input())
A1 = list(map(int,input().split()))
A2 = list(map(int,input().split()))
sums = [sum(A1[:i])+sum(A2[i-1:]) for i in range(1,N+1)]
print(max(sums))
``` | output | 1 | 89,961 | 15 | 179,923 |
Provide a correct Python 3 solution for this coding contest problem.
We have a 2 \times N grid. We will denote the square at the i-th row and j-th column (1 \leq i \leq 2, 1 \leq j \leq N) as (i, j).
You are initially in the top-left square, (1, 1). You will travel to the bottom-right square, (2, N), by repeatedly moving right or down.
The square (i, j) contains A_{i, j} candies. You will collect all the candies you visit during the travel. The top-left and bottom-right squares also contain candies, and you will also collect them.
At most how many candies can you collect when you choose the best way to travel?
Constraints
* 1 \leq N \leq 100
* 1 \leq A_{i, j} \leq 100 (1 \leq i \leq 2, 1 \leq j \leq N)
Input
Input is given from Standard Input in the following format:
N
A_{1, 1} A_{1, 2} ... A_{1, N}
A_{2, 1} A_{2, 2} ... A_{2, N}
Output
Print the maximum number of candies that can be collected.
Examples
Input
5
3 2 2 4 1
1 2 2 2 1
Output
14
Input
4
1 1 1 1
1 1 1 1
Output
5
Input
7
3 3 4 5 4 5 3
5 3 4 4 2 3 2
Output
29
Input
1
2
3
Output
5 | instruction | 0 | 89,962 | 15 | 179,924 |
"Correct Solution:
```
n = int(input())
a1 = list(map(int, input().split()))
a2 = list(map(int, input().split()))
m = 0
for i in range(n):
m = max(m, sum(a1[:i+1])+sum(a2[i:]))
print (m)
``` | output | 1 | 89,962 | 15 | 179,925 |
Provide a correct Python 3 solution for this coding contest problem.
We have a 2 \times N grid. We will denote the square at the i-th row and j-th column (1 \leq i \leq 2, 1 \leq j \leq N) as (i, j).
You are initially in the top-left square, (1, 1). You will travel to the bottom-right square, (2, N), by repeatedly moving right or down.
The square (i, j) contains A_{i, j} candies. You will collect all the candies you visit during the travel. The top-left and bottom-right squares also contain candies, and you will also collect them.
At most how many candies can you collect when you choose the best way to travel?
Constraints
* 1 \leq N \leq 100
* 1 \leq A_{i, j} \leq 100 (1 \leq i \leq 2, 1 \leq j \leq N)
Input
Input is given from Standard Input in the following format:
N
A_{1, 1} A_{1, 2} ... A_{1, N}
A_{2, 1} A_{2, 2} ... A_{2, N}
Output
Print the maximum number of candies that can be collected.
Examples
Input
5
3 2 2 4 1
1 2 2 2 1
Output
14
Input
4
1 1 1 1
1 1 1 1
Output
5
Input
7
3 3 4 5 4 5 3
5 3 4 4 2 3 2
Output
29
Input
1
2
3
Output
5 | instruction | 0 | 89,963 | 15 | 179,926 |
"Correct Solution:
```
n=int(input())
l=[list(map(int,input().split()))for _ in range(2)]
a=0
for i in range(n):
a=max(a,sum(l[0][:i+1])+sum(l[1][i:]))
print(a)
``` | output | 1 | 89,963 | 15 | 179,927 |
Provide a correct Python 3 solution for this coding contest problem.
We have a 2 \times N grid. We will denote the square at the i-th row and j-th column (1 \leq i \leq 2, 1 \leq j \leq N) as (i, j).
You are initially in the top-left square, (1, 1). You will travel to the bottom-right square, (2, N), by repeatedly moving right or down.
The square (i, j) contains A_{i, j} candies. You will collect all the candies you visit during the travel. The top-left and bottom-right squares also contain candies, and you will also collect them.
At most how many candies can you collect when you choose the best way to travel?
Constraints
* 1 \leq N \leq 100
* 1 \leq A_{i, j} \leq 100 (1 \leq i \leq 2, 1 \leq j \leq N)
Input
Input is given from Standard Input in the following format:
N
A_{1, 1} A_{1, 2} ... A_{1, N}
A_{2, 1} A_{2, 2} ... A_{2, N}
Output
Print the maximum number of candies that can be collected.
Examples
Input
5
3 2 2 4 1
1 2 2 2 1
Output
14
Input
4
1 1 1 1
1 1 1 1
Output
5
Input
7
3 3 4 5 4 5 3
5 3 4 4 2 3 2
Output
29
Input
1
2
3
Output
5 | instruction | 0 | 89,964 | 15 | 179,928 |
"Correct Solution:
```
n = int(input())
a = [list(map(int,input().split())) for i in range(2)]
get = []
for i in range(n):
get =get+ [sum(a[0][:i+1]) + sum(a[1][i:])]
print(max(get))
``` | output | 1 | 89,964 | 15 | 179,929 |
Provide a correct Python 3 solution for this coding contest problem.
We have a 2 \times N grid. We will denote the square at the i-th row and j-th column (1 \leq i \leq 2, 1 \leq j \leq N) as (i, j).
You are initially in the top-left square, (1, 1). You will travel to the bottom-right square, (2, N), by repeatedly moving right or down.
The square (i, j) contains A_{i, j} candies. You will collect all the candies you visit during the travel. The top-left and bottom-right squares also contain candies, and you will also collect them.
At most how many candies can you collect when you choose the best way to travel?
Constraints
* 1 \leq N \leq 100
* 1 \leq A_{i, j} \leq 100 (1 \leq i \leq 2, 1 \leq j \leq N)
Input
Input is given from Standard Input in the following format:
N
A_{1, 1} A_{1, 2} ... A_{1, N}
A_{2, 1} A_{2, 2} ... A_{2, N}
Output
Print the maximum number of candies that can be collected.
Examples
Input
5
3 2 2 4 1
1 2 2 2 1
Output
14
Input
4
1 1 1 1
1 1 1 1
Output
5
Input
7
3 3 4 5 4 5 3
5 3 4 4 2 3 2
Output
29
Input
1
2
3
Output
5 | instruction | 0 | 89,965 | 15 | 179,930 |
"Correct Solution:
```
n=int(input())
A_1=list(map(int,input().split()))
A_2=list(map(int,input().split()))
ans=0
for i in range(n):
c=sum(A_1[:i+1])+sum(A_2[i:])
ans=max(ans,c)
print(ans)
``` | output | 1 | 89,965 | 15 | 179,931 |
Provide a correct Python 3 solution for this coding contest problem.
We have a 2 \times N grid. We will denote the square at the i-th row and j-th column (1 \leq i \leq 2, 1 \leq j \leq N) as (i, j).
You are initially in the top-left square, (1, 1). You will travel to the bottom-right square, (2, N), by repeatedly moving right or down.
The square (i, j) contains A_{i, j} candies. You will collect all the candies you visit during the travel. The top-left and bottom-right squares also contain candies, and you will also collect them.
At most how many candies can you collect when you choose the best way to travel?
Constraints
* 1 \leq N \leq 100
* 1 \leq A_{i, j} \leq 100 (1 \leq i \leq 2, 1 \leq j \leq N)
Input
Input is given from Standard Input in the following format:
N
A_{1, 1} A_{1, 2} ... A_{1, N}
A_{2, 1} A_{2, 2} ... A_{2, N}
Output
Print the maximum number of candies that can be collected.
Examples
Input
5
3 2 2 4 1
1 2 2 2 1
Output
14
Input
4
1 1 1 1
1 1 1 1
Output
5
Input
7
3 3 4 5 4 5 3
5 3 4 4 2 3 2
Output
29
Input
1
2
3
Output
5 | instruction | 0 | 89,966 | 15 | 179,932 |
"Correct Solution:
```
n=int(input())
x=list(map(int,input().split()))
y=list(map(int,input().split()))
mx=0
for i in range(n):
ans=sum(x[:i+1])+sum(y[i:])
if ans>mx:
mx=ans
print(mx)
``` | output | 1 | 89,966 | 15 | 179,933 |
Provide a correct Python 3 solution for this coding contest problem.
We have a 2 \times N grid. We will denote the square at the i-th row and j-th column (1 \leq i \leq 2, 1 \leq j \leq N) as (i, j).
You are initially in the top-left square, (1, 1). You will travel to the bottom-right square, (2, N), by repeatedly moving right or down.
The square (i, j) contains A_{i, j} candies. You will collect all the candies you visit during the travel. The top-left and bottom-right squares also contain candies, and you will also collect them.
At most how many candies can you collect when you choose the best way to travel?
Constraints
* 1 \leq N \leq 100
* 1 \leq A_{i, j} \leq 100 (1 \leq i \leq 2, 1 \leq j \leq N)
Input
Input is given from Standard Input in the following format:
N
A_{1, 1} A_{1, 2} ... A_{1, N}
A_{2, 1} A_{2, 2} ... A_{2, N}
Output
Print the maximum number of candies that can be collected.
Examples
Input
5
3 2 2 4 1
1 2 2 2 1
Output
14
Input
4
1 1 1 1
1 1 1 1
Output
5
Input
7
3 3 4 5 4 5 3
5 3 4 4 2 3 2
Output
29
Input
1
2
3
Output
5 | instruction | 0 | 89,967 | 15 | 179,934 |
"Correct Solution:
```
n=int(input())
a=list(map(int,input().split()))
b=list(map(int, input().split()))
ans = 0
for i in range(n):
ans = max(ans,sum(a[:i+1])+sum(b[i:n]))
print(ans)
``` | output | 1 | 89,967 | 15 | 179,935 |
Provide a correct Python 3 solution for this coding contest problem.
We have a 2 \times N grid. We will denote the square at the i-th row and j-th column (1 \leq i \leq 2, 1 \leq j \leq N) as (i, j).
You are initially in the top-left square, (1, 1). You will travel to the bottom-right square, (2, N), by repeatedly moving right or down.
The square (i, j) contains A_{i, j} candies. You will collect all the candies you visit during the travel. The top-left and bottom-right squares also contain candies, and you will also collect them.
At most how many candies can you collect when you choose the best way to travel?
Constraints
* 1 \leq N \leq 100
* 1 \leq A_{i, j} \leq 100 (1 \leq i \leq 2, 1 \leq j \leq N)
Input
Input is given from Standard Input in the following format:
N
A_{1, 1} A_{1, 2} ... A_{1, N}
A_{2, 1} A_{2, 2} ... A_{2, N}
Output
Print the maximum number of candies that can be collected.
Examples
Input
5
3 2 2 4 1
1 2 2 2 1
Output
14
Input
4
1 1 1 1
1 1 1 1
Output
5
Input
7
3 3 4 5 4 5 3
5 3 4 4 2 3 2
Output
29
Input
1
2
3
Output
5 | instruction | 0 | 89,968 | 15 | 179,936 |
"Correct Solution:
```
N = int(input())
A = list(map(int, input().split()))
B= list(map(int, input().split()))
m=0
for i in range(N):
s=sum(A[:i+1])+sum(B[i:])
m=max(m, s)
print(m)
``` | output | 1 | 89,968 | 15 | 179,937 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a 2 \times N grid. We will denote the square at the i-th row and j-th column (1 \leq i \leq 2, 1 \leq j \leq N) as (i, j).
You are initially in the top-left square, (1, 1). You will travel to the bottom-right square, (2, N), by repeatedly moving right or down.
The square (i, j) contains A_{i, j} candies. You will collect all the candies you visit during the travel. The top-left and bottom-right squares also contain candies, and you will also collect them.
At most how many candies can you collect when you choose the best way to travel?
Constraints
* 1 \leq N \leq 100
* 1 \leq A_{i, j} \leq 100 (1 \leq i \leq 2, 1 \leq j \leq N)
Input
Input is given from Standard Input in the following format:
N
A_{1, 1} A_{1, 2} ... A_{1, N}
A_{2, 1} A_{2, 2} ... A_{2, N}
Output
Print the maximum number of candies that can be collected.
Examples
Input
5
3 2 2 4 1
1 2 2 2 1
Output
14
Input
4
1 1 1 1
1 1 1 1
Output
5
Input
7
3 3 4 5 4 5 3
5 3 4 4 2 3 2
Output
29
Input
1
2
3
Output
5
Submitted Solution:
```
#087_C
n=int(input())
a=[[int(j) for j in input().split()] for _ in range(2)]
print(max([sum(a[0][:(i+1)])+sum(a[1][i:]) for i in range(n)]))
``` | instruction | 0 | 89,969 | 15 | 179,938 |
Yes | output | 1 | 89,969 | 15 | 179,939 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a 2 \times N grid. We will denote the square at the i-th row and j-th column (1 \leq i \leq 2, 1 \leq j \leq N) as (i, j).
You are initially in the top-left square, (1, 1). You will travel to the bottom-right square, (2, N), by repeatedly moving right or down.
The square (i, j) contains A_{i, j} candies. You will collect all the candies you visit during the travel. The top-left and bottom-right squares also contain candies, and you will also collect them.
At most how many candies can you collect when you choose the best way to travel?
Constraints
* 1 \leq N \leq 100
* 1 \leq A_{i, j} \leq 100 (1 \leq i \leq 2, 1 \leq j \leq N)
Input
Input is given from Standard Input in the following format:
N
A_{1, 1} A_{1, 2} ... A_{1, N}
A_{2, 1} A_{2, 2} ... A_{2, N}
Output
Print the maximum number of candies that can be collected.
Examples
Input
5
3 2 2 4 1
1 2 2 2 1
Output
14
Input
4
1 1 1 1
1 1 1 1
Output
5
Input
7
3 3 4 5 4 5 3
5 3 4 4 2 3 2
Output
29
Input
1
2
3
Output
5
Submitted Solution:
```
N=int(input())
A1=list(map(int,input().split()))
A2=list(map(int,input().split()))
ans=0
for i in range(1,N+1):
ans=max(ans,sum(A1[0:i]+A2[i-1:N]))
print(ans)
``` | instruction | 0 | 89,970 | 15 | 179,940 |
Yes | output | 1 | 89,970 | 15 | 179,941 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a 2 \times N grid. We will denote the square at the i-th row and j-th column (1 \leq i \leq 2, 1 \leq j \leq N) as (i, j).
You are initially in the top-left square, (1, 1). You will travel to the bottom-right square, (2, N), by repeatedly moving right or down.
The square (i, j) contains A_{i, j} candies. You will collect all the candies you visit during the travel. The top-left and bottom-right squares also contain candies, and you will also collect them.
At most how many candies can you collect when you choose the best way to travel?
Constraints
* 1 \leq N \leq 100
* 1 \leq A_{i, j} \leq 100 (1 \leq i \leq 2, 1 \leq j \leq N)
Input
Input is given from Standard Input in the following format:
N
A_{1, 1} A_{1, 2} ... A_{1, N}
A_{2, 1} A_{2, 2} ... A_{2, N}
Output
Print the maximum number of candies that can be collected.
Examples
Input
5
3 2 2 4 1
1 2 2 2 1
Output
14
Input
4
1 1 1 1
1 1 1 1
Output
5
Input
7
3 3 4 5 4 5 3
5 3 4 4 2 3 2
Output
29
Input
1
2
3
Output
5
Submitted Solution:
```
n=int(input())
A1 = list(map(int, input().split()))
A2 = list(map(int, input().split()))
m=-1
for i in range(n):
m=max(m, sum(A1[:i+1])+sum(A2[i:]))
print(m)
``` | instruction | 0 | 89,971 | 15 | 179,942 |
Yes | output | 1 | 89,971 | 15 | 179,943 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a 2 \times N grid. We will denote the square at the i-th row and j-th column (1 \leq i \leq 2, 1 \leq j \leq N) as (i, j).
You are initially in the top-left square, (1, 1). You will travel to the bottom-right square, (2, N), by repeatedly moving right or down.
The square (i, j) contains A_{i, j} candies. You will collect all the candies you visit during the travel. The top-left and bottom-right squares also contain candies, and you will also collect them.
At most how many candies can you collect when you choose the best way to travel?
Constraints
* 1 \leq N \leq 100
* 1 \leq A_{i, j} \leq 100 (1 \leq i \leq 2, 1 \leq j \leq N)
Input
Input is given from Standard Input in the following format:
N
A_{1, 1} A_{1, 2} ... A_{1, N}
A_{2, 1} A_{2, 2} ... A_{2, N}
Output
Print the maximum number of candies that can be collected.
Examples
Input
5
3 2 2 4 1
1 2 2 2 1
Output
14
Input
4
1 1 1 1
1 1 1 1
Output
5
Input
7
3 3 4 5 4 5 3
5 3 4 4 2 3 2
Output
29
Input
1
2
3
Output
5
Submitted Solution:
```
N = int(input())
A = [list(map(int,input().split())) for k in range(2)]
ans = 0
for x in range(N):
ans = max(ans,sum(A[0][:x+1])+sum(A[1][x:]))
print(ans)
``` | instruction | 0 | 89,972 | 15 | 179,944 |
Yes | output | 1 | 89,972 | 15 | 179,945 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a 2 \times N grid. We will denote the square at the i-th row and j-th column (1 \leq i \leq 2, 1 \leq j \leq N) as (i, j).
You are initially in the top-left square, (1, 1). You will travel to the bottom-right square, (2, N), by repeatedly moving right or down.
The square (i, j) contains A_{i, j} candies. You will collect all the candies you visit during the travel. The top-left and bottom-right squares also contain candies, and you will also collect them.
At most how many candies can you collect when you choose the best way to travel?
Constraints
* 1 \leq N \leq 100
* 1 \leq A_{i, j} \leq 100 (1 \leq i \leq 2, 1 \leq j \leq N)
Input
Input is given from Standard Input in the following format:
N
A_{1, 1} A_{1, 2} ... A_{1, N}
A_{2, 1} A_{2, 2} ... A_{2, N}
Output
Print the maximum number of candies that can be collected.
Examples
Input
5
3 2 2 4 1
1 2 2 2 1
Output
14
Input
4
1 1 1 1
1 1 1 1
Output
5
Input
7
3 3 4 5 4 5 3
5 3 4 4 2 3 2
Output
29
Input
1
2
3
Output
5
Submitted Solution:
```
a, b, c, x = [int(input()) for i in range(4)]
can_count = 0
for i in range(a+1):
coins_price_500 = i * 500
for j in range(b+1):
coins_price_100 = j * 100
for k in range(c+1):
coins_price_50 = k * 50
if x == coins_price_500 * coins_price_100 * coins_price_50:
can_count += 1
print(can_count)
``` | instruction | 0 | 89,973 | 15 | 179,946 |
No | output | 1 | 89,973 | 15 | 179,947 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a 2 \times N grid. We will denote the square at the i-th row and j-th column (1 \leq i \leq 2, 1 \leq j \leq N) as (i, j).
You are initially in the top-left square, (1, 1). You will travel to the bottom-right square, (2, N), by repeatedly moving right or down.
The square (i, j) contains A_{i, j} candies. You will collect all the candies you visit during the travel. The top-left and bottom-right squares also contain candies, and you will also collect them.
At most how many candies can you collect when you choose the best way to travel?
Constraints
* 1 \leq N \leq 100
* 1 \leq A_{i, j} \leq 100 (1 \leq i \leq 2, 1 \leq j \leq N)
Input
Input is given from Standard Input in the following format:
N
A_{1, 1} A_{1, 2} ... A_{1, N}
A_{2, 1} A_{2, 2} ... A_{2, N}
Output
Print the maximum number of candies that can be collected.
Examples
Input
5
3 2 2 4 1
1 2 2 2 1
Output
14
Input
4
1 1 1 1
1 1 1 1
Output
5
Input
7
3 3 4 5 4 5 3
5 3 4 4 2 3 2
Output
29
Input
1
2
3
Output
5
Submitted Solution:
```
def main():
n = int(input())
a_lst1 = list(map(int, input().split()))
a_lst2 = list(map(int, input().split()))
candies_lst = []
tmp = 1
while tmp <= 7:
a1_tmp = a_lst1[:tmp]
a2_tmp = a_lst2[tmp-1:]
a1 = sum(a1_tmp)
a2 = sum(a2_tmp)
tmp += 1
candies = a1 + a2
candies_lst.append(candies)
maximum = max(candies_lst)
print(maximum)
if __name__ == '__main__':
main()
``` | instruction | 0 | 89,974 | 15 | 179,948 |
No | output | 1 | 89,974 | 15 | 179,949 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a 2 \times N grid. We will denote the square at the i-th row and j-th column (1 \leq i \leq 2, 1 \leq j \leq N) as (i, j).
You are initially in the top-left square, (1, 1). You will travel to the bottom-right square, (2, N), by repeatedly moving right or down.
The square (i, j) contains A_{i, j} candies. You will collect all the candies you visit during the travel. The top-left and bottom-right squares also contain candies, and you will also collect them.
At most how many candies can you collect when you choose the best way to travel?
Constraints
* 1 \leq N \leq 100
* 1 \leq A_{i, j} \leq 100 (1 \leq i \leq 2, 1 \leq j \leq N)
Input
Input is given from Standard Input in the following format:
N
A_{1, 1} A_{1, 2} ... A_{1, N}
A_{2, 1} A_{2, 2} ... A_{2, N}
Output
Print the maximum number of candies that can be collected.
Examples
Input
5
3 2 2 4 1
1 2 2 2 1
Output
14
Input
4
1 1 1 1
1 1 1 1
Output
5
Input
7
3 3 4 5 4 5 3
5 3 4 4 2 3 2
Output
29
Input
1
2
3
Output
5
Submitted Solution:
```
n = int(input())
a = [list(map(int,input().split())) for i in range(2)]
ans = a[0][0] + a[1][-1]
for i in range(n-1):
if sum(a[0][i+1:]) >= sum(a[1][i:n-1]):
ans += a[0][i+1]
else:
ans += sum(a[1][i:n-1])
break
print(ans)
``` | instruction | 0 | 89,975 | 15 | 179,950 |
No | output | 1 | 89,975 | 15 | 179,951 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a 2 \times N grid. We will denote the square at the i-th row and j-th column (1 \leq i \leq 2, 1 \leq j \leq N) as (i, j).
You are initially in the top-left square, (1, 1). You will travel to the bottom-right square, (2, N), by repeatedly moving right or down.
The square (i, j) contains A_{i, j} candies. You will collect all the candies you visit during the travel. The top-left and bottom-right squares also contain candies, and you will also collect them.
At most how many candies can you collect when you choose the best way to travel?
Constraints
* 1 \leq N \leq 100
* 1 \leq A_{i, j} \leq 100 (1 \leq i \leq 2, 1 \leq j \leq N)
Input
Input is given from Standard Input in the following format:
N
A_{1, 1} A_{1, 2} ... A_{1, N}
A_{2, 1} A_{2, 2} ... A_{2, N}
Output
Print the maximum number of candies that can be collected.
Examples
Input
5
3 2 2 4 1
1 2 2 2 1
Output
14
Input
4
1 1 1 1
1 1 1 1
Output
5
Input
7
3 3 4 5 4 5 3
5 3 4 4 2 3 2
Output
29
Input
1
2
3
Output
5
Submitted Solution:
```
n=int(input())
a_list=[list(map(int,input().split())) for _ in range(2)]
top_cnts=[]
beneath_cnts=[]
top_cnt = 0
beneath_cnt=sum(a_list[1])
for a in a_list[0]:
top_cnt += a
top_cnts.append(top_cnt)
beneath_cnts.append(beneath_cnt)
beneath_cnt-=a
max_cnt = 0
for i in range(n):
cnt = top_cnts[i]+beneath_cnts[i]
if cnt >max_cnt:
max_cnt=cnt
print(max_cnt)
``` | instruction | 0 | 89,976 | 15 | 179,952 |
No | output | 1 | 89,976 | 15 | 179,953 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Dima overslept the alarm clock, which was supposed to raise him to school.
Dima wonders if he will have time to come to the first lesson. To do this, he needs to know the minimum time it will take him to get from home to school.
The city where Dima lives is a rectangular field of n × m size. Each cell (i, j) on this field is denoted by one number a_{ij}:
* The number -1 means that the passage through the cell is prohibited;
* The number 0 means that the cell is free and Dima can walk though it.
* The number x (1 ≤ x ≤ 10^9) means that the cell contains a portal with a cost of x. A cell with a portal is also considered free.
From any portal, Dima can go to any other portal, while the time of moving from the portal (i, j) to the portal (x, y) corresponds to the sum of their costs a_{ij} + a_{xy}.
In addition to moving between portals, Dima can also move between unoccupied cells adjacent to one side in time w. In particular, he can enter a cell with a portal and not use it.
Initially, Dima is in the upper-left cell (1, 1), and the school is in the lower right cell (n, m).
Input
The first line contains three integers n, m and w (2 ≤ n, m ≤ 2 ⋅ 10^3, 1 ≤ w ≤ 10^9), where n and m are city size, w is time during which Dima moves between unoccupied cells.
The next n lines each contain m numbers (-1 ≤ a_{ij} ≤ 10^9) — descriptions of cells.
It is guaranteed that the cells (1, 1) and (n, m) are free.
Output
Output the minimum time it will take for Dima to get to school. If he cannot get to school at all, then output "-1".
Example
Input
5 5 1
0 -1 0 1 -1
0 20 0 0 -1
-1 -1 -1 -1 -1
3 0 0 0 0
-1 0 0 0 0
Output
14
Note
Explanation for the first sample:
<image> | instruction | 0 | 90,380 | 15 | 180,760 |
Tags: brute force, dfs and similar, graphs, greedy, implementation, shortest paths
Correct Solution:
```
import io,os
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
from collections import deque
m,n,w = map(int,input().split())
grid = []
for i in range(m):
temp = list(map(int,input().split()))
grid.append(temp)
#visited_s = [[0 for j in range(n)] for i in range(m)]
#visited_e = [[0 for j in range(n)] for i in range(m)]
visited_s = [0 for i in range(m*n)]
visited_e = [0 for i in range(m*n)]
visited_s[0] = 1
visited_e[-1] = 1
startp = float('inf')
endp = float('inf')
noportal = float('inf')
start = deque()
end = deque()
direc = [[0,-1],[0,1],[-1,0],[1,0]]
start.append([0,0])
end.append([m*n-1,0])
if grid[0][0]>0: startp = grid[0][0]
if grid[m-1][n-1]>0: endp = grid[m-1][n-1]
while start:
[index,cost] = start.popleft()
x = index//n
y = index%n
for d in range(4):
newx = x + direc[d][0]
newy = y + direc[d][1]
if newx<0 or newx>=m or newy<0 or newy>=n: continue
newindex = newx*n+newy
# print(newx,newy,newindex)
if grid[newx][newy]==-1: continue
if visited_s[newindex]>0: continue
visited_s[newindex] = 1
start.append([newindex,cost+w])
if grid[newx][newy]>0: startp = min(startp,cost+w+grid[newx][newy])
if newindex==m*n-1: noportal = cost+w
#print(visited_s)
while end:
[index,cost] = end.popleft()
x = index//n
y = index%n
for d in range(4):
newx = x + direc[d][0]
newy = y + direc[d][1]
if newx<0 or newx>=m or newy<0 or newy>=n: continue
newindex = newx*n+newy
if grid[newx][newy]==-1: continue
if visited_e[newindex]>0: continue
if cost+w>=endp: continue
if cost+w>=noportal-startp: continue
visited_e[newindex] = 1
end.append([newindex,cost+w])
if grid[newx][newy]>0: endp = min(endp,cost+w+grid[newx][newy])
#print(startp,endp)
#print(ds,de)
ans = min(noportal, startp+endp)
if ans==float('inf'):
ans = -1
print(ans)
``` | output | 1 | 90,380 | 15 | 180,761 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Dima overslept the alarm clock, which was supposed to raise him to school.
Dima wonders if he will have time to come to the first lesson. To do this, he needs to know the minimum time it will take him to get from home to school.
The city where Dima lives is a rectangular field of n × m size. Each cell (i, j) on this field is denoted by one number a_{ij}:
* The number -1 means that the passage through the cell is prohibited;
* The number 0 means that the cell is free and Dima can walk though it.
* The number x (1 ≤ x ≤ 10^9) means that the cell contains a portal with a cost of x. A cell with a portal is also considered free.
From any portal, Dima can go to any other portal, while the time of moving from the portal (i, j) to the portal (x, y) corresponds to the sum of their costs a_{ij} + a_{xy}.
In addition to moving between portals, Dima can also move between unoccupied cells adjacent to one side in time w. In particular, he can enter a cell with a portal and not use it.
Initially, Dima is in the upper-left cell (1, 1), and the school is in the lower right cell (n, m).
Input
The first line contains three integers n, m and w (2 ≤ n, m ≤ 2 ⋅ 10^3, 1 ≤ w ≤ 10^9), where n and m are city size, w is time during which Dima moves between unoccupied cells.
The next n lines each contain m numbers (-1 ≤ a_{ij} ≤ 10^9) — descriptions of cells.
It is guaranteed that the cells (1, 1) and (n, m) are free.
Output
Output the minimum time it will take for Dima to get to school. If he cannot get to school at all, then output "-1".
Example
Input
5 5 1
0 -1 0 1 -1
0 20 0 0 -1
-1 -1 -1 -1 -1
3 0 0 0 0
-1 0 0 0 0
Output
14
Note
Explanation for the first sample:
<image> | instruction | 0 | 90,381 | 15 | 180,762 |
Tags: brute force, dfs and similar, graphs, greedy, implementation, shortest paths
Correct Solution:
```
'''
from bisect import bisect,bisect_left
from collections import *
from heapq import *
from math import gcd,ceil,sqrt,floor,inf
from itertools import *
from operator import add,mul,sub,xor,truediv,floordiv
from functools import *
'''
#------------------------------------------------------------------------
import os
import sys
from io import BytesIO, IOBase
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#------------------------------------------------------------------------
def RL(): return map(int, sys.stdin.readline().split())
def RLL(): return list(map(int, sys.stdin.readline().split()))
def N(): return int(input())
def A(n):return [0]*n
def AI(n,x): return [x]*n
def A2(n,m): return [[0]*m for i in range(n)]
def G(n): return [[] for i in range(n)]
def GP(it): return [[ch,len(list(g))] for ch,g in groupby(it)]
#------------------------------------------------------------------------
#from random import randint
from collections import deque
from math import inf
dxy=[(-1, 0), (0, -1), (0, 1), (1, 0)]
def bfs(s):
dis=[inf]*sz
dis[s]=0
q=deque([s])
while q:
u=q.popleft()
i,j=divmod(u,m)
for dx,dy in dxy:
ni=i+dx
nj=j+dy
if 0<=ni<n and 0<=nj<m and res[ni][nj]!=-1:
v=ni*m+nj
if dis[v]>dis[u]+1:
dis[v]=dis[u]+1
q.append(v)
return dis
t=1
for i in range(t):
n,m,w=RL()
sz=n*m
res=[list(map(int, sys.stdin.readline().split())) for i in range(n)]
ds=bfs(0)
de=bfs(sz-1)
ans=ds[sz-1]*w
fi=se=inf
for i in range(n):
for j in range(m):
if res[i][j]>0:
fi=min(fi,ds[i*m+j]*w+res[i][j])
se=min(se,de[i*m+j]*w+res[i][j])
ans=min(ans,fi+se)
if ans==inf:
ans=-1
print(ans)
'''
sys.setrecursionlimit(200000)
import threading
threading.stack_size(10**8)
t=threading.Thr
ead(target=main)
t.start()
t.join()
'''
``` | output | 1 | 90,381 | 15 | 180,763 |
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