message stringlengths 2 44.5k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 42 109k | cluster float64 5 5 | __index_level_0__ int64 84 217k |
|---|---|---|---|---|---|
Provide a correct Python 3 solution for this coding contest problem.
Given are an integer N and arrays S, T, U, and V, each of length N. Construct an N×N matrix a that satisfy the following conditions:
* a_{i,j} is an integer.
* 0 \leq a_{i,j} \lt 2^{64}.
* If S_{i} = 0, the bitwise AND of the elements in the i-th row is U_{i}.
* If S_{i} = 1, the bitwise OR of the elements in the i-th row is U_{i}.
* If T_{i} = 0, the bitwise AND of the elements in the i-th column is V_{i}.
* If T_{i} = 1, the bitwise OR of the elements in the i-th column is V_{i}.
However, there may be cases where no matrix satisfies the conditions.
Constraints
* All values in input are integers.
* 1 \leq N \leq 500
* 0 \leq S_{i} \leq 1
* 0 \leq T_{i} \leq 1
* 0 \leq U_{i} \lt 2^{64}
* 0 \leq V_{i} \lt 2^{64}
Input
Input is given from Standard Input in the following format:
N
S_{1} S_{2} ... S_{N}
T_{1} T_{2} ... T_{N}
U_{1} U_{2} ... U_{N}
V_{1} V_{2} ... V_{N}
Output
If there exists a matrix that satisfies the conditions, print one such matrix in the following format:
a_{1,1} ... a_{1,N}
:
a_{N,1} ... a_{N,N}
Note that any matrix satisfying the conditions is accepted.
If no matrix satisfies the conditions, print -1.
Examples
Input
2
0 1
1 0
1 1
1 0
Output
1 1
1 0
Input
2
1 1
1 0
15 15
15 11
Output
15 11
15 11 | instruction | 0 | 16,581 | 5 | 33,162 |
"Correct Solution:
```
from functools import reduce
from operator import and_, or_
N = int(input())
S, T, U, V = [[int(a) for a in input().split()] for _ in range(4)]
X = [[U[i] & V[j] if S[i] and T[j] else U[i] if T[j] else V[j] if S[i] else U[i] | V[j] for j in range(N)] for i in range(N)]
for _ in range(2):
A1 = [[-1] * N for _ in range(N)]
A2 = [[-1] * N for _ in range(N)]
for i in range(N-1)[::-1]:
for j in range(N):
A2[i][j] = A2[i+1][j] & X[i+1][j]
for i in range(N):
a = ~U[i] | reduce(or_, X[i])
for j in range(N):
if i:
A1[i][j] = A1[i-1][j] & X[i-1][j]
if S[i] and not T[j]:
X[i][j] |= ~(A1[i][j] & A2[i][j] | a)
a |= X[i][j]
if [reduce(or_ if S[i] else and_, x) for i, x in enumerate(X)] != U:
print(-1)
exit()
X = [[X[j][i] for j in range(N)] for i in range(N)]
S, T, U, V = T, S, V, U
for x in X:
print(*x)
``` | output | 1 | 16,581 | 5 | 33,163 |
Provide a correct Python 3 solution for this coding contest problem.
Given are an integer N and arrays S, T, U, and V, each of length N. Construct an N×N matrix a that satisfy the following conditions:
* a_{i,j} is an integer.
* 0 \leq a_{i,j} \lt 2^{64}.
* If S_{i} = 0, the bitwise AND of the elements in the i-th row is U_{i}.
* If S_{i} = 1, the bitwise OR of the elements in the i-th row is U_{i}.
* If T_{i} = 0, the bitwise AND of the elements in the i-th column is V_{i}.
* If T_{i} = 1, the bitwise OR of the elements in the i-th column is V_{i}.
However, there may be cases where no matrix satisfies the conditions.
Constraints
* All values in input are integers.
* 1 \leq N \leq 500
* 0 \leq S_{i} \leq 1
* 0 \leq T_{i} \leq 1
* 0 \leq U_{i} \lt 2^{64}
* 0 \leq V_{i} \lt 2^{64}
Input
Input is given from Standard Input in the following format:
N
S_{1} S_{2} ... S_{N}
T_{1} T_{2} ... T_{N}
U_{1} U_{2} ... U_{N}
V_{1} V_{2} ... V_{N}
Output
If there exists a matrix that satisfies the conditions, print one such matrix in the following format:
a_{1,1} ... a_{1,N}
:
a_{N,1} ... a_{N,N}
Note that any matrix satisfying the conditions is accepted.
If no matrix satisfies the conditions, print -1.
Examples
Input
2
0 1
1 0
1 1
1 0
Output
1 1
1 0
Input
2
1 1
1 0
15 15
15 11
Output
15 11
15 11 | instruction | 0 | 16,582 | 5 | 33,164 |
"Correct Solution:
```
N = int(input())
S = list(map(int,input().split()))
T = list(map(int,input().split()))
U = list(map(int,input().split()))
V = list(map(int,input().split()))
def invalid_exit():
print(-1)
exit()
ans = [[0]*N for _ in range(N)]
for k in range(64):
tmp = [[-1]*N for _ in range(N)]
row_has_0 = [0] * N
col_has_0 = [0] * N
row_has_1 = [0] * N
col_has_1 = [0] * N
def fill(i,j,bit):
if bit == 1 - tmp[i][j]: invalid_exit()
tmp[i][j] = bit
if bit:
row_has_1[i] = col_has_1[j] = 1
else:
row_has_0[i] = col_has_0[j] = 1
blank_rows = set(range(N))
blank_cols = set(range(N))
for i,(s,u) in enumerate(zip(S,U)):
x = (u>>k)&1
if s==0 and x:
for j in range(N):
tmp[i][j] = 1
col_has_1[j] = 1
row_has_1[i] = 1
blank_rows.remove(i)
elif s and x==0:
for j in range(N):
tmp[i][j] = 0
col_has_0[j] = 1
row_has_0[i] = 1
blank_rows.remove(i)
for j,(t,v) in enumerate(zip(T,V)):
x = (v>>k)&1
if t==0 and x:
for i in range(N):
if tmp[i][j] == 0: invalid_exit()
tmp[i][j] = 1
row_has_1[i] = 1
col_has_1[j] = 1
blank_cols.remove(j)
elif t and x==0:
for i in range(N):
if tmp[i][j] == 1: invalid_exit()
tmp[i][j] = 0
row_has_0[i] = 1
col_has_0[j] = 1
blank_cols.remove(j)
for _ in range(2):
if blank_rows or blank_cols:
j = list(blank_cols)[0] if len(blank_cols)==1 else None
for i,(s,u) in enumerate(zip(S,U)):
x = (u>>k)&1
if s==0 and x==0:
if not row_has_0[i] and len(blank_cols) == 0: invalid_exit()
if j is not None:
y = (V[j]>>k)&1
if x==y:
fill(i,j,x)
else:
if x==0 and (not row_has_0[i]) and (not col_has_1[j]) and len(blank_rows) < 2: invalid_exit()
if x==1 and (not row_has_1[i]) and (not col_has_0[j]) and len(blank_rows) < 2: invalid_exit()
if row_has_0[i]:
fill(i,j,1)
else:
fill(i,j,0)
blank_rows.remove(i)
elif s and x:
if not row_has_1[i] and len(blank_cols) == 0: invalid_exit()
if j is not None:
y = (V[j]>>k)&1
if x==y:
fill(i,j,x)
else:
if x==0 and (not row_has_0[i]) and (not col_has_1[j]) and len(blank_rows) < 2: invalid_exit()
if x==1 and (not row_has_1[i]) and (not col_has_0[j]) and len(blank_rows) < 2: invalid_exit()
if row_has_0[i]:
fill(i,j,1)
else:
fill(i,j,0)
blank_rows.remove(i)
if j is not None:
blank_cols = set()
if blank_rows or blank_cols:
i = list(blank_rows)[0] if len(blank_rows)==1 else None
for j,(t,v) in enumerate(zip(T,V)):
x = (v>>k)&1
if t==0 and x==0:
if not col_has_0[j] and len(blank_rows) == 0: invalid_exit()
if i is not None:
y = (U[i]>>k)&1
if x==y:
fill(i,j,x)
else:
if x==0 and (not col_has_0[i]) and (not row_has_1[j]) and len(blank_cols) < 2: invalid_exit()
if x==1 and (not col_has_1[i]) and (not row_has_0[j]) and len(blank_cols) < 2: invalid_exit()
if col_has_0[i]:
fill(i,j,1)
else:
fill(i,j,0)
blank_cols.remove(j)
elif t and x:
if not col_has_1[j] and len(blank_rows) == 0: invalid_exit()
if i is not None:
y = (U[i]>>k)&1
if x==y:
fill(i,j,x)
else:
if x==0 and (not col_has_0[i]) and (not row_has_1[j]) and len(blank_cols) < 2: invalid_exit()
if x==1 and (not col_has_1[i]) and (not row_has_0[j]) and len(blank_cols) < 2: invalid_exit()
if col_has_0[i]:
fill(i,j,1)
else:
fill(i,j,0)
blank_cols.remove(j)
if i is not None:
blank_rows = set()
assert len(blank_rows) != 1
assert len(blank_cols) != 1
for i,r in enumerate(blank_rows):
for j,c in enumerate(blank_cols):
tmp[r][c] = (i%2)^(j%2)
for i,row in enumerate(tmp):
for j,c in enumerate(row):
if c == -1: invalid_exit()
if c:
ans[i][j] |= (1<<k)
for row in ans:
print(*row)
``` | output | 1 | 16,582 | 5 | 33,165 |
Provide a correct Python 3 solution for this coding contest problem.
Given are an integer N and arrays S, T, U, and V, each of length N. Construct an N×N matrix a that satisfy the following conditions:
* a_{i,j} is an integer.
* 0 \leq a_{i,j} \lt 2^{64}.
* If S_{i} = 0, the bitwise AND of the elements in the i-th row is U_{i}.
* If S_{i} = 1, the bitwise OR of the elements in the i-th row is U_{i}.
* If T_{i} = 0, the bitwise AND of the elements in the i-th column is V_{i}.
* If T_{i} = 1, the bitwise OR of the elements in the i-th column is V_{i}.
However, there may be cases where no matrix satisfies the conditions.
Constraints
* All values in input are integers.
* 1 \leq N \leq 500
* 0 \leq S_{i} \leq 1
* 0 \leq T_{i} \leq 1
* 0 \leq U_{i} \lt 2^{64}
* 0 \leq V_{i} \lt 2^{64}
Input
Input is given from Standard Input in the following format:
N
S_{1} S_{2} ... S_{N}
T_{1} T_{2} ... T_{N}
U_{1} U_{2} ... U_{N}
V_{1} V_{2} ... V_{N}
Output
If there exists a matrix that satisfies the conditions, print one such matrix in the following format:
a_{1,1} ... a_{1,N}
:
a_{N,1} ... a_{N,N}
Note that any matrix satisfying the conditions is accepted.
If no matrix satisfies the conditions, print -1.
Examples
Input
2
0 1
1 0
1 1
1 0
Output
1 1
1 0
Input
2
1 1
1 0
15 15
15 11
Output
15 11
15 11 | instruction | 0 | 16,583 | 5 | 33,166 |
"Correct Solution:
```
#!/usr/bin/env python
# -*- coding: utf-8 -*-
import sys
sys.setrecursionlimit(10**7)
from pprint import pprint as pp
from pprint import pformat as pf
#import pysnooper # @pysnooper.snoop() #TODO
import math
#from sortedcontainers import SortedList, SortedDict, SortedSet # no in atcoder
import bisect
class Solver:
def __init__(self, n, s_list, t_list, u_list, v_list):
self.n = n
self.s_list = s_list
self.t_list = t_list
self.u_list = u_list
self.v_list = v_list
self.size = 64
#self.size = 1 # debug
self.bit_key = 1
self.mode = 'l' # l or c
self.ans = self.make_ans()
def make_ans(self):
ans = [None] * self.n
for l in range(self.n):
ans[l] = [0] * self.n
return ans
def print_ans(self):
for l in range(self.n):
for c in range(self.n):
print(self.ans[l][c], end=" ")
print("")
def add_straight(self, key, value):
if not value:
return
if self.mode == 'l':
for c in range(self.n):
self.ans[key][c] |= self.bit_key
else:
for l in range(self.n):
self.ans[l][key] |= self.bit_key
def add(self, key, sub_key, value):
if not value:
return
if self.mode == 'l':
l, c = key, sub_key
else:
l, c = sub_key, key
self.ans[l][c] |= self.bit_key
def run(self):
try:
for bit in range(self.size):
self.bit_key = 1 << bit
self.solve_for_bit()
except RuntimeError as err:
#raise err # debug
return False
return True
def solve_for_bit(self):
# determine
self.mode = 'l'
filled_l, ambiguous_l = self.determine(self.s_list, self.u_list)
#print('filled_l, ambiguous_l') # debug
#print(filled_l, ambiguous_l) # debug
self.mode = 'c'
filled_c, ambiguous_c = self.determine(self.t_list, self.v_list)
#print('filled_c, ambiguous_c') # debug
#print(filled_c, ambiguous_c) # debug
#print('self.ans') # debug
#self.print_ans() # debug
# check
if filled_l[0] and filled_c[1] or filled_l[1] and filled_c[0]:
#print('hoge') # debug
raise RuntimeError
# fill 1 straight or like checker
if len(ambiguous_l) == 1:
self.handle_straight('l', filled_l, ambiguous_l, filled_c, ambiguous_c)
elif len(ambiguous_c) == 1:
self.handle_straight('c', filled_c, ambiguous_c, filled_l, ambiguous_l)
else:
self.checker_fill(ambiguous_l, ambiguous_c)
def determine(self, l1, l2):
filled = [False, False]
ambiguous = [] # (key, wanted)
for key in range(self.n):
st = l1[key]
uv = 1 if l2[key] & self.bit_key else 0
if st == uv:
ambiguous.append((key, uv))
continue
filled[uv] = True
self.add_straight(key, uv)
return filled, ambiguous
def handle_straight(self, mode, filled, ambiguous, sub_filled, sub_ambiguous):
#print('self.ans') # debug
#print(self.ans) # debug
self.mode = mode
key, wanted = ambiguous[0]
ok = False
if sub_filled[wanted]:
ok = True
for sub_key, sub_wanted in sub_ambiguous:
if filled[sub_wanted]:
# sub_wanted is satisfied, so use for wanted
self.add(key, sub_key, wanted)
ok = True
else:
# satisfy sub_wanted
self.add(key, sub_key, sub_wanted)
if wanted == sub_wanted:
ok = True
if not ok:
raise RuntimeError
def checker_fill(self, ambiguous_l, ambiguous_c):
self.mode = 'l'
for il, (l_key, _) in enumerate(ambiguous_l):
for ic, (c_key, _) in enumerate(ambiguous_c):
v = (il + ic) % 2 == 0
#print('l_key, c_key', l_key, c_key) # debug
self.add(l_key, c_key, v)
if __name__ == '__main__':
n = int(input())
s_list = list(map(int, input().split()))
t_list = list(map(int, input().split()))
u_list = list(map(int, input().split()))
v_list = list(map(int, input().split()))
#print('n, s_list, t_list, u_list, v_list') # debug
#print(n, s_list, t_list, u_list, v_list) # debug
solver = Solver(n, s_list, t_list, u_list, v_list)
ans = solver.run()
if ans is False:
print(-1)
else:
solver.print_ans()
#print('\33[32m' + 'end' + '\033[0m') # debug
``` | output | 1 | 16,583 | 5 | 33,167 |
Provide a correct Python 3 solution for this coding contest problem.
Given are an integer N and arrays S, T, U, and V, each of length N. Construct an N×N matrix a that satisfy the following conditions:
* a_{i,j} is an integer.
* 0 \leq a_{i,j} \lt 2^{64}.
* If S_{i} = 0, the bitwise AND of the elements in the i-th row is U_{i}.
* If S_{i} = 1, the bitwise OR of the elements in the i-th row is U_{i}.
* If T_{i} = 0, the bitwise AND of the elements in the i-th column is V_{i}.
* If T_{i} = 1, the bitwise OR of the elements in the i-th column is V_{i}.
However, there may be cases where no matrix satisfies the conditions.
Constraints
* All values in input are integers.
* 1 \leq N \leq 500
* 0 \leq S_{i} \leq 1
* 0 \leq T_{i} \leq 1
* 0 \leq U_{i} \lt 2^{64}
* 0 \leq V_{i} \lt 2^{64}
Input
Input is given from Standard Input in the following format:
N
S_{1} S_{2} ... S_{N}
T_{1} T_{2} ... T_{N}
U_{1} U_{2} ... U_{N}
V_{1} V_{2} ... V_{N}
Output
If there exists a matrix that satisfies the conditions, print one such matrix in the following format:
a_{1,1} ... a_{1,N}
:
a_{N,1} ... a_{N,N}
Note that any matrix satisfying the conditions is accepted.
If no matrix satisfies the conditions, print -1.
Examples
Input
2
0 1
1 0
1 1
1 0
Output
1 1
1 0
Input
2
1 1
1 0
15 15
15 11
Output
15 11
15 11 | instruction | 0 | 16,584 | 5 | 33,168 |
"Correct Solution:
```
n,*s=map(int,open(0).read().split());s,t,f,g=[s[i*n:i*n+n]for i in range(4)];r=range(n);a=[n*[0]for _ in r];e=enumerate
for b in range(64):
y=1<<b;u=[k&y for k in f];v=[k&y for k in g];l=[0]*n;m=l[:]
for i,k in e(u):
for j,h in e(v):
if(k&h)|((s[i]^1)*k)|((t[j]^1)*h):a[i][j]|=y;l[i]+=1;m[j]+=1
for i in r:
if(l[i]==0)*s[i]*u[i]:
for j in r:
if(t[j]^1)*(n-m[j]>1):a[i][j]|=y;m[j]+=1;break
if(m[i]==0)*t[i]*v[i]:
for j in r:
if(s[j]^1)*(n-l[j]>1):a[j][i]|=y;l[j]+=1;break
h=a[0][:];w=[i[0]for i in a]
for i,k in e(a):
for j,l in e(k):w[i]=(l|w[i]if s[i]else w[i]&l);h[j]=(l|h[j]if t[j]else h[j]&l)
if(g!=h)|(f!=w):exit(print(-1))
for i in a:print(*i)
``` | output | 1 | 16,584 | 5 | 33,169 |
Provide a correct Python 3 solution for this coding contest problem.
Given are an integer N and arrays S, T, U, and V, each of length N. Construct an N×N matrix a that satisfy the following conditions:
* a_{i,j} is an integer.
* 0 \leq a_{i,j} \lt 2^{64}.
* If S_{i} = 0, the bitwise AND of the elements in the i-th row is U_{i}.
* If S_{i} = 1, the bitwise OR of the elements in the i-th row is U_{i}.
* If T_{i} = 0, the bitwise AND of the elements in the i-th column is V_{i}.
* If T_{i} = 1, the bitwise OR of the elements in the i-th column is V_{i}.
However, there may be cases where no matrix satisfies the conditions.
Constraints
* All values in input are integers.
* 1 \leq N \leq 500
* 0 \leq S_{i} \leq 1
* 0 \leq T_{i} \leq 1
* 0 \leq U_{i} \lt 2^{64}
* 0 \leq V_{i} \lt 2^{64}
Input
Input is given from Standard Input in the following format:
N
S_{1} S_{2} ... S_{N}
T_{1} T_{2} ... T_{N}
U_{1} U_{2} ... U_{N}
V_{1} V_{2} ... V_{N}
Output
If there exists a matrix that satisfies the conditions, print one such matrix in the following format:
a_{1,1} ... a_{1,N}
:
a_{N,1} ... a_{N,N}
Note that any matrix satisfying the conditions is accepted.
If no matrix satisfies the conditions, print -1.
Examples
Input
2
0 1
1 0
1 1
1 0
Output
1 1
1 0
Input
2
1 1
1 0
15 15
15 11
Output
15 11
15 11 | instruction | 0 | 16,585 | 5 | 33,170 |
"Correct Solution:
```
n,*s=map(int,open(0).read().split());s,t,f,g=[s[i*n:i*n+n]for i in range(4)];r=range(n);a=[n*[0]for _ in r];e=1
for _ in[0]*64:
u=[k&e for k in f];v=[k&e for k in g];l=[0]*n;m=l[:]
for i in r:
for j in r:
if(u[i]&v[j])|((s[i]^1)*u[i])|((t[j]^1)*v[j]):a[i][j]|=e;l[i]+=1;m[j]+=1
for i in r:
if(l[i]==0)*s[i]*u[i]:
for j in r:
if(t[j]^1)*(n-m[j]>1):a[i][j]|=e;m[j]+=1;break
if(m[i]==0)*t[i]*v[i]:
for j in r:
if(s[j]^1)*(n-l[j]>1):a[j][i]|=e;l[j]+=1;break
e<<=1
h=a[0][:];w=[i[0]for i in a]
for i in r:
for j in r:w[i]=(w[i]|a[i][j]if s[i]else w[i]&a[i][j]);h[j]=(h[j]|a[i][j]if t[j]else h[j]&a[i][j])
if(g!=h)|(f!=w):print(-1);exit()
for i in a:print(*i)
``` | output | 1 | 16,585 | 5 | 33,171 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given are an integer N and arrays S, T, U, and V, each of length N. Construct an N×N matrix a that satisfy the following conditions:
* a_{i,j} is an integer.
* 0 \leq a_{i,j} \lt 2^{64}.
* If S_{i} = 0, the bitwise AND of the elements in the i-th row is U_{i}.
* If S_{i} = 1, the bitwise OR of the elements in the i-th row is U_{i}.
* If T_{i} = 0, the bitwise AND of the elements in the i-th column is V_{i}.
* If T_{i} = 1, the bitwise OR of the elements in the i-th column is V_{i}.
However, there may be cases where no matrix satisfies the conditions.
Constraints
* All values in input are integers.
* 1 \leq N \leq 500
* 0 \leq S_{i} \leq 1
* 0 \leq T_{i} \leq 1
* 0 \leq U_{i} \lt 2^{64}
* 0 \leq V_{i} \lt 2^{64}
Input
Input is given from Standard Input in the following format:
N
S_{1} S_{2} ... S_{N}
T_{1} T_{2} ... T_{N}
U_{1} U_{2} ... U_{N}
V_{1} V_{2} ... V_{N}
Output
If there exists a matrix that satisfies the conditions, print one such matrix in the following format:
a_{1,1} ... a_{1,N}
:
a_{N,1} ... a_{N,N}
Note that any matrix satisfying the conditions is accepted.
If no matrix satisfies the conditions, print -1.
Examples
Input
2
0 1
1 0
1 1
1 0
Output
1 1
1 0
Input
2
1 1
1 0
15 15
15 11
Output
15 11
15 11
Submitted Solution:
```
import sys
input = sys.stdin.readline
sys.setrecursionlimit(10**8)
mod = 10**9+7
def main():
n = int(input())
ans = [[0 for _ in range(n)] for _ in range(n)]
s = list(map(int, input().split()))
t = list(map(int, input().split()))
u = list(map(int, input().split()))
v = list(map(int, input().split()))
for i in range(n):
if not s[i]:
for j in range(n):
ans[i][j] = ans[i][j]|u[i]
if not t[i]:
for j in range(n):
ans[j][i] = ans[j][i]|v[i]
for i in range(n):
for j in range(n):
if (u[i] & v[j]):
ans[i][j] = ans[i][j]|(u[i]&v[j])
for x in range(n):
if not s[x]:
continue
x_sum = ans[x][0]
for y in range(n):
x_sum = x_sum|ans[x][y]
if x_sum == u[x]:
continue
up = u[x] - x_sum
for y in range(n):
if t[y]:
continue
y_mul = ans[0][y]
for i in range(n):
if i == x:
continue
y_mul = y_mul&ans[i][y]
up_y = (~y_mul)&up
ans[x][y] += up_y
up -= up_y
if not up:
break
for y in range(n):
if not t[y]:
continue
y_sum = ans[0][y]
for x in range(n):
y_sum = y_sum|ans[x][y]
if y_sum == v[y]:
continue
up = v[y] - y_sum
for x in range(n):
if s[x]:
continue
x_mul = ans[x][0]
for j in range(n):
if y == j:
continue
x_mul = x_mul&ans[x][j]
up_x = (~x_mul)&up
ans[x][y] += up_x
up -= up_x
if not up:
break
for i in range(n):
check_xs = ans[i][0]
check_ys = ans[0][i]
check_xm = ans[i][0]
check_ym = ans[0][i]
for j in range(n):
check_xs = check_xs|ans[i][j]
check_ys = check_ys|ans[j][i]
check_xm = check_xm&ans[i][j]
check_ym = check_ym&ans[j][i]
if s[i] and u[i] != check_xs:
print(-1)
return
if t[i] and v[i] != check_ys:
print(-1)
return
if not s[i] and u[i] != check_xm:
print(-1)
return
if not t[i] and v[i] != check_ym:
print(-1)
return
for i in range(n):
print(*ans[i])
if __name__ == "__main__":
main()
``` | instruction | 0 | 16,586 | 5 | 33,172 |
Yes | output | 1 | 16,586 | 5 | 33,173 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given are an integer N and arrays S, T, U, and V, each of length N. Construct an N×N matrix a that satisfy the following conditions:
* a_{i,j} is an integer.
* 0 \leq a_{i,j} \lt 2^{64}.
* If S_{i} = 0, the bitwise AND of the elements in the i-th row is U_{i}.
* If S_{i} = 1, the bitwise OR of the elements in the i-th row is U_{i}.
* If T_{i} = 0, the bitwise AND of the elements in the i-th column is V_{i}.
* If T_{i} = 1, the bitwise OR of the elements in the i-th column is V_{i}.
However, there may be cases where no matrix satisfies the conditions.
Constraints
* All values in input are integers.
* 1 \leq N \leq 500
* 0 \leq S_{i} \leq 1
* 0 \leq T_{i} \leq 1
* 0 \leq U_{i} \lt 2^{64}
* 0 \leq V_{i} \lt 2^{64}
Input
Input is given from Standard Input in the following format:
N
S_{1} S_{2} ... S_{N}
T_{1} T_{2} ... T_{N}
U_{1} U_{2} ... U_{N}
V_{1} V_{2} ... V_{N}
Output
If there exists a matrix that satisfies the conditions, print one such matrix in the following format:
a_{1,1} ... a_{1,N}
:
a_{N,1} ... a_{N,N}
Note that any matrix satisfying the conditions is accepted.
If no matrix satisfies the conditions, print -1.
Examples
Input
2
0 1
1 0
1 1
1 0
Output
1 1
1 0
Input
2
1 1
1 0
15 15
15 11
Output
15 11
15 11
Submitted Solution:
```
def disp(s, L):
return
print(s + " =")
for x in L:
print(*x)
print()
N = int(input())
S = [int(a) for a in input().split()]
T = [int(a) for a in input().split()]
U = [int(a) for a in input().split()]
V = [int(a) for a in input().split()]
X = [[0] * N for _ in range(N)]
m = (1 << 64) - 1
for i in range(N):
for j in range(N):
if not S[i]:
X[i][j] |= U[i]
if not T[j]:
X[i][j] |= V[j]
X[i][j] |= U[i] & V[j]
disp("X", X)
###
A1 = [[m] * N for _ in range(N)]
A2 = [[m] * N for _ in range(N)]
for i in range(N-1)[::-1]:
for j in range(N):
A2[i][j] = A2[i+1][j] & X[i+1][j]
for i in range(N):
a = 0
for j in range(N):
a |= X[i][j]
for j in range(N):
if i:
A1[i][j] = A1[i-1][j] & X[i-1][j]
if S[i] and not T[j]:
X[i][j] |= U[i] & ~(A1[i][j] & A2[i][j]) & ~a
a |= X[i][j]
disp("X", X)
disp("A1", A1)
disp("A2", A2)
###
if False:
for i in range(N):
if not S[i]: continue
A = [m] * N
for j in range(N-1)[::-1]:
A[j] = A[j+1] & X[i][j+1]
a = m
for j in range(N):
if not T[j]:
X[i][j] |= U[i] & ~(a & A[j])
a &= X[i][j]
###
A1 = [[m] * N for _ in range(N)]
A2 = [[m] * N for _ in range(N)]
for j in range(N-1)[::-1]:
for i in range(N):
A2[i][j] = A2[i][j+1] & X[i][j+1]
for j in range(N):
a = 0
for i in range(N):
a |= X[i][j]
for i in range(N):
if j:
A1[i][j] = A1[i][j-1] & X[i][j-1]
if T[j] and not S[i]:
X[i][j] |= V[j] & ~(A1[i][j] & A2[i][j]) & ~a
a |= X[i][j]
disp("X", X)
disp("A1", A1)
disp("A2", A2)
###
if False:
for j in range(N):
if not T[j]: continue
A = [m] * N
for i in range(N-1)[::-1]:
A[i] = A[i+1] & X[i+1][j]
a = m
for i in range(N):
if not S[i]:
X[i][j] |= V[j] & ~(a & A[i])
a &= X[i][j]
###
for i in range(N):
if S[i] == 0:
s = m
for j in range(N):
s &= X[i][j]
if s != U[i]:
print(-1)
exit()
else:
s = 0
for j in range(N):
s |= X[i][j]
if s != U[i]:
print(-1)
exit()
for j in range(N):
if T[j] == 0:
s = m
for i in range(N):
s &= X[i][j]
if s != V[j]:
print(-1)
exit()
else:
s = 0
for i in range(N):
s |= X[i][j]
if s != V[j]:
print(-1)
exit()
for x in X:
print(*x)
disp("X", X)
disp("A1", A1)
disp("A2", A2)
``` | instruction | 0 | 16,587 | 5 | 33,174 |
Yes | output | 1 | 16,587 | 5 | 33,175 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given are an integer N and arrays S, T, U, and V, each of length N. Construct an N×N matrix a that satisfy the following conditions:
* a_{i,j} is an integer.
* 0 \leq a_{i,j} \lt 2^{64}.
* If S_{i} = 0, the bitwise AND of the elements in the i-th row is U_{i}.
* If S_{i} = 1, the bitwise OR of the elements in the i-th row is U_{i}.
* If T_{i} = 0, the bitwise AND of the elements in the i-th column is V_{i}.
* If T_{i} = 1, the bitwise OR of the elements in the i-th column is V_{i}.
However, there may be cases where no matrix satisfies the conditions.
Constraints
* All values in input are integers.
* 1 \leq N \leq 500
* 0 \leq S_{i} \leq 1
* 0 \leq T_{i} \leq 1
* 0 \leq U_{i} \lt 2^{64}
* 0 \leq V_{i} \lt 2^{64}
Input
Input is given from Standard Input in the following format:
N
S_{1} S_{2} ... S_{N}
T_{1} T_{2} ... T_{N}
U_{1} U_{2} ... U_{N}
V_{1} V_{2} ... V_{N}
Output
If there exists a matrix that satisfies the conditions, print one such matrix in the following format:
a_{1,1} ... a_{1,N}
:
a_{N,1} ... a_{N,N}
Note that any matrix satisfying the conditions is accepted.
If no matrix satisfies the conditions, print -1.
Examples
Input
2
0 1
1 0
1 1
1 0
Output
1 1
1 0
Input
2
1 1
1 0
15 15
15 11
Output
15 11
15 11
Submitted Solution:
```
import sys
readline = sys.stdin.readline
def check(k):
res = [[None]*N for _ in range(N)]
H = []
W = []
for i in range(N):
if S[i] == 0 and U[i] & (1<<k):
res[i] = [1]*N
elif S[i] == 1 and not U[i] & (1<<k):
res[i] = [0]*N
else:
H.append(i)
for j in range(N):
if T[j] == 0 and V[j] & (1<<k):
for i in range(N):
if res[i][j] == 0:
return -1
res[i][j] = 1
elif T[j] == 1 and not V[j] & (1<<k):
for i in range(N):
if res[i][j] == 1:
return -1
res[i][j] = 0
else:
W.append(j)
for st in range(4):
if st > 1:
for i in range(len(H)):
for j in range(len(W)):
h, w = H[i], W[j]
res[h][w] = (i+j+st)%2
else:
for i in range(len(H)):
for j in range(len(W)):
h, w = H[i], W[j]
res[h][w] = st
flag = True
for i in H:
if S[i] == 0:
cnt = 1
for j in range(N):
cnt &= res[i][j]
if (cnt ^ (U[i]>>k)) & 1:
flag = False
break
else:
cnt = 0
for j in range(N):
cnt |= res[i][j]
if (cnt ^ (U[i]>>k)) & 1:
flag = False
break
else:
for j in W:
if T[j] == 0:
cnt = 1
for i in range(N):
cnt &= res[i][j]
if (cnt ^ (V[j]>>k)) & 1:
flag = False
break
else:
cnt = 0
for i in range(N):
cnt |= res[i][j]
if (cnt ^ (V[j]>>k)) & 1:
flag = False
break
if flag:
return res
return -1
N = int(readline())
S = list(map(int, readline().split()))
T = list(map(int, readline().split()))
U = list(map(int, readline().split()))
V = list(map(int, readline().split()))
ans = [[0]*N for _ in range(N)]
for k in range(64):
res = check(k)
if res == -1:
ans = -1
break
for i in range(N):
for j in range(N):
ans[i][j] |= res[i][j]<<k
if ans == -1:
print(-1)
else:
print('\n'.join([' '.join(map(str, a)) for a in ans]))
``` | instruction | 0 | 16,588 | 5 | 33,176 |
Yes | output | 1 | 16,588 | 5 | 33,177 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given are an integer N and arrays S, T, U, and V, each of length N. Construct an N×N matrix a that satisfy the following conditions:
* a_{i,j} is an integer.
* 0 \leq a_{i,j} \lt 2^{64}.
* If S_{i} = 0, the bitwise AND of the elements in the i-th row is U_{i}.
* If S_{i} = 1, the bitwise OR of the elements in the i-th row is U_{i}.
* If T_{i} = 0, the bitwise AND of the elements in the i-th column is V_{i}.
* If T_{i} = 1, the bitwise OR of the elements in the i-th column is V_{i}.
However, there may be cases where no matrix satisfies the conditions.
Constraints
* All values in input are integers.
* 1 \leq N \leq 500
* 0 \leq S_{i} \leq 1
* 0 \leq T_{i} \leq 1
* 0 \leq U_{i} \lt 2^{64}
* 0 \leq V_{i} \lt 2^{64}
Input
Input is given from Standard Input in the following format:
N
S_{1} S_{2} ... S_{N}
T_{1} T_{2} ... T_{N}
U_{1} U_{2} ... U_{N}
V_{1} V_{2} ... V_{N}
Output
If there exists a matrix that satisfies the conditions, print one such matrix in the following format:
a_{1,1} ... a_{1,N}
:
a_{N,1} ... a_{N,N}
Note that any matrix satisfying the conditions is accepted.
If no matrix satisfies the conditions, print -1.
Examples
Input
2
0 1
1 0
1 1
1 0
Output
1 1
1 0
Input
2
1 1
1 0
15 15
15 11
Output
15 11
15 11
Submitted Solution:
```
n,*s=map(int,open(0).read().split());s,t,f,g=[s[i*n:i*n+n]for i in range(4)];r=range(n);a=[n*[0]for _ in r];e=enumerate
for b in range(64):
y=1<<b;u=[k&y for k in f];v=[k&y for k in g];l=[0]*n;m=l[:]
for i,k in e(u):
for j,h in e(v):
if(k&h)|((s[i]^1)*k)|((t[j]^1)*h):a[i][j]|=y;l[i]+=1;m[j]+=1
for i in r:
if(l[i]==0)*s[i]*u[i]:
for j in r:
if(t[j]^1)*(n-m[j]>1):a[i][j]|=y;m[j]+=1;break
if(m[i]==0)*t[i]*v[i]:
for j in r:
if(s[j]^1)*(n-l[j]>1):a[j][i]|=y;l[j]+=1;break
h=a[0][:];w=[i[0]for i in a]
for i,k in e(a):
for j,l in e(k):w[i]=(l|w[i]if s[i]else w[i]&l);h[j]=(l|h[j]if t[j]else h[j]&l)
if g+f!=h+w:exit(print(-1))
for i in a:print(*i)
``` | instruction | 0 | 16,589 | 5 | 33,178 |
Yes | output | 1 | 16,589 | 5 | 33,179 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given are an integer N and arrays S, T, U, and V, each of length N. Construct an N×N matrix a that satisfy the following conditions:
* a_{i,j} is an integer.
* 0 \leq a_{i,j} \lt 2^{64}.
* If S_{i} = 0, the bitwise AND of the elements in the i-th row is U_{i}.
* If S_{i} = 1, the bitwise OR of the elements in the i-th row is U_{i}.
* If T_{i} = 0, the bitwise AND of the elements in the i-th column is V_{i}.
* If T_{i} = 1, the bitwise OR of the elements in the i-th column is V_{i}.
However, there may be cases where no matrix satisfies the conditions.
Constraints
* All values in input are integers.
* 1 \leq N \leq 500
* 0 \leq S_{i} \leq 1
* 0 \leq T_{i} \leq 1
* 0 \leq U_{i} \lt 2^{64}
* 0 \leq V_{i} \lt 2^{64}
Input
Input is given from Standard Input in the following format:
N
S_{1} S_{2} ... S_{N}
T_{1} T_{2} ... T_{N}
U_{1} U_{2} ... U_{N}
V_{1} V_{2} ... V_{N}
Output
If there exists a matrix that satisfies the conditions, print one such matrix in the following format:
a_{1,1} ... a_{1,N}
:
a_{N,1} ... a_{N,N}
Note that any matrix satisfying the conditions is accepted.
If no matrix satisfies the conditions, print -1.
Examples
Input
2
0 1
1 0
1 1
1 0
Output
1 1
1 0
Input
2
1 1
1 0
15 15
15 11
Output
15 11
15 11
Submitted Solution:
```
"""
おもしろそうｙ！
桁ごとに独立に考えてよい
積で桁が1→ ALL 1 W
積で桁が0→ 少なくとも一つ0 X
和で桁が1→ 少なくとも1つ 1 Y
和で桁が0→ ALL 0 Z
積1,和で0は簡単に埋まる
→ここでぶつかったら当然1
それ以外は列と行が抜けた感じになるので座圧できる
01を1こずつ入れてあげれば安泰
2*2以上なら
01
10…にすればおわり
そうでない=1行しかないなら
1じゃない方を構築して、もう片方を調べればよい
"""
import sys
N = int(input())
S = list(map(int,input().split()))
T = list(map(int,input().split()))
U = list(map(int,input().split()))
V = list(map(int,input().split()))
ans = [ [0] * N for i in range(N) ]
for dig in range(64):
nd = 2**dig
XlisR = []
YlisR = []
R = []
Wexist = False
Zexist = False
for i in range(N):
if S[i] == 0:
if U[i] & nd == 1:
Wexist = True
for j in range(N):
ans[i][j] |= nd
else:
XlisR.append(i)
R.append(i)
else:
if U[i] & nd == 0:
Zexist = True
else:
YlisR.append(i)
R.append(i)
WexC = False
ZexC = False
XlisC = []
YlisC = []
C = []
for j in range(N):
if T[j] == 0:
if V[j] & nd == 1:
WexC = True
if Zexist:
print (-1)
sys.exit()
for i in range(N):
ans[i][j] |= nd
else:
XlisC.append(j)
C.append(j)
else:
if V[j] & nd == 0:
ZexC = True
if Wexist:
print (-1)
sys.exit()
else:
YlisC.append(j)
C.append(j)
if len(C) == 0:
if len(XlisR) > 0 and not ZexC:
print (-1)
sys.exit()
if len(YlisR) > 0 and not WexC:
print (-1)
sys.exit()
elif len(R) == 0:
if len(XlisC) > 0 and not Zexist:
print (-1)
sys.exit()
if len(YlisC) > 0 and not Wexist:
print (-1)
sys.exit()
elif len(C) == 1:
for i in YlisR:
for j in C:
ans[i][j] |= nd
if len(XlisC) == 1:
zex = False
for i in range(N):
for j in C:
if nd & ans[i][j] == 0:
zex = True
if not zex:
print (-1)
sys.exit()
else:
oex = False
for i in range(N):
for j in C:
if nd & ans[i][j] > 0:
oex = True
if not oex:
print (-1)
sys.exit()
elif len(R) == 1:
for i in R:
for j in YlisC:
ans[i][j] |= nd
if len(XlisR) == 1:
zex = False
for i in R:
for j in range(N):
if nd & ans[i][j] == 0:
zex = True
if not zex:
print (-1)
sys.exit()
else:
oex = False
for i in R:
for j in range(N):
if nd & ans[i][j] > 0:
oex = True
if not oex:
print (-1)
sys.exit()
else:
for i in range(len(R)):
for j in range(len(C)):
ans[R[i]][C[j]] |= nd * ((i+j)%2)
for i in range(N):
print (*ans[i])
``` | instruction | 0 | 16,590 | 5 | 33,180 |
No | output | 1 | 16,590 | 5 | 33,181 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given are an integer N and arrays S, T, U, and V, each of length N. Construct an N×N matrix a that satisfy the following conditions:
* a_{i,j} is an integer.
* 0 \leq a_{i,j} \lt 2^{64}.
* If S_{i} = 0, the bitwise AND of the elements in the i-th row is U_{i}.
* If S_{i} = 1, the bitwise OR of the elements in the i-th row is U_{i}.
* If T_{i} = 0, the bitwise AND of the elements in the i-th column is V_{i}.
* If T_{i} = 1, the bitwise OR of the elements in the i-th column is V_{i}.
However, there may be cases where no matrix satisfies the conditions.
Constraints
* All values in input are integers.
* 1 \leq N \leq 500
* 0 \leq S_{i} \leq 1
* 0 \leq T_{i} \leq 1
* 0 \leq U_{i} \lt 2^{64}
* 0 \leq V_{i} \lt 2^{64}
Input
Input is given from Standard Input in the following format:
N
S_{1} S_{2} ... S_{N}
T_{1} T_{2} ... T_{N}
U_{1} U_{2} ... U_{N}
V_{1} V_{2} ... V_{N}
Output
If there exists a matrix that satisfies the conditions, print one such matrix in the following format:
a_{1,1} ... a_{1,N}
:
a_{N,1} ... a_{N,N}
Note that any matrix satisfying the conditions is accepted.
If no matrix satisfies the conditions, print -1.
Examples
Input
2
0 1
1 0
1 1
1 0
Output
1 1
1 0
Input
2
1 1
1 0
15 15
15 11
Output
15 11
15 11
Submitted Solution:
```
# coding: utf-8
import sys
sysread = sys.stdin.readline
from heapq import heappop, heappush
from itertools import combinations
sys.setrecursionlimit(10**6)
import numpy as np
def run():
# import sys
# A = list(map(int, sys.stdin.readline().split()))
# all = sys.stdin.readlines()
#N, *A = map(int, open(0))
#N = int(input())
process_length = lambda x:'0' * (max_n - len(str(x))) + str(x)
N = int(sysread())
S = list(map(int, sysread().split()))
T = list(map(int, sysread().split()))
U = list(map(int, sysread().split()))
V = list(map(int, sysread().split()))
max_n = len(bin(max(U+V))[2:])
def to_bin(x):
x = bin(x)[2:]
return "0" * (max_n - len(x)) + x
U = list(map(to_bin, U))
V = list(map(to_bin, V))
Z = (np.zeros((N,N,max_n), dtype=np.int32) - 1).astype(str)
# fill absolutely determined columns
for i, (s, val) in enumerate(zip(S,U)):
if s == 0:# all 1
for idx, k in enumerate(val):
if k=="1":
for j in range(N):
Z[i, j, idx] = '1'
else:# all 0
for idx, k in enumerate(val):
if k=="0":
for j in range(N):
if Z[i, j, idx] == '1':
print(-1)
return
Z[i, j, idx] = '0'
for j, (t, val) in enumerate(zip(T,V)):
if t == 0:# all 1
for idx, k in enumerate(val):
if k=="1":
for i in range(N):
if Z[i, j, idx] == '0':
print(-1)
return
Z[i, j, idx] = '1'
else:# all 0
for idx, k in enumerate(val):
if k=="0":
print(idx, k)
for i in range(N):
if Z[i, j, idx] == '1':
print(-1)
return
Z[i, j, idx] = '0'
# fill empty columns, according to sum-criteria
for i, (s, val) in enumerate(zip(S,U)):
if s == 1:# all 1
for idx, v in enumerate(val):
if v == '1':
if '1' in Z[i, :, idx]:continue
is_empty = [k for k,_v in enumerate(Z[i, :, idx]) if _v=='-1']
if is_empty == []:
return print(-1)
for e in is_empty:
if T[e] == 0:
if (Z[:, e, idx] == '0').sum() > 0:
Z[i, e, idx] = '1'
break
else:
Z[i, e, idx] = '1'
break
for j, (t, val) in enumerate(zip(T, V)):
if t == 1: # all 1
for idx, v in enumerate(val):
if v == '1':
if '1' in Z[:, j, idx]: continue
is_empty = [k for k,_v in enumerate(Z[:, j, idx]) if _v == '-1']
if is_empty == []:
return print(-1)
for e in is_empty:
if S[e] == 0:
if (Z[e, :, idx] == '0').sum() > 0:
Z[e, j, idx] = '1'
break
else:
Z[e, j, idx] = '1'
break
Z[Z=='-1'] = 0
A = [[0] * N for _ in range(N)]
for i in range(N):
for j in range(N):
tmp = str(int(''.join(Z[i,j,:].tolist()), 2))
A[i][j] = tmp
for s in A:
print(' '.join(s))
if __name__ == "__main__":
run()
``` | instruction | 0 | 16,591 | 5 | 33,182 |
No | output | 1 | 16,591 | 5 | 33,183 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given are an integer N and arrays S, T, U, and V, each of length N. Construct an N×N matrix a that satisfy the following conditions:
* a_{i,j} is an integer.
* 0 \leq a_{i,j} \lt 2^{64}.
* If S_{i} = 0, the bitwise AND of the elements in the i-th row is U_{i}.
* If S_{i} = 1, the bitwise OR of the elements in the i-th row is U_{i}.
* If T_{i} = 0, the bitwise AND of the elements in the i-th column is V_{i}.
* If T_{i} = 1, the bitwise OR of the elements in the i-th column is V_{i}.
However, there may be cases where no matrix satisfies the conditions.
Constraints
* All values in input are integers.
* 1 \leq N \leq 500
* 0 \leq S_{i} \leq 1
* 0 \leq T_{i} \leq 1
* 0 \leq U_{i} \lt 2^{64}
* 0 \leq V_{i} \lt 2^{64}
Input
Input is given from Standard Input in the following format:
N
S_{1} S_{2} ... S_{N}
T_{1} T_{2} ... T_{N}
U_{1} U_{2} ... U_{N}
V_{1} V_{2} ... V_{N}
Output
If there exists a matrix that satisfies the conditions, print one such matrix in the following format:
a_{1,1} ... a_{1,N}
:
a_{N,1} ... a_{N,N}
Note that any matrix satisfying the conditions is accepted.
If no matrix satisfies the conditions, print -1.
Examples
Input
2
0 1
1 0
1 1
1 0
Output
1 1
1 0
Input
2
1 1
1 0
15 15
15 11
Output
15 11
15 11
Submitted Solution:
```
N = int(input())
S = list(map(int,input().split()))
T = list(map(int,input().split()))
U = list(map(int,input().split()))
V = list(map(int,input().split()))
ans = [[0]*N for _ in range(N)]
possible = True
for n in range(64):
pn = 2**n
ud = {(0,0):0,(0,1):0,(1,0):0,(1,1):0}
vd = {(0,0):0,(0,1):0,(1,0):0,(1,1):0}
for i in range(N):
s = S[i]
u = U[i]&pn
if u:
u = 1
ud[(s,u)] += 1
for j in range(N):
t = T[j]
v = V[j]&pn
if v:
v = 1
vd[(t,v)] += 1
if ud[(0,1)]*vd[(1,0)] or ud[(1,0)]*vd[(0,1)]:
possible = False
break
if ud[(1,0)]+ud[(0,0)]==N:
if ud[(0,0)]<=1 and vd[(1,1)]==N:
possible = False
break
elif ud[(0,0)]==0 and vd[(1,1)]>=1:
possible = False
break
if ud[(1,1)]+ud[(0,1)]==N:
if ud[(1,1)]<=1 and vd[(0,0)]==N:
possible = False
break
elif ud[(1,1)]==0 and vd[(0,0)]>=1:
possible = False
break
if vd[(1,0)]+vd[(0,0)]==N:
if vd[(0,0)]<=1 and ud[(1,1)]==N:
possible = False
break
elif vd[(0,0)]==0 and ud[(1,1)]>=1:
possible = False
break
if vd[(1,1)]+vd[(0,1)]==N:
if vd[(1,1)]<=1 and ud[(0,0)]==N:
possible = False
break
elif vd[(1,1)]==0 and ud[(0,0)]>=1:
possible = False
break
for i in range(N):
s = S[i]
u = U[i]&pn
if u:
u = 1
for j in range(N):
t = T[j]
v = V[j]&pn
if v:
v = 1
if ud[(0,1)]*vd[(0,1)]:
if (s,u) != (0,1) and (t,v) != (0,1):
ans[i][j] += pn
elif ud[(1,0)]*vd[(1,0)]:
if (s,u) == (0,1) or (t,v) == (0,1):
ans[i][j] += pn
elif ud[(0,1)] or ud[(1,0)]:
if u:
ans[i][j] += pn
elif vd[(0,1)] or vd[(1,0)]:
if v:
ans[i][j] += pn
if ud[(0,0)]+ud[(1,0)]==N and vd[(1,1)]==N:
c = 0
for i in range(N):
s = S[i]
u = U[i]&pn
if (s,u)==(0,0):
if c==1:
ans[i][0] += pn
break
if c==0:
for j in range(1,N):
ans[i][j] += pn
c+=1
if ud[(0,0)]+ud[(1,0)]==N and vd[(1,1)]+vd[(0,0)]==N:
for i in range(N):
s = S[i]
u = U[i]&pn
if (s,u)==(0,0):
for j in range(N):
t = T[j]
v = V[j]&pn
if v:
ans[i][j] += pn
break
if ud[(1,1)]+ud[(0,1)]==N and vd[(0,0)]==N:
c = 0
for i in range(N):
s = S[i]
u = U[i]&pn
if (s,u)==(1,1):
if c==1:
ans[i][0] -= pn
break
if c==0:
for j in range(1,N):
ans[i][j] -= pn
c+=1
if ud[(1,1)]+ud[(0,1)]==N and vd[(0,0)]+vd[(1,1)]==N:
for i in range(N):
s = S[i]
u = U[i]&pn
if (s,u)==(1,1):
for j in range(N):
t = T[j]
v = V[j]&pn
if v:
ans[i][j] -= pn
break
if vd[(0,0)]+vd[(1,0)]==N and ud[(1,1)]==N:
c = 0
for j in range(N):
t = T[j]
v = V[j]&pn
if (t,v)==(0,0):
if c==1:
ans[0][j] += pn
break
if c==0:
for i in range(1,N):
ans[i][j] += pn
c+=1
if vd[(0,0)]+vd[(1,0)]==N and ud[(1,1)]+ud[(0,0)]==N:
for j in range(N):
t = T[j]
v = V[j]&pn
if (t,v)==(0,0):
for i in range(N):
s = S[i]
u = U[i]&pn
if u:
ans[i][j] += pn
break
if vd[(1,1)]+vd[(0,1)]==N and ud[(0,0)]==N:
c = 0
for j in range(N):
t = T[j]
v = V[j]&pn
if (t,v)==(1,1):
if c==1:
ans[0][j] -= pn
break
if c==0:
for i in range(1,N):
ans[i][j] -= pn
c+=1
if vd[(1,1)]+vd[(0,1)]==N and ud[(0,0)]+ud[(1,1)]==N:
for j in range(N):
t = T[j]
v = V[j]&pn
if (t,v)==(1,1):
for i in range(N):
s = S[i]
u = U[i]&pn
if u:
ans[i][j] -= pn
break
if possible:
for i in range(N):
print(*ans[i])
else:
print(-1)
``` | instruction | 0 | 16,592 | 5 | 33,184 |
No | output | 1 | 16,592 | 5 | 33,185 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given are an integer N and arrays S, T, U, and V, each of length N. Construct an N×N matrix a that satisfy the following conditions:
* a_{i,j} is an integer.
* 0 \leq a_{i,j} \lt 2^{64}.
* If S_{i} = 0, the bitwise AND of the elements in the i-th row is U_{i}.
* If S_{i} = 1, the bitwise OR of the elements in the i-th row is U_{i}.
* If T_{i} = 0, the bitwise AND of the elements in the i-th column is V_{i}.
* If T_{i} = 1, the bitwise OR of the elements in the i-th column is V_{i}.
However, there may be cases where no matrix satisfies the conditions.
Constraints
* All values in input are integers.
* 1 \leq N \leq 500
* 0 \leq S_{i} \leq 1
* 0 \leq T_{i} \leq 1
* 0 \leq U_{i} \lt 2^{64}
* 0 \leq V_{i} \lt 2^{64}
Input
Input is given from Standard Input in the following format:
N
S_{1} S_{2} ... S_{N}
T_{1} T_{2} ... T_{N}
U_{1} U_{2} ... U_{N}
V_{1} V_{2} ... V_{N}
Output
If there exists a matrix that satisfies the conditions, print one such matrix in the following format:
a_{1,1} ... a_{1,N}
:
a_{N,1} ... a_{N,N}
Note that any matrix satisfying the conditions is accepted.
If no matrix satisfies the conditions, print -1.
Examples
Input
2
0 1
1 0
1 1
1 0
Output
1 1
1 0
Input
2
1 1
1 0
15 15
15 11
Output
15 11
15 11
Submitted Solution:
```
n=int(input())
s=list(map(int,input().split()))
t=list(map(int,input().split()))
u=list(map(lambda x:'0'*64+bin(int(x))[2:],input().split()))
v=list(map(lambda x:'0'*64+bin(int(x))[2:],input().split()))
ans=[[0]*n for _ in range(n)]
for i in range(65): # 2^iの位で考える
ansi=[[-1]*n for _ in range(n)] # -1で残ったものは0とみなす
for r in range(n): # 和で0、積で1の行と列に値を入れる。
if s[r]==1 and u[r][-i-1]=='0':
for c in range(n):
ansi[r][c]=0
elif s[r]==0 and u[r][-i-1]=='1':
for c in range(n):
ansi[r][c]=1
for c in range(n):
if t[c]==1 and v[c][-i-1]=='0':
for r in range(n):
if ansi[r][c]==-1:
ansi[r][c]=0
elif ansi[r][c]==1:
print(-1)
exit(0)
elif t[c]==0 and v[c][-i-1]=='1':
for r in range(n):
if ansi[r][c]==-1:
ansi[r][c]=1
elif ansi[r][c]==0:
print(-1)
exit(0)
for r in range(n):
for c in range(n):
if ansi[r][c]==-1 and u[r][-i-1]==v[c][-i-1]:
ansi[r][c]=int(v[c][-i-1])
for r in range(n): #行に対する操作
if s[r]==1 and u[r][-i-1]=='1' and 1>max(ansi[r]): #和で1の行だが、まだ1がない
ok=False
for c in range(n):
if ansi[r][c]<1 and t[c]==0 and u[c][-1-i]=='0' and 1>min([ansi[l][c] for l in range(n) if l!=r]): #積で0の列の0を1にする。ただし他の要素に0がある場合のみ。
ansi[r][c]=1
ok=True
break
if ok==False:
print(-1)
exit(0)
if s[r]==0 and u[r][-i-1]=='0' and 0<min(ansi[r]): #積で0の行だが、まだ0がない
ok=False
for c in range(n):
if ansi[r][c]==1 and t[c]==1 and u[c][-1-i]=='1' and 0<max([ansi[l][c] for l in range(n) if l!=r]): #和で1の列の1を0にする。ただし他の要素に1がある場合のみ。
ansi[r][c]=0
ok=True
break
if ok==False:
print(-1)
exit(0)
for c in range(n): #列に対する操作
if t[c]==1 and v[c][-i-1]=='1' and 1>max([ansi[l][c] for l in range(n)]): #和で1の列だが、まだ1がない
ok=False
for r in range(n):
if ansi[r][c]<1 and s[r]==0 and v[r][-1-i]=='0' and 1>min([ansi[r][l] for l in range(n) if l!=c]): #積で0の行の0を1にする。ただし他の要素に0がある場合のみ。
ansi[r][c]=1
ok=True
break
if ok==False:
print(-1)
exit(0)
if t[c]==0 and v[c][-i-1]=='0' and 0<min([ansi[l][c] for l in range(n)]): #積で0の列だが、まだ0がない
ok=False
for r in range(n):
if ansi[r][c]==1 and s[c]==1 and v[c][-1-i]=='1' and 0<max([ansi[r][l] for l in range(n) if l!=c]): #和で1の行の1を0にする。ただし他の要素に1がある場合のみ。
ansi[r][c]=1
ok=True
break
if ok==False:
print(-1)
exit(0)
for r in range(n):
for c in range(n):
ans[r][c]+=pow(2,i)*max(0,ansi[r][c])
for a in ans:
print(' '.join(map(str,a)))
``` | instruction | 0 | 16,593 | 5 | 33,186 |
No | output | 1 | 16,593 | 5 | 33,187 |
Provide a correct Python 3 solution for this coding contest problem.
There are N apple trees in a row. People say that one of them will bear golden apples.
We want to deploy some number of inspectors so that each of these trees will be inspected.
Each inspector will be deployed under one of the trees. For convenience, we will assign numbers from 1 through N to the trees. An inspector deployed under the i-th tree (1 \leq i \leq N) will inspect the trees with numbers between i-D and i+D (inclusive).
Find the minimum number of inspectors that we need to deploy to achieve the objective.
Constraints
* All values in input are integers.
* 1 \leq N \leq 20
* 1 \leq D \leq 20
Input
Input is given from Standard Input in the following format:
N D
Output
Print the minimum number of inspectors that we need to deploy to achieve the objective.
Examples
Input
6 2
Output
2
Input
14 3
Output
2
Input
20 4
Output
3 | instruction | 0 | 16,610 | 5 | 33,220 |
"Correct Solution:
```
N,D = map(int, input().split())
print(-(-N//(1+D*2)))
``` | output | 1 | 16,610 | 5 | 33,221 |
Provide a correct Python 3 solution for this coding contest problem.
There are N apple trees in a row. People say that one of them will bear golden apples.
We want to deploy some number of inspectors so that each of these trees will be inspected.
Each inspector will be deployed under one of the trees. For convenience, we will assign numbers from 1 through N to the trees. An inspector deployed under the i-th tree (1 \leq i \leq N) will inspect the trees with numbers between i-D and i+D (inclusive).
Find the minimum number of inspectors that we need to deploy to achieve the objective.
Constraints
* All values in input are integers.
* 1 \leq N \leq 20
* 1 \leq D \leq 20
Input
Input is given from Standard Input in the following format:
N D
Output
Print the minimum number of inspectors that we need to deploy to achieve the objective.
Examples
Input
6 2
Output
2
Input
14 3
Output
2
Input
20 4
Output
3 | instruction | 0 | 16,611 | 5 | 33,222 |
"Correct Solution:
```
N, D = map(int, input().split())
L = 2*D+1
print((N+L-1)//L)
``` | output | 1 | 16,611 | 5 | 33,223 |
Provide a correct Python 3 solution for this coding contest problem.
There are N apple trees in a row. People say that one of them will bear golden apples.
We want to deploy some number of inspectors so that each of these trees will be inspected.
Each inspector will be deployed under one of the trees. For convenience, we will assign numbers from 1 through N to the trees. An inspector deployed under the i-th tree (1 \leq i \leq N) will inspect the trees with numbers between i-D and i+D (inclusive).
Find the minimum number of inspectors that we need to deploy to achieve the objective.
Constraints
* All values in input are integers.
* 1 \leq N \leq 20
* 1 \leq D \leq 20
Input
Input is given from Standard Input in the following format:
N D
Output
Print the minimum number of inspectors that we need to deploy to achieve the objective.
Examples
Input
6 2
Output
2
Input
14 3
Output
2
Input
20 4
Output
3 | instruction | 0 | 16,612 | 5 | 33,224 |
"Correct Solution:
```
n,d=map(int,input().split())
d=2*d+1
print(-(-n//d))
``` | output | 1 | 16,612 | 5 | 33,225 |
Provide a correct Python 3 solution for this coding contest problem.
There are N apple trees in a row. People say that one of them will bear golden apples.
We want to deploy some number of inspectors so that each of these trees will be inspected.
Each inspector will be deployed under one of the trees. For convenience, we will assign numbers from 1 through N to the trees. An inspector deployed under the i-th tree (1 \leq i \leq N) will inspect the trees with numbers between i-D and i+D (inclusive).
Find the minimum number of inspectors that we need to deploy to achieve the objective.
Constraints
* All values in input are integers.
* 1 \leq N \leq 20
* 1 \leq D \leq 20
Input
Input is given from Standard Input in the following format:
N D
Output
Print the minimum number of inspectors that we need to deploy to achieve the objective.
Examples
Input
6 2
Output
2
Input
14 3
Output
2
Input
20 4
Output
3 | instruction | 0 | 16,613 | 5 | 33,226 |
"Correct Solution:
```
N,D = map(int,input().split())
a = D*2+1
ans = -(-N//a)
print(ans)
``` | output | 1 | 16,613 | 5 | 33,227 |
Provide a correct Python 3 solution for this coding contest problem.
There are N apple trees in a row. People say that one of them will bear golden apples.
We want to deploy some number of inspectors so that each of these trees will be inspected.
Each inspector will be deployed under one of the trees. For convenience, we will assign numbers from 1 through N to the trees. An inspector deployed under the i-th tree (1 \leq i \leq N) will inspect the trees with numbers between i-D and i+D (inclusive).
Find the minimum number of inspectors that we need to deploy to achieve the objective.
Constraints
* All values in input are integers.
* 1 \leq N \leq 20
* 1 \leq D \leq 20
Input
Input is given from Standard Input in the following format:
N D
Output
Print the minimum number of inspectors that we need to deploy to achieve the objective.
Examples
Input
6 2
Output
2
Input
14 3
Output
2
Input
20 4
Output
3 | instruction | 0 | 16,614 | 5 | 33,228 |
"Correct Solution:
```
n,d=map(int,input().split());print(-(-n//(2*d+1)))
``` | output | 1 | 16,614 | 5 | 33,229 |
Provide a correct Python 3 solution for this coding contest problem.
There are N apple trees in a row. People say that one of them will bear golden apples.
We want to deploy some number of inspectors so that each of these trees will be inspected.
Each inspector will be deployed under one of the trees. For convenience, we will assign numbers from 1 through N to the trees. An inspector deployed under the i-th tree (1 \leq i \leq N) will inspect the trees with numbers between i-D and i+D (inclusive).
Find the minimum number of inspectors that we need to deploy to achieve the objective.
Constraints
* All values in input are integers.
* 1 \leq N \leq 20
* 1 \leq D \leq 20
Input
Input is given from Standard Input in the following format:
N D
Output
Print the minimum number of inspectors that we need to deploy to achieve the objective.
Examples
Input
6 2
Output
2
Input
14 3
Output
2
Input
20 4
Output
3 | instruction | 0 | 16,615 | 5 | 33,230 |
"Correct Solution:
```
N,D = map(int,input().split())
A = D*2 + 1
print(N//A + (N%A != 0))
``` | output | 1 | 16,615 | 5 | 33,231 |
Provide a correct Python 3 solution for this coding contest problem.
There are N apple trees in a row. People say that one of them will bear golden apples.
We want to deploy some number of inspectors so that each of these trees will be inspected.
Each inspector will be deployed under one of the trees. For convenience, we will assign numbers from 1 through N to the trees. An inspector deployed under the i-th tree (1 \leq i \leq N) will inspect the trees with numbers between i-D and i+D (inclusive).
Find the minimum number of inspectors that we need to deploy to achieve the objective.
Constraints
* All values in input are integers.
* 1 \leq N \leq 20
* 1 \leq D \leq 20
Input
Input is given from Standard Input in the following format:
N D
Output
Print the minimum number of inspectors that we need to deploy to achieve the objective.
Examples
Input
6 2
Output
2
Input
14 3
Output
2
Input
20 4
Output
3 | instruction | 0 | 16,616 | 5 | 33,232 |
"Correct Solution:
```
a = list(map(int,input().split()))
print(-(-a[0]//(2*a[1]+1)))
``` | output | 1 | 16,616 | 5 | 33,233 |
Provide a correct Python 3 solution for this coding contest problem.
There are N apple trees in a row. People say that one of them will bear golden apples.
We want to deploy some number of inspectors so that each of these trees will be inspected.
Each inspector will be deployed under one of the trees. For convenience, we will assign numbers from 1 through N to the trees. An inspector deployed under the i-th tree (1 \leq i \leq N) will inspect the trees with numbers between i-D and i+D (inclusive).
Find the minimum number of inspectors that we need to deploy to achieve the objective.
Constraints
* All values in input are integers.
* 1 \leq N \leq 20
* 1 \leq D \leq 20
Input
Input is given from Standard Input in the following format:
N D
Output
Print the minimum number of inspectors that we need to deploy to achieve the objective.
Examples
Input
6 2
Output
2
Input
14 3
Output
2
Input
20 4
Output
3 | instruction | 0 | 16,617 | 5 | 33,234 |
"Correct Solution:
```
n,d=list(map(int, input().split()))
a=d*2+1
print((n-1)//a+1)
``` | output | 1 | 16,617 | 5 | 33,235 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N apple trees in a row. People say that one of them will bear golden apples.
We want to deploy some number of inspectors so that each of these trees will be inspected.
Each inspector will be deployed under one of the trees. For convenience, we will assign numbers from 1 through N to the trees. An inspector deployed under the i-th tree (1 \leq i \leq N) will inspect the trees with numbers between i-D and i+D (inclusive).
Find the minimum number of inspectors that we need to deploy to achieve the objective.
Constraints
* All values in input are integers.
* 1 \leq N \leq 20
* 1 \leq D \leq 20
Input
Input is given from Standard Input in the following format:
N D
Output
Print the minimum number of inspectors that we need to deploy to achieve the objective.
Examples
Input
6 2
Output
2
Input
14 3
Output
2
Input
20 4
Output
3
Submitted Solution:
```
N,D=map(int,input().split())
z=2*D+1
while N>z:
z=z+1
print(z)
``` | instruction | 0 | 16,625 | 5 | 33,250 |
No | output | 1 | 16,625 | 5 | 33,251 |
Provide a correct Python 3 solution for this coding contest problem.
You are given positive integers N and M.
How many sequences a of length N consisting of positive integers satisfy a_1 \times a_2 \times ... \times a_N = M? Find the count modulo 10^9+7.
Here, two sequences a' and a'' are considered different when there exists some i such that a_i' \neq a_i''.
Constraints
* All values in input are integers.
* 1 \leq N \leq 10^5
* 1 \leq M \leq 10^9
Input
Input is given from Standard Input in the following format:
N M
Output
Print the number of the sequences consisting of positive integers that satisfy the condition, modulo 10^9 + 7.
Examples
Input
2 6
Output
4
Input
3 12
Output
18
Input
100000 1000000000
Output
957870001 | instruction | 0 | 16,645 | 5 | 33,290 |
"Correct Solution:
```
n,m=map(int,input().split())
l={}
p=2
while m>=p*p:
if m%p==0:
if not p in l.keys():
l[p]=0
l[p]+=1
m//=p
else:
p+=1
if not m in l.keys():
l[m]=0
l[m] += 1
ans=1
for p,q in l.items():
if p==1:
continue
for i in range(q):
ans*=n+i
for i in range(q):
ans//=i+1
print(ans%1000000007)
``` | output | 1 | 16,645 | 5 | 33,291 |
Provide a correct Python 3 solution for this coding contest problem.
You are given positive integers N and M.
How many sequences a of length N consisting of positive integers satisfy a_1 \times a_2 \times ... \times a_N = M? Find the count modulo 10^9+7.
Here, two sequences a' and a'' are considered different when there exists some i such that a_i' \neq a_i''.
Constraints
* All values in input are integers.
* 1 \leq N \leq 10^5
* 1 \leq M \leq 10^9
Input
Input is given from Standard Input in the following format:
N M
Output
Print the number of the sequences consisting of positive integers that satisfy the condition, modulo 10^9 + 7.
Examples
Input
2 6
Output
4
Input
3 12
Output
18
Input
100000 1000000000
Output
957870001 | instruction | 0 | 16,648 | 5 | 33,296 |
"Correct Solution:
```
N,M = map(int,input().split())
div = []
for i in range(2,int(M**(1/2))+2):
cur = 0
while M%i==0:
M//=i
cur += 1
if cur >=1:
div.append(cur)
if M>1:div.append(1)
div.sort()
mod = 10**9+7
frac = [1]*(N+50)
num=len(frac)
for i in range(num-1):
frac[i+1] = frac[i]*(i+1)%mod
finv = [1]*(N+50)
finv[-1] = pow(frac[-1],mod-2,mod)
ans = 1
for i in range(1,num):
finv[num-1-i] = finv[num-i]*(num-i)%mod
for i in div:
ans =ans*frac[N+i-1]*finv[N-1]*finv[i]%mod
print(ans)
``` | output | 1 | 16,648 | 5 | 33,297 |
Provide a correct Python 3 solution for this coding contest problem.
You are given positive integers N and M.
How many sequences a of length N consisting of positive integers satisfy a_1 \times a_2 \times ... \times a_N = M? Find the count modulo 10^9+7.
Here, two sequences a' and a'' are considered different when there exists some i such that a_i' \neq a_i''.
Constraints
* All values in input are integers.
* 1 \leq N \leq 10^5
* 1 \leq M \leq 10^9
Input
Input is given from Standard Input in the following format:
N M
Output
Print the number of the sequences consisting of positive integers that satisfy the condition, modulo 10^9 + 7.
Examples
Input
2 6
Output
4
Input
3 12
Output
18
Input
100000 1000000000
Output
957870001 | instruction | 0 | 16,649 | 5 | 33,298 |
"Correct Solution:
```
n, m = map(int, input().split())
MOD = 1000000007
def comb(x, y):
ret = 1
for a in range(x, x - y, -1):
ret *= a
for b in range(y, 0, -1):
ret //= b
return ret
ans = 1
for i in range(2, 100000):
if i > m:break
cnt = 0
while m % i == 0:
cnt += 1
m //= i
ans *= comb(cnt + n - 1, cnt)
if m != 1:
ans *= comb(n, 1)
print(ans % MOD)
``` | output | 1 | 16,649 | 5 | 33,299 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given positive integers N and M.
How many sequences a of length N consisting of positive integers satisfy a_1 \times a_2 \times ... \times a_N = M? Find the count modulo 10^9+7.
Here, two sequences a' and a'' are considered different when there exists some i such that a_i' \neq a_i''.
Constraints
* All values in input are integers.
* 1 \leq N \leq 10^5
* 1 \leq M \leq 10^9
Input
Input is given from Standard Input in the following format:
N M
Output
Print the number of the sequences consisting of positive integers that satisfy the condition, modulo 10^9 + 7.
Examples
Input
2 6
Output
4
Input
3 12
Output
18
Input
100000 1000000000
Output
957870001
Submitted Solution:
```
from math import factorial
N,M = map(int,input().split())
pdic={}
p=2
while M>1:
c=0
while M%p==0:
c+=1
M//=p
if c>0:
pdic[p]=c
p+=1
ans=1
tmp=factorial(N-1)
for p in pdic.keys():
e=pdic[p]
ans*=factorial(N+e-1)//(factorial(e)*tmp)
print(ans%(10**9+7))
``` | instruction | 0 | 16,650 | 5 | 33,300 |
Yes | output | 1 | 16,650 | 5 | 33,301 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given positive integers N and M.
How many sequences a of length N consisting of positive integers satisfy a_1 \times a_2 \times ... \times a_N = M? Find the count modulo 10^9+7.
Here, two sequences a' and a'' are considered different when there exists some i such that a_i' \neq a_i''.
Constraints
* All values in input are integers.
* 1 \leq N \leq 10^5
* 1 \leq M \leq 10^9
Input
Input is given from Standard Input in the following format:
N M
Output
Print the number of the sequences consisting of positive integers that satisfy the condition, modulo 10^9 + 7.
Examples
Input
2 6
Output
4
Input
3 12
Output
18
Input
100000 1000000000
Output
957870001
Submitted Solution:
```
n,m=map(int,input().split())
mod=10**9+7
def combination(n, r):
a = 1
b = 1
for i in range(r):
a *= (n - i)
b *= (i + 1)
return a // b
def get_prime_dic(n):
dic = {}
i = 2
while i * i <= n:
while n % i == 0:
n //= i
if i in dic:
dic[i] += 1
else:
dic[i] = 1
i += 1
if n > 1:
dic[n] = 1
return dic
d=get_prime_dic(m)
ans=1
for i,j in d.items():
ans*=combination(n-1+j, j)
ans%=mod
print(ans)
``` | instruction | 0 | 16,651 | 5 | 33,302 |
Yes | output | 1 | 16,651 | 5 | 33,303 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given positive integers N and M.
How many sequences a of length N consisting of positive integers satisfy a_1 \times a_2 \times ... \times a_N = M? Find the count modulo 10^9+7.
Here, two sequences a' and a'' are considered different when there exists some i such that a_i' \neq a_i''.
Constraints
* All values in input are integers.
* 1 \leq N \leq 10^5
* 1 \leq M \leq 10^9
Input
Input is given from Standard Input in the following format:
N M
Output
Print the number of the sequences consisting of positive integers that satisfy the condition, modulo 10^9 + 7.
Examples
Input
2 6
Output
4
Input
3 12
Output
18
Input
100000 1000000000
Output
957870001
Submitted Solution:
```
#imsosorryididntwanttoimplementmodpow
import math
n,m = [int(x) for x in input().split()]
def comb(n, r):
return math.factorial(n) // (math.factorial(n-r)*math.factorial(r))
i = 2
ans = 1
while (i<=m):
if m%i==0:
mult = 0
while (m%i==0):
mult += 1
m //= i
ans *= comb(n+mult-1,n-1)
i += 1
if m!=1:
ans *= comb(n,n-1)
print(ans%int(1e9+7))
``` | instruction | 0 | 16,652 | 5 | 33,304 |
Yes | output | 1 | 16,652 | 5 | 33,305 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given positive integers N and M.
How many sequences a of length N consisting of positive integers satisfy a_1 \times a_2 \times ... \times a_N = M? Find the count modulo 10^9+7.
Here, two sequences a' and a'' are considered different when there exists some i such that a_i' \neq a_i''.
Constraints
* All values in input are integers.
* 1 \leq N \leq 10^5
* 1 \leq M \leq 10^9
Input
Input is given from Standard Input in the following format:
N M
Output
Print the number of the sequences consisting of positive integers that satisfy the condition, modulo 10^9 + 7.
Examples
Input
2 6
Output
4
Input
3 12
Output
18
Input
100000 1000000000
Output
957870001
Submitted Solution:
```
N,M = list(map(int,input().split()))
P = 10**9+7
def get_soinsu(M):
for i in range(2,1+int(M**0.5)):
if M%i == 0:
return [i] + get_soinsu(M//i)
else:
return [M]
soinsu = {}
for n in get_soinsu(M):
soinsu[n] = 1 + soinsu.get(n,0)
def combi(n,r):
tmp1,tmp2 = (1,1)
for i in range(n-r+1,n+1):
tmp1 *= i
for i in range(2,r+1):
tmp2 *= i
return (tmp1//tmp2)%P
out = 1
soinsu.pop(1,None)
for n in soinsu.values():
out *= combi(N-1+n,n)
out %= P
print(out)
``` | instruction | 0 | 16,653 | 5 | 33,306 |
Yes | output | 1 | 16,653 | 5 | 33,307 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given positive integers N and M.
How many sequences a of length N consisting of positive integers satisfy a_1 \times a_2 \times ... \times a_N = M? Find the count modulo 10^9+7.
Here, two sequences a' and a'' are considered different when there exists some i such that a_i' \neq a_i''.
Constraints
* All values in input are integers.
* 1 \leq N \leq 10^5
* 1 \leq M \leq 10^9
Input
Input is given from Standard Input in the following format:
N M
Output
Print the number of the sequences consisting of positive integers that satisfy the condition, modulo 10^9 + 7.
Examples
Input
2 6
Output
4
Input
3 12
Output
18
Input
100000 1000000000
Output
957870001
Submitted Solution:
```
def prime(M):
lst = []
i = 2
while M != 1:
if M % i == 0:
lst.append(i)
M = M // i
else:
i += 1
return lst
def counter(lst):
dic = {}
for i in lst:
if i not in dic:
dic[i] = 1
else:
dic[i] += 1
return dic
def cmb(n, r):
if n - r < r: r = n - r
if r == 0: return 1
if r == 1: return n
numerator = [n - r + k + 1 for k in range(r)]
denominator = [k + 1 for k in range(r)]
for p in range(2, r+1):
pivot = denominator[p - 1]
if pivot > 1:
offset = (n - r) % p
for k in range(p-1, r, p):
numerator[k - offset] /= pivot
denominator[k] /= pivot
result = 1
for k in range(r):
if numerator[k] > 1:
result *= int(numerator[k])
return result
N, M = map(int, input().split())
val = 1
for i in counter(prime(M)).values():
val *= cmb(N + i - 1, i)
if val >= 10 * 9 + 7:
val = val % (10 ** 9 + 7)
print(val)
``` | instruction | 0 | 16,654 | 5 | 33,308 |
No | output | 1 | 16,654 | 5 | 33,309 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given positive integers N and M.
How many sequences a of length N consisting of positive integers satisfy a_1 \times a_2 \times ... \times a_N = M? Find the count modulo 10^9+7.
Here, two sequences a' and a'' are considered different when there exists some i such that a_i' \neq a_i''.
Constraints
* All values in input are integers.
* 1 \leq N \leq 10^5
* 1 \leq M \leq 10^9
Input
Input is given from Standard Input in the following format:
N M
Output
Print the number of the sequences consisting of positive integers that satisfy the condition, modulo 10^9 + 7.
Examples
Input
2 6
Output
4
Input
3 12
Output
18
Input
100000 1000000000
Output
957870001
Submitted Solution:
```
from scipy.misc import comb
n,m=map(int,input().split())
a=1
p=2
while m>1:
if m%p:
p+=1
continue
c=0
while not m%p:
m//=p
c+=1
a=a*comb(n+c-1,c,exact=1)%(10**9+7)
print(a)
``` | instruction | 0 | 16,656 | 5 | 33,312 |
No | output | 1 | 16,656 | 5 | 33,313 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given positive integers N and M.
How many sequences a of length N consisting of positive integers satisfy a_1 \times a_2 \times ... \times a_N = M? Find the count modulo 10^9+7.
Here, two sequences a' and a'' are considered different when there exists some i such that a_i' \neq a_i''.
Constraints
* All values in input are integers.
* 1 \leq N \leq 10^5
* 1 \leq M \leq 10^9
Input
Input is given from Standard Input in the following format:
N M
Output
Print the number of the sequences consisting of positive integers that satisfy the condition, modulo 10^9 + 7.
Examples
Input
2 6
Output
4
Input
3 12
Output
18
Input
100000 1000000000
Output
957870001
Submitted Solution:
```
a,b=map(int,input().split())
if a==1:
print(1)
exit()
from collections import Counter
from operator import mul
from functools import reduce
def cmb(n,r):
r = min(n-r,r)
if r == 0: return 1
over = reduce(mul, range(n, n - r, -1))
under = reduce(mul, range(1,r + 1))
return over // under
def factorize(n):
b = 2
fct = []
while b * b <= n:
while n % b == 0:
n //= b
fct.append(b)
b = b + 1
if n > 1:
fct.append(n)
return fct
mod=10**9+7
print(reduce(mul,map(lambda x:cmb(a-1+x,a-1),Counter(factorize(b)).values()))%mod)
``` | instruction | 0 | 16,657 | 5 | 33,314 |
No | output | 1 | 16,657 | 5 | 33,315 |
Provide a correct Python 3 solution for this coding contest problem.
In an electric circuit, when two resistors R_1 and R_2 are connected in parallel, the equivalent resistance R_3 can be derived from the following formula:
* \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{R_3}
Given R_1 and R_2, find R_3.
Constraints
* 1 \leq R_1, R_2 \leq 100
* R_1 and R_2 are integers.
Input
The input is given from Standard Input in the following format:
R_1 R_2
Output
Print the value of R_3.
The output is considered correct if the absolute or relative error is at most 10^{-6}.
Examples
Input
2 3
Output
1.2000000000
Input
100 99
Output
49.7487437186 | instruction | 0 | 16,695 | 5 | 33,390 |
"Correct Solution:
```
#ライブラリインポート
from collections import defaultdict
#入力受け取り
def getlist():
return list(map(int, input().split()))
#処理内容
def main():
R1, R2 = getlist()
ans = R1 * R2 / (R1 + R2)
print(ans)
if __name__ == '__main__':
main()
``` | output | 1 | 16,695 | 5 | 33,391 |
Provide a correct Python 3 solution for this coding contest problem.
In an electric circuit, when two resistors R_1 and R_2 are connected in parallel, the equivalent resistance R_3 can be derived from the following formula:
* \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{R_3}
Given R_1 and R_2, find R_3.
Constraints
* 1 \leq R_1, R_2 \leq 100
* R_1 and R_2 are integers.
Input
The input is given from Standard Input in the following format:
R_1 R_2
Output
Print the value of R_3.
The output is considered correct if the absolute or relative error is at most 10^{-6}.
Examples
Input
2 3
Output
1.2000000000
Input
100 99
Output
49.7487437186 | instruction | 0 | 16,696 | 5 | 33,392 |
"Correct Solution:
```
# -*- coding: utf-8 -*-
import sys
def input(): return sys.stdin.readline().strip()
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
def list4d(a, b, c, d, e): return [[[[e] * d for j in range(c)] for j in range(b)] for i in range(a)]
def ceil(x, y=1): return int(-(-x // y))
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(N=None): return list(MAP()) if N is None else [INT() for i in range(N)]
def Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
sys.setrecursionlimit(10 ** 9)
INF = 10 ** 18
MOD = 10 ** 9 + 7
R1, R2 = MAP()
ans = 1 / (1/R1 + 1/R2)
print(ans)
``` | output | 1 | 16,696 | 5 | 33,393 |
Provide a correct Python 3 solution for this coding contest problem.
In an electric circuit, when two resistors R_1 and R_2 are connected in parallel, the equivalent resistance R_3 can be derived from the following formula:
* \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{R_3}
Given R_1 and R_2, find R_3.
Constraints
* 1 \leq R_1, R_2 \leq 100
* R_1 and R_2 are integers.
Input
The input is given from Standard Input in the following format:
R_1 R_2
Output
Print the value of R_3.
The output is considered correct if the absolute or relative error is at most 10^{-6}.
Examples
Input
2 3
Output
1.2000000000
Input
100 99
Output
49.7487437186 | instruction | 0 | 16,697 | 5 | 33,394 |
"Correct Solution:
```
print(eval('1/'+input().replace(' ','+1/'))**-1)
``` | output | 1 | 16,697 | 5 | 33,395 |
Provide a correct Python 3 solution for this coding contest problem.
In an electric circuit, when two resistors R_1 and R_2 are connected in parallel, the equivalent resistance R_3 can be derived from the following formula:
* \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{R_3}
Given R_1 and R_2, find R_3.
Constraints
* 1 \leq R_1, R_2 \leq 100
* R_1 and R_2 are integers.
Input
The input is given from Standard Input in the following format:
R_1 R_2
Output
Print the value of R_3.
The output is considered correct if the absolute or relative error is at most 10^{-6}.
Examples
Input
2 3
Output
1.2000000000
Input
100 99
Output
49.7487437186 | instruction | 0 | 16,698 | 5 | 33,396 |
"Correct Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**15
mod = 10**9+7
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def main():
a,b = LI()
return 1 / (1 / a + 1 / b)
print(main())
``` | output | 1 | 16,698 | 5 | 33,397 |
Provide a correct Python 3 solution for this coding contest problem.
In an electric circuit, when two resistors R_1 and R_2 are connected in parallel, the equivalent resistance R_3 can be derived from the following formula:
* \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{R_3}
Given R_1 and R_2, find R_3.
Constraints
* 1 \leq R_1, R_2 \leq 100
* R_1 and R_2 are integers.
Input
The input is given from Standard Input in the following format:
R_1 R_2
Output
Print the value of R_3.
The output is considered correct if the absolute or relative error is at most 10^{-6}.
Examples
Input
2 3
Output
1.2000000000
Input
100 99
Output
49.7487437186 | instruction | 0 | 16,699 | 5 | 33,398 |
"Correct Solution:
```
a=list(map(int,input().split()))
print(1/(1/a[0]+1/a[1]))
``` | output | 1 | 16,699 | 5 | 33,399 |
Provide a correct Python 3 solution for this coding contest problem.
In an electric circuit, when two resistors R_1 and R_2 are connected in parallel, the equivalent resistance R_3 can be derived from the following formula:
* \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{R_3}
Given R_1 and R_2, find R_3.
Constraints
* 1 \leq R_1, R_2 \leq 100
* R_1 and R_2 are integers.
Input
The input is given from Standard Input in the following format:
R_1 R_2
Output
Print the value of R_3.
The output is considered correct if the absolute or relative error is at most 10^{-6}.
Examples
Input
2 3
Output
1.2000000000
Input
100 99
Output
49.7487437186 | instruction | 0 | 16,700 | 5 | 33,400 |
"Correct Solution:
```
R1, R2 = map(int, input().split())
print(R1*R2/(R1+R2))
``` | output | 1 | 16,700 | 5 | 33,401 |
Provide a correct Python 3 solution for this coding contest problem.
In an electric circuit, when two resistors R_1 and R_2 are connected in parallel, the equivalent resistance R_3 can be derived from the following formula:
* \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{R_3}
Given R_1 and R_2, find R_3.
Constraints
* 1 \leq R_1, R_2 \leq 100
* R_1 and R_2 are integers.
Input
The input is given from Standard Input in the following format:
R_1 R_2
Output
Print the value of R_3.
The output is considered correct if the absolute or relative error is at most 10^{-6}.
Examples
Input
2 3
Output
1.2000000000
Input
100 99
Output
49.7487437186 | instruction | 0 | 16,701 | 5 | 33,402 |
"Correct Solution:
```
n,m=map(float,input().split())
print(n*m/(n+m))
``` | output | 1 | 16,701 | 5 | 33,403 |
Provide a correct Python 3 solution for this coding contest problem.
In an electric circuit, when two resistors R_1 and R_2 are connected in parallel, the equivalent resistance R_3 can be derived from the following formula:
* \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{R_3}
Given R_1 and R_2, find R_3.
Constraints
* 1 \leq R_1, R_2 \leq 100
* R_1 and R_2 are integers.
Input
The input is given from Standard Input in the following format:
R_1 R_2
Output
Print the value of R_3.
The output is considered correct if the absolute or relative error is at most 10^{-6}.
Examples
Input
2 3
Output
1.2000000000
Input
100 99
Output
49.7487437186 | instruction | 0 | 16,702 | 5 | 33,404 |
"Correct Solution:
```
R1,R2 = map(int,input().split())
R = 1/(1/R1 + 1/R2)
print(R)
``` | output | 1 | 16,702 | 5 | 33,405 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In an electric circuit, when two resistors R_1 and R_2 are connected in parallel, the equivalent resistance R_3 can be derived from the following formula:
* \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{R_3}
Given R_1 and R_2, find R_3.
Constraints
* 1 \leq R_1, R_2 \leq 100
* R_1 and R_2 are integers.
Input
The input is given from Standard Input in the following format:
R_1 R_2
Output
Print the value of R_3.
The output is considered correct if the absolute or relative error is at most 10^{-6}.
Examples
Input
2 3
Output
1.2000000000
Input
100 99
Output
49.7487437186
Submitted Solution:
```
from fractions import *
r1,r2=map(int,input().split())
R1=Fraction(1,r1)
R2=Fraction(1,r2)
R3=R1+R2
print(R3.denominator/R3.numerator)
``` | instruction | 0 | 16,703 | 5 | 33,406 |
Yes | output | 1 | 16,703 | 5 | 33,407 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In an electric circuit, when two resistors R_1 and R_2 are connected in parallel, the equivalent resistance R_3 can be derived from the following formula:
* \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{R_3}
Given R_1 and R_2, find R_3.
Constraints
* 1 \leq R_1, R_2 \leq 100
* R_1 and R_2 are integers.
Input
The input is given from Standard Input in the following format:
R_1 R_2
Output
Print the value of R_3.
The output is considered correct if the absolute or relative error is at most 10^{-6}.
Examples
Input
2 3
Output
1.2000000000
Input
100 99
Output
49.7487437186
Submitted Solution:
```
x,y = map(int,input().split())
print(x*y/(x+y))
``` | instruction | 0 | 16,704 | 5 | 33,408 |
Yes | output | 1 | 16,704 | 5 | 33,409 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In an electric circuit, when two resistors R_1 and R_2 are connected in parallel, the equivalent resistance R_3 can be derived from the following formula:
* \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{R_3}
Given R_1 and R_2, find R_3.
Constraints
* 1 \leq R_1, R_2 \leq 100
* R_1 and R_2 are integers.
Input
The input is given from Standard Input in the following format:
R_1 R_2
Output
Print the value of R_3.
The output is considered correct if the absolute or relative error is at most 10^{-6}.
Examples
Input
2 3
Output
1.2000000000
Input
100 99
Output
49.7487437186
Submitted Solution:
```
A,B=map(int,input().split())
print((A*B)/(A+B))
``` | instruction | 0 | 16,705 | 5 | 33,410 |
Yes | output | 1 | 16,705 | 5 | 33,411 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In an electric circuit, when two resistors R_1 and R_2 are connected in parallel, the equivalent resistance R_3 can be derived from the following formula:
* \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{R_3}
Given R_1 and R_2, find R_3.
Constraints
* 1 \leq R_1, R_2 \leq 100
* R_1 and R_2 are integers.
Input
The input is given from Standard Input in the following format:
R_1 R_2
Output
Print the value of R_3.
The output is considered correct if the absolute or relative error is at most 10^{-6}.
Examples
Input
2 3
Output
1.2000000000
Input
100 99
Output
49.7487437186
Submitted Solution:
```
a,b=map(int,input().split())
print((a**-1+b**-1)**-1)
``` | instruction | 0 | 16,706 | 5 | 33,412 |
Yes | output | 1 | 16,706 | 5 | 33,413 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In an electric circuit, when two resistors R_1 and R_2 are connected in parallel, the equivalent resistance R_3 can be derived from the following formula:
* \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{R_3}
Given R_1 and R_2, find R_3.
Constraints
* 1 \leq R_1, R_2 \leq 100
* R_1 and R_2 are integers.
Input
The input is given from Standard Input in the following format:
R_1 R_2
Output
Print the value of R_3.
The output is considered correct if the absolute or relative error is at most 10^{-6}.
Examples
Input
2 3
Output
1.2000000000
Input
100 99
Output
49.7487437186
Submitted Solution:
```
a,b = map(int,input().split())
print(1/a + 1/b)
``` | instruction | 0 | 16,707 | 5 | 33,414 |
No | output | 1 | 16,707 | 5 | 33,415 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In an electric circuit, when two resistors R_1 and R_2 are connected in parallel, the equivalent resistance R_3 can be derived from the following formula:
* \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{R_3}
Given R_1 and R_2, find R_3.
Constraints
* 1 \leq R_1, R_2 \leq 100
* R_1 and R_2 are integers.
Input
The input is given from Standard Input in the following format:
R_1 R_2
Output
Print the value of R_3.
The output is considered correct if the absolute or relative error is at most 10^{-6}.
Examples
Input
2 3
Output
1.2000000000
Input
100 99
Output
49.7487437186
Submitted Solution:
```
a, b, c = map(int, input().split())
print((a * b / (a + b)))
``` | instruction | 0 | 16,708 | 5 | 33,416 |
No | output | 1 | 16,708 | 5 | 33,417 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In an electric circuit, when two resistors R_1 and R_2 are connected in parallel, the equivalent resistance R_3 can be derived from the following formula:
* \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{R_3}
Given R_1 and R_2, find R_3.
Constraints
* 1 \leq R_1, R_2 \leq 100
* R_1 and R_2 are integers.
Input
The input is given from Standard Input in the following format:
R_1 R_2
Output
Print the value of R_3.
The output is considered correct if the absolute or relative error is at most 10^{-6}.
Examples
Input
2 3
Output
1.2000000000
Input
100 99
Output
49.7487437186
Submitted Solution:
```
s = input()
rs = ''.join(['d' if(c=='b') else 'b' if(c=='d') else 'q' if(c=='p') else 'p' for c in s[::-1]])
ans = 'Yes' if(s == rs) else 'No'
print(ans)
``` | instruction | 0 | 16,709 | 5 | 33,418 |
No | output | 1 | 16,709 | 5 | 33,419 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In an electric circuit, when two resistors R_1 and R_2 are connected in parallel, the equivalent resistance R_3 can be derived from the following formula:
* \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{R_3}
Given R_1 and R_2, find R_3.
Constraints
* 1 \leq R_1, R_2 \leq 100
* R_1 and R_2 are integers.
Input
The input is given from Standard Input in the following format:
R_1 R_2
Output
Print the value of R_3.
The output is considered correct if the absolute or relative error is at most 10^{-6}.
Examples
Input
2 3
Output
1.2000000000
Input
100 99
Output
49.7487437186
Submitted Solution:
```
n,m = map(input().split())
print((n*m)/(n+m))
``` | instruction | 0 | 16,710 | 5 | 33,420 |
No | output | 1 | 16,710 | 5 | 33,421 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let s(x) be sum of digits in decimal representation of positive integer x. Given two integers n and m, find some positive integers a and b such that
* s(a) ≥ n,
* s(b) ≥ n,
* s(a + b) ≤ m.
Input
The only line of input contain two integers n and m (1 ≤ n, m ≤ 1129).
Output
Print two lines, one for decimal representation of a and one for decimal representation of b. Both numbers must not contain leading zeros and must have length no more than 2230.
Examples
Input
6 5
Output
6
7
Input
8 16
Output
35
53
Note
In the first sample, we have n = 6 and m = 5. One valid solution is a = 6, b = 7. Indeed, we have s(a) = 6 ≥ n and s(b) = 7 ≥ n, and also s(a + b) = s(13) = 4 ≤ m.
Submitted Solution:
```
class CodeforcesTask1028BSolution:
def __init__(self):
self.result = ''
def read_input(self):
pass
def process_task(self):
self.result = "{0}\n{1}".format("9" * 200 + "0" * 199 + "1", "9" * 200)
def get_result(self):
return self.result
if __name__ == "__main__":
Solution = CodeforcesTask1028BSolution()
Solution.read_input()
Solution.process_task()
print(Solution.get_result())
``` | instruction | 0 | 16,838 | 5 | 33,676 |
Yes | output | 1 | 16,838 | 5 | 33,677 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let s(x) be sum of digits in decimal representation of positive integer x. Given two integers n and m, find some positive integers a and b such that
* s(a) ≥ n,
* s(b) ≥ n,
* s(a + b) ≤ m.
Input
The only line of input contain two integers n and m (1 ≤ n, m ≤ 1129).
Output
Print two lines, one for decimal representation of a and one for decimal representation of b. Both numbers must not contain leading zeros and must have length no more than 2230.
Examples
Input
6 5
Output
6
7
Input
8 16
Output
35
53
Note
In the first sample, we have n = 6 and m = 5. One valid solution is a = 6, b = 7. Indeed, we have s(a) = 6 ≥ n and s(b) = 7 ≥ n, and also s(a + b) = s(13) = 4 ≤ m.
Submitted Solution:
```
n, m = map(int, input().split())
i = 1
j = 1
while sum(map(int, str(i))) < n and sum(map(int, str(j))) < n and sum(map(int, str(i + j))) <= m:
if sum(map(int, str(i))) < n:
i += 1
if sum(map(int, str(j))) < n or sum(map(int, str(i + j))) != m:
j += 1
if sum(map(int, str(i))) >= n and sum(map(int, str(j))) >= n and sum(map(int, str(i + j))) <= m:
print(i)
print(j)
``` | instruction | 0 | 16,842 | 5 | 33,684 |
No | output | 1 | 16,842 | 5 | 33,685 |
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