message stringlengths 2 433k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 113 108k | cluster float64 12 12 | __index_level_0__ int64 226 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has two arrays A and B of lengths n and m, respectively.
He can perform the following operation arbitrary number of times (possibly zero): he takes some consecutive subsegment of the array and replaces it with a single element, equal to the sum of all elements on this subsegment. For example, from the array [1, 10, 100, 1000, 10000] Vasya can obtain array [1, 1110, 10000], and from array [1, 2, 3] Vasya can obtain array [6].
Two arrays A and B are considered equal if and only if they have the same length and for each valid i A_i = B_i.
Vasya wants to perform some of these operations on array A, some on array B, in such a way that arrays A and B become equal. Moreover, the lengths of the resulting arrays should be maximal possible.
Help Vasya to determine the maximum length of the arrays that he can achieve or output that it is impossible to make arrays A and B equal.
Input
The first line contains a single integer n~(1 β€ n β€ 3 β
10^5) β the length of the first array.
The second line contains n integers a_1, a_2, β
β
β
, a_n~(1 β€ a_i β€ 10^9) β elements of the array A.
The third line contains a single integer m~(1 β€ m β€ 3 β
10^5) β the length of the second array.
The fourth line contains m integers b_1, b_2, β
β
β
, b_m~(1 β€ b_i β€ 10^9) - elements of the array B.
Output
Print a single integer β the maximum length of the resulting arrays after some operations were performed on arrays A and B in such a way that they became equal.
If there is no way to make array equal, print "-1".
Examples
Input
5
11 2 3 5 7
4
11 7 3 7
Output
3
Input
2
1 2
1
100
Output
-1
Input
3
1 2 3
3
1 2 3
Output
3 | instruction | 0 | 87,546 | 12 | 175,092 |
Tags: greedy, two pointers
Correct Solution:
```
n = int(input())
a = list(map(int, input().split()))
m = int(input())
b = list(map(int, input().split()))
ptra = 1
ptrb = 1
sa = a[0]
sb = b[0]
ans = 0
while ptra != n and ptrb != m:
if sa == sb:
ans += 1
sa = a[ptra]
sb = b[ptrb]
ptra += 1
ptrb += 1
continue
if sa < sb:
sa += a[ptra]
ptra += 1
else:
sb += b[ptrb]
ptrb += 1
while ptra != n:
sa += a[ptra]
ptra += 1
while ptrb != m:
sb += b[ptrb]
ptrb += 1
if sa != sb:
print(-1)
exit(0)
print(ans + 1)
``` | output | 1 | 87,546 | 12 | 175,093 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has two arrays A and B of lengths n and m, respectively.
He can perform the following operation arbitrary number of times (possibly zero): he takes some consecutive subsegment of the array and replaces it with a single element, equal to the sum of all elements on this subsegment. For example, from the array [1, 10, 100, 1000, 10000] Vasya can obtain array [1, 1110, 10000], and from array [1, 2, 3] Vasya can obtain array [6].
Two arrays A and B are considered equal if and only if they have the same length and for each valid i A_i = B_i.
Vasya wants to perform some of these operations on array A, some on array B, in such a way that arrays A and B become equal. Moreover, the lengths of the resulting arrays should be maximal possible.
Help Vasya to determine the maximum length of the arrays that he can achieve or output that it is impossible to make arrays A and B equal.
Input
The first line contains a single integer n~(1 β€ n β€ 3 β
10^5) β the length of the first array.
The second line contains n integers a_1, a_2, β
β
β
, a_n~(1 β€ a_i β€ 10^9) β elements of the array A.
The third line contains a single integer m~(1 β€ m β€ 3 β
10^5) β the length of the second array.
The fourth line contains m integers b_1, b_2, β
β
β
, b_m~(1 β€ b_i β€ 10^9) - elements of the array B.
Output
Print a single integer β the maximum length of the resulting arrays after some operations were performed on arrays A and B in such a way that they became equal.
If there is no way to make array equal, print "-1".
Examples
Input
5
11 2 3 5 7
4
11 7 3 7
Output
3
Input
2
1 2
1
100
Output
-1
Input
3
1 2 3
3
1 2 3
Output
3 | instruction | 0 | 87,547 | 12 | 175,094 |
Tags: greedy, two pointers
Correct Solution:
```
# import atexit
# import io
# import sys
#
# _INPUT_LINES = sys.stdin.read().splitlines()
# input = iter(_INPUT_LINES).__next__
# _OUTPUT_BUFFER = io.StringIO()
# sys.stdout = _OUTPUT_BUFFER
#
#
# @atexit.register
# def write():
# sys.__stdout__.write(_OUTPUT_BUFFER.getvalue())
from itertools import accumulate
def main():
n = int(input())
arr1 = list(accumulate(map(int, input().split())))
m = int(input())
arr2 = list(accumulate(map(int, input().split())))
if arr1[-1] != arr2[-1]:
print(-1)
exit(0)
s1 = set(arr1)
cnt = 0
for v in arr2:
if v in s1:
cnt += 1
print(cnt)
if __name__ == "__main__":
main()
``` | output | 1 | 87,547 | 12 | 175,095 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has two arrays A and B of lengths n and m, respectively.
He can perform the following operation arbitrary number of times (possibly zero): he takes some consecutive subsegment of the array and replaces it with a single element, equal to the sum of all elements on this subsegment. For example, from the array [1, 10, 100, 1000, 10000] Vasya can obtain array [1, 1110, 10000], and from array [1, 2, 3] Vasya can obtain array [6].
Two arrays A and B are considered equal if and only if they have the same length and for each valid i A_i = B_i.
Vasya wants to perform some of these operations on array A, some on array B, in such a way that arrays A and B become equal. Moreover, the lengths of the resulting arrays should be maximal possible.
Help Vasya to determine the maximum length of the arrays that he can achieve or output that it is impossible to make arrays A and B equal.
Input
The first line contains a single integer n~(1 β€ n β€ 3 β
10^5) β the length of the first array.
The second line contains n integers a_1, a_2, β
β
β
, a_n~(1 β€ a_i β€ 10^9) β elements of the array A.
The third line contains a single integer m~(1 β€ m β€ 3 β
10^5) β the length of the second array.
The fourth line contains m integers b_1, b_2, β
β
β
, b_m~(1 β€ b_i β€ 10^9) - elements of the array B.
Output
Print a single integer β the maximum length of the resulting arrays after some operations were performed on arrays A and B in such a way that they became equal.
If there is no way to make array equal, print "-1".
Examples
Input
5
11 2 3 5 7
4
11 7 3 7
Output
3
Input
2
1 2
1
100
Output
-1
Input
3
1 2 3
3
1 2 3
Output
3 | instruction | 0 | 87,548 | 12 | 175,096 |
Tags: greedy, two pointers
Correct Solution:
```
n = int(input())
arr = list(map(int, input().split()))
m = int(input())
brr = list(map(int, input().split()))
if sum(arr) != sum(brr):
print(-1)
exit()
ai = 0
bi = 0
asm = 0
bsm = 0
cnt = 0
while ai < len(arr) and bi < len(brr):
if asm < bsm:
asm += arr[ai]
ai += 1
elif bsm < asm:
bsm += brr[bi]
bi += 1
else:
cnt += 1
asm = arr[ai]
bsm = brr[bi]
ai += 1
bi += 1
print(cnt)
``` | output | 1 | 87,548 | 12 | 175,097 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has two arrays A and B of lengths n and m, respectively.
He can perform the following operation arbitrary number of times (possibly zero): he takes some consecutive subsegment of the array and replaces it with a single element, equal to the sum of all elements on this subsegment. For example, from the array [1, 10, 100, 1000, 10000] Vasya can obtain array [1, 1110, 10000], and from array [1, 2, 3] Vasya can obtain array [6].
Two arrays A and B are considered equal if and only if they have the same length and for each valid i A_i = B_i.
Vasya wants to perform some of these operations on array A, some on array B, in such a way that arrays A and B become equal. Moreover, the lengths of the resulting arrays should be maximal possible.
Help Vasya to determine the maximum length of the arrays that he can achieve or output that it is impossible to make arrays A and B equal.
Input
The first line contains a single integer n~(1 β€ n β€ 3 β
10^5) β the length of the first array.
The second line contains n integers a_1, a_2, β
β
β
, a_n~(1 β€ a_i β€ 10^9) β elements of the array A.
The third line contains a single integer m~(1 β€ m β€ 3 β
10^5) β the length of the second array.
The fourth line contains m integers b_1, b_2, β
β
β
, b_m~(1 β€ b_i β€ 10^9) - elements of the array B.
Output
Print a single integer β the maximum length of the resulting arrays after some operations were performed on arrays A and B in such a way that they became equal.
If there is no way to make array equal, print "-1".
Examples
Input
5
11 2 3 5 7
4
11 7 3 7
Output
3
Input
2
1 2
1
100
Output
-1
Input
3
1 2 3
3
1 2 3
Output
3
Submitted Solution:
```
n=int(input())
a=[int(i) for i in input().split(' ')]
m=int(input())
b=[int(i) for i in input().split(' ')]
aa=[a[0]]
bb=[b[0]]
for i in range(1,n):
aa.append(aa[-1]+a[i])
for i in range(1,m):
bb.append(bb[-1]+b[i])
a1=set(aa)
a2=set(bb)
if sum(a)!=sum(b):
print(-1)
else:
print(len((a1.intersection(a2))))
``` | instruction | 0 | 87,549 | 12 | 175,098 |
Yes | output | 1 | 87,549 | 12 | 175,099 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has two arrays A and B of lengths n and m, respectively.
He can perform the following operation arbitrary number of times (possibly zero): he takes some consecutive subsegment of the array and replaces it with a single element, equal to the sum of all elements on this subsegment. For example, from the array [1, 10, 100, 1000, 10000] Vasya can obtain array [1, 1110, 10000], and from array [1, 2, 3] Vasya can obtain array [6].
Two arrays A and B are considered equal if and only if they have the same length and for each valid i A_i = B_i.
Vasya wants to perform some of these operations on array A, some on array B, in such a way that arrays A and B become equal. Moreover, the lengths of the resulting arrays should be maximal possible.
Help Vasya to determine the maximum length of the arrays that he can achieve or output that it is impossible to make arrays A and B equal.
Input
The first line contains a single integer n~(1 β€ n β€ 3 β
10^5) β the length of the first array.
The second line contains n integers a_1, a_2, β
β
β
, a_n~(1 β€ a_i β€ 10^9) β elements of the array A.
The third line contains a single integer m~(1 β€ m β€ 3 β
10^5) β the length of the second array.
The fourth line contains m integers b_1, b_2, β
β
β
, b_m~(1 β€ b_i β€ 10^9) - elements of the array B.
Output
Print a single integer β the maximum length of the resulting arrays after some operations were performed on arrays A and B in such a way that they became equal.
If there is no way to make array equal, print "-1".
Examples
Input
5
11 2 3 5 7
4
11 7 3 7
Output
3
Input
2
1 2
1
100
Output
-1
Input
3
1 2 3
3
1 2 3
Output
3
Submitted Solution:
```
n = int(input())
a = [int(x) for x in input().split()]
m = int(input())
b = [int(x) for x in input().split()]
i = j = 0
x = a[0]
y = b[0]
length = 0
while True:
if x == y and i < (n - 1) and j < (m - 1):
length += 1
i += 1
j += 1
# if i < n and j < m:
x = a[i]
y = b[j]
elif x < y and i < (n - 1):
i += 1
x += a[i]
elif x > y and j < (m - 1):
j += 1
y += b[j]
else:
break
if i == (n - 1) and j == (m - 1) and x == y:
print(length + 1)
else:
print(-1)
``` | instruction | 0 | 87,550 | 12 | 175,100 |
Yes | output | 1 | 87,550 | 12 | 175,101 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has two arrays A and B of lengths n and m, respectively.
He can perform the following operation arbitrary number of times (possibly zero): he takes some consecutive subsegment of the array and replaces it with a single element, equal to the sum of all elements on this subsegment. For example, from the array [1, 10, 100, 1000, 10000] Vasya can obtain array [1, 1110, 10000], and from array [1, 2, 3] Vasya can obtain array [6].
Two arrays A and B are considered equal if and only if they have the same length and for each valid i A_i = B_i.
Vasya wants to perform some of these operations on array A, some on array B, in such a way that arrays A and B become equal. Moreover, the lengths of the resulting arrays should be maximal possible.
Help Vasya to determine the maximum length of the arrays that he can achieve or output that it is impossible to make arrays A and B equal.
Input
The first line contains a single integer n~(1 β€ n β€ 3 β
10^5) β the length of the first array.
The second line contains n integers a_1, a_2, β
β
β
, a_n~(1 β€ a_i β€ 10^9) β elements of the array A.
The third line contains a single integer m~(1 β€ m β€ 3 β
10^5) β the length of the second array.
The fourth line contains m integers b_1, b_2, β
β
β
, b_m~(1 β€ b_i β€ 10^9) - elements of the array B.
Output
Print a single integer β the maximum length of the resulting arrays after some operations were performed on arrays A and B in such a way that they became equal.
If there is no way to make array equal, print "-1".
Examples
Input
5
11 2 3 5 7
4
11 7 3 7
Output
3
Input
2
1 2
1
100
Output
-1
Input
3
1 2 3
3
1 2 3
Output
3
Submitted Solution:
```
# import sys
# sys.stdin = open('input.txt', 'r')
# sys.stdout = open('output.txt', 'w')
n = int(input())
a = list(map(int,input().split()))
m = int(input())
b = list(map(int,input().split()))
aNew = [0]*n
bNew = [0]*m
aNew[0] = a[0]
bNew[0] = b[0]
for i in range(1,n):
aNew[i] = aNew[i-1] + a[i]
for i in range(1,m):
bNew[i] = bNew[i-1] + b[i]
count = 0
i = 0
j = 0
while (i<n and j<m):
if (aNew[i] > bNew[j]):
j += 1
elif (aNew[i] < bNew[j]):
i += 1
else:
count += 1
i += 1
j += 1
if (i==n and j==m):
print(count)
else:
print(-1)
``` | instruction | 0 | 87,551 | 12 | 175,102 |
Yes | output | 1 | 87,551 | 12 | 175,103 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has two arrays A and B of lengths n and m, respectively.
He can perform the following operation arbitrary number of times (possibly zero): he takes some consecutive subsegment of the array and replaces it with a single element, equal to the sum of all elements on this subsegment. For example, from the array [1, 10, 100, 1000, 10000] Vasya can obtain array [1, 1110, 10000], and from array [1, 2, 3] Vasya can obtain array [6].
Two arrays A and B are considered equal if and only if they have the same length and for each valid i A_i = B_i.
Vasya wants to perform some of these operations on array A, some on array B, in such a way that arrays A and B become equal. Moreover, the lengths of the resulting arrays should be maximal possible.
Help Vasya to determine the maximum length of the arrays that he can achieve or output that it is impossible to make arrays A and B equal.
Input
The first line contains a single integer n~(1 β€ n β€ 3 β
10^5) β the length of the first array.
The second line contains n integers a_1, a_2, β
β
β
, a_n~(1 β€ a_i β€ 10^9) β elements of the array A.
The third line contains a single integer m~(1 β€ m β€ 3 β
10^5) β the length of the second array.
The fourth line contains m integers b_1, b_2, β
β
β
, b_m~(1 β€ b_i β€ 10^9) - elements of the array B.
Output
Print a single integer β the maximum length of the resulting arrays after some operations were performed on arrays A and B in such a way that they became equal.
If there is no way to make array equal, print "-1".
Examples
Input
5
11 2 3 5 7
4
11 7 3 7
Output
3
Input
2
1 2
1
100
Output
-1
Input
3
1 2 3
3
1 2 3
Output
3
Submitted Solution:
```
n = int(input())
arr1 = [int(element) for element in input().split()]
m = int(input())
arr2 = [int(element) for element in input().split()]
res = 0
p1 = 0
p2 = 0
sum1 = arr1[p1]
sum2 = arr2[p2]
arr1.append(0)
arr2.append(0)
while (p1 < n and p2 < m):
if (sum1 == sum2):
res += 1
sum1 = arr1[p1 + 1]
sum2 = arr2[p2 + 1]
p1 += 1
p2 += 1
elif (sum1 < sum2):
sum1 += arr1[p1 + 1]
p1 += 1
elif (sum1 > sum2):
sum2 += arr2[p2 + 1]
p2 += 1
if (p1 == n and p2 == m):
print(res)
else:
print(-1)
``` | instruction | 0 | 87,552 | 12 | 175,104 |
Yes | output | 1 | 87,552 | 12 | 175,105 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has two arrays A and B of lengths n and m, respectively.
He can perform the following operation arbitrary number of times (possibly zero): he takes some consecutive subsegment of the array and replaces it with a single element, equal to the sum of all elements on this subsegment. For example, from the array [1, 10, 100, 1000, 10000] Vasya can obtain array [1, 1110, 10000], and from array [1, 2, 3] Vasya can obtain array [6].
Two arrays A and B are considered equal if and only if they have the same length and for each valid i A_i = B_i.
Vasya wants to perform some of these operations on array A, some on array B, in such a way that arrays A and B become equal. Moreover, the lengths of the resulting arrays should be maximal possible.
Help Vasya to determine the maximum length of the arrays that he can achieve or output that it is impossible to make arrays A and B equal.
Input
The first line contains a single integer n~(1 β€ n β€ 3 β
10^5) β the length of the first array.
The second line contains n integers a_1, a_2, β
β
β
, a_n~(1 β€ a_i β€ 10^9) β elements of the array A.
The third line contains a single integer m~(1 β€ m β€ 3 β
10^5) β the length of the second array.
The fourth line contains m integers b_1, b_2, β
β
β
, b_m~(1 β€ b_i β€ 10^9) - elements of the array B.
Output
Print a single integer β the maximum length of the resulting arrays after some operations were performed on arrays A and B in such a way that they became equal.
If there is no way to make array equal, print "-1".
Examples
Input
5
11 2 3 5 7
4
11 7 3 7
Output
3
Input
2
1 2
1
100
Output
-1
Input
3
1 2 3
3
1 2 3
Output
3
Submitted Solution:
```
n = int(input())
a = [int(i) for i in input().split()]
m = int(input())
b = [int(i) for i in input().split()]
if sum(a) != sum(b):
print(-1)
else:
i = 0
j = 0
c1 = a[0]
c2 = b[0]
count = 0
while i < n and j < m:
if c1 == c2:
count += 1
i += 1
j += 1
if i < n and j < n:
c1 += a[i]
c2 += b[j]
elif c1 < c2:
i += 1
if i < n:
c1 += a[i]
elif c2 < c1:
j += 1
if j < n:
c2 += b[j]
print(count)
``` | instruction | 0 | 87,553 | 12 | 175,106 |
No | output | 1 | 87,553 | 12 | 175,107 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has two arrays A and B of lengths n and m, respectively.
He can perform the following operation arbitrary number of times (possibly zero): he takes some consecutive subsegment of the array and replaces it with a single element, equal to the sum of all elements on this subsegment. For example, from the array [1, 10, 100, 1000, 10000] Vasya can obtain array [1, 1110, 10000], and from array [1, 2, 3] Vasya can obtain array [6].
Two arrays A and B are considered equal if and only if they have the same length and for each valid i A_i = B_i.
Vasya wants to perform some of these operations on array A, some on array B, in such a way that arrays A and B become equal. Moreover, the lengths of the resulting arrays should be maximal possible.
Help Vasya to determine the maximum length of the arrays that he can achieve or output that it is impossible to make arrays A and B equal.
Input
The first line contains a single integer n~(1 β€ n β€ 3 β
10^5) β the length of the first array.
The second line contains n integers a_1, a_2, β
β
β
, a_n~(1 β€ a_i β€ 10^9) β elements of the array A.
The third line contains a single integer m~(1 β€ m β€ 3 β
10^5) β the length of the second array.
The fourth line contains m integers b_1, b_2, β
β
β
, b_m~(1 β€ b_i β€ 10^9) - elements of the array B.
Output
Print a single integer β the maximum length of the resulting arrays after some operations were performed on arrays A and B in such a way that they became equal.
If there is no way to make array equal, print "-1".
Examples
Input
5
11 2 3 5 7
4
11 7 3 7
Output
3
Input
2
1 2
1
100
Output
-1
Input
3
1 2 3
3
1 2 3
Output
3
Submitted Solution:
```
from bisect import bisect_left as bl
from bisect import bisect_right as br
from heapq import heappush,heappop
import math
from collections import *
from functools import reduce,cmp_to_key, lru_cache
import sys
input = sys.stdin.readline
M = mod = 10 ** 9 + 7
def factors(n):return sorted(set(reduce(list.__add__, ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0))))
def inv_mod(n):return pow(n, mod - 2, mod)
def li():return [int(i) for i in input().rstrip('\n').split()]
def st():return input().rstrip('\n')
def val():return int(input().rstrip('\n'))
def li2():return [i for i in input().rstrip('\n')]
def li3():return [int(i) for i in input().rstrip('\n')]
n = val()
l1 = li()
m = val()
l2 = li()
if sum(l1) != sum(l2):
print(-1)
exit()
curr1 = ans = i = j = curr2 = 0
i = j = 1
for itr in range(n):
if l1[itr] != l2[itr]:
ans = itr
i = itr + 1
curr1 = l1[itr]
j = itr
break
ans += 1
i = itr + 1
j = itr + 1
if i == n:
print(ans)
exit()
while i < n and j < m:
if curr1 == curr2:
curr1 = curr2 = 0
i += 1
j += 1
ans += 1
if i < n:
curr1 = l1[i]
i += 1
elif curr1 < curr2:
curr1 += l1[i]
i += 1
else:
curr2 += l2[j]
j += 1
# print(i, j, curr1, curr2, ans)
# print(ans)
ans += 1
print(ans)
``` | instruction | 0 | 87,554 | 12 | 175,108 |
No | output | 1 | 87,554 | 12 | 175,109 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has two arrays A and B of lengths n and m, respectively.
He can perform the following operation arbitrary number of times (possibly zero): he takes some consecutive subsegment of the array and replaces it with a single element, equal to the sum of all elements on this subsegment. For example, from the array [1, 10, 100, 1000, 10000] Vasya can obtain array [1, 1110, 10000], and from array [1, 2, 3] Vasya can obtain array [6].
Two arrays A and B are considered equal if and only if they have the same length and for each valid i A_i = B_i.
Vasya wants to perform some of these operations on array A, some on array B, in such a way that arrays A and B become equal. Moreover, the lengths of the resulting arrays should be maximal possible.
Help Vasya to determine the maximum length of the arrays that he can achieve or output that it is impossible to make arrays A and B equal.
Input
The first line contains a single integer n~(1 β€ n β€ 3 β
10^5) β the length of the first array.
The second line contains n integers a_1, a_2, β
β
β
, a_n~(1 β€ a_i β€ 10^9) β elements of the array A.
The third line contains a single integer m~(1 β€ m β€ 3 β
10^5) β the length of the second array.
The fourth line contains m integers b_1, b_2, β
β
β
, b_m~(1 β€ b_i β€ 10^9) - elements of the array B.
Output
Print a single integer β the maximum length of the resulting arrays after some operations were performed on arrays A and B in such a way that they became equal.
If there is no way to make array equal, print "-1".
Examples
Input
5
11 2 3 5 7
4
11 7 3 7
Output
3
Input
2
1 2
1
100
Output
-1
Input
3
1 2 3
3
1 2 3
Output
3
Submitted Solution:
```
"""import math
n,k=map(int,input().split())
#s=[int(x) for x in input().split()]
#for i in range(0,len(s)):
ans=k//n
if(k%n!=0):
ans=ans+1
print(ans)
#ans=math.ceil(k/n)
#print(ans)
"""
"""q=int(input())
for i in range(0,q):
n,m,k=map(int,input().split())
if(k<n or k<m):
print('-1')
else:
#ct=min(n,m)
x=max(n,m)-min(n,m)
y=k-min(n,m)
if(x%2==0 and (y-x)%2==0):
print(k)
elif(x%2==0 and (y-x)%2!=0):
print(k-2)
else:
print(k-1)"""
n=int(input())
s1=[int(x) for x in input().split()]
m=int(input())
s2=[int(x) for x in input().split()]
if(n==1 or m==1):
if(n==1):
r=0
for j in range(0,m):
r=r+s2[j]
if(r==s1[0]):
print('1')
else:
print('-1')
else:
r=0
for j in range(0,n):
r=r+s1[j]
if(r==s2[0]):
print('1')
else:
print('-1')
else:
pt1=1
pt2=1
sm1=s1[0]
sm2=s2[0]
L1=[]
L2=[]
c=0
for i in range(0,n+m+1):
if(sm1==sm2):
c=c+1
L1.append(sm1)
L2.append(sm2)
sm1=s1[pt1]
sm2=s2[pt2]
pt1=pt1+1
pt2=pt2+1
elif(sm1>sm2):
sm2=sm2+s2[pt2]
pt2=pt2+1
else:
sm1=sm1+s1[pt1]
pt1=pt1+1
if(pt1==n or pt2==m):
break
if(sm1==sm2):
c=c+1
L1.append(sm1)
L2.append(sm2)
#sm1=s1[pt1]
#sm2=s2[pt2]
pt1=pt1+1
pt2=pt2+1
if(len(L1)==0 or len(L2)==0):
print('-1')
else:
if(len(L1)==len(L2)):
print(len(L1))
``` | instruction | 0 | 87,555 | 12 | 175,110 |
No | output | 1 | 87,555 | 12 | 175,111 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has two arrays A and B of lengths n and m, respectively.
He can perform the following operation arbitrary number of times (possibly zero): he takes some consecutive subsegment of the array and replaces it with a single element, equal to the sum of all elements on this subsegment. For example, from the array [1, 10, 100, 1000, 10000] Vasya can obtain array [1, 1110, 10000], and from array [1, 2, 3] Vasya can obtain array [6].
Two arrays A and B are considered equal if and only if they have the same length and for each valid i A_i = B_i.
Vasya wants to perform some of these operations on array A, some on array B, in such a way that arrays A and B become equal. Moreover, the lengths of the resulting arrays should be maximal possible.
Help Vasya to determine the maximum length of the arrays that he can achieve or output that it is impossible to make arrays A and B equal.
Input
The first line contains a single integer n~(1 β€ n β€ 3 β
10^5) β the length of the first array.
The second line contains n integers a_1, a_2, β
β
β
, a_n~(1 β€ a_i β€ 10^9) β elements of the array A.
The third line contains a single integer m~(1 β€ m β€ 3 β
10^5) β the length of the second array.
The fourth line contains m integers b_1, b_2, β
β
β
, b_m~(1 β€ b_i β€ 10^9) - elements of the array B.
Output
Print a single integer β the maximum length of the resulting arrays after some operations were performed on arrays A and B in such a way that they became equal.
If there is no way to make array equal, print "-1".
Examples
Input
5
11 2 3 5 7
4
11 7 3 7
Output
3
Input
2
1 2
1
100
Output
-1
Input
3
1 2 3
3
1 2 3
Output
3
Submitted Solution:
```
from sys import stdin, stdout
import cProfile
printHeap = str()
test = False
memory_constrained = False
def display(string_to_print):
stdout.write(str(string_to_print) + "\n")
def test_print(output):
if test:
stdout.write(str(output) + "\n")
def display_list(list1, sep=" "):
stdout.write(sep.join(map(str, list1)) + "\n")
def get_int():
return int(stdin.readline())
def get_tuple():
return map(int, stdin.readline().split())
def get_list():
return list(map(int, stdin.readline().split()))
memory = dict()
def clear_cache():
global memory
memory = dict()
def cached_fn(fn, *args):
global memory
if args in memory:
return memory[args]
else:
result = fn(*args)
memory[args] = result
return result
# ----------------------------------------------------------------------------------- MAIN PROGRAM
def main():
n = get_int()
a = get_list()
m = get_int()
b = get_list()
if sum(a)!=sum(b):
print("-1")
return
i =0
j = 0
sa = 0
sb = 0
count = 0
while True:
test_print(str(j)+"=JI="+str(i))
if (i==n or j==m):
break
if a[i]==b[j]:
count +=1
i+=1
j+=1
sa,sb = 0,0
continue
elif sa>sb:
sb+=b[j]
j+=1
if sa==sb:
count+=1
sa, sb = 0, 0
elif sa<sb:
sa+=a[i]
i+=1
if sa==sb:
count+=1
sa, sb = 0, 0
else:
if a[i]>b[j]:
sb = b[j]
j+=1
else:
sa = a[i]
i+=1
display(count)
# --------------------------------------------------------------------------------------------- END
cProfile.run('main()') if test else main()
``` | instruction | 0 | 87,556 | 12 | 175,112 |
No | output | 1 | 87,556 | 12 | 175,113 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This problem differs from one which was on the online contest.
The sequence a1, a2, ..., an is called increasing, if ai < ai + 1 for i < n.
The sequence s1, s2, ..., sk is called the subsequence of the sequence a1, a2, ..., an, if there exist such a set of indexes 1 β€ i1 < i2 < ... < ik β€ n that aij = sj. In other words, the sequence s can be derived from the sequence a by crossing out some elements.
You are given two sequences of integer numbers. You are to find their longest common increasing subsequence, i.e. an increasing sequence of maximum length that is the subsequence of both sequences.
Input
The first line contains an integer n (1 β€ n β€ 500) β the length of the first sequence. The second line contains n space-separated integers from the range [0, 109] β elements of the first sequence. The third line contains an integer m (1 β€ m β€ 500) β the length of the second sequence. The fourth line contains m space-separated integers from the range [0, 109] β elements of the second sequence.
Output
In the first line output k β the length of the longest common increasing subsequence. In the second line output the subsequence itself. Separate the elements with a space. If there are several solutions, output any.
Examples
Input
7
2 3 1 6 5 4 6
4
1 3 5 6
Output
3
3 5 6
Input
5
1 2 0 2 1
3
1 0 1
Output
2
0 1 | instruction | 0 | 87,577 | 12 | 175,154 |
Tags: dp
Correct Solution:
```
def get_lcis_length(X, Y):
len_Y = len(Y)
# maintain current lcis length
# curr_lcis_len: len lcis(X, Y[0..i])
curr_lcis_len = [0] * len_Y
for u in X:
lcis_prefix_max = []
curr_max = 0
# copy current lcis length table and create prefix max from it
# to make access in next step be O(1)
for i, v in enumerate(Y):
if v < u:
curr_max = max(curr_max, curr_lcis_len[i])
lcis_prefix_max.append(curr_max)
# loop through all elements in Y
# update current lcis length if the element matches with current element in X
for i, v in enumerate(Y):
if u == v:
curr_lcis_len[i] = max(curr_lcis_len[i], (0 if i == 0 else lcis_prefix_max[i - 1]) + 1)
# find maximum lcis length
max_lcis_length = curr_lcis_len[0]
for i, u in enumerate(curr_lcis_len):
max_lcis_length = max(max_lcis_length, u)
# Backtrack to get the LCIS
# Redo all previous step, but break when we found lcis that has the same length
# with max_lcis_length
# Need to break the loop to stop backtracking table (bt) being updated
# No need to backtrack if there is no LCIS
if max_lcis_length == 0:
print(0)
return max_lcis_length
curr_lcis_len = [0] * len_Y
bt = [-1] * len_Y # backtracking table
for u in X:
lcis_prefix_max = []
bt_id = []
curr_max = 0
curr_id_max = -1;
# copy current lcis length table and create prefix max from it
# to make access in next step be O(1)
# but, we also keep where should we backtrack if we use the lcis_pref_max
for i, v in enumerate(Y):
if v < u:
if curr_max < curr_lcis_len[i]:
curr_max = curr_lcis_len[i]
curr_id_max = i
lcis_prefix_max.append(curr_max)
bt_id.append(curr_id_max)
# print("bt_id", bt_id)
# print("lcis pref max", lcis_prefix_max)
# print("curr lcis len", curr_lcis_len)
# loop through all elements in Y
# update current lcis length if the element matches with current element in X
for i, v in enumerate(Y):
if u == v:
# print(u, v)
if curr_lcis_len[i] < (0 if i == 0 else lcis_prefix_max[i - 1]) + 1:
curr_lcis_len[i] = (0 if i == 0 else lcis_prefix_max[i - 1]) + 1
bt[i] = (-1 if i == 0 else bt_id[i - 1]) # also update where should we backtrack if using this index
# if this is one of the lcis, use the backtrack table to track the values
if curr_lcis_len[i] == max_lcis_length:
lcis_list = []
now = i
while now != -1:
lcis_list.append(Y[now]);
now = bt[now]
lcis_list.reverse()
print(max_lcis_length)
for i, u in enumerate(lcis_list):
print(u, end=('\n' if i == max_lcis_length - 1 else ' '))
return max_lcis_length, lcis_list
return max_lcis_length
x = []
try:
while True:
y = input()
x.append(y)
except EOFError:
pass
x = ' '.join(x)
x = x.split()
x = [int(i) for i in x]
X = []
Y = []
num_x = x[0]
for i in range(1, num_x + 1):
X.append(x[i])
for i in range(x[num_x + 1]):
Y.append(x[num_x + 2 + i])
get_lcis_length(X, Y)
``` | output | 1 | 87,577 | 12 | 175,155 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This problem differs from one which was on the online contest.
The sequence a1, a2, ..., an is called increasing, if ai < ai + 1 for i < n.
The sequence s1, s2, ..., sk is called the subsequence of the sequence a1, a2, ..., an, if there exist such a set of indexes 1 β€ i1 < i2 < ... < ik β€ n that aij = sj. In other words, the sequence s can be derived from the sequence a by crossing out some elements.
You are given two sequences of integer numbers. You are to find their longest common increasing subsequence, i.e. an increasing sequence of maximum length that is the subsequence of both sequences.
Input
The first line contains an integer n (1 β€ n β€ 500) β the length of the first sequence. The second line contains n space-separated integers from the range [0, 109] β elements of the first sequence. The third line contains an integer m (1 β€ m β€ 500) β the length of the second sequence. The fourth line contains m space-separated integers from the range [0, 109] β elements of the second sequence.
Output
In the first line output k β the length of the longest common increasing subsequence. In the second line output the subsequence itself. Separate the elements with a space. If there are several solutions, output any.
Examples
Input
7
2 3 1 6 5 4 6
4
1 3 5 6
Output
3
3 5 6
Input
5
1 2 0 2 1
3
1 0 1
Output
2
0 1 | instruction | 0 | 87,578 | 12 | 175,156 |
Tags: dp
Correct Solution:
```
n = int(input())
a = [0] + list(map(int, input().split()))
m = int(input())
b = [0] + list(map(int, input().split()))
d = [[0 for j in range(m+1)] for i in range(n+1)]
prev = [0] * (n+1)
for i in range(1, n+1):
for j in range(1, m+1):
if a[i] == b[j]:
d[i][j] = 1
for k in range(1, i):
if a[k] < a[i] and d[i][j] < d[k][j] + 1:
d[i][j] = d[k][j] + 1
prev[i] = k
else:
d[i][j] = d[i][j - 1]
pos = 0
for i in range(1, n+1):
if d[pos][m] < d[i][m]:
pos = i
ans = []
while pos != 0:
ans.append(a[pos])
pos = prev[pos]
print(len(ans))
print(*ans[::-1])
# Fri Oct 09 2020 18:06:27 GMT+0300 (ΠΠΎΡΠΊΠ²Π°, ΡΡΠ°Π½Π΄Π°ΡΡΠ½ΠΎΠ΅ Π²ΡΠ΅ΠΌΡ)
``` | output | 1 | 87,578 | 12 | 175,157 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This problem differs from one which was on the online contest.
The sequence a1, a2, ..., an is called increasing, if ai < ai + 1 for i < n.
The sequence s1, s2, ..., sk is called the subsequence of the sequence a1, a2, ..., an, if there exist such a set of indexes 1 β€ i1 < i2 < ... < ik β€ n that aij = sj. In other words, the sequence s can be derived from the sequence a by crossing out some elements.
You are given two sequences of integer numbers. You are to find their longest common increasing subsequence, i.e. an increasing sequence of maximum length that is the subsequence of both sequences.
Input
The first line contains an integer n (1 β€ n β€ 500) β the length of the first sequence. The second line contains n space-separated integers from the range [0, 109] β elements of the first sequence. The third line contains an integer m (1 β€ m β€ 500) β the length of the second sequence. The fourth line contains m space-separated integers from the range [0, 109] β elements of the second sequence.
Output
In the first line output k β the length of the longest common increasing subsequence. In the second line output the subsequence itself. Separate the elements with a space. If there are several solutions, output any.
Examples
Input
7
2 3 1 6 5 4 6
4
1 3 5 6
Output
3
3 5 6
Input
5
1 2 0 2 1
3
1 0 1
Output
2
0 1 | instruction | 0 | 87,579 | 12 | 175,158 |
Tags: dp
Correct Solution:
```
__author__ = 'Darren'
# dp(i,j): length of the LCIS between f[0:i+1] and s[0:j+1] that ends at s[j]
# if f[i] == s[j]: dp(i,j) = max_len + 1
# else: dp(i,j) = dp(i-1,j)
def solve():
n = int(input())
first = list(map(int, input().split()))
m = int(input())
second = list(map(int, input().split()))
if m > n:
n, m = m, n
first, second = second, first
# dp[j]: LCIS ending at second[j]
# prev[j]: index of the second-to-last number for the LCIS ending at second[j]
dp, prev = [0] * m, [-1] * m
for i in range(n):
# max_len: length of the LCIS whose largest number is smaller than first[i]
max_len, last_index = 0, -1
for j in range(m):
if first[i] == second[j] and dp[j] < max_len + 1:
dp[j] = max_len + 1
prev[j] = last_index
elif first[i] > second[j] and max_len < dp[j]:
max_len = dp[j]
last_index = j
# Find the length of LCIS between the first and second sequences, and
# the index of the last common number in the second sequence
max_value, index_max = 0, -1
for index, value in enumerate(dp):
if value > max_value:
max_value = value
index_max = index
print(max_value)
if max_value > 0:
seq = []
index = index_max
while index >= 0:
seq.append(str(second[index]))
index = prev[index]
seq.reverse()
print(' '.join(seq))
if __name__ == '__main__':
solve()
``` | output | 1 | 87,579 | 12 | 175,159 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This problem differs from one which was on the online contest.
The sequence a1, a2, ..., an is called increasing, if ai < ai + 1 for i < n.
The sequence s1, s2, ..., sk is called the subsequence of the sequence a1, a2, ..., an, if there exist such a set of indexes 1 β€ i1 < i2 < ... < ik β€ n that aij = sj. In other words, the sequence s can be derived from the sequence a by crossing out some elements.
You are given two sequences of integer numbers. You are to find their longest common increasing subsequence, i.e. an increasing sequence of maximum length that is the subsequence of both sequences.
Input
The first line contains an integer n (1 β€ n β€ 500) β the length of the first sequence. The second line contains n space-separated integers from the range [0, 109] β elements of the first sequence. The third line contains an integer m (1 β€ m β€ 500) β the length of the second sequence. The fourth line contains m space-separated integers from the range [0, 109] β elements of the second sequence.
Output
In the first line output k β the length of the longest common increasing subsequence. In the second line output the subsequence itself. Separate the elements with a space. If there are several solutions, output any.
Examples
Input
7
2 3 1 6 5 4 6
4
1 3 5 6
Output
3
3 5 6
Input
5
1 2 0 2 1
3
1 0 1
Output
2
0 1 | instruction | 0 | 87,580 | 12 | 175,160 |
Tags: dp
Correct Solution:
```
n = int(input())
a = [0] + list(map(int, input().split()))
m = int(input())
b = [0] + list(map(int, input().split()))
d = [[0 for j in range(m+1)] for i in range(n+1)]
prev = [0] * (n+1)
for i in range(1, n+1):
for j in range(1, m+1):
if a[i] == b[j]:
d[i][j] = 1
for k in range(1, i):
if a[k] < a[i] and d[i][j] < d[k][j] + 1:
d[i][j] = d[k][j] + 1
prev[i] = k
else:
d[i][j] = d[i][j - 1]
pos = 0
for i in range(1, n+1):
if d[pos][m] < d[i][m]:
pos = i
ans = []
while pos != 0:
ans.append(a[pos])
pos = prev[pos]
print(len(ans))
print(*ans[::-1])
# Tue Oct 15 2019 10:46:29 GMT+0300 (MSK)
``` | output | 1 | 87,580 | 12 | 175,161 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This problem differs from one which was on the online contest.
The sequence a1, a2, ..., an is called increasing, if ai < ai + 1 for i < n.
The sequence s1, s2, ..., sk is called the subsequence of the sequence a1, a2, ..., an, if there exist such a set of indexes 1 β€ i1 < i2 < ... < ik β€ n that aij = sj. In other words, the sequence s can be derived from the sequence a by crossing out some elements.
You are given two sequences of integer numbers. You are to find their longest common increasing subsequence, i.e. an increasing sequence of maximum length that is the subsequence of both sequences.
Input
The first line contains an integer n (1 β€ n β€ 500) β the length of the first sequence. The second line contains n space-separated integers from the range [0, 109] β elements of the first sequence. The third line contains an integer m (1 β€ m β€ 500) β the length of the second sequence. The fourth line contains m space-separated integers from the range [0, 109] β elements of the second sequence.
Output
In the first line output k β the length of the longest common increasing subsequence. In the second line output the subsequence itself. Separate the elements with a space. If there are several solutions, output any.
Examples
Input
7
2 3 1 6 5 4 6
4
1 3 5 6
Output
3
3 5 6
Input
5
1 2 0 2 1
3
1 0 1
Output
2
0 1 | instruction | 0 | 87,581 | 12 | 175,162 |
Tags: dp
Correct Solution:
```
n=int(input())
a=list(map(int,input().split()))
m=int(input())
b=list(map(int,input().split()))
dp=[[0]*(m + 1) for i in range(n + 1)]
p=[[0]*(m + 1) for i in range(n + 1)]
v1=0
mx=0
for i in range(1,n+1):
bst=0
v=0
for j in range(1,m+1):
dp[i][j]=dp[i-1][j]
p[i][j]=p[i-1][j]
if(a[i-1]==b[j-1] and bst+1>dp[i][j]):
dp[i][j]=bst+1
p[i][j]=v
if(a[i - 1]>b[j - 1] and dp[i-1][j]>bst):
v=b[j-1]
bst=dp[i-1][j]
if(dp[i][j]>mx):
mx=dp[i][j]
v1 = b[j - 1]
print(mx)
i=n-1
j=m-1
arr=[0]*mx
for k in range(mx):
while(a[i]!=v1):
i=i-1
while(b[j]!=v1):
j=j-1
arr[mx-1-k]=a[i]
v1=p[i+1][j+1]
for k in range(mx):
print(arr[k],end=' ')
# Tue Aug 04 2020 01:02:44 GMT+0300 (ΠΠΎΡΠΊΠ²Π°, ΡΡΠ°Π½Π΄Π°ΡΡΠ½ΠΎΠ΅ Π²ΡΠ΅ΠΌΡ)
``` | output | 1 | 87,581 | 12 | 175,163 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This problem differs from one which was on the online contest.
The sequence a1, a2, ..., an is called increasing, if ai < ai + 1 for i < n.
The sequence s1, s2, ..., sk is called the subsequence of the sequence a1, a2, ..., an, if there exist such a set of indexes 1 β€ i1 < i2 < ... < ik β€ n that aij = sj. In other words, the sequence s can be derived from the sequence a by crossing out some elements.
You are given two sequences of integer numbers. You are to find their longest common increasing subsequence, i.e. an increasing sequence of maximum length that is the subsequence of both sequences.
Input
The first line contains an integer n (1 β€ n β€ 500) β the length of the first sequence. The second line contains n space-separated integers from the range [0, 109] β elements of the first sequence. The third line contains an integer m (1 β€ m β€ 500) β the length of the second sequence. The fourth line contains m space-separated integers from the range [0, 109] β elements of the second sequence.
Output
In the first line output k β the length of the longest common increasing subsequence. In the second line output the subsequence itself. Separate the elements with a space. If there are several solutions, output any.
Examples
Input
7
2 3 1 6 5 4 6
4
1 3 5 6
Output
3
3 5 6
Input
5
1 2 0 2 1
3
1 0 1
Output
2
0 1 | instruction | 0 | 87,582 | 12 | 175,164 |
Tags: dp
Correct Solution:
```
n1=int(input())
arr1=list(map(int,input().split()))
n2=int(input())
arr2=list(map(int,input().split()))
if(n1>n2):
temp=n1
n1=n2
n2=temp
temp1=arr1.copy()
arr1=arr2.copy()
arr2=temp1
table=[]
count=[0]*n2
table1=[]
for i in range(n2):
table.append([])
for i in range(n1):
table1.append([])
for i in range(n1):
ct=0
for j in range(n2):
if(arr1[i]==arr2[j]):
if(ct+1>count[j]):
count[j]=ct+1
table[j]=table1[i].copy()
table[j].append(arr1[i])
if(arr1[i]>arr2[j]):
if(count[j]>ct):
ct=count[j]
table1[i]=table[j].copy()
mx=0
res=[]
for i in range(n2):
if(count[i]>mx):
mx=count[i]
res=table[i]
print(mx)
if(mx!=0):
for i in res:
print(i,end=" ")
else:
print()
``` | output | 1 | 87,582 | 12 | 175,165 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This problem differs from one which was on the online contest.
The sequence a1, a2, ..., an is called increasing, if ai < ai + 1 for i < n.
The sequence s1, s2, ..., sk is called the subsequence of the sequence a1, a2, ..., an, if there exist such a set of indexes 1 β€ i1 < i2 < ... < ik β€ n that aij = sj. In other words, the sequence s can be derived from the sequence a by crossing out some elements.
You are given two sequences of integer numbers. You are to find their longest common increasing subsequence, i.e. an increasing sequence of maximum length that is the subsequence of both sequences.
Input
The first line contains an integer n (1 β€ n β€ 500) β the length of the first sequence. The second line contains n space-separated integers from the range [0, 109] β elements of the first sequence. The third line contains an integer m (1 β€ m β€ 500) β the length of the second sequence. The fourth line contains m space-separated integers from the range [0, 109] β elements of the second sequence.
Output
In the first line output k β the length of the longest common increasing subsequence. In the second line output the subsequence itself. Separate the elements with a space. If there are several solutions, output any.
Examples
Input
7
2 3 1 6 5 4 6
4
1 3 5 6
Output
3
3 5 6
Input
5
1 2 0 2 1
3
1 0 1
Output
2
0 1 | instruction | 0 | 87,583 | 12 | 175,166 |
Tags: dp
Correct Solution:
```
def sub_seq(initial,lis1,lis2,sus):
total=[]
inc_initial=initial
c_lis2=lis2
if sus==[]:
subs=[initial]
else:
subs=sus
recur_sub=[initial]
ind=0
for i in lis1:
if int(i)>int(inc_initial):
if i in c_lis2:
subs.append(i)
inc_initial=i
c_lis2=c_lis2[c_lis2.index(i)+1:]
ind+=1
else:
if int(i)>int(initial):
if i in lis2:
temp_subs=subs.copy()
for h in range (len(subs)-1,-1,-1):
if int(subs[h]) < int(i):
temp_subs.append(int(i))
break
else:
temp_subs.pop()
recur_sub=sub_seq(i,lis1[ind+1:],lis2[lis2.index(i):],temp_subs)
total.append(recur_sub)
ind+=1
else:
ind+=1
min=0
su=[]
for z in total:
if len(z)>min:
min=len(z)
su=z
if min > len(subs):
return su
else:
return subs
is1=[]
ns1=int(input())
s1=input()
s1=s1.split()
ns2=int(input())
s2=input()
s2=s2.split()
if ns1>ns2:
indexi=0
for i in s2:
if i in s1:
if indexi+1<len(s2):
is1.append(sub_seq(i,s2[indexi+1:],s1[s1.index(i):],[]))
else:
is1.append([i])
indexi+=1
else:
indexi=0
for i in s1:
if i in s2:
if indexi+1<len(s1):
is1.append(sub_seq(i,s1[indexi+1:],s2[s2.index(i):],[]))
else:
is1.append([i])
indexi+=1
min=0
su=[]
for i in is1:
if len(i)>min:
min=len(i)
su=i
print(min)
print(*su)
``` | output | 1 | 87,583 | 12 | 175,167 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This problem differs from one which was on the online contest.
The sequence a1, a2, ..., an is called increasing, if ai < ai + 1 for i < n.
The sequence s1, s2, ..., sk is called the subsequence of the sequence a1, a2, ..., an, if there exist such a set of indexes 1 β€ i1 < i2 < ... < ik β€ n that aij = sj. In other words, the sequence s can be derived from the sequence a by crossing out some elements.
You are given two sequences of integer numbers. You are to find their longest common increasing subsequence, i.e. an increasing sequence of maximum length that is the subsequence of both sequences.
Input
The first line contains an integer n (1 β€ n β€ 500) β the length of the first sequence. The second line contains n space-separated integers from the range [0, 109] β elements of the first sequence. The third line contains an integer m (1 β€ m β€ 500) β the length of the second sequence. The fourth line contains m space-separated integers from the range [0, 109] β elements of the second sequence.
Output
In the first line output k β the length of the longest common increasing subsequence. In the second line output the subsequence itself. Separate the elements with a space. If there are several solutions, output any.
Examples
Input
7
2 3 1 6 5 4 6
4
1 3 5 6
Output
3
3 5 6
Input
5
1 2 0 2 1
3
1 0 1
Output
2
0 1 | instruction | 0 | 87,584 | 12 | 175,168 |
Tags: dp
Correct Solution:
```
n = int(input())
a = [0] + list(map(int, input().split()))
m = int(input())
b = [0] + list(map(int, input().split()))
n += 1
m += 1
dp = [[0] * max(m,n) for i in range(max(m,n))]
pr = [0] * (max(n, m) + 1)
ans = []
for i in range(1, n):
for j in range(1, m):
if a[i] == b[j]:
dp[i][j] = 1 #max(1, dp[i][j])
for k in range(1, i): #for k in range(1, i + 1):
if (a[k] < a[i]):
if dp[i][j] < dp[k][j] + 1:
pr[i] = k
dp[i][j] = dp[k][j] + 1
else:
dp[i][j] = dp[i][j - 1]
mi = 1
mj = 1
for i in range(1, n):
if dp[i][m-1] > dp[mi][m-1]:
mi = i
print(dp[mi][m-1])
k = dp[mi][m-1]
s = 0
while (mi != 0 and k > s):
ans.append(a[mi])
s += 1
mi = pr[mi]
if ans:
ans.reverse()
print(*ans)
'''
5
-12 0 2 5 10
8
0 2 5 10 -12 -15 0 2
6
-12 0 2 5 10 -12
8
0 2 5 10 -12 -15 0 2
'''
``` | output | 1 | 87,584 | 12 | 175,169 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This problem differs from one which was on the online contest.
The sequence a1, a2, ..., an is called increasing, if ai < ai + 1 for i < n.
The sequence s1, s2, ..., sk is called the subsequence of the sequence a1, a2, ..., an, if there exist such a set of indexes 1 β€ i1 < i2 < ... < ik β€ n that aij = sj. In other words, the sequence s can be derived from the sequence a by crossing out some elements.
You are given two sequences of integer numbers. You are to find their longest common increasing subsequence, i.e. an increasing sequence of maximum length that is the subsequence of both sequences.
Input
The first line contains an integer n (1 β€ n β€ 500) β the length of the first sequence. The second line contains n space-separated integers from the range [0, 109] β elements of the first sequence. The third line contains an integer m (1 β€ m β€ 500) β the length of the second sequence. The fourth line contains m space-separated integers from the range [0, 109] β elements of the second sequence.
Output
In the first line output k β the length of the longest common increasing subsequence. In the second line output the subsequence itself. Separate the elements with a space. If there are several solutions, output any.
Examples
Input
7
2 3 1 6 5 4 6
4
1 3 5 6
Output
3
3 5 6
Input
5
1 2 0 2 1
3
1 0 1
Output
2
0 1
Submitted Solution:
```
def get_lcis_length(X, Y):
len_Y = len(Y)
# maintain current lcis length
# curr_lcis_len: len lcis(X, Y[0..i])
curr_lcis_len = [0] * len_Y
for u in X:
lcis_prefix_max = []
curr_max = 0
# copy current lcis length table and create prefix max from it
# to make access in next step be O(1)
for i, v in enumerate(Y):
if v < u:
curr_max = max(curr_max, curr_lcis_len[i])
lcis_prefix_max.append(curr_max)
# loop through all elements in Y
# update current lcis length if the element matches with current element in X
for i, v in enumerate(Y):
if u == v:
curr_lcis_len[i] = max(curr_lcis_len[i], (0 if i == 0 else lcis_prefix_max[i - 1]) + 1)
# find maximum lcis length
max_lcis_length = curr_lcis_len[0]
for i, u in enumerate(curr_lcis_len):
max_lcis_length = max(max_lcis_length, u)
# Backtrack to get the LCIS
# Redo all previous step, but break when we found lcis that has the same length
# with max_lcis_length
# Need to break the loop to stop backtracking table (bt) being updated
# No need to backtrack if there is no LCIS
if max_lcis_length == 0:
print(0)
return max_lcis_length
curr_lcis_len = [0] * len_Y
bt = [-1] * len_Y # backtracking table
for u in X:
lcis_prefix_max = []
bt_id = []
curr_max = 0
curr_id_max = -1;
# copy current lcis length table and create prefix max from it
# to make access in next step be O(1)
# but, we also keep where should we backtrack if we use the lcis_pref_max
for i, v in enumerate(Y):
if v < u:
if curr_max < curr_lcis_len[i]:
curr_max = curr_lcis_len[i]
curr_id_max = i
lcis_prefix_max.append(curr_max)
bt_id.append(curr_id_max)
# loop through all elements in Y
# update current lcis length if the element matches with current element in X
for i, v in enumerate(Y):
if u == v:
if curr_lcis_len[i] < (0 if i == 0 else lcis_prefix_max[i - 1]) + 1:
curr_lcis_len[i] = (0 if i == 0 else lcis_prefix_max[i - 1]) + 1
bt[i] = (-1 if i == 0 else bt_id[i - 1]) # also update where should we backtrack if using this index
# if this is one of the lcis, use the backtrack table to track the values
if curr_lcis_len[i] == max_lcis_length:
lcis_list = []
now = i
while now != -1:
lcis_list.append(Y[now]);
now = bt[now]
lcis_list.reverse()
print(max_lcis_length)
for i, u in enumerate(lcis_list):
print(u, end=('\n' if i == max_lcis_length - 1 else ' '))
return max_lcis_length, lcis_list
return max_lcis_length
x = []
try:
while True:
y = input()
x.append(y)
except EOFError:
pass
x = ' '.join(x)
x = x.split()
x = [int(i) for i in x]
X = []
Y = []
num_x = x[0]
for i in range(1, num_x + 1):
X.append(x[i])
for i in range(x[num_x + 1]):
Y.append(x[num_x + 2 + i])
get_lcis_length(X, Y)
``` | instruction | 0 | 87,585 | 12 | 175,170 |
Yes | output | 1 | 87,585 | 12 | 175,171 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This problem differs from one which was on the online contest.
The sequence a1, a2, ..., an is called increasing, if ai < ai + 1 for i < n.
The sequence s1, s2, ..., sk is called the subsequence of the sequence a1, a2, ..., an, if there exist such a set of indexes 1 β€ i1 < i2 < ... < ik β€ n that aij = sj. In other words, the sequence s can be derived from the sequence a by crossing out some elements.
You are given two sequences of integer numbers. You are to find their longest common increasing subsequence, i.e. an increasing sequence of maximum length that is the subsequence of both sequences.
Input
The first line contains an integer n (1 β€ n β€ 500) β the length of the first sequence. The second line contains n space-separated integers from the range [0, 109] β elements of the first sequence. The third line contains an integer m (1 β€ m β€ 500) β the length of the second sequence. The fourth line contains m space-separated integers from the range [0, 109] β elements of the second sequence.
Output
In the first line output k β the length of the longest common increasing subsequence. In the second line output the subsequence itself. Separate the elements with a space. If there are several solutions, output any.
Examples
Input
7
2 3 1 6 5 4 6
4
1 3 5 6
Output
3
3 5 6
Input
5
1 2 0 2 1
3
1 0 1
Output
2
0 1
Submitted Solution:
```
"""
s1=input()
s2=input()
n=len(s1)
m=len(s2)
dp=[]
for i in range(n+1):
h=[]
for j in range(m+1):
h.append(0)
dp.append(h)
for i in range(1,n+1):
for j in range(1,m+1):
if(s1[i-1]==s2[j-1]):
dp[i][j]=dp[i-1][j-1]+1
else:
dp[i][j]=max(dp[i-1][j],dp[i][j-1])
print(dp[n][m])"""
"""
n=int(input())
s1=[int(x) for x in input().split()]
m=int(input())
s2=[int(x) for x in input().split()]
table=[0]*m
for i in range(n):
curr=0
for j in range(m):
if(s1[i]==s2[j]):
table[j]=max(table[j],curr+1)
if(s1[i]>s2[j]):
curr=max(curr,table[j])
print(max(table))
"""
n=int(input())
s1=[0]+[int(x) for x in input().split()]
m=int(input())
s2=[0]+[int(x) for x in input().split()]
table=[0]*(m+1)
pos=[0]*(m+1)
for i in range(1,n+1):
curr=0
for j in range(1,m+1):
if(s1[i]==s2[j]):
table[j]=table[curr]+1
pos[j]=curr
if(s1[i]>s2[j] and table[curr]<table[j]):
curr=j
p=0
for i in range(1,m+1):
if(table[i]>table[p]):
p=i
print(table[p])
def out(p):
if(not p):
return
out(pos[p])
print(s2[p],end=" ")
out(p)
print(" ")
``` | instruction | 0 | 87,586 | 12 | 175,172 |
Yes | output | 1 | 87,586 | 12 | 175,173 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This problem differs from one which was on the online contest.
The sequence a1, a2, ..., an is called increasing, if ai < ai + 1 for i < n.
The sequence s1, s2, ..., sk is called the subsequence of the sequence a1, a2, ..., an, if there exist such a set of indexes 1 β€ i1 < i2 < ... < ik β€ n that aij = sj. In other words, the sequence s can be derived from the sequence a by crossing out some elements.
You are given two sequences of integer numbers. You are to find their longest common increasing subsequence, i.e. an increasing sequence of maximum length that is the subsequence of both sequences.
Input
The first line contains an integer n (1 β€ n β€ 500) β the length of the first sequence. The second line contains n space-separated integers from the range [0, 109] β elements of the first sequence. The third line contains an integer m (1 β€ m β€ 500) β the length of the second sequence. The fourth line contains m space-separated integers from the range [0, 109] β elements of the second sequence.
Output
In the first line output k β the length of the longest common increasing subsequence. In the second line output the subsequence itself. Separate the elements with a space. If there are several solutions, output any.
Examples
Input
7
2 3 1 6 5 4 6
4
1 3 5 6
Output
3
3 5 6
Input
5
1 2 0 2 1
3
1 0 1
Output
2
0 1
Submitted Solution:
```
__author__ = 'Darren'
def solve():
n = int(input())
first = list(map(int, input().split()))
m = int(input())
second = list(map(int, input().split()))
if m > n:
n, m = m, n
first, second = second, first
dp, prev = [0] * m, [-1] * m
for i in range(n):
max_len, last_index = 0, -1
for j in range(m):
if first[i] == second[j] and dp[j] < max_len + 1:
dp[j] = max_len + 1
prev[j] = last_index
elif first[i] > second[j] and max_len < dp[j]:
max_len = dp[j]
last_index = j
max_value, index_max = 0, -1
for index, value in enumerate(dp):
if value > max_value:
max_value = value
index_max = index
print(max_value)
if max_value > 0:
seq = []
index = index_max
while index >= 0:
seq.append(str(second[index]))
index = prev[index]
seq.reverse()
print(' '.join(seq))
if __name__ == '__main__':
solve()
``` | instruction | 0 | 87,587 | 12 | 175,174 |
Yes | output | 1 | 87,587 | 12 | 175,175 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This problem differs from one which was on the online contest.
The sequence a1, a2, ..., an is called increasing, if ai < ai + 1 for i < n.
The sequence s1, s2, ..., sk is called the subsequence of the sequence a1, a2, ..., an, if there exist such a set of indexes 1 β€ i1 < i2 < ... < ik β€ n that aij = sj. In other words, the sequence s can be derived from the sequence a by crossing out some elements.
You are given two sequences of integer numbers. You are to find their longest common increasing subsequence, i.e. an increasing sequence of maximum length that is the subsequence of both sequences.
Input
The first line contains an integer n (1 β€ n β€ 500) β the length of the first sequence. The second line contains n space-separated integers from the range [0, 109] β elements of the first sequence. The third line contains an integer m (1 β€ m β€ 500) β the length of the second sequence. The fourth line contains m space-separated integers from the range [0, 109] β elements of the second sequence.
Output
In the first line output k β the length of the longest common increasing subsequence. In the second line output the subsequence itself. Separate the elements with a space. If there are several solutions, output any.
Examples
Input
7
2 3 1 6 5 4 6
4
1 3 5 6
Output
3
3 5 6
Input
5
1 2 0 2 1
3
1 0 1
Output
2
0 1
Submitted Solution:
```
def lcis(a, b):
n, m = len(a), len(b)
dp = [0]*m
best_seq, max_len = [], 0
for i in range(n):
current = 0
seq = []
for j in range(m):
if a[i] == b[j]:
f = seq + [a[i]]
if len(f) > max_len:
max_len = len(f)
best_seq = f
dp[j] = max(current+1, dp[j])
if a[i] > b[j]:
if dp[j] > current:
current = dp[j]
seq.append(b[j])
return max_len, best_seq
n = int(input())
a = input().split(' ')
a = [int(x) for x in a]
m = int(input())
b = input().split(' ')
b = [int(x) for x in b]
x, y = lcis(a, b)
print(x)
y = [str(x) for x in y]
print(' '.join(y))
``` | instruction | 0 | 87,588 | 12 | 175,176 |
Yes | output | 1 | 87,588 | 12 | 175,177 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This problem differs from one which was on the online contest.
The sequence a1, a2, ..., an is called increasing, if ai < ai + 1 for i < n.
The sequence s1, s2, ..., sk is called the subsequence of the sequence a1, a2, ..., an, if there exist such a set of indexes 1 β€ i1 < i2 < ... < ik β€ n that aij = sj. In other words, the sequence s can be derived from the sequence a by crossing out some elements.
You are given two sequences of integer numbers. You are to find their longest common increasing subsequence, i.e. an increasing sequence of maximum length that is the subsequence of both sequences.
Input
The first line contains an integer n (1 β€ n β€ 500) β the length of the first sequence. The second line contains n space-separated integers from the range [0, 109] β elements of the first sequence. The third line contains an integer m (1 β€ m β€ 500) β the length of the second sequence. The fourth line contains m space-separated integers from the range [0, 109] β elements of the second sequence.
Output
In the first line output k β the length of the longest common increasing subsequence. In the second line output the subsequence itself. Separate the elements with a space. If there are several solutions, output any.
Examples
Input
7
2 3 1 6 5 4 6
4
1 3 5 6
Output
3
3 5 6
Input
5
1 2 0 2 1
3
1 0 1
Output
2
0 1
Submitted Solution:
```
is1=[]
is2=[]
ns1=int(input())
s1=input()
s1=s1.split()
ns2=int(input())
s2=input()
s2=s2.split()
if ns1>ns2:
seq=[]
for i in s2:
subs=[]
inde=0
if i in s1:
position=s1.index(i)
subs.append(i)
if position+1<len(s1):
temp_s1=s1[position+1:]
if (s2.index(i)+1)<len(s2):
for j in s2[s2.index(i)+1:]:
if int(j)>int(subs[inde]):
if j in temp_s1:
subs.append(j)
inde+=1
if temp_s1.index(j)+1<len(temp_s1):
temp_s1=temp_s1[temp_s1.index(j)+1:]
else:
break
seq.append(subs)
else:
seq=[]
for i in s1:
subs=[]
inde=0
if i in s2:
position=s2.index(i)
subs.append(i)
if position+1<len(s2):
temp_s2=s2[position+1:]
if (s1.index(i)+1)<len(s1):
for j in s1[s1.index(i)+1:]:
if int(j)>int(subs[inde]):
if j in temp_s2:
subs.append(j)
inde+=1
if temp_s2.index(j)+1<len(temp_s2):
temp_s2=temp_s2[temp_s2.index(j)+1:]
else:
break
seq.append(subs)
min=0
su=[]
for i in seq:
if len(i)>min:
min=len(i)
su=i
print(min)
print(*su)
``` | instruction | 0 | 87,589 | 12 | 175,178 |
No | output | 1 | 87,589 | 12 | 175,179 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This problem differs from one which was on the online contest.
The sequence a1, a2, ..., an is called increasing, if ai < ai + 1 for i < n.
The sequence s1, s2, ..., sk is called the subsequence of the sequence a1, a2, ..., an, if there exist such a set of indexes 1 β€ i1 < i2 < ... < ik β€ n that aij = sj. In other words, the sequence s can be derived from the sequence a by crossing out some elements.
You are given two sequences of integer numbers. You are to find their longest common increasing subsequence, i.e. an increasing sequence of maximum length that is the subsequence of both sequences.
Input
The first line contains an integer n (1 β€ n β€ 500) β the length of the first sequence. The second line contains n space-separated integers from the range [0, 109] β elements of the first sequence. The third line contains an integer m (1 β€ m β€ 500) β the length of the second sequence. The fourth line contains m space-separated integers from the range [0, 109] β elements of the second sequence.
Output
In the first line output k β the length of the longest common increasing subsequence. In the second line output the subsequence itself. Separate the elements with a space. If there are several solutions, output any.
Examples
Input
7
2 3 1 6 5 4 6
4
1 3 5 6
Output
3
3 5 6
Input
5
1 2 0 2 1
3
1 0 1
Output
2
0 1
Submitted Solution:
```
n1=int(input())
arr1=list(map(int,input().split()))
n2=int(input())
arr2=list(map(int,input().split()))
arr3=[]
for i in range(n1+1):
c=[]
for j in range(n2+1):
c.append([])
arr3.append(c.copy())
for i in range(1,n1+1):
for j in range(1,n2+1):
if(arr1[i-1]==arr2[j-1]):
arr3[i][j].extend(arr3[i-1][j-1].copy())
arr3[i][j].append(arr1[i-1])
else:
x=arr3[i-1][j]
y=arr3[i][j-1]
if(len(x)>=len(y)):
arr3[i][j].extend(x.copy())
else:
arr3[i][j].extend(y.copy())
# print(arr3[n1][n2])
arr4=arr3[n1][n2].copy()
l=len(arr4)
arr1=[]
arr2=[1]*l
for i in range(l):
arr1.append([arr4[i]])
for i in range(1,l):
for j in range(i):
if(arr4[i]>arr4[j]):
if(arr2[i]<arr2[j]+1):
arr2[i]=arr2[j]+1
arr1[i]=arr1[j].copy()
arr1[i].append(arr4[i])
mx=-1
res=[]
for i in range(l):
if(mx<arr2[i]):
mx=arr2[i]
res=arr1[i]
print(mx)
for i in res:
print(i,end=" ")
``` | instruction | 0 | 87,590 | 12 | 175,180 |
No | output | 1 | 87,590 | 12 | 175,181 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This problem differs from one which was on the online contest.
The sequence a1, a2, ..., an is called increasing, if ai < ai + 1 for i < n.
The sequence s1, s2, ..., sk is called the subsequence of the sequence a1, a2, ..., an, if there exist such a set of indexes 1 β€ i1 < i2 < ... < ik β€ n that aij = sj. In other words, the sequence s can be derived from the sequence a by crossing out some elements.
You are given two sequences of integer numbers. You are to find their longest common increasing subsequence, i.e. an increasing sequence of maximum length that is the subsequence of both sequences.
Input
The first line contains an integer n (1 β€ n β€ 500) β the length of the first sequence. The second line contains n space-separated integers from the range [0, 109] β elements of the first sequence. The third line contains an integer m (1 β€ m β€ 500) β the length of the second sequence. The fourth line contains m space-separated integers from the range [0, 109] β elements of the second sequence.
Output
In the first line output k β the length of the longest common increasing subsequence. In the second line output the subsequence itself. Separate the elements with a space. If there are several solutions, output any.
Examples
Input
7
2 3 1 6 5 4 6
4
1 3 5 6
Output
3
3 5 6
Input
5
1 2 0 2 1
3
1 0 1
Output
2
0 1
Submitted Solution:
```
n1=int(input())
arr1=list(map(int,input().split()))
n2=int(input())
arr2=list(map(int,input().split()))
arr3=[]
for i in range(n1+1):
c=[]
for j in range(n2+1):
c.append([])
arr3.append(c.copy())
for i in range(1,n1+1):
for j in range(1,n2+1):
if(arr1[i-1]==arr2[j-1]):
arr3[i][j].extend(arr3[i-1][j-1].copy())
arr3[i][j].append(arr1[i-1])
else:
x=arr3[i-1][j]
y=arr3[i][j-1]
if(len(x)>=len(y)):
arr3[i][j].extend(x.copy())
else:
arr3[i][j].extend(y.copy())
# print(arr3[n1][n2])
arr4=arr3[n1][n2].copy()
l=len(arr4)
arr1=[]
arr2=[1]*l
for i in range(l):
arr1.append([arr4[i]])
for i in range(1,l):
for j in range(i):
if(arr4[i]>arr4[j]):
if(arr2[i]<arr2[j]+1):
arr2[i]=arr2[j]+1
arr1[i]=arr1[j].copy()
arr1[i].append(arr4[i])
mx=0
res=[]
for i in range(l):
if(mx<arr2[i]):
mx=arr2[i]
res=arr1[i]
print(mx)
if(mx!=0):
for i in res:
print(i,end=" ")
else:
print()
``` | instruction | 0 | 87,591 | 12 | 175,182 |
No | output | 1 | 87,591 | 12 | 175,183 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This problem differs from one which was on the online contest.
The sequence a1, a2, ..., an is called increasing, if ai < ai + 1 for i < n.
The sequence s1, s2, ..., sk is called the subsequence of the sequence a1, a2, ..., an, if there exist such a set of indexes 1 β€ i1 < i2 < ... < ik β€ n that aij = sj. In other words, the sequence s can be derived from the sequence a by crossing out some elements.
You are given two sequences of integer numbers. You are to find their longest common increasing subsequence, i.e. an increasing sequence of maximum length that is the subsequence of both sequences.
Input
The first line contains an integer n (1 β€ n β€ 500) β the length of the first sequence. The second line contains n space-separated integers from the range [0, 109] β elements of the first sequence. The third line contains an integer m (1 β€ m β€ 500) β the length of the second sequence. The fourth line contains m space-separated integers from the range [0, 109] β elements of the second sequence.
Output
In the first line output k β the length of the longest common increasing subsequence. In the second line output the subsequence itself. Separate the elements with a space. If there are several solutions, output any.
Examples
Input
7
2 3 1 6 5 4 6
4
1 3 5 6
Output
3
3 5 6
Input
5
1 2 0 2 1
3
1 0 1
Output
2
0 1
Submitted Solution:
```
n = int(input())
S = input().split()
m = int(input())
T = input().split()
x = len(S);y = len(T)
d = [[ 0 for i in range(y+1)] for r in range(x+1) ]
prev = [0]*(max(x+1,y+1))
for i in range(1,x+1):
for j in range(1,y+1):
if S[i-1] == T[j-1] :
d[i][j] = 1
for k in range(1,i):
if S[k-1] < S[i-1] and d[i][j] < 1 + d[k][j]:
d[i][j] = 1 + d[k][j]
prev[i]=k
else:
d[i][j] = d[i][j-1]
p = 0
for i in range(1,n+1):
if d[p][m] < d[i][m]:
p = i
ans = []
while p !=0:
ans.append(S[p-1])
p = prev[p]
print(len(ans))
print(*ans[::-1])
``` | instruction | 0 | 87,592 | 12 | 175,184 |
No | output | 1 | 87,592 | 12 | 175,185 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two matrices A and B. Each matrix contains exactly n rows and m columns. Each element of A is either 0 or 1; each element of B is initially 0.
You may perform some operations with matrix B. During each operation, you choose any submatrix of B having size 2 Γ 2, and replace every element in the chosen submatrix with 1. In other words, you choose two integers x and y such that 1 β€ x < n and 1 β€ y < m, and then set B_{x, y}, B_{x, y + 1}, B_{x + 1, y} and B_{x + 1, y + 1} to 1.
Your goal is to make matrix B equal to matrix A. Two matrices A and B are equal if and only if every element of matrix A is equal to the corresponding element of matrix B.
Is it possible to make these matrices equal? If it is, you have to come up with a sequence of operations that makes B equal to A. Note that you don't have to minimize the number of operations.
Input
The first line contains two integers n and m (2 β€ n, m β€ 50).
Then n lines follow, each containing m integers. The j-th integer in the i-th line is A_{i, j}. Each integer is either 0 or 1.
Output
If it is impossible to make B equal to A, print one integer -1.
Otherwise, print any sequence of operations that transforms B into A in the following format: the first line should contain one integer k β the number of operations, and then k lines should follow, each line containing two integers x and y for the corresponding operation (set B_{x, y}, B_{x, y + 1}, B_{x + 1, y} and B_{x + 1, y + 1} to 1). The condition 0 β€ k β€ 2500 should hold.
Examples
Input
3 3
1 1 1
1 1 1
0 1 1
Output
3
1 1
1 2
2 2
Input
3 3
1 0 1
1 0 1
0 0 0
Output
-1
Input
3 2
0 0
0 0
0 0
Output
0
Note
The sequence of operations in the first example:
\begin{matrix} 0 & 0 & 0 & & 1 & 1 & 0 & & 1 & 1 & 1 & & 1 & 1 & 1 \\\ 0 & 0 & 0 & β & 1 & 1 & 0 & β & 1 & 1 & 1 & β & 1 & 1 & 1 \\\ 0 & 0 & 0 & & 0 & 0 & 0 & & 0 & 0 & 0 & & 0 & 1 & 1 \end{matrix} | instruction | 0 | 87,637 | 12 | 175,274 |
Tags: constructive algorithms, greedy, implementation
Correct Solution:
```
n,m=map(int,input().split())
li=[]
for i in range(n):
a=list(map(int,input().split()))
li.append(a)
li1=[[0 for i in range(m)] for _ in range(n)]
ans=[]
ans1=0
for i in range(n-1):
for j in range(m-1):
if li[i][j]+li[i][j+1]+li[i+1][j]+li[i+1][j+1]==4:
li1[i][j]=1
li1[i][j+1]=1
li1[i+1][j]=1
li1[i+1][j+1]=1
ans.append([i+1,j+1])
ans1+=1
if li1==li:
print(ans1)
for i in ans:
print(*i)
else:
print(-1)
``` | output | 1 | 87,637 | 12 | 175,275 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two matrices A and B. Each matrix contains exactly n rows and m columns. Each element of A is either 0 or 1; each element of B is initially 0.
You may perform some operations with matrix B. During each operation, you choose any submatrix of B having size 2 Γ 2, and replace every element in the chosen submatrix with 1. In other words, you choose two integers x and y such that 1 β€ x < n and 1 β€ y < m, and then set B_{x, y}, B_{x, y + 1}, B_{x + 1, y} and B_{x + 1, y + 1} to 1.
Your goal is to make matrix B equal to matrix A. Two matrices A and B are equal if and only if every element of matrix A is equal to the corresponding element of matrix B.
Is it possible to make these matrices equal? If it is, you have to come up with a sequence of operations that makes B equal to A. Note that you don't have to minimize the number of operations.
Input
The first line contains two integers n and m (2 β€ n, m β€ 50).
Then n lines follow, each containing m integers. The j-th integer in the i-th line is A_{i, j}. Each integer is either 0 or 1.
Output
If it is impossible to make B equal to A, print one integer -1.
Otherwise, print any sequence of operations that transforms B into A in the following format: the first line should contain one integer k β the number of operations, and then k lines should follow, each line containing two integers x and y for the corresponding operation (set B_{x, y}, B_{x, y + 1}, B_{x + 1, y} and B_{x + 1, y + 1} to 1). The condition 0 β€ k β€ 2500 should hold.
Examples
Input
3 3
1 1 1
1 1 1
0 1 1
Output
3
1 1
1 2
2 2
Input
3 3
1 0 1
1 0 1
0 0 0
Output
-1
Input
3 2
0 0
0 0
0 0
Output
0
Note
The sequence of operations in the first example:
\begin{matrix} 0 & 0 & 0 & & 1 & 1 & 0 & & 1 & 1 & 1 & & 1 & 1 & 1 \\\ 0 & 0 & 0 & β & 1 & 1 & 0 & β & 1 & 1 & 1 & β & 1 & 1 & 1 \\\ 0 & 0 & 0 & & 0 & 0 & 0 & & 0 & 0 & 0 & & 0 & 1 & 1 \end{matrix} | instruction | 0 | 87,638 | 12 | 175,276 |
Tags: constructive algorithms, greedy, implementation
Correct Solution:
```
a,b=list(map(int,input().split()))
array=[]
array_b=[]
answer=[]
flag=5
for x in range(a):
row=list(map(int,input().split()))
array.append(row)
row=[0]*b
array_b.append(row)
for z in range(len(array_b)):
row=array_b[z]
for x in range(len(row)):
if row[x]==0:
if array[z][x]==0:
pass
else:
if x+1<b and z+1<a:
if array[z][x+1]==1 and array[z+1][x]==1 and array[z+1][x+1]==1:
array_b[z][x]=1
array_b[z][x+1]=1
array_b[z+1][x]=1
array_b[z+1][x+1]=1
answer.append([z+1,x+1])
else:
flag=6
break
else:
flag=6
break
else:
if array[z][x]==0:
pass
else:
if x+1<b and z+1<a:
if array[z][x+1]==1 and array[z+1][x]==1 and array[z+1][x+1]==1:
array_b[z][x]=1
array_b[z][x+1]=1
array_b[z+1][x]=1
array_b[z+1][x+1]=1
answer.append([z+1,x+1])
else:
pass
else:
pass
if flag==6:
break
if flag==6:
print(-1)
else:
print(len(answer))
for it in answer:
print(*it)
``` | output | 1 | 87,638 | 12 | 175,277 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two matrices A and B. Each matrix contains exactly n rows and m columns. Each element of A is either 0 or 1; each element of B is initially 0.
You may perform some operations with matrix B. During each operation, you choose any submatrix of B having size 2 Γ 2, and replace every element in the chosen submatrix with 1. In other words, you choose two integers x and y such that 1 β€ x < n and 1 β€ y < m, and then set B_{x, y}, B_{x, y + 1}, B_{x + 1, y} and B_{x + 1, y + 1} to 1.
Your goal is to make matrix B equal to matrix A. Two matrices A and B are equal if and only if every element of matrix A is equal to the corresponding element of matrix B.
Is it possible to make these matrices equal? If it is, you have to come up with a sequence of operations that makes B equal to A. Note that you don't have to minimize the number of operations.
Input
The first line contains two integers n and m (2 β€ n, m β€ 50).
Then n lines follow, each containing m integers. The j-th integer in the i-th line is A_{i, j}. Each integer is either 0 or 1.
Output
If it is impossible to make B equal to A, print one integer -1.
Otherwise, print any sequence of operations that transforms B into A in the following format: the first line should contain one integer k β the number of operations, and then k lines should follow, each line containing two integers x and y for the corresponding operation (set B_{x, y}, B_{x, y + 1}, B_{x + 1, y} and B_{x + 1, y + 1} to 1). The condition 0 β€ k β€ 2500 should hold.
Examples
Input
3 3
1 1 1
1 1 1
0 1 1
Output
3
1 1
1 2
2 2
Input
3 3
1 0 1
1 0 1
0 0 0
Output
-1
Input
3 2
0 0
0 0
0 0
Output
0
Note
The sequence of operations in the first example:
\begin{matrix} 0 & 0 & 0 & & 1 & 1 & 0 & & 1 & 1 & 1 & & 1 & 1 & 1 \\\ 0 & 0 & 0 & β & 1 & 1 & 0 & β & 1 & 1 & 1 & β & 1 & 1 & 1 \\\ 0 & 0 & 0 & & 0 & 0 & 0 & & 0 & 0 & 0 & & 0 & 1 & 1 \end{matrix} | instruction | 0 | 87,639 | 12 | 175,278 |
Tags: constructive algorithms, greedy, implementation
Correct Solution:
```
def make_equal_matrix(n, m, a, cmd):
b = [['0'] * m for _ in range(n)]
for i in range(n - 1):
for j in range(m - 1):
summ = sum(map(int, (a[i][j], a[i + 1][j], a[i][j + 1], a[i + 1][j + 1])))
if summ == 4:
cmd.append((str(i + 1), str(j + 1)))
b[i][j], b[i + 1][j], b[i][j + 1], b[i + 1][j + 1] = '1', '1', '1', '1'
return a == b
n, m = map(int, input().split())
a = [input().split() for _ in range(n)]
cmd = []
if not make_equal_matrix(n, m, a, cmd):
print(-1)
else:
print(len(cmd))
print(*[p[0] + ' ' + p[1] for p in cmd], sep='\n')
``` | output | 1 | 87,639 | 12 | 175,279 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two matrices A and B. Each matrix contains exactly n rows and m columns. Each element of A is either 0 or 1; each element of B is initially 0.
You may perform some operations with matrix B. During each operation, you choose any submatrix of B having size 2 Γ 2, and replace every element in the chosen submatrix with 1. In other words, you choose two integers x and y such that 1 β€ x < n and 1 β€ y < m, and then set B_{x, y}, B_{x, y + 1}, B_{x + 1, y} and B_{x + 1, y + 1} to 1.
Your goal is to make matrix B equal to matrix A. Two matrices A and B are equal if and only if every element of matrix A is equal to the corresponding element of matrix B.
Is it possible to make these matrices equal? If it is, you have to come up with a sequence of operations that makes B equal to A. Note that you don't have to minimize the number of operations.
Input
The first line contains two integers n and m (2 β€ n, m β€ 50).
Then n lines follow, each containing m integers. The j-th integer in the i-th line is A_{i, j}. Each integer is either 0 or 1.
Output
If it is impossible to make B equal to A, print one integer -1.
Otherwise, print any sequence of operations that transforms B into A in the following format: the first line should contain one integer k β the number of operations, and then k lines should follow, each line containing two integers x and y for the corresponding operation (set B_{x, y}, B_{x, y + 1}, B_{x + 1, y} and B_{x + 1, y + 1} to 1). The condition 0 β€ k β€ 2500 should hold.
Examples
Input
3 3
1 1 1
1 1 1
0 1 1
Output
3
1 1
1 2
2 2
Input
3 3
1 0 1
1 0 1
0 0 0
Output
-1
Input
3 2
0 0
0 0
0 0
Output
0
Note
The sequence of operations in the first example:
\begin{matrix} 0 & 0 & 0 & & 1 & 1 & 0 & & 1 & 1 & 1 & & 1 & 1 & 1 \\\ 0 & 0 & 0 & β & 1 & 1 & 0 & β & 1 & 1 & 1 & β & 1 & 1 & 1 \\\ 0 & 0 & 0 & & 0 & 0 & 0 & & 0 & 0 & 0 & & 0 & 1 & 1 \end{matrix} | instruction | 0 | 87,640 | 12 | 175,280 |
Tags: constructive algorithms, greedy, implementation
Correct Solution:
```
n,m=map(int,input().split())
a=[0]*n
b= [[0 for i in range(m)] for j in range(n)]
c=[]
p=0
for i in range(n):
a[i]=list(map(int,input().split()))
for i in range(n-1):
for j in range(m-1):
if a[i][j]==1 and a[i][j+1]==1 and a[i+1][j]==1 and a[i+1][j+1]==1:
b[i][j]=1
b[i][j+1]=1
b[i+1][j]=1
b[i+1][j+1]=1
c.append(i+1)
c.append(j+1)
#print(c,a,b)
if a==b:
if len(c)==0:
print(0)
else:
x=len(c)//2
print(x)
for i in range(0,x*2,2):
print(c[i],c[i+1])
else:
print(-1)
``` | output | 1 | 87,640 | 12 | 175,281 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two matrices A and B. Each matrix contains exactly n rows and m columns. Each element of A is either 0 or 1; each element of B is initially 0.
You may perform some operations with matrix B. During each operation, you choose any submatrix of B having size 2 Γ 2, and replace every element in the chosen submatrix with 1. In other words, you choose two integers x and y such that 1 β€ x < n and 1 β€ y < m, and then set B_{x, y}, B_{x, y + 1}, B_{x + 1, y} and B_{x + 1, y + 1} to 1.
Your goal is to make matrix B equal to matrix A. Two matrices A and B are equal if and only if every element of matrix A is equal to the corresponding element of matrix B.
Is it possible to make these matrices equal? If it is, you have to come up with a sequence of operations that makes B equal to A. Note that you don't have to minimize the number of operations.
Input
The first line contains two integers n and m (2 β€ n, m β€ 50).
Then n lines follow, each containing m integers. The j-th integer in the i-th line is A_{i, j}. Each integer is either 0 or 1.
Output
If it is impossible to make B equal to A, print one integer -1.
Otherwise, print any sequence of operations that transforms B into A in the following format: the first line should contain one integer k β the number of operations, and then k lines should follow, each line containing two integers x and y for the corresponding operation (set B_{x, y}, B_{x, y + 1}, B_{x + 1, y} and B_{x + 1, y + 1} to 1). The condition 0 β€ k β€ 2500 should hold.
Examples
Input
3 3
1 1 1
1 1 1
0 1 1
Output
3
1 1
1 2
2 2
Input
3 3
1 0 1
1 0 1
0 0 0
Output
-1
Input
3 2
0 0
0 0
0 0
Output
0
Note
The sequence of operations in the first example:
\begin{matrix} 0 & 0 & 0 & & 1 & 1 & 0 & & 1 & 1 & 1 & & 1 & 1 & 1 \\\ 0 & 0 & 0 & β & 1 & 1 & 0 & β & 1 & 1 & 1 & β & 1 & 1 & 1 \\\ 0 & 0 & 0 & & 0 & 0 & 0 & & 0 & 0 & 0 & & 0 & 1 & 1 \end{matrix} | instruction | 0 | 87,641 | 12 | 175,282 |
Tags: constructive algorithms, greedy, implementation
Correct Solution:
```
n, m = map(int, input().split())
A = list()
B = list()
for i in range(n):
row = list(map(int, input().split()))
A.append(row)
B.append([0 for _ in range(m)])
ops = list()
for i in range(n - 1):
for j in range(m - 1):
if A[i][j] == 1 and A[i + 1][j] == 1 and A[i][j + 1] == 1 and A[i + 1][j + 1] == 1:
B[i][j] = 1
B[i + 1][j] = 1
B[i][j + 1] = 1
B[i + 1][j + 1] = 1
ops.append((i, j))
if A == B:
print(len(ops))
for p in ops:
print(p[0] + 1, p[1] + 1)
else:
print(-1)
``` | output | 1 | 87,641 | 12 | 175,283 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two matrices A and B. Each matrix contains exactly n rows and m columns. Each element of A is either 0 or 1; each element of B is initially 0.
You may perform some operations with matrix B. During each operation, you choose any submatrix of B having size 2 Γ 2, and replace every element in the chosen submatrix with 1. In other words, you choose two integers x and y such that 1 β€ x < n and 1 β€ y < m, and then set B_{x, y}, B_{x, y + 1}, B_{x + 1, y} and B_{x + 1, y + 1} to 1.
Your goal is to make matrix B equal to matrix A. Two matrices A and B are equal if and only if every element of matrix A is equal to the corresponding element of matrix B.
Is it possible to make these matrices equal? If it is, you have to come up with a sequence of operations that makes B equal to A. Note that you don't have to minimize the number of operations.
Input
The first line contains two integers n and m (2 β€ n, m β€ 50).
Then n lines follow, each containing m integers. The j-th integer in the i-th line is A_{i, j}. Each integer is either 0 or 1.
Output
If it is impossible to make B equal to A, print one integer -1.
Otherwise, print any sequence of operations that transforms B into A in the following format: the first line should contain one integer k β the number of operations, and then k lines should follow, each line containing two integers x and y for the corresponding operation (set B_{x, y}, B_{x, y + 1}, B_{x + 1, y} and B_{x + 1, y + 1} to 1). The condition 0 β€ k β€ 2500 should hold.
Examples
Input
3 3
1 1 1
1 1 1
0 1 1
Output
3
1 1
1 2
2 2
Input
3 3
1 0 1
1 0 1
0 0 0
Output
-1
Input
3 2
0 0
0 0
0 0
Output
0
Note
The sequence of operations in the first example:
\begin{matrix} 0 & 0 & 0 & & 1 & 1 & 0 & & 1 & 1 & 1 & & 1 & 1 & 1 \\\ 0 & 0 & 0 & β & 1 & 1 & 0 & β & 1 & 1 & 1 & β & 1 & 1 & 1 \\\ 0 & 0 & 0 & & 0 & 0 & 0 & & 0 & 0 & 0 & & 0 & 1 & 1 \end{matrix} | instruction | 0 | 87,642 | 12 | 175,284 |
Tags: constructive algorithms, greedy, implementation
Correct Solution:
```
from itertools import product
def main():
n,m = map(int,input().split())
aa = []
for _ in range(n):
aa.append(list(map(int, input().split())))
bb = [[0]*m for _ in range(n)]
ops = []
for i, j in product(range(n-1), range(m-1)):
for k,l in product(range(2), range(2)):
if aa[i+k][j+l] == 0:
break
else:
for k, l in product(range(2), range(2)):
bb[i + k][j + l] =1
ops.append((i+1,j+1))
if aa==bb:
print(len(ops))
for o in ops:
print(*o)
else:
print(-1)
if __name__ == "__main__":
main()
``` | output | 1 | 87,642 | 12 | 175,285 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two matrices A and B. Each matrix contains exactly n rows and m columns. Each element of A is either 0 or 1; each element of B is initially 0.
You may perform some operations with matrix B. During each operation, you choose any submatrix of B having size 2 Γ 2, and replace every element in the chosen submatrix with 1. In other words, you choose two integers x and y such that 1 β€ x < n and 1 β€ y < m, and then set B_{x, y}, B_{x, y + 1}, B_{x + 1, y} and B_{x + 1, y + 1} to 1.
Your goal is to make matrix B equal to matrix A. Two matrices A and B are equal if and only if every element of matrix A is equal to the corresponding element of matrix B.
Is it possible to make these matrices equal? If it is, you have to come up with a sequence of operations that makes B equal to A. Note that you don't have to minimize the number of operations.
Input
The first line contains two integers n and m (2 β€ n, m β€ 50).
Then n lines follow, each containing m integers. The j-th integer in the i-th line is A_{i, j}. Each integer is either 0 or 1.
Output
If it is impossible to make B equal to A, print one integer -1.
Otherwise, print any sequence of operations that transforms B into A in the following format: the first line should contain one integer k β the number of operations, and then k lines should follow, each line containing two integers x and y for the corresponding operation (set B_{x, y}, B_{x, y + 1}, B_{x + 1, y} and B_{x + 1, y + 1} to 1). The condition 0 β€ k β€ 2500 should hold.
Examples
Input
3 3
1 1 1
1 1 1
0 1 1
Output
3
1 1
1 2
2 2
Input
3 3
1 0 1
1 0 1
0 0 0
Output
-1
Input
3 2
0 0
0 0
0 0
Output
0
Note
The sequence of operations in the first example:
\begin{matrix} 0 & 0 & 0 & & 1 & 1 & 0 & & 1 & 1 & 1 & & 1 & 1 & 1 \\\ 0 & 0 & 0 & β & 1 & 1 & 0 & β & 1 & 1 & 1 & β & 1 & 1 & 1 \\\ 0 & 0 & 0 & & 0 & 0 & 0 & & 0 & 0 & 0 & & 0 & 1 & 1 \end{matrix} | instruction | 0 | 87,643 | 12 | 175,286 |
Tags: constructive algorithms, greedy, implementation
Correct Solution:
```
n, m = list(map(int, input().split()))
matrix_a = list()
matrix_b = list()
for i in range(n):
matrix_a.append(list(map(int, input().split())))
matrix_b.append([0] * len(matrix_a[i]))
k = 0
transforms = list()
def transform_matrix_b(x, y):
matrix_b[x][y] = 1
matrix_b[x + 1][y] = 1
matrix_b[x][y + 1] = 1
matrix_b[x + 1][y + 1] = 1
global k
k += 1
transforms.append((x + 1, y + 1))
for i in range(n - 1):
for j in range(m - 1):
if matrix_a[i][j] == 1 and matrix_a[i + 1][j] == 1 and matrix_a[i][
j + 1] == 1 and matrix_a[i + 1][j + 1] == 1:
transform_matrix_b(i, j)
if matrix_a == matrix_b:
if matrix_b.count(0) == n * m:
print(0)
else:
print(k)
for i in range(k):
print(*transforms[i])
else:
print(-1)
``` | output | 1 | 87,643 | 12 | 175,287 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two matrices A and B. Each matrix contains exactly n rows and m columns. Each element of A is either 0 or 1; each element of B is initially 0.
You may perform some operations with matrix B. During each operation, you choose any submatrix of B having size 2 Γ 2, and replace every element in the chosen submatrix with 1. In other words, you choose two integers x and y such that 1 β€ x < n and 1 β€ y < m, and then set B_{x, y}, B_{x, y + 1}, B_{x + 1, y} and B_{x + 1, y + 1} to 1.
Your goal is to make matrix B equal to matrix A. Two matrices A and B are equal if and only if every element of matrix A is equal to the corresponding element of matrix B.
Is it possible to make these matrices equal? If it is, you have to come up with a sequence of operations that makes B equal to A. Note that you don't have to minimize the number of operations.
Input
The first line contains two integers n and m (2 β€ n, m β€ 50).
Then n lines follow, each containing m integers. The j-th integer in the i-th line is A_{i, j}. Each integer is either 0 or 1.
Output
If it is impossible to make B equal to A, print one integer -1.
Otherwise, print any sequence of operations that transforms B into A in the following format: the first line should contain one integer k β the number of operations, and then k lines should follow, each line containing two integers x and y for the corresponding operation (set B_{x, y}, B_{x, y + 1}, B_{x + 1, y} and B_{x + 1, y + 1} to 1). The condition 0 β€ k β€ 2500 should hold.
Examples
Input
3 3
1 1 1
1 1 1
0 1 1
Output
3
1 1
1 2
2 2
Input
3 3
1 0 1
1 0 1
0 0 0
Output
-1
Input
3 2
0 0
0 0
0 0
Output
0
Note
The sequence of operations in the first example:
\begin{matrix} 0 & 0 & 0 & & 1 & 1 & 0 & & 1 & 1 & 1 & & 1 & 1 & 1 \\\ 0 & 0 & 0 & β & 1 & 1 & 0 & β & 1 & 1 & 1 & β & 1 & 1 & 1 \\\ 0 & 0 & 0 & & 0 & 0 & 0 & & 0 & 0 & 0 & & 0 & 1 & 1 \end{matrix} | instruction | 0 | 87,644 | 12 | 175,288 |
Tags: constructive algorithms, greedy, implementation
Correct Solution:
```
n,m=map(int,input().split())
a=[]
for j in range(n):
a.append(list(map(int,input().split())))
an=[]
f=0
for j in range(n):
for k in range(m):
if a[j][k]==1:
if j+1<n and k+1<m and a[j+1][k+1]==1 and a[j][k+1]==1 and a[j+1][k]==1:
an.append((j+1,k+1))
elif j-1>=0 and k-1>=0 and a[j-1][k-1]==1 and a[j][k-1]==1 and a[j-1][k]==1 :
continue
elif j-1>=0 and k+1<m and a[j-1][k+1]==1 and a[j-1][k]==1 and a[j][k+1]==1:
continue
elif j+1<n and k-1>=0 and a[j+1][k-1]==1 and a[j+1][k]==1 and a[j][k-1]==1:
continue
else:
f=1
break
if f==1:
print(-1)
else:
print(len(an))
for u,v in an:
print(u,v)
``` | output | 1 | 87,644 | 12 | 175,289 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Once again, Boris needs the help of Anton in creating a task. This time Anton needs to solve the following problem:
There are two arrays of integers a and b of length n. It turned out that array a contains only elements from the set \{-1, 0, 1\}.
Anton can perform the following sequence of operations any number of times:
1. Choose any pair of indexes (i, j) such that 1 β€ i < j β€ n. It is possible to choose the same pair (i, j) more than once.
2. Add a_i to a_j. In other words, j-th element of the array becomes equal to a_i + a_j.
For example, if you are given array [1, -1, 0], you can transform it only to [1, -1, -1], [1, 0, 0] and [1, -1, 1] by one operation.
Anton wants to predict if it is possible to apply some number (zero or more) of these operations to the array a so that it becomes equal to array b. Can you help him?
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 β€ t β€ 10000). The description of the test cases follows.
The first line of each test case contains a single integer n (1 β€ n β€ 10^5) β the length of arrays.
The second line of each test case contains n integers a_1, a_2, ..., a_n (-1 β€ a_i β€ 1) β elements of array a. There can be duplicates among elements.
The third line of each test case contains n integers b_1, b_2, ..., b_n (-10^9 β€ b_i β€ 10^9) β elements of array b. There can be duplicates among elements.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, output one line containing "YES" if it's possible to make arrays a and b equal by performing the described operations, or "NO" if it's impossible.
You can print each letter in any case (upper or lower).
Example
Input
5
3
1 -1 0
1 1 -2
3
0 1 1
0 2 2
2
1 0
1 41
2
-1 0
-1 -41
5
0 1 -1 1 -1
1 1 -1 1 -1
Output
YES
NO
YES
YES
NO
Note
In the first test-case we can choose (i, j)=(2, 3) twice and after that choose (i, j)=(1, 2) twice too. These operations will transform [1, -1, 0] β [1, -1, -2] β [1, 1, -2]
In the second test case we can't make equal numbers on the second position.
In the third test case we can choose (i, j)=(1, 2) 41 times. The same about the fourth test case.
In the last lest case, it is impossible to make array a equal to the array b. | instruction | 0 | 87,716 | 12 | 175,432 |
Tags: greedy, implementation
Correct Solution:
```
printn = lambda x: print(x,end='')
inn = lambda : int(input())
inl = lambda: list(map(int, input().split()))
inm = lambda: map(int, input().split())
ins = lambda : input().strip()
DBG = True # and False
BIG = 10**18
R = 10**9 + 7
def ddprint(x):
if DBG:
print(x)
t = inn()
for tt in range(t):
n = inn()
a = inl()
b = inl()
m = p = n
for i in range(n):
if a[i] == -1:
m = i
break
for i in range(n):
if a[i] == 1:
p = i
break
if m<p:
ok = True
for i in range(n):
if i<m and b[i]!=0 or i==m and b[i] != -1 or \
m<i<=p and b[i]>a[i]:
ok = False
break
print('YES' if ok else 'NO')
else: # p<m
ok = True
for i in range(n):
if i<p and b[i]!=0 or i==p and b[i] != 1 or \
p<i<=m and b[i]<a[i]:
ok = False
break
print('YES' if ok else 'NO')
``` | output | 1 | 87,716 | 12 | 175,433 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Once again, Boris needs the help of Anton in creating a task. This time Anton needs to solve the following problem:
There are two arrays of integers a and b of length n. It turned out that array a contains only elements from the set \{-1, 0, 1\}.
Anton can perform the following sequence of operations any number of times:
1. Choose any pair of indexes (i, j) such that 1 β€ i < j β€ n. It is possible to choose the same pair (i, j) more than once.
2. Add a_i to a_j. In other words, j-th element of the array becomes equal to a_i + a_j.
For example, if you are given array [1, -1, 0], you can transform it only to [1, -1, -1], [1, 0, 0] and [1, -1, 1] by one operation.
Anton wants to predict if it is possible to apply some number (zero or more) of these operations to the array a so that it becomes equal to array b. Can you help him?
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 β€ t β€ 10000). The description of the test cases follows.
The first line of each test case contains a single integer n (1 β€ n β€ 10^5) β the length of arrays.
The second line of each test case contains n integers a_1, a_2, ..., a_n (-1 β€ a_i β€ 1) β elements of array a. There can be duplicates among elements.
The third line of each test case contains n integers b_1, b_2, ..., b_n (-10^9 β€ b_i β€ 10^9) β elements of array b. There can be duplicates among elements.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, output one line containing "YES" if it's possible to make arrays a and b equal by performing the described operations, or "NO" if it's impossible.
You can print each letter in any case (upper or lower).
Example
Input
5
3
1 -1 0
1 1 -2
3
0 1 1
0 2 2
2
1 0
1 41
2
-1 0
-1 -41
5
0 1 -1 1 -1
1 1 -1 1 -1
Output
YES
NO
YES
YES
NO
Note
In the first test-case we can choose (i, j)=(2, 3) twice and after that choose (i, j)=(1, 2) twice too. These operations will transform [1, -1, 0] β [1, -1, -2] β [1, 1, -2]
In the second test case we can't make equal numbers on the second position.
In the third test case we can choose (i, j)=(1, 2) 41 times. The same about the fourth test case.
In the last lest case, it is impossible to make array a equal to the array b. | instruction | 0 | 87,717 | 12 | 175,434 |
Tags: greedy, implementation
Correct Solution:
```
def check(a,b):
neg,pos=0,0
for j in range(len(a)):
if(b[j]<a[j] and neg!=1):
print("NO")
return
elif(b[j]>a[j]):
if(pos!=1):
print("NO")
return
if(a[j]==-1):
neg=1
if(a[j]==1):
pos=1
print("YES")
t=int(input())
for i in range(t):
n=int(input())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
check(a,b)
``` | output | 1 | 87,717 | 12 | 175,435 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Once again, Boris needs the help of Anton in creating a task. This time Anton needs to solve the following problem:
There are two arrays of integers a and b of length n. It turned out that array a contains only elements from the set \{-1, 0, 1\}.
Anton can perform the following sequence of operations any number of times:
1. Choose any pair of indexes (i, j) such that 1 β€ i < j β€ n. It is possible to choose the same pair (i, j) more than once.
2. Add a_i to a_j. In other words, j-th element of the array becomes equal to a_i + a_j.
For example, if you are given array [1, -1, 0], you can transform it only to [1, -1, -1], [1, 0, 0] and [1, -1, 1] by one operation.
Anton wants to predict if it is possible to apply some number (zero or more) of these operations to the array a so that it becomes equal to array b. Can you help him?
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 β€ t β€ 10000). The description of the test cases follows.
The first line of each test case contains a single integer n (1 β€ n β€ 10^5) β the length of arrays.
The second line of each test case contains n integers a_1, a_2, ..., a_n (-1 β€ a_i β€ 1) β elements of array a. There can be duplicates among elements.
The third line of each test case contains n integers b_1, b_2, ..., b_n (-10^9 β€ b_i β€ 10^9) β elements of array b. There can be duplicates among elements.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, output one line containing "YES" if it's possible to make arrays a and b equal by performing the described operations, or "NO" if it's impossible.
You can print each letter in any case (upper or lower).
Example
Input
5
3
1 -1 0
1 1 -2
3
0 1 1
0 2 2
2
1 0
1 41
2
-1 0
-1 -41
5
0 1 -1 1 -1
1 1 -1 1 -1
Output
YES
NO
YES
YES
NO
Note
In the first test-case we can choose (i, j)=(2, 3) twice and after that choose (i, j)=(1, 2) twice too. These operations will transform [1, -1, 0] β [1, -1, -2] β [1, 1, -2]
In the second test case we can't make equal numbers on the second position.
In the third test case we can choose (i, j)=(1, 2) 41 times. The same about the fourth test case.
In the last lest case, it is impossible to make array a equal to the array b. | instruction | 0 | 87,718 | 12 | 175,436 |
Tags: greedy, implementation
Correct Solution:
```
# cook your dish here
for _ in range(int(input())):
n=int(input())
l1=list(map(int,input().split()))
l2=list(map(int,input().split()))
x=-1
y=-1
z=-1
for i in range(n):
if(l1[i]==0):
x=i
break
for i in range(n):
if(l1[i]==1):
y=i
break
for i in range(n):
if(l1[i]==-1):
z=i
break
if(l1[0]!=l2[0]):
print("NO")
else:
f=0
for i in range(1,n):
if(l1[i]!=l2[i]):
if(l1[i]>l2[i]):
if(z==-1 or z>=i):
f=1
break
elif(l1[i]<l2[i]):
if(y==-1 or y>=i):
f=1
break
if(f==1):
print("NO")
else:
print("YES")
``` | output | 1 | 87,718 | 12 | 175,437 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Once again, Boris needs the help of Anton in creating a task. This time Anton needs to solve the following problem:
There are two arrays of integers a and b of length n. It turned out that array a contains only elements from the set \{-1, 0, 1\}.
Anton can perform the following sequence of operations any number of times:
1. Choose any pair of indexes (i, j) such that 1 β€ i < j β€ n. It is possible to choose the same pair (i, j) more than once.
2. Add a_i to a_j. In other words, j-th element of the array becomes equal to a_i + a_j.
For example, if you are given array [1, -1, 0], you can transform it only to [1, -1, -1], [1, 0, 0] and [1, -1, 1] by one operation.
Anton wants to predict if it is possible to apply some number (zero or more) of these operations to the array a so that it becomes equal to array b. Can you help him?
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 β€ t β€ 10000). The description of the test cases follows.
The first line of each test case contains a single integer n (1 β€ n β€ 10^5) β the length of arrays.
The second line of each test case contains n integers a_1, a_2, ..., a_n (-1 β€ a_i β€ 1) β elements of array a. There can be duplicates among elements.
The third line of each test case contains n integers b_1, b_2, ..., b_n (-10^9 β€ b_i β€ 10^9) β elements of array b. There can be duplicates among elements.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, output one line containing "YES" if it's possible to make arrays a and b equal by performing the described operations, or "NO" if it's impossible.
You can print each letter in any case (upper or lower).
Example
Input
5
3
1 -1 0
1 1 -2
3
0 1 1
0 2 2
2
1 0
1 41
2
-1 0
-1 -41
5
0 1 -1 1 -1
1 1 -1 1 -1
Output
YES
NO
YES
YES
NO
Note
In the first test-case we can choose (i, j)=(2, 3) twice and after that choose (i, j)=(1, 2) twice too. These operations will transform [1, -1, 0] β [1, -1, -2] β [1, 1, -2]
In the second test case we can't make equal numbers on the second position.
In the third test case we can choose (i, j)=(1, 2) 41 times. The same about the fourth test case.
In the last lest case, it is impossible to make array a equal to the array b. | instruction | 0 | 87,719 | 12 | 175,438 |
Tags: greedy, implementation
Correct Solution:
```
for _ in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
left = set()
res = 'YES'
for i in range(n):
if b[i] != a[i]:
if b[i] > a[i] and 1 not in left:
res = 'NO'
break
elif b[i] < a[i] and -1 not in left:
res = 'NO'
break
left.add(a[i])
print(res)
``` | output | 1 | 87,719 | 12 | 175,439 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Once again, Boris needs the help of Anton in creating a task. This time Anton needs to solve the following problem:
There are two arrays of integers a and b of length n. It turned out that array a contains only elements from the set \{-1, 0, 1\}.
Anton can perform the following sequence of operations any number of times:
1. Choose any pair of indexes (i, j) such that 1 β€ i < j β€ n. It is possible to choose the same pair (i, j) more than once.
2. Add a_i to a_j. In other words, j-th element of the array becomes equal to a_i + a_j.
For example, if you are given array [1, -1, 0], you can transform it only to [1, -1, -1], [1, 0, 0] and [1, -1, 1] by one operation.
Anton wants to predict if it is possible to apply some number (zero or more) of these operations to the array a so that it becomes equal to array b. Can you help him?
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 β€ t β€ 10000). The description of the test cases follows.
The first line of each test case contains a single integer n (1 β€ n β€ 10^5) β the length of arrays.
The second line of each test case contains n integers a_1, a_2, ..., a_n (-1 β€ a_i β€ 1) β elements of array a. There can be duplicates among elements.
The third line of each test case contains n integers b_1, b_2, ..., b_n (-10^9 β€ b_i β€ 10^9) β elements of array b. There can be duplicates among elements.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, output one line containing "YES" if it's possible to make arrays a and b equal by performing the described operations, or "NO" if it's impossible.
You can print each letter in any case (upper or lower).
Example
Input
5
3
1 -1 0
1 1 -2
3
0 1 1
0 2 2
2
1 0
1 41
2
-1 0
-1 -41
5
0 1 -1 1 -1
1 1 -1 1 -1
Output
YES
NO
YES
YES
NO
Note
In the first test-case we can choose (i, j)=(2, 3) twice and after that choose (i, j)=(1, 2) twice too. These operations will transform [1, -1, 0] β [1, -1, -2] β [1, 1, -2]
In the second test case we can't make equal numbers on the second position.
In the third test case we can choose (i, j)=(1, 2) 41 times. The same about the fourth test case.
In the last lest case, it is impossible to make array a equal to the array b. | instruction | 0 | 87,720 | 12 | 175,440 |
Tags: greedy, implementation
Correct Solution:
```
t = int(input())
for x in range(t):
n = int(input())
a = list(map(int,input().split()))
b = list(map(int,input().split()))
v = [999999999999999]*3
deff = 0
v0 = 0
v2 = 0
v1 = 0
for i in range(n):
if a[i] == -1:
if v0 == 0:
v[0] = i
v0 = 1
elif a[i] == 0:
if v1 == 0:
v[1] = i
v1 = 1
elif a[i] == 1:
if v2 == 0:
v[2] = i
v2 = 1
for i in range(1,n+1):
if a[-i] - b[-i] > 0:
if v[0] < n - i:
pass
else:
if deff == 0:
print('NO')
deff = 1
if a[-i] - b[-i] < 0:
if v[2] < n - i:
pass
else:
if deff == 0:
print('NO')
deff = 1
if deff == 0:
print('YES')
``` | output | 1 | 87,720 | 12 | 175,441 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Once again, Boris needs the help of Anton in creating a task. This time Anton needs to solve the following problem:
There are two arrays of integers a and b of length n. It turned out that array a contains only elements from the set \{-1, 0, 1\}.
Anton can perform the following sequence of operations any number of times:
1. Choose any pair of indexes (i, j) such that 1 β€ i < j β€ n. It is possible to choose the same pair (i, j) more than once.
2. Add a_i to a_j. In other words, j-th element of the array becomes equal to a_i + a_j.
For example, if you are given array [1, -1, 0], you can transform it only to [1, -1, -1], [1, 0, 0] and [1, -1, 1] by one operation.
Anton wants to predict if it is possible to apply some number (zero or more) of these operations to the array a so that it becomes equal to array b. Can you help him?
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 β€ t β€ 10000). The description of the test cases follows.
The first line of each test case contains a single integer n (1 β€ n β€ 10^5) β the length of arrays.
The second line of each test case contains n integers a_1, a_2, ..., a_n (-1 β€ a_i β€ 1) β elements of array a. There can be duplicates among elements.
The third line of each test case contains n integers b_1, b_2, ..., b_n (-10^9 β€ b_i β€ 10^9) β elements of array b. There can be duplicates among elements.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, output one line containing "YES" if it's possible to make arrays a and b equal by performing the described operations, or "NO" if it's impossible.
You can print each letter in any case (upper or lower).
Example
Input
5
3
1 -1 0
1 1 -2
3
0 1 1
0 2 2
2
1 0
1 41
2
-1 0
-1 -41
5
0 1 -1 1 -1
1 1 -1 1 -1
Output
YES
NO
YES
YES
NO
Note
In the first test-case we can choose (i, j)=(2, 3) twice and after that choose (i, j)=(1, 2) twice too. These operations will transform [1, -1, 0] β [1, -1, -2] β [1, 1, -2]
In the second test case we can't make equal numbers on the second position.
In the third test case we can choose (i, j)=(1, 2) 41 times. The same about the fourth test case.
In the last lest case, it is impossible to make array a equal to the array b. | instruction | 0 | 87,721 | 12 | 175,442 |
Tags: greedy, implementation
Correct Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
a = [int(s) for s in input().split()]
b = [int(s) for s in input().split()]
pos = False
neg = False
for x, y in zip(a, b):
if y > x and not pos:
print('NO')
break
elif y < x and not neg:
print('NO')
break
if x == 1:
pos = True
elif x == -1:
neg = True
else:
print('YES')
``` | output | 1 | 87,721 | 12 | 175,443 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Once again, Boris needs the help of Anton in creating a task. This time Anton needs to solve the following problem:
There are two arrays of integers a and b of length n. It turned out that array a contains only elements from the set \{-1, 0, 1\}.
Anton can perform the following sequence of operations any number of times:
1. Choose any pair of indexes (i, j) such that 1 β€ i < j β€ n. It is possible to choose the same pair (i, j) more than once.
2. Add a_i to a_j. In other words, j-th element of the array becomes equal to a_i + a_j.
For example, if you are given array [1, -1, 0], you can transform it only to [1, -1, -1], [1, 0, 0] and [1, -1, 1] by one operation.
Anton wants to predict if it is possible to apply some number (zero or more) of these operations to the array a so that it becomes equal to array b. Can you help him?
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 β€ t β€ 10000). The description of the test cases follows.
The first line of each test case contains a single integer n (1 β€ n β€ 10^5) β the length of arrays.
The second line of each test case contains n integers a_1, a_2, ..., a_n (-1 β€ a_i β€ 1) β elements of array a. There can be duplicates among elements.
The third line of each test case contains n integers b_1, b_2, ..., b_n (-10^9 β€ b_i β€ 10^9) β elements of array b. There can be duplicates among elements.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, output one line containing "YES" if it's possible to make arrays a and b equal by performing the described operations, or "NO" if it's impossible.
You can print each letter in any case (upper or lower).
Example
Input
5
3
1 -1 0
1 1 -2
3
0 1 1
0 2 2
2
1 0
1 41
2
-1 0
-1 -41
5
0 1 -1 1 -1
1 1 -1 1 -1
Output
YES
NO
YES
YES
NO
Note
In the first test-case we can choose (i, j)=(2, 3) twice and after that choose (i, j)=(1, 2) twice too. These operations will transform [1, -1, 0] β [1, -1, -2] β [1, 1, -2]
In the second test case we can't make equal numbers on the second position.
In the third test case we can choose (i, j)=(1, 2) 41 times. The same about the fourth test case.
In the last lest case, it is impossible to make array a equal to the array b. | instruction | 0 | 87,722 | 12 | 175,444 |
Tags: greedy, implementation
Correct Solution:
```
n = input
v = lambda: map(int, n().split())
def f():
s = 0
for a, b in zip(v(), v()):
if s == 0:
if a != b: return 'NO'
s = a
elif s > 0:
if a > b: return 'NO'
if a < 0: return 'YES'
elif s < 0:
if a < b: return 'NO'
if a > 0: return 'YES'
return 'YES'
for i in range(int(n())): n(), print(f())
``` | output | 1 | 87,722 | 12 | 175,445 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Once again, Boris needs the help of Anton in creating a task. This time Anton needs to solve the following problem:
There are two arrays of integers a and b of length n. It turned out that array a contains only elements from the set \{-1, 0, 1\}.
Anton can perform the following sequence of operations any number of times:
1. Choose any pair of indexes (i, j) such that 1 β€ i < j β€ n. It is possible to choose the same pair (i, j) more than once.
2. Add a_i to a_j. In other words, j-th element of the array becomes equal to a_i + a_j.
For example, if you are given array [1, -1, 0], you can transform it only to [1, -1, -1], [1, 0, 0] and [1, -1, 1] by one operation.
Anton wants to predict if it is possible to apply some number (zero or more) of these operations to the array a so that it becomes equal to array b. Can you help him?
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 β€ t β€ 10000). The description of the test cases follows.
The first line of each test case contains a single integer n (1 β€ n β€ 10^5) β the length of arrays.
The second line of each test case contains n integers a_1, a_2, ..., a_n (-1 β€ a_i β€ 1) β elements of array a. There can be duplicates among elements.
The third line of each test case contains n integers b_1, b_2, ..., b_n (-10^9 β€ b_i β€ 10^9) β elements of array b. There can be duplicates among elements.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, output one line containing "YES" if it's possible to make arrays a and b equal by performing the described operations, or "NO" if it's impossible.
You can print each letter in any case (upper or lower).
Example
Input
5
3
1 -1 0
1 1 -2
3
0 1 1
0 2 2
2
1 0
1 41
2
-1 0
-1 -41
5
0 1 -1 1 -1
1 1 -1 1 -1
Output
YES
NO
YES
YES
NO
Note
In the first test-case we can choose (i, j)=(2, 3) twice and after that choose (i, j)=(1, 2) twice too. These operations will transform [1, -1, 0] β [1, -1, -2] β [1, 1, -2]
In the second test case we can't make equal numbers on the second position.
In the third test case we can choose (i, j)=(1, 2) 41 times. The same about the fourth test case.
In the last lest case, it is impossible to make array a equal to the array b. | instruction | 0 | 87,723 | 12 | 175,446 |
Tags: greedy, implementation
Correct Solution:
```
t=int(input())
for i in range(t):
n=int(input())
ar=list(map(int,input().split()))
br=list(map(int,input().split()))
d={}
d[1]=0
d[-1]=0
for i in range(n):
if ar[i]==1 or ar[i]==-1:
d[ar[i]]=d.get(ar[i])+1
fl=False
for i in range(n-1,-1,-1):
if ar[i]==-1 or ar[i]==1:
d[ar[i]]=d.get(ar[i])-1
if ar[i]!=br[i]:
t=br[i]-ar[i]
if t<0:
if d.get(-1)==0:
fl=True
elif t>0:
if d.get(1)==0:
fl=True
if fl:
print("NO")
else:
print("YES")
``` | output | 1 | 87,723 | 12 | 175,447 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once again, Boris needs the help of Anton in creating a task. This time Anton needs to solve the following problem:
There are two arrays of integers a and b of length n. It turned out that array a contains only elements from the set \{-1, 0, 1\}.
Anton can perform the following sequence of operations any number of times:
1. Choose any pair of indexes (i, j) such that 1 β€ i < j β€ n. It is possible to choose the same pair (i, j) more than once.
2. Add a_i to a_j. In other words, j-th element of the array becomes equal to a_i + a_j.
For example, if you are given array [1, -1, 0], you can transform it only to [1, -1, -1], [1, 0, 0] and [1, -1, 1] by one operation.
Anton wants to predict if it is possible to apply some number (zero or more) of these operations to the array a so that it becomes equal to array b. Can you help him?
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 β€ t β€ 10000). The description of the test cases follows.
The first line of each test case contains a single integer n (1 β€ n β€ 10^5) β the length of arrays.
The second line of each test case contains n integers a_1, a_2, ..., a_n (-1 β€ a_i β€ 1) β elements of array a. There can be duplicates among elements.
The third line of each test case contains n integers b_1, b_2, ..., b_n (-10^9 β€ b_i β€ 10^9) β elements of array b. There can be duplicates among elements.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, output one line containing "YES" if it's possible to make arrays a and b equal by performing the described operations, or "NO" if it's impossible.
You can print each letter in any case (upper or lower).
Example
Input
5
3
1 -1 0
1 1 -2
3
0 1 1
0 2 2
2
1 0
1 41
2
-1 0
-1 -41
5
0 1 -1 1 -1
1 1 -1 1 -1
Output
YES
NO
YES
YES
NO
Note
In the first test-case we can choose (i, j)=(2, 3) twice and after that choose (i, j)=(1, 2) twice too. These operations will transform [1, -1, 0] β [1, -1, -2] β [1, 1, -2]
In the second test case we can't make equal numbers on the second position.
In the third test case we can choose (i, j)=(1, 2) 41 times. The same about the fourth test case.
In the last lest case, it is impossible to make array a equal to the array b.
Submitted Solution:
```
t = int(input())
al = []
for i in range(t):
n = int(input())
a = list(map(int,input().split()))
b = list(map(int,input().split()))
s = 0
f = 0
z = 0
for j in range(n):
if a[j] == 0:
z += 1
elif a[j] == 1:
s += 1
else:
f += 1
ans = "YES"
for j in range(n-1,-1,-1):
if a[j] == 0:
z -= 0
elif a[j] == 1:
s -= 1
else:
f -= 1
if b[j] > a[j]:
if s == 0:
ans = "NO"
break
elif b[j] < a[j]:
if f == 0:
ans = "NO"
al.append(ans)
for i in al:
print(i)
``` | instruction | 0 | 87,724 | 12 | 175,448 |
Yes | output | 1 | 87,724 | 12 | 175,449 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once again, Boris needs the help of Anton in creating a task. This time Anton needs to solve the following problem:
There are two arrays of integers a and b of length n. It turned out that array a contains only elements from the set \{-1, 0, 1\}.
Anton can perform the following sequence of operations any number of times:
1. Choose any pair of indexes (i, j) such that 1 β€ i < j β€ n. It is possible to choose the same pair (i, j) more than once.
2. Add a_i to a_j. In other words, j-th element of the array becomes equal to a_i + a_j.
For example, if you are given array [1, -1, 0], you can transform it only to [1, -1, -1], [1, 0, 0] and [1, -1, 1] by one operation.
Anton wants to predict if it is possible to apply some number (zero or more) of these operations to the array a so that it becomes equal to array b. Can you help him?
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 β€ t β€ 10000). The description of the test cases follows.
The first line of each test case contains a single integer n (1 β€ n β€ 10^5) β the length of arrays.
The second line of each test case contains n integers a_1, a_2, ..., a_n (-1 β€ a_i β€ 1) β elements of array a. There can be duplicates among elements.
The third line of each test case contains n integers b_1, b_2, ..., b_n (-10^9 β€ b_i β€ 10^9) β elements of array b. There can be duplicates among elements.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, output one line containing "YES" if it's possible to make arrays a and b equal by performing the described operations, or "NO" if it's impossible.
You can print each letter in any case (upper or lower).
Example
Input
5
3
1 -1 0
1 1 -2
3
0 1 1
0 2 2
2
1 0
1 41
2
-1 0
-1 -41
5
0 1 -1 1 -1
1 1 -1 1 -1
Output
YES
NO
YES
YES
NO
Note
In the first test-case we can choose (i, j)=(2, 3) twice and after that choose (i, j)=(1, 2) twice too. These operations will transform [1, -1, 0] β [1, -1, -2] β [1, 1, -2]
In the second test case we can't make equal numbers on the second position.
In the third test case we can choose (i, j)=(1, 2) 41 times. The same about the fourth test case.
In the last lest case, it is impossible to make array a equal to the array b.
Submitted Solution:
```
def solve(a,b,n):
s,p=0,0
for i in range(n):
if p==0 and (b[i]-a[i])>0:
return "NO"
if s==0 and (b[i]-a[i])<0:
return "NO"
if a[i]<0:
s=1
if a[i]>0:
p=1
return "YES"
for i in range(int(input())):
n=int(input())
a=[int(i) for i in input().split()]
b=[int(i) for i in input().split()]
print(solve(a,b,n))
``` | instruction | 0 | 87,725 | 12 | 175,450 |
Yes | output | 1 | 87,725 | 12 | 175,451 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once again, Boris needs the help of Anton in creating a task. This time Anton needs to solve the following problem:
There are two arrays of integers a and b of length n. It turned out that array a contains only elements from the set \{-1, 0, 1\}.
Anton can perform the following sequence of operations any number of times:
1. Choose any pair of indexes (i, j) such that 1 β€ i < j β€ n. It is possible to choose the same pair (i, j) more than once.
2. Add a_i to a_j. In other words, j-th element of the array becomes equal to a_i + a_j.
For example, if you are given array [1, -1, 0], you can transform it only to [1, -1, -1], [1, 0, 0] and [1, -1, 1] by one operation.
Anton wants to predict if it is possible to apply some number (zero or more) of these operations to the array a so that it becomes equal to array b. Can you help him?
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 β€ t β€ 10000). The description of the test cases follows.
The first line of each test case contains a single integer n (1 β€ n β€ 10^5) β the length of arrays.
The second line of each test case contains n integers a_1, a_2, ..., a_n (-1 β€ a_i β€ 1) β elements of array a. There can be duplicates among elements.
The third line of each test case contains n integers b_1, b_2, ..., b_n (-10^9 β€ b_i β€ 10^9) β elements of array b. There can be duplicates among elements.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, output one line containing "YES" if it's possible to make arrays a and b equal by performing the described operations, or "NO" if it's impossible.
You can print each letter in any case (upper or lower).
Example
Input
5
3
1 -1 0
1 1 -2
3
0 1 1
0 2 2
2
1 0
1 41
2
-1 0
-1 -41
5
0 1 -1 1 -1
1 1 -1 1 -1
Output
YES
NO
YES
YES
NO
Note
In the first test-case we can choose (i, j)=(2, 3) twice and after that choose (i, j)=(1, 2) twice too. These operations will transform [1, -1, 0] β [1, -1, -2] β [1, 1, -2]
In the second test case we can't make equal numbers on the second position.
In the third test case we can choose (i, j)=(1, 2) 41 times. The same about the fourth test case.
In the last lest case, it is impossible to make array a equal to the array b.
Submitted Solution:
```
t=int(input())
for _ in range(t):
n=int(input())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
flag1=False
flag2=False
ans=False
for i in range(n):
if a[i]<b[i]:
if flag1==False:
ans=True
break
elif a[i]>b[i]:
if flag2==False:
ans=True
break
if a[i]==1:
flag1=True
elif a[i]==-1:
flag2=True
if ans:
print("NO")
else:
print("YES")
``` | instruction | 0 | 87,726 | 12 | 175,452 |
Yes | output | 1 | 87,726 | 12 | 175,453 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once again, Boris needs the help of Anton in creating a task. This time Anton needs to solve the following problem:
There are two arrays of integers a and b of length n. It turned out that array a contains only elements from the set \{-1, 0, 1\}.
Anton can perform the following sequence of operations any number of times:
1. Choose any pair of indexes (i, j) such that 1 β€ i < j β€ n. It is possible to choose the same pair (i, j) more than once.
2. Add a_i to a_j. In other words, j-th element of the array becomes equal to a_i + a_j.
For example, if you are given array [1, -1, 0], you can transform it only to [1, -1, -1], [1, 0, 0] and [1, -1, 1] by one operation.
Anton wants to predict if it is possible to apply some number (zero or more) of these operations to the array a so that it becomes equal to array b. Can you help him?
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 β€ t β€ 10000). The description of the test cases follows.
The first line of each test case contains a single integer n (1 β€ n β€ 10^5) β the length of arrays.
The second line of each test case contains n integers a_1, a_2, ..., a_n (-1 β€ a_i β€ 1) β elements of array a. There can be duplicates among elements.
The third line of each test case contains n integers b_1, b_2, ..., b_n (-10^9 β€ b_i β€ 10^9) β elements of array b. There can be duplicates among elements.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, output one line containing "YES" if it's possible to make arrays a and b equal by performing the described operations, or "NO" if it's impossible.
You can print each letter in any case (upper or lower).
Example
Input
5
3
1 -1 0
1 1 -2
3
0 1 1
0 2 2
2
1 0
1 41
2
-1 0
-1 -41
5
0 1 -1 1 -1
1 1 -1 1 -1
Output
YES
NO
YES
YES
NO
Note
In the first test-case we can choose (i, j)=(2, 3) twice and after that choose (i, j)=(1, 2) twice too. These operations will transform [1, -1, 0] β [1, -1, -2] β [1, 1, -2]
In the second test case we can't make equal numbers on the second position.
In the third test case we can choose (i, j)=(1, 2) 41 times. The same about the fourth test case.
In the last lest case, it is impossible to make array a equal to the array b.
Submitted Solution:
```
def noob():
return 5+10
t = int(input())
while t!=0:
t-=1
n = int(input())
a = [int(ele) for ele in input().split()]
b = [int(elem) for elem in input().split()]
on = [-1]*n
non = [-1]*n
pronoob = -1
for i in range(n):
on[i] = pronoob
if a[i]==1: pronoob=i
pronoob = -1
for i in range(n):
non[i] = pronoob
if a[i]==-1: pronoob=i
flag = 0
for i in range(n):
t1 = a[i]-b[i]
if t1<0 and on[i]==-1:
flag = 1
break
if t1>0 and non[i]==-1:
flag = 1
break
if flag==1: print("NO")
else: print("YES")
``` | instruction | 0 | 87,727 | 12 | 175,454 |
Yes | output | 1 | 87,727 | 12 | 175,455 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once again, Boris needs the help of Anton in creating a task. This time Anton needs to solve the following problem:
There are two arrays of integers a and b of length n. It turned out that array a contains only elements from the set \{-1, 0, 1\}.
Anton can perform the following sequence of operations any number of times:
1. Choose any pair of indexes (i, j) such that 1 β€ i < j β€ n. It is possible to choose the same pair (i, j) more than once.
2. Add a_i to a_j. In other words, j-th element of the array becomes equal to a_i + a_j.
For example, if you are given array [1, -1, 0], you can transform it only to [1, -1, -1], [1, 0, 0] and [1, -1, 1] by one operation.
Anton wants to predict if it is possible to apply some number (zero or more) of these operations to the array a so that it becomes equal to array b. Can you help him?
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 β€ t β€ 10000). The description of the test cases follows.
The first line of each test case contains a single integer n (1 β€ n β€ 10^5) β the length of arrays.
The second line of each test case contains n integers a_1, a_2, ..., a_n (-1 β€ a_i β€ 1) β elements of array a. There can be duplicates among elements.
The third line of each test case contains n integers b_1, b_2, ..., b_n (-10^9 β€ b_i β€ 10^9) β elements of array b. There can be duplicates among elements.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, output one line containing "YES" if it's possible to make arrays a and b equal by performing the described operations, or "NO" if it's impossible.
You can print each letter in any case (upper or lower).
Example
Input
5
3
1 -1 0
1 1 -2
3
0 1 1
0 2 2
2
1 0
1 41
2
-1 0
-1 -41
5
0 1 -1 1 -1
1 1 -1 1 -1
Output
YES
NO
YES
YES
NO
Note
In the first test-case we can choose (i, j)=(2, 3) twice and after that choose (i, j)=(1, 2) twice too. These operations will transform [1, -1, 0] β [1, -1, -2] β [1, 1, -2]
In the second test case we can't make equal numbers on the second position.
In the third test case we can choose (i, j)=(1, 2) 41 times. The same about the fourth test case.
In the last lest case, it is impossible to make array a equal to the array b.
Submitted Solution:
```
#!/usr/bin/env python
import os
import sys
from io import BytesIO, IOBase
import threading
from heapq import heapify,heappush,heappop
def can_win(s,e):
if e%2==1:
return 1 if s%2==0 else 0
else:
if e//2<s<=e:
return 1 if s%2==1 else 0
if e//4<s<=e//2:
return 1
else:
return can_win(s,e//4)
def main():
for _ in range(int(input())):
n=int(input())
arr=list(map(int,input().split()))
brr=list(map(int,input().split()))
is1=is_1=-1
b=True
is1=is_1=-1
for i in range(n):
if brr[i]>arr[i] and is1==-1:
b=False
break
elif brr[i]<arr[i] and is_1==-1:
b=False
break
elif arr[i]==1:
is1=1
else:
is_1=1
if b:
print('YES')
else:
print('NO')
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
main()
``` | instruction | 0 | 87,728 | 12 | 175,456 |
No | output | 1 | 87,728 | 12 | 175,457 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once again, Boris needs the help of Anton in creating a task. This time Anton needs to solve the following problem:
There are two arrays of integers a and b of length n. It turned out that array a contains only elements from the set \{-1, 0, 1\}.
Anton can perform the following sequence of operations any number of times:
1. Choose any pair of indexes (i, j) such that 1 β€ i < j β€ n. It is possible to choose the same pair (i, j) more than once.
2. Add a_i to a_j. In other words, j-th element of the array becomes equal to a_i + a_j.
For example, if you are given array [1, -1, 0], you can transform it only to [1, -1, -1], [1, 0, 0] and [1, -1, 1] by one operation.
Anton wants to predict if it is possible to apply some number (zero or more) of these operations to the array a so that it becomes equal to array b. Can you help him?
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 β€ t β€ 10000). The description of the test cases follows.
The first line of each test case contains a single integer n (1 β€ n β€ 10^5) β the length of arrays.
The second line of each test case contains n integers a_1, a_2, ..., a_n (-1 β€ a_i β€ 1) β elements of array a. There can be duplicates among elements.
The third line of each test case contains n integers b_1, b_2, ..., b_n (-10^9 β€ b_i β€ 10^9) β elements of array b. There can be duplicates among elements.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, output one line containing "YES" if it's possible to make arrays a and b equal by performing the described operations, or "NO" if it's impossible.
You can print each letter in any case (upper or lower).
Example
Input
5
3
1 -1 0
1 1 -2
3
0 1 1
0 2 2
2
1 0
1 41
2
-1 0
-1 -41
5
0 1 -1 1 -1
1 1 -1 1 -1
Output
YES
NO
YES
YES
NO
Note
In the first test-case we can choose (i, j)=(2, 3) twice and after that choose (i, j)=(1, 2) twice too. These operations will transform [1, -1, 0] β [1, -1, -2] β [1, 1, -2]
In the second test case we can't make equal numbers on the second position.
In the third test case we can choose (i, j)=(1, 2) 41 times. The same about the fourth test case.
In the last lest case, it is impossible to make array a equal to the array b.
Submitted Solution:
```
for _ in range(int(input())):
n=int(input())
a=[int(x) for x in input().split()]
b=[int(x) for x in input().split()]
f=0
pos,neg=0,0
if a[0]!=b[0]:
print('NO')
continue
for i in range(1,n):
if a[i-1]==1:
pos=1
elif a[i-1]==-1:
neg=1
if a[i]==b[i]:
continue
elif b[i]>0 and pos:
continue
elif b[i]<0 and neg:
continue
elif b[i]==0:
if a[i]>0 and neg:
continue
elif a[i]<0 and pos:
continue
else:
f=1
break
if f:
print('NO')
else:
print('YES')
``` | instruction | 0 | 87,729 | 12 | 175,458 |
No | output | 1 | 87,729 | 12 | 175,459 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once again, Boris needs the help of Anton in creating a task. This time Anton needs to solve the following problem:
There are two arrays of integers a and b of length n. It turned out that array a contains only elements from the set \{-1, 0, 1\}.
Anton can perform the following sequence of operations any number of times:
1. Choose any pair of indexes (i, j) such that 1 β€ i < j β€ n. It is possible to choose the same pair (i, j) more than once.
2. Add a_i to a_j. In other words, j-th element of the array becomes equal to a_i + a_j.
For example, if you are given array [1, -1, 0], you can transform it only to [1, -1, -1], [1, 0, 0] and [1, -1, 1] by one operation.
Anton wants to predict if it is possible to apply some number (zero or more) of these operations to the array a so that it becomes equal to array b. Can you help him?
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 β€ t β€ 10000). The description of the test cases follows.
The first line of each test case contains a single integer n (1 β€ n β€ 10^5) β the length of arrays.
The second line of each test case contains n integers a_1, a_2, ..., a_n (-1 β€ a_i β€ 1) β elements of array a. There can be duplicates among elements.
The third line of each test case contains n integers b_1, b_2, ..., b_n (-10^9 β€ b_i β€ 10^9) β elements of array b. There can be duplicates among elements.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, output one line containing "YES" if it's possible to make arrays a and b equal by performing the described operations, or "NO" if it's impossible.
You can print each letter in any case (upper or lower).
Example
Input
5
3
1 -1 0
1 1 -2
3
0 1 1
0 2 2
2
1 0
1 41
2
-1 0
-1 -41
5
0 1 -1 1 -1
1 1 -1 1 -1
Output
YES
NO
YES
YES
NO
Note
In the first test-case we can choose (i, j)=(2, 3) twice and after that choose (i, j)=(1, 2) twice too. These operations will transform [1, -1, 0] β [1, -1, -2] β [1, 1, -2]
In the second test case we can't make equal numbers on the second position.
In the third test case we can choose (i, j)=(1, 2) 41 times. The same about the fourth test case.
In the last lest case, it is impossible to make array a equal to the array b.
Submitted Solution:
```
def artem(n,m):
if (m*n)%2 == 1:
return ('BW'*(m//2) + 'B\n' + 'WB'*(m//2) + 'W\n')*(n//2) + 'BW'*(m//2) + 'B\n'
else:
return ('BW'*((m-1)//2) + 'B\n'*(m%2) + 'BW\n'*((m-1)%2) + 'WB'*((m-1)//2) + 'W\n'*(m%2) + 'WB\n'*((m-1)%2))*((n-1)//2) + 'BW'*((m-1)//2) + 'BB\n'*((m-1)%2) + 'B\n'*(m%2) + ('WB'*((m-1)//2) + 'B\n'*(m%2) + 'WB\n'*((m-1)%2))*((n-1)%2)
# cases = int(input())
# for _ in range(cases):
# n,m = list(map(int,input().split()))
# print(artem(n,m)[:-1])
def anton(a,b):
found = [False,False]
for i in range(len(a)):
if a[i] == b[i]:
if a[i] == 1 and not found[1]:
found[1] = True
if a[i] == -1 and not found[0]:
found[0] = True
if found[0] and found[1]:
return 'YES'
elif a[i] > b[i] and not found[0]:
return 'NO'
elif a[i] < b[i] and not found[1]:
return 'NO'
return 'YES'
cases = int(input())
for _ in range(cases):
l = int(input())
a = list(map(int,input().split()))
b = list(map(int,input().split()))
print(anton(a,b))
``` | instruction | 0 | 87,730 | 12 | 175,460 |
No | output | 1 | 87,730 | 12 | 175,461 |
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