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 |
|---|---|---|---|---|---|
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 ≤ i < j ≤ 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 ≤ t ≤ 10000). The description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the length of arrays.
The second line of each test case contains n integers a_1, a_2, ..., a_n (-1 ≤ a_i ≤ 1) — 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 ≤ b_i ≤ 10^9) — 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] → [1, -1, -2] → [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:
```
testCases = int(input())
def convertArrayToInt(A):
for i in range(0, len(A), 1):
A[i] = int(A[i])
return A
def canBeConverted(A, B):
for i in range(len(A)):
if A[i]!=B[i]:
diff = B[i]-A[i]
flag = 0
for x in range(0, i, 1):
if A[x]!=0 and diff%A[x]==0:
flag=1
break
if flag==0:
return "NO"
return "YES"
while testCases:
length = int(input())
A = input().split()
B = input().split()
A = convertArrayToInt(A)
B = convertArrayToInt(B)
ans = canBeConverted(A, B)
print (ans)
testCases-=1
``` | instruction | 0 | 87,731 | 12 | 175,462 |
No | output | 1 | 87,731 | 12 | 175,463 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n integers. Indices of the array start from zero (i. e. the first element is a_0, the second one is a_1, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of a with borders l and r is a[l; r] = a_l, a_{l + 1}, ..., a_{r}.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements a_0, a_2, ..., a_{2k} for integer k = ⌊(n-1)/(2)⌋ should be maximum possible).
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of a. The second line of the test case contains n integers a_0, a_1, ..., a_{n-1} (1 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of a.
Example
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | instruction | 0 | 87,748 | 12 | 175,496 |
Tags: divide and conquer, dp, greedy, implementation
Correct Solution:
```
from sys import stdin, setrecursionlimit, stdout
#setrecursionlimit(1000000)
from collections import deque
from math import sqrt, floor, ceil, log, log2, log10, pi, gcd, sin, cos, asin
from heapq import heapify, heappop, heappush, heappushpop, heapreplace
def ii(): return int(stdin.readline())
def fi(): return float(stdin.readline())
def mi(): return map(int, stdin.readline().split())
def fmi(): return map(float, stdin.readline().split())
def li(): return list(mi())
def si(): return stdin.readline().rstrip()
def lsi(): return list(si())
#mod=1000000007
res=['NET', 'DA']
############# CODE STARTS HERE #############
test_case=ii()
while test_case:
test_case-=1
n=ii()
a=li()
s=sum([a[i] for i in range(0, n, 2)])
s1=s2=mx1=mx2=0
for i in range(1, n, 2):
x=a[i]-a[i-1]
if s1+x>0:
s1+=x
mx1=max(mx1, s1)
else:
s1=0
for i in range(2, n, 2):
y=a[i-1]-a[i]
if s2+y>0:
s2+=y
mx2=max(mx2, s2)
else:
s2=0
print(s+max(mx1, mx2))
``` | output | 1 | 87,748 | 12 | 175,497 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n integers. Indices of the array start from zero (i. e. the first element is a_0, the second one is a_1, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of a with borders l and r is a[l; r] = a_l, a_{l + 1}, ..., a_{r}.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements a_0, a_2, ..., a_{2k} for integer k = ⌊(n-1)/(2)⌋ should be maximum possible).
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of a. The second line of the test case contains n integers a_0, a_1, ..., a_{n-1} (1 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of a.
Example
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | instruction | 0 | 87,749 | 12 | 175,498 |
Tags: divide and conquer, dp, greedy, implementation
Correct Solution:
```
for t in range(int(input())):
n = int(input())
lst= list(map(int, input().split()))
r = s = rs = 0
x=n//2
for i in range(x):
s =s+ (lst[2 * i + 1] - lst[2 * i])
rs = rs+ lst[2 * i]
s = max(s, 0)
if s > r:
r = s
if n % 2:
rs += lst[-1]
s = 0
y=(n + 1) // 2
for i in range(1,y):
s += (lst[2 * i - 1] - lst[2 * i])
s = max(s, 0)
if s > r:
r =s
print(rs + r)
``` | output | 1 | 87,749 | 12 | 175,499 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n integers. Indices of the array start from zero (i. e. the first element is a_0, the second one is a_1, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of a with borders l and r is a[l; r] = a_l, a_{l + 1}, ..., a_{r}.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements a_0, a_2, ..., a_{2k} for integer k = ⌊(n-1)/(2)⌋ should be maximum possible).
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of a. The second line of the test case contains n integers a_0, a_1, ..., a_{n-1} (1 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of a.
Example
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | instruction | 0 | 87,750 | 12 | 175,500 |
Tags: divide and conquer, dp, greedy, implementation
Correct Solution:
```
import sys
t=int(sys.stdin.readline())
for i in range (t):
num=0
parr=0
imparrr=0
n=int(sys.stdin.readline())
a=list(map(int,input().split()))
suma=0
for i in range(len(a)):
if i%2==0:
suma+=a[i]
for i in range (0,n-1,2):
num += a[i+1] - a[i]
parr=max(parr,num)
if num<0: num=0
num=0
for i in range (1,n-1,2):
num+= a[i] - a[i+1]
parr=max(parr,num)
if num < 0: num = 0
print(parr+suma)
``` | output | 1 | 87,750 | 12 | 175,501 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n integers. Indices of the array start from zero (i. e. the first element is a_0, the second one is a_1, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of a with borders l and r is a[l; r] = a_l, a_{l + 1}, ..., a_{r}.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements a_0, a_2, ..., a_{2k} for integer k = ⌊(n-1)/(2)⌋ should be maximum possible).
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of a. The second line of the test case contains n integers a_0, a_1, ..., a_{n-1} (1 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of a.
Example
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | instruction | 0 | 87,751 | 12 | 175,502 |
Tags: divide and conquer, dp, greedy, implementation
Correct Solution:
```
def main():
t = int(input())
for _ in range(t):
n = int(input())
a = list(map(int, input().split()))
print(f(n, a))
def f(n, a):
odd_sum = 0
for i in range(0, n, 2): odd_sum += a[i]
prefix = 0
min_prefix_even = 0
min_prefix_odd = float('inf')
max_shift = 0
for i in range(n):
if i % 2 == 0:
prefix -= a[i]
else:
prefix += a[i]
if i % 2 == 1:
max_shift = max(max_shift, prefix - min_prefix_even)
min_prefix_even = min(min_prefix_even, prefix)
else:
max_shift = max(max_shift, prefix - min_prefix_odd)
min_prefix_odd = min(min_prefix_odd, prefix)
return max(odd_sum, odd_sum + max_shift)
if __name__ == '__main__':
main()
``` | output | 1 | 87,751 | 12 | 175,503 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n integers. Indices of the array start from zero (i. e. the first element is a_0, the second one is a_1, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of a with borders l and r is a[l; r] = a_l, a_{l + 1}, ..., a_{r}.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements a_0, a_2, ..., a_{2k} for integer k = ⌊(n-1)/(2)⌋ should be maximum possible).
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of a. The second line of the test case contains n integers a_0, a_1, ..., a_{n-1} (1 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of a.
Example
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | instruction | 0 | 87,752 | 12 | 175,504 |
Tags: divide and conquer, dp, greedy, implementation
Correct Solution:
```
import sys
import random
from math import *
def input():
return sys.stdin.readline().strip()
def iinput():
return int(input())
def finput():
return float(input())
def tinput():
return input().split()
def linput():
return list(input())
def rinput():
return map(int, tinput())
def fiinput():
return map(float, tinput())
def rlinput():
return list(map(int, input().split()))
def trinput():
return tuple(rinput())
def srlinput():
return sorted(list(map(int, input().split())))
def NOYES(fl):
if fl:
print("NO")
else:
print("YES")
def YESNO(fl):
if fl:
print("YES")
else:
print("NO")
def main():
n = iinput()
#k = iinput()
#m = iinput()
#n = int(sys.stdin.readline().strip())
#n, k = rinput()
#n, m = rinput()
#m, k = rinput()
#n, k, m = rinput()
#n, m, k = rinput()
#k, n, m = rinput()
#k, m, n = rinput()
#m, k, n = rinput()
#m, n, k = rinput()
#q = srlinput()
#q = linput()
s, mn, m, res, fp = 0, 0, 0, 0, 0
q = rlinput()
for i in range(n):
f = fp * (i > 0)
if i % 2 == 0:
f -= q[i]
s += q[i]
res = max(res, f - mn)
mn = min(mn, f)
else:
f += q[i]
res = max(res, f - m)
m = min(m, f)
fp = f
print(s + res)
for inytd in range(iinput()):
main()
``` | output | 1 | 87,752 | 12 | 175,505 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n integers. Indices of the array start from zero (i. e. the first element is a_0, the second one is a_1, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of a with borders l and r is a[l; r] = a_l, a_{l + 1}, ..., a_{r}.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements a_0, a_2, ..., a_{2k} for integer k = ⌊(n-1)/(2)⌋ should be maximum possible).
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of a. The second line of the test case contains n integers a_0, a_1, ..., a_{n-1} (1 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of a.
Example
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | instruction | 0 | 87,753 | 12 | 175,506 |
Tags: divide and conquer, dp, greedy, implementation
Correct Solution:
```
import sys
input = sys.stdin.buffer.readline
Q = int(input())
Query = []
for _ in range(Q):
N = int(input())
A = list(map(int, input().split()))
Query.append((N, A))
def solve(A, ans):
nowmin = 0
w = 0
for p1 in A:
w += p1
nowmin = min(nowmin, w)
ans = max(ans, T + w - nowmin)
return ans
for N, A in Query:
T = 0
P1 = []
P2 = []
for i in range((N-1)//2+1):
T += A[2*i]
if 2*i+1 <= N-1:
d = A[2*i+1] - A[2*i]
P1.append(d)
if 0 <= 2*i-1:
d = A[2*i-1] - A[2*i]
P2.append(d)
ans = solve(P1, T)
ans = solve(P2, ans)
print(ans)
``` | output | 1 | 87,753 | 12 | 175,507 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n integers. Indices of the array start from zero (i. e. the first element is a_0, the second one is a_1, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of a with borders l and r is a[l; r] = a_l, a_{l + 1}, ..., a_{r}.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements a_0, a_2, ..., a_{2k} for integer k = ⌊(n-1)/(2)⌋ should be maximum possible).
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of a. The second line of the test case contains n integers a_0, a_1, ..., a_{n-1} (1 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of a.
Example
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | instruction | 0 | 87,754 | 12 | 175,508 |
Tags: divide and conquer, dp, greedy, implementation
Correct Solution:
```
for _ in range(int(input())):
n = int(input())
a = list(map(int,input().split()))
tot = 0
dif1 = 0
extra = 0
for i in range(0,n,2):
tot+=a[i]
for i in range(0,n-1,2):
dif1-=a[i]
dif1+=a[i+1]
extra = max(extra,dif1)
if dif1 < 0:
dif1 = 0
dif2 = 0
for i in range(1,n-1,2):
dif2+=a[i]
dif2-=a[i+1]
extra = max(extra,dif2)
if dif2 < 0:
dif2 = 0
tot = tot+extra
print(tot)
``` | output | 1 | 87,754 | 12 | 175,509 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n integers. Indices of the array start from zero (i. e. the first element is a_0, the second one is a_1, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of a with borders l and r is a[l; r] = a_l, a_{l + 1}, ..., a_{r}.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements a_0, a_2, ..., a_{2k} for integer k = ⌊(n-1)/(2)⌋ should be maximum possible).
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of a. The second line of the test case contains n integers a_0, a_1, ..., a_{n-1} (1 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of a.
Example
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5 | instruction | 0 | 87,755 | 12 | 175,510 |
Tags: divide and conquer, dp, greedy, implementation
Correct Solution:
```
def maxSubArraySum(a,size):
max_so_far =a[0]
curr_max = a[0]
for i in range(1,size):
curr_max = max(a[i], curr_max + a[i])
max_so_far = max(max_so_far,curr_max,0)
return max(max_so_far,0)
for _ in range(int(input())):
N=int(input())
A=list(map(int,input().split()))
temp1=[]
temp2=[]
if(N==1):
print(A[0])
else:
if(N%2==0):
suma=0
for i in range(0,N,2):
temp1.append(A[i+1]-A[i])
suma+=A[i]
for i in range(1,N-1,2):
temp2.append(A[i]-A[i+1])
# print(temp1)
t=maxSubArraySum(temp1,len(temp1))
if(len(temp2)==0):
t2=0
else:
t2=maxSubArraySum(temp2,len(temp2))
r=max(t,t2)
print(r+suma)
else:
suma=0
for i in range(0,N-1,2):
temp1.append(A[i+1]-A[i])
suma+=A[i]
suma+=A[N-1]
# print(suma)
for i in range(1,N,2):
temp2.append(A[i]-A[i+1])
t1=maxSubArraySum(temp1,len(temp1))
t2=maxSubArraySum(temp2,len(temp2))
r=max(t1,t2)
# print(r)
print(suma+r)
``` | output | 1 | 87,755 | 12 | 175,511 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n integers. Indices of the array start from zero (i. e. the first element is a_0, the second one is a_1, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of a with borders l and r is a[l; r] = a_l, a_{l + 1}, ..., a_{r}.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements a_0, a_2, ..., a_{2k} for integer k = ⌊(n-1)/(2)⌋ should be maximum possible).
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of a. The second line of the test case contains n integers a_0, a_1, ..., a_{n-1} (1 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of a.
Example
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5
Submitted Solution:
```
from sys import stdin
inp = lambda : stdin.readline().strip()
t = int(inp())
for _ in range(t):
n = int(inp())
a = [int(x) for x in inp().split()]
even = 0
for i in range(0,n,2):
even += a[i]
x = [0]*n
p = [0]*n
for i in range(0,n-1,2):
x[i//2] = a[i+1]-a[i]
for i in range(1,n-1,2):
p[i//2] = a[i]-a[i+1]
y = x[0]
ans = x[0]
for i in x[1:]:
y = max(y + i, i)
ans = max(ans,y)
y = p[0]
ans = max(ans,y)
for i in p[1:]:
y = max(y + i, i)
ans = max(ans,y)
print(ans + even)
``` | instruction | 0 | 87,756 | 12 | 175,512 |
Yes | output | 1 | 87,756 | 12 | 175,513 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n integers. Indices of the array start from zero (i. e. the first element is a_0, the second one is a_1, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of a with borders l and r is a[l; r] = a_l, a_{l + 1}, ..., a_{r}.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements a_0, a_2, ..., a_{2k} for integer k = ⌊(n-1)/(2)⌋ should be maximum possible).
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of a. The second line of the test case contains n integers a_0, a_1, ..., a_{n-1} (1 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of a.
Example
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5
Submitted Solution:
```
import io,os
input=io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
import sys
def solve(n,A):
k=n//2
DP=[[0]*3 for _ in range(k+1)]
for i in range(k):
l,r=2*i,2*i+1
DP[i+1][0]=DP[i][0]+A[l]
DP[i+1][1]=max(DP[i][0]+A[r],DP[i][1]+A[r])
DP[i+1][2]=max(DP[i][1]+A[l],DP[i][2]+A[l])
ans=max(DP[-1])
return ans
def main():
t=int(input())
for _ in range(t):
n=int(input())
A=list(map(int,input().split()))
if n%2:
A.append(0)
n+=1
ans1=solve(n,A)
B=A[1:-1]
B.reverse()
ans2=A[0]+solve(n-2,B)
#print(ans1,ans2)
ans=max(ans1,ans2)
sys.stdout.write(str(ans)+'\n')
if __name__=='__main__':
main()
``` | instruction | 0 | 87,757 | 12 | 175,514 |
Yes | output | 1 | 87,757 | 12 | 175,515 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n integers. Indices of the array start from zero (i. e. the first element is a_0, the second one is a_1, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of a with borders l and r is a[l; r] = a_l, a_{l + 1}, ..., a_{r}.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements a_0, a_2, ..., a_{2k} for integer k = ⌊(n-1)/(2)⌋ should be maximum possible).
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of a. The second line of the test case contains n integers a_0, a_1, ..., a_{n-1} (1 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of a.
Example
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5
Submitted Solution:
```
t=int(input())
for i in range(t):
n=int(input())
a=[int(v) for v in input().split()]
v1=[0]
v2=[0]
m=0
s=sum(a[::2])
for j in range(0,n-(n%2),2):
v1.append(max(v1[-1]+a[j+1]-a[j],0))
for j in range(1,n-(1-(n%2)),2):
v2.append(max(a[j]-a[j+1]+v2[-1],0))
m=max(max(v1),max(v2))
print(s+m)
``` | instruction | 0 | 87,758 | 12 | 175,516 |
Yes | output | 1 | 87,758 | 12 | 175,517 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n integers. Indices of the array start from zero (i. e. the first element is a_0, the second one is a_1, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of a with borders l and r is a[l; r] = a_l, a_{l + 1}, ..., a_{r}.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements a_0, a_2, ..., a_{2k} for integer k = ⌊(n-1)/(2)⌋ should be maximum possible).
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of a. The second line of the test case contains n integers a_0, a_1, ..., a_{n-1} (1 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of a.
Example
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5
Submitted Solution:
```
for _ in range(int(input())):
n=int(input())
l=list(map(int,input().split()))
o=[0]*n
e=[0]*n
e[0]=l[0]
dp=[0]*n
dp[0]=o[0]-e[0]
ans=0
emin=min(0,dp[0])
omin=0
for i in range(1,n):
o[i]=o[i-1]
e[i]=e[i-1]
if i%2==0:
e[i]+=l[i]
else:
o[i]+=l[i]
dp[i]=o[i]-e[i]
if i%2==0:
temp = dp[i] - emin
ans=max(ans,temp)
emin = min(emin,dp[i])
else:
temp = dp[i] - omin
ans = max(ans,temp)
omin = min(omin,dp[i])
# print(dp)
ans+=e[-1]
print(ans)
``` | instruction | 0 | 87,759 | 12 | 175,518 |
Yes | output | 1 | 87,759 | 12 | 175,519 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n integers. Indices of the array start from zero (i. e. the first element is a_0, the second one is a_1, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of a with borders l and r is a[l; r] = a_l, a_{l + 1}, ..., a_{r}.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements a_0, a_2, ..., a_{2k} for integer k = ⌊(n-1)/(2)⌋ should be maximum possible).
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of a. The second line of the test case contains n integers a_0, a_1, ..., a_{n-1} (1 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of a.
Example
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5
Submitted Solution:
```
t=int(input())
for _ in range(t):
n=int(input())
l=list(map(int,input().split()))
o=0
e=0
m=0
s=0
for i in range(n):
f=1
if i%2==0:
f=0
s+=l[i]
e+=l[i]
else:
o+=l[i]
if o>=e:
e=0
o=l[i]
if o-e>m and (f==0 or i==n-1):
m=o-e
print(s+m)
``` | instruction | 0 | 87,760 | 12 | 175,520 |
No | output | 1 | 87,760 | 12 | 175,521 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n integers. Indices of the array start from zero (i. e. the first element is a_0, the second one is a_1, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of a with borders l and r is a[l; r] = a_l, a_{l + 1}, ..., a_{r}.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements a_0, a_2, ..., a_{2k} for integer k = ⌊(n-1)/(2)⌋ should be maximum possible).
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of a. The second line of the test case contains n integers a_0, a_1, ..., a_{n-1} (1 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of a.
Example
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5
Submitted Solution:
```
def calc_01(n, a):
# 01
end = n if n % 2 == 0 else n - 1
cp = 0
P = []
for i in range(0, end, 2):
p = a[i + 1] - a[i]
#print(p)
if p > 0:
if cp > 0:
cp += p
else:
P.append(cp)
cp = p
elif p == 0:
continue
else:
if cp < 0:
cp += p
else:
P.append(cp)
cp = p
#print(P, "dldl")
if cp > 0:
P.append(cp)
while P and P[0] <= 0:
P.pop(0)
while P and P[-1] <= 0:
P.pop()
if P == []:
return 0
#print(P)
M = max(P)
z = 0
csum = P[0]
while z < len(P) - 2:
add = P[z + 2] + P[z + 1]
csum += add
#print(csum, add)
M = max(M, csum)
z += 2
if csum < 0:
csum = P[z]
return M
def calc_12(n, a):
# 12
end = n
cp = 0
P = []
for i in range(2, end, 2):
p = a[i - 1] - a[i]
# print(p)
if p > 0:
if cp > 0:
cp += p
else:
P.append(cp)
cp = p
elif p == 0:
continue
else:
if cp < 0:
cp += p
else:
P.append(cp)
cp = p
#print(P, "dk")
if cp > 0:
P.append(cp)
while P and P[0] <= 0:
P.pop(0)
while P and P[-1] <= 0:
P.pop()
if P == []:
return 0
#print(P, "dl")
M = max(P)
z = 0
csum = P[0]
while z < len(P) - 2:
add = P[z + 2] + P[z + 1]
csum += add
M = max(M, csum)
z += 2
if csum < 0:
csum = P[z]
return M
def sum_evens(n, a):
total = 0
for i in range(0, n, 2):
total += a[i]
return total
def main():
t = int(input())
for _ in range(t):
n = int(input())
a = list(map(int, input().split()))
M01 = calc_01(n, a)
M12 = calc_12(n, a)
#print(M01, M12)
max_profit = max([M01, M12, 0])
even_sum = sum_evens(n, a)
print(even_sum + max_profit)
main()
``` | instruction | 0 | 87,761 | 12 | 175,522 |
No | output | 1 | 87,761 | 12 | 175,523 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n integers. Indices of the array start from zero (i. e. the first element is a_0, the second one is a_1, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of a with borders l and r is a[l; r] = a_l, a_{l + 1}, ..., a_{r}.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements a_0, a_2, ..., a_{2k} for integer k = ⌊(n-1)/(2)⌋ should be maximum possible).
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of a. The second line of the test case contains n integers a_0, a_1, ..., a_{n-1} (1 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of a.
Example
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5
Submitted Solution:
```
for test in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
if len(a) < 3:
print(a[0])
continue
zero = [a[i + 1] - a[i] for i in range(0, n - n % 2, 2)]
one = [a[i] - a[i + 1] for i in range(1, n - (1 - n % 2), 2)]
ans_zero = zero[0]
summ = 0
for i in zero:
summ += i
ans_zero = max(ans_zero, summ)
summ = max(0, summ)
ans_one = one[0]
summ = 0
for i in one:
summ += i
ans_one = max(ans_one, summ)
summ = max(0, summ)
summ = sum(a[::2])
ans = summ + max(ans_one, ans_zero)
print(max(summ, ans))
``` | instruction | 0 | 87,762 | 12 | 175,524 |
No | output | 1 | 87,762 | 12 | 175,525 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n integers. Indices of the array start from zero (i. e. the first element is a_0, the second one is a_1, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of a with borders l and r is a[l; r] = a_l, a_{l + 1}, ..., a_{r}.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements a_0, a_2, ..., a_{2k} for integer k = ⌊(n-1)/(2)⌋ should be maximum possible).
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of a. The second line of the test case contains n integers a_0, a_1, ..., a_{n-1} (1 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of a.
Example
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5
Submitted Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
a = list(map(int, input().split()))
if n == 1:
print(a[0])
exit()
if n == 2:
print(max(a))
exit()
even, odd = [], []
for i in range(n):
if i % 2 == 0:
even.append(a[i])
else:
odd.append(a[i])
ans1, ans2 = 0, 0
dif1 = []
for i in range(len(odd)):
dif1.append(odd[i] - even[i])
tmp = 0
for i in range(len(dif1)):
tmp += dif1[i]
if tmp < 0:
tmp = 0
if ans1 < tmp:
ans1 = tmp
#print(ans)
dif2 = []
for i in range(len(even)-1):
dif2.append(odd[i] - even[i+1])
tmp = 0
for i in range(len(dif2)):
tmp += dif2[i]
if tmp < 0:
tmp = 0
if ans2 < tmp:
ans2 = tmp
#print(ans)
print(sum(even) + max(ans1, ans2))
``` | instruction | 0 | 87,763 | 12 | 175,526 |
No | output | 1 | 87,763 | 12 | 175,527 |
Evaluate the correctness of the submitted Python 2 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n integers. Indices of the array start from zero (i. e. the first element is a_0, the second one is a_1, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of a with borders l and r is a[l; r] = a_l, a_{l + 1}, ..., a_{r}.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements a_0, a_2, ..., a_{2k} for integer k = ⌊(n-1)/(2)⌋ should be maximum possible).
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of a. The second line of the test case contains n integers a_0, a_1, ..., a_{n-1} (1 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of a.
Example
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5
Submitted Solution:
```
from sys import stdin, stdout
from collections import Counter, defaultdict
from itertools import permutations, combinations
raw_input = stdin.readline
pr = stdout.write
mod=10**9+7
def ni():
return int(raw_input())
def li():
return map(int,raw_input().split())
def pn(n):
stdout.write(str(n)+'\n')
def pa(arr):
pr(' '.join(map(str,arr))+'\n')
# fast read function for total integer input
def inp():
# this function returns whole input of
# space/line seperated integers
# Use Ctrl+D to flush stdin.
return map(int,stdin.read().split())
range = xrange # not for python 3.0+
for t in range(ni()):
n=ni()
l=li()
ans=0
for i in range(0,n,2):
ans+=l[i]
sm1=0
a1=0
for i in range(0,n-1,2):
val=l[i+1]-l[i]
if val<0:
sm1=0
continue
sm1+=val
a1=max(sm1,a1)
sm2=0
a2=0
for i in range(1,n-1,2):
val=l[i]-l[i+1]
if val<0:
sm2=0
continue
sm2+=val
a2=max(a2,sm2)
pn(ans+max(a1,a2))
``` | instruction | 0 | 87,764 | 12 | 175,528 |
No | output | 1 | 87,764 | 12 | 175,529 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Yurii is sure he can do everything. Can he solve this task, though?
He has an array a consisting of n positive integers. Let's call a subarray a[l...r] good if the following conditions are simultaneously satisfied:
* l+1 ≤ r-1, i. e. the subarray has length at least 3;
* (a_l ⊕ a_r) = (a_{l+1}+a_{l+2}+…+a_{r-2}+a_{r-1}), where ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
In other words, a subarray is good if the bitwise XOR of the two border elements is equal to the sum of the rest of the elements.
Yurii wants to calculate the total number of good subarrays. What is it equal to?
An array c is a subarray of an array d if c can be obtained from d by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.
Input
The first line contains a single integer n (3 ≤ n ≤ 2⋅ 10^5) — the length of a.
The second line contains n integers a_1,a_2,…,a_n (1 ≤ a_i < 2^{30}) — elements of a.
Output
Output a single integer — the number of good subarrays.
Examples
Input
8
3 1 2 3 1 2 3 15
Output
6
Input
10
997230370 58052053 240970544 715275815 250707702 156801523 44100666 64791577 43523002 480196854
Output
2
Note
There are 6 good subarrays in the example:
* [3,1,2] (twice) because (3 ⊕ 2) = 1;
* [1,2,3] (twice) because (1 ⊕ 3) = 2;
* [2,3,1] because (2 ⊕ 1) = 3;
* [3,1,2,3,1,2,3,15] because (3 ⊕ 15) = (1+2+3+1+2+3). | instruction | 0 | 87,785 | 12 | 175,570 |
Tags: binary search, bitmasks, brute force, constructive algorithms, divide and conquer, two pointers
Correct Solution:
```
n = int(input())
a = [int(x) for x in input().split()]
ans = set([])
for i in range(n):
x = 1
while x <= a[i]:
x *= 2
j = i+1
sum = 0
while j < n-1 and sum < x:
sum += a[j]
if a[i] ^ a[j+1] == sum:
ans.add(n * i + j + 1)
j += 1
sum = 0
j = i-1
while j>0 and sum < x:
sum += a[j]
if a[i] ^ a[j-1] == sum:
ans.add(n * (j-1) + i)
j -= 1
print(len(ans))
``` | output | 1 | 87,785 | 12 | 175,571 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Yurii is sure he can do everything. Can he solve this task, though?
He has an array a consisting of n positive integers. Let's call a subarray a[l...r] good if the following conditions are simultaneously satisfied:
* l+1 ≤ r-1, i. e. the subarray has length at least 3;
* (a_l ⊕ a_r) = (a_{l+1}+a_{l+2}+…+a_{r-2}+a_{r-1}), where ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
In other words, a subarray is good if the bitwise XOR of the two border elements is equal to the sum of the rest of the elements.
Yurii wants to calculate the total number of good subarrays. What is it equal to?
An array c is a subarray of an array d if c can be obtained from d by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.
Input
The first line contains a single integer n (3 ≤ n ≤ 2⋅ 10^5) — the length of a.
The second line contains n integers a_1,a_2,…,a_n (1 ≤ a_i < 2^{30}) — elements of a.
Output
Output a single integer — the number of good subarrays.
Examples
Input
8
3 1 2 3 1 2 3 15
Output
6
Input
10
997230370 58052053 240970544 715275815 250707702 156801523 44100666 64791577 43523002 480196854
Output
2
Note
There are 6 good subarrays in the example:
* [3,1,2] (twice) because (3 ⊕ 2) = 1;
* [1,2,3] (twice) because (1 ⊕ 3) = 2;
* [2,3,1] because (2 ⊕ 1) = 3;
* [3,1,2,3,1,2,3,15] because (3 ⊕ 15) = (1+2+3+1+2+3). | instruction | 0 | 87,786 | 12 | 175,572 |
Tags: binary search, bitmasks, brute force, constructive algorithms, divide and conquer, two pointers
Correct Solution:
```
import sys
try:sys.stdin,sys.stdout=open('in.txt','r'),open('out.txt','w')
except:pass
ii1=lambda:int(sys.stdin.readline().strip()) # for interger
is1=lambda:sys.stdin.readline().strip() # for str
iia=lambda:list(map(int,sys.stdin.readline().strip().split())) # for List[int]
isa=lambda:sys.stdin.readline().strip().split() # for List[str]
mod=int(1e9 + 7);from collections import *;from math import *
from itertools import *
from functools import *
###################### Start Here ######################
n = ii1()
arr = iia()
ans = 0
for l in range(30):
for i in range(n):
if arr[i]&(1<<l):
currsum = 0
for j in range(i+2,n):
currsum+=arr[j-1]
if currsum>=(2<<l):break
if currsum<(1<<l):continue
if arr[i]^arr[j]==currsum:ans+=1
currsum = 0
for j in range(i-2,-1,-1):
currsum+=arr[j+1]
if currsum>=(2<<l):break
if currsum<(1<<l):continue
if arr[i]^arr[j]==currsum:ans+=1
print(ans)
``` | output | 1 | 87,786 | 12 | 175,573 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Yurii is sure he can do everything. Can he solve this task, though?
He has an array a consisting of n positive integers. Let's call a subarray a[l...r] good if the following conditions are simultaneously satisfied:
* l+1 ≤ r-1, i. e. the subarray has length at least 3;
* (a_l ⊕ a_r) = (a_{l+1}+a_{l+2}+…+a_{r-2}+a_{r-1}), where ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
In other words, a subarray is good if the bitwise XOR of the two border elements is equal to the sum of the rest of the elements.
Yurii wants to calculate the total number of good subarrays. What is it equal to?
An array c is a subarray of an array d if c can be obtained from d by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.
Input
The first line contains a single integer n (3 ≤ n ≤ 2⋅ 10^5) — the length of a.
The second line contains n integers a_1,a_2,…,a_n (1 ≤ a_i < 2^{30}) — elements of a.
Output
Output a single integer — the number of good subarrays.
Examples
Input
8
3 1 2 3 1 2 3 15
Output
6
Input
10
997230370 58052053 240970544 715275815 250707702 156801523 44100666 64791577 43523002 480196854
Output
2
Note
There are 6 good subarrays in the example:
* [3,1,2] (twice) because (3 ⊕ 2) = 1;
* [1,2,3] (twice) because (1 ⊕ 3) = 2;
* [2,3,1] because (2 ⊕ 1) = 3;
* [3,1,2,3,1,2,3,15] because (3 ⊕ 15) = (1+2+3+1+2+3). | instruction | 0 | 87,787 | 12 | 175,574 |
Tags: binary search, bitmasks, brute force, constructive algorithms, divide and conquer, two pointers
Correct Solution:
```
def solve(a):
seen = set()
for i in range(len(a)):
c = 0
for j in range(i+2,len(a)):
c += a[j-1]
if a[i]^a[j] == c: seen.add((i,j))
if c >= 2*a[i]: break
for i in range(len(a)-1,-1,-1):
c = 0
for j in range(i-2,-1,-1):
c += a[j+1]
if a[i]^a[j] == c: seen.add((j,i))
if c >= 2 *a[i]: break
print(len(seen))
n = int(input());solve(list(map(int,input().split())))
``` | output | 1 | 87,787 | 12 | 175,575 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Yurii is sure he can do everything. Can he solve this task, though?
He has an array a consisting of n positive integers. Let's call a subarray a[l...r] good if the following conditions are simultaneously satisfied:
* l+1 ≤ r-1, i. e. the subarray has length at least 3;
* (a_l ⊕ a_r) = (a_{l+1}+a_{l+2}+…+a_{r-2}+a_{r-1}), where ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
In other words, a subarray is good if the bitwise XOR of the two border elements is equal to the sum of the rest of the elements.
Yurii wants to calculate the total number of good subarrays. What is it equal to?
An array c is a subarray of an array d if c can be obtained from d by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.
Input
The first line contains a single integer n (3 ≤ n ≤ 2⋅ 10^5) — the length of a.
The second line contains n integers a_1,a_2,…,a_n (1 ≤ a_i < 2^{30}) — elements of a.
Output
Output a single integer — the number of good subarrays.
Examples
Input
8
3 1 2 3 1 2 3 15
Output
6
Input
10
997230370 58052053 240970544 715275815 250707702 156801523 44100666 64791577 43523002 480196854
Output
2
Note
There are 6 good subarrays in the example:
* [3,1,2] (twice) because (3 ⊕ 2) = 1;
* [1,2,3] (twice) because (1 ⊕ 3) = 2;
* [2,3,1] because (2 ⊕ 1) = 3;
* [3,1,2,3,1,2,3,15] because (3 ⊕ 15) = (1+2+3+1+2+3). | instruction | 0 | 87,788 | 12 | 175,576 |
Tags: binary search, bitmasks, brute force, constructive algorithms, divide and conquer, two pointers
Correct Solution:
```
n = int(input())
a = list(map(int, input().split()))
def calc(a):
ans = 0
for i in range(2, len(a)):
sum = 0
for j in reversed(range(0, i-1)):
sum += a[j+1]
ans += a[i] > a[j] and a[i]^a[j] == sum
if sum > 2*a[i] or a[j].bit_length() > a[i].bit_length():
break
return ans
print(calc(a) + calc(a[::-1]))
``` | output | 1 | 87,788 | 12 | 175,577 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Yurii is sure he can do everything. Can he solve this task, though?
He has an array a consisting of n positive integers. Let's call a subarray a[l...r] good if the following conditions are simultaneously satisfied:
* l+1 ≤ r-1, i. e. the subarray has length at least 3;
* (a_l ⊕ a_r) = (a_{l+1}+a_{l+2}+…+a_{r-2}+a_{r-1}), where ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
In other words, a subarray is good if the bitwise XOR of the two border elements is equal to the sum of the rest of the elements.
Yurii wants to calculate the total number of good subarrays. What is it equal to?
An array c is a subarray of an array d if c can be obtained from d by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.
Input
The first line contains a single integer n (3 ≤ n ≤ 2⋅ 10^5) — the length of a.
The second line contains n integers a_1,a_2,…,a_n (1 ≤ a_i < 2^{30}) — elements of a.
Output
Output a single integer — the number of good subarrays.
Examples
Input
8
3 1 2 3 1 2 3 15
Output
6
Input
10
997230370 58052053 240970544 715275815 250707702 156801523 44100666 64791577 43523002 480196854
Output
2
Note
There are 6 good subarrays in the example:
* [3,1,2] (twice) because (3 ⊕ 2) = 1;
* [1,2,3] (twice) because (1 ⊕ 3) = 2;
* [2,3,1] because (2 ⊕ 1) = 3;
* [3,1,2,3,1,2,3,15] because (3 ⊕ 15) = (1+2+3+1+2+3). | instruction | 0 | 87,789 | 12 | 175,578 |
Tags: binary search, bitmasks, brute force, constructive algorithms, divide and conquer, two pointers
Correct Solution:
```
import itertools, math
n = int(input())
A = list(map(int, input().split()))
acc = [0] + list(itertools.accumulate(A))
ans = 0
seen = set()
for i in range(n - 2):
a = int(math.log2(A[i]))
for j in range(i + 2, n):
cur = acc[j] - acc[i + 1]
b = int(math.log2(cur))
if b > a:
break
if A[i] ^ A[j] == cur and (i, j) not in seen:
ans += 1
seen.add((i, j))
for j in range(n - 1, 1, -1):
a = int(math.log2(A[j]))
for i in range(j - 2, -1, -1):
cur = acc[j] - acc[i + 1]
b = int(math.log2(cur))
if b > a:
break
if A[i] ^ A[j] == cur and (i, j) not in seen:
ans += 1
seen.add((i, j))
print(ans)
``` | output | 1 | 87,789 | 12 | 175,579 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Yurii is sure he can do everything. Can he solve this task, though?
He has an array a consisting of n positive integers. Let's call a subarray a[l...r] good if the following conditions are simultaneously satisfied:
* l+1 ≤ r-1, i. e. the subarray has length at least 3;
* (a_l ⊕ a_r) = (a_{l+1}+a_{l+2}+…+a_{r-2}+a_{r-1}), where ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
In other words, a subarray is good if the bitwise XOR of the two border elements is equal to the sum of the rest of the elements.
Yurii wants to calculate the total number of good subarrays. What is it equal to?
An array c is a subarray of an array d if c can be obtained from d by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.
Input
The first line contains a single integer n (3 ≤ n ≤ 2⋅ 10^5) — the length of a.
The second line contains n integers a_1,a_2,…,a_n (1 ≤ a_i < 2^{30}) — elements of a.
Output
Output a single integer — the number of good subarrays.
Examples
Input
8
3 1 2 3 1 2 3 15
Output
6
Input
10
997230370 58052053 240970544 715275815 250707702 156801523 44100666 64791577 43523002 480196854
Output
2
Note
There are 6 good subarrays in the example:
* [3,1,2] (twice) because (3 ⊕ 2) = 1;
* [1,2,3] (twice) because (1 ⊕ 3) = 2;
* [2,3,1] because (2 ⊕ 1) = 3;
* [3,1,2,3,1,2,3,15] because (3 ⊕ 15) = (1+2+3+1+2+3). | instruction | 0 | 87,790 | 12 | 175,580 |
Tags: binary search, bitmasks, brute force, constructive algorithms, divide and conquer, two pointers
Correct Solution:
```
from bisect import *
from collections import *
from math import gcd,ceil,sqrt,floor,inf
from heapq import *
from itertools import *
from operator import add,mul,sub,xor,truediv,floordiv
from functools import *
#------------------------------------------------------------------------
import os
import sys
from io import BytesIO, IOBase
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#------------------------------------------------------------------------
def RL(): return map(int, sys.stdin.readline().rstrip().split())
def RLL(): return list(map(int, sys.stdin.readline().rstrip().split()))
def N(): return int(input())
#------------------------------------------------------------------------
from types import GeneratorType
def bootstrap(f, stack=[]):
def wrappedfunc(*args, **kwargs):
if stack:
return f(*args, **kwargs)
else:
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
else:
stack.pop()
if not stack:
break
to = stack[-1].send(to)
return to
return wrappedfunc
farr=[1]
ifa=[]
def fact(x,mod=0):
if mod:
while x>=len(farr):
farr.append(farr[-1]*len(farr)%mod)
else:
while x>=len(farr):
farr.append(farr[-1]*len(farr))
return farr[x]
def ifact(x,mod):
global ifa
ifa.append(pow(farr[-1],mod-2,mod))
for i in range(x,0,-1):
ifa.append(ifa[-1]*i%mod)
ifa=ifa[::-1]
def per(i,j,mod=0):
if i<j: return 0
if not mod:
return fact(i)//fact(i-j)
return farr[i]*ifa[i-j]%mod
def com(i,j,mod=0):
if i<j: return 0
if not mod:
return per(i,j)//fact(j)
return per(i,j,mod)*ifa[j]%mod
def catalan(n):
return com(2*n,n)//(n+1)
def linc(f,t,l,r):
while l<r:
mid=(l+r)//2
if t>f(mid):
l=mid+1
else:
r=mid
return l
def rinc(f,t,l,r):
while l<r:
mid=(l+r+1)//2
if t<f(mid):
r=mid-1
else:
l=mid
return l
def ldec(f,t,l,r):
while l<r:
mid=(l+r)//2
if t<f(mid):
l=mid+1
else:
r=mid
return l
def rdec(f,t,l,r):
while l<r:
mid=(l+r+1)//2
if t>f(mid):
r=mid-1
else:
l=mid
return l
def isprime(n):
for i in range(2,int(n**0.5)+1):
if n%i==0:
return False
return True
def binfun(x):
c=0
for w in arr:
c+=ceil(w/x)
return c
def lowbit(n):
return n&-n
def inverse(a,m):
a%=m
if a<=1: return a
return ((1-inverse(m,a)*m)//a)%m
class BIT:
def __init__(self,arr):
self.arr=arr
self.n=len(arr)-1
def update(self,x,v):
while x<=self.n:
self.arr[x]+=v
x+=x&-x
def query(self,x):
ans=0
while x:
ans+=self.arr[x]
x&=x-1
return ans
'''
class SMT:
def __init__(self,arr):
self.n=len(arr)-1
self.arr=[0]*(self.n<<2)
self.lazy=[0]*(self.n<<2)
def Build(l,r,rt):
if l==r:
self.arr[rt]=arr[l]
return
m=(l+r)>>1
Build(l,m,rt<<1)
Build(m+1,r,rt<<1|1)
self.pushup(rt)
Build(1,self.n,1)
def pushup(self,rt):
self.arr[rt]=self.arr[rt<<1]+self.arr[rt<<1|1]
def pushdown(self,rt,ln,rn):#lr,rn表区间数字数
if self.lazy[rt]:
self.lazy[rt<<1]+=self.lazy[rt]
self.lazy[rt<<1|1]=self.lazy[rt]
self.arr[rt<<1]+=self.lazy[rt]*ln
self.arr[rt<<1|1]+=self.lazy[rt]*rn
self.lazy[rt]=0
def update(self,L,R,c,l=1,r=None,rt=1):#L,R表示操作区间
if r==None: r=self.n
if L<=l and r<=R:
self.arr[rt]+=c*(r-l+1)
self.lazy[rt]+=c
return
m=(l+r)>>1
self.pushdown(rt,m-l+1,r-m)
if L<=m: self.update(L,R,c,l,m,rt<<1)
if R>m: self.update(L,R,c,m+1,r,rt<<1|1)
self.pushup(rt)
def query(self,L,R,l=1,r=None,rt=1):
if r==None: r=self.n
#print(L,R,l,r,rt)
if L<=l and R>=r:
return self.arr[rt]
m=(l+r)>>1
self.pushdown(rt,m-l+1,r-m)
ans=0
if L<=m: ans+=self.query(L,R,l,m,rt<<1)
if R>m: ans+=self.query(L,R,m+1,r,rt<<1|1)
return ans
'''
class DSU:#容量+路径压缩
def __init__(self,n):
self.c=[-1]*n
def same(self,x,y):
return self.find(x)==self.find(y)
def find(self,x):
if self.c[x]<0:
return x
self.c[x]=self.find(self.c[x])
return self.c[x]
def union(self,u,v):
u,v=self.find(u),self.find(v)
if u==v:
return False
if self.c[u]<self.c[v]:
u,v=v,u
self.c[u]+=self.c[v]
self.c[v]=u
return True
def size(self,x): return -self.c[self.find(x)]
class UFS:#秩+路径
def __init__(self,n):
self.parent=[i for i in range(n)]
self.ranks=[0]*n
def find(self,x):
if x!=self.parent[x]:
self.parent[x]=self.find(self.parent[x])
return self.parent[x]
def union(self,u,v):
pu,pv=self.find(u),self.find(v)
if pu==pv:
return False
if self.ranks[pu]>=self.ranks[pv]:
self.parent[pv]=pu
if self.ranks[pv]==self.ranks[pu]:
self.ranks[pu]+=1
else:
self.parent[pu]=pv
def Prime(n):
c=0
prime=[]
flag=[0]*(n+1)
for i in range(2,n+1):
if not flag[i]:
prime.append(i)
c+=1
for j in range(c):
if i*prime[j]>n: break
flag[i*prime[j]]=prime[j]
if i%prime[j]==0: break
return prime
def dij(s,graph):
d={}
d[s]=0
heap=[(0,s)]
seen=set()
while heap:
dis,u=heappop(heap)
if u in seen:
continue
for v in graph[u]:
if v not in d or d[v]>d[u]+graph[u][v]:
d[v]=d[u]+graph[u][v]
heappush(heap,(d[v],v))
return d
def GP(it): return [(ch,len(list(g))) for ch,g in groupby(it)]
class DLN:
def __init__(self,val):
self.val=val
self.pre=None
self.next=None
t=1
for i in range(t):
n=N()
a=RLL()
ans=0
for j in range(2):
pre=[0]
for i in range(n):
pre.append(pre[-1]+a[i])
for i in range(n-2):
k=len(bin(a[i]))-2
k=1<<k
for r in range(i+2,n):
if pre[r]-pre[i+1]>k:
break
if a[i]>a[r] and a[i]^a[r]==pre[r]-pre[i+1]:
ans+=1
a=a[::-1]
print(ans)
'''
sys.setrecursionlimit(200000)
import threading
threading.stack_size(10**8)
t=threading.Thread(target=main)
t.start()
t.join()
'''
``` | output | 1 | 87,790 | 12 | 175,581 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Yurii is sure he can do everything. Can he solve this task, though?
He has an array a consisting of n positive integers. Let's call a subarray a[l...r] good if the following conditions are simultaneously satisfied:
* l+1 ≤ r-1, i. e. the subarray has length at least 3;
* (a_l ⊕ a_r) = (a_{l+1}+a_{l+2}+…+a_{r-2}+a_{r-1}), where ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
In other words, a subarray is good if the bitwise XOR of the two border elements is equal to the sum of the rest of the elements.
Yurii wants to calculate the total number of good subarrays. What is it equal to?
An array c is a subarray of an array d if c can be obtained from d by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.
Input
The first line contains a single integer n (3 ≤ n ≤ 2⋅ 10^5) — the length of a.
The second line contains n integers a_1,a_2,…,a_n (1 ≤ a_i < 2^{30}) — elements of a.
Output
Output a single integer — the number of good subarrays.
Examples
Input
8
3 1 2 3 1 2 3 15
Output
6
Input
10
997230370 58052053 240970544 715275815 250707702 156801523 44100666 64791577 43523002 480196854
Output
2
Note
There are 6 good subarrays in the example:
* [3,1,2] (twice) because (3 ⊕ 2) = 1;
* [1,2,3] (twice) because (1 ⊕ 3) = 2;
* [2,3,1] because (2 ⊕ 1) = 3;
* [3,1,2,3,1,2,3,15] because (3 ⊕ 15) = (1+2+3+1+2+3). | instruction | 0 | 87,791 | 12 | 175,582 |
Tags: binary search, bitmasks, brute force, constructive algorithms, divide and conquer, two pointers
Correct Solution:
```
from sys import *
input = stdin.readline
def solve(n, a, t):
ans = 0
for i in range(n):
sum = 0
high1 = a[i].bit_length()-1
for j in range(i+1, n-1):
high2 = a[j+1].bit_length()-1
sum += a[j]
if(sum >= (1<<(high1+1))):
break
if((a[i]^a[j+1]) == sum and (t == 0 or high1 != high2)):
ans += 1
return ans
n = int(input())
a = list(map(int, input().split()))
ans = solve(n, a, 0)
a = a[::-1]
ans += solve(n, a, 1)
print(ans)
``` | output | 1 | 87,791 | 12 | 175,583 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Yurii is sure he can do everything. Can he solve this task, though?
He has an array a consisting of n positive integers. Let's call a subarray a[l...r] good if the following conditions are simultaneously satisfied:
* l+1 ≤ r-1, i. e. the subarray has length at least 3;
* (a_l ⊕ a_r) = (a_{l+1}+a_{l+2}+…+a_{r-2}+a_{r-1}), where ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
In other words, a subarray is good if the bitwise XOR of the two border elements is equal to the sum of the rest of the elements.
Yurii wants to calculate the total number of good subarrays. What is it equal to?
An array c is a subarray of an array d if c can be obtained from d by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.
Input
The first line contains a single integer n (3 ≤ n ≤ 2⋅ 10^5) — the length of a.
The second line contains n integers a_1,a_2,…,a_n (1 ≤ a_i < 2^{30}) — elements of a.
Output
Output a single integer — the number of good subarrays.
Examples
Input
8
3 1 2 3 1 2 3 15
Output
6
Input
10
997230370 58052053 240970544 715275815 250707702 156801523 44100666 64791577 43523002 480196854
Output
2
Note
There are 6 good subarrays in the example:
* [3,1,2] (twice) because (3 ⊕ 2) = 1;
* [1,2,3] (twice) because (1 ⊕ 3) = 2;
* [2,3,1] because (2 ⊕ 1) = 3;
* [3,1,2,3,1,2,3,15] because (3 ⊕ 15) = (1+2+3+1+2+3). | instruction | 0 | 87,792 | 12 | 175,584 |
Tags: binary search, bitmasks, brute force, constructive algorithms, divide and conquer, two pointers
Correct Solution:
```
def solve(a):
seen = set()
for i in range(len(a)):
c = 0
for j in range(i+2,len(a)):
c += a[j-1]
if a[i]^a[j] == c:
seen.add((i,j))
if c >= 2*a[i]:
break
for i in range(len(a)-1,-1,-1):
c = 0
for j in range(i-2,-1,-1):
c += a[j+1]
if a[i]^a[j] == c:
seen.add((j,i))
if c >= 2 *a[i]:
break
print(len(seen))
n = int(input())
a = list(map(int,input().split()))
solve(a)
``` | output | 1 | 87,792 | 12 | 175,585 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Yurii is sure he can do everything. Can he solve this task, though?
He has an array a consisting of n positive integers. Let's call a subarray a[l...r] good if the following conditions are simultaneously satisfied:
* l+1 ≤ r-1, i. e. the subarray has length at least 3;
* (a_l ⊕ a_r) = (a_{l+1}+a_{l+2}+…+a_{r-2}+a_{r-1}), where ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
In other words, a subarray is good if the bitwise XOR of the two border elements is equal to the sum of the rest of the elements.
Yurii wants to calculate the total number of good subarrays. What is it equal to?
An array c is a subarray of an array d if c can be obtained from d by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.
Input
The first line contains a single integer n (3 ≤ n ≤ 2⋅ 10^5) — the length of a.
The second line contains n integers a_1,a_2,…,a_n (1 ≤ a_i < 2^{30}) — elements of a.
Output
Output a single integer — the number of good subarrays.
Examples
Input
8
3 1 2 3 1 2 3 15
Output
6
Input
10
997230370 58052053 240970544 715275815 250707702 156801523 44100666 64791577 43523002 480196854
Output
2
Note
There are 6 good subarrays in the example:
* [3,1,2] (twice) because (3 ⊕ 2) = 1;
* [1,2,3] (twice) because (1 ⊕ 3) = 2;
* [2,3,1] because (2 ⊕ 1) = 3;
* [3,1,2,3,1,2,3,15] because (3 ⊕ 15) = (1+2+3+1+2+3).
Submitted Solution:
```
import math
n = int(input())
a = list(map(int, input().split()))
ans = 0
for i in range(2, n):
sum = 0
for j in reversed(range(0, i-1)):
sum += a[j+1]
ans += a[i] > a[j] and a[i]^a[j] == sum
if sum > 2*a[i] or a[j].bit_length() > a[i].bit_length():
break
a.reverse()
for i in range(2, n):
sum = 0
for j in reversed(range(0, i-1)):
sum += a[j+1]
ans += a[i] >= a[j] and a[i]^a[j] == sum
if sum > 2*a[i] or a[j].bit_length() > a[i].bit_length():
break
print(ans)
``` | instruction | 0 | 87,793 | 12 | 175,586 |
Yes | output | 1 | 87,793 | 12 | 175,587 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Yurii is sure he can do everything. Can he solve this task, though?
He has an array a consisting of n positive integers. Let's call a subarray a[l...r] good if the following conditions are simultaneously satisfied:
* l+1 ≤ r-1, i. e. the subarray has length at least 3;
* (a_l ⊕ a_r) = (a_{l+1}+a_{l+2}+…+a_{r-2}+a_{r-1}), where ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
In other words, a subarray is good if the bitwise XOR of the two border elements is equal to the sum of the rest of the elements.
Yurii wants to calculate the total number of good subarrays. What is it equal to?
An array c is a subarray of an array d if c can be obtained from d by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.
Input
The first line contains a single integer n (3 ≤ n ≤ 2⋅ 10^5) — the length of a.
The second line contains n integers a_1,a_2,…,a_n (1 ≤ a_i < 2^{30}) — elements of a.
Output
Output a single integer — the number of good subarrays.
Examples
Input
8
3 1 2 3 1 2 3 15
Output
6
Input
10
997230370 58052053 240970544 715275815 250707702 156801523 44100666 64791577 43523002 480196854
Output
2
Note
There are 6 good subarrays in the example:
* [3,1,2] (twice) because (3 ⊕ 2) = 1;
* [1,2,3] (twice) because (1 ⊕ 3) = 2;
* [2,3,1] because (2 ⊕ 1) = 3;
* [3,1,2,3,1,2,3,15] because (3 ⊕ 15) = (1+2+3+1+2+3).
Submitted Solution:
```
def f(a):
ans = 0
for i in range(n - 2):
s = 0
for j in range(i + 2, n):
s += a[j - 1]
ans += a[i] > a[j] and a[i] ^ a[j] == s
if s > 2 * a[i]: break
return ans
read = lambda: map(int, input().split())
n = int(input())
a = list(read())
print(f(a) + f(a[::-1]))
``` | instruction | 0 | 87,794 | 12 | 175,588 |
Yes | output | 1 | 87,794 | 12 | 175,589 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Yurii is sure he can do everything. Can he solve this task, though?
He has an array a consisting of n positive integers. Let's call a subarray a[l...r] good if the following conditions are simultaneously satisfied:
* l+1 ≤ r-1, i. e. the subarray has length at least 3;
* (a_l ⊕ a_r) = (a_{l+1}+a_{l+2}+…+a_{r-2}+a_{r-1}), where ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
In other words, a subarray is good if the bitwise XOR of the two border elements is equal to the sum of the rest of the elements.
Yurii wants to calculate the total number of good subarrays. What is it equal to?
An array c is a subarray of an array d if c can be obtained from d by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.
Input
The first line contains a single integer n (3 ≤ n ≤ 2⋅ 10^5) — the length of a.
The second line contains n integers a_1,a_2,…,a_n (1 ≤ a_i < 2^{30}) — elements of a.
Output
Output a single integer — the number of good subarrays.
Examples
Input
8
3 1 2 3 1 2 3 15
Output
6
Input
10
997230370 58052053 240970544 715275815 250707702 156801523 44100666 64791577 43523002 480196854
Output
2
Note
There are 6 good subarrays in the example:
* [3,1,2] (twice) because (3 ⊕ 2) = 1;
* [1,2,3] (twice) because (1 ⊕ 3) = 2;
* [2,3,1] because (2 ⊕ 1) = 3;
* [3,1,2,3,1,2,3,15] because (3 ⊕ 15) = (1+2+3+1+2+3).
Submitted Solution:
```
n = int(input())
a = list(map(int, input().split(' ')))
ans = 0
for i in range(n):
sum = 0
for j in range(i + 2, n, 1):
sum += a[j - 1]
if (sum >= a[i] + a[i]):
break
if ((a[i] ^ a[j]) == sum and a[i] >= a[j]):
ans += 1
for i in range(n):
sum = 0
for j in range(i - 2, -1, -1):
sum += a[j + 1]
if (sum >= a[i] + a[i]):
break
if ((a[i] ^ a[j]) == sum and a[i] > a[j]):
ans += 1
print(ans)
``` | instruction | 0 | 87,795 | 12 | 175,590 |
Yes | output | 1 | 87,795 | 12 | 175,591 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Yurii is sure he can do everything. Can he solve this task, though?
He has an array a consisting of n positive integers. Let's call a subarray a[l...r] good if the following conditions are simultaneously satisfied:
* l+1 ≤ r-1, i. e. the subarray has length at least 3;
* (a_l ⊕ a_r) = (a_{l+1}+a_{l+2}+…+a_{r-2}+a_{r-1}), where ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
In other words, a subarray is good if the bitwise XOR of the two border elements is equal to the sum of the rest of the elements.
Yurii wants to calculate the total number of good subarrays. What is it equal to?
An array c is a subarray of an array d if c can be obtained from d by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.
Input
The first line contains a single integer n (3 ≤ n ≤ 2⋅ 10^5) — the length of a.
The second line contains n integers a_1,a_2,…,a_n (1 ≤ a_i < 2^{30}) — elements of a.
Output
Output a single integer — the number of good subarrays.
Examples
Input
8
3 1 2 3 1 2 3 15
Output
6
Input
10
997230370 58052053 240970544 715275815 250707702 156801523 44100666 64791577 43523002 480196854
Output
2
Note
There are 6 good subarrays in the example:
* [3,1,2] (twice) because (3 ⊕ 2) = 1;
* [1,2,3] (twice) because (1 ⊕ 3) = 2;
* [2,3,1] because (2 ⊕ 1) = 3;
* [3,1,2,3,1,2,3,15] because (3 ⊕ 15) = (1+2+3+1+2+3).
Submitted Solution:
```
import math
n = int(input())
a = list(map(int, input().split()))
ans = 0
for i in range(2, n):
sum = 0
for j in reversed(range(0, i-1)):
sum += a[j+1]
ans += a[i] < a[j] and a[i]^a[j] == sum
if sum > 2*a[i] or int(math.log2(a[j])) > int(math.log2(a[i])):
break
a.reverse()
for i in range(2, n):
sum = 0
for j in reversed(range(0, i-1)):
sum += a[j+1]
ans += a[i] >= a[j] and a[i]^a[j] == sum
if sum > 2*a[i] or int(math.log2(a[j])) > int(math.log2(a[i])):
break
print(ans)
``` | instruction | 0 | 87,796 | 12 | 175,592 |
No | output | 1 | 87,796 | 12 | 175,593 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Yurii is sure he can do everything. Can he solve this task, though?
He has an array a consisting of n positive integers. Let's call a subarray a[l...r] good if the following conditions are simultaneously satisfied:
* l+1 ≤ r-1, i. e. the subarray has length at least 3;
* (a_l ⊕ a_r) = (a_{l+1}+a_{l+2}+…+a_{r-2}+a_{r-1}), where ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
In other words, a subarray is good if the bitwise XOR of the two border elements is equal to the sum of the rest of the elements.
Yurii wants to calculate the total number of good subarrays. What is it equal to?
An array c is a subarray of an array d if c can be obtained from d by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.
Input
The first line contains a single integer n (3 ≤ n ≤ 2⋅ 10^5) — the length of a.
The second line contains n integers a_1,a_2,…,a_n (1 ≤ a_i < 2^{30}) — elements of a.
Output
Output a single integer — the number of good subarrays.
Examples
Input
8
3 1 2 3 1 2 3 15
Output
6
Input
10
997230370 58052053 240970544 715275815 250707702 156801523 44100666 64791577 43523002 480196854
Output
2
Note
There are 6 good subarrays in the example:
* [3,1,2] (twice) because (3 ⊕ 2) = 1;
* [1,2,3] (twice) because (1 ⊕ 3) = 2;
* [2,3,1] because (2 ⊕ 1) = 3;
* [3,1,2,3,1,2,3,15] because (3 ⊕ 15) = (1+2+3+1+2+3).
Submitted Solution:
```
n = int(input())
a = list(map(int,input().split()))
cnt = 0
for i in range(n-2):
s = 0
for j in range(i+1,n-1):
s += a[j]
if (a[i] ^ a[j] == s):
cnt += 1
k = len(bin(a[i])[2:])
if (s >= (1 << (k+1))):
break
a = list(reversed(a))
for i in range(n-2):
s = 0
for j in range(i+1,n-1):
s += a[j]
if (a[i] ^ a[j+1] == s):
cnt += 1
k = len(bin(a[i])[2:])
if (s >= (1 << (k+1))):
break
print(cnt)
``` | instruction | 0 | 87,797 | 12 | 175,594 |
No | output | 1 | 87,797 | 12 | 175,595 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Yurii is sure he can do everything. Can he solve this task, though?
He has an array a consisting of n positive integers. Let's call a subarray a[l...r] good if the following conditions are simultaneously satisfied:
* l+1 ≤ r-1, i. e. the subarray has length at least 3;
* (a_l ⊕ a_r) = (a_{l+1}+a_{l+2}+…+a_{r-2}+a_{r-1}), where ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
In other words, a subarray is good if the bitwise XOR of the two border elements is equal to the sum of the rest of the elements.
Yurii wants to calculate the total number of good subarrays. What is it equal to?
An array c is a subarray of an array d if c can be obtained from d by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.
Input
The first line contains a single integer n (3 ≤ n ≤ 2⋅ 10^5) — the length of a.
The second line contains n integers a_1,a_2,…,a_n (1 ≤ a_i < 2^{30}) — elements of a.
Output
Output a single integer — the number of good subarrays.
Examples
Input
8
3 1 2 3 1 2 3 15
Output
6
Input
10
997230370 58052053 240970544 715275815 250707702 156801523 44100666 64791577 43523002 480196854
Output
2
Note
There are 6 good subarrays in the example:
* [3,1,2] (twice) because (3 ⊕ 2) = 1;
* [1,2,3] (twice) because (1 ⊕ 3) = 2;
* [2,3,1] because (2 ⊕ 1) = 3;
* [3,1,2,3,1,2,3,15] because (3 ⊕ 15) = (1+2+3+1+2+3).
Submitted Solution:
```
def checkgood(a,n):
if len(a)>=3:
if (a[0]^a[-1])==sum(a[1:n-1]):
return True
else:
return False
else:
return False
t = int(input())
a = (input().split())
a = [int(i) for i in a]
x = 0
for i in range(t):
for j in range(i,t+1):
p = a[i:j]
if checkgood(p,len(p)):
print(p)
x += 1
print(x)
``` | instruction | 0 | 87,798 | 12 | 175,596 |
No | output | 1 | 87,798 | 12 | 175,597 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Yurii is sure he can do everything. Can he solve this task, though?
He has an array a consisting of n positive integers. Let's call a subarray a[l...r] good if the following conditions are simultaneously satisfied:
* l+1 ≤ r-1, i. e. the subarray has length at least 3;
* (a_l ⊕ a_r) = (a_{l+1}+a_{l+2}+…+a_{r-2}+a_{r-1}), where ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
In other words, a subarray is good if the bitwise XOR of the two border elements is equal to the sum of the rest of the elements.
Yurii wants to calculate the total number of good subarrays. What is it equal to?
An array c is a subarray of an array d if c can be obtained from d by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.
Input
The first line contains a single integer n (3 ≤ n ≤ 2⋅ 10^5) — the length of a.
The second line contains n integers a_1,a_2,…,a_n (1 ≤ a_i < 2^{30}) — elements of a.
Output
Output a single integer — the number of good subarrays.
Examples
Input
8
3 1 2 3 1 2 3 15
Output
6
Input
10
997230370 58052053 240970544 715275815 250707702 156801523 44100666 64791577 43523002 480196854
Output
2
Note
There are 6 good subarrays in the example:
* [3,1,2] (twice) because (3 ⊕ 2) = 1;
* [1,2,3] (twice) because (1 ⊕ 3) = 2;
* [2,3,1] because (2 ⊕ 1) = 3;
* [3,1,2,3,1,2,3,15] because (3 ⊕ 15) = (1+2+3+1+2+3).
Submitted Solution:
```
l = int(input())
data = [int(x) for x in input().split()]
fl = False
def func(num):
a = []
while True:
a.append(num % 2)
num //= 2
if num < 1:
return len(a)
def main_f(data):
count = 0
for i in range(l):
s = 0
a = data[i]
func_a = func(a)
for j in range(i + 1, l):
b = data[j]
if s == a ^ b and (not fl or fl and a > b) and j != i + 1:
count += 1
elif func(s + data[j]) <= func_a:
s += data[j]
else:
break
return count
f_res = main_f(data)
fl = True
s_res = main_f(data[-1::-1])
print(f_res + s_res)
``` | instruction | 0 | 87,799 | 12 | 175,598 |
No | output | 1 | 87,799 | 12 | 175,599 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have 2n integers 1, 2, ..., 2n. You have to redistribute these 2n elements into n pairs. After that, you choose x pairs and take minimum elements from them, and from the other n - x pairs, you take maximum elements.
Your goal is to obtain the set of numbers \\{b_1, b_2, ..., b_n\} as the result of taking elements from the pairs.
What is the number of different x-s (0 ≤ x ≤ n) such that it's possible to obtain the set b if for each x you can choose how to distribute numbers into pairs and from which x pairs choose minimum elements?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the integer n (1 ≤ n ≤ 2 ⋅ 10^5).
The second line of each test case contains n integers b_1, b_2, ..., b_n (1 ≤ b_1 < b_2 < ... < b_n ≤ 2n) — the set you'd like to get.
It's guaranteed that the sum of n over test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case, print one number — the number of different x-s such that it's possible to obtain the set b.
Example
Input
3
1
1
5
1 4 5 9 10
2
3 4
Output
1
3
1
Note
In the first test case, x = 1 is the only option: you have one pair (1, 2) and choose the minimum from this pair.
In the second test case, there are three possible x-s. If x = 1, then you can form the following pairs: (1, 6), (2, 4), (3, 5), (7, 9), (8, 10). You can take minimum from (1, 6) (equal to 1) and the maximum elements from all other pairs to get set b.
If x = 2, you can form pairs (1, 2), (3, 4), (5, 6), (7, 9), (8, 10) and take the minimum elements from (1, 2), (5, 6) and the maximum elements from the other pairs.
If x = 3, you can form pairs (1, 3), (4, 6), (5, 7), (2, 9), (8, 10) and take the minimum elements from (1, 3), (4, 6), (5, 7).
In the third test case, x = 0 is the only option: you can form pairs (1, 3), (2, 4) and take the maximum elements from both of them. | instruction | 0 | 87,800 | 12 | 175,600 |
Tags: binary search, constructive algorithms, greedy, two pointers
Correct Solution:
```
import sys
from bisect import bisect_left
input = sys.stdin.buffer.readline
T = int(input())
for _ in range(T):
n, b = int(input()), list(map(int, input().split()))
setb = set(b)
rb = [ u for u in range(1, 2*n+1) if not u in setb ]
sm = [0]*(n+2)
for i, u in enumerate(rb):
p = bisect_left(b, u)
p1, p2 = max(p-i, 0), n-1-i
p3, p4 = 0+n-i, min(p-1+n-i, n)
if p1 <= p2: sm[p1] += 1; sm[p2+1] -= 1
if p3 <= p4: sm[p3] += 1; sm[p4+1] -= 1
for u in range(1, n+2):
sm[u] += sm[u-1]
cc = sum([ 1 for u in sm if u == n ])
print(cc)
``` | output | 1 | 87,800 | 12 | 175,601 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have 2n integers 1, 2, ..., 2n. You have to redistribute these 2n elements into n pairs. After that, you choose x pairs and take minimum elements from them, and from the other n - x pairs, you take maximum elements.
Your goal is to obtain the set of numbers \\{b_1, b_2, ..., b_n\} as the result of taking elements from the pairs.
What is the number of different x-s (0 ≤ x ≤ n) such that it's possible to obtain the set b if for each x you can choose how to distribute numbers into pairs and from which x pairs choose minimum elements?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the integer n (1 ≤ n ≤ 2 ⋅ 10^5).
The second line of each test case contains n integers b_1, b_2, ..., b_n (1 ≤ b_1 < b_2 < ... < b_n ≤ 2n) — the set you'd like to get.
It's guaranteed that the sum of n over test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case, print one number — the number of different x-s such that it's possible to obtain the set b.
Example
Input
3
1
1
5
1 4 5 9 10
2
3 4
Output
1
3
1
Note
In the first test case, x = 1 is the only option: you have one pair (1, 2) and choose the minimum from this pair.
In the second test case, there are three possible x-s. If x = 1, then you can form the following pairs: (1, 6), (2, 4), (3, 5), (7, 9), (8, 10). You can take minimum from (1, 6) (equal to 1) and the maximum elements from all other pairs to get set b.
If x = 2, you can form pairs (1, 2), (3, 4), (5, 6), (7, 9), (8, 10) and take the minimum elements from (1, 2), (5, 6) and the maximum elements from the other pairs.
If x = 3, you can form pairs (1, 3), (4, 6), (5, 7), (2, 9), (8, 10) and take the minimum elements from (1, 3), (4, 6), (5, 7).
In the third test case, x = 0 is the only option: you can form pairs (1, 3), (2, 4) and take the maximum elements from both of them. | instruction | 0 | 87,801 | 12 | 175,602 |
Tags: binary search, constructive algorithms, greedy, two pointers
Correct Solution:
```
t=int(input())
for _ in range(t):
n=int(input())
b=list(map(int,input().split()))
s=set(b)
arr=[1]*(2*n)
pos=0
mini=0
for i in range(n):
pos=max(pos,b[i])
temp=pos
for j in range(pos,2*n):
if(((j+1) not in s) and arr[j]==1):
mini+=1
arr[j]=0
arr[b[i]-1]=0
pos=j+1
break
if(pos==temp):
pos=2*n
arr=[1]*(2*n)
pos=2*n-1
maxi=0
for i in range(n-1,-1,-1):
pos=min(pos,b[i]-2)
temp=pos
for j in range(pos,-1,-1):
if(((j+1) not in s) and arr[j]==1):
maxi+=1
arr[j]=0
arr[b[i]-1]=0
pos=j-1
break
if(pos==temp):
pos=-1
maxi=n-maxi
print(mini-maxi+1)
``` | output | 1 | 87,801 | 12 | 175,603 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have 2n integers 1, 2, ..., 2n. You have to redistribute these 2n elements into n pairs. After that, you choose x pairs and take minimum elements from them, and from the other n - x pairs, you take maximum elements.
Your goal is to obtain the set of numbers \\{b_1, b_2, ..., b_n\} as the result of taking elements from the pairs.
What is the number of different x-s (0 ≤ x ≤ n) such that it's possible to obtain the set b if for each x you can choose how to distribute numbers into pairs and from which x pairs choose minimum elements?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the integer n (1 ≤ n ≤ 2 ⋅ 10^5).
The second line of each test case contains n integers b_1, b_2, ..., b_n (1 ≤ b_1 < b_2 < ... < b_n ≤ 2n) — the set you'd like to get.
It's guaranteed that the sum of n over test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case, print one number — the number of different x-s such that it's possible to obtain the set b.
Example
Input
3
1
1
5
1 4 5 9 10
2
3 4
Output
1
3
1
Note
In the first test case, x = 1 is the only option: you have one pair (1, 2) and choose the minimum from this pair.
In the second test case, there are three possible x-s. If x = 1, then you can form the following pairs: (1, 6), (2, 4), (3, 5), (7, 9), (8, 10). You can take minimum from (1, 6) (equal to 1) and the maximum elements from all other pairs to get set b.
If x = 2, you can form pairs (1, 2), (3, 4), (5, 6), (7, 9), (8, 10) and take the minimum elements from (1, 2), (5, 6) and the maximum elements from the other pairs.
If x = 3, you can form pairs (1, 3), (4, 6), (5, 7), (2, 9), (8, 10) and take the minimum elements from (1, 3), (4, 6), (5, 7).
In the third test case, x = 0 is the only option: you can form pairs (1, 3), (2, 4) and take the maximum elements from both of them. | instruction | 0 | 87,802 | 12 | 175,604 |
Tags: binary search, constructive algorithms, greedy, two pointers
Correct Solution:
```
import os,io
input=io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
INF=10**9+1
t=int(input())
for _ in range(t):
n=int(input())
b=list(map(int,input().split()))
maxofmins=0
minofmaxs=0
maxofmaxs=0
minofmins=0
pointer=1
grid=[-1]*(2*n)
for i in b:
grid[i-1]=1
tmpmax=0
tmpmin=0
curr=0
for i in range(2*n):
curr+=grid[i]
tmpmax=max(tmpmax,curr)
tmpmin=min(tmpmin,curr)
print(n+tmpmin-tmpmax+1)
``` | output | 1 | 87,802 | 12 | 175,605 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have 2n integers 1, 2, ..., 2n. You have to redistribute these 2n elements into n pairs. After that, you choose x pairs and take minimum elements from them, and from the other n - x pairs, you take maximum elements.
Your goal is to obtain the set of numbers \\{b_1, b_2, ..., b_n\} as the result of taking elements from the pairs.
What is the number of different x-s (0 ≤ x ≤ n) such that it's possible to obtain the set b if for each x you can choose how to distribute numbers into pairs and from which x pairs choose minimum elements?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the integer n (1 ≤ n ≤ 2 ⋅ 10^5).
The second line of each test case contains n integers b_1, b_2, ..., b_n (1 ≤ b_1 < b_2 < ... < b_n ≤ 2n) — the set you'd like to get.
It's guaranteed that the sum of n over test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case, print one number — the number of different x-s such that it's possible to obtain the set b.
Example
Input
3
1
1
5
1 4 5 9 10
2
3 4
Output
1
3
1
Note
In the first test case, x = 1 is the only option: you have one pair (1, 2) and choose the minimum from this pair.
In the second test case, there are three possible x-s. If x = 1, then you can form the following pairs: (1, 6), (2, 4), (3, 5), (7, 9), (8, 10). You can take minimum from (1, 6) (equal to 1) and the maximum elements from all other pairs to get set b.
If x = 2, you can form pairs (1, 2), (3, 4), (5, 6), (7, 9), (8, 10) and take the minimum elements from (1, 2), (5, 6) and the maximum elements from the other pairs.
If x = 3, you can form pairs (1, 3), (4, 6), (5, 7), (2, 9), (8, 10) and take the minimum elements from (1, 3), (4, 6), (5, 7).
In the third test case, x = 0 is the only option: you can form pairs (1, 3), (2, 4) and take the maximum elements from both of them. | instruction | 0 | 87,803 | 12 | 175,606 |
Tags: binary search, constructive algorithms, greedy, two pointers
Correct Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
b = list(map(int, input().split()))
max_force = 0
min_force = 0
legacy = 0
possibility = 0
for i in b:
jump = i - legacy - 1
possibility += jump
if possibility > 0:
possibility -= 1
else:
min_force += 1
legacy = i
possibility = 0
legacy = 2 * n + 1
for i in reversed(b):
jump = legacy - i - 1
possibility += jump
if possibility > 0:
possibility -= 1
else:
max_force += 1
legacy = i
print(n - max_force - min_force + 1)
``` | output | 1 | 87,803 | 12 | 175,607 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have 2n integers 1, 2, ..., 2n. You have to redistribute these 2n elements into n pairs. After that, you choose x pairs and take minimum elements from them, and from the other n - x pairs, you take maximum elements.
Your goal is to obtain the set of numbers \\{b_1, b_2, ..., b_n\} as the result of taking elements from the pairs.
What is the number of different x-s (0 ≤ x ≤ n) such that it's possible to obtain the set b if for each x you can choose how to distribute numbers into pairs and from which x pairs choose minimum elements?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the integer n (1 ≤ n ≤ 2 ⋅ 10^5).
The second line of each test case contains n integers b_1, b_2, ..., b_n (1 ≤ b_1 < b_2 < ... < b_n ≤ 2n) — the set you'd like to get.
It's guaranteed that the sum of n over test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case, print one number — the number of different x-s such that it's possible to obtain the set b.
Example
Input
3
1
1
5
1 4 5 9 10
2
3 4
Output
1
3
1
Note
In the first test case, x = 1 is the only option: you have one pair (1, 2) and choose the minimum from this pair.
In the second test case, there are three possible x-s. If x = 1, then you can form the following pairs: (1, 6), (2, 4), (3, 5), (7, 9), (8, 10). You can take minimum from (1, 6) (equal to 1) and the maximum elements from all other pairs to get set b.
If x = 2, you can form pairs (1, 2), (3, 4), (5, 6), (7, 9), (8, 10) and take the minimum elements from (1, 2), (5, 6) and the maximum elements from the other pairs.
If x = 3, you can form pairs (1, 3), (4, 6), (5, 7), (2, 9), (8, 10) and take the minimum elements from (1, 3), (4, 6), (5, 7).
In the third test case, x = 0 is the only option: you can form pairs (1, 3), (2, 4) and take the maximum elements from both of them. | instruction | 0 | 87,804 | 12 | 175,608 |
Tags: binary search, constructive algorithms, greedy, two pointers
Correct Solution:
```
for _ in range(int(input())):
n = int(input())
b = list(map(int, input().split()))
used = [0] * (2*n+1)
for x in b:
used[x] = 1
a = [x for x in range(1, 2*n+1) if used[x]==0]
max_ = 0
l, r = 0, len(a) - 1
for x in b[::-1]:
if x < a[r]:
max_ += 1
r -= 1
else:
l += 1
min_ = 0
l, r = 0, len(b) - 1
for x in a[::-1]:
if x < b[r]:
r -= 1
else:
min_ += 1
l += 1
print(max_ - min_ + 1)
``` | output | 1 | 87,804 | 12 | 175,609 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have 2n integers 1, 2, ..., 2n. You have to redistribute these 2n elements into n pairs. After that, you choose x pairs and take minimum elements from them, and from the other n - x pairs, you take maximum elements.
Your goal is to obtain the set of numbers \\{b_1, b_2, ..., b_n\} as the result of taking elements from the pairs.
What is the number of different x-s (0 ≤ x ≤ n) such that it's possible to obtain the set b if for each x you can choose how to distribute numbers into pairs and from which x pairs choose minimum elements?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the integer n (1 ≤ n ≤ 2 ⋅ 10^5).
The second line of each test case contains n integers b_1, b_2, ..., b_n (1 ≤ b_1 < b_2 < ... < b_n ≤ 2n) — the set you'd like to get.
It's guaranteed that the sum of n over test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case, print one number — the number of different x-s such that it's possible to obtain the set b.
Example
Input
3
1
1
5
1 4 5 9 10
2
3 4
Output
1
3
1
Note
In the first test case, x = 1 is the only option: you have one pair (1, 2) and choose the minimum from this pair.
In the second test case, there are three possible x-s. If x = 1, then you can form the following pairs: (1, 6), (2, 4), (3, 5), (7, 9), (8, 10). You can take minimum from (1, 6) (equal to 1) and the maximum elements from all other pairs to get set b.
If x = 2, you can form pairs (1, 2), (3, 4), (5, 6), (7, 9), (8, 10) and take the minimum elements from (1, 2), (5, 6) and the maximum elements from the other pairs.
If x = 3, you can form pairs (1, 3), (4, 6), (5, 7), (2, 9), (8, 10) and take the minimum elements from (1, 3), (4, 6), (5, 7).
In the third test case, x = 0 is the only option: you can form pairs (1, 3), (2, 4) and take the maximum elements from both of them. | instruction | 0 | 87,805 | 12 | 175,610 |
Tags: binary search, constructive algorithms, greedy, two pointers
Correct Solution:
```
tests = int(input())
for t in range(tests):
n = int(input())
ls = list(map(int, input().split()))
if min(ls) == (n+1) or max(ls) == (n):
print(1)
else:
#right to left
mini = 0
curr = 2*n
idx = n-1
remain = 0
while idx >= 0:
remain += curr - ls[idx]
if remain > 0:
remain -= 1
mini += 1
curr = ls[idx]-1
idx -= 1
#left to right
maxi = 0
curr = 1
idx = 0
remain = 0
while idx < n:
remain += ls[idx] - curr
if remain > 0:
remain -= 1
maxi += 1
curr = ls[idx]+1
idx += 1
print(mini - (n-maxi)+1)
``` | output | 1 | 87,805 | 12 | 175,611 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have 2n integers 1, 2, ..., 2n. You have to redistribute these 2n elements into n pairs. After that, you choose x pairs and take minimum elements from them, and from the other n - x pairs, you take maximum elements.
Your goal is to obtain the set of numbers \\{b_1, b_2, ..., b_n\} as the result of taking elements from the pairs.
What is the number of different x-s (0 ≤ x ≤ n) such that it's possible to obtain the set b if for each x you can choose how to distribute numbers into pairs and from which x pairs choose minimum elements?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the integer n (1 ≤ n ≤ 2 ⋅ 10^5).
The second line of each test case contains n integers b_1, b_2, ..., b_n (1 ≤ b_1 < b_2 < ... < b_n ≤ 2n) — the set you'd like to get.
It's guaranteed that the sum of n over test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case, print one number — the number of different x-s such that it's possible to obtain the set b.
Example
Input
3
1
1
5
1 4 5 9 10
2
3 4
Output
1
3
1
Note
In the first test case, x = 1 is the only option: you have one pair (1, 2) and choose the minimum from this pair.
In the second test case, there are three possible x-s. If x = 1, then you can form the following pairs: (1, 6), (2, 4), (3, 5), (7, 9), (8, 10). You can take minimum from (1, 6) (equal to 1) and the maximum elements from all other pairs to get set b.
If x = 2, you can form pairs (1, 2), (3, 4), (5, 6), (7, 9), (8, 10) and take the minimum elements from (1, 2), (5, 6) and the maximum elements from the other pairs.
If x = 3, you can form pairs (1, 3), (4, 6), (5, 7), (2, 9), (8, 10) and take the minimum elements from (1, 3), (4, 6), (5, 7).
In the third test case, x = 0 is the only option: you can form pairs (1, 3), (2, 4) and take the maximum elements from both of them. | instruction | 0 | 87,806 | 12 | 175,612 |
Tags: binary search, constructive algorithms, greedy, two pointers
Correct Solution:
```
#! /usr/bin/env python3
import sys
t = int(sys.stdin.readline())
for ti in range(t):
n = int(sys.stdin.readline())
b = [int(a) for a in sys.stdin.readline().split()]
max_min = -1
min_max = n
prev = 0
E = n
U = 0
for i in range(n):
leaved = b[i] - prev - 1
prev = b[i]
U = max(0, U - leaved) + 1
E = E - leaved
if E > U:
max_min = i
elif E == U:
max_min = i
break
elif E < U:
break
E = n
U = 0
prev = 2 * n + 1
for i in range(n):
leaved = prev - b[n - i - 1] - 1
prev = b[n - i - 1]
U = max(0, U - leaved) + 1
E = E - leaved
if E > U:
min_max = n - i - 1
elif E == U:
min_max = n - i - 1
break
elif E < U:
break
print(max_min - min_max + 2)
``` | output | 1 | 87,806 | 12 | 175,613 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have 2n integers 1, 2, ..., 2n. You have to redistribute these 2n elements into n pairs. After that, you choose x pairs and take minimum elements from them, and from the other n - x pairs, you take maximum elements.
Your goal is to obtain the set of numbers \\{b_1, b_2, ..., b_n\} as the result of taking elements from the pairs.
What is the number of different x-s (0 ≤ x ≤ n) such that it's possible to obtain the set b if for each x you can choose how to distribute numbers into pairs and from which x pairs choose minimum elements?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the integer n (1 ≤ n ≤ 2 ⋅ 10^5).
The second line of each test case contains n integers b_1, b_2, ..., b_n (1 ≤ b_1 < b_2 < ... < b_n ≤ 2n) — the set you'd like to get.
It's guaranteed that the sum of n over test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case, print one number — the number of different x-s such that it's possible to obtain the set b.
Example
Input
3
1
1
5
1 4 5 9 10
2
3 4
Output
1
3
1
Note
In the first test case, x = 1 is the only option: you have one pair (1, 2) and choose the minimum from this pair.
In the second test case, there are three possible x-s. If x = 1, then you can form the following pairs: (1, 6), (2, 4), (3, 5), (7, 9), (8, 10). You can take minimum from (1, 6) (equal to 1) and the maximum elements from all other pairs to get set b.
If x = 2, you can form pairs (1, 2), (3, 4), (5, 6), (7, 9), (8, 10) and take the minimum elements from (1, 2), (5, 6) and the maximum elements from the other pairs.
If x = 3, you can form pairs (1, 3), (4, 6), (5, 7), (2, 9), (8, 10) and take the minimum elements from (1, 3), (4, 6), (5, 7).
In the third test case, x = 0 is the only option: you can form pairs (1, 3), (2, 4) and take the maximum elements from both of them. | instruction | 0 | 87,807 | 12 | 175,614 |
Tags: binary search, constructive algorithms, greedy, two pointers
Correct Solution:
```
for _ in range(int(input())):
n = int(input())
b = [int(x) for x in input().split()]
j, count = 0, 0
bad_left, bad_right = 0, 0
for i in range(1, 2*n+1):
if j < n and b[j] == i:
j += 1
if count == 0:
bad_left += 1
else:
count -= 1
else:
count += 1
j, count = n-1, 0
for i in range(2*n, 0, -1):
if j > -1 and b[j] == i:
j -= 1
if count == 0:
bad_right += 1
else:
count -= 1
else:
count += 1
print(n+1 - bad_left - bad_right)
``` | output | 1 | 87,807 | 12 | 175,615 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have 2n integers 1, 2, ..., 2n. You have to redistribute these 2n elements into n pairs. After that, you choose x pairs and take minimum elements from them, and from the other n - x pairs, you take maximum elements.
Your goal is to obtain the set of numbers \\{b_1, b_2, ..., b_n\} as the result of taking elements from the pairs.
What is the number of different x-s (0 ≤ x ≤ n) such that it's possible to obtain the set b if for each x you can choose how to distribute numbers into pairs and from which x pairs choose minimum elements?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the integer n (1 ≤ n ≤ 2 ⋅ 10^5).
The second line of each test case contains n integers b_1, b_2, ..., b_n (1 ≤ b_1 < b_2 < ... < b_n ≤ 2n) — the set you'd like to get.
It's guaranteed that the sum of n over test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case, print one number — the number of different x-s such that it's possible to obtain the set b.
Example
Input
3
1
1
5
1 4 5 9 10
2
3 4
Output
1
3
1
Note
In the first test case, x = 1 is the only option: you have one pair (1, 2) and choose the minimum from this pair.
In the second test case, there are three possible x-s. If x = 1, then you can form the following pairs: (1, 6), (2, 4), (3, 5), (7, 9), (8, 10). You can take minimum from (1, 6) (equal to 1) and the maximum elements from all other pairs to get set b.
If x = 2, you can form pairs (1, 2), (3, 4), (5, 6), (7, 9), (8, 10) and take the minimum elements from (1, 2), (5, 6) and the maximum elements from the other pairs.
If x = 3, you can form pairs (1, 3), (4, 6), (5, 7), (2, 9), (8, 10) and take the minimum elements from (1, 3), (4, 6), (5, 7).
In the third test case, x = 0 is the only option: you can form pairs (1, 3), (2, 4) and take the maximum elements from both of them.
Submitted Solution:
```
import sys
def main():
def modst(a, s):
ret = 1
while s:
if s % 2:
ret = ret * a % mod
a = a * a % mod
s //= 2
return ret
def Cnk(n, k):
return (k <= n and n >= 0) * ((f[n] * modst((f[k] * f[n - k]) % mod, mod - 2)) % mod)
#x, w = map(int, sys.stdin.readline().split())
#a,b,c = map(int, sys.stdin.readline().split())
n = int(sys.stdin.readline().strip())
#a, w = map(int, sys.stdin.readline().split())
q = list(map(int, sys.stdin.readline().split()))
#q = sorted(list(map(int, sys.stdin.readline().split())), reverse=True)
'''mod = 998244353
f = [1, 1]
for i in range(2, 200007):
f.append((f[-1] * i) % mod)
a = 0
for i in range(2 - n % 2, n + 1, 2):
a = (a + Cnk(max(((n - i) // 2) + i - 1, 0), i - 1)) % mod
print((a * modst(modst(2, n), mod - 2)) % mod)'''
s = set(q)
f, ans, res, l, r = 2 * n, 0, 0, 0, 0
for i in range(1, f + 1):
if i in s:
if l:
l -= 1
ans += 1
else:
l += 1
for i in range(f, 0, -1):
if i in s:
if r:
r -= 1
res += 1
else:
r += 1
print(abs(ans + res - n) + 1)
for i in range(int(input())):
main()
``` | instruction | 0 | 87,808 | 12 | 175,616 |
Yes | output | 1 | 87,808 | 12 | 175,617 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have 2n integers 1, 2, ..., 2n. You have to redistribute these 2n elements into n pairs. After that, you choose x pairs and take minimum elements from them, and from the other n - x pairs, you take maximum elements.
Your goal is to obtain the set of numbers \\{b_1, b_2, ..., b_n\} as the result of taking elements from the pairs.
What is the number of different x-s (0 ≤ x ≤ n) such that it's possible to obtain the set b if for each x you can choose how to distribute numbers into pairs and from which x pairs choose minimum elements?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the integer n (1 ≤ n ≤ 2 ⋅ 10^5).
The second line of each test case contains n integers b_1, b_2, ..., b_n (1 ≤ b_1 < b_2 < ... < b_n ≤ 2n) — the set you'd like to get.
It's guaranteed that the sum of n over test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case, print one number — the number of different x-s such that it's possible to obtain the set b.
Example
Input
3
1
1
5
1 4 5 9 10
2
3 4
Output
1
3
1
Note
In the first test case, x = 1 is the only option: you have one pair (1, 2) and choose the minimum from this pair.
In the second test case, there are three possible x-s. If x = 1, then you can form the following pairs: (1, 6), (2, 4), (3, 5), (7, 9), (8, 10). You can take minimum from (1, 6) (equal to 1) and the maximum elements from all other pairs to get set b.
If x = 2, you can form pairs (1, 2), (3, 4), (5, 6), (7, 9), (8, 10) and take the minimum elements from (1, 2), (5, 6) and the maximum elements from the other pairs.
If x = 3, you can form pairs (1, 3), (4, 6), (5, 7), (2, 9), (8, 10) and take the minimum elements from (1, 3), (4, 6), (5, 7).
In the third test case, x = 0 is the only option: you can form pairs (1, 3), (2, 4) and take the maximum elements from both of them.
Submitted Solution:
```
# Write your code here
import sys, math
import io, os
#data = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
from bisect import bisect_left as bl, bisect_right as br, insort
from heapq import heapify, heappush, heappop
from collections import defaultdict as dd, deque, Counter
from itertools import permutations,combinations
def data(): return sys.stdin.readline().strip()
def mdata(): return list(map(int, data().split()))
def outl(var) : sys.stdout.write(' '.join(map(str, var))+'\n')
def out(var) : sys.stdout.write(str(var)+'\n')
from decimal import Decimal
#from fractions import Fraction
sys.setrecursionlimit(100000)
INF = float('inf')
mod = 10**9 + 7
for _ in range(int(data())):
n=int(data())
b=mdata()
l,s,ol,os=0,0,0,0
for i in range(n):
if b[i]-i-1>ol:
ol+=1
l+=1
if 2*n-b[n-i-1]-i>os:
os+=1
s+=1
out(os - n + ol + 1)
``` | instruction | 0 | 87,809 | 12 | 175,618 |
Yes | output | 1 | 87,809 | 12 | 175,619 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have 2n integers 1, 2, ..., 2n. You have to redistribute these 2n elements into n pairs. After that, you choose x pairs and take minimum elements from them, and from the other n - x pairs, you take maximum elements.
Your goal is to obtain the set of numbers \\{b_1, b_2, ..., b_n\} as the result of taking elements from the pairs.
What is the number of different x-s (0 ≤ x ≤ n) such that it's possible to obtain the set b if for each x you can choose how to distribute numbers into pairs and from which x pairs choose minimum elements?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the integer n (1 ≤ n ≤ 2 ⋅ 10^5).
The second line of each test case contains n integers b_1, b_2, ..., b_n (1 ≤ b_1 < b_2 < ... < b_n ≤ 2n) — the set you'd like to get.
It's guaranteed that the sum of n over test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case, print one number — the number of different x-s such that it's possible to obtain the set b.
Example
Input
3
1
1
5
1 4 5 9 10
2
3 4
Output
1
3
1
Note
In the first test case, x = 1 is the only option: you have one pair (1, 2) and choose the minimum from this pair.
In the second test case, there are three possible x-s. If x = 1, then you can form the following pairs: (1, 6), (2, 4), (3, 5), (7, 9), (8, 10). You can take minimum from (1, 6) (equal to 1) and the maximum elements from all other pairs to get set b.
If x = 2, you can form pairs (1, 2), (3, 4), (5, 6), (7, 9), (8, 10) and take the minimum elements from (1, 2), (5, 6) and the maximum elements from the other pairs.
If x = 3, you can form pairs (1, 3), (4, 6), (5, 7), (2, 9), (8, 10) and take the minimum elements from (1, 3), (4, 6), (5, 7).
In the third test case, x = 0 is the only option: you can form pairs (1, 3), (2, 4) and take the maximum elements from both of them.
Submitted Solution:
```
import sys
input = sys.stdin.readline
t=int(input())
for tests in range(t):
n=int(input())
B=list(map(int,input().split()))
SETB=set(B)
MINOK=[0]*n
MAXOK=[0]*n
minpair=0
for i in range(n):
b=B[i]
while minpair<=b or minpair in SETB:
minpair+=1
MINOK[i]=minpair
minpair+=1
maxpair=n*2
for i in range(n-1,-1,-1):
b=B[i]
while maxpair>=b or maxpair in SETB:
maxpair-=1
MAXOK[i]=maxpair
maxpair-=1
#print(MINOK,MAXOK)
for i in range(n):
if 1<=MINOK[i]<=n*2:
MINOK[i]=1
else:
MINOK[i]=0
if 1<=MAXOK[i]<=n*2:
MAXOK[i]=1
else:
MAXOK[i]=0
#print(MINOK,MAXOK)
ANS=0
if MAXOK[0]==1:
ANS+=1
for i in range(n-1):
if MINOK[i]==1 and MAXOK[i+1]==1:
ANS+=1
if MINOK[-1]==1:
ANS+=1
print(ANS)
``` | instruction | 0 | 87,810 | 12 | 175,620 |
Yes | output | 1 | 87,810 | 12 | 175,621 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have 2n integers 1, 2, ..., 2n. You have to redistribute these 2n elements into n pairs. After that, you choose x pairs and take minimum elements from them, and from the other n - x pairs, you take maximum elements.
Your goal is to obtain the set of numbers \\{b_1, b_2, ..., b_n\} as the result of taking elements from the pairs.
What is the number of different x-s (0 ≤ x ≤ n) such that it's possible to obtain the set b if for each x you can choose how to distribute numbers into pairs and from which x pairs choose minimum elements?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the integer n (1 ≤ n ≤ 2 ⋅ 10^5).
The second line of each test case contains n integers b_1, b_2, ..., b_n (1 ≤ b_1 < b_2 < ... < b_n ≤ 2n) — the set you'd like to get.
It's guaranteed that the sum of n over test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case, print one number — the number of different x-s such that it's possible to obtain the set b.
Example
Input
3
1
1
5
1 4 5 9 10
2
3 4
Output
1
3
1
Note
In the first test case, x = 1 is the only option: you have one pair (1, 2) and choose the minimum from this pair.
In the second test case, there are three possible x-s. If x = 1, then you can form the following pairs: (1, 6), (2, 4), (3, 5), (7, 9), (8, 10). You can take minimum from (1, 6) (equal to 1) and the maximum elements from all other pairs to get set b.
If x = 2, you can form pairs (1, 2), (3, 4), (5, 6), (7, 9), (8, 10) and take the minimum elements from (1, 2), (5, 6) and the maximum elements from the other pairs.
If x = 3, you can form pairs (1, 3), (4, 6), (5, 7), (2, 9), (8, 10) and take the minimum elements from (1, 3), (4, 6), (5, 7).
In the third test case, x = 0 is the only option: you can form pairs (1, 3), (2, 4) and take the maximum elements from both of them.
Submitted Solution:
```
def main(n, d):
o = []
i = 0
c = d[0]
for j in range(1, 2 * n + 1):
if i < n and c == j:
i += 1
if i == n: continue
c = d[i]
else:
o.append(j)
f, e = 0, 0
i = 0
c = o[i]
for j in d:
if c < j:
i += 1
if i == n: break
c = o[i]
else:
f += 1
i = n - 1
c = o[i]
for j in range(n - 1, -1, -1):
if c > d[j]:
i -= 1
if i < 0: break
c = o[i]
else:
e += 1
return n - e - f + 1
t = int(input())
for i in range(t):
n =int(input())
*d, = map(int, input().split())
print(main(n, d))
``` | instruction | 0 | 87,811 | 12 | 175,622 |
Yes | output | 1 | 87,811 | 12 | 175,623 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have 2n integers 1, 2, ..., 2n. You have to redistribute these 2n elements into n pairs. After that, you choose x pairs and take minimum elements from them, and from the other n - x pairs, you take maximum elements.
Your goal is to obtain the set of numbers \\{b_1, b_2, ..., b_n\} as the result of taking elements from the pairs.
What is the number of different x-s (0 ≤ x ≤ n) such that it's possible to obtain the set b if for each x you can choose how to distribute numbers into pairs and from which x pairs choose minimum elements?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the integer n (1 ≤ n ≤ 2 ⋅ 10^5).
The second line of each test case contains n integers b_1, b_2, ..., b_n (1 ≤ b_1 < b_2 < ... < b_n ≤ 2n) — the set you'd like to get.
It's guaranteed that the sum of n over test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case, print one number — the number of different x-s such that it's possible to obtain the set b.
Example
Input
3
1
1
5
1 4 5 9 10
2
3 4
Output
1
3
1
Note
In the first test case, x = 1 is the only option: you have one pair (1, 2) and choose the minimum from this pair.
In the second test case, there are three possible x-s. If x = 1, then you can form the following pairs: (1, 6), (2, 4), (3, 5), (7, 9), (8, 10). You can take minimum from (1, 6) (equal to 1) and the maximum elements from all other pairs to get set b.
If x = 2, you can form pairs (1, 2), (3, 4), (5, 6), (7, 9), (8, 10) and take the minimum elements from (1, 2), (5, 6) and the maximum elements from the other pairs.
If x = 3, you can form pairs (1, 3), (4, 6), (5, 7), (2, 9), (8, 10) and take the minimum elements from (1, 3), (4, 6), (5, 7).
In the third test case, x = 0 is the only option: you can form pairs (1, 3), (2, 4) and take the maximum elements from both of them.
Submitted Solution:
```
from sys import stdin, stdout
from math import floor, gcd, fabs, factorial, fmod, sqrt, inf, log
from bisect import bisect_left as bl, bisect_right as br, bisect
mod = pow(10, 9) + 7
mod2 = 998244353
def inp(): return stdin.readline().strip()
def out(var, end="\n"): stdout.write(str(var)+"\n")
def outa(*var, end="\n"): stdout.write(' '.join(map(str, var)) + end)
def lmp(): return list(mp())
def mp(): return map(int, inp().split())
def smp(): return map(str, inp().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(n, val) for j in range(m)]
def remadd(x, y): return 1 if x%y else 0
def ceil(a,b): return (a+b-1)//b
def chkprime(x):
if x<=1: return False
if x in (2, 3): return True
if x%2 == 0: return False
for i in range(3, int(sqrt(x))+1, 2):
if x%i == 0: return False
return True
for _ in range(int(inp())):
n = int(inp())
arr = lmp()
# c1, c2 = 0, 0
# for i in range(n):
# if arr[i] <= n: c1+=1
# else: c2 += 1
# print(min(c1, c2)+1)
md = {}
for i in range(n):
if arr[i] not in md: md[arr[i]]=1
c1, c2 = 0, 2*n+1
for i in range(n):
if arr[i] <= n: c1 = max(c1, arr[i])
else: c2 = min(c2, arr[i])
i = 1
k1, k2 = 0, 0
while(i<c1):
if i not in md:
k1 += 1
i += 1
i = 2*n
while(i>c2):
if i not in md:
k2 += 1
i -= 1
print(k1+k2+1)
``` | instruction | 0 | 87,812 | 12 | 175,624 |
No | output | 1 | 87,812 | 12 | 175,625 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have 2n integers 1, 2, ..., 2n. You have to redistribute these 2n elements into n pairs. After that, you choose x pairs and take minimum elements from them, and from the other n - x pairs, you take maximum elements.
Your goal is to obtain the set of numbers \\{b_1, b_2, ..., b_n\} as the result of taking elements from the pairs.
What is the number of different x-s (0 ≤ x ≤ n) such that it's possible to obtain the set b if for each x you can choose how to distribute numbers into pairs and from which x pairs choose minimum elements?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the integer n (1 ≤ n ≤ 2 ⋅ 10^5).
The second line of each test case contains n integers b_1, b_2, ..., b_n (1 ≤ b_1 < b_2 < ... < b_n ≤ 2n) — the set you'd like to get.
It's guaranteed that the sum of n over test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case, print one number — the number of different x-s such that it's possible to obtain the set b.
Example
Input
3
1
1
5
1 4 5 9 10
2
3 4
Output
1
3
1
Note
In the first test case, x = 1 is the only option: you have one pair (1, 2) and choose the minimum from this pair.
In the second test case, there are three possible x-s. If x = 1, then you can form the following pairs: (1, 6), (2, 4), (3, 5), (7, 9), (8, 10). You can take minimum from (1, 6) (equal to 1) and the maximum elements from all other pairs to get set b.
If x = 2, you can form pairs (1, 2), (3, 4), (5, 6), (7, 9), (8, 10) and take the minimum elements from (1, 2), (5, 6) and the maximum elements from the other pairs.
If x = 3, you can form pairs (1, 3), (4, 6), (5, 7), (2, 9), (8, 10) and take the minimum elements from (1, 3), (4, 6), (5, 7).
In the third test case, x = 0 is the only option: you can form pairs (1, 3), (2, 4) and take the maximum elements from both of them.
Submitted Solution:
```
#import io,os
#input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
T = int(input())
r = 1
while r<=T:
n = int(input())
b = list(map(int,input().split()))
ans = 0
occu = [False]*(2*n+2)
for i in range(n):
occu[b[i]] = True
s = 1
l = 2*n
maxbig = 0
minbig = n
for i in range(n):
if b[i]>s:
s += 1
while occu[s]: s+= 1
maxbig += 1
else:
while occu[l]: l -= 1
l -= 1
s += 1
s = 1
l = 2*n
for i in range(n-1,-1,-1):
if b[i]<l:
l -= 1
while occu[l]: l -= 1
minbig -= 1
else:
while occu[s]: s += 1
s += 1
l -= 1
# print(minbig)
# print(maxbig)
print(maxbig-minbig+1)
r += 1
``` | instruction | 0 | 87,813 | 12 | 175,626 |
No | output | 1 | 87,813 | 12 | 175,627 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have 2n integers 1, 2, ..., 2n. You have to redistribute these 2n elements into n pairs. After that, you choose x pairs and take minimum elements from them, and from the other n - x pairs, you take maximum elements.
Your goal is to obtain the set of numbers \\{b_1, b_2, ..., b_n\} as the result of taking elements from the pairs.
What is the number of different x-s (0 ≤ x ≤ n) such that it's possible to obtain the set b if for each x you can choose how to distribute numbers into pairs and from which x pairs choose minimum elements?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the integer n (1 ≤ n ≤ 2 ⋅ 10^5).
The second line of each test case contains n integers b_1, b_2, ..., b_n (1 ≤ b_1 < b_2 < ... < b_n ≤ 2n) — the set you'd like to get.
It's guaranteed that the sum of n over test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case, print one number — the number of different x-s such that it's possible to obtain the set b.
Example
Input
3
1
1
5
1 4 5 9 10
2
3 4
Output
1
3
1
Note
In the first test case, x = 1 is the only option: you have one pair (1, 2) and choose the minimum from this pair.
In the second test case, there are three possible x-s. If x = 1, then you can form the following pairs: (1, 6), (2, 4), (3, 5), (7, 9), (8, 10). You can take minimum from (1, 6) (equal to 1) and the maximum elements from all other pairs to get set b.
If x = 2, you can form pairs (1, 2), (3, 4), (5, 6), (7, 9), (8, 10) and take the minimum elements from (1, 2), (5, 6) and the maximum elements from the other pairs.
If x = 3, you can form pairs (1, 3), (4, 6), (5, 7), (2, 9), (8, 10) and take the minimum elements from (1, 3), (4, 6), (5, 7).
In the third test case, x = 0 is the only option: you can form pairs (1, 3), (2, 4) and take the maximum elements from both of them.
Submitted Solution:
```
#!/usr/bin/env python
# -*- coding: utf-8 -*-
# @oj: codeforces
# @id: hitwanyang
# @email: 296866643@qq.com
# @date: 2020/12/18 11:12
# @url: https://codeforc.es/contest/1463/problem/D
import sys, os
from io import BytesIO, IOBase
import collections, itertools, bisect, heapq, math, string
from decimal import *
# region fastio
BUFSIZE = 8192
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")
# ------------------------------
## 注意嵌套括号!!!!!!
## 先有思路,再写代码,别着急!!!
## 先有朴素解法,不要有思维定式,试着换思路解决
## 精度 print("%.10f" % ans)
## sqrt:int(math.sqrt(n))+1
## 字符串拼接不要用+操作，会超时
## 二进制转换:bin(1)[2:].rjust(32,'0')
## array copy:cur=array[::]
## oeis:example 1, 3, _, 1260, _, _, _, _, _, 12164510040883200
## sqrt:Decimal(x).sqrt()避免精度误差
## 无穷大表示:float('inf')
## py 10**6 排序+双指针 3秒可能TLE
## 按区间右端点排序,current.left>pre.right,贪心求不相交区间的最大个数
def main():
t = int(input())
for i in range(t):
n = int(input())
b = list(map(int, input().split()))
sb = sorted(b)
x,y=0,0
for j in sb:
if j<=n:
x+=1
else:
y+=1
print (min(n-x,n-y)+1)
if __name__ == "__main__":
main()
``` | instruction | 0 | 87,814 | 12 | 175,628 |
No | output | 1 | 87,814 | 12 | 175,629 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have 2n integers 1, 2, ..., 2n. You have to redistribute these 2n elements into n pairs. After that, you choose x pairs and take minimum elements from them, and from the other n - x pairs, you take maximum elements.
Your goal is to obtain the set of numbers \\{b_1, b_2, ..., b_n\} as the result of taking elements from the pairs.
What is the number of different x-s (0 ≤ x ≤ n) such that it's possible to obtain the set b if for each x you can choose how to distribute numbers into pairs and from which x pairs choose minimum elements?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the integer n (1 ≤ n ≤ 2 ⋅ 10^5).
The second line of each test case contains n integers b_1, b_2, ..., b_n (1 ≤ b_1 < b_2 < ... < b_n ≤ 2n) — the set you'd like to get.
It's guaranteed that the sum of n over test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case, print one number — the number of different x-s such that it's possible to obtain the set b.
Example
Input
3
1
1
5
1 4 5 9 10
2
3 4
Output
1
3
1
Note
In the first test case, x = 1 is the only option: you have one pair (1, 2) and choose the minimum from this pair.
In the second test case, there are three possible x-s. If x = 1, then you can form the following pairs: (1, 6), (2, 4), (3, 5), (7, 9), (8, 10). You can take minimum from (1, 6) (equal to 1) and the maximum elements from all other pairs to get set b.
If x = 2, you can form pairs (1, 2), (3, 4), (5, 6), (7, 9), (8, 10) and take the minimum elements from (1, 2), (5, 6) and the maximum elements from the other pairs.
If x = 3, you can form pairs (1, 3), (4, 6), (5, 7), (2, 9), (8, 10) and take the minimum elements from (1, 3), (4, 6), (5, 7).
In the third test case, x = 0 is the only option: you can form pairs (1, 3), (2, 4) and take the maximum elements from both of them.
Submitted Solution:
```
import sys
import math,bisect,operator
inf,m = float('inf'),10**9+7
sys.setrecursionlimit(10 ** 5)
from itertools import groupby,accumulate
from heapq import heapify,heappop,heappush
from collections import deque,Counter,defaultdict
I = lambda : int(sys.stdin.readline())
neo = lambda : map(int, sys.stdin.readline().split())
Neo = lambda : list(map(int, sys.stdin.readline().split()))
for _ in range(I()):
n = I()
A = Neo()
B = [1]*(2*n+1)
B[0] = 0
for i in A:
B[i] = 0
C = list(accumulate(B[::-1]))[::-1]
D = list(accumulate(B))
A.sort()
Ans = 0
for i in A:
if C[i]-Ans > 0:
Ans += 1
t = 0
f = 1
for i in A:
if D[i]-t > 0:
t += 1
else:
f = 0
break
if f:
Ans += 1
print(Ans)
``` | instruction | 0 | 87,815 | 12 | 175,630 |
No | output | 1 | 87,815 | 12 | 175,631 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A median in an array with the length of n is an element which occupies position number <image> after we sort the elements in the non-decreasing order (the array elements are numbered starting with 1). A median of an array (2, 6, 1, 2, 3) is the number 2, and a median of array (0, 96, 17, 23) — the number 17.
We define an expression <image> as the integer part of dividing number a by number b.
One day Vasya showed Petya an array consisting of n integers and suggested finding the array's median. Petya didn't even look at the array and said that it equals x. Petya is a very honest boy, so he decided to add several numbers to the given array so that the median of the resulting array would be equal to x.
Petya can add any integers from 1 to 105 to the array, including the same numbers. Of course, he can add nothing to the array. If a number is added multiple times, then we should consider it the number of times it occurs. It is not allowed to delete of change initial numbers of the array.
While Petya is busy distracting Vasya, your task is to find the minimum number of elements he will need.
Input
The first input line contains two space-separated integers n and x (1 ≤ n ≤ 500, 1 ≤ x ≤ 105) — the initial array's length and the required median's value. The second line contains n space-separated numbers — the initial array. The elements of the array are integers from 1 to 105. The array elements are not necessarily different.
Output
Print the only integer — the minimum number of elements Petya needs to add to the array so that its median equals x.
Examples
Input
3 10
10 20 30
Output
1
Input
3 4
1 2 3
Output
4
Note
In the first sample we can add number 9 to array (10, 20, 30). The resulting array (9, 10, 20, 30) will have a median in position <image>, that is, 10.
In the second sample you should add numbers 4, 5, 5, 5. The resulting array has median equal to 4. | instruction | 0 | 87,849 | 12 | 175,698 |
Tags: greedy, math, sortings
Correct Solution:
```
def median(n,m,li):
count = 0
li.sort()
while True:
if li[(n+1)//2 -1]== m:
return count
li.append(m)
li.sort()
n+=1
count+=1
n,m = input().split()
li = [int(x) for x in input().split()]
print(median(int(n),int(m),li))
``` | output | 1 | 87,849 | 12 | 175,699 |
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