time limit per test
1 second
memory limit per test
256 megabytes
Luntik came out for a morning stroll and found an array a of length n. He calculated the sum s of the elements of the array (s=∑ni=1ai). Luntik calls a subsequence of the array a nearly full if the sum of the numbers in that subsequence is equal to s−1.
Luntik really wants to know the number of nearly full subsequences of the array a. But he needs to come home so he asks you to solve that problem!
A sequence x is a subsequence of a sequence y if x can be obtained from y by deletion of several (possibly, zero or all) elements.
Input
The first line contains a single integer t (1≤t≤1000) --- the number of test cases. The next 2⋅t lines contain descriptions of test cases. The description of each test case consists of two lines.
The first line of each test case contains a single integer n (1≤n≤60) --- the length of the array.
The second line contains n integers a1,a2,...,an (0≤ai≤109) --- the elements of the array a.
Output
For each test case print the number of nearly full subsequences of the array.
Example
Input
Copy
5
5
1 2 3 4 5
2
1000 1000
2
1 0
5
3 0 2 1 1
5
2 1 0 3 0Output
Copy
1
0
2
4
4Note
In the first test case, s=1+2+3+4+5=15, only (2,3,4,5) is a nearly full subsequence among all subsequences, the sum in it is equal to 2+3+4+5=14=15−1.
In the second test case, there are no nearly full subsequences.
In the third test case, s=1+0=1, the nearly full subsequences are (0) and () (the sum of an empty subsequence is 0).
解题说明：此题是一道模拟题，其实就是找出数列中1和0的个数，找规律能发现结果就是1的个数乘以2^（0的个数）。
cpp
#include<stdio.h>
int main()
{
long long int t, a[65], i;
scanf("%lld", &t);
a[0] = 0;
for (i = 1; i <= 61; i++)
{
a[i] = 1 + a[i - 1] * 2;
}
while (t--)
{
long long int s = 0, ans = 0, b = 0, c = 0, n, x;
scanf("%lld", &n);
for (i = 0; i < n; i++)
{
scanf("%lld", &x);
if (x == 1)
{
b++;
}
if (x == 0)
{
c++;
}
}
if (b)
{
ans = b + a[c] * b;
}
printf("%lld\n", ans);
}
return 0;
}