Por Costel the Pig has received a royal invitation to the palace of the Egg-Emperor of Programming, Azerah. Azerah had heard of the renowned pig and wanted to see him with his own eyes. Por Costel, having arrived at the palace, tells the Egg-Emperor that he looks "tasty". Azerah feels insulted (even though Por Costel meant it as a compliment) and, infuratied all the way to his yolk, threatens to kill our round friend if he doesn't get the answer to a counting problem that he's been struggling with for a while
Given an array of numbers, how many non-empty subsequences of this array have the sum of their numbers even ? Calculate this value mod (Azerah won't tell the difference anyway)
Help Por Costel get out of this mischief!
Input
The file azerah.in will contain on its first line an integer , the number of test cases. Each test has the following format: the first line contains an integer
(the size of the array) and the second line will contain
integers
(
), separated by single spaces.
It is guaranteed that the sum of over all test cases is at most
Output
The file azerah.out should contain lines. The
-th line should contain a single integer, the answer to the
-th test case.
Example
Input
2 3 3 10 1 2 4 2
Output
3 3
题意：给你一个序列，问你有多少个子序列的元素和为偶数。
直接看代码就懂，用到了全排的问题
#include<iostream>
#include<stdio.h>
#include<math.h>
#include<algorithm>
using namespace std;
const int inf=0x3f3f3f3f;
int mod=1e9+7;
int main()
{
freopen("azerah.in","r",stdin);
freopen("azerah.out","w",stdout);
int t;
scanf("%d",&t);
while(t--)
{
long long n;
cin>>n;
long long l=0,r=0,x;
for(int i=1; i<=n; i++)
{
cin>>x;
if(x%2==0)
l++;
else r++;
}
long long ans=1,sum=1;
for(int i=1; i<=l; i++)
{
ans*=2;
ans%=mod;
}
for(int i=1; i<r; i++)
{
sum*=2;
sum%=mod;
}
ans--;
sum--;
cout<<(sum+ans+sum*ans)%mod<<endl;
}
return 0;
}