Gym - 100923A Por Costel and Azerah 新手dp水题

azerah.in / azerah.out

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

题解：这是一道dp水题，对于每一位可以选择取或不取，对于奇数dp结果为上一位的奇数情况和偶数情况之和，对于偶数，则是偶数情况是偶数情况*2和奇数情况是上一位奇数情况*2，非常水，适合新手

代码如下

#include <iostream>
#include <bits/stdc++.h>

using namespace std;

long long mod=1e9+7;

long long dp[1000005][2];

int main()
{
    freopen("azerah.in","r",stdin);
    freopen("azerah.out","w",stdout);
    int t;
    cin>>t;
    while(t--){
    long long n1=0,n2=0;
    long long n;
        cin>>n;
        dp[0][0]=1;
        dp[0][1]=0;
        for(long long i=1;i<=n;i++)
        {
            long long l;
            scanf("%lld",&l);
            if(l&1){
                n1++;
                dp[i][0]=(dp[i-1][1]+dp[i-1][0])%mod;
                dp[i][1]=(dp[i-1][0]+dp[i-1][1])%mod;
            }
            else
            {
                n2++;
                dp[i][0]=2*dp[i-1][0]%mod;
                dp[i][1]=2*dp[i-1][1]%mod;
            }
            //cout<<dp[i][0]<<' '<<dp[i][1]<<endl;
        }
            cout<<dp[n][0]-1<<endl;
    }
    //cout << "Hello world!" << endl;
    return 0;
}