【Gym

Por Costel and Azerah

Descriptions

给你n个数 问你，有多少个子序列 的和是偶数 

Example

Input

2
3
3 10 1
2
4 2

Output

3
3

题目链接

https://vjudge.net/problem/Gym-100923A

 

恶心死了  

freopen("azerah.in","r",stdin);
freopen("azerah.out","w",stdout);

必须加上 不然一直错  卡了我1小时

直接看代码吧 挺水的

AC代码

#include <iostream>
#include <cstdio>
#include <fstream>
#include <algorithm>
#include <cmath>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <cstring>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#define IOS ios_base::sync_with_stdio(0); cin.tie(0);
#define Mod 1000000007
#define eps 1e-6
#define ll long long
#define INF 0x3f3f3f3f
#define MEM(x,y) memset(x,y,sizeof(x))
#define Maxn 1000010
using namespace std;
ll T,n;
int main()
{
    freopen("azerah.in","r",stdin);
    freopen("azerah.out","w",stdout);
    cin>>T;
    while(T--)
    {
        cin>>n;
        ll x;
        ll n1=0;//奇数个数
        ll n2=0;//偶数个数
        for(ll i=1; i<=n; i++)
        {
            cin>>x;
            if(x%2)
                n1++;
            else
                n2++;
        }
        ll s1=1;//奇数中的偶数子序列个数=2^(s1-1)-1
        ll s2=1;//偶数中的偶数子序列个数=2^s2-1
        //怕溢出，就一步一步循环吧
        for(ll i=1; i<=n2; i++)
        {
            s1*=2;
            s1%=Mod;
        }
        for(ll i=1; i<n1; i++)
        {
            s2*=2;
            s2%=Mod;
        }
        s1--;
        s2--;
        cout<<(s1+s2+s1*s2)%Mod<<endl;
    }
    return 0;
}