多校6-Key Set 2015-08-09 20:35 2人阅读 评论(0) 收藏

Key Set
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 722 Accepted Submission(s): 442

Problem Description
soda has a set S with n integers {1,2,…,n}. A set is called key set if the sum of integers in the set is an even number. He wants to know how many nonempty subsets of S are key set.

Input
There are multiple test cases. The first line of input contains an integer T (1≤T≤105), indicating the number of test cases. For each test case:

The first line contains an integer n (1≤n≤109), the number of integers in the set.

Output
For each test case, output the number of key sets modulo 1000000007.

Sample Input

4
1
2
3
4

Sample Output

0
1
3
7

Source
2015 Multi-University Training Contest 6

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快速幂

#include <map>
#include <list>
#include <climits>
#include <cmath>
#include <queue>
#include <stack>
#include <string>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
#define eps 1e-9
#define LL long long
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
#define CRR fclose(stdin)
#define CWW fclose(stdout)
#define RR freopen("input.txt","r",stdin)
#define WW freopen("output.txt","w",stdout)

const int Max = 10010;

const int Mod = 1000000007;



LL pow_mod(LL n)
{
    LL res=1;
    LL a=2;
    while(n)
    {
        if(n&1)
        {
            res=(res*a)%Mod;
        }
        a=(a*a)%Mod;
        n>>=1;
    }
    return res;
}

int main()
{
    int T;

    scanf("%d",&T);
    LL n;
    while(T--)
    {
        scanf("%I64d",&n);
        printf("%I64d
",pow_mod(n-1)-1);
    }

    return 0;
}

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