标签：acm   hdu   

Key Set

4
1
2
3
4

0
1
3
7

//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cmath>
#include<stdlib.h>
#include<map>
#include<set>
#include<time.h>
#include<vector>
#include<queue>
#include<string>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
#define eps 1e-8
#define INF 0x3f3f3f3f
#define LL long long
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
typedef pair<int , int> P;
#define mod 1000000007

int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        int n;
        scanf("%d", &n);
        int ans = 1;
        int a = 2;
        n--;
        while(n)
        {
            if(n & 1) ans = ((LL)ans * a) % mod;
            a = ((LL)a * a) % mod;
            n >>= 1;
        }
        ans--;
        printf("%d\n", ans % mod);
    }
    return 0;
}

2015 multiply 6 1011

标签：acm   hdu   

原文地址：http://blog.csdn.net/dojintian/article/details/47322089