HDU 5363 Key Set(快速幂取余)

最新推荐文章于 2020-03-27 22:51:12 发布

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最新推荐文章于 2020-03-27 22:51:12 发布

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分类专栏： ---- HDU ---- ---- 数论or数学 ----

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本博客讨论了如何计算给定整数集合中形成关键子集的数量，关键子集的定义为元素之和为偶数的子集。通过输入整数n表示集合大小，并输出所有非空关键子集的数量取模1000000007的结果。

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Key Set

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 362 Accepted Submission(s): 238

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

(1≤T≤105) , indicating the number of test cases. For each test case:

The first line contains an integer

(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

Sample Output

Source

2015 Multi-University Training Contest 6

题目链接：点击打开链接

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;
__int64 f( __int64 n,__int64 m,__int64 s)
{
    __int64 t;
    n=n%s;
    t=1;
    if(n==0)return 0;
    while(m)
    {
        if(m%2==1)
        {
            t*=n;
            t%=s;
        }
        n*=n;
        n%=s;
        m/=2;
    }
    return t;
}
int main()
{
    __int64 n,i,x;
    scanf("%I64d",&n);
    for(i=1;i<=n;i++)
    {
        scanf("%I64d",&x);
        if(x==1)
        {
            printf("0\n");
            continue;
        }

       __int64 t=f(2,x-1,1000000007);
       printf("%I64d\n",t-1);
    }

}