快速幂

最新推荐文章于 2023-02-07 10:46:33 发布

原创最新推荐文章于 2023-02-07 10:46:33 发布 · 169 阅读

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本文深入解析了快速幂算法的实现原理与模板代码，通过一个具体的编程问题展示了如何运用快速幂解决数学运算中大规模指数计算的问题。文章包含了完整的代码示例，并详细解释了算法的每一步操作，为读者提供了理解和掌握快速幂算法的有效途径。

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快速幂模板


long long quickpow(int x,int N)
{
    long long res = x;
    long long ans = 1;
    while(N)
    {
        if(N&1)
        {
            ans=ans*res;
        }
        res=res*res;
        N=N>>1;
    }
    return ans;
}

需要取余时自行取余

下面是例题

soda has a set SS with nn integers {1,2,…,n}{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 SS are key set.

Input

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

The first line contains an integer nn (1≤n≤109)(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

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<stack>
#include<cstring> 
using namespace std;
#define M 1000000007
long long quickpow(int x,int N)
{
    long long res = x;
    long long ans = 1;
    while(N)
    {
        if(N&1)
        {
            ans=(ans*res)%M;
        }
        res=(res*res)%M;
        N=N>>1;
    }
    return ans;
}
int main()
{
	int n;
	cin>>n;
	while(n--)
	{
		int m;
		cin>>m;
		long long ans=quickpow(2,m-1)-1;
		cout<<ans<<endl;
	}
	
	
	return 0;
}