Easy Summation

最新推荐文章于 2021-10-13 18:45:58 发布

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最新推荐文章于 2021-10-13 18:45:58 发布

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分类专栏： ACM. 数论算法文章标签：数论快速幂

本文链接：https://blog.youkuaiyun.com/gjs935219/article/details/81394830

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本文介绍了一种解决大数幂求和问题的算法，并提供了一个具体的实现案例。该问题要求计算从1到n的整数k次幂的和并对结果取模。文章通过示例解释了算法的工作原理及其实现细节。

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You are encountered with a traditional problem concerning the sums of powers.
Given two integers nn and kk. Let f(i)=ikf(i)=ik, please evaluate the sum f(1)+f(2)+...+f(n)f(1)+f(2)+...+f(n). The problem is simple as it looks, apart from the value of nnin this question is quite large.
Can you figure the answer out? Since the answer may be too large, please output the answer modulo 109+7109+7.

Input

The first line of the input contains an integer T(1≤T≤20)T(1≤T≤20), denoting the number of test cases.
Each of the following TT lines contains two integers n(1≤n≤10000)n(1≤n≤10000) and k(0≤k≤5)k(0≤k≤5).

Output

For each test case, print a single line containing an integer modulo 109+7109+7.

Sample Input

Sample Output

33
30
10

#include<stdio.h>
long long dabiao(long long x,long long m) {  //long long 型，否则会报错
	long long i,j,y,sum=0,r=1;
		y=m;
		while(y!=0) {
			if(1&y)r=r*x%1000000007;
			x=x*x%1000000007;
			y=y>>1;
		}
		return r;
}
int main() {
	int s;
	scanf("%d",&s);
	while(s--) {
		long long n,i,k,sum;
		scanf("%lld%lld",&n,&k);
		sum=0;
		for(i=1;i<=n;i++)
		sum=(sum+dabiao(i,k))%1000000007;
		printf("%d\n",sum);
		}
}