求i^k之和——HDU - 6027

最新推荐文章于 2025-07-09 17:34:59 发布

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最新推荐文章于 2025-07-09 17:34:59 发布

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分类专栏：快速幂 XYNUOJ 数论

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本文介绍了一种解决大数幂求和问题的方法，通过快速幂运算计算一系列整数的k次幂之和，并对结果取模以避免溢出。采用C语言实现，适用于处理较大范围内的整数。

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You are encountered with a traditional problem concerning the sums of powers.
Given two integers n and k. Let f(i)= $i^{k}$ , please evaluate the sum= f(1)+f(2)+...+f(n). The problem is simple as it looks, apart from the value of nn in this question is quite large.
Can you figure the answer out? Since the answer may be too large, please output the answer modulo $10^{9}$ +7.

Input

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

Output

For each test case, print a single line containing an integer modulo $10^{9}$ +7.

Sample Input

Sample Output

33
30
10

题解：

快速幂求a^n的值，再在主函数中调用。

由于数值比较大，用 long long 型数

代码：

#include<stdio.h>
long long pow(int a,int n)
{//返回a^n
    long long result=1,flag=a;
	while(n)
	{
		if(n&1)
		result=result*flag%1000000007;
		flag=flag*flag%1000000007;
		n/=2;
	 } 
	 return result%1000000007;
}
int main()
{
	int t,n,k;
	scanf("%d",&t);
	while(t--)
	{
		int i;
		long long sum=0;
		scanf("%d %d",&n,&k);
		for(i=1;i<=n;i++)
		{
			sum+=pow(i,k)%1000000007;
		 } 
		 printf("%lld\n",sum%1000000007);
	}
	return 0;
}