E - Easy Summation

最新推荐文章于 2024-04-27 17:20:14 发布

原创最新推荐文章于 2024-04-27 17:20:14 发布 · 446 阅读

0 ·

CC 4.0 BY-SA版权

简单题专栏收录该内容

36 篇文章

订阅专栏

解决一个经典的幂次求和问题，对于给定的n和k，计算从1到n的所有整数的k次幂之和，并输出结果模10^9+7。此问题涉及到大数运算，采用自定义的幂次函数来避免溢出。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

这个题的数据量是很大的，因此每求一次a^b,都要进行取模，所以我们把pow()函数在这儿重新写 一次，同时，最后的输出结果也要取模。


#include <iostream>
#include<cstdio>
#include<cmath>
using namespace std;
const int N=1000000000+7;
long long pow1(int a,int b)
{
   long long p=1;
   for(int i=1;i<=b;i++)
   {
       p*=a;
       p%=N;
   }
   return p;
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
      long long a,b,sum=0;
      scanf("%lld%lld",&a,&b);
      for(int i=1;i<=a;i++)
        sum+=pow1(i,b);
      printf("%lld\n",sum%N);
    }
    return 0;
}