HDU 6027 Easy Summation

Easy Summation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 54    Accepted Submission(s): 32

Problem Description

You are encountered with a traditional problem concerning the sums of powers.
Given two integers n and k. Let f(i)=ik, please evaluate the sum f(1)+f(2)+...+f(n). The problem is simple as it looks, apart from the value of n in this question is quite large.
Can you figure the answer out? Since the answer may be too large, please output the answer modulo109+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 modulo109+7.

 

Sample Input

32 54 24 1

 

Sample Output

333010

题意：给出n，k，求n以内的所有数的k次方之和。对1000000007取模。

很简单直接照着算，不会超时的。

#include<cstdio>#include<cstring>#include<algorithm>#define N 1000000007using namespace std;int main(){    int i,j,n,t,a,b;    __int64 s,sum;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&a,&b);        sum=0;        for(i=1;i<=a;i++)        {            s=1;            for(j=0;j<b;j++)            {                s*=i;                if(s>N)                s=s%N;            }            sum+=s;            if(sum>N)                sum=sum%N;        }        printf("%d\n",sum);    }    return 0;}