山东省第八届 ACM 省赛 sum of power（SDUT 3899）

Problem Description

Calculate ∑ni=1im mod (1000000000+7) for given n，m.

Input

Input contains two integers n,m(1≤n≤1000,0≤m≤10).

Output

Output the answer in a single line.

Example Input

10 0

Example Output

10

方法：快速幂和大数求和

#include <iostream>using namespace std;const int d=1000000000+7;long long poww(long a,long b){    long long ans=1,base=a;    while(b!=0)    {        if(b&1)            ans=(ans*base)%d;        base=(base*base)%d;        b>>=1;    }    return ans;}int main(){    long long n,m;    cin>>n>>m;    long long sum = 0;    for(int i=1;i<=n;i++)        sum = (sum+poww(i,m))%d;    cout<<sum<<endl;    return 0;}