hdu6027Easy Summation(快速幂取模)

Easy Summation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 4112    Accepted Submission(s): 1676

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 modulo 109+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 109+7.

 

Sample Input

3

2 5

4 2

4 1

 

Sample Output

33

30

10

题意：给出n,k，定义f(i)=i^k,求出[f(1)+f(2)+……+f(n)]mod 1e9+7

题解：直接套快速幂，注意用long long

#include<bits/stdc++.h>
using namespace std;
const long long mod=1e9+7;
long long PowerMod(long long a, long long b,long long c) {
    long long ans = 1;
    a = a % c;
    while(b>0) {

        if(b % 2 == 1)
            ans = (ans * a) % c;
        b = b/2;
        a = (a * a) % c;
    }
    return ans;
}
int main()
{
    int t;
    while(~scanf("%d",&t))
    {
        while(t--)
        {
            long long n,k,sum=0;
            scanf("%lld %lld",&n,&k);
            for(int i=1;i<=n;i++)
            {
                sum+=PowerMod(i,k,mod);
                sum=sum%mod;
            }
            printf("%lld
",sum);
        }
    }
    return 0;
}