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

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 const long long mod=1e9+7;
 4 long long PowerMod(long long a, long long b,long long c) {
 5     long long ans = 1;
 6     a = a % c;
 7     while(b>0) {
 8
 9         if(b % 2 == 1)
10             ans = (ans * a) % c;
11         b = b/2;
12         a = (a * a) % c;
13     }
14     return ans;
15 }
16 int main()
17 {
18     int t;
19     while(~scanf("%d",&t))
20     {
21         while(t--)
22         {
23             long long n,k,sum=0;
24             scanf("%lld %lld",&n,&k);
25             for(int i=1;i<=n;i++)
26             {
27                 sum+=PowerMod(i,k,mod);
28                 sum=sum%mod;
29             }
30             printf("%lld\n",sum);
31         }
32     }
33     return 0;
34 }

原文地址：https://www.cnblogs.com/fqfzs/p/9858893.html