Easy Summation
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 0 Accepted Submission(s): 0
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：给出一个n和k，求从1^k到n^k的和
解题思路：暴力，注意取模即可
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <cmath>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <functional>
using namespace std;
#define LL long long
const int INF=0x3f3f3f3f;
const int mod=1e9+7;
int n,m;
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&m,&n);
LL ans=0;
for(int i=1; i<=m; i++)
{
LL x=1;
for(int j=1; j<=n; j++)
{
x*=i;
x%=mod;
}
ans+=x;
ans%=mod;
}
printf("%lld\n",ans);
}
return 0;
}
