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
当时一激动没考虑到直接pow超出了longlong范围wa了两发
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<string>
#include<queue>
#include<stack>
#include<vector>
#include<algorithm>
using namespace std;
#define ll long long
#define ms(a,b) memset(a,b,sizeof(a))
const int M=1e5+10;
const int inf=0x3f3f3f3f;
int i,j,k,n,m;
const ll mod=1e9+7;
int main()
{
int T;
scanf("%d",&T);
while(T--){
scanf("%d%d",&n,&k);
ll ans=0;
for(int i=1;i<=n;i++){
ll sum=1;
for(int j=1;j<=k;j++)
sum*=i,sum%=mod;
ans+=sum;
ans%=mod;
}
printf("%lld\n",ans);
}
return 0;
}
