Easy Summation
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 54 Accepted Submission(s): 32
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，求n以内的所有数的k次方之和。对1000000007取模。
很简单直接照着算，不会超时的。
#include<cstdio>
#include<cstring>
#include<algorithm>
#define N 1000000007
using namespace std;
int main()
{
int i,j,n,t,a,b;
__int64 s,sum;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&a,&b);
sum=0;
for(i=1;i<=a;i++)
{
s=1;
for(j=0;j<b;j++)
{
s*=i;
if(s>N)
s=s%N;
}
sum+=s;
if(sum>N)
sum=sum%N;
}
printf("%d\n",sum);
}
return 0;
}
