【CSU 1556】Pseudoprime numbers

题

Description

 Jerry is caught by Tom. He was penned up in one room with a door, which only can be opened by its code. The code is the answer of the sum of the sequence of number written on the door. The type of the sequence of number is

1^m + 2^m + 3^m + …… + n^m

But Jerry’s mathematics is poor, help him to escape from the room.

Input

 There are some cases (about 500). For each case, there are two integer numbers n, m describe as above ( 1 <= n < 1 000 000, 1 <= m < 1000).

Output

 For each case, you program will output the answer of the sum of the sequence of number (mod 1e9+7).

Sample Input

4 1
5 1
4 2
5 2
4 3

Sample Output

10
15
30
55
100

题意：求1^m + 2^m + 3^m + …… + n^m  (mod 1e9+7)

注意：每个加起来的时候和加起来后也要mod

#include<stdio.h>
#define M 1000000007
long long qpow(long long a,long long m)//快速幂
{
    long long ans=1,k=a;
    while(m)
    {
        if(m&1)
            ans=(ans*k)%M;
        k=(k*k)%M;
        m>>=1;
    }
    return ans;
}
int main()
{
    long long n,m,ans;
    while(scanf("%lld%lld",&n,&m)!=EOF)
    {
        ans=0;
        for(int i=1; i<=n; i++)
            ans+=qpow(i,m)%M;//这里mod一下
        printf("%lld
",ans%M);//这里mod一下
    }
    return 0;
}