csu 1556(快速幂)

1556: Jerry's trouble

Time Limit: 10 Sec  Memory Limit: 256 MB
Submit: 787  Solved: 317
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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

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:求解 (1^m+2^m+...+n^m)%mod
题解:快速幂

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <math.h>
#include <algorithm>
using namespace std;
typedef long long LL;
const LL mod = ;
LL pow_mod(LL a,LL n){
    LL ans = ;
    while(n){
        if(n&) ans = ans*a%mod;
        a = a*a%mod;
        n>>=;
    }
    return ans;
}
int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF){
        LL res = ;
        for(int i=;i<=n;i++){
            res = (res+pow_mod(i,m))%mod;
        }
        printf("%lld\n",res);
    }
}