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.
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).
For each case, you program will output the answer of the sum of the sequence of number (mod 1e9+7).
4 1
5 1
4 2
5 2
4 3
10
15
30
55
100
HINT
求1到n之间，所有数的m次幂之和。
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<cmath>
#define Mod 1000000007
#define ll long long
using namespace std;
ll pow_mod(ll a,ll b) {
ll res=1;
while(b) {
if(b&1)
res*=a,res%=Mod;
a*=a;
a%=Mod;
b>>=1;
}
return res;
}
ll n,m;
int main() {
while(~scanf("%lld%lld",&n,&m)) {
ll ans=0;
for(int i=1; i<=n; i++) {
ans+=pow_mod(i,m);
ans%=Mod;
}
printf("%lld\n",ans);
}
}csu 1556: Jerry's trouble(快速幂)
原文地址：http://blog.csdn.net/acm_baihuzi/article/details/45801639
