| Time Limit: 10000MS | Memory Limit: 262144KB | 64bit IO Format: %lld & %llu |
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：给出两个数，n 和 m， 求 1~n 之间每个数的 m 次幂的和，再取模
思路：快速幂
<span style="font-size:18px;">#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <string>
#include <algorithm>
#include <queue>
#include <stack>
using namespace std;
#define ll long long
const double PI = acos(-1.0);
const double e = 2.718281828459;
const double eps = 1e-8;
const ll MOD = 1e9+7;
ll pow_mod(ll n, ll m) {
ll res = 1;
while(m) {
if(m&1)
res = res*n%MOD;
n = n*n%MOD;
m >>= 1;
}
return res;
}
int main() {
//freopen("in.txt", "r", stdin);
//freopen("out.txt", "w", stdout);
ll n, m;
ll ans;
while(cin>>n>>m) {
ans = 0;
for(ll i = 1; i <= n; i++) {
ans += pow_mod(i, m);
ans %= MOD;
}
cout<<ans<<endl;
}
return 0;
}
</span>CSU - 1556 Jerry's trouble(快速幂)
原文地址：http://blog.csdn.net/u014028317/article/details/46573301
