RXD and math

题目链接  http://acm.pdsu.edu.cn/problem.php?cid=1027&pid=7

RXD and math

时间限制: 2 Sec   内存限制: 128 MB 提交: 25   解决: 21 [ 提交][ 状态][ 讨论版]

题目描述

RXD is a good mathematician. One day he wants to calculate: ∑i=1nkμ2(i)×⌊nki−−−√⌋ output the answer module  109+7 . 1≤n,k≤1018 μ(n)=1(n=1) μ(n)=(−1)k(n=p1p2…pk) μ(n)=0(otherwise) p1,p2,p3…pk  are different prime numbers

输入

There are several test cases, please keep reading until EOF. There are exact 10000 cases. For each test case, there are 2 numbers  n,k .

输出

For each test case, output "Case #x: y", which means the test case number and the answer.

样例输入

10 10

样例输出

Case #1: 999999937 解题思路：

公式化简后，就是n^k。也就是说该题是求n^k.。是快速幂的应用，如果不会快速幂，且想快速学会快速幂的代码，那么请点击链接  http://blog.csdn.net/jiyi_xiaoli/article/details/76587519   如果想详细的理解快速幂，那么请点击链接  http://blog.csdn.net/jiyi_xiaoli/article/details/76578376

代码：

#include<iostream> using namespace std; const long long int mod=1e9+7; int pow(long long int a,long long int n) { long long int t=1; while(n) { if(n&1) t=(t%mod)*(a%mod)%mod; n=n>>1; a=(a%mod)*(a%mod); } return t; } int main() { long long int n,k; int i=0; while(cin>>n>>k) { i++; cout<<"Case #"<<i<<": "; cout<<pow(n,k)<<endl; } return 0; }