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;}