HDU6063-RXD and math

RXD and math

                                                                    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
                                                                                            Total Submission(s): 1035    Accepted Submission(s): 570

Problem Description

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

 

Input

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.

 

Output

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

 

Sample Input

10 10

 

Sample Output

Case #1: 999999937

 

Source

2017 Multi-University Training Contest - Team 3

 

题意：给出一个n和k，求出题目给出的式子

解题思路：其实就是一个莫比乌斯函数，打表找规律后发现就是求n^k

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <map>#include <set>#include <stack>#include <queue>#include <vector>#include <bitset>#include <functional>using namespace std;#define LL long longconst int INF = 0x3f3f3f3f;const LL mod=1000000007;LL n,k;LL mypow(LL a,LL b){    LL ans=1;    a%=mod;    while(b)    {        if(b&1) {ans*=a;ans%=mod;}        a=(a*a)%mod;        b>>=1;    }    return ans;}int main(){    int cas=0;    while(~scanf("%lld%lld",&n,&k))    {        printf("Case #%d: %lld\n",++cas,mypow(n,k));    }    return 0;}