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 long
const 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;
}