Hdu6063 RXD and math(2017多校第3场)

RXD and math

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 532    Accepted Submission(s): 287

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

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题目的意思是给你这个公式，给出n，k计算公式和

思路：比赛时u没给出来，队友研究了下莫比乌斯函数告诉我，我尝试达标找规律，发现答案就是n^k,快速幂解决

感慨高中生的不讲道理，一个简单的式子可以包装到这么复杂

#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;
}