2017多校联合三／hdu6063 ( RXD and math )快速幂＋思维

RXD and math

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 310    Accepted Submission(s): 160

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

注意到一个数字$x$必然会被唯一表示成a^2\times ba​2​​×b的形式.其中$|\mu (b)|=1$。 所以这个式子会把[1, n^k][1,n​k​​]的每个整数恰好算一次. 所以答案就是n^kn​k​​,快速幂即可. 时间复杂度$O(\log k)$.

#include <iostream>#include <algorithm>#include <cstdio>#include <cmath>#include <cstring>#include <string>#include <string.h>#include <map>#include <set>#include <queue>#include <deque>#include <list>#include <bitset>#include <stack>#include <stdlib.h>#define lowbit(x) (x&-x)#define e exp(1.0)//ios::sync_with_stdio(false);//    auto start = clock();//    cout << (clock() - start) / (double)CLOCKS_PER_SEC;typedef long long ll;typedef long long LL;const int mod=1e9+7;using namespace std;ll qmod(ll n,ll k){    ll ans=1;    n=n%mod;    while(k)    {        if(k%2==0)        {            n=n*n%mod;            k/=2;        }        else        {            ans=ans*n%mod;            k--;        }    }    return ans;}int main(){    ll n,k;    int cas=1;    while(cin>>n>>k)    {        cout<<"Case #"<<cas++<<": "<<qmod(n,k)<<endl;    }    return 0;}