17 多校 3

RXD and math

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 984    Accepted Submission(s): 540

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

 

  找规律是个好方法！

可以打表发现公式可转换为 n^k

#include<iostream>#include<string.h>#include<stdio.h>using namespace std;#define ll long longll n,k,mod = 1e9 + 7;ll qkm(ll n,ll m){    ll base = n;    n = 1;    while(m)    {        if(m%2)            n = (n * base) % mod;        base = (base * base) % mod;        m/=2;    }    return n;}int main(){    int cas = 0;    while(scanf("%lld%lld",&n,&k)!=EOF)        printf("Case #%d: %lld\n",++cas,qkm(n%mod,k));    return 0;}