17暑假多校联赛3.8 HDU 6063 RXD and math-优快云博客

RXD and math

Time Limit: 2000/1000 MS (Java/Others)
Memory Limit: 524288/524288 K (Java/Others)

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

题目网址：http://acm.hdu.edu.cn/showproblem.php?pid=6063

分析

题意：根据给定的公式，求解
比较迷的一道题，推公式推不出来，然后打表发现，就是求n的k次方，直接用整数快速幂，不会的话点超链接看看，记得不断取模

代码

#include <bits/stdc++.h>
using namespace std;
long long MOD=1e9+7;
long long poww(long long a,long long b)
{
    long long ans=1,res=a;
    while(b!=0)
    {
        if(b&1!=0)ans=(ans*(res%MOD))%MOD;
        res=((res%MOD)*(res%MOD))%MOD;
        b>>=1;
    }
    return ans;
}
int main()
{
    long long n,k,i=0;
    while(~scanf("%lld%lld",&n,&k))
    {
        printf("Case #%lld: %lld\n",++i,poww(n,k));
    }
}