HDU 6063 17多校3 RXD and math（暴力打表题）

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

 打表大法好啊！打表之后发现就是求n^k%MOD

记得对n先做预处理取模，否则快速幂也救不了啊

#include<cstdio>
#include<iostream>
#include<cmath>
#include<cstring>
using namespace std;
const long long MOD=1e9+7;

long long quickmod(long long a,long long b,long long m) 
{ 
    long long ans = 1; 
    while(b)//用一个循环从右到左遍历b的所有二进制位 
    { 
        if(b&1)//判断此时b[i]的二进制位是否为1 
        { 
            ans = (ans*a)%m;//乘到结果上，这里a是a^(2^i)%m 
            b--;//把该为变0 
        } 
        b/=2; 
        a = a*a%m; 
    } 
    return ans; 
} 

int main()
{
    long long n,k;
    int t=1;
    while(~scanf("%lld%lld",&n,&k))
    {
        printf("Case #%d: ",t++);
        n%=MOD;
        printf("%lld
",quickmod(n,k,MOD));
    }
    return 0;
}

打表程序如下：

#include<cstdio>
#include<iostream>
#include<cmath>
using namespace std;

#define MOD 1000000000+7

bool panduan (long long num)
{
    long long i;
    for(i=2;i<=sqrt((double)num)+1;i++)
    {
        if(num%(i*i)==0)
            return true;
    }
    return false;
}

int main()
{
    int n,k;
    long long num;
    long long res=0;
    for(int n=1;n<=10;n++)
    for(int k=1;k<=10;k++)
    {
        res=0;
        num=pow((double)n,(double)k);
        for(int i=1;i<=num;i++)
        {
            if(!panduan(i))
                res+=(long long)(sqrt((double)(num/i)));
            res%=MOD;
        }
        printf("%d %d %lld
",n,k,res);
    }
    return 0;
}

 每一行三个数字分别表示n,k,res