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+;

 long long quickmod(long long a,long long b,long long m)
 {
     long long ans = ;
     while(b)//用一个循环从右到左遍历b的所有二进制位
     {
         if(b&)//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/=;
         a = a*a%m;
     }
     return ans;
 } 

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

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=;i<=sqrt((double)num)+;i++)
     {
         if(num%(i*i)==)
             return true;
     }
     return false;
 }

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

每一行三个数字分别表示n,k,res