hdu 3826

Squarefree number

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1984    Accepted Submission(s):
526

Problem Description

In mathematics, a squarefree number is one which is
divisible by no perfect squares, except 1. For example, 10 is square-free but 18
is not, as it is divisible by 9 = 3^2. Now you need to determine whether an
integer is squarefree or not.

 

Input

The first line contains an integer T indicating the
number of test cases.
For each test case, there is a single line contains an
integer N.

Technical Specification

1. 1 <= T <= 20
2. 2
<= N <= 10^18

 

Output

For each test case, output the case number first. Then
output "Yes" if N is squarefree, "No" otherwise.

 

Sample Input

2

30

75

 

Sample Output

Case 1: Yes

Case 2: No

 

Author

hanshuai

采用如果素因子超过2次，NO。这个很好理解。

这样还是会超时。10^18次方，最坏的情况下10^9次方。

但是，我们只需要枚举10^6内的素数就可以。为什么？

 

假如n 有(10^6,10^9)的一个素因子平方构成 ,不在10^6范围，但是可能直接开方判断即可。

还有假如是(10^,10^9)的一个素因子平方*a构成，则a必在10^6内。不然数字就超过10^18次方了。

我们用[1,10^6]内的素数，刷一遍的时候，就可能除去a，然后又可以开方，直接判断就行。

 #include<iostream>
 #include<stdio.h>
 #include<cstring>
 #include<cstdlib>
 #include<math.h>
 using namespace std;
 typedef __int64 LL;

 bool s[];
 LL prime[],len;
 void init()
 {
     int i,j;
     len=;
     memset(s,false,sizeof(s));
     for(i=;i<=;i++)
     {
         if(s[i]==true)continue;
         prime[++len]=i;
         for(j=i*;j<=;j=j+i)
             s[j]=true;
     }
 }
 LL Euler(LL n)
 {
     LL i,num;
     for(i=;prime[i]*prime[i]<=n && i<=len;i++)
     {
         if(n%prime[i]==)
         {
             num=;
             while(n%prime[i]==)
             {
                 n=n/prime[i];
                 num++;
                 if(num>=) return ;
             }
         }
     }
     return n;
 }
 int main()
 {
     init();
     int T,t;
     LL n;
     scanf("%d",&T);
     for(t=;t<=T;t++)
     {
         scanf("%I64d",&n);
         LL ans = Euler(n);
         if(ans==)
         {
             printf("Case %d: No\n",t);
         }
         else
         {
             LL cur=(LL)sqrt(ans*1.0);
             if(cur*cur==ans)printf("Case %d: No\n",t);
             else printf("Case %d: Yes\n",t);
         }
     }
     return ;
 }