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。

 1 #include<iostream>

 2 #include<stdio.h>

 3 #include<cstring>

 4 #include<cstdlib>

 5 #include<math.h>

 6 using namespace std;

 7 typedef __int64 LL;

 8 

 9 bool s[1000002];

10 LL prime[78500],len;

11 void init()

12 {

13     int i,j;

14     len=0;

15     memset(s,false,sizeof(s));

16     for(i=2;i<=1000000;i++)

17     {

18         if(s[i]==true)continue;

19         prime[++len]=i;

20         for(j=i*2;j<=1000000;j=j+i)

21             s[j]=true;

22     }

23 }

24 LL Euler(LL n)

25 {

26     LL i,num;

27     for(i=1;prime[i]*prime[i]<=n && i<=len;i++)

28     {

29         if(n%prime[i]==0)

30         {

31             num=0;

32             while(n%prime[i]==0)

33             {

34                 n=n/prime[i];

35                 num++;

36                 if(num>=2) return 0;

37             }

38         }

39     }

40     return n;

41 }

42 int main()

43 {

44     init();

45     int T,t;

46     LL n;

47     scanf("%d",&T);

48     for(t=1;t<=T;t++)

49     {

50         scanf("%I64d",&n);

51         LL ans = Euler(n);

52         if(ans==0)

53         {

54             printf("Case %d: No\n",t);

55         }

56         else

57         {

58             LL cur=(LL)sqrt(ans*1.0);

59             if(cur*cur==ans)printf("Case %d: No\n",t);

60             else printf("Case %d: Yes\n",t);

61         }

62     }

63     return 0;

64 }