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
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 ",t); 55 } 56 else 57 { 58 LL cur=(LL)sqrt(ans*1.0); 59 if(cur*cur==ans)printf("Case %d: No ",t); 60 else printf("Case %d: Yes ",t); 61 } 62 } 63 return 0; 64 }