Moscow subregional problem H. hometask conversion, prime number screening

ACM ICPC 2010-2011 Neerc Moscow subregional Contest Moscow, October 24, 2010

Problem H. hometask
Time Limit:1 Second
Memory limit:256 Megabytes

Kolya is still trying to pass a test on Numbers theory. The lecturer is so desperate about Kolya ' s
Knowledge that she gives him the same task every time.
The problem is to check if N! is divisible by n^2.

Input
The first line of input contains the only integer N (1≤n≤10^9).

Output
Please, print to output ' YES ' provided that N! is divisible by N2, otherwise print "NO".

Examples
stdin stdout
3 NO
9 YES

Note
N! = 1 · 2 · . . . · N

Source

Moscow Subregional 2010

My Solution

Prime number Screening

First, n^2 is divisible by n!.

i.e. N is (N-1)! Divisible

Arbitrarily basically as long as N is not a prime number can have a constraint other than N and 1,

So as long as the prime number is NO

Also for 4 because only 1 2 4 so the number of pages does not meet N is (N-1)! divisible, so 4 special treatment no//good caution, by the way, or 4 of special treatment may have to WA and later found

0! is 1 so N = = 1 also satisfies 1^1 is divisible by 1 YES

Complexity O (sqrt (n))

The code was implemented by the teammates, so the AC code was cs_lyj1997 to the teammate.

In addition we have also had another 2 questions, but that 2 questions, the author did not participate, so do not tidy up

#include <iostream> #include <cmath>using namespace Std;int main () {    int n,i;    bool F;    F=false;    cin>>n;    for (I=2;I<=SQRT (n); i++)        if (n%i==0) f=true;    if (n==1) f=true;    if (!f | | n==4) cout<< "NO" <<endl;    else cout<< "YES" <<endl;    return 0;}

Thank you!

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Moscow subregional problem H. hometask conversion, prime number screening