3

13121

12131

Sample Output

YES!

YES!

NO!

OJ-ID： 
HDU-1719

author：
Caution_X

date of submission：
20190930

tags：
数学推导

description modelling：
Friend number are defined recursively as follows.
(1) numbers 1 and 2 are friend number;
(2) if a and b are friend numbers, so is ab+a+b;
(3) only the numbers defined in (1) and (2) are friend number.
Now your task is to judge whether an integer is a friend number. 

major steps to solve it：
1.设n是Friend数，则n=ab+a+b=(a+1)*(b+1)-1   =>   n+1=(a+1)*(b+1)
2.a,b都是Friend数,所以a+1=(a'+1)*(b'+1),b+1=(a''+1)*(b''+1)直到a,b=1,2
3.那么：n+1= 2^x * 3^y

warnings：
0需要特判

AC CODE:

#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

int main()

{

    ll n;

    while(~scanf("%lld",&n)){

        if(n==){

            printf("NO!\n");

            continue;

        }

        n++;

        while(n%==)    n/=;

        while(n%==)    n/=;

        if(n==)    printf("YES!\n");

        else    printf("NO!\n");

    }

    return ;

}