（周日赛1）Friend

自己尝试推 但是没有推出来 太不机智了！后来提交的时候还忘记加‘！’ wa了好几次也没发现=。=

///转

首先，原式ab+a+b = ab+a+b+1-1 = (a+1)*(b+1)-1

令a = (a1+1)*(a2+1)-1;

b = (b1+1)*(b2+1)-1;

代入原式中可得：n = (a1+1)*(b1+1)*(a2+1)*(b2+1)-1;

因为原式的朋友数都是由1,2推到出来的

所以递推到最底层，那么(a1+1)*(b1+1)*(a2+1)*(b2+1)肯定是2,3的倍数（即是1+1,2+1）

所以最后就是要解决n+1得到的数到底是不是只有2,3这些因子

 

 

Problem Description

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.

 

 

Input

There are several lines in input, each line has a nunnegative integer a, 0<=a<=2^30.

 

 

Output

For the number a on each line of the input, if a is a friend number, output “YES!”, otherwise output “NO!”.

 

 

Sample Input

3

13121

12131

 

 

Sample Output

YES!

YES!

NO!

 

 

#include<stdio.h>

int main()

{

    __int64 n;

    while(~scanf("%I64d",&n))

    {

        if(!n) 

        {

        puts("NO!");

        continue;

        }

        n++;

        while(n%2==0||n%3==0)

        {

        if(n%2==0)  n/=2;

        else if(n%3==0)    n/=3;

        }

        if(n==1)

        puts("YES！");

        else puts("NO!");

    }

    return 0;

}