Friend HDU杭电1719 【规律题】

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!
   
   
   
   

friend numbers = 2^x + 3^y -1
解释：

变形可得到  x = ab+a+b+1 = (a+1)(b+1)

如果x为friend数，a，b一定得是friend数a就可以写成

a = a1b1+a1+b1-1   b = a2b2+a2+b2-1;就这样一直递推就

一开始只有1和2是朋友数。故可以得到一个规律：
friend numbers + = 2^x + 3^y 

#include <stdio.h>
int main(){
	int n;
	while(~scanf("%d", &n))
	{
		if(!n)
		{
			printf("NO!\n");
			continue;
		}
		n+=1;
		while(n%3==0 || n%2==0)
		{
			if(n%3==0) n/=3;
			if(n%2==0) n/=2;
		}
		n==1?printf("YES!\n"):printf("NO!\n");
		
	}
	return 0;
}