hdoj 1719 Friend

Friend

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2234    Accepted Submission(s): 1121

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!
 
思路： 设c是Friend数，题中给出公式ab+a+b。则c=ab+a+b。c+1=ab+a+b+1=(a+1)*(b+1),所以推断n+1是否满足这个公式就可以。又由于1,2是最小的Friend数。
所以推断n+1=(a+1)^x*(b+1)^y即n+1=2^x*3^y是否成立。
 
 
代码：
 
#include<stdio.h>
int main()
{
	int n,x,y;
	while(scanf("%d",&n)!=EOF)
	{
		n+=1;//n先加1； 
		x=0;y=0;
		while(n%2==0)
		{
			n=n/2;
			x++;
		}
		while(n%3==0)
		{
			n=n/3;
			y++;
		}
		if(n==1&&(x>0||y>0))//n=0时不是Friend数。 
		printf("YES!
");
		else
		printf("NO!
");
	}
	return 0;
}