hdu1719-Friend（规律）

Problem C

Time Limit : 1000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 53   Accepted Submission(s) : 16

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

 

这个题主要是看个人观察力了！

假如  n=a*b+a+b

那么n+1=a*b+a+b+1=(a+1)(b+1)

而所有的友好数都有1和2演化而来，那么所有的友好数一定能被2或者3整除

所以：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int main()
{
	int n;
	while(scanf("%d",&n)!=EOF)
	{
		if(n>0)
		n+=1;
		if(n>0)//这里必须要排除n==0的情况！
		while(n%2==0||n%3==0)
		{
			if(n%2==0)
			n/=2;
			if(n%3==0)
			n/=3;
		}
		if(n==1)
		printf("YES!\n");
		else
		printf("NO!\n");
	}
	return 0;
}