HDU1719

Friend

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1689    Accepted Submission(s): 835

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!
  
  

 

题解来自： http://972169909-qq-com.iteye.com/blog/1436549

题意：

①1,2都是friend数

②如果a，b都是friend数，那么ab+a+b也是friend数

任务：判断一个数n是不是friend数 (0<=n<=2^30)

 

设a, b都是friend数，

那么可以生成一个friend数 x = ab+a+b = (a+1)(b+1)-1

设c, d都是friend数，

那么可以生成一个friend数 y = (c+1)(d+1)-1

由x，y又可以生成friend数n = (x+1)(y+1)-1

代入得：n = [(a+1)(b+1)][(c+1)(d+1)]-1

1,2生成的是 (1+1)(2+1)-1;

1,1生成的是 (1+1)^2 - 1;

2,2生成的是 (2+1)^2 - 1;

由递归理解可知friend数n = [(1+1)^x * (2+1)^y] - 1;

AC code：

#include <iostream>
using namespace std;
int main()
{
	int n;
	while(cin>>n)
	{
		if(n==0){
			cout<<"NO!"<<endl;
			continue;
		}
		n++;
		while(n%2==0||n%3==0)
		{
			if(n%2==0)
				n/=2;
			if(n%3==0)
				n/=3;
		}
		if(n==1)
			cout<<"YES!"<<endl;
		else
			cout<<"NO!"<<endl;
	}
	return 0;
}