HDU 1719 Friend 【數學題】

 

 

Friend

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2005    Accepted Submission(s): 1009

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!

 

 

Source

 

2007省賽集訓隊練習賽（2）

 

 

第一次寫硬是沒有寫出來，一直以爲是打表寫的，發現也沒啥規律可循的，唯一找出來是一個遞推方程，可是感覺太麻煩，不想寫。後來看了別人寫的，感覺自己在數學方面還是有點欠缺的，所以有必要解釋解釋，給自己看。

 

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;

因爲原式的朋友數都是由1,2推到出來的

所以遞推到最底層，那麼(a1+1)*(b1+1)*(a2+1)*(b2+1)肯定是2,3的倍數（即是1+1,2+1）

所以最後就是要解決n+1得到的數到底是不是隻有2,3這些因子

 

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