hdu 1719 Friend

Friend

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1292    Accepted Submission(s): 630

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!

 

数学推论题：此题题意是一个friend number是最终由1和2这对友好数字演化而来的，且这个friend number必须满足由两个其他friend number这样构成ab+a+b；

假设这个friend number=ab+a+b；

那么：

friend number=ab+a+b+1-1；

friend number=(a+1)*(b+1)-1;

因为a和b也是其他friend number,那么可设a=(c+1)*(d+1)-1;b=(e+1)*(f+1)-1;代入得：

friend number=(c+1)*(d+1)*(e+1)*(f+1)-1;

又因为c,d,e,f也是其他friend number，可以在设在代入，一直带到原始的friend number，1和2为止，可以得到friend number=（(1+1)^x）*((1+2)^y)-1;

friend number=(2^x)*(3^y)-1;(x,y>=0)；

也就是只要这个数 N 满足 N++后能被2整数得到一个整数X后继续能被3整数得到一个Y，那么这个数 N 就是friend number了；

C++代码如下：(比较灵活，这里换了另外一种思路去避开找X和Y直接求结果)；

 

#include<iostream>
#include<cstdio>
using namespace std;

int main()
{
	int n;
	while (scanf("%d",&n) == 1)
	{
		if (n == 0)
		{
			printf("NO!\n");
			continue;
		}
		n++;
		while(n)
		{
			if (n % 2 == 0)
			    n /= 2;
			else 
			    break;
		}
		while(n)
		{
			if (n % 3 == 0)
			    n /= 3;
			else 
			    break;
		}
		if (n == 1)
			printf("YES!\n");
		else 
			printf("NO!\n");		
	}
	return 0;
}