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!

题目大意：

问输入一个数是否为朋友数。下面是朋友数的定义： 
1. 1和2为朋友数 
2. 如果有一个朋友数为，另一个为b，那么a*b+a+b也是朋友数 
3. 满足上面的两个条件的所有情况都为朋友数 
是朋友数输出“YES!”，不是输出“NO!”。

C++

/*&同为1为1，|同为0为0。假如a和b都为朋友数，那么可以通过a*b+a+b得到另一个朋友数c。我们可以将其列出来：
a*b+a+b = c; -   (a+1)(b+1) = c+1;
a和b起初只有1和2。那么就可以看出来，一个朋友数c，加上1，肯定是2或3的倍数。输入一个数n，通过不断地对(n+1)除等3和2，最后如果得到的结果为1，那么就可以证明c为朋友数。
注意：0不为朋友数。*/
#include <iostream>
#include <stdio.h>
using namespace std;
int main()
{
    int a;
    while(~scanf("%d", &a))
    {
        if(a == 0)
        {
            cout << "NO!" << endl;
            continue;
        }

        int t = a + 1;
        while(t%2 == 0 || t%3 == 0)
        {
            if(t % 2 == 0)
            {
                t /= 2;
                continue;
            }
            if(t % 3 == 0)
            {
                t /= 3;
                continue;
            }
        }
        if(t == 1)
            cout << "YES!" << endl;
        else
            cout << "NO!" << endl;
    }
    return 0;
}