HDU1719：Friend

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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

31312112131

Sample Output

YES!YES!NO!

这道题就是以道纯数学题，我记得第一次看到这道题还是在刚刚搞ACM不久的时候，当时硬是没有想出解决的方法

直到昨天训练赛的时候又出了这道题，仍然是没有弄出来，果然我的数学思维还是不行啊，在看了别人的题解之后，终于算是看懂意思了，在此有必要解释一下

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;

因为原式的朋友数都是由1,2推到出来的

所以递推到最底层，那么(a1+1)*(b1+1)*(a2+1)*(b2+1)肯定是2,3的倍数（即是1+1,2+1）

所以最后就是要解决n+1得到的数到底是不是只有2,3这些因子

#include <stdio.h>int main(){    __int64 n;    while(~scanf("%I64d",&n))    {        if(!n)        {            printf("NO!\n");            continue;        }        n++;        while(n%2==0)            n/=2;        while(n%3 == 0)            n/=3;        if(n==1)            printf("YES!\n");        else            printf("NO!\n");    }    return 0;}