HDU 1719 Friend(规律)

Friend

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2838 Accepted Submission(s): 1472

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省赛集训队练习赛（2）

n=a*b+a+b 那么n+1=a*b+a+b+1=(a+1)(b+1) 而所有的友好数都有1和2演化而来，那么所有的友好数一定能被2或者3整除

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;

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