hdoj 1719 Friend（公式）

Friend（链接）

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!
   
   

注： （1）1和2是友好数

（2）如果 a 和 b 都是友好数的话  a*b+a+b 也是友好数

嗯嗯嗯：

当 c = 0 是直接输出 NO！！！！！！！！别忘记感叹号

令 c = a*b+a+b；注意到（a + 1）*（b + 1）= c + 1；

所以 c + 1 不断除以2 然后除以3 得到的结果如果是1的话 c 就是完美数

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

int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        if(n==0)
        {
            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;
}