杭电1719 简单数学题

       这道题经过推导之后可以得到n=2^x*3^y-1,凡是形如这样的数都是friend数，，之后程序就简单了，需要注意的是0不是friend数。。。。题目：

Friend

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 998    Accepted Submission(s): 481

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!

 

ac代码：

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