hdoj 1719 Friend

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Friend

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2022 Accepted Submission(s): 1019

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!

好巧的题：题目中所给判断式 a*b+a+b = (a+1)*(b+1)-1。

一开始只有1，2两个friend number，所以对于一个数 n ，我们只需判断 n+1 能否写成 2* x + 3 * y的形式即可（x , y均为整数 )

代码：

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

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