friend（hdoj 1719）

Friend

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2207    Accepted Submission(s): 1108

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!

 

Source

2007省赛集训队练习赛（2）

 

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/*刚刚看到这道题的时候一心只想着打表，但是试了几次都超时，没想到有这种规律....c=ab+b+a=b(a+1)+b=(b+1)(a+1)-1,c+1=(b+1)(a+1),所以输入的数加一之后，它只要是2或者3的倍数就可以*/#include<stdio.h>#include<string.h> int main(){int n;while(scanf("%d",&n)!=EOF){if(n<1){printf("NO!\n");}else{n=n+1;while(n%2==0)n=n/2;while(n%3==0)n=n/3;if(n==1)printf("YES!\n");else printf("NO!\n");}}}