HDU1719(找规律)

Friend

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1444    Accepted Submission(s): 718

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!
 
friend的数满足一下要求之一：
1.1、2是friend 数
2.若a、b是friend数，则a*b+a+b是friend数
 
a*b+a+b=a*b+a+b+1-1=(a+1)*(b+1)-1
其中a、b是friend数，满足a=(c+1)*(d+1)-1,b=(e+1)*(f+1)-1
直到推到a*b+a+b=(1+1)^x*(1+2)^y-1,其中x,y为整数
 
 
#include<cstdio>int main(){int n;while(~scanf("%d",&n)){n++;if(n==1){printf("NO!\n");continue;}while(n){if(n%2==0)n/=2;elsebreak;}while(n){if(n%3==0)n/=3;else                break;}if(n==1)printf("YES!\n");elseprintf("NO!\n");}return 0;}