Friend
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2200 Accepted Submission(s): 1103
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!
//看了半天没看懂，最后看了一下题解才知道规律
//a=1;b=2;
//=a*b+a+b=(a+1)(b+1)-1 === >>c+1=(a+1)(b+1);
//所以输入的数+1，只要是2或者3的倍数就是了
#include<stdio.h>
#include<string.h>
#include<math.h>
int main(){
int n,m,i,j;
while(scanf("%d",&n)!=EOF)
{
if(n<1)
printf("NO!\n");
else
{
n=n+1;
while(n%2==0)
n/=2;
while(n%3==0)
n/=3;
if(n==1)
printf("YES!\n");
else
printf("NO!\n");
}
}
return 0;
}
