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.
(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!
Source
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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!
");
}
else
{
n=n+1;
while(n%2==0)
n=n/2;
while(n%3==0)
n=n/3;
if(n==1)
printf("YES!
");
else printf("NO!
");
}
}
}