杭电1719--Friend(找规律)

Friend

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2131    Accepted Submission(s): 1074

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!

 

 

Source

2007省赛集训队练习赛（2）

 

 

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friend数： n = (a + 1)(b + 1) - 1;

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
int main(){
    int n;
    while(~scanf("%d", &n)){
        if(!n){
            printf("NO!\n");
            continue;
        }
        n += 1;
        while(n % 2 == 0 || n % 3 == 0){
            if(n % 2 == 0)
                n /= 2;
            else
                n /= 3;
        }
        if(n == 1)
            printf("YES!\n");
        else
            printf("NO!\n");
    }
    return 0;
}