Friend

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2073 Accepted Submission(s): 1046

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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lcy | We have carefully selected several similar problems for you: 1905 2082 1073 1715 1257

心塞的一题。。。。。。。

friend 数为ab+a+b，那么可以变成(a+1)(b+1)-1,只要给一个数加上1，然后判断他是否是由一系列2和一系列3组成的，

因为所有的数都是有(a+1)(b+1) ,2,3递推而来的。

#include <iostream>

#include <cstdio>

#define LL int

using namespace std;

void solve(LL x)

{

    while((x%)==)

    {

        x/=;

        if(x==)

            break;

    }

    while((x%)==)

    {

        x/=;

        if(x==)

            break;

    }

    if(x==)

       printf("YES!\n");

    else

        printf("NO!\n");

}

int main()

{

    LL x,T;

    while(~scanf("%d",&x))

    {

        if(x== || x==)

            printf("YES!\n");

        if(x==)

             printf("NO!\n");

        if(x!= && x!= && x!=)

        solve(x+);

    }

    return ;

}

巴特西

hdu 1719

Friend

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