HDOJ---1719 Friend[数论]

Friend

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1139    Accepted Submission(s): 558

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）

 

 

Recommend

lcy

 

 

 

 

 

 

 

题意：

①1,2都是friend数

②如果a，b都是friend数，那么ab+a+b也是friend数

任务：判断一个数n是不是friend数 (0<=n<=2^30)

 

设a, b都是friend数，

那么可以生成一个friend数 x = ab+a+b = (a+1)(b+1)-1

设c, d都是friend数，

那么可以生成一个friend数 y = (c+1)(d+1)-1

由x，y又可以生成friend数n = (x+1)(y+1)-1

代入得：n = [(a+1)(b+1)][(c+1)(d+1)]-1

1,2生成的是 (1+1)(2+1)-1;

1,1生成的是 (1+1)^2 - 1;

2,2生成的是 (2+1)^2 - 1;

由递归理解可知friend数n = [(1+1)^x * (2+1)^y] - 1;

 

 

 

 

 

code:

#include <iostream>   
#include <iomanip>   
#include <fstream>   
#include <sstream>   
#include <algorithm>   
#include <string>   
#include <set>   
#include <utility>   
#include <queue>   
#include <stack>   
#include <list>   
#include <vector>   
#include <cstdio>   
#include <cstdlib>   
#include <cstring>   
#include <cmath>   
#include <ctime>   
#include <ctype.h> 
using namespace std;

int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        n++;
        int x,y;
        x=y=0;
        while(n%2==0)
        {
            n/=2;
            x++;
        }
        while(n%3==0)
        {
            n/=3;
            y++;
        }
        if(n==1&&(x>0||y>0))
            printf("YES!\n");
        else
            printf("NO!\n");
    }
    return 0;
}