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:

 1 #include <iostream>   

 2 #include <iomanip>   

 3 #include <fstream>   

 4 #include <sstream>   

 5 #include <algorithm>   

 6 #include <string>   

 7 #include <set>   

 8 #include <utility>   

 9 #include <queue>   

10 #include <stack>   

11 #include <list>   

12 #include <vector>   

13 #include <cstdio>   

14 #include <cstdlib>   

15 #include <cstring>   

16 #include <cmath>   

17 #include <ctime>   

18 #include <ctype.h> 

19 using namespace std;

20 

21 int main()

22 {

23     int n;

24     while(~scanf("%d",&n))

25     {

26         n++;

27         int x,y;

28         x=y=0;

29         while(n%2==0)

30         {

31             n/=2;

32             x++;

33         }

34         while(n%3==0)

35         {

36             n/=3;

37             y++;

38         }

39         if(n==1&&(x>0||y>0))

40             printf("YES!\n");

41         else

42             printf("NO!\n");

43     }

44     return 0;

45 }