Friend
http://acm.hdu.edu.cn/showproblem.php?pid=1719
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2002 Accepted Submission(s): 1006
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
Sample Output
Source
/*
这道题之前我用模拟，类似斐波那契数列那样的模拟，发现有问题，后来发现找规律才是真理
思路：
假设：x 时friend number;
x=a*b+a+b
x+1=a*b+a+b+1
x+1=(a+1)(b+1);
又因为 1,2均是friend number；
所以 我们就看x+1 能否被2 3整除；
如果能被2和3整除，那么就是friend number
*/
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<algorithm>
using namespace std;
int main()
{
int a;
while(~scanf("%d",&a))
{
if(a==0)
printf("NO!\n");
else
if(a==1||a==2)
printf("YES!\n");
else
{
int n=a;
n+=1;
while(n!=1)
{
if(n%2==0)
n/=2;
else if(n%3==0)
n/=3;
else break;
}
if(n==1)
printf("YES!\n");
else
printf("NO!\n");
}
}
return 0;
}hdoj 1719 Friend(找规律)
原文：http://blog.csdn.net/lh__huahuan/article/details/44539279
