hdu1719-Friend（规律）

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Problem C

Time Limit : 1000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 53 Accepted Submission(s) : 16

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

31312112131

这个题主要是看个人观察力了！

假如 n=a*b+a+b

那么n+1=a*b+a+b+1=(a+1)(b+1)

而所有的友好数都有1和2演化而来，那么所有的友好数一定能被2或者3整除

所以：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int main(){int n;while(scanf("%d",&n)!=EOF){if(n>0)n+=1;if(n>0)//这里必须要排除n==0的情况！while(n%2==0||n%3==0){if(n%2==0)n/=2;if(n%3==0)n/=3;}if(n==1)printf("YES!\n");elseprintf("NO!\n");}return 0;}

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