HDU 1719 Friend 【规律题】

来源：互联网发布：淘宝信用贷款入口编辑：程序博客网时间：2024/04/28 14:09

Friend

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2215 Accepted Submission(s): 1111

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

Sample Output

YES!YES!NO!

Source

2007省赛集训队练习赛（2）

思路：

刚开始就只想到把给出来的那个式子进行变形n=a*b+a+b=(a+1)(b+1)-1;然后（n+1）=(a+1)*(b+1),想了想也没有什么规律，然后在网上看到人家的推导过程，原来就差了一点！哎，无语了！

设a,b是朋友数，则x=(a+1)*(b+1)-1;

设c,d是朋友数，则y=(c+1)*(d+1)-1;

则n=(x+1)*(y+1)-1,n也是朋友数；

n=(x+1)*(y+1)-1=[(a+1)*(b+1)]*[(c+1)*(d+1)]-1;

...................

如果往后继续算，我们能够得到一个规律，就是任何一个朋友数都是由最基本的朋友数构成的，也就是任何一个朋友数n+1=（1+1）^p+(1+2)^q,(其中p,q>=0),这一点需要注意的是在判断的时候，并不是那个数加1是2或3的倍数就是朋友数，而是由多个2和多个3相乘得到的！具体看代码：

代码：

#include <stdio.h>#include <string.h>int main(){int n;while(scanf("%d",&n)!=EOF){if(n==0){printf("NO!\n");continue;}n=n+1;while(n%2==0){n=n/2;}while(n%3==0){n=n/3;}if(n==1)printf("YES!\n");else printf("NO!\n");}return 0;}

搜索

0 0