HDOJ 题目1719 Friend（数学）

Friend

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1904    Accepted Submission(s): 949

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）

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 当n=a+b+a*b时是friend数即n+1=（a+1）*（b+1）所以n+1可以表示2^x*3^y（n==0时不符合）

ac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++;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");}}