hdu---1719Friend

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//是一个由题设条件，推导公式的题目，一开始有点想法，但知道肯定超时。

//后来参看了傅总的结题报告，下面是其转载。学习学习！

hdu-1719 Friend

2011-07-25 14:50:47| 分类： ACM-hdu|举报|字号订阅

Friend

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 757 Accepted Submission(s): 353

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

13121

12131

Sample Output

YES!

NO!

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解题思路：

这是一道想法题！因为任意一个friend number可以由两个friend numbera，b构成，且为ab+b+a;

想法就是在这个表达式中!

因为 ab+a+b=ab+a+b+1-1=(a+1)*(b+1)-1

假设 a=(c+1)*(d+1)-1;b=(e+1)*(f+1)-1;

得到 ab+a+b=(c+1)*(d+1)*(e+1)*(f+1)-1

带入数值化简就得到，ab+a+b=((1+1)^x)*((1+2)^y)-1(x,y >=0)【也就是说c,d,e,f也可以看成是a和b，找到他们的c,d,e,f，找到最后，c,d,e,f一定能够用a和b来表示，所以代入1和2，得到这样的等式】

就把题目转化为一个数如果过能写成 (2^x)*(3^y)-1 那他就是一个friend number！

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Code：

#include<iostream>using namespace std; int main(){ #ifndef ONLINE_JUDGEfreopen("testin.txt","r",stdin); #endifint a; while(scanf("%d",&a)!=EOF){ if(!a) printf("NO!\n"); else { a++; while(a%2==0) a=a/2; while(a%3==0) a=a/3; if(a==1) printf("YES!\n"); else printf("NO!\n");}} return 0;}

------------------转载自 fudq

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