HDU1719
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Friend
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1689 Accepted Submission(s): 835
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.
(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!
题解来自:http://972169909-qq-com.iteye.com/blog/1436549
题意:
①1,2都是friend数
②如果a,b都是friend数,那么ab+a+b也是friend数
任务:判断一个数n是不是friend数 (0<=n<=2^30)
设a, b都是friend数,
那么可以生成一个friend数 x = ab+a+b = (a+1)(b+1)-1
设c, d都是friend数,
那么可以生成一个friend数 y = (c+1)(d+1)-1
由x,y又可以生成friend数n = (x+1)(y+1)-1
代入得:n = [(a+1)(b+1)][(c+1)(d+1)]-1
1,2生成的是 (1+1)(2+1)-1;
1,1生成的是 (1+1)^2 - 1;
2,2生成的是 (2+1)^2 - 1;
由递归理解可知friend数n = [(1+1)^x * (2+1)^y] - 1;
AC code:
#include <iostream>using namespace std;int main(){int n;while(cin>>n){if(n==0){cout<<"NO!"<<endl;continue;}n++;while(n%2==0||n%3==0){if(n%2==0)n/=2;if(n%3==0)n/=3;}if(n==1)cout<<"YES!"<<endl;elsecout<<"NO!"<<endl;}return 0;}
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