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hdu 1719

时间:2015-05-26 18:09:09      阅读:221      评论:0      收藏:0      [点我收藏+]

Friend

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2073    Accepted Submission(s): 1046

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
3 13121 12131
 
Sample Output
YES! YES! NO!
 
Source
 
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心塞的一题。。。。。。。
friend 数为ab+a+b,那么可以变成(a+1)(b+1)-1,只要给一个数加上1,然后判断他是否是由一系列2和一系列3组成的,
因为所有的数都是有(a+1)(b+1) ,2,3递推而来的。
#include <iostream>
#include <cstdio>
#define LL int
using namespace std;

void solve(LL x)
{
    while((x%2)==0)
    {
        x/=2;
        if(x==1)
            break;
    }
    while((x%3)==0)
    {
        x/=3;
        if(x==1)
            break;
    }
    if(x==1)
       printf("YES!\n");
    else
        printf("NO!\n");
}
int main()
{
    LL x,T;
    while(~scanf("%d",&x))
    {
        if(x==1 || x==2)
            printf("YES!\n");
        if(x==0)
             printf("NO!\n");
        if(x!=1 && x!=2 && x!=0)
        solve(x+1);
    }
    return 0;
}

 

hdu 1719

原文:http://www.cnblogs.com/xianbin7/p/4530879.html

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