poj 1426 深搜dfs

最新推荐文章于 2020-06-28 21:11:27 发布

活着就是为了自己想要的

最新推荐文章于 2020-06-28 21:11:27 发布

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分类专栏：深搜

本文链接：https://blog.youkuaiyun.com/qq_34289848/article/details/51992304

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本文介绍了一个编程问题，即寻找一个正整数n的倍数m，该倍数仅由0和1组成，且不超过100位。文章提供了一种递归深度优先搜索的方法，并附带了一个C++实现示例。

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Find The Multiple

Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%lld & %llu

Submit Status

Description

Given a positive integer n, write a program to find out a nonzero multiple m of n whose decimal representation contains only the digits 0 and 1. You may assume that n is not greater than 200 and there is a corresponding m containing no more than 100 decimal digits.

Input

The input file may contain multiple test cases. Each line contains a value of n (1 <= n <= 200). A line containing a zero terminates the input.

Output

For each value of n in the input print a line containing the corresponding value of m. The decimal representation of m must not contain more than 100 digits. If there are multiple solutions for a given value of n, any one of them is acceptable.

Sample Input

Sample Output

10
100100100100100100
111111111111111111

这道题唯一的坑点就是任意值都可以，你输出来就行，我输出6的倍数与样例的不同，然后交了下试试然后就对了。要好好学英语了！！！


  
   #include
   <stdio.h>
  

  
   int
    state
   ;///定义是否已找到那个倍数
  

  
   long
    
   long
    ans
   ;
  

  
   void
    
   dfs
   (
   long
    
   long
    a
   ,
    
   long
    
   long
    b
   ,
    
   int
    k
   )
  

  
   {
  
  
       
   if
   (
   state
   )
  
  
           
   return
    
   ;
  
  
       
   if
   (
   k
   ==
   19
   )
  
  
           
   return
    
   ;
  
  
       
   if
   (
   b
   %
   a
   ==
   0
    
   &&
    state
   ==
   0
   )//这个加不加state==0都行 我害怕出意外就加上了
  
  
       
   {
  
  
           state
   =
   1
   ;
  
  
           ans
   =
   b
   ;
  
  
           
   return
    
   ;
  
  
       
   }
  
  
   

  
  
       
   dfs
   (
   a
   ,
   10
   *
   b
   ,
   k
   +1
   );
  
  
       
   dfs
   (
   a
   ,
   10
   *
   b
   +1
   ,
   k
   +1
   );
  

  
   }
  

  
   int
    
   main
   ()
  

  
   {
  
  
       
   long
    
   long
    n
   ;
  
  
       
   while
   (
   scanf
   (
   "%lld"
   ,&
   n
   ),
   n
   )
  
  
       
   {
  
  
           state
   =
   0
   ;
  
  
           
   dfs
   (
   n
   ,
   1
   ,
   0
   );
  
  
           
   printf
   (
   "%lld
   \n
   "
   ,
   ans
   );
  
  
       
   }
  
  
       
   return
    
   0
   ;
  

  
   }