【原】 POJ 1426 Find The Multiple BFS搜索 解题报告

 

http://poj.org/problem?id=1426

方法：

找到n的倍数m，m中只能有0或1
0，1可构造出二叉树进行BFS搜索
该题的重点就在于如何构造出BFS进行搜索。
所有0，1组成的数可以构造出一颗二叉树从而进行BFS。每层的数字长度相同，每个节点的左右孩子为其末尾分别添加0，1
BFS搜索一颗二叉树。二叉树的节点按层序遍历依次为(1)，(10，11)，(100，101，110，111)....
即每个节点vertex的邻接点为10*vertex和10*vertex+1

               1
              /  \
             /    \
            /      \
          10        11
          / \       /\
         /   \     /  \
       100  101  110  111

 

注意：
1、使用STL的queue会TLE，改用自己手写的才126MS左右，差别很大
2、自己手写的queue的capacity是测试出来的，即使得200以内的答案正确即可
3、虽然题目中说m可能有200位那么长，似乎要用到bignum，但是经测试可知__int64足够了。

 

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

2

6

19

0

Sample Output

10

100100100100100100

111111111111111111

 

   1: void run1426_2()

   2: {

   3:     int n ;

   4:     __int64 m ;

   5:     const int capacity = 9999999 ;

   6:     __int64 *myQueue = new __int64[capacity] ; 

   7:     int front,rear ;

   8:  

   9:     //for( n=1 ;n<=200; ++n )

  10:     while( scanf("%d",&n) && n!=0 )

  11:     {

  12:         rear = 0 ;

  13:         front = 1 ;

  14:  

  15:         myQueue[++rear] = 1 ;

  16:         while( true )

  17:         {

  18:             m = myQueue[front] ;

  19:             if(++front == capacity)

  20:                 front = 0 ;

  21:  

  22:             if( m%n==0 )

  23:                 break ;

  24:  

  25:             m = 10*m ;

  26:             if( m%n==0 )

  27:                 break ;

  28:             if(++rear == capacity)

  29:                 rear = 0 ;

  30:             myQueue[rear]=m ;

  31:  

  32:             m += 1 ;

  33:             if( m%n==0 )

  34:                 break ;

  35:             if(++rear == capacity)

  36:                 rear = 0 ;

  37:             myQueue[rear]=m ;

  38:         }

  39:         //printf("%d\t%d\n",n,m) ;

  40:         printf("%I64d\n",m) ;

  41:     }

  42:     delete []myQueue ;

  43: }