Find The Multiple

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 43294		Accepted: 18175		Special Judge

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

10100100100100100100111111111111111111

Source

大致题意：

给你一个n，让你算出一个每一位数都是1或0的n的倍数。

代码：

#include 
   
    #include 
    
     #include 
     
      #include 
      
       #include 
       
        #include 
        #include 
         
          #include 
          
           #include 
           
            using namespace std;bool ok;void DFS(int a,unsigned long long b, int k){ if (ok) { return; } if (b % a == 0) { ok = 1; cout << b << endl; return; } if (k == 19) return;//到了19层就回溯回去，走另一条路 DFS (a, b * 10, k + 1);//b有两种方式可以获取，一个是b*10，一个是b*10+1 DFS (a, b * 10 + 1, k + 1);}int main(){ int n; while (cin>>n) { if (n == 0) break;//别忘了终止条件 ok = false; DFS (n, 1, 0); } return 0;}

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