POJ 1426 Find The Multiple ——————DFS

雨点打透心脏的1/2处 2022-05-13 05:34 106阅读 0赞

**Find The Multiple**

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Find The Multiple

<table> 
 <tbody> 
  <tr> 
   <td><b>Time Limit:</b> 1000MS</td> 
   <td></td> 
   <td colspan="3"><b>Memory Limit:</b> 10000K</td> 
  </tr> 
  <tr> 
   <td><b>Total Submissions:</b> 42225</td> 
   <td></td> 
   <td><b>Accepted:</b> 17735</td> 
   <td></td> 
   <td>Special Judge</td> 
  </tr> 
 </tbody> 
</table>

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

Source

Dhaka 2002

—– 这道题让找 n 的 01 十进制倍数， 也就是说，找出一个01序列，它是 n 的倍数 第一位只能是1 所以就从 1 开始DFS 下一位只能是0或者1 所以说复杂度是  O(2n)  O ( 2 n )  的级别，有点儿不敢写，但是后来发现这个最大也就是  218  2 18  ,能过得 —

#include<cstdio>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    typedef long long ll;
    ll n,k,flag;
    void dfs(ll t,ll i)
    {
        if(i==19)    return ;
        if(flag)    return ;
        if(t%n==0)
        {
            flag=1;
            printf("%lld\n",t);
            return ;
        }
    
        dfs(t*10,i+1);
        dfs(t*10+1,i+1);
    }
    int main()
    {
        while(~scanf("%lld",&n)&&n)
        {
            flag=0;
            dfs(1,0);
        }
        return 0;
    }