E - Find The Multiple

Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%I64d & %I64u

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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

26190

 

Sample Output

10100100100100100100111111111111111111

题意：给出一个数n，然后求得一个数字k，数字满足：能被n整除，每一位只有0,1。这样的数字k会有很多个，然以输出一个就可以。

思路：n的最大值为200，用dfs从数字k的个位开始往高位搜索，每一位只有0或1。找到能被n整除的时候输出就可以了。

ps:(一开始不知道到底应该搜到多少位时可以return，后来看到网上说最多到19位就可以)

代码如下：

 

#include 
      
       using namespace std;int n,flag;void dfs(int k,long long cur){      //k:位数 cur:当前数字    if(k == 19 || flag) return ;    if(cur%n == 0){        cout<
       
        <
        
         >n,n != 0){        flag = 0;        dfs(0,1);           }    return 0;}