POJ1426 Find The Multiple

                                                                                  Find The Multiple

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 12097		Accepted: 4968		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

2
6
19
0

Sample Output

10
100100100100100100
111111111111111111

Source

Dhaka 2002

 

 

思路：寻找能整除N的，由0,1组成的数。本题采用DFS，虽然题目要求最后的结果可能在100digits。但是实际情况下，在unsigned long long  范围内完全可以找到相应的答案。

 

采用 unsigned long long 的版本：

 

#include <cstdlib>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <map>
#include <string>
#include <algorithm>




using namespace std;

unsigned long long answer;



int dfs(unsigned long long prenum,unsigned long long lev,int n)
{
   if(prenum>=lev)
   return 0;
    
    if(lev%n==0)
    {answer=lev;return 1;}
    
    if(dfs(lev,lev*10,n))
    {return 1;}
    
    return dfs(lev,lev*10+1,n);
    
    
}






int main(int argc, char *argv[])
{
    int n;
    int i,j,k;
    
    while(scanf("%d",&n)!=EOF)
    {
                              if(!n)
                              break;
                              
                              if(dfs(0,1,n))
                              cout<<answer<<endl;
    }
                              
                              
                              
                              
   // system("PAUSE");
    return EXIT_SUCCESS;
}