Find The Multiple

Time Limit: 1000MS

 

Memory Limit: 10000K

Total Submissions: 32261

 

Accepted: 13470

 

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


【分析】

比较简单的一道搜索题...*10和*10+1两种情况，long long能吃的下最终情况..所以直接搜索就可以了

不过这里有一种，非常简洁的方式，因为只有两种情况*10或者*10+1，非常像二叉树，所以这里可以用哈夫曼思想，直接在数组中模拟搜索

【代码】


#include<iostream>  
#include<stdio.h>  
#include<string.h>  
#include<queue>  
using namespace std;  
  
long long f[1000000];  
  
int main()  
{  
    int n;f[0]=0;
    while(~scanf("%d",&n)&&n)  
    {  
        int i;  
        for(i=1;;i++)  
        {  
            f[i]=f[i/2]*10+i%2;  
            if(f[i]%n==0) break;  
        }  
        printf("%lld\n",f[i]);  
    }  
    return 0;  
}

dfs 


#include <stdio.h>
int n;
long long ans=-1;
void dfs(long long now,int deep)
{
  if (ans!=-1) return;
  if (deep>19) return;
  if (now%n==0)
  {
    ans=now;return;
  }
  
  dfs(now*10,deep+1);
  if (ans!=-1) return;
  dfs(now*10+1,deep+1);
}

int main()
{
  while (~scanf("%d",&n)&&n)
  {
    ans=-1;
    dfs(1,1);
    printf("%lld\n",ans);
  }
  return 0;
}