Find The Multiple



C - Find The Multiple

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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

2
6
19
0

Sample Output

10
100100100100100100
111111111111111111

题目大意：

输入一个正整数n，寻找只由0和1组成的能够将n整除的十进制数

代码如下

#include <iostream>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<cstdlib>
#include<stack>
#include<queue>

using namespace std;

int n,m;

void dfs(long long t,int y)
{
    if(y > 18||m==1)//long long 最多存储十九位数，第二十位时已经超出数据范围，当找到结果时，要return，设置m=1，使递归结束
      return;
    if( t % n == 0)
    {
        m=1;
        printf("%lld\n",t);
        return;
    }
        dfs(t*10,y+1);//只由0,1组成的二进制数一定是由1为第一位数字，如不是1，增大数字则乘十或者乘十再加一
       dfs(t*10+1,y+1);//不能用y++，会改变y的值

}
int main()
{
    while (scanf("%d",&n)!=EOF&&n)
        {
            m=0;//标记是否找到答案
            dfs(1,0);
        }

    return 0;
}