Find The Multiple

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 27297 Accepted: 11361 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

26190

Sample Output

10100100100100100100111111111111111111

#include<iostream>#include<cstdlib>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<queue>using namespace std; bool found; void DFS(unsigned __int64 t ,int n,int k) {     if(found)         return ;//如果已经发现了答案就没搜的必要了     if(t%n==0)     {//发现答案，输出，标记变量该true         printf("%I64u\n",t);        found=true;         return ;     }     if(k==19)//到第19层，回溯         return ;     DFS(t*10,n,k+1);    //搜索×10     DFS(t*10+1,n,k+1);    //搜索×10+1 } int main() {     int n;     while(cin>>n,n)     {         found=false;//标记变量，当为true代表搜到了题意第一的m         DFS(1,n,0);    //从1开始搜n的倍数，第三个参数代表搜的层数，当到第19层时返回（因为第20层64位整数存不下）    }     return 0;}