Find The Multiple（深搜找倍数）

Find The Multiple

 

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

 

题目介绍：（这个题我是真的没翻译懂，很尴尬。。。。。）

题目大意是给出一个数n，找出一个数要求是n的倍数，并且这个数的十进制只由1和0组成，明显这样的数不止一个（如果，满足条件一定会有m×10也满足，故不止一种），题目要求输出任意一个满足该条件的m

对于数据1，可知2×5=10，故答案可以得出是10（当然，100，1000...也满足，但是special judge，只用输出一个满足条件的解），其他数据也同理。此题中，最后需要输出的数据很大，因此需要用 _int64，关于这方面的知识，我也是第一次了解，详见http://blog.csdn.net/zhangxiaoduoduo/article/details/78196982

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int n;int flag;void dfs(unsigned __int64 t,int k){if(flag==1)return ;          //找到答案后就可以停止了 if(t%n==0){printf("%I64u\n",t);      flag=1;return ;}if(k==19)return;         //19位以后就超出了可表示范围，所以要返回 dfs(10*t,k+1);dfs(10*t+1,k+1);}int main(){while(scanf("%d",&n)&&n){flag=0;dfs(1,0);}return 0;}