Find The Multiple
| Time Limit: 1000MS | Memory Limit: 10000K | |||
| Total Submissions: 27518 | Accepted: 11455 | 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
题意：
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
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int n, ls;
void dfs(unsigned long long ans, int c)
{
if(ls == 1)
return ;
if(ans % n == 0)
{
cout<<ans<<endl;
ls = 1;
return ;
}
if(c == 19)
return ;
dfs(ans*10, c+1);
dfs(ans * 10 + 1, c + 1);
}
int main()
{
while(scanf("%d", &n) != EOF)
{
if(n == 0)
break;
ls = 0;
dfs(1, 0);
}
return 0;
}