poj 1426 Find The Multiple

poj 1426 的传送门

Language:
Find The Multiple
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 21735 Accepted: 8939 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

题目大意：给你一个数，让你找可以整除这个数的并且仅仅含有0和1 的数，比方说
input :3;
output : 111;
可能有多个答案。仅仅须要输出一个就可以；
解题思路：用dfs搜索。仅仅搜关于0和1 的数，详情见代码。，，，

上代码：

#include <iostream>

using namespace std;
bool fo;
void dfs(unsigned long long t, int k, int m)//unsigned一定要有，这是一个无符号字符类型,
{
    if(fo)//一定要有这句话，要不然会有非常多个答案的。
        return ;
    if(t % m == 0)
    {
        cout<<t<<endl;
        fo=1;//标记
        return;
    }
    if(k == 19)//long long 最多有19位，
        return;
    dfs(t*10, k+1, m);//搜*10的
    dfs(t*10+1, k+1, m);//搜*10+1的
}
int main()
{
    int m;
    while(cin>>m,m)
    {
        fo=0;
        dfs(1, 0, m);
    }
    return 0;
}