poj-1426-Find The Multiple(dfs)

Find The Multiple

Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 40713 Accepted: 17088 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

给定一个正整数 n，请编写一个程序来寻找 n 的一个非零的倍数 m，这个 m 应当在十进制表示时每一位上只包含 0 或者 1。你可以假定 n 不大于 200 且 m 不多于 100 位。
提示：本题采用 Special Judge，你无需输出所有符合条件的 m，你只需要输出任一符合条件的 m 即可。

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#include"iostream"
using namespace std;
typedef unsigned long long ll;
int n;
bool sign;

void dfs(ll x,int count)
{
    if(sign)
        return ;
    if(x%n==0){
        sign=true;
        cout<<x<<endl;
        return ;
    }
    if(count==19)//m 最多 200 位
        return ;
    dfs(x*10,count+1);
    dfs(x*10+1,count+1);
    //每两位数后两位有两种情况，10 或 11，深搜所有情况，找到一种就返回，找不到找另外一颗子树
}
int main()
{
    while(cin>>n&&n)
    {
        sign=false;
        dfs(1,0);
    }
    return 0;
}