POJ 1426 Find The Multiple（DFS，BFS）

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：
　　对于一个数字n,找到n的倍数且这个数的每一位只能是0和1。
题解:
　　因为是special judge，所以只要输出任何一个满足条件的就可以了。因为n<=200，所以最小满足条件的答案在18位以内（long long就足够了）。
两种解法：DFS和BFS都行。
DFS：

#include<iostream>
using namespace std;
int n;
bool flag;
void dfs(long long x,int k)
{
    if(flag)
        return;
    if(x%n==0)
    {
        flag=true;
        cout<<x<<endl;
        return ;
    }
    if(k==18)//搜索到18位的时候还没找到就返回
        return ;
    dfs(x*10,k+1);
    dfs(x*10+1,k+1);
}
int main()
{
    while(cin>>n&&n)
    {
        flag=false;
        dfs(1,0);
    }
    return 0;
}

 BFS：

#include<iostream>
#include<cstring>
#include<queue>
using namespace std;
typedef long long LL;
LL bfs(int n)
{
    queue<LL> que;
    que.push(1);//从1开始搜索
    while(que.size())
    {
        LL cur=que.front();
        que.pop();
        if(cur%n==0)
            return cur;
        que.push(cur*10);
        que.push(cur*10+1);
    }
    return -1;
}
int main()
{
    int n;
    while(cin>>n,n)
    {
        cout<<bfs(n)<<endl;
    }
    return 0;
}