Find The Multiple--POJ1426

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

题意是求出n任意一个倍数而这个数是只有0和1构成的十进制整数，输出任意符合题意的答案即可；
我们可以输出最小的符合条件的；这个数第一位一定为1；接着判断10,11,101,100,110,111......;可以用bfs来写；

代码如下：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
using namespace std;
__int64 bfs(int n)
{
    queue<__int64>Q;
    Q.push(1);
    long long q;
    while(!Q.empty())
    {
        q=Q.front();
        Q.pop();
        if(q%n==0)
            return q;
        Q.push(q*10);
        Q.push(q*10+1);
    }
    return -1;
}

int main()
{
    int n;
    __int64 ans;
    while(scanf("%d",&n),n)
    {
        ans=bfs(n);
        printf("%I64d
",ans);
    }
    return 0;
}