POJ 1426 Find The Multiple

题目链接：POJ 1426

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

多组输入，输入一个数n，然后寻找到一个只包含0和1的十进制数，并且这个数可以被n整除，然后输出它

题解：

刚开始想复杂了，以为有什么巧妙的公式，后来发现这是完全的暴力题好吧？？枚举从0,1,10,11开始依次类推的所有数，因为题目的长度限制，不用担心超时的问题，而且在找到的数不唯一时，输出任意一个就可以了。用bfs，从0,1开始，每次*10入队，*10+1入队，然后一直找下去，符合条件就输出。

代码

#include<iostream>
#include<queue>
using namespace std;
queue<long long>P;
int main() {
	int n;
	long long ans;
	while(cin>>n,n){
		while (!P.empty())
			P.pop();
		P.push(1);
		long long t;
		while (!P.empty()) {
			t = P.front();
			P.pop();
			if (t%n == 0)
			{
				ans = t;
				break;
			}
			P.push(t * 10);
			P.push(t * 10 + 1);
		}
		cout << ans << endl;
	}
	return  0;
}