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POJ1426 Find The Multiple(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

 

題意:
給出一個數n,找出任意一個數m,m是n的倍數且只有0與1組成。

解題思路:

BFS。從1開始,依次添0和1,即

1

10 11

100 101 110 111

......

只要找到一個滿足條件的m即可結束搜尋。

 

AC程式碼:
 

#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;

queue<long long> q;

void bfs(int n)
{
	while(!q.empty())
	{
		q.pop();
	}
	q.push(1);
	while(!q.empty())
	{
		if(q.front()%n==0)
		{
			printf("%lld\n",q.front());
			return;
		}
		q.push(q.front()*10);
		q.push(q.front()*10+1);
		q.pop();
	}
}

int main()
{
	int n;
	while(~scanf("%d",&n)&&n)
	{
		bfs(n);
	}
	return 0;
}