寒假前刷题（5）搜索系列 bfs poj 1426

Find The Multiple

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 13192		Accepted: 5389		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

这是一道非常简单的bfs题目，其实如果不是放在搜索这一类里，我都不敢相信这是一道搜索的题目，首先题目所给的数据非常之大，所以如果想要用暴力去解题的话，肯定会超，所以可以用bfs，将其建成一颗树的样子，如下图：

然后按照层次去遍历他，一旦符合条件即可输出

我的这个程序是用数组来代替队列模拟的，写的比较挫，大牛请绕道

代码如下：

#include<stdio.h>
#include<stdlib.h>
long long q[9999999];
int n;int bfs()
{
int front,rear;
long long top;
front=rear=0;
q[rear]=1;
rear++;
while(rear>front)
{
top=q[front];
if(top%n==0)
break;
top*=10;
front++;
q[rear++]=top;
q[rear++]=top+1;
}
printf("%lld\n",top);
}int main()
{
while(scanf("%ld",&n)&&n)
{
bfs();
}
return 0;
}