POJ 1426 广搜BFS

DescriptionGiven 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.InputThe 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.OutputFor 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 Input2
6
19
0

Sample Output

10
100100100100100100
111111111111111111

题意：给出一个数n输出一个n的倍数m，m只能由0和1组成的10进制

一开始想不出是搜索，看到1010可以看成队列从而进行搜索得出。

 

思路，从0,1,10,11，一个一个进队开始搜索，知道求出解。

同时考虑剪枝。网上还看到可以用同余求模定理来做。打表太扯淡了。。。

看到最基本的广搜：

广搜思路：队列，每次进行队列的头，将最上层先完成，再慢慢往下。

广搜一般可利用queue队列或者数组。

循环while（head<tail）；

进行操作，操作中控制tail，每次操作head++；

 

#include<stdio.h>#include<string.h>#include<math.h>int n;long long now,q[1000000];void bfs(){        int head,tail;        head=0;        tail=1;        q[tail]=1;        while(head<tail)        {            head++;            now=q[head];            now=now*10;            if(now%n==0)            {                break;            }            tail++;            q[tail]=now;            tail++;            q[tail]=now+1;        }        printf("%I64d\n",now);}int main(){    while(scanf("%d",&n)!=EOF&&n!=0)        {            bfs();        }        return 0;}