How can I use BFS to solve the multiple finding problem in POJ 1426?

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描述：给定一个正整数n，编写一个程序找出一个非零的n的倍数m，其十进制表示只包含数字0和1。假设n不超过200，并且存在相应的m。

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

求任意一个数是n的倍数，且该数的每一位只能是0或1。

思路

利用bfs，从最低的1开始枚举，对于每一位的两种情况进行判断，合理即输出。

AC 代码

#include <iostream>
#include<stdio.h>
#include<queue>
#include<math.h>
using namespace std;
typedef __int64 LL;

LL bfs(LL n)
{
queue<LL>sk;
sk.push(1);
while(!sk.empty())
{
LL s=sk.front();
sk.pop();
if(s%n==0)return s;
sk.push(s*10);
sk.push(s*10+1);
}
return 0;
}
int main()
{
LL n;
while(~scanf("%I64d",&n)&&n)
printf("%I64d\n",bfs(n));
}