POJ 1426 Find The Multiple

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本文探讨了一道编程题目，要求找到给定整数n的倍数t，使得t仅由0和1组成。通过深度优先搜索(DFS)和广度优先搜索(BFS)两种算法实现，展示了不同算法解决同一问题的效率对比。

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

Sample Output

10
100100100100100100
111111111111111111

题目大意：找给定的n的倍数t，且t的十进制表示只含0、1。

思路：bfs，贴一个别人的做法。顺带吐槽一下这题的数据也太水了。

dfs居然都能过：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#define INF 0x3f3f3f3f
using namespace std;
typedef long long ll;

int n,flag;

void dfs(unsigned long long t,int k)
{
    if(flag)
        return;
    if(t%n==0)
    {
        printf("%llu\n",t);
        flag=1;
        return;
    }
    if(k==19)
        return;
    dfs(t*10,k+1);
    dfs(t*10+1,k+1);
}

int main()
{
    while(~scanf("%d",&n)&&n)
    {
        flag=0;
        dfs(1,0);
    }
    return 0;
}

正解是bfs：

https://blog.youkuaiyun.com/lyy289065406/article/details/6647917