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最新推荐文章于 2025-06-04 22:15:46 发布

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本文介绍了一个算法问题：给定一个正整数n，找出一个由0和1组成的非零倍数m，使得m可以被n整除。通过递归深度优先搜索的方法，在不超过100位数的情况下找到符合条件的最小m值。示例输入输出展示了如何使用该算法解决具体实例。

原题

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
思路解析
寻找能整除N的，由0,1组成的数。
AC代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
bool vis;
int n;
void dfs(unsigned __int64 x,int n,int k){
    if(vis) return;
    if(x%n==0){
        printf("%I64u\n",x);
        vis=true;
        return;
    }
    if(k==19) return;
    dfs(x*10,n,k+1);
    dfs(x*10+1,n,k+1);
}
int main(){
    while(scanf("%d",&n)){
        if(n==0) break;
        vis=false;
        dfs(1,n,0);
    }
    return 0;
}