E - Find The Multiple POJ - 1426_given a positive integer n, write a program to fin-优快云博客

本文链接：https://blog.youkuaiyun.com/qwqwdqwqwe/article/details/80471853

本文介绍了一个编程挑战：对于给定的正整数n（n≤200），找到一个只包含0和1的非零倍数m，且m的十进制表示不超过100位。文章提供了一个使用深度优先搜索算法的示例程序，该程序能够解决这个问题，并展示了一些输入输出示例。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <vector>

using namespace std;
typedef long long ll;
int t, n, m, flag;
void DFS(int cur, ll num) {
    if(cur == 19 || flag) {
        return;
    }
    if(num % n == 0) {
        cout << num << endl;
        flag = 1;
        return;
    }
    DFS(cur + 1, num * 10);
    DFS(cur + 1, num * 10 + 1);
}
int main() {
    ios::sync_with_stdio(false);
    while(cin >> n && n) {
        flag = 0 ;
        DFS(0, 1);
    }
    return 0;
}