POJ 1426 Find The Multiple

最新推荐文章于 2021-02-08 19:07:26 发布

原创最新推荐文章于 2021-02-08 19:07:26 发布 · 168 阅读

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本文介绍了一个编程问题，即给定一个正整数n，如何通过递归深度优先搜索找到一个只包含0和1的非零倍数m。文章提供了一个C++实现示例，该算法能够解决输入限制内的问题，并给出了示例输入输出。

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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<cstring>
using namespace std;
int found;
void DFS(unsigned long long t,int n,int k)
{
    if(found) return ;
    if(t%n==0)
    {
        printf("%llu\n",t);
        found=1;
        return ;
    }
    if(k==19) return ;
    DFS(t*10,n,k+1);
    DFS(t*10+1,n,k+1);
}
int main()
{
    int n;
    while(cin>>n)
    {
        if(n==0) break;
        found=0;
        DFS(1,n,0);
    }
    return 0;
}

unsigned long long 要用%llu来输出

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