POJ - 1426 Find The Multiple-优快云博客

本文链接：https://blog.youkuaiyun.com/acm_du/article/details/77855767

本文介绍了一个编程问题：给定一个正整数n，如何找出一个只包含0和1的数，该数为n的倍数。文章提供了一个使用C++实现的解决方案，通过递归深度优先搜索的方法来寻找符合条件的数。

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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
2
6
19
0
Sample Output
10
100100100100100100
111111111111111111
这道题题意很好理解，给一个n，找到一个只有0和1组成的数可以整除n。
一直long long不能输出结果，百度看到都用的unsigned long long。

#include <iostream>
#include <cstdio>
using namespace std;

int flag;
void dfs(unsigned long long num,int step,int n)
{
    if(flag)return;
    if(num%n==0)
    {
        flag=1;
        printf("%I64u\n",num);
        return;
    }
    if(step==19)return;//防止越界
    dfs(num*10,step+1,n);
    dfs(num*10+1,step+1,n);
    return;
}

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