POJ1426 Find The Multiple (DFS)

                                     

E - Find The Multiple

  POJ - 1426

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

         题目：找到一个能被n整除得并且只由0和1组成得十进制数

        利用dfs搜索

        代码：

      

#include <iostream>
#include <cstdio>

using namespace std;

typedef long long ll;

int n,flag;

void dfs(int num,ll ans)
{
    if(flag==1||num>19) return;

    if( ans%n==0)
    {
        flag = 1;
        printf("%lld\n",ans);
        return ;
    }

    dfs(num+1,ans*10);
    dfs(num+1,ans*10+1);
}



int main()
{

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

另一种大佬得方法：

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;

long long mod[600001];

int main()
{
    int n;
    while(~scanf("%d",&n)&&n)
    {
        int i;
        for(i=1;;i++)
        {
            mod[i]=mod[i/2]*10+i%2;
            // x=(x/10)%n*10+i%2;
            // mod[i]表示十进制里长成二进制i这样的数
            if(mod[i]%n==0)
            {
                break;
            }
        }
        printf("%I64d\n",mod[i]);
    }
    return 0;
}