poj1426 Find The Multiple (DFS)

题目：

Find The Multiple

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 41845		Accepted: 17564		Special Judge

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

Source

Dhaka 2002

题意：找出一个最小的由1或者0组成的十进制数能整除n

题解：dfs，一开始看题目说十进制数会达到100位，其实不会，在longlong的范围内，下次遇到这种可以直接打表找一下规律

代码:

#include<iostream>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PII;
const ll mod=;
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
//head
#define MAX 1000
int n;
int flag;
void dfs(int cur,ll sum)
{
    if(cur==)return ;
    if(flag) return ;
   if(sum%n==&&sum!=)
   {
       flag=;
       cout<<sum<<endl;
       return ;
   }
   else
    dfs(cur+,sum*),dfs(cur+,sum*+);
}
int main()
{
      cin.tie();
    cout.tie();
    ios::sync_with_stdio();
    while(cin>>n)
    {
        if(n==)break;
        flag=;
        dfs(,);
    }
    return ;
}