POJ 1426 Find The Multiple 【简单搜索】【dfs】

Find The Multiple

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 42382		Accepted: 17805		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

从1开始，每次递归递归两条线路

x*10和x*10+1

1可以变成 10 和 11

直到它可以整除n，递归19层左右。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
#define LL long long
#define M(a,b) memset(a,b,sizeof(a))
const int MAXN = 205;
const int INF =  0x3f3f3f3f;
LL vis[205];
LL n;
int flag;
void dfs(int x,LL num)
{
    if(x>=19) return ;
    if(flag==1) return;
    if(num%n==0)
    {
        flag = 1;
        printf("%lld\n",num);
        return ;
    }
    dfs(x+1,num*10);
    dfs(x+1,num*10 + 1);
}
int main()
{
    while(scanf("%lld",&n)&&n!=0)
    {
        flag = 0;
        M(vis,0);
        dfs(0,1);
    }
    return 0;
}