POJ--1426.Find The Multiple（DFS）

Find The Multiple

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 72588		Accepted: 28677		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

连样本数据都没过，但提交了就是对的，就离谱。

看网上大神都是这么写的，思路是没错 

比较疑惑的是if(k==19)return;//剪枝  ，因为m最大可以到100位数，但是我们直接就取n<19位的....直接剪枝，咱也不知道为啥。

#include
#include
#include
using namespace std;
typedef long long ll;
int n;
int f;

void dfs(ll x,int k)
{
	if(f)return;
	if(k==19)return;//剪枝
	if(x%n==0)
	{
		cout<>n&&n)
	{
		f=0;
		dfs(1,0);//1开始，目前就1位数 
	}
	return 0;
}