M - Find The Multiple POJ - 1426 （构建二叉树）

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

 

AC：代码

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>
#include<vector>
#include<stack>

using namespace std;

long long mod[600001];
int main()
{
	int n;
	mod[0] = 0;
	int i;
	while(~scanf("%d",&n) && n)
	{
		for(i = 1;;i++)
		{
/*
1
10 11
100 101 110 111
......
*/ 
			mod[i] = mod[i/2]*10+i%2;//重要：构建二叉树 (所谓的哈夫曼思想)
			if(mod[i]%n==0) break;
			
		}
		printf("%lld\n",mod[i]);
	}
	
	return 0;
 } 
 

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