poj1426 Find The Multiple（哈夫曼思想）

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

http://poj.org/problem?id=1426

找一个比N大于等于的数，使这个数每一位只有0和1，且能被N整除。

看网上有人说用了哈夫曼思想，发现在把数字存入数组的时候用到了，原来并不连续的数字可以通过一定的联系放入数组。

#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;
}