POJ 1426 Find The Multiple

Find The Multiple

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 46927		Accepted: 19608		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

 

思路： 使用dfs枚举所有可能的结果(01串)，当该结果可以整除n时，就输出。由于long long最多可以表示19位数字，所以当该数的位数大于19时，return。

#include <iostream>

using namespace std;

int n, flag;

void dfs(int k, long long cur)  // k:位数 cur:当前数字
{     
    if (k == 19 || flag)        // long long最多只能表示19位数
        return;

    if (cur % n == 0) {
        cout << cur << endl;
        flag = 1;
        return;
    }
    for (int i = 0; i <= 1; ++i)
    {
        dfs(k + 1, cur * 10 + i);
    }
}


int main()
{
    while (cin >> n, n != 0) {
        flag = 0;
        dfs(0, 1);
    }
    return 0;
}