POJ - 1426 Find The Multiple

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

题目大意：找出任意一个只包含1和0的十进制数保证他能整除n

从1开始 乘10和乘10+1分别做到在其后方加1和加0的操作，

需要注意的是，样例输出位数大于longlong最大范围，因此用无符号数来增加正数范围。

#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cstring>
#include<queue>
#include<algorithm>
#include<cmath>
using namespace std;

bool found;
void dfs(unsigned long long int  t, int n, int k)
{
    if (found)
        return;
    if (t%n == 0)
    {
        printf("%I64u\n", t);
        found = true;
        return;
    }
    if (k == 19)
        return;
    dfs(t * 10, n, k + 1);
    dfs(t * 10 + 1, n, k + 1);
}
int main()
{
    int n;
    while (cin >> n)
    {
        if(n==0)
            break;
        found = false;
        dfs(1, n, 0);
    }
    return 0;
}