POJ 1426 Find The Multiple —————

POJ 1426 Find The Multiple ——————DFS

Find The Multiple

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Find The Multiple

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 42225		Accepted: 17735		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

Sample Output

10
100100100100100100
111111111111111111

Source

Dhaka 2002

—– 这道题让找 n 的 01 十进制倍数，也就是说，找出一个01序列，它是 n 的倍数第一位只能是1 所以就从 1 开始DFS 下一位只能是0或者1 所以说复杂度是 O(2n) O ( 2 n ) 的级别，有点儿不敢写，但是后来发现这个最大也就是 218 2 18 ,能过得 —

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
ll n,k,flag;
void dfs(ll t,ll i)
{
    if(i==19)    return ;
    if(flag)    return ;
    if(t%n==0)
    {
        flag=1;
        printf("%lld\n",t);
        return ;
    }
    dfs(t*10,i+1);
    dfs(t*10+1,i+1);
}
int main()
{
    while(~scanf("%lld",&n)&&n)
    {
        flag=0;
        dfs(1,0);
    }
    return 0;
}