POJ-1426 Find The Multiple

Find The Multiple

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 31553 Accepted: 13158 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

26190

Sample Output

10100100100100100100111111111111111111

Source

Dhaka 2002

一道不是数水坑也不是走迷宫的DFS，题意：给一个数，让你找到一个数字对所给数取模，结果为0，但是这个数只能由数字0和1构成答案也许不唯一，输出一个结果即可。

思路：既然寻找数只能由0,1构成，那么1,10,11,100,110这样的数字满足第一个条件，设a=1，那么满足条件的数可以表示为a*10和a*10+1，递归就可以把a越滚越大且只由0,1构成，对每个构造出来的数取模检查，如果结果为0，输出结果，退出函数。由于数字可能会很大且为整数，所以使用unsigned long long类型定义变量。step是为了限制递归深度，超过20之后速度会很慢，但貌似这里在递归到19层之前就能找到结果。

#include<iostream>#define ull unsigned long longusing namespace std;int n;int flag=0;void dfs(ull a,int step){    if(flag || step==19)        return ;    if(a%n==0)    {        cout << a << endl;        flag=1;        return ;    }    else    {       dfs(a*10,step+1);       dfs(a*10+1,step+1);    }}int main(){    while(cin >> n && n)    {        flag=0;        dfs(1,0);    }    return 0;}