Poj 1426 Find The Multiple 【DFS】

Find The Multiple

Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 19815 Accepted: 8048 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
题意：输入一个n，找出一个是由0,1组成的并且是n的整数倍的数。
用usigned long long 就可以。
代码：

#include <cstdio>
#include <iostream>
#include <cstring>
#define LL unsigned long long
using namespace std;

bool flag;
LL n;

void dfs(LL cur, int step){
    if(flag|| step == 19) return ;
    if(cur%n == 0){
            printf("%llu\n", cur);
        //cout << cur<<endl;
        flag = 1; return ;
    }
    dfs(cur*10, step+1);
    dfs(cur*10+1, step+1);


}

int main(){
    LL s = 1;
    while(scanf("%llu", &n), n){
        flag = 0;
        dfs(s, 0);
    }
    return 0;
}