POJ1426 Find The Multiple(DFS)

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

这道题大意就是输入一个n,然后求n的倍数，但是这个倍数只能由0或1构成。因为这个代码有借鉴他人的，加了小编的一点思想，在这里详细解释一下

#include<iostream>#include<cstdio>using namespace std;bool  found;void dfs(long long k,int n,int m){if (found)   return;if (k%n == 0){printf("%lld\n", k);found = 1;return;}if (m == 18)    return;else{dfs(k * 10,n,m+1);dfs(k * 10 + 1,n,m+1);}}int main(){int n;while (cin>>n,n){found = 0; dfs(1,n,0);}return 0;}

这个代码首先第一点要加一个判断bool类型的作为最后输出，因为找到了结果就没有再找下去的必要了。然后有另一点注意的是，要大致算一下范围，long long类型范围大小是-2^63 到 2^63-1，代码中dfs必须要有两种清况，就是如果第一个dfs找不到结果会进行第二个dfs，所以必须加一个上限，就是乘到第18层的时候就得跳有一个跳出条件，从而执行第二个dfs.