文章标题 POJ 1426 ：Find The Multiple （dfs）

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

题意：给一个1~200的数n，找出能将n整出的一个数，这个数只含有1和0
分析：一开始看题目说这个数不超过100个数，其实这个数在unsigned long long 内，所以直接dfs深搜一下。
代码：

#include<iostream>#include<string>#include<cstdio>#include<cstring>#include<vector>#include<math.h>#include<map>#include<queue> #include<algorithm>using namespace std;const int inf = 0x3f3f3f3f;int n;unsigned long long answer;int flag;void dfs(int n,unsigned long long tmp,int deep){    //找到答案就结束    if (flag) return;    if (tmp%n==0){        flag=1;        cout<<tmp<<endl;        return;    }    //当深度到达19，则回溯    if (deep>=19) return;    dfs(n,tmp*10,deep+1);    dfs(n,tmp*10+1,deep+1);}int main (){    while (cin>>n){        if (n==0) break;        flag=0;        dfs(n,1,0);    }       return 0;}