poj 1426 Find The Multiple （简单搜索dfs）

题目：

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：

        本题要找出数m，m是只有0和1构成的十进制数，并且是n的倍数，若有多个答案，输出任意一个就可以。

题解：

        经过思考会发现m最大不会超过unsigned long long 的范围，所以用unsigned long long保存就可以，接下来就是深搜就ok。

代码：

#include <iostream>

using namespace std;
unsigned long long ans;
bool f;

void dfs(unsigned long long s,int n,int k)
{
    if(f) return ;
    if(s%n==0) {ans=s;f=true; return ;}
    if(k==19) return ;        //如果k超过19就不在unsigned long long的范围内了
    dfs(s*10,n,k+1);
    dfs(s*10+1,n,k+1);
    return ;
}

int main()
{
    int n;
    while(cin>>n,n)
    {
        ans=0;
        f=false;
        dfs(1,n,0);
        cout<<ans<<endl;
    }
    return 0;
}