Find The Multiple ——深搜POJ

Find The Multiple

 POJ - 1426 

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

题意：找出任意一个是n的倍数且只含0和1的十进制数，这里说道n不大于200，long long int范围为19位，且long long int中只包含0和1的十进制数满足所有200之内任意一个数都有整数倍（这里需要自己想出）。

题解：开头不能为0，直接从1开始深搜，因为只能包含0和1，所以（x*10,step+1）和（x*10+1,step+1）；不能超出long long范围，所以step最多是18。

代码：

#include<stdio.h>int n,flag;void dfs(long long int x,int step)//x定义为long long型{    if(flag||step>18)         return ;    if(x%n==0)    {        flag=1;        printf("%lld\n",x);        return ;    }    dfs(x*10,step+1);//满足只含0和1    dfs(x*10+1,step+1);}int main(){    while(scanf("%d",&n)&&n)    {        flag=0;//标记        dfs(1,0);    }    return 0;}