Find The Multiple

Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 32524 Accepted: 13602 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

10100100100100100100

111111111111111111

题意：

输入一个数,n,寻求一个数x,使x,只由0和1组成，位数不超过十八位，并且是输入数的倍数；

思路：

由题知输出数是long long型的，可以用搜索dfs,每次搜x*10    x*10+1 从而保证输入数只由0和1，同时注意若输入是0或者是输出数超过十八位的，直接return

       当时，知道以前做过，但是再读不知到啥意思，还有输出数的龙long long 型都没想到，以前学长讲过觉得挺简单的，但过后就忘了，不能记代码，或者本来不会做，别人讲过，看着代码会就过，应该看题中信息，知道怎么从哪看出意思，知道思路。下面是参考代码：

include<stdio.h>int n,f;void dfs(long long int x,int step)   //x,long long 型{    if(f || step > 18)return ;       //x,不能超过18位，找一个符合题意就行，f,标记已经找到输出，用在往下了    if(x % n == 0)    {        f=1;        printf("%lld\n",x);        return ;    }    dfs(x*10,step+1);    dfs(x*10+1,step+1);           //保证只有0和1组成}int main(){    while(scanf("%d",&n) && n)    {        f=0;        dfs(1,0);    }}