poj1426Find The Multiple

文章大意就是：给你一个正整数N，让你用 1 和 0 组成一个不超过100位的数字，问这个数字是多少恰好能整除N，可能这样的数不唯一，输入其中一个就行

当时没做出来确实不应该，因为是英文题先看了眼数据，发现数据这么大，没仔细读题都以为肯定会超时。如果当时在勇敢一点吧题好好读一遍，把数据在测试一遍，应该是能做出来的。其实你能输入来的数最大也不超过19位！！！（做题也考验人的胆量，吼吼）

解题思路：其实就是一个简单的dfs，注意最大不超过19位，如果超过19位或者能整除直接return就行了，上代码吧，一看就一目了然了

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

#include<stdio.h>#include<string.h>int n,f;void dfs(long long int x,int step){    if(f||step>18)return ;   //被标记或者大于18 就返回    if(x%n==0)    {        f=1;        printf("%lld\n",x);        return ;    }    dfs(x*10,step+1);     //要么×10，要么×10加一，因为只有0和1    dfs(x*10+1,step+1);}int main(){    while(~scanf("%d",&n),n)    {        f=0;        dfs(1,0);    //肯定从1开始搜    }    return 0;}