Find The Multiple

C -Find The MultipleCrawling in process...Crawling failedTime Limit:1000MSMemory Limit:10000KB 64bit IO Format:%I64d & %I64uSubmitStatusDescriptionGiven 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 thereis a corresponding m containing no more than 100 decimal digits.InputThe 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.OutputFor 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 givenvalue of n, any one of them is acceptable.Sample Input

2
6
19
0

Sample Output

10
100100100100100100
111111111111111111

题目大意：

输入一个正整数n，寻找只由0和1组成的能够将n整除的十进制数

代码如下

#include <iostream>#include<algorithm>#include<cstdio>#include<cmath>#include<cstring>#include<cstdlib>#include<stack>#include<queue>

using namespace std;

int n,m;

void dfs(long long t,int y){if(y > 18||m==1)//long long 最多存储十九位数，第二十位时已经超出数据范围，当找到结果时，要return，设置m=1，使递归结束return;if( t % n == 0){m=1;printf("%lld\n",t);return;}dfs(t*10,y+1);//只由0,1组成的二进制数一定是由1为第一位数字，如不是1，增大数字则乘十或者乘十再加一dfs(t*10+1,y+1);//不能用y++，会改变y的值

}int main(){while (scanf("%d",&n)!=EOF&&n){m=0;//标记是否找到答案dfs(1,0);}

    return 0;}