POJ

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
这道题题意很好理解，给一个n，找到一个只有0和1组成的数可以整除n。
一直long long不能输出结果，百度看到都用的unsigned long long。

#include <iostream>#include <cstdio>using namespace std;int flag;void dfs(unsigned long long num,int step,int n){    if(flag)return;    if(num%n==0)    {        flag=1;        printf("%I64u\n",num);        return;    }    if(step==19)return;//防止越界    dfs(num*10,step+1,n);    dfs(num*10+1,step+1,n);    return;}int main(){    int n;    while(scanf("%d",&n)!=EOF&&n)    {        flag=0;        dfs(1,0,n);    }    return 0;}