E - Find The Multiple
Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64u
Submit Status
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
10 100100100100100100 111111111111111111
#include<iostream>
#include<algorithm>
#include<stdio.h>
using namespace std;
int n, flag;void DFS(unsigned long long result, int n, int digit);//result是我们最后要求的结果是n的倍数，n是读入的数据，digit是result的位数；unsigned long long result=1844674473709551616LL，其实就是一个大数，因为本题说只要任意一组数据就行，所以要对数的大小进行一个限制。此程序限制就是19位，
int main()
{
while(cin >> n, n)
{
flag=0;
DFS(1, n, 0);//因为任何一个由0和1构成的数据开头都为1，且位数是从零开始的；
}
return 0;
}
void DFS(unsigned long long result, int n, int digit)
{
if(flag==1)return;//这是代表着找到第一个由0和1构成的是n的倍数的result，是为了防止无限递归，因为有很多result；
if(result%n==0)
{
flag=1;
printf("%I64u\n", result);
return;
}
if(digit==19)//因为unsigned long long 是19位
return;
//搜索的两个方向，result的后一位可能是0也可能是1
DFS(result*10, n, digit+1);//后一位是0
DFS(result*10+1, n, digit+1);//后一位是1
}