Find The Multiple
| Time Limit: 1000MS | Memory Limit: 10000K | |||
| Total Submissions: 22234 | Accepted: 9128 | 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
10 100100100100100100 111111111111111111
Source
题目大意：
就是给你一个数 m，让你找只含有 0 和 1 的数而且是 m 的倍数的数
解题思路：
从网上看到的说最大的数不会超过18位，（m<=200），所以，可以直接用无符号位的数就行了，不涉及大数问题，
直接上dfs搞一下子，就行了，具体详见我代码：
/*
2015 - 09 - 09
Author: ITAK
Motto:
今日的我要超越昨日的我，明日的我要胜过今日的我，
以创作出更好的代码为目标，不断地超越自己。
*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
bool ok;
/// t:表示0和1组成的数， k:位数，m:给定的数
void dfs(unsigned long long t, int k, int m)
{
if(ok)///退出的标志
return;
if(t%m == 0)
{
cout<<t<<endl;
ok = 1;
return;
}
if(k == 19)
return;
dfs(t*10, k+1, m);///有可能是0
dfs(t*10+1, k+1, m);///有可能是1
}
int main()
{
int m;
while(cin>>m,m)
{
ok = 0;
dfs(1, 0, m);///从1开始搜，因为最高位是1
}
return 0;
}
