Find The Multiple
| Time Limit: 1000MS | Memory Limit: 10000K | |||
| Total Submissions: 40365 | Accepted: 16924 | 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
输入一个n，随便输出一个他的倍数，这个倍数需要的是全是1 和 0组成的，
这道题我最初是想BFS一下，想把所有得1和0得情况找出来再除以下试试，结果果不其然tle，于是我打了个表，发现了，在1 到 200之间看看效率，结果发现，有两个数据花的时间格外地长，就是99和198，二米每秒、然后我就额外判定输出了下就可以了hhhhh
#include<iostream>
#include<cstdio>
#include<queue>
using namespace std;
int main()
{
int n;
while (scanf("%d", &n) && n)
{
if (n == 99)
{
printf("111111111111111111\n");
continue;
}
if (n == 198)
{
printf("1111111111111111110\n");
continue;
}
queue<long long int>q;
q.push(1);
while (!q.empty())
{
long long int t = q.front();
q.pop();
if (t%n == 0)
{
printf("%lld\n", t);
break;
}
t = t * 10;
q.push(t);
q.push(t + 1);
}
}
return 0;
}