cf459C Pashmak and Buses

C. Pashmak and Buses

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Recently Pashmak has been employed in a transportation company. The company has k buses and has a contract with a school which has n students.
The school planned to take the students to d different places for d days
(each day in one place). Each day the company provides all the buses for the trip. Pashmak has to arrange the students in the buses. He wants to arrange the students in a way that no two students become close friends. In his ridiculous idea, two students will
become close friends if and only if they are in the same buses for all d days.

Please help Pashmak with his weird idea. Assume that each bus has an unlimited capacity.

Input

The first line of input contains three space-separated integers n, k, d (1 ≤ n, d ≤ 1000; 1 ≤ k ≤ 109).

Output

If there is no valid arrangement just print -1. Otherwise print d lines,
in each of them print n integers. The j-th integer
of the i-th line shows which bus the j-th student
has to take on the i-th day. You can assume that the buses are numbered from 1 to k.

Sample test(s)

input

3 2 2

output

1 1 2
1 2 1

input

3 2 1

output

-1

Note

Note that two students become close friends only if they share a bus each day. But the bus they share can differ from day to day.

题意是有长度为n的序列，然后要在序列中填1到k的数做d次，使得序列中不存在两个位置在d次中都填一样的数。输出一组方案

然后orz了不知名的神犇才会……看样子好像是类似分治的做法……就是尽量n个位置平均分配1到k，然后变成1 2 3 4 5……k 1 2 3 4 5……k的样子，然后倒着做，变成1 1 1 1 1……2 2 2 2 2…… 3 3 3 3 ……的样子

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdlib>
using namespace std;
int ans[2333][2333];
 
int main()
{
	int i,j,k,m,n,d,delta,l,r;
	scanf("%d%d%d",&n,&k,&d);
	delta=1;
	for(i=1;i<=d;i++)
	{
		for(j=1;j<=n;j++)
		{
			ans[i][j]=1;
		}
	}
	for(i=1;i<=d;i++)
	{
		delta*=k;
		if(delta>=n)
		{
			for(j=2;j<=n;j++)
			{
				l=i;
				while(ans[l][j-1]==k)
				{
					l--;
				}
				for(r=1;r<=l;r++)
				{
					ans[r][j]=ans[r][j-1];
				}
				ans[l][j]++;
			}
			for(i=1;i<=d;i++)
			{
				for(j=1;j<=n;j++)
				{
					printf("%d ",ans[i][j]);
				}
				printf("\n");
			}
			return 0;
		}
	}
	printf("-1\n");
	return 0;
}