codeforces #261 C题 Pashmak and Buses（瞎搞）

题目地址：http://codeforces.com/contest/459/problem/C

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.

这题就是求一全排列。由于要求每一天的都不同样，所以最多是k^d种。假设要输出的话，最简单的方法就是进行全排列呗。。。。

要注意。。

假设用的跟我的求全排列方法一样的话。那须要注意中间的cnt值是会非常大的。可是由于最多仅仅须要输出n种，所以假设大于n的话就直接让他等于n+1。

代码例如以下：

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <stdlib.h>
#include <math.h>
#include <ctype.h>
#include <queue>
#include <map>
#include <set>
#include <algorithm>

using namespace std;
#define LL __int64
LL mp[3100][3100];
int main()
{
    LL n, k, d, i, j, cnt, h, flag=0, x, y;
    scanf("%I64d%I64d%I64d",&n,&k,&d);
        memset(mp,0,sizeof(mp));
        cnt=1;
        for(i=1;i<=d;i++)
        {
            for(j=1;j<=n;j++)
            {
                if((j-1)%cnt==0)
                {
                    mp[i][j]=mp[i][j-1]+1;
                    if(mp[i][j]>k)
                    {
                        mp[i][j]=1;
                        if(i==d)
                        {
                            flag=1;
                            break;
                        }
                    }
                }
                else
                    mp[i][j]=mp[i][j-1];
            }
            if(j!=n+1)
                break;
            cnt*=k;
            if(cnt>1000)
                cnt=1001;
        }
        if(flag)
        {
            printf("-1
");
        }
        else
        {
            for(i=d;i>=1;i--)
            {
                for(j=1;j<=n;j++)
                {
                    printf("%I64d ",mp[i][j]);
                }
                printf("
");
            }
        }
    return 0;
}