Codeforces #261 (Div. 2) C. Pashmak and Buses（数学）

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.

给定n, k, d，即n个学生，k俩车, d天

问是否存在d天中任两个学生都不一直在一张车上的情况

如果存在输出d天，k俩车上的对应的学生

学生的编号为1-n

这个题首先判断对于k辆车，d天最多可以使k^d个学生满足条件（从全1到全n的排列）

所以只需判断k^d与n大小即可

如果存在怎么输出结果呢？

我的方法是DFS，但是数据量太大超时了

今天看了下别人用数学的做法，感觉很巧妙

证明用语言不太好描述，直接放图和代码运行结果吧，一看就明白了

上图对应k=3， d=3的情况，树上每一条路径对应输出结果中的一列

我的DFS也是按照这样的策略搜索的

但是用数学方法也可以做

这是代码运行结果，可能这样看不容易看出来，但是如果从下往上看呢

是不是发现结果和上图是一样的

其实这种方法就是从叶子向跟打印而已

下面贴代码吧：

#include <cstdio>#include <vector>#include <iostream>#define MAXN 1010using namespace std;int ans[MAXN][MAXN];int main() {    int n, k, d;    scanf("%d%d%d", &n, &k, &d);    //vector< vector<int> > ans(d, vector<int>(n, -1));    int x = 1;    for(int i=0; i<d; ++i) {        for(int j=0; j<n; ++j)             ans[i][j] = j / x % k;        if((long long)x * k >= n)            x = n;        else            x *= k;    }    if(x < n) {        puts("-1");    }else {        for(int i=0; i<d; ++i) {            for(int j=0; j<n; ++j) {                if(j != 0) putchar(' ');                printf("%d", ans[i][j]+1);            }            puts("");        }    }    return 0;}