Recently Pashmak has been employed in a transportation company. The company has kbuses 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.

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

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

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：有K台公交，n个人，d天，任意两个人全部d天都不能做同一辆公交，输出这种安排

解法：K台公交安排d天，自然是K^d种方法，Kⁱ大于等于n说明符合要求，再将1~n变成k进制保证题目要求，然后按照题目要求输出

 #include<bits/stdc++.h>

 using namespace std;

 #define ll long long

 ll Pow(ll a,ll b)

 {

     ll ans=;

     ll base=a;

     while(b)

     {

         if(b&)

         {

             ans*=base;

         }

         base*=base;

         b>>=;

     }

     return ans;

 }

 ll n,k,d;

 ll solve[][];

 int main()

 {

     int flag=;

     cin>>n>>k>>d;

     for(int i=;i<=d;i++)

     {

         if(Pow(k,i)>=n)

         {

             flag=;

             break;

         }

     }

     if(flag)

     {

         for(int i=;i<=n;i++)

         {

             ll num=i;

             for(int j=;j<=d;j++)

             {

                 solve[i][j]=num%k+;

                 num/=k;

             }

         }

         for(int i=;i<=d;i++)

         {

             for(int j=;j<=n;j++)

             {

                 cout<<solve[j][i]<<" ";

             }

             cout<<endl;

         }

     }

     else

     {

         cout<<"-1"<<endl;

     }

     return ;

 }