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构造数列Codeforces Round #261 (Div. 2)C
2015-07-20 17:53:40 来源: 作者: 【 】 浏览:1
Tags:构造 数列 Codeforces Round #261 Div.

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 nstudents. 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:n个学生,k张车(容量无限),d天,要求每个学生每天所坐的车,要求不能有学生d天全部坐同一张车.

当成k进制来搞就行,当成n个d位的k进制数,至少有一位不同就行.代码如下:

#include
  
   
#include
   
     #include
    
      #include
      #include
      
        #include
       
         #include
        
          using namespace std; typedef long long LL; int A[1005][1005]; int main() { int n,d,k; while(~scanf("%d%d%d",&n,&k,&d)) { bool f1=false; LL tmp=1; for(int i=1;i<=d;i++) { tmp*=k; if(tmp>=n) { f1=true; break; } } if(!f1){ puts("-1"); continue; } for(int i=0;i
         
          


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