cf 459c Pashmak and Buses

E - Pashmak and Buses

Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

Description

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 Input

Input

3 2 2

Output

1 1 2 
1 2 1

Input

3 2 1

Output

-1

#include<iostream>
#include<stdio.h>
#include<math.h>
#include<fstream>
#include<cstring>
using namespace std;
int code[1005][1005];
int main()
{
    int n,k,d;
    //cout<<1000*log10(1000000000+0.0)<<endl;
    //cout<<pow(5,4)<<endl;
    while(~scanf("%d%d%d",&n,&k,&d))
    {
        int tmp;
        int flag=k;
        for(int i=0;i<d-1;i++)
            {
                if(flag>n) break;
                flag*=k;

            }
        //cout<<flag;
        if(flag<n)
        {
            printf("-1
");
            continue;
        }
        tmp=  (pow(n+0.0,1.0/d)>(int)pow(n+0.0,1.0/d))?(int)pow(n+0.0,1.0/d)+1:(int)pow(n+0.0,1.0/d);
        for(int j=0;j<d;j++)
            code[j][1]=code[j][0]=0;
        for(int i=1; i<n; i++)
        {
            int j=d-1;

            //cout<<"ss";
            code[j][i]++;
            int now=code[j][i];
            int row=j;
            int col=i;
            while(now>tmp-1&&row>-1)
            {
                code[row][col]=now%(tmp);
                code[row-1][col]+=(now/(tmp));
                now=code[row-1][col];
                row--;
            }
            for(int j=0;j<d;j++)
                code[j][i+1]=code[j][i];
        }
        for(int i=0; i<d; i++)
        {
            for(int j=0; j<n; j++)
            {
                cout<<code[i][j]+1<<" ";
            }
            cout<<endl;
        }
    }
    return 0;
}

View Code

这里

for(int i=0;i<d-1;i++)
{
if(flag>n) break;
flag*=k;

}

如果用pow的话，存在两个问题：1 pow（1e9,1000)过大 2 由于精度问题，有些数的比较是失败的