Codeforces Round #342 (Div. 2) C. K-special Tables 构造

C. K-special Tables

题目连接：

http://www.codeforces.com/contest/625/problem/C

Description

People do many crazy things to stand out in a crowd. Some of them dance, some learn by heart rules of Russian language, some try to become an outstanding competitive programmers, while others collect funny math objects.

Alis is among these collectors. Right now she wants to get one of k-special tables. In case you forget, the table n × n is called k-special if the following three conditions are satisfied:

every integer from 1 to n2 appears in the table exactly once;

in each row numbers are situated in increasing order;

the sum of numbers in the k-th column is maximum possible.

Your goal is to help Alice and find at least one k-special table of size n × n. Both rows and columns are numbered from 1 to n, with rows numbered from top to bottom and columns numbered from left to right.

Input

The first line of the input contains two integers n and k (1 ≤ n ≤ 500, 1 ≤ k ≤ n) — the size of the table Alice is looking for and the column that should have maximum possible sum.

Output

First print the sum of the integers in the k-th column of the required table.

Next n lines should contain the description of the table itself: first line should contains n elements of the first row, second line should contain n elements of the second row and so on.

If there are multiple suitable table, you are allowed to print any.

Sample Input

4 1

Sample Output

28

1 2 3 4

5 6 7 8

9 10 11 12

13 14 15 16

Hint

题意

构造题，你需要构造出n*n的矩阵，然后使得这个矩阵的第k列的和尽量大

并且这个矩阵每一行都是递增的

题解：

构造题，小于k列的，我们从小到大放就好了，大于等于k列的，我们倒着放，从最后一个位置，最后一个数开始放，然后放放放就好了

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 505;
int n,k;
int mp[maxn][maxn];
int main()
{
    scanf("%d%d",&n,&k);
    int L = 1,R = n*n;
    for(int i=1;i<=n;i++)
        for(int j=1;j<k;j++)
            mp[i][j]=L++;
    for(int i=n;i>=1;i--)
        for(int j=n;j>=k;j--)
            mp[i][j]=R--;
    long long sum = 0;
    for(int i=1;i<=n;i++)
        sum+=mp[i][k];
    cout<<sum<<endl;
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=n;j++)
        {
            printf("%d ",mp[i][j]);
        }
        printf("\n");
    }
}