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

本文介绍了一种算法问题，目标是找到满足特定条件的k-special表格。这类表格要求包含从1到n²的所有整数各出现一次，每行数字递增排序，且第k列的总和达到最大值。文章提供了实现思路及示例代码。

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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 test(s)

input
4 1

output
28
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16

input
5 3

output
85
5 6 17 18 19
9 10 23 24 25
7 8 20 21 22
3 4 14 15 16
1 2 11 12 13这题理解了不难，其实就是要你找出第k列最大的序列。这题一开始没什么思路，最后看着数据突然有灵感，发现是根据第k列来划分为两部分，这样就能使得k列最大了。AC代码：#include<iostream>
#include<functional>
#include<algorithm>
#include<cstring>
#include<string>
#include<vector>
#include<cstdio>
#include<cmath>
#include<map>
using namespace std;
#define CRL(a) memset(a,0,sizeof(a))
typedef unsigned __int64 LL;
typedef  __int64 ll;
#define CMP bool cmp(const node& a,const node& b){	return a.R<b.R||(a.R==b.R&&a.L<b.L); }
const int T = 503000;
const int mod = 1000000007;

int v[550][550];

int main()
{
#ifdef zsc
    freopen("input.txt","r",stdin);
#endif

	int n,m,i,j,k;
	
	while(~scanf("%d%d",&n,&m))
	{
		k = 0;
		int t = 0;
		for(i=0;i<n;++i){
			for(j=0;j<n&&j<m-1;++j){
				v[i][j]=++t;
			}
		}
		for(i=0;i<n;++i){
			for(j=m-1;j<n;++j){
				if(j==m-1)k+=t+1;
				v[i][j]=++t;
			}
		}
			printf("%d\n",k);
			for(i=0;i<n;++i){
				for(j=0;j<n;++j){
					printf("%d ",v[i][j]);
				}
				printf("\n");
			}
		
	}

    return 0;
}