Codeforces 625C K-special Tables 【贪心】

C. K-special Tables

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

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 containn 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

题意：让你用1-n*n这些数填充n*n矩阵，要求每行的数字严格递增、每个数字只能用一次且保证第k列的数字和最大。

思路：贪心选数，随便搞搞就过了。

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <string>
#define INF 1000000
#define eps 1e-8
#define MAXN (100000+10)
#define MAXM (10000+10)
#define Ri(a) scanf("%d", &a)
#define Rl(a) scanf("%lld", &a)
#define Rf(a) scanf("%lf", &a)
#define Rs(a) scanf("%s", a)
#define Pi(a) printf("%d\n", (a))
#define Pf(a) printf("%.2lf\n", (a))
#define Pl(a) printf("%lld\n", (a))
#define Ps(a) printf("%s\n", (a))
#define W(a) while((a)--)
#define CLR(a, b) memset(a, (b), sizeof(a))
#define MOD 1000000007
#define LL long long
#define lson o<<1, l, mid
#define rson o<<1|1, mid+1, r
#define ll o<<1
#define rr o<<1|1
#define PI acos(-1.0)
#pragma comment(linker, "/STACK:102400000,102400000")
#define fi first
#define se second
using namespace std;
typedef pair<int, int> pii;
int Map[600][600];
int main()
{
int n, k;
Ri(n); Ri(k); int Max = n * n;
int Min = 1;
for(int i = 1; i <= n; i++)
{
Max -= (n - k);
Map[i][k] = Max;
for(int j = k+1; j <= n; j++)
Map[i][j] = Map[i][j-1] + 1;
for(int j = 1; j < k; j++)
Map[i][j] = Min++;
Max--;
}
int sum = 0;
for(int i = 1; i <= n; i++) sum += Map[i][k];
Pi(sum);
for(int i = 1; i <= n; i++)
{
for(int j = 1; j <= n; j++)
{
if(j > 1) printf(" ");
printf("%d", Map[i][j]);
}
printf("\n");
}
return 0;
}