Codeforces Round #342 (Div. 2) 625C K-special Tables(脑洞)

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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
281 2 3 45 6 7 89 10 11 1213 14 15 16
input
5 3
output
855 6 17 18 199 10 23 24 257 8 20 21 223 4 14 15 161 2 11 12 13

题目链接:点击打开链接

给出n, k构造一个n * n的矩阵, 元素从1到n ^ 2, 使得第k列和尽量大, 每行递增.

找规律, 发现前k - 1列按行递增, 同时后面的列按行递增, 即可构造矩阵成功.

AC代码:

#include "iostream"#include "cstdio"#include "cstring"#include "algorithm"#include "queue"#include "stack"#include "cmath"#include "utility"#include "map"#include "set"#include "vector"#include "list"#include "string"#include "cstdlib"using namespace std;typedef long long ll;#define X first#define Y secondconst int MOD = 1e9 + 7;const int INF = 0x3f3f3f3f;const int MAXN = 505;int n, k, num, ans[MAXN][MAXN], sum;int main(int argc, char const *argv[]){scanf("%d%d", &n, &k);for(int i = 1; i <= n; ++i)for(int j = 1; j <= k - 1; ++j)ans[i][j] = ++num;for(int i = 1; i <= n; ++i)for(int j = k; j <= n; ++j)ans[i][j] = ++num;for(int i = 1; i <= n; ++i)sum += ans[i][k];printf("%d\n", sum);for(int i = 1; i <= n; ++i) {for(int j = 1; j <= n; ++j)printf("%d ", ans[i][j]);printf("\n");}return 0;}

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