矩阵中通过(x，y)的左右对角线上的单元数
给定四个整数 row、col、x 和 y ，其中 row 和 col 是二维矩阵的行数和列数， x 和 y 是同一矩阵中一个单元格的坐标，任务是找出与矩阵的单元格(x，y)相关联的左对角线和右对角线中的单元格数。 举例:
输入:行= 4，列= 3，x = 2，y = 2 T3】输出: 3 3
(2，2)的左对角线和右对角线上的单元格数分别为 3 和 3。 输入:行= 4，列= 5，x = 2，y = 2 输出: 4 3
进场:
- 分别计算单元格(x，y)左对角线左上角和右下角的单元格数。然后将它们相加，得到左对角线上的单元格数。
- 同样，分别计算单元格(x，y)右对角线的右上角和左下角的单元格数。
以下是上述方法的实现:
C++
// C++ implementation of the approach
#include<bits/stdc++.h>
using namespace std;
// Function to return the number of cells
// in the left and the right diagonal of
// the matrix for a cell (x, y)
void count_left_right(int n, int m, int x, int y)
{
int left = 0, right = 0;
// number of cells in the left diagonal
int left_upper_part = min(x-1, y-1);
int left_lower_part = min(n-x, m-y);
left = left_upper_part + left_lower_part + 1;
// number of cells in the right diagonal
int right_upper_part = min(m-y, x-1);
int right_lower_part = min(y-1, n-x);
right = right_upper_part + right_lower_part + 1;
cout<<(left)<<" "<<(right);
}
// Driver code
int main()
{
int row = 4;
int col = 3;
int x = 2;
int y = 2;
count_left_right(row, col, x, y);
}
// This code is contributed by
// Sanjit_Prasad
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
class GFG {
// Function to return the number of cells
// in the left and the right diagonal of
// the matrix for a cell (x, y)
static void count_left_right(int n, int m
, int x, int y)
{
int left = 0, right = 0;
// number of cells in the left diagonal
int left_upper_part = Math.min(x - 1, y - 1);
int left_lower_part = Math.min(n - x, m - y);
left = left_upper_part + left_lower_part + 1;
// number of cells in the right diagonal
int right_upper_part = Math.min(m - y, x - 1);
int right_lower_part = Math.min(y - 1, n - x);
right = right_upper_part + right_lower_part + 1;
System.out.println(left + " " + right);
}
// Driver code
public static void main(String[] args)
{
int row = 4;
int col = 3;
int x = 2;
int y = 2;
count_left_right(row, col, x, y);
}
}
Python 3
# Python 3 implementation of the approach
# Function to return the number of cells
# in the left and the right diagonal of
# the matrix for a cell (x, y)
def count_left_right(n, m, x, y):
left = 0
right = 0
# number of cells in the left diagonal
left_upper_part = min(x - 1, y - 1)
left_lower_part = min(n - x, m - y)
left = left_upper_part + left_lower_part + 1
# number of cells in the right diagonal
right_upper_part = min(m - y, x - 1)
right_lower_part = min(y - 1, n - x)
right = right_upper_part + right_lower_part + 1
print(left, right)
# Driver code
if __name__ == "__main__":
row = 4
col = 3
x = 2
y = 2
count_left_right(row, col, x, y)
# This code is contributed by ChitraNayal
C
// C# implementation of the above approach
using System;
class Program
{
// Function to return the number of cells
// in the left and the right diagonal of
// the matrix for a cell (x, y)
static void count_left_right(int n, int m
, int x, int y)
{
int left = 0, right = 0;
// number of cells in the left diagonal
int left_upper_part = Math.Min(x - 1, y - 1);
int left_lower_part = Math.Min(n - x, m - y);
left = left_upper_part + left_lower_part + 1;
// number of cells in the right diagonal
int right_upper_part = Math.Min(m - y, x - 1);
int right_lower_part = Math.Min(y - 1, n - x);
right = right_upper_part + right_lower_part + 1;
Console.WriteLine(left + " " + right);
}
//Driver code
static void Main()
{
int row = 4;
int col = 3;
int x = 2;
int y = 2;
count_left_right(row, col, x, y);
}
// This code is contributed by ANKITRAI1
}
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP implementation of the approach
// Function to return the number of cells
// in the left and the right diagonal of
// the matrix for a cell (x, y)
function count_left_right($n, $m, $x, $y)
{
$left = 0;
$right = 0;
// number of cells in the left diagonal
$left_upper_part = min($x - 1, $y - 1);
$left_lower_part = min($n - $x, $m - $y);
$left = $left_upper_part +
$left_lower_part + 1;
// number of cells in the right diagonal
$right_upper_part = min($m - $y, $x - 1);
$right_lower_part = min($y - 1, $n - $x);
$right = $right_upper_part +
$right_lower_part + 1;
echo $left, " " , $right;
}
// Driver code
$row = 4;
$col = 3;
$x = 2;
$y = 2;
count_left_right($row, $col, $x, $y);
// This code is contributed by jit_t
?>
java 描述语言
<script>
// Javascript implementation of the approach
// Function to return the number of cells
// in the left and the right diagonal of
// the matrix for a cell (x, y)
function count_left_right(n, m, x, y)
{
let left = 0, right = 0;
// number of cells in the left diagonal
let left_upper_part = Math.min(x-1, y-1);
let left_lower_part = Math.min(n-x, m-y);
left = left_upper_part + left_lower_part + 1;
// number of cells in the right diagonal
let right_upper_part = Math.min(m-y, x-1);
let right_lower_part = Math.min(y-1, n-x);
right = right_upper_part + right_lower_part + 1;
document.write((left) + " " + (right));
}
// Driver code
let row = 4;
let col = 3;
let x = 2;
let y = 2;
count_left_right(row, col, x, y);
// This code is contributed by souravmahato348.
</script>
Output:
3 3
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