计算可以攻击 N * N 棋盘上给定棋子的骑士数量
原文:https://www . geesforgeks . org/count-可以攻击给定棋子的骑士-in-n-n-board/
给定一个大小为 N * 2 的 2D 阵 骑士【】【】，表格中的每一行 { X，Y } 代表骑士的坐标，一个阵卒【】代表一个 N * N 棋盘中一个卒的坐标，任务是找出棋盘中正在攻击该卒的骑士数量。
示例:
输入:骑士[][] = { { 0，4 }，{ 4，5 }，{ 1，4 }，{ 3，1 } }，棋子[] = { 2，3 } 输出: 2 说明: 坐标{ { 0，4 }，{ 3，1 } }处出现的骑士正在攻击棋子。 因此，要求的输出为 2。
输入:骑士[][] = { { 4，6 }，{ 7，5 }，{ 5，5 } }，棋子[] = { 6，7 } 输出: 3 说明: 坐标{ { 4，6 }，{ 7，5 }，{ 5，5 } }处出现的骑士正在攻击棋子。 因此，要求的输出为 3。
方法:按照下面给出的步骤解决问题
- 初始化一个变量，比如 cntKnights ，来存储攻击棋子的骑士数量。
- 使用变量 i 遍历骑士[][] 数组，对于每个数组元素骑士[i] ，检查数组 {(骑士[I][0]–棋子[0])，(骑士[I][1]–棋子[1]) } 是否等于 { 1，2 } 或 { 2，1 } 。如果发现为真，则将的数值增加1。
- 最后，打印 cntKnights 的值。
下面是上述方法的实现:
C++
// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to count the knights that are
// attacking the pawn in an M * M board
int cntKnightsAttackPawn(int knights[][2],
int pawn[], int M)
{
// Stores count of knights that
// are attacking the pawn
int cntKnights = 0;
// Traverse the knights[][] array
for (int i = 0; i < M; i++) {
// Stores absolute difference of X
// co-ordinate of i-th knight and pawn
int X = abs(knights[i][0]
- pawn[0]);
// Stores absolute difference of Y
// co-ordinate of i-th knight and pawn
int Y = abs(knights[i][1]
- pawn[1]);
// If X is 1 and Y is 2 or
// X is 2 and Y is 1
if ((X == 1 && Y == 2)
|| (X == 2 && Y == 1)) {
// Update cntKnights
cntKnights++;
}
}
return cntKnights;
}
// Driver Code
int main()
{
int knights[][2] = { { 0, 4 }, { 4, 5 },
{ 1, 4 }, { 3, 1 } };
int pawn[] = { 2, 3 };
// Stores total count of knights
int M = sizeof(knights)
/ sizeof(knights[0]);
cout << cntKnightsAttackPawn(
knights, pawn, M);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to implement
// the above approach
import java.io.*;
import java.lang.Math;
class GFG{
// Function to count the knights that are
// attacking the pawn in an M * M board
static int cntKnightsAttackPawn(int knights[][],
int pawn[], int M)
{
// Stores count of knights that
// are attacking the pawn
int cntKnights = 0;
// Traverse the knights[][] array
for(int i = 0; i < M; i++)
{
// Stores absolute difference of X
// co-ordinate of i-th knight and pawn
int X = Math.abs(knights[i][0] - pawn[0]);
// Stores absolute difference of Y
// co-ordinate of i-th knight and pawn
int Y = Math.abs(knights[i][1] - pawn[1]);
// If X is 1 and Y is 2 or
// X is 2 and Y is 1
if ((X == 1 && Y == 2) ||
(X == 2 && Y == 1))
{
// Update cntKnights
cntKnights++;
}
}
return cntKnights;
}
// Driver code
public static void main(String[] args)
{
int[][] knights = { { 0, 4 }, { 4, 5 },
{ 1, 4 }, { 3, 1 } };
int[] pawn = new int[]{2, 3};
// Stores total count of knights
int M = knights.length;
System.out.println(cntKnightsAttackPawn(
knights, pawn, M));
}
}
// This code is contributed by vandanakillari54935
Python 3
# Python program to implement
# the above approach
# Function to count the knights that are
# attacking the pawn in an M * M board
def cntKnightsAttackPawn(knights, pawn, M):
# Stores count of knights that
# are attacking the pawn
cntKnights = 0;
# Traverse the knights array
for i in range(M):
# Stores absolute difference of X
# co-ordinate of i-th knight and pawn
X = abs(knights[i][0] - pawn[0]);
# Stores absolute difference of Y
# co-ordinate of i-th knight and pawn
Y = abs(knights[i][1] - pawn[1]);
# If X is 1 and Y is 2 or
# X is 2 and Y is 1
if ((X == 1 and Y == 2) or (X == 2 and Y == 1)):
# Update cntKnights
cntKnights += 1;
return cntKnights;
# Driver code
if __name__ == '__main__':
knights = [[0, 4], [4, 5], [1, 4], [3, 1]];
pawn = [2, 3];
# Stores total count of knights
M = len(knights);
print(cntKnightsAttackPawn(knights, pawn, M));
# This code is contributed by Amit Katiyar
C
// C# program to implement
// the above approach
using System;
class GFG
{
// Function to count the knights that are
// attacking the pawn in an M * M board
static int cntKnightsAttackPawn(int[,] knights, int[] pawn, int M)
{
// Stores count of knights that
// are attacking the pawn
int cntKnights = 0;
// Traverse the knights[][] array
for (int i = 0; i < M; i++) {
// Stores absolute difference of X
// co-ordinate of i-th knight and pawn
int X = Math.Abs(knights[i, 0] - pawn[0]);
// Stores absolute difference of Y
// co-ordinate of i-th knight and pawn
int Y = Math.Abs(knights[i, 1] - pawn[1]);
// If X is 1 and Y is 2 or
// X is 2 and Y is 1
if ((X == 1 && Y == 2)
|| (X == 2 && Y == 1)) {
// Update cntKnights
cntKnights++;
}
}
return cntKnights;
}
// Driver code
static void Main()
{
int[,] knights = {{ 0, 4 }, { 4, 5 }, { 1, 4 }, { 3, 1 }};
int[] pawn = {2, 3};
// Stores total count of knights
int M = knights.GetLength(0);
Console.WriteLine(cntKnightsAttackPawn(knights, pawn, M));
}
}
// This code is contributed by divyeshrabadiya07
java 描述语言
<script>
// javascript program for the above approach
// Function to count the knights that are
// attacking the pawn in an M * M board
function cntKnightsAttackPawn(knights,
pawn, M)
{
// Stores count of knights that
// are attacking the pawn
let cntKnights = 0;
// Traverse the knights[][] array
for(let i = 0; i < M; i++)
{
// Stores absolute difference of X
// co-ordinate of i-th knight and pawn
let X = Math.abs(knights[i][0] - pawn[0]);
// Stores absolute difference of Y
// co-ordinate of i-th knight and pawn
let Y = Math.abs(knights[i][1] - pawn[1]);
// If X is 1 and Y is 2 or
// X is 2 and Y is 1
if ((X == 1 && Y == 2) ||
(X == 2 && Y == 1))
{
// Update cntKnights
cntKnights++;
}
}
return cntKnights;
}
// Driver Code
let knights = [[ 0, 4 ], [ 4, 5 ],
[ 1, 4 ], [ 3, 1 ]];
let pawn = [2, 3];
// Stores total count of knights
let M = knights.length;
document.write(cntKnightsAttackPawn(
knights, pawn, M));
</script>
Output
2
时间复杂度:* O(M)，其中 M 为骑士总数 辅助空间:* O(1)
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