求给定方格内的方格数
给定一个边 N * N 的网格，任务是找出其中存在的方块总数。所有选定的方块可以是任意长度。 例:
输入:N = 1 T3】输出: 1
输入:N = 2 T3】输出: 5
输入:N = 4 T3】输出: 30
进场 1: 举几个例子，可以观察到对于一个大小为 N * N 的格子，里面的方块数会是12+22+32+…+N2 以下是上述进场的实施:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the number
// of squares inside an n*n grid
int cntSquares(int n)
{
int squares = 0;
for (int i = 1; i <= n; i++) {
squares += pow(i, 2);
}
return squares;
}
// Driver code
int main()
{
int n = 4;
cout << cntSquares(4);
return 0;
}
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 squares inside an n*n grid
static int cntSquares(int n)
{
int squares = 0;
for (int i = 1; i <= n; i++) {
squares += Math.pow(i, 2);
}
return squares;
}
// Driver code
public static void main(String args[])
{
int n = 4;
System.out.print(cntSquares(4));
}
}
Python 3
# Python3 implementation of the approach
# Function to return the number
# of squares inside an n*n grid
def cntSquares(n) :
squares = 0;
for i in range(1, n + 1) :
squares += i ** 2;
return squares;
# Driver code
if __name__ == "__main__" :
n = 4;
print(cntSquares(4));
# This code is contributed by AnkitRai01
C
// C# implementation of the approach
using System;
class GFG
{
// Function to return the number
// of squares inside an n*n grid
static int cntSquares(int n)
{
int squares = 0;
for (int i = 1; i <= n; i++)
{
squares += (int)Math.Pow(i, 2);
}
return squares;
}
// Driver code
public static void Main(String []args)
{
int n = 4;
Console.Write(cntSquares(n));
}
}
// This code is contributed by 29AjayKumar
java 描述语言
<script>
// JavaScript implementation of the approach
// Function to return the number
// of squares inside an n*n grid
function cntSquares(n)
{
let squares = 0;
for (let i = 1; i <= n; i++)
{
squares += Math.pow(i, 2);
}
return squares;
}
let n = 4;
document.write(cntSquares(n));
</script>
Output:
30
方法 2: 通过使用直接公式。
然而，和
具有封闭形式(直接公式)
。因此，我们可以用它来计算
时间内的总和。
以下是上述方法的实现:
C++
// C++ implementation of the approach
#include <iostream>
using namespace std;
int cnt_squares (int n)
{
/* Function to return the number
of squares inside an n*n grid */
return n * (n + 1) * (2 * n + 1) / 6;
}
// Driver code
int main()
{
cout << cnt_squares (4) << endl;
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
class GFG {
static int cntSquares (int n) {
/* Function to return the number
of squares inside an n*n grid */
return n * (n + 1) * (2 * n + 1) / 6;
}
// Driver code
public static void main(String args[]) {
System.out.println (cntSquares(4));
}
}
Python 3
# Python3 implementation of the approach
"""
Function to return the number
of squares inside an n*n grid
"""
def cntSquares(n) :
return int (n * (n + 1) * (2 * n + 1) / 6)
# Driver code
if __name__ == "__main__" :
print (cntSquares (4));
C
// C# implementation of the approach
using System;
class GFG
{
/* Function to return the number
of squares inside an n*n grid */
static int cntSquares (int n)
{
return n * (n + 1) * (2 * n + 1) / 6;
}
// Driver code
public static void Main (String[] args)
{
Console.Write (cntSquares (4));
}
}
java 描述语言
<script>
// Javascript implementation of the approach
/* Function to return the number
of squares inside an n*n grid */
function cntSquares (n)
{
return n * (n + 1) * (2 * n + 1) / 6;
}
document.write(cntSquares(4));
// This code is contributed by divyeshrabadiya07.
</script>
Output:
30
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