算法题（如何计算矩形中的正方形数（））

本文概述

建议：在继续解决方案之前, 请先在"实践"上解决它。
C ++
Java
Python3
C#
的PHP
C ++
Java
Python3
C#

给定一个m x n的矩形, 其中有多少个正方形？
例子：

Input:m = 2, n = 2Output: 5There are 4 squares of size 1x1 + 1 square of size 2x2.Input: m = 4, n = 3Output: 20There are 12 squares of size 1x1 + 6 squares of size 2x2 + 2 squares of size 3x3.

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推荐：请在"实践首先, 在继续解决方案之前。
让我们首先解决m = n的问题, 即正方形的问题：
对于m = n = 1, 输出：1
对于m = n = 2, 输出：4 + 1 [4的大小为1×1 + 1的大小为2×2]
对于m = n = 3, 输出：9 + 4 + 1 [4尺寸为1×1 + 4尺寸为2×2 + 1尺寸为3×3]
对于m = n = 4, 输出16 + 9 + 4 +1 [16的大小为1×1 + 9的大小为2×2 + 4的大小为3×3 + 1的大小为4×4]
通常, 它似乎是n ^ 2 +(n-1)^ 2 +…1 = n(n + 1)(2n + 1)/ 6
当m不等于n时, 让我们解决这个问题：
让我们假设m < = n
通过以上解释, 我们知道m x m矩阵中的平方数为m(m + 1)(2m + 1)/ 6
当我们添加一列时, 即m x(m + 1)矩阵中的平方数是多少？
当我们添加一列时, 增加的平方数为m +(m-1)+…+ 3 + 2 +1
[尺寸为1×1的m平方+尺寸为2×2的(m-1)平方+…+尺寸为m x m的1平方]
等于m(m + 1)/ 2
因此, 当我们添加(n-m)列时, 增加的平方总数为(n-m)* m(m + 1)/ 2。
因此, 平方总数为m(m + 1)(2m + 1)/ 6 +(n-m)* m(m + 1)/ 2。
使用相同的逻辑, 我们可以证明n < = m。
所以总的来说

Total number of squares = m x (m+1) x (2m+1)/6 + (n-m) x m x (m+1)/2 when n is larger dimension

.
对矩形使用上述逻辑, 我们还可以证明一个正方形中的正方形数为n(n + 1)(2n + 1)/ 6
下面是上述公式的实现。
C ++

// C++ program to count squares // in a rectangle of size m x n #include< iostream> using namespace std; // Returns count of all squares // in a rectangle of size m x n int countSquares( int m, int n) { // If n is smaller, swap m and n if (n < m) swap(m, n); // Now n is greater dimension, // apply formula return m * (m + 1) * (2 * m + 1) / 6 + (n - m) * m *(m + 1) / 2; }// Driver Code int main() { int m = 4, n = 3; cout < < "Count of squares is " < < countSquares(m, n); }

Java

// Java program to count squares // in a rectangle of size m x nclass GFG { // Returns count of all squares // in a rectangle of size m x n static int countSquares( int m, int n) { // If n is smaller, swap m and n if (n < m) { // swap(m, n) int temp = m; m = n; n = temp; }// Now n is greater dimension, // apply formula return m * (m + 1 ) * ( 2 * m + 1 ) / 6 + (n - m) * m * (m + 1 ) / 2 ; }// Driver Code public static void main(String[] args) { int m = 4 , n = 3 ; System.out.println( "Count of squares is " + countSquares(m, n)); } }

Python3

# Python3 program to count squares # in a rectangle of size m x n# Returns count of all squares # in a rectangle of size m x n def countSquares(m, n):# If n is smaller, swap m and n if (n < m): temp = m m = n n = temp# Now n is greater dimension, # apply formula return ((m * (m + 1 ) * ( 2 * m + 1 ) / 6 + (n - m) * m * (m + 1 ) / 2 ))# Driver Code if __name__ = = '__main__' : m = 4 n = 3 print ( "Count of squares is " , countSquares(m, n))# This code is contributed by mits.

// C# program to count squares in a rectangle // of size m x n using System; class GFG {// Returns count of all squares in a // rectangle of size m x n static int countSquares( int m, int n) { // If n is smaller, swap m and n if (n < m) { // swap(m, n) int temp = m; m = n; n = temp; }// Now n is greater dimension, apply // formula return m * (m + 1) * (2 * m + 1) / 6 + (n - m) * m * (m + 1) / 2; }// Driver method public static void Main() { int m = 4, n = 3; Console.WriteLine( "Count of squares is " + countSquares(m, n)); } }//This code is contributed by vt_m.

的PHP

< ?php // PHP program to count squares // in a rectangle of size m x n// Returns count of all squares // in a rectangle of size m x n function countSquares( $m , $n ) { // If n is smaller, swap m and n if ( $n < $m ) list( $m , $n ) = array ( $n , $m ); // Now n is greater dimension, // apply formula return $m * ( $m + 1) * (2 * $m + 1) / 6 + ( $n - $m ) * $m * ( $m + 1) / 2; }// Driver Code $m = 4; $n = 3; echo ( "Count of squares is " . countSquares( $m , $n )); // This code is contributed by Ajit. ?>

输出：

Count of Squares is 20

【算法题（如何计算矩形中的正方形数（））】替代解决方案：
让我们取m = 2, n = 3；
边1的平方数将是6, 因为将存在两种情况, 一种是沿水平(2)的1单元边的正方形, 第二种是沿垂直(3)的1单元边的正方形。给我们2 * 3 = 6平方
当边为2个单位时, 一种情况将是仅沿一个位置水平放置2个单位的边的正方形, 而第二个情况将为垂直放置两个位置的情况。所以平方数= 2
所以我们可以推断出
大小为1 * 1的平方数将为m * n
大小2 * 2的平方数将为(n-1)(m-1)
因此, 大小为n的平方数将为1 *(m-n + 1)
总平方数的最终公式为
n *(n + 1)(3m-n + 1)/ 6
C ++

// C++ program to count squares // in a rectangle of size m x n #include< iostream> using namespace std; // Returns count of all squares // in a rectangle of size m x n int countSquares( int m, int n) {// If n is smaller, swap m and n if (n < m) { int temp = m; m = n; n = temp; }// Now n is greater dimension, // apply formula return n * (n + 1) * (3 * m - n + 1) / 6; }// Driver Code int main() { int m = 4, n = 3; cout < < "Count of squares is " < < countSquares(m, n); }// This code is contributed by 29AjayKumar

Java

// Java program to count squares // in a rectangle of size m x n import java.util.*; class GFG {// Returns count of all squares // in a rectangle of size m x n static int countSquares( int m, int n) {// If n is smaller, swap m and n if (n < m) { int temp = m; m = n; n = temp; }// Now n is greater dimension, // apply formula return n * (n + 1 ) * ( 3 * m - n + 1 ) / 6 ; }// Driver Code public static void main(String[] args) { int m = 4 ; int n = 3 ; System.out.print( "Count of squares is " + countSquares(m, n)); } }// This code is contributed by 29AjayKumar

Python3

# Python3 program to count squares # in a rectangle of size m x n # Returns count of all squares # in a rectangle of size m x n def countSquares(m, n): # If n is smaller, swap m and n if (n < m): temp = m m = n n = temp # Now n is greater dimension, # apply formula return n * (n + 1 ) * ( 3 * m - n + 1 ) / / 6# Driver Code if __name__ = = '__main__' : m = 4 n = 3 print ( "Count of squares is" , countSquares(m, n)) # This code is contributed by AnkitRai01

// C# program to count squares // in a rectangle of size m x n using System; class GFG {// Returns count of all squares // in a rectangle of size m x n static int countSquares( int m, int n) {// If n is smaller, swap m and n if (n < m) { int temp = m; m = n; n = temp; }// Now n is greater dimension, // apply formula return n * (n + 1) * (3 * m - n + 1) / 6; }// Driver Code public static void Main(String[] args) { int m = 4; int n = 3; Console.Write( "Count of squares is " + countSquares(m, n)); } }// This code is contributed by Rajput-Ji

输出：