给定大小为N (1≤N≤10 3 )的2D数组A [] [2] ,其中A [i] [0]和A [i] [1]分别表示矩形i的长度和宽度。
Two rectangle i and j where (i < j) are similar if the ratio of their length and breadth is equal
A[i][0] / A[i][1] = A[j][0] / A[j][1]
任务是计算几乎相似的一对矩形。
例子:
Input : A[][2] = {{4, 8}, {15, 30}, {3, 6}, {10, 20}}
Output: 6
Explanation: Pairs of similar rectangles are (0, 1), {0, 2), (0, 3), (1, 2), (1, 3), (2, 3). For every rectangle, ratio of length : breadth is 1 : 2
Input : A[][2] = {{2, 3}, {4, 5}, {7, 8}}
Output: 0
Explanation: No pair of similar rectangles exists.
方法:按照步骤解决问题
- 遍历数组。
- 对于每个( i
)对,通过检查条件A [i] [0] / A [i] [1] = A [j] [0] / A [ j] [1]是否满足。 - 如果发现为真,则增加计数。
- 最后,打印获得的计数。
下面是上述方法的实现:
C++
// C++ Program for the above approach
#include
using namespace std;
// Function to calculate the count
// of similar rectangles
int getCount(int rows,
int columns, int A[][2])
{
int res = 0;
for (int i = 0; i < rows; i++) {
for (int j = i + 1; j < rows; j++) {
if (A[i][0] * 1LL * A[j][1]
== A[i][1] * 1LL * A[j][0]) {
res++;
}
}
}
return res;
}
// Driver Code
int main()
{
// Input
int A[][2]
= { { 4, 8 }, { 10, 20 }, { 15, 30 }, { 3, 6 } };
int columns = 2;
int rows = sizeof(A) / sizeof(A[0]);
cout << getCount(rows, columns, A);
return 0;
}
Java
// Java program for the above approach
import java.util.*;
class GFG{
// Function to calculate the count
// of similar rectangles
static int getCount(int rows, int columns,
int[][] A)
{
int res = 0;
for(int i = 0; i < rows; i++)
{
for(int j = i + 1; j < rows; j++)
{
if (A[i][0] * A[j][1] ==
A[i][1] * A[j][0])
{
res++;
}
}
}
return res;
}
// Driver Code
public static void main(String[] args)
{
// Input
int[][] A = { { 4, 8 }, { 10, 20 },
{ 15, 30 }, { 3, 6 } };
int columns = 2;
int rows = 4;
System.out.print(getCount(rows, columns, A));
}
}
// This code is contributed by code_hunt.
Python3
# Python3 program for the above approach
# Function to calculate the count
# of similar rectangles
def getCount(rows, columns, A):
res = 0
for i in range(rows):
for j in range(i + 1, rows, 1):
if (A[i][0] * A[j][1] ==
A[i][1] * A[j][0]):
res += 1
return res
# Driver Code
if __name__ == '__main__':
# Input
A = [ [ 4, 8 ], [ 10, 20 ],
[ 15, 30 ], [ 3, 6 ] ]
columns = 2
rows = len(A)
print(getCount(rows, columns, A))
# This code is contributed by SURENDRA_GANGWAR
C#
// C# program for the above approach
using System;
class GFG{
// Function to calculate the count
// of similar rectangles
static int getCount(int rows, int columns,
int[,] A)
{
int res = 0;
for(int i = 0; i < rows; i++)
{
for(int j = i + 1; j < rows; j++)
{
if (A[i, 0] * A[j, 1] ==
A[i, 1] * A[j, 0])
{
res++;
}
}
}
return res;
}
// Driver code
static void Main()
{
// Input
int[,] A = { { 4, 8 }, { 10, 20 },
{ 15, 30 }, { 3, 6 } };
int columns = 2;
int rows = 4;
Console.Write(getCount(rows, columns, A));
}
}
// This code is contributed by divyesh072019
Javascript
输出:
6
时间复杂度:O(N 2 )
辅助空间:O(1)