给定三个正整数N ， M和A ，任务是计算M * N网格中面积等于A的矩形的数量。
例子：
Input: N = 2, M = 2, A = 2
Output: 4
Explanation:
In the given grid of size 2 × 2, 2 rectangles of dimension 2 × 1 and 2 rectangles of dimension 1 × 2 can be inscribed.
Therefore, the required output is 4.
Input: N = 2, M = 2, A = 3
Output: 0
Explanation:
The possible rectangles with area A (= 3) are of dimensions either 1 × 3 or 3 × 1.
But, the maximum length of a side in the grid can only be 2. Therefore, no rectangles can be inscribed within the grid.
方法：可以根据以下观察结果解决问题：
The total number of ways to select a segment of length X on the segment of length M is equal to (M – X + 1).
Therefore, the total count of rectangles of size X * Y in the rectangle of size M * N is equal to (M – X + 1) * (N – Y + 1).
请按照以下步骤解决问题：
- 在[1，√A]范围内迭代。对于第i次迭代，在给定网格内找到矩形的长度和宽度的所有可能值，例如{i，(A / i)}或{(A / i)，i} 。
- 遍历所有可能的长度值X和宽度Y，并将矩形的计数增加(M – X + 1)*(N – Y + 1) 。
- 最后，打印获得的计数。
下面是上述方法的实现：
C++
// C++ program of the above approach
#include
using namespace std;
// Function to find the count of rectangles
// in an M * N grid such that the area of
// the rectangles is equal to A
int count_number(int N, int M, int A)
{
// Stores all possible values of length
// and breadth whose area equal to A
vector > v;
// Calculate all divisors of A
for (int i = 1; i * i <= A; i++) {
// If N is divisible by i
if (N % i == 0) {
// Stores length of the rectangle
int length = i;
// Stores breadth of the rectangle
int breadth = A / i;
// If length of rectangle is not
// equal to breadth of rectangle
if (length != breadth) {
// Insert { length, breadth }
v.push_back({ length, breadth });
// Insert { breadth, length }
v.push_back({ breadth, length });
}
else {
// Insert { length, breadth}
// because both are equal
v.push_back({ length, breadth });
}
}
}
// Stores the count of rectangles
// in a grid whose area equal to A
long long total = 0;
// Iterate over all possible
// values of { length, breadth }
for (auto it : v) {
// Stores total count of ways to
// select a segment of length it.first
// on the segment of length M
int num1 = (max(0, M - it.first + 1));
// Stores total count of ways to
// select a segment of length it.second
// on the segment of length N
int num2 = (max(0, N - it.second + 1));
// Update total
total += (num1 * num2);
}
return total;
}
// Drivers Code
int main()
{
// Input
int N = 2, M = 2, A = 2;
// Print the result
cout << count_number(N, M, A) << endl;
} Java
// Java program of the above approach
import java.util.*;
class GFG
{
static class pair
{
int first, second;
public pair(int first, int second)
{
this.first = first;
this.second = second;
}
}
// Function to find the count of rectangles
// in an M * N grid such that the area of
// the rectangles is equal to A
static int count_number(int N, int M, int A)
{
// Stores all possible values of length
// and breadth whose area equal to A
Vector v = new Vector();
// Calculate all divisors of A
for (int i = 1; i * i <= A; i++)
{
// If N is divisible by i
if (N % i == 0)
{
// Stores length of the rectangle
int length = i;
// Stores breadth of the rectangle
int breadth = A / i;
// If length of rectangle is not
// equal to breadth of rectangle
if (length != breadth)
{
// Insert { length, breadth }
v.add(new pair(length, breadth));
// Insert { breadth, length }
v.add(new pair(breadth, length));
}
else
{
// Insert { length, breadth}
// because both are equal
v.add(new pair(length, breadth));
}
}
}
// Stores the count of rectangles
// in a grid whose area equal to A
int total = 0;
// Iterate over all possible
// values of { length, breadth }
for (pair it : v)
{
// Stores total count of ways to
// select a segment of length it.first
// on the segment of length M
int num1 = (Math.max(0, M - it.first + 1));
// Stores total count of ways to
// select a segment of length it.second
// on the segment of length N
int num2 = (Math.max(0, N - it.second + 1));
// Update total
total += (num1 * num2);
}
return total;
}
// Drivers Code
public static void main(String[] args)
{
// Input
int N = 2, M = 2, A = 2;
// Print the result
System.out.print(count_number(N, M, A) +"\n");
}
}
// This code is contributed by 29AjayKumar Python3
# Python3 program of the above approach
# Function to find the count of rectangles
# in an M * N grid such that the area of
# the rectangles is equal to A
def count_number(N, M, A):
# Stores all possible values of length
# and breadth whose area equal to A
v = []
# Calculate all divisors of A
for i in range(1, A + 1):
if i * i > A:
break
# If N is divisible by i
if (N % i == 0):
# Stores length of the rectangle
length = i
# Stores breadth of the rectangle
breadth = A // i
# If length of rectangle is not
# equal to breadth of rectangle
if (length != breadth):
# Insert { length, breadth }
v.append([length, breadth ])
# Insert { breadth, length }
v.append([breadth, length ])
else:
# Insert { length, breadth}
# because both are equal
v.append([length, breadth ])
# Stores the count of rectangles
# in a grid whose area equal to A
total = 0
# Iterate over all possible
# values of { length, breadth }
for it in v:
# Stores total count of ways to
# select a segment of length it.first
# on the segment of length M
num1 = (max(0, M - it[0] + 1))
# Stores total count of ways to
# select a segment of length it.second
# on the segment of length N
num2 = (max(0, N - it[1] + 1))
# Update total
total += (num1 * num2)
return total
# Drivers Code
if __name__ == '__main__':
# Input
N, M, A = 2, 2, 2
# Prthe result
print(count_number(N, M, A))
# This code is contributed by mohit kumar 29.C#
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG
{
public class pair
{
public int first, second;
public pair(int first, int second)
{
this.first = first;
this.second = second;
}
}
// Function to find the count of rectangles
// in an M * N grid such that the area of
// the rectangles is equal to A
static int count_number(int N, int M, int A)
{
// Stores all possible values of length
// and breadth whose area equal to A
List v = new List();
// Calculate all divisors of A
for (int i = 1; i * i <= A; i++)
{
// If N is divisible by i
if (N % i == 0)
{
// Stores length of the rectangle
int length = i;
// Stores breadth of the rectangle
int breadth = A / i;
// If length of rectangle is not
// equal to breadth of rectangle
if (length != breadth)
{
v.Add(new pair(length, breadth));
// Insert { breadth, length }
v.Add(new pair(breadth, length));
}
else
{
// Insert { length, breadth}
// because both are equal
v.Add(new pair(length, breadth));
}
}
}
// Stores the count of rectangles
// in a grid whose area equal to A
int total = 0;
// Iterate over all possible
// values of { length, breadth }
foreach (pair it in v)
{
// Stores total count of ways to
// select a segment of length it.first
// on the segment of length M
int num1 = (Math.Max(0, M - it.first + 1));
// Stores total count of ways to
// select a segment of length it.second
// on the segment of length N
int num2 = (Math.Max(0, N - it.second + 1));
// Update total
total += (num1 * num2);
}
return total;
}
// Driver code
public static void Main(String[] args)
{
// Input
int N = 2, M = 2, A = 2;
// Print the result
Console.Write(count_number(N, M, A) +"\n");
}
}
// This code is contributed by susmitakundugoaldang 输出：
4时间复杂度： O(sqrt(N))
辅助空间： O(sqrt(N))