矩阵元素之和，其中每个元素是行和列的整数除法

考虑一个NXN矩阵，其中每个元素除以列号（整数除法），即 mat[i][j] = floor((i+1)/(j+1)) 其中 0 <= i < n 和 0 <= j < n。任务是找到所有矩阵元素的总和。

例子 ：

Input  : N = 2
Output : 4
2 X 2 matrix with given constraint:
1 0
2 1
Sum of matrix element: 4

Input  : N = 3
Output : 9

方法1（蛮力）：
运行两个循环，一个用于行，另一个用于列，并找到 (i / j) 的整数部分并添加到答案中。
下面是这种方法的实现：

C++

// C++ program to find sum of matrix element
// where each element is integer division of
// row and column.
#include
using namespace std;
 
// Return sum of matrix element where each element
// is division of its corresponding row and column.
int findSum(int n)
{
    int ans = 0;
    for (int i = 1; i <= n; i++)   // for rows
        for (int j = 1; j <= n; j++) // for columns
            ans += (i/j);
    return ans;
}
 
// Driven Program
int main()
{
    int N = 2;
    cout << findSum(N) << endl;
    return 0;
}

Java

// java program to find sum of matrix
// element where each element is integer
// division of row and column.
 
import java.io.*;
 
class GFG {
     
    // Return sum of matrix element
    // where each element is division
    // of its corresponding row and
    // column.
    static int findSum(int n)
    {
        int ans = 0;
         
        // for rows
        for (int i = 1; i <= n; i++)
         
            // for columns
            for (int j = 1; j <= n; j++)
                ans += (i/j);
                 
        return ans;
    }
     
    // Driven Program
    public static void main (String[] args)
    {
        int N = 2;
        System.out.println( findSum(N));
    }
}
 
// This code is contributed by anuj_67.

Python3

# Python 3 program to find sum of
# matrix element where each element
# is integer division of row and column.
 
# Return sum of matrix element
# where each element is division
# of its corresponding row and column.
def findSum(N):
    ans = 0
    for i in range(1, N + 1):
        for j in range(1, N + 1):
            ans += i // j
    return ans
 
# Driver code
N = 2
print(findSum(N))
 
# This code is contributed
# by Shrikant13

C#

// C# program to find the sum of matrix
// element where each element is an integer
// division of row and column.
using System;
 
class GFG {
     
    // Return sum of matrix element
    // where each element is division
    // of its corresponding row and
    // column.
    static int findSum(int n)
    {
        int ans = 0;
         
        // for rows
        for (int i = 1; i <= n; i++)
         
            // for columns
            for (int j = 1; j <= n; j++)
                ans += (i/j);
                 
        return ans;
    }
     
    // Driven Program
    public static void Main ()
    {
        int N = 2;
        Console.WriteLine( findSum(N));
    }
}
 
// This code is contributed by anuj_67.

Javascript

    // Javascript program to find the sum of matrix
    // element where each element is an integer
    // division of row and column.
     
    // Return sum of matrix element
    // where each element is division
    // of its corresponding row and
    // column.
    function findSum(n)
    {
        let ans = 0;
           
        // for rows
        for (let i = 1; i <= n; i++)
           
            // for columns
            for (let j = 1; j <= n; j++)
                ans += parseInt(i/j, 10);
                   
        return ans;
    }
     
    let N = 2;
    document.write(findSum(N));
     

输出：

4

方法2（高效）：
让 N = 9，矩阵将是

观察，对于每 j 列
mat[i][k] = 0，对于 1 <= k < j，1 <= i <= N
mat[i][k] = 1，对于 j <= k < 2*j，1 <= i <= N
mat[i][k] = 2，对于 2*j <= k < 3*j，1 <= i <= N
等等。
因此，在每列i 中，有 i – 1 个零，然后是 i 乘以 1，然后是 i 乘以 2，依此类推。
我们逐列遍历矩阵并对元素求和。
下面是这种方法的实现。

C++

// C++ program to find sum of matrix element
// where each element is integer division of
// row and column.
#include
using namespace std;
 
// Return sum of matrix element where each
// element is division of its corresponding
// row and column.
int findSum(int n)
{
    int ans = 0, temp = 0, num;
 
    // For each column.
    for (int i = 1; i <= n && temp < n; i++)
    {
        // count the number of elements of
        // each column. Initialize to i -1
        // because number of zeroes are i - 1.
        temp = i - 1;
 
        // For multiply
        num = 1;
 
        while (temp < n)
        {
            if (temp + i <= n)
                ans += (i * num);
            else
                ans += ((n - temp) * num);
 
            temp += i;
            num ++;
        }
    }
 
    return ans;
}
 
// Driven Program
int main()
{
    int N = 2;
    cout << findSum(N) << endl;
    return 0;
}

Java

// java program to find sum of matrix element
// where each element is integer division of
// row and column.
 
import java.io.*;
 
class GFG {
     
    // Return sum of matrix element where each
    // element is division of its corresponding
    // row and column.
    static int findSum(int n)
    {
        int ans = 0, temp = 0, num;
     
        // For each column.
        for (int i = 1; i <= n && temp < n; i++)
        {
             
            // count the number of elements of
            // each column. Initialize to i -1
            // because number of zeroes are i - 1.
            temp = i - 1;
     
            // For multiply
            num = 1;
     
            while (temp < n)
            {
                if (temp + i <= n)
                    ans += (i * num);
                else
                    ans += ((n - temp) * num);
     
                temp += i;
                num ++;
            }
        }
     
        return ans;
    }
     
    // Driven Program
    public static void main (String[] args)
    {
        int N = 2;
        System.out.println(findSum(N));
    }
}
 
// This code is contributed by anuj_67.

Python3

# Program to find sum of matrix element
# where each element is integer division 
# of row and column.
 
# Return sum of matrix element where each
# element is division of its corresponding
# row and column.
def findSum(n):
    ans = 0; temp = 0;
 
    for i in range(1, n + 1):
 
        # count the number of elements of
        # each column. Initialize to i -1
        # because number of zeroes are i - 1.
        if temp < n:
            temp = i - 1
 
            # For multiply
            num = 1
            while temp < n:
                if temp + i <= n:
                    ans += i * num
                else:
                    ans += (n - temp) * num
                temp += i
                num += 1
    return ans
 
# Driver Code
N = 2
print(findSum(N))
 
# This code is contributed by Shrikant13

C＃

// C# program to find sum of matrix
// element where each element is
// integer division of row and column.
using System;
 
class GFG
{
     
    // Return sum of matrix element
    // where each element is division
    // of its corresponding row and column.
    static int findSum(int n)
    {
        int ans = 0, temp = 0, num;
     
        // For each column.
        for (int i = 1; i <= n && temp < n; i++)
        {
             
            // count the number of elements
            // of each column. Initialize
            // to i -1 because number of
            // zeroes are i - 1.
            temp = i - 1;
     
            // For multiply
            num = 1;
     
            while (temp < n)
            {
                if (temp + i <= n)
                    ans += (i * num);
                else
                    ans += ((n - temp) * num);
     
                temp += i;
                num ++;
            }
        }
     
        return ans;
    }
     
    // Driver Code
    public static void Main ()
    {
        int N = 2;
        Console.WriteLine(findSum(N));
    }
}
 
// This code is contributed by anuj_67.

Javascript

// java Script program to find sum of matrix element
// where each element is integer division of
// row and column.
 
 
     
// Return sum of matrix element where each
// element is division of its corresponding
// row and column.
function findSum(n) {
    let ans = 0, temp = 0, num;
     
    // For each column.
    for (let i = 1; i <= n && temp < n; i++)
    {
             
        // count the number of elements of
        // each column. Initialize to i -1
        // because number of zeroes are i - 1.
        temp = i - 1;
     
        // For multiply
        num = 1;
     
        while (temp < n)
        {
            if (temp + i <= n)
                ans += (i * num);
            else
                ans += ((n - temp) * num);
     
            temp += i;
            num ++;
        }
    }
     
    return ans;
}
     
// Driven Program
 
    let N = 2;
    document.write(findSum(N));
     
 
 
// This code is contributed by sravan kumar G

输出 ：

3