# 每個元素是其行號和列號的絕對差的矩陣總和 > 原文： [https://www.geeksforgeeks.org/sum-matrix-element-absolute-difference-row-column-numbers/](https://www.geeksforgeeks.org/sum-matrix-element-absolute-difference-row-column-numbers/) 給定正整數 **n** 。 考慮一個 **n** 行和 **n** 列的矩陣，其中每個元素包含其行號和數字的絕對差。 任務是計算矩陣中每個元素的和。 **示例**： ``` Input : n = 2 Output : 2 Matrix formed with n = 2 with given constraint: 0 1 1 0 Sum of matrix = 2. Input : n = 3 Output : 8 Matrix formed with n = 3 with given constraint: 0 1 2 1 0 1 2 1 0 Sum of matrix = 8. ``` **方法 1（暴力）**： 只需構造一個 n 行 n 列的矩陣，并以其對應的行號和列號的絕對差初始化每個單元格。 現在，找到每個單元格的總和。 以下是上述想法的實現： ## C++ ```cpp // C++ program to find sum of matrix in which each // element is absolute difference of its corresponding // row and column number row. #include<bits/stdc++.h> using namespace std; // Retuen the sum of matrix in which each element // is absolute difference of its corresponding row // and column number row int findSum(int n) { ????// Generate matrix ????int arr[n][n]; ????for (int i = 0; i < n; i++) ????????for (int j = 0; j < n; j++) ????????????arr[i][j] = abs(i - j); ????// Compute sum ????int sum = 0; ????for (int i = 0; i < n; i++) ????????for (int j = 0; j < n; j++) ????????????sum += arr[i][j]; ????return sum; } // Driven Program int main() { ????int n = 3; ????cout << findSum(n) << endl; ????return 0; } ```