# 矩陣的中心元素等于對角線的一半 > 原文： [https://www.geeksforgeeks.org/center-element-of-matrix-equals-sums-of-half-diagonals/](https://www.geeksforgeeks.org/center-element-of-matrix-equals-sums-of-half-diagonals/) 給定一個奇數階矩陣，即（5 * 5）。 任務是檢查矩陣的中心元素是否等于所有半對角線的總和。  **示例**： ``` Input : mat[][] = { 2 9 1 4 -2 6 7 2 11 4 4 2 9 2 4 1 9 2 4 4 0 2 4 2 5 } Output :Yes Explanation : Sum of Half Diagonal 1 = 2 + 7 = 9 Sum of Half Diagonal 2 = 9 + 0 = 9 Sum of Half Diagonal 3 = 11 + -2 = 9 Sum of Half Diagonal 4 = 5 + 4 = 9 Here, All the sums equal to the center element that is mat[2][2] = 9 ``` **簡單方法**： 迭代兩個循環，找到所有對角線總和，然后檢查所有總和是否等于矩陣的中心元素。 如果其中任何一個不等于中心元素，則打印“否”，否則打印“是”。 時間復雜度：O（N * N） **有效方法**：基于[有效方法來找到`O(n)`](https://www.geeksforgeeks.org/efficiently-compute-sums-of-diagonals-of-a-matrix/)中的對角線總和。 以下是此方法的實現 ## C++ ```cpp // C++ Program to check if the center // element is equal to the individual? // sum of all the half diagonals #include <stdio.h> #include<bits/stdc++.h> using namespace std; const int MAX = 100; // Function to Check center element? // is equal to the individual? // sum of all the half diagonals bool HalfDiagonalSums(int mat[][MAX], int n) {???? ????// Find sums of half diagonals ????int diag1_left = 0, diag1_right = 0; ????int diag2_left = 0, diag2_right = 0;???? ????for (int i = 0, j = n - 1; i < n; i++, j--) { ????????if (i < n/2) { ????????????diag1_left += mat[i][i]; ????????????diag2_left += mat[j][i];??????????? ????????} ????????else if (i > n/2) { ????????????diag1_right += mat[i][i]; ????????????diag2_right += mat[j][i];??????????? ????????} ????} ????return (diag1_left == diag2_right &&? ????????????diag2_right == diag2_left && ????????????diag1_right == diag2_left && ????????????diag2_right == mat[n/2][n/2]); } // Driver code int main() { ????int a[][MAX] = { { 2, 9, 1, 4, -2}, ?????????????????????{ 6, 7, 2, 11, 4},? ?????????????????????{ 4, 2, 9, 2, 4}, ?????????????????????{ 1, 9, 2, 4, 4}, ?????????????????????{ 0, 2, 4, 2, 5} }; ????????cout << ( HalfDiagonalSums(a, 5) ? "Yes" : "No" ); ????return 0; } ```