求一个方阵两条对角线之和的乘积
给定一个由大小为 NxN 的整数组成的正方形矩阵 mat ，任务是计算其对角线之和之间的乘积。 例:
Input: mat[][] = {{5, 8, 1},
{5, 10, 3},
{-6, 17, -9}}
Output: 30
Sum of primary diagonal = 5 + 10 + (-9) = 6.
Sum of secondary diagonal = 1 + 10 + (-6) = 5.
Product = 6 * 5 = 30.
Input: mat[][] = {{22, -8, 11},
{55, 87, -1},
{-61, 69, 19}}
Output: 4736
天真的做法: 遍历整个矩阵找到对角元素。计算正方形矩阵两条对角线的和。然后，取两个和的乘积。 时间复杂度:O(N2) 朴素方法:**通过观察对角元素索引中的模式，只遍历对角元素，而不是整个矩阵。 以下是本办法的实施情况:
卡片打印处理机（Card Print Processor 的缩写）
// C++ program to find the product
// of the sum of diagonals.
#include <bits/stdc++.h>
using namespace std;
// Function to find the product
// of the sum of diagonals.
long long product(vector<vector<int>> &mat, int n)
{
// Initialize sums of diagonals
long long d1 = 0, d2 = 0;
for (int i = 0; i < n; i++)
{
d1 += mat[i][i];
d2 += mat[i][n - i - 1];
}
// Return the answer
return 1LL * d1 * d2;
}
// Driven code
int main()
{
vector<vector<int>> mat = {{ 5, 8, 1},
{ 5, 10, 3},
{ -6, 17, -9}};
int n = mat.size();
// Function call
cout << product(mat, n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to find the product
// of the sum of diagonals.
class GFG{
// Function to find the product
// of the sum of diagonals.
static long product(int [][]mat, int n)
{
// Initialize sums of diagonals
long d1 = 0, d2 = 0;
for (int i = 0; i < n; i++)
{
d1 += mat[i][i];
d2 += mat[i][n - i - 1];
}
// Return the answer
return 1L * d1 * d2;
}
// Driven code
public static void main(String[] args)
{
int [][]mat = {{ 5, 8, 1},
{ 5, 10, 3},
{ -6, 17, -9}};
int n = mat.length;
// Function call
System.out.print(product(mat, n));
}
}
// This code is contributed by 29AjayKumar
Python 3
# Python3 program to find the product
# of the sum of diagonals.
# Function to find the product
# of the sum of diagonals.
def product(mat,n):
# Initialize sums of diagonals
d1 = 0
d2 = 0
for i in range(n):
d1 += mat[i][i]
d2 += mat[i][n - i - 1]
# Return the answer
return d1 * d2
# Driven code
if __name__ == '__main__':
mat = [[5, 8, 1],
[5, 10, 3],
[-6, 17, -9]]
n = len(mat)
# Function call
print(product(mat, n))
# This code is contributed by mohit kumar 29
C
// C# program to find the product
// of the sum of diagonals.
using System;
class GFG{
// Function to find the product
// of the sum of diagonals.
static long product(int [,]mat, int n)
{
// Initialize sums of diagonals
long d1 = 0, d2 = 0;
for (int i = 0; i < n; i++)
{
d1 += mat[i, i];
d2 += mat[i, n - i - 1];
}
// Return the answer
return 1L * d1 * d2;
}
// Driven code
public static void Main(String[] args)
{
int [,]mat = {{ 5, 8, 1},
{ 5, 10, 3},
{ -6, 17, -9}};
int n = mat.GetLength(0);
// Function call
Console.Write(product(mat, n));
}
}
// This code is contributed by Princi Singh
java 描述语言
<script>
// Javascript program to find the product
// of the sum of diagonals.
// Function to find the product
// of the sum of diagonals.
function product(mat, n)
{
// Initialize sums of diagonals
let d1 = 0, d2 = 0;
for (let i = 0; i < n; i++)
{
d1 += mat[i][i];
d2 += mat[i][n - i - 1];
}
// Return the answer
return d1 * d2;
}
// Driven code
let mat = [[ 5, 8, 1],
[ 5, 10, 3],
[ -6, 17, -9]];
let n = mat.length;
// Function call
document.write(product(mat, n));
// This code is contributed by rishavmahato348.
</script>
Output:
30
时间复杂度:O(N)T4】
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