构造一个和等于对角元素之和的矩阵
原文:https://www . geeksforgeeks . org/construct-a-sum 等于对角线元素之和的矩阵/
给定一个整数 N ，任务是使用正负整数并排除 0 来构造一个大小为 N 2 的矩阵，使得矩阵的和等于矩阵对角线的和。
示例:
输入: N = 2 输出: 1 -2 2 4 说明: 对角线总和= (1 + 4) = 5 矩阵总和=(1–2+2+4)= 5
输入: N = 5 输出: 1 2 3 5 10 3 1 4-9 1 -19 6 1 5-8 4-7 2 1 12 -17 1 1 1 解释: 对角线总和= (1 + 1 + 1 + 1 + 1) = 5 矩阵总和= 5
方法: 解决问题的方法是遍历矩阵的所有索引，在 N 对角位置打印一个正元素(比如说 y ，在剩余的N2–N位置平均分配一个单值的正负整数(比如说 x 和 -x )。
下面是上述方法的实现:
C++
// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to construct matrix with
// diagonal sum equal to matrix sum
void constructmatrix(int N)
{
bool check = true;
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
// If diagonal position
if (i == j) {
cout << 1 << " ";
}
else if (check) {
// Positive element
cout << 2 << " ";
check = false;
}
else {
// Negative element
cout << -2 << " ";
check = true;
}
}
cout << endl;
}
}
// Driver Code
int main()
{
int N = 5;
constructmatrix(5);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to implement
// the above approach
public class Main {
// Function to construct matrix with
// diagonal sum equal to matrix sum
public static void constructmatrix(int N)
{
boolean check = true;
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
// If diagonal position
if (i == j) {
System.out.print("1 ");
}
else if (check) {
// Positive element
System.out.print("2 ");
check = false;
}
else {
// Negative element
System.out.print("-2 ");
check = true;
}
}
System.out.println();
}
}
// Driver Code
public static void main(String[] args)
{
int N = 5;
constructmatrix(5);
}
}
Python 3
# Python3 program to implement
# the above approach
# Function to construct matrix with
# diagonal sum equal to matrix sum
def constructmatrix(N):
check = bool(True)
for i in range(N):
for j in range(N):
# If diagonal position
if (i == j):
print(1, end = " ")
elif (check):
# Positive element
print(2, end = " ")
check = bool(False)
else:
# Negative element
print(-2, end = " ")
check = bool(True)
print()
# Driver code
N = 5
constructmatrix(5)
# This code is contributed by divyeshrabadiya07
C
// C# program to implement
// the above approach
using System;
class GFG{
// Function to construct matrix with
// diagonal sum equal to matrix sum
public static void constructmatrix(int N)
{
bool check = true;
for(int i = 0; i < N; i++)
{
for(int j = 0; j < N; j++)
{
// If diagonal position
if (i == j)
{
Console.Write("1 ");
}
else if (check)
{
// Positive element
Console.Write("2 ");
check = false;
}
else
{
// Negative element
Console.Write("-2 ");
check = true;
}
}
Console.WriteLine();
}
}
// Driver Code
static public void Main ()
{
int N = 5;
constructmatrix(N);
}
}
// This code is contributed by piyush3010
java 描述语言
<script>
// Javascript program to implement
// the above approach
// Function to construct matrix with
// diagonal sum equal to matrix sum
function constructmatrix(N)
{
let check = true;
for (let i = 0; i < N; i++) {
for (let j = 0; j < N; j++) {
// If diagonal position
if (i == j) {
document.write("1 ");
}
else if (check) {
// Positive element
document.write("2 ");
check = false;
}
else {
// Negative element
document.write("-2 ");
check = true;
}
}
document.write("<br/>");
}
}
// Driver Code
let N = 5;
constructmatrix(5);
</script>
Output:
1 2 -2 2 -2
2 1 -2 2 -2
2 -2 1 2 -2
2 -2 2 1 -2
2 -2 2 -2 1
时间复杂度:O(N2) 辅助空间: O(1)
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