编写一个程序,输入两个矩阵AN×N与BN×N(2<N<10)进行如下运算: 1. 加 2. 减 3. 乘
时间: 2023-04-10 07:01:28 浏览: 204
好的,这是一个计算问题,我可以回答。以下是程序的伪代码:
输入矩阵 AN×N 和 BN×N
选择运算类型(加、减、乘)
如果选择加或减:
对应位置的元素相加或相减,得到新的矩阵 CN×N
如果选择乘:
对于矩阵 AN×N 中的每个元素 aij,乘以矩阵 BN×N 中对应列的每个元素 bj1, bj2, ..., bjN,得到新的矩阵 CN×N 中的元素 cij
输出矩阵 CN×N
希望这可以帮到你!
相关问题
编写一个程序,输入两个矩阵AN×N与BN×N(2<N<10)进行如下运算: 1. 加 2. 减 3. 乘
好的,我可以回答这个编程类的问题。以下是一个示例程序,可以实现输入两个矩阵并进行加、减、乘运算:
```python
N = int(input("请输入矩阵的维度N:"))
# 输入矩阵A
print("请输入矩阵A:")
A = []
for i in range(N):
row = list(map(int, input().split()))
A.append(row)
# 输入矩阵B
print("请输入矩阵B:")
B = []
for i in range(N):
row = list(map(int, input().split()))
B.append(row)
# 加法
C = []
for i in range(N):
row = []
for j in range(N):
row.append(A[i][j] + B[i][j])
C.append(row)
print("A + B = ")
for row in C:
print(row)
# 减法
C = []
for i in range(N):
row = []
for j in range(N):
row.append(A[i][j] - B[i][j])
C.append(row)
print("A - B = ")
for row in C:
print(row)
# 乘法
C = []
for i in range(N):
row = []
for j in range(N):
s = 0
for k in range(N):
s += A[i][k] * B[k][j]
row.append(s)
C.append(row)
print("A * B = ")
for row in C:
print(row)
```
希望这个程序能够帮到你!
用java编写一个程序,输入两个矩阵AN×N与BN×N(2<N<10)进行如下运算: 1. 加 2. 减 3. 乘
好的,这是一个计算问题,我可以回答。以下是Java代码示例:
import java.util.Scanner;
public class MatrixOperation {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
System.out.print("Enter the size of the matrix (N): ");
int n = input.nextInt();
int[][] a = new int[n][n];
int[][] b = new int[n][n];
System.out.println("Enter the elements of matrix A:");
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
a[i][j] = input.nextInt();
}
}
System.out.println("Enter the elements of matrix B:");
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
b[i][j] = input.nextInt();
}
}
System.out.println("Choose an operation:");
System.out.println("1. Addition");
System.out.println("2. Subtraction");
System.out.println("3. Multiplication");
int choice = input.nextInt();
int[][] result = new int[n][n];
switch (choice) {
case 1:
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
result[i][j] = a[i][j] + b[i][j];
}
}
break;
case 2:
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
result[i][j] = a[i][j] - b[i][j];
}
}
break;
case 3:
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
for (int k = 0; k < n; k++) {
result[i][j] += a[i][k] * b[k][j];
}
}
}
break;
default:
System.out.println("Invalid choice!");
return;
}
System.out.println("Result:");
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
System.out.print(result[i][j] + " ");
}
System.out.println();
}
}
}