C语言：编写程序，求一个n×n矩阵对角线元素之和。

求一个n×n矩阵对角答脊仔线元素之和的源代码如下

public class lianxi28 {

public static void main(String[] args) {

System.out.println("请输入九个整数：")

Scanner in = new Scanner(System.in)

int[][] a = new int[3][3]

for(int i=0i<3i++){

for(int j=0j<3j++){

a[i][j]=in.nextInt()

}

}

System.out.println("你输入的n×n矩阵是：")

for(int i=0i<3i++){

for(int j=0j<3j++){

System.out.print(a[i][j]+" ")

}

System.out.println()

}

System.out.println()

int sum=0

for(int i=0i<3i++){

for(int j=0j<3j++){

if(i==j){

sum=sum+a[i][j]

}

System.out.println(sum)

}

扩展资料

1、用arrange生成一个数组，元素为0,1,...,8，然后利用reshape生成一个n×n的矩阵。

2、清汪利用for循环将该矩阵的野磨主对角线上的元素相加，然后输出结果。

command1 是显示上三角，command2是显示下三角，command3 是显示前面两个picturebox中的主对角线亏游和销伏销 以及 次对角线和

Private Sub Command2_Click()

Randomize

For i = 1 To 4

For j = 1 To i

sb(i, j) = Int(Rnd * 34 + 36)

Picture2.Print Tab((j - 1) * 5 + 1)sb(i, j)

Next j

Print

Next i

End Sub

Private Sub Command1_Click()

Randomize

For m = 1 To 4

For n = m To 4

sa(m, n) = Int(Rnd * 37 + 103)

Picture1.Print Tab((n - 1) * 5 + 1)sa(m, n)

Next n

Print

Next m

End Sub

Private Sub Command3_Click()

Dim sum1%, sum2%

For m = 1 To 4

For n = 1 To 4

If m = n Then

sum1 = sum1 + sa(m, n)

End If

Next n

Next m

For i = 1 To 4

For j = 1 To 4

If i + j = 5 Then

sum2 = sum2 + sb(i, j)

End If

Next j

Next i

Picture3.Print "A数厅册组主对角线元素和为:" &sum1

Picture3.Print "B数组副对角线元素和为:" &sum2

End Sub

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