Question

显示方阵的非对角线元素必须满足的条件是什么。例如，

L=123
  456
  789

答案应该是以下形式：

   3
  5
 7

Answer 1

>>> l = len(matrix[0])

所有行都有相同的长度：

>>> [matrix[l-1-i][i] for i in range(l-1,-1,-1)]
[3, 5, 7]

您希望遍历列表，从第一行开始并获取其第三个元素，然后是第二行的第二个元素，依此类推：

// Create Controller

var popoverContent = CustomPop()

// Get NavController
var nav = UINavigationController(rootViewController: popoverContent)
nav.modalPresentationStyle = UIModalPresentationStyle.Popover

// Get PopoverPresentationController 
var popover = nav.popoverPresentationController as UIPopoverPresentationController
popover.sourceView = self.view
popover.sourceRect = CGRectMake(100,100,0,0)

// Show Popover
self.presentViewController(nav, animated: true, completion: nil)

Answer 2

我已经从上面的答案中得到了提示并且说得对，我猜。

n=input('Enter the no. of rows=')
a=[[0 for j in range(n)]for i in range(n)]
for i in range(n):
    for j in range(n):
         a[i][j]=input()
print'\nInput Matrix is->'
for i in range(n):
    for j in range(n):
         print a[i][j],' ',
    print

print'\nOff-diagonal elements are->'
for i in range(n):
    for j in range(n):
        if i+j==n-1:
            print a[i][j],
        else: print ' ',
    print

在方形矩阵中打印非对角线元素要满足的条件

2 个答案: