n = int(input()) matrix = [] for i in range(n): row = list(map(float, input().split())) matrix.append(row) sum1 = sum(matrix[i][i] for i in range(n)) # 主对角线上的元素之和 sum2 = sum(matrix[i][n-i-1] for i in range(n)) # 副对角线上的元素之和 print("sum1={:.2f}, sum2={:.2f}".format(sum1, sum2))将这串代码改为求sum1sum2d 的和

n = int(input())
matrix = []
for i in range(n):
row = list(map(float, input().split()))
matrix.append(row)

sum1 = sum(matrix[i][i] for i in range(n)) # 主对角线上的元素之和
sum2 = sum(matrix[i][n-i-1] for i in range(n)) # 副对角线上的元素之和
d = sum(matrix[i][j] for i in range(n) for j in range(n) if i != j and i+j != n-1) # 其它元素之和

print("sum1={:.2f}, sum2={:.2f}, d={:.2f}".format(sum1, sum2, d))
# 注意：这里直接将求和的表达式放入格式化字符串中，不需要另外定义变量

相关问题

Solve the problem with code:Ack Country has discovered large quantities of crude oil under its territory. The king decided that all n industrial facilities in the country should be connected with crude oil to keep production working. There are two ways to supply crude oil to industrial facilities, either by laying pipelines to other industrial facilities that are already supplied with crude oil, or by exploring for oil at the plant site.The cost of crude oil exploration is w, and the cost of connecting two industrial facilities is c, given in matrix form, and the main diagonal of matrix is 0. Calculate the minimum cost of supplying crude oil to all industrial facilities.

Sure, here is a Python code that solves the problem using the Floyd-Warshall algorithm:

INF = float('inf')

def find_min_cost(n, c, w):
    for k in range(n):
        for i in range(n):
            for j in range(n):
                c[i][j] = min(c[i][j], c[i][k] + c[k][j])
                w[i][j] = min(w[i][j], c[i][j] + min(w[i][k], w[k][j]))
    return sum(sum(row) for row in w)

n = int(input())
c = []
w = []
for i in range(n):
    c_row = list(map(int, input().split()))
    c.append(c_row)
    w_row = [INF if x == 0 else x for x in c_row]
    w.append(w_row)

print(find_min_cost(n, c, w))

This code takes the input of `n`, the size of the matrix, `c`, the cost of connecting two industrial facilities, and `w`, the cost of crude oil exploration. It then initializes the `INF` constant to represent infinity and sets the diagonal of `w` to infinity since exploration is not required for facilities that already have oil.

The algorithm then uses three nested loops to calculate the minimum cost of connecting all the facilities. It first calculates the shortest path between two facilities using the `c` matrix with the Floyd-Warshall algorithm. It then calculates the cost of supplying oil to each facility by adding the cost of connecting the facility to the nearest facility with oil and the cost of exploration if necessary.

Finally, the algorithm returns the total minimum cost of supplying oil to all facilities by summing up the values in the `w` matrix.

Please note that this code assumes that the input is valid and follows the problem statement. It also assumes that the input is formatted correctly.

Input 在第一行输入N（2≤N≤1000）代表矩阵的行数和列数。 接下来N行，每行输入N个数字。对于每个数字num（1≤num≤109) Output 输出答案，矩阵四条边中最小的数。用Python写给我

Input部分描述的是一个程序的输入格式，它涉及到一个二维整数数组（矩阵）。你需要读取的变量有：

1. `N`：表示矩阵的行数和列数，其范围是从2到1000。
2. 矩阵数据：接下来的N行，每行包含N个整数，数值范围从1到10^9。

例如，如果输入是一个3x3的矩阵：

3
1 2 3
4 5 6
7 8 9

这表示矩阵的第一行是[1, 2, 3]，第二行是[4, 5, 6]，第三行是[7, 8, 9]。

Output部分要求你计算并输出这个矩阵四条边上所有元素中的最小值。在Python中，你可以使用嵌套循环遍历矩阵，并维护四个边界（左上、右上、左下、右下）的最小值。这是一个简单的示例：

def min_edge(matrix):
    rows = len(matrix)
    cols = len(matrix[0])
    
    # 初始化边缘最小值
    min_top_left = matrix[0][0]
    min_top_right = float('inf') if rows > 1 else matrix[0][cols - 1]
    min_bot_left = float('inf') if cols > 1 else matrix[rows - 1][0]
    min_bot_right = float('inf')
    
    # 更新边界最小值
    for i in range(1, rows):
        min_top_right = min(min_top_right, matrix[i][cols - 1])
        min_bot_left = min(min_bot_left, matrix[i][0])
        
    for j in range(1, cols):
        min_top_left = min(min_top_left, matrix[0][j])
        min_bot_right = min(min_bot_right, matrix[rows - 1][j])
        
    return min(min_top_left, min_top_right, min_bot_left, min_bot_right)

# 示例输入处理
n = int(input())
matrix = []
for _ in range(n):
    row = list(map(int, input().split()))
    matrix.append(row)

# 计算并输出结果
min_num = min_edge(matrix)
print(min_num)