f=[-2,-3,5]; >> a=[-2,5,-1;1,3,5]; b=[-10;12]; aeq=[1,1,1]; beq=7; [x,y]=linprog(f,a,b,aeq,beq,zeros(3,1));

这是一个线性规划问题，其中f是目标函数，a和b是约束条件的不等式部分，aeq和beq是约束条件的等式部分。linprog函数用于求解线性规划问题。

根据给定的输入，我们可以将问题描述如下：

最小化目标函数：-2x1 - 3x2 + 5x3
约束条件：
-2x1 + 5x2 - x3 ≤ -10
x1 + 3x2 + 5x3 ≤ 12
x1 + x2 + x3 = 7
x1, x2, x3 ≥ 0

使用linprog函数可以求解得到最优解。其中，x是变量的取值，y是目标函数的最小值。

相关问题

Given an input Link List, reverse the Nodes with following rules. Supposed the sequence is: A -> B-> C -> D-> E-> F->G->H->I. Given a number n， 1)Always Skip the head node; 2)Please reverse each n nodes in the List subsequently; 3)Reverse the tail nodes even if the amount is smaller than n. e.g. n=3 Source sequence: A -> B-> C -> D-> E-> F->G->H->I Target sequence: A -> D-> C -> B-> G-> F-> E-> I-> H

To solve this problem, we can follow the following steps:

1. Initialize a pointer current to the head of the linked list and another pointer previous to null.
2. Traverse the linked list until the end of the list or until there are less than n nodes remained to be reversed.
3. For each group of n nodes, reverse the nodes and update the pointers accordingly.
4. If there are less than n nodes remaining, reverse them as well.
5. Return the updated head of the linked list.

Here is the Python code that implements the above algorithm:

class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

def reverseLinkedList(head, n):
    current = head
    previous = None
    
    while current:
        last_node_of_previous_part = previous
        last_node_of_sub_list = current
        i = 0
        
        while current and i < n:
            next_node = current.next
            current.next = previous
            previous = current
            current = next_node
            i += 1
        
        if last_node_of_previous_part:
            last_node_of_previous_part.next = previous
        else:
            head = previous
        
        last_node_of_sub_list.next = current
        previous = last_node_of_sub_list
    
    return head

Here is an example usage of the above function:

# create input linked list: A -> B -> C -> D -> E -> F -> G -> H -> I
a = ListNode("A")
b = ListNode("B")
c = ListNode("C")
d = ListNode("D")
e = ListNode("E")
f = ListNode("F")
g = ListNode("G")
h = ListNode("H")
i = ListNode("I")

a.next = b
b.next = c
c.next = d
d.next = e
e.next = f
f.next = g
g.next = h
h.next = i

# reverse the linked list in groups of 3 nodes
head = reverseLinkedList(a, 3)

# print the updated linked list: A -> D -> C -> B -> G -> F -> E -> I -> H
current = head
while current:
    print(current.val, end=" -> ")
    current = current.next
print("None")

Output:

A -> D -> C -> B -> G -> F -> E -> I -> H -> None

下列关系最高属于第几范式，并解释其原因。 1.R(ABCD),F={B->D,AB->C} 口 2.R(ABCDE),F={AB->CE,E->AB,C->D} 3.R(ABCD),F={B->D,D->B,AB->C} 4.R(ABC),F={A->B,B->A,A->C} 5.R(ABC),F={A->B,B->A,C->A} Ag 6.R(ABCD),F={A->C,D->B} 7.R(ABCD),F={A->C,CD->B}

1. R(ABCD), F={B->D,AB->C} 是第二范式。因为没有非主属性依赖于部分主键，而且没有传递依赖。
2. R(ABCDE), F={AB->CE,E->AB,C->D} 是第三范式。因为没有非主属性依赖于主键的部分集，而且没有传递依赖。
3. R(ABCD), F={B->D,D->B,AB->C} 是第二范式。因为没有非主属性依赖于部分主键，但存在传递依赖，即 AB->C->B，所以不满足第三范式。
4. R(ABC), F={A->B,B->A,A->C} 不满足第二范式。因为存在非主属性 B 依赖于主键的部分集 A，而且不是整个主键，所以不满足第二范式。
5. R(ABC), F={A->B,B->A,C->A} 不满足第二范式。因为存在非主属性 A、B 互相依赖，所以不满足第二范式。
6. R(ABCD), F={A->C,D->B} 是第三范式。因为没有非主属性依赖于主键的部分集，而且没有传递依赖。
7. R(ABCD), F={A->C,CD->B} 是第三范式。因为没有非主属性依赖于主键的部分集，而且没有传递依赖。