编程语言：Python

解题思想：

1.T,E统计在输入的时候就进行

2.注意E统计的是发生掉落的苹果棵数，而不是发生掉落的次数

3.使用T来判断E的情况

4.因为使用列表，需要对E额外进行两次首尾相连的特殊情况判断。

时间复杂度：O(n^2)

IDE代码如下：

NOT = int(input())  # 输入苹果树的棵数
Operate = list()  # 存放每一行输入值的列表
Continuous = [0 for i in range(NOT)]  # 存放掉落苹果情况的苹果树编号
T, D, E = 0, 0, 0  # 分别代表疏果操作结束后总的苹果个数，掉落苹果的苹果棵数，连续三棵树发生掉落苹果的情况的组数
 
# 完成T和D的统计
for i in range(NOT):
    Operate = list(map(int, input().split()))  # 将输入的一行全部存入列表中
    Initial = Operate[1]  # 每棵苹果树记录的初始苹果个数为列表下标的第1个元素
    DJudge = False  # 该苹果树是否掉落
    for j in range(Operate.__len__() - 2):  # 对列表的最后（列表-2）个元素进行遍历
        if Operate[j + 2] > 0:  # 发现更新的苹果个数
            if Operate[j + 2] < Initial and DJudge == False:  # 该苹果树首次发生苹果掉落的情况
                D += 1
                DJudge = True
                Continuous[i] = 1  # 将该苹果树的掉落标记改为1
            Initial = Operate[j + 2]  # 更新记录的苹果个数
        else:  # 疏果个数
            Initial += Operate[j + 2]
    T += Initial
 
# 完成E的统计
if T < 3:  # 不可能存在连续3棵掉落的情况
    E = 0
else:
    for i in range(Continuous.__len__() - 2):  # 对掉落苹果树列表进行遍历
        if Continuous[i] == Continuous[i + 1] == Continuous[i + 2] == 1:
            E += 1
    if Continuous[Continuous.__len__() - 2] == Continuous[Continuous.__len__() - 1] == Continuous[0] == 1:  # 对特殊情况进行判断
        E += 1
    if Continuous[Continuous.__len__() - 1] == Continuous[0] == Continuous[1] == 1:  # 对特殊情况进行判断
        E += 1
print(T, D, E)

< >

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CCF-CSP_20190902:小明种苹果（续）

编程语言：Python

解题思想：

时间复杂度：O(n^2)

IDE代码如下：

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共有 11 条评论

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