《Python程序设计》 第五章 编程题题解_7-5 分析活动投票情况

目录

7-1 输出星期名缩写

7-2 图的字典表示

7-3 四则运算（用字典实现）

7-4 分析活动投票情况

7-5 统计字符出现次数

7-6 统计工龄

7-7 列表去重

7-8 能被3,5和7整除的数的个数（用集合实现）

7-9 求矩阵鞍点的个数

7-10 两数之和

7-11 字典合并

 

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7-1 输出星期名缩写

box = {1: 'Mon', 2: 'Tue', 3: 'Wed', 4: 'Thu', 5: 'Fri', 6: 'Sat', 7: 'Sun'}
key = int(input())
print(box[key])

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7-2 图的字典表示

N = int(input())
roadNum = 0
sumLength = 0
for index in range(0, N):
    dict1 = eval(input())       # 读取第一层字典
    for z in dict1:
        dict2 = dict1[z]        # 读取第二层字典
        for z1 in dict2:        # 遍历字典
            # print("%c %d" % (z1,dict2[z1]))
            sumLength += dict2[z1]
            roadNum += 1
        
        
print("%d %d %d" % (N, roadNum, sumLength))

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7-3 四则运算（用字典实现）

dict1 = {'+': 'a+b', '-': 'a-b', '*': 'a*b', '/': 'a/b'}
a = int(input())
op = input()
b = int(input())
if op == '/' and b == 0:
    print("divided by zero")
else:
    result = eval(dict1[op])
    print("%.2f" % result)

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7-4 分析活动投票情况

nums = [int(x) for x in input().split(',')]
ticket = [0 for i in range(0, 11)]
for x in nums:
    ticket[x] += 1
 
output = 0
for index in range(6, 11):
    if ticket[index] == 0:
        if output == 1:
            print(' ', end='')
        output = 1
        print(index, end='')

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7-5 统计字符出现次数

str1 = input()
key = input()
result = 0
for index in str1:
    if index == key:
        result += 1
print(result)

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7-6 统计工龄

N = int(input())
key = [0 for z in range(0, 51)]
nums = [int(x) for x in input().split()]
for index in nums:
    key[index] += 1
 
for index in range(0, 51):
    if key[index] > 0:
        print("%d:%d" % (index, key[index]))

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7-7 列表去重

list1 = eval(input())
dict1 = {}
for index in list1:
    if index not in dict1:
        dict1[index] = 0
 
output = 0
for index in dict1:
    if output == 1:
        print(' ', end='')
    print(index, end='')
    output = 1
 
 

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7-8 能被3,5和7整除的数的个数（用集合实现）

n, m = map(int, input().split())
print(int(m/105)-int(n/105))

a, b = map(int, input().split())
f3 = set()
f5 = set()
f7 = set()
for num in range(a, b + 1):
    if num % 3 == 0:
        f3.add(num)
    if num % 5 == 0:
        f5.add(num)
    if num % 7 == 0:
        f7.add(num)
 
print(len(f3 & f5 & f7))

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7-9 求矩阵鞍点的个数

n = int(input())
box = []
for z in range(0, n):
    line = [int(x) for x in input().split()]
    box.append(line)
 
 
# 找到改行最大的数
def lineMAX(lineNum):
    maxNum = 0
    for index1 in range(1, n):
        if box[lineNum][index1] > box[lineNum][maxNum]:
            maxNum = index1
 
    return box[lineNum][maxNum]
 
 
# 找到该列最小的数
def rankMin(rankNum):
    minNum = 0
    for index2 in range(1, n):
        if box[index2][rankNum] < box[minNum][rankNum]:
            minNum = index2
 
    return box[minNum][rankNum]
 
 
result = 0
for z1 in range(0, n):
    for z2 in range(0, n):
        if box[z1][z2] == lineMAX(z1) == rankMin(z2):
            result += 1
 
print(result)

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7-10 两数之和

nums = [int(x) for x in input().split(',')]
key = int(input())
 
 
flag = 0
for z in range(0,len(nums)):
    for z1 in range(z,len(nums)):
        if nums[z] + nums[z1] == key:
            print("%d %d" % (z,z1))
            flag = 1
 
if flag == 0:
    print("no answer")

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7-11 字典合并

dict1 = eval(input())
dict2 = eval(input())
 
for index in dict2:
    if index in dict1:
        dict1[index] += dict2[index]
    else:
        dict1[index] = dict2[index]
 
result = sorted(dict1.items(), key=lambda item: item[0] if type(item[0]) == int else ord(item[0]))
 
print(str(dict(result)).replace(' ', '').replace("'", '"'))
 
# 网上找到的说明,说实话这题如果用队列+字典模拟这个过程实在是太麻烦了
# for key,value in b.items():
#     a[key] = a.get(key,0)+value
#     #get方法语法：dict.get(key,default=None),key-字典中要查找的值，default-如果指定键的值不存在时，返回该默认值
# d = sorted(a.items(),key = lambda item:item[0] if type(item[0])==int else ord(item[0]))
# #lambda是匿名函数，lambda函数实现的主要功能是：如果是数字正常比较合并即可，如果是字母的话要转ascii码值然后再比较
# out = str(dict(d)).replace(' ','').replace("'",'"')
# #将得到的字典d按照指定格式进行改造
# print(out)

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