《Python程序设计》 第六章 函数+编程题解
目录
6-1 使用函数求特殊a串数列和
6-2 使用函数求素数和
6-3 使用函数统计指定数字的个数
6-4 使用函数输出指定范围内Fibonacci数的个数
6-5 使用函数求余弦函数的近似值
6-6 缩写词
7-1 输入列表，求列表元素和(eval输入应用）
7-2 一帮一
7-3 验证“哥德巴赫猜想”
7-4 列表或元组的数字元素求和
7-5 列表数字元素加权和(1)
7-6 列表元素个数的加权和(2)
7-7 求指定层的元素个数
7-8 输出全排列
此解法为"动态规划",没算法底子可能比较难看懂
6-1 使用函数求特殊a串数列和
def fn(a, n):
result = a
key = a
for index in range(1, n):
key = key * 10 + a
result += key
return result
6-2 使用函数求素数和
import math
def prime(num):
if num < 2:
return False
if num < 4:
return True
for index in range(2, int(math.sqrt(num)) + 1):
if num % index == 0:
return False
return True
def PrimeSum(m, n):
result = 0
for index in range(m, n + 1):
if prime(index):
result += index
return result
6-3 使用函数统计指定数字的个数
def CountDigit(number, digit):
result = 0
num = int(abs(number))
while num > 0:
if num % 10 == digit:
result += 1
num = int(num/10)
return result
6-4 使用函数输出指定范围内Fibonacci数的个数
dict1 = {0: 1, 1: 1} # 记录每一次的斐波那契数
dict2 = [1] # 记录出现过的斐波那契数
a = b = key = 1
while b < 100001:
key += 1
a, b = b, a + b
dict1[key] = b
dict2.append(b)
def fib(n):
return dict1[n]
def PrintFN(m, n):
result = []
for index in range(m, n + 1):
if index in dict2:
result.append(index)
return result6-5 使用函数求余弦函数的近似值
def Jie(num):
if num == 1:
return 1
return num * Jie(num - 1)
def funcos(eps, x): # 返回cos(x)的值
result = 1
flag = 2
key = pow(x, flag) / 2
while abs(key) >= eps:
result += key if flag % 4 == 0 else -key
flag += 2
key = pow(x, flag) / Jie(flag)
return result
6-6 缩写词
def acronym(str1):
list1 = str1.split()
result = ""
for index in range(len(list1)):
ch = list1[index][0]
if 'a' <= ch <= 'z':
ch = chr(ord(ch) - 32)
result += ch
return result
7-1 输入列表，求列表元素和(eval输入应用）
box = eval(input())
print(sum(box))
7-2 一帮一
n = int(input())
box = []
for index in range(n):
sex, name = input().split()
box.append([sex, name])
while len(box) > 0:
for index in range(len(box) - 1, 0, -1):
if box[index][0] != box[0][0]:
print("%s %s" % (box[0][1], box[index][1]))
box.pop(index)
box.pop(0)
break
7-3 验证“哥德巴赫猜想”
from math import sqrt
def SU(num):
if num < 2:
return False
if num < 4:
return True
for index in range(3, int(sqrt(num) + 3), 2):
if num % index == 0:
return False
return True
N = int(input())
key = 2
for z in range(3, N, 2):
if SU(z) and SU(N - z):
key = z
break
print("%d = %d + %d" % (N, key, N - key))
7-4 列表或元组的数字元素求和
def SUM(str1):
if type(str1) == int:
return str1
if type(str1) != str:
return sum(SUM(x) for x in str1)
return 0
box = eval(input())
print(SUM(box))
7-5 列表数字元素加权和(1)
def SUM(str1, deep):
if type(str1) == int:
# print("%d %d" % (str1, deep))
return str1 * deep
if type(str1) != str:
return sum(SUM(x, deep+1) for x in str1)
return 0
box = eval(input())
print(SUM(box, 0))
7-6 列表元素个数的加权和(2)
def SUM(str1, deep):
if type(str1) == int:
# print("%d %d" % (str1, deep))
return deep
if type(str1) != str:
return sum(SUM(x, deep+1) for x in str1)
return 0
box = eval(input())
print(SUM(box, 0))
7-7 求指定层的元素个数
result = [0 for index in range(101)] # 记录各层出现元素的次数(数组)
def SUM(str1, deep):
if type(str1) == int:
# print("%d %d" % (str1, deep))
result[deep] += 1
elif type(str1) != str:
for index in str1:
SUM(index, deep + 1)
box = eval(input())
SUM(box, 0)
command = int(input())
print(result[command])
7-8 输出全排列
appear = [0 for cdl in range(520)]
N = int(input())
result = []
def printNum(num, deep):
if appear[num] == 1:
return
else:
result.append(num)
appear[num] = 1
for index in range(1, N + 1):
printNum(index, deep + 1)
if deep == N:
for index in result:
print(index, end='')
print()
appear[num] = 0
result.pop(deep-1)
for z in range(1, N + 1):
printNum(z, 1)