Python 入门编程题（一）答案

缺乏、安全感 2022-04-04 11:26 236阅读 0赞

# Python 入门编程题：1~10（答案） #

提示：最好还是先思考，先编写，再看答案哦 ^\_^

for i in range(1, 5):
    	for j in range(1,5):
    		for k in range(1, 5):
    			if i != j and j!= k and i != k:
    				print(i,j,k)

--------------------

i = int(input('净利润:'))
    arr = [1000000,600000,400000,200000,100000,0]
    rat = [0.01,0.015,0.03,0.05,0.075,0.1]
    r = 0
    for idx in range(0,6):
        if i > arr[idx]:
            r += (i - arr[idx]) * rat[idx]
            print((i-arr[idx]) * rat[idx])
            i = arr[idx]
    print(r)

--------------------

import math
    for i in range(10001):
    	if math.sqrt(i + 100) == int(math.sqrt(i + 100)) and math.sqrt(i + 268) == int(math.sqrt(i + 268)):
    		print(i)

--------------------

def leapyear(n):
        return True if (n % 4 == 0 and n % 100 != 0) or n % 400 == 0 else False
    
    days = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30 ]
    year, month, day = [int(x) for x in input('input year/month/day: ').split('/')]
    day2 = sum(days[:month - 1]) + day
    if leapyear(year) and month > 2:
        day2 += 1
    print(day2)

--------------------

L = []
    a = int(input('请输入第一个整数：'))
    b = int(input('请输入第二个整数：'))
    c = int(input('请输入第三个整数：'))
    L.append(a)
    L.append(b)
    L.append(c)
    L.sort()
    print(L)

--------------------

#斐波那契数列。
    def fib(n):
            a,b = 1,1
            for i in range(n-1):
                    a,b = b,a+b
            return a

--------------------

#输出 9*9 乘法口诀表
    for i in range(1, 10):
    	for j in range(1, i+1):
    		print('%s * %s = %s' %(i, j, i * j), end = '    ')
    	print()

--------------------

import math
    leap = 1
    h = 0
    for m in range(101, 201):
    	
    	k = int(math.sqrt(m + 1))
    	for i in range(2, k+1):
    		if  m % i == 0:
    			leap = 0
    			break
    	if leap == 1:
    		print('%-4d' %m, end = '')
    		h += 1
    		if h % 10 == 0:
    			print()
    	leap = 1
    print('The total is %d' % h)

--------------------

from math import sqrt
    #判断n是否为素数
    def isprime(n):
        if n <= 1:
            return 0
        m = int(sqrt(n))+1
        for x in range(2,m):
            if n%x == 0:
                return 0
        return 1
    #利用递归分解n并打印质因数
    def bprime(n):
        if isprime(n):
            print(n)
        else:
            x = 2
            while x <= int(n/2):
                if n%x == 0:
                    print(x)
                    return bprime(n/x)
                x = x + 1

--------------------

10.

x2 = 1
    for day in range(9,0,-1):
        x1 = (x2 + 1) * 2
        x2 = x1
    print(x1)

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*（未完待续，持续更新）*