基础

读程序，总结程序的功能：

1

numbers = 1
for i in range(0,20):
    numbers *= 2
print(numbers)

功能：求2的20次方

2

summation = 0
num = 1 
while num <= 100:
    if (num % 3 == 0 or num % 7 == 0) and num % 21 != 0:
        summation += 1
    num += 1
print(summation)

功能：找出在1~100中，能被3或7整除，且不能被21整除的数字个数。

编程实现(for和while各写一遍)：

1

for

sum = 0
for x in range(1,101):
    sum += x
print(sum,sum/100)

while

num = 0
sum = 0
while num <= 100:
    sum += num
    num += 1
print(sum,sum/100)
print(type(sum and sum/100))

2

for

sum = 0
for x in range(1,101):
    if x % 3 == 0:
        sum += x
print(sum)

while

x = 1
sum = 0
while x <= 100:
    if x % 3 == 0:
        sum += x
    x += 1
print(sum)

3

for

sum = 0
for x in range(1,101):
    if x % 7 != 0:
        sum += x
print(sum)

while

x = 1
sum = 0
while x <= 100:
    if x % 7 != 0:
        sum += x
    x += 1
print(sum)

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求斐波那契数列中的第n个数是多少？1 1 2 3 5 8 13 21...

n = 6
pre_1 = 1
pre_2 = 1
current = 0
for x in range(1,n+1):
    if x == 1 or x == 2:
        # print(1)
        continue
    current = pre_1 + pre_2
    # pre_1 = current 
    # pre_2 = pre_1
    pre_2 = pre_1   
    pre_1 = current
print(current)

判断101-200之间有多少个素数，并输出所有素数。

for num in range(101,201):
    count = 0
    for x in range(2,num):
        if num % x == 0:
            count += 1
            break # print('%d不是素数'%num)
    if count == 0:
        print('%d是素数'%num)

for x in (2,3,4):
    print(x)

number = 101
x=(2-100)
x=2 101%2==0
x=3 101%3==0
x=4 101%4==0
x=5
x=6
x=100 101%100==0
number=102
x=(2,101)
x=2 102%2==0 102不是素数
number=103
x=(2,102)

打印出所有的水仙花数,所谓水仙花数是指一个三位数，其各位数字立方和等于该数本身。

例如：153是一个水仙花数,因为153 = 1^3 + 5^3 + 3^3

for num in range(100,999):
    ge_wei = num % 10
    shi_wei = num // 10 % 10
    bai_wei = num // 100
    if num == ge_wei**3 + shi_wei**3 + bai_wei**3:
        print('%d是水仙数'%num)