1.讀程序，總結(jié)程序的功能:
(1).

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

答：求2^20的值

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)

答：求100以內(nèi)能被3，7整除不能被21整除的數(shù)的個(gè)數(shù)。
2.編程實(shí)現(xiàn)(for和while各寫?遍):
(1). 求1到100之間所有數(shù)的和、平均值

sum1 = 0
for num in range(1, 101):
    sum1 += num
ave = sum1 / 100
print(sum1, ave)

sum1 = 0
num = 1
while num in range(1, 101):
    sum1 += num
    num += 1
ave = sum1 / 100
print(sum1, ave)

(2). 計(jì)算1-100之間能3整除的數(shù)的和

sum1 = 0
for num in range(1, 101):
    if num % 3 == 0:
        sum1 += num
print(sum1)

sum1 = 0
num = 1
while num in range(1, 101):
    if num % 3 == 0:
        sum1 += num
    num += 1
print(sum1)

(3). 計(jì)算1-100之間不能被7整除的數(shù)的和

sum1 = 0
for num in range(1, 101):
    if num % 7 != 0:
        sum1 += num
print(sum1)

sum1 = 0
num = 1
while num in range(1, 101):
    if num % 7 != 0:
        sum1 += num
    num += 1
print(sum1)

1. 求斐波那契數(shù)列中第n個(gè)數(shù)的值：1，1，2，3，5，8，13，21，34....

a = 1
b = 1
sum1 = 0
n = int(input('請(qǐng)輸入一個(gè)數(shù)'))
for sum1 in range(0, n-2):
    x = a +b
    a = b
    b = x
    sum1 += 1
print(b)

2. 判斷101-200之間有多少個(gè)素?cái)?shù)，并輸出所有素?cái)?shù)。判斷素?cái)?shù)的?法：??個(gè)數(shù)分別除2到sqrt(這個(gè)
   數(shù))，如果能被整除，則表明此數(shù)不是素?cái)?shù)，反之是素?cái)?shù)

import math
for i in range(101, 201):
    # for j in range(2, i):
    for j in range(2, int(math.sqrt(i))):
        if i % j == 0:
            print(i, '不是素?cái)?shù)')
            break
    else:
        print(i, '是素?cái)?shù)')


# 方法2
import math
for i in range(101, 201):
    flag = True     #假設(shè)當(dāng)前數(shù)是素?cái)?shù)
    for j in range(2, int(math.sqrt(i))):
         if i % j == 0:
             flag = False # 如果在2-i之間遇到一個(gè)能整除的假設(shè)不成立
             break
    if flag:
        print(i, '是素?cái)?shù)')

3. 打印出所有的?仙花數(shù),所謂?仙花數(shù)是指?個(gè)三位數(shù)，其各位數(shù)字??和等于該數(shù)本身。例如：153是
   ?個(gè)?仙花數(shù),因?yàn)?53 = 1^3 + 5^3 + 3^3

for n in range(100,1000):
    if n == (n // 100) ** 3 + (n % 100 // 10)**3 \
            + (n % 100 % 10) ** 3:
        print(n)

4. 有?分?jǐn)?shù)序列：2/1,3/2,5/3,8/5,13/8,21/13...求出這個(gè)數(shù)列的第20個(gè)分?jǐn)?shù)
   分?：上?個(gè)分?jǐn)?shù)的分?加分? 分?: 上?個(gè)分?jǐn)?shù)的分? fz = 2 fm = 1 fz+fm / fz

fz = 1
fm = 1
for n in range(20):
    fz, fm = fz + fm, fz
    # sum1 = fz + fm
    # fm = fz
    # fz = sum1
print(fz, '/', fm)

5. 給?個(gè)正整數(shù)，要求：1、求它是?位數(shù) 2.逆序打印出各位數(shù)字

m = int(input("請(qǐng)輸入："))
string = ""
count = 0
while True:
        string += str(m % 10) + " "
        m = int(m / 10)
        count += 1
        if m == 0:
            break
print("m是" + str(count) + "位數(shù)")
print(string)