1.解釋下列代碼的含義:
a.
numbers = 1
for i in range(0,20):
numbers *= 2
print(numbers)
解釋：打印 2**20的值
b.
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)
解釋：求 100 以內(nèi) 有多少個(gè) 能被 3或者7 整除但是不能同時(shí)被3和7整除的數(shù)？
編程實(shí)現(xiàn)(for 和 while 都寫一遍)
1.求1-100之間所有數(shù)的和，平均數(shù)！
num = 0 # for語(yǔ)句
for n in range(1,101):
num += n
average = num / 100
print('和：',num,'平均值：',average)
num = 0 # while語(yǔ)句
n=1
while n <=100:
num+=n
n+=1
average = num / 100
print('和：',num,'平均值：',average)
2.計(jì)算1-100之間能被3整除的數(shù)的和
sum = 0 # for語(yǔ)句
for num in range(1,100):
if num %3 == 0:
sum+= num
print(sum)
sum = 0 # while語(yǔ)句
num = 1
while num <=100:
if num % 3 == 0:
sum+=num
num+=1
print(sum)
3.計(jì)算1-100之間不能被7整除的數(shù)的和
sum = 0 # for語(yǔ)句
for num in range(1,101):
if num %7 != 0:
sum+= num
print(sum)
sum = 0 # while語(yǔ)句
num = 1
while num <=100:
if num % 7 != 0:
sum+=num
num+=1
print(sum)
1.求婓波那契數(shù)列中第n個(gè)數(shù)的值:1,1,2,3,5,8......
n = int(input('第幾位數(shù):'))
s2= 1
s1 = 1
x = 2
num = 1
while x < n:
num = s1+s2
s1 = s2
s2 = num
x+=1
print('數(shù)值:',num)
2.判斷101-200之間有多少個(gè)素?cái)?shù)，并輸出所有素?cái)?shù)。
什么是素?cái)?shù)(質(zhì)數(shù)): 除了1和它本身，不能被其他的數(shù)整除！
方法一
for num in range(101,201):
for x in range(2,num):
if num % x == 0:
break
else:
print(num,'是素?cái)?shù)')
方法二
for num in range(101,200):
isprime = True
for x in range(2,num):
if num % x == 0:
isprime = False
break
if isprime:
print(num,'是素?cái)?shù)')
