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

求2的20次方

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)

求1到100中能被3或者7整除但不能被3和7同时整除的数的个数

1. 求1到100之间所有数的和、平均值
   for循环

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

while循环

sum1 = 0 
num = 1   
while num<=100:
    sum1+=num
    num+=1
print(sum1)
average=sum1/100
print(average)

2. 计算1-100之间能3整除的数的和

for循环

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

while循环

num=1
sum1=0
while num<=100:
    if num%3==0:
        sum1+=num
    num+=1
print(sum1)

计算1-100之间不能被7整除的数的和

for循环

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

while循环

num=1
sum1=0
while num<=100:
    if num%7!=0:
        sum1+=num
    num+=1
print(sum1)