• 读程序，总结程序功能

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

求出2的20次方，结果为1048576

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)

找出1~100之间，能够被3或者7整除，并且不被21整除的数，输出符合要求的数字的个数。

• 编程实现(for和while各写一遍):
  求1到100之间所有数的和、平均值

for 循环
sum = 0
for x in range(1,101):
    sum += x
sum1 = sum / 100
print(sum,sum1)

while 循环
sum = 0
x = 1
while  x <= 100:
    sum += x
    x += 1
sum1 = sum / 100
print(sum,sum1)

结果

5050 50.5

计算1-100之间能3整除的数的和

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

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

结果

1683