读程序，总结程序的功能:

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

程序执行过程：
numbers=1；i=0；true；numbers=12=2; print2;
numbers=2; i=1; true; numbers=22=4; print4;
numbers=4; i=2; true; numbers=42=8; print8;
...
number=262144, i=19; true; numbers=2621442=524288;print524288;
number=262144, i=20; false
执行结束；
功能：求2的19次方的值

 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)

执行过程：
summation=0; num=1; 1<=100; true; 1%3==0; false; 1%7==0; false; or语句; false; and语句; false; num+=1; num=2;
summation=0; num=2; 2<=100; true; 2%3==0; false; 2%7==0; false; or语句; false; and语句; false; num+=1; num=3;
summation=0; num=3; 3<=100; ture; 3%3==0; true; or语句; true; 3%21!=0; true; and语句; true; summation=1; num+=1; num=4;
...
summation=0; num=100; 100<=100; true; 100%3==0; false; or语句; false; and语句; false; num+=1; num=101;
summation=0; num=101; 101<=100; false;
执行结束;
功能：求1-100中能被3或者7整除，且同时不能被21整除的数的个数

编程实现(for和while各写一遍):

1. 求1到100之间所有数的和、平均值

sum = 0
count = 1
for sum1 in range(1, 101):
    sum += sum1
    avg = sum / count
    count += 1
print(sum)
print(avg)

sum1 = 0
sum = 1
count = 1
while sum < 101:
    sum1 += sum
    vag = sum1 / count
    sum += 1
    count += 1
print(sum1)
print(vag)

2. 计算1-100之间能3整除的数的和

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

sum = 0
x = 1
while x < 101:
    if x % 3 == 0:
        sum += x
    x += 1
print(sum)

3. 计算1-100之间不不能被7整除的数的和

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

sum = 0
x = 1
while x < 101:
    if x % 7 != 0:
        sum += x
    x += 1
print(sum)