Day3作业

2018-07-19  本文已影响0人  zhazhaK丶
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。

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
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