for i in range(20):
    number *= 2
print(number)

20个2相乘

num = 1
while num<100:
    if (num%3==0 or num%7==0) and num% 21!=0:
        summation += 1
    num += 1
print(summation) 

找到1-100 不能被21整除且3和7的倍数的个数和

1.求出1到100所有数的和，平均值

for  i in range(101):
    sum1 += i

avr = sum1/len(range(100))
print(sum1)
print(avr)

number=1
sum2= 0 
while number<101:
    sum2 += number
    number += 1
avr = sum1/len(range(100))
print(sum2)
print(avr)

2.计算1-100之间能3整除的数的和

sum3 =0
for  i in range(3,101,3):
    sum3 += i
print(sum3)

sum4 = 0
for  i in range(1,101):
    if i%3==0:
        sum4 += i

print(sum4)

number =0
total= 0
while number<101:
    
    total += number
    number+=3

print(total)

number =0
total= 0
while number<101:
    
    if number%3 ==0:
        total += number

    number+=1


print(total)

3.计算1-100之间不能被7整除的数的和

total2=0
for i in range(1,101):
    if i%7==1:
        total2 += i
print(total2)

number2 = 0
total3 = 0
while number2<101:
    
    if number2%7==1:
        total3 += number2
    number2 += 1
print(total3)

1.png 2.png 3.png 4.png 5.png 6.png