numbers=1
for i in range(0,20):
numbers*=2
print(numbers)
2**20
2.
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)
# 1到100之内的能被3整除或能被7整除,但是不能同时被3或7整除的数有多少个
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
# for循环
sum1 = 0
average = 0
n = 0
for n in range(1, 101):
sum1 += n
average = sum1 / n
print('和为%d,平均值为%.2f' % (sum1, average))
# while循环
sum1 = 0
average = 0
n = 1
while n <= 100:
sum1 += n
n += 1
average = sum1 / (n - 1)
print('和为%d,平均值为%.2f' % (sum1, average))
2. 计算1-100之间能3整除的数的和
# for循环
sum2 = 0
for n in range(1, 101):
if not n % 3:
sum2 += n
print('1-100能被3整除的数的和为%d' % sum2)
# while循环
sum2 = 0
n = 1
while n <= 100:
if n % 3 == 0:
sum2 += n
n += 1
print('1-100能被3整除的数的和为%d' % sum2)
3. 计算1-100之间不能被7整除的数的和
# for循环
sum3 = 0
for n in range(1, 101):
if n % 7 == 0:
continue
sum3 += n
print('1-100不能被7整除的数的和为%d' % sum3)
# while循环
sum3 = 0
n = 1
while n <= 100:
if n % 7:
sum3 += n
n += 1
print('1-100不能被7整除的数的和为%d' % sum3)
二、稍微困难
1. 求斐波那契数列中第n个数的值:1,1,2,3,5,8,13,21,34....
num1, num2 = 0, 1
n = int(input('请输入你想得到第几位斐波那契数:'))
for _ in range(n):
num1, num2 = num2, (num1 + num2)
print('第%d位斐波那契数是%d' % (n, num1))
count = 0
for x in range(101, 201):
# 把这个数从2除到它本身都不能整除则是素数,也可以用math.sqrt来判断
for n in range(2, x):
if x % n == 0:
break
else:
count += 1
print(x, end=' ')
print('101-200之间共有%d个素数' % count)