Python 阶段编程练习(九)

编程练习

计算1到100以内能被3或者7整除，但不能同时被3和7整除的数的个数，运行结果为39

任务

定义变量num用来存放数值，count用来存放个数
使用while实现循环
通过if来设置符合的条件，符合条件count计数加1
输出符合条件的总个数count

任务提示

同时被3和7整除，即为21的倍数；
除以21不等于0即不能同时被3和7整除

初始代码

num = 1
count = 0
# 循环条件
 
	# 循环体，设置条件
	# 补全代码
		
	num = num + 1
print(count)
hosted with ❤ by GitHub

代码提交区

# coding:utf-8
 
num = 1
count = 0
# 循环条件
while num < 100:
    # 循环体，设置条件
    # 补全代码
    if num % 3 == 0 or num % 7 == 0:
        count += 1
    if num % 21 == 0:
        count -= 1
    num = num + 1
print(count)
 
hosted with ❤ by GitHub

# coding:utf-8
 
num = 1
count = 0
# 循环条件
while num < 100:
    # 循环体，设置条件
    # 补全代码
    if num % 3 == 0 or num % 7 == 0:
        if num % 21 != 0:
            count += 1
    num = num + 1
print(count)
hosted with ❤ by GitHub

	num = 1
	count = 0
	# 循环条件

	# 循环体，设置条件
	# 补全代码

	num = num + 1
	print(count)

	# coding:utf-8

	num = 1
	count = 0
	# 循环条件
	while num < 100:
	# 循环体，设置条件
	# 补全代码
	if num % 3 == 0 or num % 7 == 0:
	count += 1
	if num % 21 == 0:
	count -= 1
	num = num + 1
	print(count)