python 在一个范围内,寻找另一个数字的所有整数倍数,并计算一共有多少个倍数

python 在一个范围内,寻找另一个数字的所有整数倍数,并计算一共有多少个倍数
这个是问题,真心不会做.我用的是python 2.7.要用 for loop
1) 建立程序 count_multiples() which takes 三个非负整数:base,start and stop,prints each integer multiple of base which
occurs between start and stop (including start but not including stop) on a separate line,
and returns the number of multiples found.假如 base = 3,那在start = 9 和stop = 15之间就有2个整倍数,9 和 12,但不包括15.the easiest way to test whether one number is an integer multiple of another is with the % operator.
\x05\x05\x05\x05\x05\x05
\x05\x05\x05\x05\x05
\x05\x05\x05\x05
\x05\x05\x05
\x05\x05
\x05\x05\x05
\x05\x05\x05\x05
\x05\x05\x05\x05\x05
\x05\x05\x05\x05\x05\x05
\x05\x05\x05\x05\x05\x05\x05
2）.Write a function user_input_multiples() which takes a single integer input base.This
function will get start and stop values from the user with two calls to raw_input(),call
count_multiples() to determine the number of integer multiples of base between the user
specified start and stop,and then ask again for new start and stop values.The function will
continue asking for new start and stop values until at least one of the following cases occurs:
\x05\x05\x05\x05\x05\x05\x05
\x05\x05\x05\x05\x05\x05
\x05\x05\x05\x05\x05\x05\x05\x05
\x05\x05\x05\x05\x05\x05\x05\x05\x05
 The user enters a negative value for start or stop.
\x05\x05\x05\x05\x05\x05\x05\x05
\x05\x05\x05\x05\x05\x05\x05\x05
\x05\x05\x05\x05\x05\x05\x05\x05\x05
 The user enters a value for stop which is less than the value for start.
\x05\x05\x05\x05\x05\x05\x05\x05
\x05\x05\x05\x05\x05\x05\x05\x05
\x05\x05\x05\x05\x05\x05\x05\x05\x05
 The function count_multiples() returns zero (eg:there were no multiples between start and stop).
Once the function stops asking for input,it will return the total number of multiples found (the total
over all calls to count_multiples()).Hint:You will
want to use a while loop for this function.
英语有点多,看着有点烦,请见谅.第一部分我已经尽量翻译最主要的举例了.
如果没有时间,给我一个详细的思路或者方向也行.:)

def count_multiples(base, start, stop):
    result=[]
    for item in range(start,stop):
        if item % base ==0:
            result.append(item)