Python 3.x python返回语句的主要问题

函数，该函数用于查找x或y的倍数为N（包括N）的所有数字之和

我试着把返回语句放在不同的缩进处，但还是没有希望

def multipleSum(N,x,y):
    z=0
    a=0
    for i in range(0,N+1):
        if i%x==0:
            z=z+i   
        elif i%y==0:
            a=a+i
        else:
            s=z+a
    return s

测试用例：

def multipleSum(N,x,y):
    total=0
    for i in range(N+1): # No need to mention 0(zero), as range starts from '0'
        if i % x == 0 or i % y == 0: # No need to use if and elif, better to use 'or' instead
            total+=i
    return total

def multipleSum(N,x,y):
    return sum(map(lambda v: v if v%x==0 or v%y==0 else 0, range(N+1)))

>>> multipleSum(10, 2, 3)
42

倍数（10,2,3）==42

预期结果为42，实际结果为15

代码：

def multipleSum(N,x,y):
    total=0
    for i in range(N+1): # No need to mention 0(zero), as range starts from '0'
        if i % x == 0 or i % y == 0: # No need to use if and elif, better to use 'or' instead
            total+=i
    return total

def multipleSum(N,x,y):
    return sum(map(lambda v: v if v%x==0 or v%y==0 else 0, range(N+1)))

>>> multipleSum(10, 2, 3)
42

还有一种方法：

def multipleSum(N,x,y):
    total=0
    for i in range(N+1): # No need to mention 0(zero), as range starts from '0'
        if i % x == 0 or i % y == 0: # No need to use if and elif, better to use 'or' instead
            total+=i
    return total

def multipleSum(N,x,y):
    return sum(map(lambda v: v if v%x==0 or v%y==0 else 0, range(N+1)))

>>> multipleSum(10, 2, 3)
42

---

输出：

def multipleSum(N,x,y):
    total=0
    for i in range(N+1): # No need to mention 0(zero), as range starts from '0'
        if i % x == 0 or i % y == 0: # No need to use if and elif, better to use 'or' instead
            total+=i
    return total

def multipleSum(N,x,y):
    return sum(map(lambda v: v if v%x==0 or v%y==0 else 0, range(N+1)))

>>> multipleSum(10, 2, 3)
42