Question

我想实现一个函数，它接受三个参数（x，low，high）作为输入 - 所有整数，并查找x和{{之间low因子的整数数1}}。例如，50到100之间有多少个整数有4个因子？

我的代码如下：

high

当我尝试运行此操作时，我得到的结果不正确，例如

def n_factors(x, lower, upper):
""" Find how many integers have x (user-specified)
    factors from - lower to upper - (user-specified) """
    int_counter = 0 # integer counter
    div_count = 0 # divisor counter 
    for i in range(lower, upper+1):
        for j in range(2, i):
            if (i%j)==0:
                div_count += 1
        if (div_count == x):
            print i
            int_counter += 1
     return int_counter

这应列出10到20之间的四个素数功能有效，但没有。任何帮助非常感谢！

Answer 1

您需要在每次迭代结束时将div_count设置为零：

def n_factors(x, lower, upper):
    int_counter = 0 # integer counter
    div_count = 0 # divisor counter
    for i in range(lower, upper+1):
        for j in range(1, i + 1):
            if (i%j)==0:
                div_count += 1
        if div_count == x:
            print i
            int_counter += 1
        div_count = 0
    return int_counter

Answer 2

def n_factors(x, lower, upper):
    facs = {}
    for f in range(1, int(upper**0.5)+1):
        for n in range(lower+f-(lower%f), upper+1, f):
            if n not in facs: facs[n] = []
            facs[n].append(f)

    for n, numFacs in facs.items():
        if len(numFacs) != x: continue
        print(n)

Answer 3

def n_factors(x, lower, upper):
""" Find how many integers have x (user-specified)
    factors from - lower to upper - (user-specified) """
    int_counter = 0
    div_count = 0
    for i in range(lower, upper+1):
        for j in range(1, i+1):
            if (i%j)==0:
                div_count += 1
        if (div_count == x):
            print i
            int_counter += 1
        div_count = 0 
    return int_counter

Answer 4

您可以通过略高的编码轻松避免这样的错误。

如果您需要计算这种情况发生的次数＆＃34;，您不需要设置手动计数器而忘记重置它 - 您可以将该句子直接翻译成蟒蛇;这里的条件是＆＃34; j除i而没有余数＆＃34;：

for i in range(lower, upper+1):
    div_count = sum(i%j == 0 for j in range(1,i+1))

确定两个数之间具有一定数量因子的整数数

4 个答案: