用户输入nmber时确定大量数字

Question

I am try to determine if a number is abundant when a user inputs the number. But I am so lost to why it isn't working. Not really good at coding since i just started. Any help would be appreciated. This is what I have so far :

def is_abundant():
    n = raw_input("Enter a number: ")
    max_divisor = int(n / 2) + 1
    sum = 0
    for x in range(1, max_divisor):
        if n % x == 0:
            sum += x
            print ("your number isn't Abundant")
is_abundant()

It isn't giving me anything when I enter it. Please help!

Answer 1

Just looked up the definition of an abundant number:

In number theory, an abundant number or excessive number is a number for which the sum of its proper divisors is greater than the number itself. The integer 12 is the first abundant number. Its proper divisors are 1, 2, 3, 4 and 6 for a total of 16. — Wikipedia

Came up with this:

def isAbundant(n):
    factors = filter(lambda j: n % j == 0, range(1, n/2 + 1));
    return sum(factors) > n;

Or equally, if you like one-liners, this:

isAbdn = lambda n: sum(filter(lambda j: n % j == 0, range(1, n/2 +1))) > n;

Now, if you insist on using raw_input :

def isInputAbundant():
    n = int(raw_input("Enter a number: "));
    if isAbundant(n): print "YEAH! It's abundant.";
    else: print "Sorry! It's not abundant.";

Hope this helps.

Answer 2

A couple of things. First, you probably want to turn your input from a string into an integer as soon as possible, by replacing:

n = raw_input("Enter a number: ")

with:

import sys
:
try:
    n = int(raw_input("Enter a number: "))
except ValueError:
    print "Invalid number"
    sys.exit(1)

---

Second, you can't tell a number is abundant until after you've summed all its integral factors. So your code will need to sum them first, then check:

for x in range(1, max_divisor):
    if n % x == 0:
        sum += x
if sum > n:
    print ("Your number is abundant")
else:
    print ("Your number isn't abundant")

---

So, in toto:

import sys

def is_abundant():
    try:
        n = int(raw_input("Enter a number: "))
    except ValueError:
        print "Invalid number"
        sys.exit(1)
    max_divisor = int(n / 2) + 1
    sum = 0
    for x in range(1, max_divisor):
        if n % x == 0:
            sum += x
    if sum > n:
        print ("Your number is abundant")
    else:
        print ("Your number isn't abundant")

is_abundant()

---

One other thing to keep in mind though this is more of a stylistic thing. Many people would consider the function is_abundant() as one taking an integer and returning a Boolean value, rather than one also responsible for the input of the number and output of the results.

You may want to keep that in mind in future, in order to "properly" structure/name your code so as to make more sense, such as:

import sys

def is_abundant(num):
    max_divisor = int(num / 2) + 1
    sum = 0
    for x in range(1, max_divisor):
        if num % x == 0:
            sum += x
    return sum > num

try:
    n = int(raw_input("Enter a number: "))
except ValueError:
    print "Invalid number"
    sys.exit(1)

if is_abundant(n):
    print ("Your number is abundant")
else:
    print ("Your number isn't abundant")

Answer 3

“ n”变量是字符串，您应该在第3行和第6行将其转换为整数:::> int（n）祝您好运