Question

我在python中编写了这个程序：

num=51

if (num % 3 == 1):
    if (num%4 == 2):
        if (num%5 == 3):
            if (num%6 ==4):
                print num
            else:
                print "not right number, try again - error 1"
        else:
            print "not right number, try again - error 2"
    else:
        print "not right number, try again - error 3"
else: 
    print "not right number, try again - error 4"

哪种方法效果很好，除非我真的不想手动迭代num，直到我得到我想要的答案（我写这个来解决我想要解决的数学问题 - 这不是虽然做作业。如果有人可以详细说明要更改所有else语句以添加一个语句，将num递增1并返回到for循环的开头，那就太棒了。

谢谢！

Answer 1

您可以使用break语句终止循环

num=1

while True:
    if (num % 3 == 1):
        if (num%4 == 2):
            if (num%5 == 3):
                if (num%6 ==4):
                    print num
                    break
                else:
                    print "not right number, try again - error 1"
            else:
                print "not right number, try again - error 2"
        else:
            print "not right number, try again - error 3"
    else: 
        print "not right number, try again - error 4"
    num += 1

Answer 2

这个怎么样？

def f(n):
    for (a, b) in [(3, 1), (4, 2), (5, 3), (6, 4)]:
        if(num % a) != b:
            return (False, b)

    return (True, n)

for num in range(100):
    print '-' * 80
    v = f(num)

    if not v[0]:
        print "{0} is not the right number, try again - error {1}".format(num, v[1])
    else:
        print "The first number found is --> {0}".format(v[1])
        break


N = 1000000
numbers = [num for num in range(N) if f(num)[0]]
print "There are {0} numbers satisfying the condition below {1}".format(
    len(numbers), N)

Answer 3

我认为代码的结构是错误的，你可以试试这个：

num=51

def test(num):
    # keep all the tests in a list
    # same as tests = [num % 3 == 1, num % 4 == 2, ...]
    tests = [num % x == y for x,y in zip(range(3,7), range(1,5))]

    if all(tests):    # if all the tests are True
        return False  # this while exit the loop 
    else:
        # message to be formatted
        msg = "{n} is not the right number, try again - error {err}"

        # I tried to keep your error numbers
        err = len(tests) - tests.count(False) + 1

        # format the message with the number and the error
        print msg.format(n=num, err=err)

        return True

while test(num):
    num += 1  # increment the number

print num, "is the right number"

while循环测试每次迭代时的数字，当数字正确时它将退出

Answer 4

你可以把支票放在一个函数中来清理它：

def good_number(num):
    if num % 3 == 1:
        if num % 4 == 2:
            if num % 5 == 3:
                if num % 6 == 4:
                    return True
    # Put your elses/prints here

# Replace 100 with your max
for num in range(100): 
    if good_number(num):
        print('{} is a good number'.format(num))

# Or use a while loop:
num = 0
while not good_num(num):
    num += 1

print('{} is a good number'.format(num))

if语句出现时迭代变量false

4 个答案: