如果if语句为false，则迭代变量

Question

I wrote this program in python:

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"

Which works well, except I really don't want to have to hand iterate num until I get the answer I want (I wrote this to solve a mathematics problem I wanted to solve - this is not homework, though). If anyone could detail to change all of the else statements to add a statement incrementing num by one and return to the beginning of the for loop, that'd be great.

Thanks!

Answer 1

You can use the break statement to terminate the loop

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

What about this one?

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

I think that the code's structure is wrong, you could try this instead:

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"

The while loop tests the number on each iteration and it will exit when the number is right

Answer 4

You could clean it up by putting your checks in a function:

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))