Question

This is for an entry level C++ course. Having trouble understanding what the question is really asking. Thank a ton to whoever can help.

Given an analog clock (displayed in a circle) with minutes on the big hand, and hour on the small hand. Take in as input a number between 0 and 11 (including zero and 11). Then the second number determines how many minutes have passed. Print out where the big hand will be after that many minutes going clockwise, then print out where the big hand will be after that many minutes going counter-clockwise. Ex: If the input is 2 5, the output is: If you start at 2 minutes and add 5 (clockwise), you will be at minute 7. To go counter clockwise, we have to realize that we are already 2 ahead of 12 (so 14). Now we subtract the number of minutes (assuming that does not go over 12) and we have where it lands Hint: Remember that the modulus operator gives the remainder after division so 14 mod 12 be determined with 14 divided by 12 is 1 with a remainder of 2. Don't forget your parenthesis for your order of operations!

Answer 1

See in this problem statement.. they didnt took actual minutes when they mention minutes. they took the numbers didplay on clock.. think it as this way..

you have numbers 1 to 12..

1 2 3 4 5 6 7 8 9 10 11 12

now take the input suppose its 5 2

if we go clock wise we end up at 7

assume that clock wise means in above list of numbers we have to go left to right.

or else you can just think it as this way... 5+2 = 7

if we go counter clock wise we end up at 3

assume that counter clock wise means in above list of numbers we have to go right to left.

or else you can just think it as this way... 5-2 = 3

let see example 2:

now if we get the input like 11 2

clock wise : 11+2 =13

see whenever our sum is over 12 . then you just have to do sum-12 . which is 13-12 = 1

so our out put will be 1

and in counter clock wise : 11-2 = 9

example 3:

lets take input 2 5

clock wise : 2+5 = 7

counter clock wise 2-5 = -3

see when our difference is less than 0 then you have to subratct it from 12(just take the number not sign)

12 - 3 = 9

sumup:

clock wise.. add the both inputs: and if the sum is greater than 12 then subtract 12 from it...( sum -12 )

counter clock wise.. subtract the both inputs .. and if the difference is less than 0 then subtract it from 12....( 12 - diff )