code is showing error. what is error..help me to fix it..

Question

https://code.sololearn.com/cRK9J585hH37/?ref=app

Answer 1

__n I have already attached the fix code. Please check! :)

Mmm. 👇
https://code.sololearn.com/cpQUTl9WC35v/?ref=app

Answer 2

__n

You are taking three input but your code consider only two.
just because of loop it's just printing 2 times.
with the same value.

for example you entered :

2
1901
2000

but when it reaches the code:
it's becomes like

2
1901

And due to for loop :
it's printing

no
no

because the value is 1901.

How to fix ?

so you have to take input inside for loop.
please check this code. it will help you.

https://code.sololearn.com/cpQUTl9WC35v/?ref=app

Answer 3

__n Your scanf is not present in loop so.

It's ignoring your 2012.

You are providing value of N only one time.
But it's should be two times. if T is 2.
Or you can override value of N
Within loop.

Answer 4

You can't chain the condition like 1 <= T <= 10.

It will be read as (1 <= T) <= 10.
Saying T is 0. 1 <= T returns 0(false)
0<=10 returns 1(true).

You have to separate it with &&.
if(1<=T && T <=10)

Answer 5

__n Pleasure of mine. :)

Answer 6

CarrieForle
yes but if we takeT that is test cases as 2 then ???
take input
2
1900
2012

Answer 7

P∆WAN M∆URY∆ then how i do it??

Answer 8

P∆WAN M∆URY∆ thank u so much☺👍

Answer 9

yes,but there is also one problem that for year which is not a leap its showing it is leap year(Yes)..

Answer 10

A leap year is divisible by 4, not 2.

Answer 11

still its showing same

Answer 12

How do you input?
I input 1 2020 and it prints yes.
If I input 1 2019 it prints no.

Answer 13

P∆WAN M∆URY∆ i think there was problem in ur code...in conditions of if ..
i uploaded this.. now this is correct..
https://code.sololearn.com/cluz4AO80mKG/?ref=app

Answer 14

__n ... What was the problem ?

Answer 15

if(1<=T && 10>=T && 1<=N && 2500>=N)

if(N%4==0 && (N%100!=0 || N%400==0))
this conditions ...

Answer 16

Hi,

This may work....

https://code.sololearn.com/c0Aw5h3Hep3X/?ref=app