GUESS - Editorial

editorial
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21

Yeah…it was just because of not using long long int. Thanks :slight_smile:

22

same solution when i am executing it is showing AC check this link CodeChef: Practical coding for everyone

23

you have used too many loops i think you will get TLE …try to use pen and paper … just a little thinking and boom you will get the answer or try to follow editorial note…

24

i think there is little problem in logic
try to replace “num=(2nm) + m -n;” with “num=(2nm) + m + n;” in your code and then try again.

25

how could You say answer will be (floor(N / 2) * ceiling(N / 2)) / (N * M) ???
for n=10 and m=10 answer is 1/4 while correct answer is 1/2.

26

you haven’t use return 0 cause your return type is int.
put return 0 and try.

27

and this is O(1)

28

willnt case 1 and case 4 interchange …??

29

Very Nice Explanation, Thank you.

30

#include<bits/stdc++.h>

#include

#include<string.h>

#define ll long long

using namespace std;

int main()

{

int t=1;

cin>>t;

while(t--)

{

   ll  int n,m;

   cin>>n>>m;

   ll int a=n/2;

   ll int b=(n+1)/2;

   ll int gcd=__gcd(a*b,n*m);

   cout<<(a*b)/gcd<<"/"<<(m*n)/gcd<<endl;

}

return 0;

}

what is error in this

31

You have not taken all possible cases into account. The sum of the 2 numbers chosen by Alice and Bob can be odd only in 2 cases :

  1. When Alice chooses an odd number and Bob chooses an even number
  2. When Alice chooses an even number and Bob chooses an odd number
    Since Alice has total n choices and Bob has total m choices, the answer will be -
    P(OA)*P(EB) + P(EA)*P(EB) where,
    P(OA) = Probability of Alice choosing an Odd Number = (n+1)/2
    P(EB) = Probability of Bob choosing an Even Number = (m/2)
    P(EA) = Probability of Alice choosing an Even Number = (n/2)
    P(OB) = Probability of Bob choosing an Odd Number = (m+1)/2
    You can also refer to your modified solution here
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32

Thank you very very much.
I clearly understood from your very nice explanation.

1 Like