Subsequence product !!! NMNMX - NO MINIMUM NO MAXIMUM
@utk911 yes i am pretty sure....test cases for subtask 1 must be weak
Last edited on
@blackmamba
What is logic to reduce tle in no strings attached.
I have done following thing.
1. Change every element from a to z , then after changing checking how many elements are greater than it in succeeding array..
This make me to score 30 only...
What is logic to reduce tle in no strings attached.
I have done following thing.
1. Change every element from a to z , then after changing checking how many elements are greater than it in succeeding array..
This make me to score 30 only...
.@coder
Ur approach is correct to handle large nCr use dp..
I scored 100 using same...
Ur approach is correct to handle large nCr use dp..
I scored 100 using same...
@coder yup that was the tricky part.
try a little harder.
try printing different variables in your code to debug your code..
try a little harder.
try printing different variables in your code to debug your code..
@decode
for(d = 1; d < n-1; d++)
{
result = (result%m * power(arr[d], c[arr[d]])%m)%m;
}
printf("%lld\n",result);
this is somewhat how used the multiplication and get 100......
for(d = 1; d < n-1; d++)
{
result = (result%m * power(arr[d], c[arr[d]])%m)%m;
}
printf("%lld\n",result);
this is somewhat how used the multiplication and get 100......
see:
we know that
a^(m-n) = (a^m)/(a^n)
here 1/(a^n) can be written as inverse of a^n.
so , a^(m-n) = (a^m)*inverse( (a^n)).
hope this helps many of u.
we know that
a^(m-n) = (a^m)/(a^n)
here 1/(a^n) can be written as inverse of a^n.
so , a^(m-n) = (a^m)*inverse( (a^n)).
hope this helps many of u.
@chelpme because he ncr part goes in power!!! by euler totient (a^b)%m=(a^(b%(m-1)))%m...so that is wht take modulo with m-1 for nCr part
@all guyz dont contact me through PM....i have posted all the necessary details to solve NSA and NMNMX problem in the respective forums....go through them and i wont be of any help more than that