I am trying hard to do better but I am failing please can any one help me in solving problems after contests so that I can also grow better.
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I am trying hard to do better but I am failing please can any one help me in solving problems after contests so that I can also grow better.
lets say we have an array a of length n.
we are given q queries with indices l,r.
we have to locate the position of minimum or maximum value within that range....of l,r.
the constraints are
.........1<=n<=10^5 .........1<=l<=r<=n .........1<=ai<=10^9 .........1<=q<=10^5
Hello everyone,
problem link:- https://mirror.codeforces.com/problemset/problem/1979/C
I was solving the problem and got stuck,
and in the editorial it was mentioned their that we have to use LCM I am not getting how the LCM is intuted.
what is the role of LCM...... is it observational or there is some proof for the use of LCM
please clarify about it..
problem link:- https://mirror.codeforces.com/problemset/problem/1881/D
here we have to choose two element in the array let ai,aj and choose a divisor of ai let x and replace ai=ai/x ans aj=aj*x;
after some operation we have to make all the element equal in that array
now let consider an array of two elements a1,a2 now let ai=a1 and aj=a2 x the divisor a1
==>let say a1=a1/x and a2=a2*x; ==>now a1/x=a2*x now multiply these two we get (a1/x)*(a2*x)=a1*a2;
we can generalise after all the operation when we multiply all the elements we get a1*a2*a3*a4...... so we can generalise if the nth root of a1*a2*a3....*an is a whole number then return yes otherwise return no; I don't no how to implement this idea..... also please clarify that the idea is right or wrong.
hello, I got stuck in this problem can any one help me in figuring out what is the mistake. submission link:- https://mirror.codeforces.com/contest/1790/submission/296512970
can anyone help me in figuring out the problem I am not getting the editorial and how it is working. please some one explain me. the problem link:- https://mirror.codeforces.com/problemset/problem/1979/C
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