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Problems have been added to practice.
We will be be adding the editorial soon. Until then you can try joining the pieces by these short description and tags:

Utkarsh and LCM: Adhoc/cake walk/simple combinatorics

Shil and the Toy Factory:easy/ Binary Search (on time)

String Match: easy-med/ Z-Algorithm/ KMP Failure Function(String Automata)

Shil and Triplets: easy-med/ DP, Tree, Combinatorics

Utkarsh Lost in Cities: Med-hard/ Graph Theory, Maths, Matrix Exponentiation, Divide and conquer(for calculating GP of Matrix)
Complexity =O(N^3 * Log^2 K). Also O(N^4 * Log K) without Divide and Conquer by karanaggarwal

Shil and Suffix Palindromes:Hard/ Palindromic tree, Link 1, Link 2

Utkarsh and Long Roads:Hard/ Polynomial Exponentiation, Divide and Conquer.
Complexity =O(L*L*Log^2 K) by xorfire. Faster O(L*Log(L)*Log^2 K) with FFT but I thought it would have been an overkill for a short contest.

If you have different solutions you can discuss here. We would be very happy to know them.

One more way of solving Utkarsh and Long roads

let wt[i] indicate the number of occ of i

observation 1:

you can get stack size to be greater than 400 let us say x if you have atleast x-400 zeros so now let dp[i][j][k] -> using the first i values(1 to 400) , have the sum to be j and use k elements. Fill till dp[400][400][400]; can be done in O(400*400*400*log(400)) time..Runs within a sec in my vm.

Now when given a query what you need to do is add the number of ways such that size <= k that are created due to zeros let y=wt[0];

iterate i from 0 to min(k,400) -> indicates the stack size formed from elements other than 0

ans=ans+f(i)*dp[400][x][i];

let n=k here f(0) = nc0*(y^n)+ n-1c0*(y^(n-1)) +.... f(1) = nc1*(y^(n-1))+n-1c1*(y^(n-2)) +.... we get f(i)=((n+1)ci*(y^(n-i+1))-f(i-1))/(y-1);

Please add questions to practice

Please add editorials of contest ASAP.