I wrote this post to help my friend , Iman Movahhedi , complete this one.

Round #40 was my first contest here in Codeforces and I feel I fell in love with CF just after that.

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Many languages have built-in reverse() function for strings. we can reverse one of the strings and check if it's equal to the other one , or we can check it manually. I prefer the second.

B-Martian Dollar: (C# code)

Since number of days is very small (2 × 10³) we can just iterate over all possibilities of buy-day and sell-day. This will take θ (n²) time which is OK.

C-Email Address: (C++ code)

The first letter of the input string can not be part of an "at" or a "dot" so we start from the second character. Greedily put "." wherever you reached "dot" and put "@" the first time you reached "at". This will take θ(n) time , where n is the length of input.

For each cell of the table store k + 1 values. Where ith value is the maximum number of peas the pawn can take while he is at that cell and this number mod k + 1 is i.

This makes a O(n² × m × k) dynamic programming which fits perfectly in the time.

E-3-cycles: (C++ code)

The road map with most edges is a complete bipartite graph with equal number of vertices on each side. (Prove it by yourself :D ). We can make such a graph by putting the first vertices on one side and the other on the other side.For sure , number of edges is .