c++ - I have solved the 1st of three phases of a panhellenic competition (it is now over) -


I am the first solution (now that it's finished) in three phases of a panhellenic competition, but I'm interested in knowing that Whether there is a simple complexity algorithm there or not 7 9

5 2

3 6

2 3

1 7

>

6 2

4 6

1 5

3 

  int main () {ifstream in ("Domes.in"); Offscreen Out ("Dome."); Int oreo, Zed; In & gt; & Gt; Orio; In & gt; & Gt; Z; Ent Dome [Oriio]; For (int i = 0; i or lt; orio; i ++) {domes [i] = 0;} int k; For (Int i = 0; I <2 * z; i ++) {In & gt; & Gt; Of; Domes [k-1] ++; } Int c = 0; For (int i = 0; i

This is about some places (represented by numbers), the number of first two numbers (orio) and the number of matches (z). Places are "connected" in some way (meaningless). You should know how many places have fewer than 2 connections and the output (in this case C) is the number of places that have fewer than 2 connections. Kashmir has temporarily added each number and "see Plasplus" Gone "is used for time. If this is seen then it means that it is connected to some other place. I do not think this is a simple solution, but some of my programs require less time to run and bother me


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