Codeforces Round #Pi (Div. 2) C

Test instructions: give you a sequence, and K, choose 3 numbers, subscript strictly increment, meet the increment of geometric series, equal to K

Idea: Count the number of all numbers first, enumerate geometric series of the middle number A, calculate the number of a*k after a, the number of a/k before a, multiply

(Playing a game of brain residue, but also think of what, love Disabled)

#include <iostream>#include<cstdio>#include<cstring>#include<string>#include<cmath>#include<algorithm>#include<map>#include<Set>#include<vector>#include<queue>#include<stack>//#include <bits/std c++.h>using namespaceStd;typedefLong Longll;typedef unsignedLong LongULL;ConstLL MOD = 1e7 +7;ConstLL MAXN = 1e5 +131; Map<LL,LL>Num1;map<LL,LL>Num2; LL LINE[MAXN<<1]; LL n,k;intMain () { while(Cin >> N >>K) {LL Sum=0; Num1.clear ();        Num2.clear ();  for(inti =1; I <= N; ++i) Cin>> Line[i], Num1[line[i]] + +;  for(inti = N; I >=1; --i) {Num2[line[i]]++; if(i = =1|| i = = N)Continue; if(Num2.count (line[i] * K) &&! (Line[i]%K)) {LL l,r; if(Line[i] = = line[i]* K) R = Num2[line[i]*k]-1; ElseR = num2[line[i]*J]; L= Num1[line[i]/K]-num2[line[i]/K]; Sum+ = (L *R); }} cout<< Sum <<Endl; }}

Codeforces Round #Pi (Div. 2) C