Codeforces Round #Pi (Div. 2) C. Geometric Progression map

C. Geometric Progression

Time Limit: 2 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/567/problem/C

Description

Polycarp loves geometric progressions very much. Since he was only three years old, he loves only the progressions of length three. He also has a favorite integer k and a sequence a, consisting of n integers.

He wants to know how many subsequences of length three can be selected from a, so that they form a geometric progression with common ratio k.

A subsequence of length three is a combination of three such indexes i1, i2, i3, that 1 ≤ i1 < i2 < i3 ≤ n. That is, a subsequence of length three are such groups of three elements that are not necessarily consecutive in the sequence, but their indexes are strictly increasing.

A geometric progression with common ratio k is a sequence of numbers of the form b·k0, b·k1, ..., b·kr - 1.

Polycarp is only three years old, so he can not calculate this number himself. Help him to do it.

Input

The first line of the input contains two integers, n and k (1 ≤ n, k ≤ 2·105), showing how many numbers Polycarp's sequence has and his favorite number.

The second line contains n integers a1, a2, ..., an ( - 109 ≤ ai ≤ 109) — elements of the sequence.

Output

Output a single number — the number of ways to choose a subsequence of length three, such that it forms a geometric progression with a common ratio k.

Sample Input

5 2
1 1 2 2 4

Sample Output

4

HINT

题意

找出 给定序列中 三个数 使得 成等比数列

题解：

枚举中值，map暴力就可以

代码

 #include <cstdio>
 #include <cmath>
 #include <cstring>
 #include <ctime>
 #include <iostream>
 #include <algorithm>
 #include <set>
 #include <vector>
 #include <queue>
 #include <typeinfo>
 #include <map>
 #include <stack>
 typedef __int64 ll;
 #define inf 1000000000000
 using namespace std;
 inline ll read()
 {
     ll x=,f=;
     char ch=getchar();
     while(ch<''||ch>'')
     {
         if(ch=='-')f=-;
         ch=getchar();
     }
     while(ch>=''&&ch<='')
     {
         x=x*+ch-'';
         ch=getchar();
     }
     return x*f;
 }

 //**************************************************************************************
 map<ll ,ll >mp;
 map<ll ,ll >mp2;
 ll a[];
 int main()
 {
     ll n,k;
     scanf("%I64d%I64d",&n,&k);
     ll sum=;
     ll f1,f2;
     for(int i=; i<=n; i++)
     {
         scanf("%I64d",&a[i]);
         if(mp.count(a[i])==)
             mp[a[i]]=;
         else mp[a[i]]++;
     }
      if(mp2.count(a[n])==)
             mp2[a[n]]=;
         else mp2[a[n]]++;
     for(int i=n-; i>; i--)
     {
         if(mp2.count(a[i])==)
             mp2[a[i]]=;
         else mp2[a[i]]++;
         ll t;
         if(a[i]==a[i]*k)t=mp2[a[i]*k]-;
         else t=mp2[a[i]*k];
       if(a[i]%k==)
         sum+=((mp[a[i]/k]-mp2[a[i]/k])*(t));
     }
     cout<<sum<<endl;
     return ;
 }