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

C. Geometric Progression
Time Limit: 20 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

要求你从一个长度为n的序列中，选出3个数，要求l1<l2<l3，使得构成一个等比为k的等比数列，问你有多少个

题解：

开两个map维护一下，维护这个数前面的每类数有多少个，维护这个数后面的每类数有多少个

然后ans+=H[a[i]/k]*H[a[i]*k]就好了

代码

#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define test freopen("test.txt","r",stdin)
#define maxn 2000001
#define mod 1000000007
#define eps 1e-9
const int inf=0x3f3f3f3f;
const ll infll = 0x3f3f3f3f3f3f3f3fLL;
inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
//**************************************************************************************

map<ll,ll> H1;
map<ll,ll> H2;
ll a[maxn];
int main()
{
    int n=read();
    ll k=read();
    for(int i=0;i<n;i++)
    {
        a[i]=read();
        H2[a[i]]++;
    }
    ll ans=0;
    for(int i=0;i<n;i++)
    {
        H2[a[i]]--;
        if(a[i]%k==0)
            ans+=H1[a[i]/k]*H2[a[i]*k];
        H1[a[i]]++;
    }
    cout<<ans<<endl;
}