cf 567 C. Geometric Progression

C. Geometric Progression

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 test(s)

input

5 2
1 1 2 2 4

output

4

input

3 1
1 1 1

output

1

input

10 3
1 2 6 2 3 6 9 18 3 9

output

6

Note

In the first sample test the answer is four, as any of the two 1s can be chosen as the first element, the second element can be any of the 2s, and the third element of the subsequence must be equal to 4.

给定一个数列,从中按照顺序挑出3个数,使这三个数构成以k为公比的等比数列.问一共有多少种方法.

STL大法好.

所以说大概map的一个作用就是当数组的下标太大?

我还是太年轻了...

做法是枚举中间的数,具体见代码注释.

/*************************************************************************
    > File Name: code/cf/#314/CC.cpp
    > Author: 111qqz
    > Email: rkz2013@126.com 
    > Created Time: 2015年08月06日 星期四 15时00分46秒
 ************************************************************************/

#include<iostream>
#include<iomanip>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<map>
#include<set>
#include<queue>
#include<vector>
#include<stack>
#define y0 abc111qqz
#define y1 hust111qqz
#define yn hez111qqz
#define j1 cute111qqz
#define tm crazy111qqz
#define lr dying111qqz
using namespace std;
#define REP(i, n) for (int i=0;i<int(n);++i)  
typedef long long LL;
typedef unsigned long long ULL;
const int inf = 0x7fffffff;
LL ans  = 0;
int n,k;
map<LL,LL>m[2];
int main()
{
    cin>>n>>k;
    for ( int i  = 1 ; i <= n ; i ++ )
    {
    int a;
    scanf("%d",&a);
    if (a%k==0)                //枚举中间的数
    {
        ans = ans + m[1][a/k];
        m[1][a] = m[1][a]+m[0][a/k];//记录的是以a为中间数的序列的数
                        // 因为有多个a,而且对于第i+1个a,第i个a前面的a%k一样可以对序列有贡献,所以是累加.
    }
    m[0][a]++; //记录a的个数
//    cout<<i<<" "<<ans<<endl;
    }
    cout<<ans<<endl;
  
    return 0;
}