CF Geometric Progression （DP和map）

题目链接

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·k^r - 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, ..., a_n ( - 10⁹ ≤ a_i ≤ 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.

输入：第一行给你两个数n和k，代表有n个数，k代表公比。

第二行有n个数。

输出：叫你选出三个数 ，要求这三个数成等比数列，且公比为k。

下面附上上决大神的做法，大神博客链接

#include<stdio.h>
#include<string.h>
#include<map>
#include<algorithm>
#define ll __int64
using namespace std;
ll dp[200002],a[200002];
int main()
{
    ll n,k;
    scanf("%I64d%I64d",&n,&k);
    map<ll,ll> m;
    for(ll i=1;i<=n;i++)
    {
        scanf("%I64d",&a[i]);
        if(a[i]%k==0) dp[i]=m[a[i]/k];//记录a[i]/k的数量
        else dp[i]=0;
        m[a[i]]++;
    }
    m.clear();
    ll ans=0;
    for(ll i=1;i<=n;i++)
    {
        if(a[i]%k==0) ans+=m[a[i]/k];
        m[a[i]]+=dp[i];
    }
    printf("%I64d\n",ans);
    return 0;
}

加入选择了i,j,k三个位置的数，dp[i]是等于0的，dp[j]不等于0，于是可以通过dp[j]算出dp[k]，有一个k满足条件就会加上dp[j]。