CodeForces 567C Geometric Progression【思维+map】

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探讨了在给定整数序列中寻找长度为三且公比为特定整数k的几何数列子序列的方法。介绍了如何通过前后扫描并利用哈希映射技术来高效解决该问题。

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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.

Examples

Input

5 2
1 1 2 2 4

Output

Input

3 1
1 1 1

Output

Input

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

Output

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.

题目大意：

一共给你N个数，再给你一个等比公项K.让你找三个数，使得其满足递增的同时，要满足其是一个以K为等比公项的等比数列。

问你能够找到几个符合条件的序列。

思路：

考虑第i个数，我们查询其前边有多少个a【i】/m记做sum1，查询其后边有多少个a【i】*m记做sum2.对应当前第i个数作为中间的数的时候，能够找到的序列个数为：sum1*sum2；

那么我们此时的任务就是从后向前map计数一波，设定数组b【i】，b【i】表示的是第i个数后边有多少个a【i】*m。

接下来再从前向后map计数一波，对应第i个数的贡献度就是：对应当前第i个数前边有多少个（a【i】/m）*b【i】；

过程累加即可。

时间复杂度O（n）；

不过我的代码好搓啊。（我做的是正数搞一波，再负数搞一波啊...）

Ac代码：

#include<stdio.h>
#include<string.h>
#include<map>
using namespace std;
#define ll __int64
ll c[6000000];
ll a[6000000];
ll b[6000000];
int m;
ll Slove(int n)
{
    map<ll ,ll >s;
    ll output=0;
    for(int i=n-1; i>=0; i--)
    {
        b[i]=s[a[i]*m];
        s[a[i]]++;
    }
    s.clear();
    for(int i=0; i<n; i++)
    {
        if(a[i]%m==0)
        {
            output+=s[a[i]/m]*b[i];
        }
        s[a[i]]++;
    }
    return output;
}
int main()
{
    int n;
    while(~scanf("%d%d",&n,&m))
    {
        ll output=0;
        for(int i=0; i<n; i++)scanf("%I64d",&c[i]);
        int cont=0;
        memset(b,0,sizeof(b));
        memset(a,0,sizeof(a));
        for(int i=0; i<n; i++)
        {
            if(c[i]>=0)
            {
                a[cont++]=c[i];
            }
        }
        output+=Slove(cont);
        memset(b,0,sizeof(b));
        memset(a,0,sizeof(a));
        cont=0;
        for(int i=0;i<n;i++)
        {
            if(c[i]<0)
            {
                a[cont++]=-c[i];
            }
        }
        output+=Slove(cont);
        printf("%I64d\n",output);
    }
}