CF Geometric Progression(DP 및 맵)

제목 링크
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.
입력: 첫 번째 줄은 n과 k를 두 개 세어 줍니다. n개의 수가 있고 k는 공비를 나타냅니다.
두 번째 줄에는 n개의 수가 있다.
출력: 세 개의 수를 선택하라고 합니다. 이 세 개의 수는 등비 수열이고 공비는 k입니다.
다음은 상결대신의 방법, 대신블로그 링크를 동봉합니다.

#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];//  a[i]/k   
        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
",ans);
    return 0;
}

에 i, j, k 세 자리를 선택한 수를 추가했다. dp[i]는 0이고 dp[j]는 0이 아니다. 그래서 dp[j]를 통해 dp[k]를 계산할 수 있다. k가 만족하면 dp[j]를 추가한다.