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

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 21 1 2 2 4

output

4

input

3 11 1 1

output

1

input

10 31 2 6 2 3 6 9 18 3 9

output

6

题意

     找出 给定序列中 三个数 使得 成等比数列

题解：

      枚举中值，map暴力就可以

//提议：在一个串中找三个等比数字。//从中间开始枚举 #include<iostream>#include<cstdio>#include<cstring>#include<map>using namespace std;int main(){int n,m,i,j;long long ans;while(scanf("%d%d",&n,&m)!=EOF){ans=0;map<long long,long long>a;map<long long,long long>pre;map<long long,long long>count;for(i=1;i<=n;i++) cin>>a[i];for(i=1;i<=n;i++) count[a[i]]++;for(i=1;i<=n;i++){count[a[i]]--;if(a[i]%m==0 && count[a[i]*m]) ans+=pre[a[i]/m]*count[a[i]*m];pre[a[i]]++;}printf("%lld\n",ans);}return 0;}

DP思想

#include <bits/stdc++.h>using namespace std;int n;long long k;long long tab[1000007];long long wyn;map <long long,long long> mapa1;map <long long,long long> mapa2;int main(){    scanf("%d%lld", &n, &k);    for (int i=1; i<=n; i++)    {        scanf("%lld", &tab[i]);        if (!(tab[i]%(k*k)))        wyn+=mapa2[tab[i]/k];        if (!(tab[i]%k))        mapa2[tab[i]]+=mapa1[tab[i]/k];        mapa1[tab[i]]++;    }    printf("%lld\n", wyn);    return 0;}