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

C. Geometric Progression

Time Limit: 2 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/567/problem/C

Description

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 Input

5 2
1 1 2 2 4

Sample Output

4

HINT

 

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

代码

 1 #include <cstdio>
 2 #include <cmath>
 3 #include <cstring>
 4 #include <ctime>
 5 #include <iostream>
 6 #include <algorithm>
 7 #include <set>
 8 #include <vector>
 9 #include <queue>
10 #include <typeinfo>
11 #include <map>
12 #include <stack>
13 typedef __int64 ll;
14 #define inf 1000000000000
15 using namespace std;
16 inline ll read()
17 {
18     ll x=0,f=1;
19     char ch=getchar();
20     while(ch<'0'||ch>'9')
21     {
22         if(ch=='-')f=-1;
23         ch=getchar();
24     }
25     while(ch>='0'&&ch<='9')
26     {
27         x=x*10+ch-'0';
28         ch=getchar();
29     }
30     return x*f;
31 }
32 
33 //**************************************************************************************
34 map<ll ,ll >mp;
35 map<ll ,ll >mp2;
36 ll a[200005];
37 int main()
38 {
39     ll n,k;
40     scanf("%I64d%I64d",&n,&k);
41     ll sum=0;
42     ll f1,f2;
43     for(int i=1; i<=n; i++)
44     {
45         scanf("%I64d",&a[i]);
46         if(mp.count(a[i])==0)
47             mp[a[i]]=1;
48         else mp[a[i]]++;
49     }
50      if(mp2.count(a[n])==0)
51             mp2[a[n]]=1;
52         else mp2[a[n]]++;
53     for(int i=n-1; i>1; i--)
54     {
55         if(mp2.count(a[i])==0)
56             mp2[a[i]]=1;
57         else mp2[a[i]]++;
58         ll t;
59         if(a[i]==a[i]*k)t=mp2[a[i]*k]-1;
60         else t=mp2[a[i]*k];
61       if(a[i]%k==0)
62         sum+=((mp[a[i]/k]-mp2[a[i]/k])*(t));
63     }
64     cout<<sum<<endl;
65     return 0;
66 }

 

posted @ 2015-08-06 03:31  meekyan  阅读(427)  评论(0)  编辑  收藏  举报