codeforces(567C)-- Geometric Progression

Geometric Progression

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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

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：给定一个序列的定义，序列包含三个数，三个数之间为等比数列，等比为k，现在给出n个数，和k的值，要求有几个这样的序列（不需要连续）。

问n个数中包含多少三个数的序列，等比为k，那么我们可以枚举从1到n，认为ai是三个数的中间那个数，那么我们可以计算出前面的一个数ai/k和后面的一个数ai*k（如果ai%k有值，那么ai就不可能是第二个数）。

现在我们需要统计的就是在ai之前，和ai之后每个数出现的次数，

因为数-10^9<=10^9，所以对数进行离散化，重新给出序列。

然后开一个数组记录所有的数出现的次数，和在ai之前的数出现的次数。计算对于以ai为第二个数来说可以组成多少序列。

最后统计总的和。

#include <cstdio>#include <cstring>#include <vector>#include <queue>#include <cmath>#include <algorithm>using namespace std ;#define LL __int64#define INF 0x3f3f3f3f#define PI acos(-1.0)LL a[200010] , b[200010] , id[200010] ;LL l[200010] , r[200010] ;LL n , m , k ;LL search1(LL x) {    int low = 0 , mid , high = m-1 ;    while( low <= high ) {        mid = (low + high) / 2 ;        if( b[mid] == x ) return mid ;        else if( b[mid] < x )            low = mid+1 ;        else            high = mid-1 ;    }    return -1 ;}int main() {    LL i , j , x , y ;    LL ans = 0 ;    scanf("%I64d %I64d", &n, &k) ;    for(i = 0 ; i < n ; i++) {        scanf("%I64d", &a[i]) ;        b[i] = a[i] ;    }    sort(b,b+n) ;    m = unique(b,b+n) - b ;    memset(l,0,sizeof(l)) ;    memset(r,0,sizeof(r)) ;    for(i = 0 ; i < n ; i++) {        id[i] = search1(a[i]) ;        r[ id[i] ]++ ;    }    for(i = 0 ; i < n ; i++) {        r[ id[i] ]-- ;        if( a[i]%k == 0 ) {            x = search1(a[i]/k) ;            y = search1(a[i]*k) ;            if( x != -1 && y != -1 ) {                ans += l[x]*r[y] ;            }        }        l[ id[i] ]++ ;    }    printf("%I64d\n", ans) ;    return 0 ;}