题目链接：http://codeforces.com/problemset/problem/567/C
C. 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 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.
题意：
给定一个数列，要求所有的子序列（顺序一定），是长度为3的等比数列(公比为K)的个数。
PS:
map！
代码如下：
#include <cstdio>
#include <cstring>
#include <map>
#include <iostream>
using namespace std;
typedef long long LL;
map<int, LL> s1, s2;
int main()
{
LL n, k;
LL ans, x;
cin >> n >> k;
ans = 0;
for(int i = 0; i < n; i++)
{
scanf("%I64d",&x);
if(x%k == 0)
{
LL t1 = s2[x/k];
ans += t1;
LL t2 = s1[x/k];
s2[x] += t2;
}
s1[x]++;
}
cout << ans << endl;
return 0;
}
/*
10 3
1 2 6 2 3 6 9 18 3 9
*/
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