時間限制：C/C++ 1秒，其他語言2秒
空間限制：C/C++ 131072K，其他語言262144K
64bit IO Format: %lld
題目描述
Given an array A with length n a[1],a[2],…,a[n] where a[i] (1<=i<=n) is positive integer.
Count the number of pair (l,r) such that a[l],a[l+1],…,a[r] can be rearranged to form a geometric sequence.
Geometric sequence is an array where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. i.e. A = [1,2,4,8,16] is a geometric sequence.
輸入描述:
The first line is an integer n (1 <= n <= 100000).
The second line consists of n integer a[1],a[2],…,a[n] where a[i] <= 100000 for 1<=i<=n.
輸出描述:
An integer answer for the problem.
示例1
輸入

5
1 1 2 4 1
輸出

11
說明

The 11 pairs of (l,r) are (1,1),(1,2),(2,2),(2,3),(2,4),(3,3),(3,4),(3,5),(4,4),(4,5),(5,5).
示例2
輸入

10
3 1 1 1 5 2 2 5 3 3
輸出

20
備註:
The answer can be quite large that you may use long long in C++ or the similar in other languages.

題意：讓你找出所有的等比序列。

分析：逆向思維便可以優雅的暴力。由於資料是小於十萬的，公比為2的序列長度最長為17，所以我們可以列舉序列長度。
公比不能為小數嗎？當然可以，但是如果是小數的話，可以轉化成分數，那麼必須是分母的整數冪。然而分母最小是2，所以證畢！

注意：比賽的時候被精度卡住，卡在90%的資料上過不了。以後能用long long int解決的絕不用double。

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
#define ll long long int
const double eps = 1e-8;
const int maxn = 1e5 + 10;
double a[maxn];
double b[maxn];
int main()
{
    int
 
 n;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
        scanf("%lf", &a[i]);
        b[i] = a[i];
    }
    ll ans = 2*n-1;
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= 17; j++) {
            if (i + j - 1 > n) break;
            if (j == 1 || j == 2) { continue; }
            sort(a + i, a + i + j);
            double cs = a[i + 1] / a[i];
            int flag = 1;
            for (int k = i + 1; k <= i + j - 1; k++) {
                double ct = a[k] / a[k - 1];
                if (fabs(ct - cs) > eps) {
                    flag = 0; break;
                }
            }
            if (fabs(cs - 1) < eps) flag = 0;
            ans += 1ll * flag;
            //cout << flag << endl;
            for (int k = i; k <= i + j - 1; k++) {
                a[k] = b[k];
            }
        }
    }
    ll sum = 1;
    for (int i = 2; i <= n; i++) {
        if (a[i] == a[i - 1]) sum++;
        else sum = 1;
        if(sum>=3)
            ans = ans + sum - 2;
    }
    printf("%lld\n", ans);
    return 0;
}