湖南大学ACM程序设计新生杯大赛（同步赛）A-Array 【证明+暴力】

时间限制：C/C++ 1秒，其他语言2秒

空间限制：C/C++ 131072K，其他语言262144K

64bit IO Format: %lld

题目描述

Given an array A with length n a[1],a[2],…,a
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.

输入描述:

The first line is an integer n (1 <= n <= 100000).

The second line consists of n integer a[1],a[2],…,a
where a[i] <= 100000 for 1<=i<=n.

输出描述:

An integer answer for the problem.

示例1

输入

5

1 1 2 4 1

输出

11

说明

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

输入

10

3 1 1 1 5 2 2 5 3 3

输出

20

备注:

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

题意：让你找出所有的等比序列。

分析：逆向思维便可以优雅的暴力。由于数据是小于十万的，公比为2的序列长度最长为17，所以我们可以枚举序列长度。

公比不能为小数吗？当然可以，但是如果是小数的话，可以转化成分数，那么必须是分母的整数幂。然而分母最小是2，所以证毕！

注意：比赛的时候被精度卡住，卡在90%的数据上过不了。以后能用long long int解决的绝不用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;
}