Multivariate function (第三次积分赛)

Multivariate function
描述

输入
第一行一个整数T，表示T组测试数据 (1≤T≤10).
每组数据第一行一个整数n(4≤n≤1000).
第二行 n个浮点数: $X_{1} , X_{2}X_{3}......X_{n} (1<=X_{i}<=1000000)$

输出
输出最大值 y，保留三位小数.
输入样例 1
2
4
1.0 2.0 3.0 4.0
5
1.6 2.6 7.1 2.3 2.6

输出样例 1
0.167
1.530
首先我们将题意进行转化

$$y = max(\frac{X_{j}*\frac{X_{p}}{X_{k}}-X_{i}}{X_{i}+X_{j}})$$
然后 $O(n^{2})预处理\frac{X_{p}}{X_{k}}$ ，最后遍历 $X_{i},X_{j}$，就可以了时间复杂度
$O(n^{2})$ 题意的数据为1000没有超
AC code：

#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <queue>

using namespace std;

const int maxn = 1e3+5;
const double INF = 1.0*0x7f7f7f7f;

double v[maxn],vis[maxn],per[maxn];

int main(){
    int t; cin>>t;
    while(t--)
    {
        int n; scanf("%d",&n);
        for (int i = 1;i<=n;i++) {
            scanf("%lf",&v[i]);
            //printf("v[%d] = %lf\n",i,v[i]);
        }
        for (int i = n-1;i>=1;i--) {
            double minn = INF;
            for (int j = i+1;j<=n;j++) {
                minn = min(minn,v[j]);
            }
            //printf("minn = %lf\n",minn);
            vis[i] = v[i]/minn;
            //printf("vis[%d] = %lf\n",i,vis[i]);
        }
        double maxx = -INF;
        for (int i = n-1;i>=1;i--) {
            maxx = max(maxx,vis[i]);
            per[i] = maxx;
            //printf("per[%d] = %lf\n",i,per[i]);
        }
        double ans = -INF;
        for (int i = 1;i<=n-3;i++) {
            for (int j = i+1;j<=n-2;j++) {
                ans = max(ans,(v[j]*per[j+1]-v[i])/(v[i]+v[j]));
            }
        }
        printf("%.3lf\n",ans);
    }
    return 0;
}