首先每次买卖一定是在某天 k 以当时的最大收入买入，再到第 i 天卖出，那么易得方程：

fi=max{AiRatekfkAkRatek+Bk+BifkAkRatek+Bk}

再令

{xk=RatekfkAkRatek+Bkyk=fkAkRatek+Bk

则有

fi=max{Aixk+Biyk}yk=−AiBixk+fiBi

那么现在需要找到一个点 (xk,yk) 使得直线的截距最大

由于斜率和横坐标皆不满足单调性，可以用平衡树等维护，这里使用CDQ分治实现

实现过程如下：

Ⅰ 将数据按照斜率AiBi降序排序

Ⅱ 将区间按照操作顺序分为左右两部分处理

Ⅲ 先处理左半部分，维护左半边凸包（注意，此时左半边已按照 x 排序）

Ⅳ 处理左半边对右半边的影响，由于已按照斜率降序排序，所以普通斜率优化即可

Ⅴ 将区间按照 x 排序

代码

cpp
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#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>

using namespace std;

const int MAXN = 1e05 + 10;

const double INF = 1e60;
const double eps = 1e-08;

int Dcmp (double p) {
    if (fabs (p) < eps)
        return 0;
    return p < 0 ? - 1 : 1;
}

struct CashSt {
    double a, b, rate;
    double k, x, y;
    int index;

    CashSt () {}

    bool operator < (const CashSt& p) const {
        return Dcmp (x - p.x) == 0 ? Dcmp (y - p.y) < 0 : Dcmp (x - p.x) < 0;
    }
} ;
CashSt Cash[MAXN];
bool comp (const CashSt& a, const CashSt& b) {
    return Dcmp (a.k - b.k) > 0;
}

int N;

double slope (CashSt a, CashSt b) {
    if (Dcmp (b.x - a.x) == 0)
        return INF;
    return (b.y - a.y) / (b.x - a.x);
}
double f[MAXN]= {0};
CashSt Que[MAXN];
int l = 1, r = 0;
CashSt temp[MAXN];
void CDQ (int left, int right) {
    if (left == right) {
        f[left] = max (f[left], f[left - 1]);
        Cash[left].y = f[left] / (Cash[left].a * Cash[left].rate + Cash[left].b);
        Cash[left].x = Cash[left].y * Cash[left].rate;
        return ;
    }
    int mid = (left + right) >> 1;
    int p1 = left - 1, p2 = mid;
    for (int i = left; i <= right; i ++)
        Cash[i].index <= mid ? temp[++ p1] = Cash[i] : temp[++ p2] = Cash[i];
    for (int i = left; i <= right; i ++)
        Cash[i] = temp[i];
    CDQ (left, mid);
    l = 1, r = 0;
    for (int i = left; i <= mid; i ++) {
        while (l < r && Dcmp (slope (Que[r - 1], Que[r]) - slope (Que[r], Cash[i])) < 0)
            r --;
        Que[++ r] = Cash[i];
    }
    for (int i = mid + 1; i <= right; i ++) {
        while (l < r && Dcmp (slope (Que[l], Que[l + 1]) - Cash[i].k) > 0)
            l ++;
        f[Cash[i].index] = max (f[Cash[i].index], Cash[i].a * Que[l].x + Cash[i].b * Que[l].y);
    }
    CDQ (mid + 1, right);
    l = left, r = mid + 1;
    int p = 0;
    while (l <= mid && r <= right) {
        // if (Dcmp (Cash[l].x - Cash[r].x) < 0 || (Dcmp (Cash[l].x - Cash[r].x) == 0 && Dcmp (Cash[l].y - Cash[r].y) < 0))
        if (Cash[l] < Cash[r])
            temp[++ p] = Cash[l], l ++;
        else
            temp[++ p] = Cash[r], r ++;
    }
    while (l <= mid)
        temp[++ p] = Cash[l], l ++;
    while (r <= right)
        temp[++ p] = Cash[r], r ++;
    for (int i = 1; i <= p; i ++)
        Cash[i + left - 1] = temp[i];
}

double getnum () {
    double num = 0.0;
    char ch = getchar ();
    double T = 1.0;

    while (! isdigit (ch))
        ch = getchar ();
    while (isdigit (ch))
        num = num * 10.0 + (ch - '0') * 1.0, ch = getchar ();
    if (ch == '.') {
        ch = getchar ();
        while (isdigit (ch))
            num = num + (T /= 10.0) * (ch - '0'), ch = getchar ();
    }

    return num;
}

int main () {
    // freopen ("Input.txt", "r", stdin);

    scanf ("%d%lf", & N, & f[0]);
    for (int i = 1; i <= N; i ++) {
        double a = getnum (), b = getnum (), rate = getnum ();
        Cash[i].a = a, Cash[i].b = b, Cash[i].rate = rate;
        Cash[i].index = i;
        Cash[i].k = - a / b;
    }
    sort (Cash + 1, Cash + N + 1, comp);
    CDQ (1, N);
    printf ("%.3f\n", f[N]);

    return 0;
}

/*
3 100
1 1 1
1 2 2
2 2 3
*/