朴素积分

\int_L^R\frac{cx+d}{ax+b}\mathrm{d}x

#include <bits/stdc++.h>
using namespace std;
typedef double db;
typedef db func(db);
db a, b, c, d, L, R;

db f(db x) {
    return (c * x + d) / (a * x + b);
}
db simpson(db l, db r, func f) {
    db mid = (l + r) / 2;
    return (f(l) + 4 * f(mid) + f(r)) * (r - l) / 6;
}
db asr(db l, db r, db eps, db A, func f) {
    db mid = (l + r) / 2;
    db L = simpson(l, mid, f), R = simpson(mid, r, f);
    if (fabs(L + R - A) <= 15 * eps) return L + R + (L + R - A) / 15;
    return asr(l, mid, eps / 2, L, f) + asr(mid, r, eps / 2, R, f);
}
db asr(db l, db r, db eps) {
    return asr(l, r, eps, simpson(l, r, f), f);
}

int main() {
    cin >> a >> b >> c >> d >> L >> R;
    printf("%.6lf\n", asr(L, R, 1e-6));
    return 0;
}

高精度积分（分段）

#include <bits/stdc++.h>
using namespace std;
typedef long double db;
typedef db func(db);
db a, b, c, d, L, R;

db f(db x) {
    return (c * x + d) / (a * x + b);
}
db simpson(db l, db r, func f) {
    db mid = (l + r) / 2;
    return (f(l) + 4 * f(mid) + f(r)) * (r - l) / 6;
}
db asr(db l, db r, db eps, db A, func f) {
    db mid = (l + r) / 2;
    db L = simpson(l, mid, f), R = simpson(mid, r, f);
    if (fabs(L + R - A) <= 15 * eps) return L + R + (L + R - A) / 15;
    return asr(l, mid, eps / 2, L, f) + asr(mid, r, eps / 2, R, f);
}
db asr(db l, db r, db eps, func f, int x = 5) {
    db dx = (r - l) / x, ret = 0;
    for (db i = l; i < r - eps; i += dx) {
        ret += asr(i, i + dx, eps, simpson(i, i + dx, f), f);
    }
    return ret;
}

双重积分

例题：gym101987G Secret Code（2019ICPC 首尔）

题目大意：

有 A,B,C 三个人要见面，每个人在 [0,S] 随机选择一个时间点作为见面时间，先到的那个人要等下一个人来了之后和他确认信息，然后马上就走。例如，假如 A 先到，B 其次，C 最后到，那么 A 要等 B 到了之后和B确认完信息，然后 A 走，B 再等 C 到了和 C 确认完信息，这样任务就完成了。

现给出 A,B,C 三人的最长等待时间 wa,wb,wc，求任务能完成的概率。

解题思路：

首先，A 到的时间可以是 [0, S]；B 到的时间是 [A, \min(A+wa,S)]；C 可以在 [B, \min(B+wb, S)] 到。因此我们枚举 3 个人来的顺序，答案为：

\frac{1}{S^3}\cdot \int_0^{S}\left(\int_A^{\min(A+wa,S)}\left(\min(B+wb,S)-B\right)\mathrm{d}B\right)\mathrm{d}A

#include <bits/stdc++.h>
using namespace std;
const double eps = 1e-9;
int S, wa, wb, wc, w[3];

typedef long double db;
typedef db func(db);

db simpson(db l, db r, func f) {
    db mid = (l + r) / 2;
    return (f(l) + 4 * f(mid) + f(r)) * (r - l) / 6;
}
db asr(db l, db r, db eps, db A, func f) {
    db mid = (l + r) / 2;
    db L = simpson(l, mid, f), R = simpson(mid, r, f);
    if (fabs(L + R - A) <= 15 * eps) return L + R + (L + R - A) / 15;
    return asr(l, mid, eps / 2, L, f) + asr(mid, r, eps / 2, R, f);
}
db asr(db l, db r, db eps, func f) {
    db dx = (r - l) / 5, ret = 0;
    for (db i = l; i < r - eps; i += dx) {
        ret += asr(i, i + dx, eps, simpson(i, i + dx, f), f);
    }
    return ret;
}

db f(db x) {
    return min(x + w[1], (db)S) - x;
}
db g(db x) {
    return asr(x, min(x + w[0], (db)S), eps, f);
}
db work() {
    return asr(0.0, (db)S, eps, g);
}

int n;

bool equ(db a, db b) {
    return fabs(a - b) < 1e-10;
}
struct Data {
    db ans;
    int i;
    bool operator < (const whnb &x) const {
        return equ(ans, x.ans) ? i < x.i : ans < x.ans;
    }
};
vector< Data > rec;

int main()
{
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
        scanf("%d %d %d %d", &S, &wa, &wb, &wc);
        double ans = 0;
        w[0] = wa, w[1] = wb, w[2] = wc; ans += work();
        w[0] = wa, w[1] = wc, w[2] = wb; ans += work();
        w[0] = wb, w[1] = wa, w[2] = wc; ans += work();
        w[0] = wb, w[1] = wc, w[2] = wa; ans += work();
        w[0] = wc, w[1] = wa, w[2] = wb; ans += work();
        w[0] = wc, w[1] = wb, w[2] = wa; ans += work();
        ans /= 1.0 * S * S * S;
        rec.push_back({ans, i});
    }
    sort(rec.begin(), rec.end());
    for (auto x : rec) printf("%d ", x.i); puts("");
    return 0;
}