思路：

相似

红色的长度等于（y - s） /  y 倍的 A' 和 B' 之间的 fence的长度

A' 是 p 和 A 连线和 x 轴交点， B'同理

交点也可以用相似求，然后lower_bound找到交点在哪里，然后通过预处理的fence长度的前缀和就可以求了，处理好边界

#pragma GCC optimize(2)#pragma GCC optimize(3)#pragma GCC optimize(4)#include
     
      using namespace std;#define fi first#define se second#define pi acos(-1.0)#define LL long long//#define mp make_pair#define pb push_back#define ls rt<<1, l, m#define rs rt<<1|1, m+1, r#define ULL unsigned LL#define pll pair
      
       #define pii pair
       
        #define piii pair
        
         #define pdd pair
         
          #define mem(a, b) memset(a, b, sizeof(a))#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);#define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stdout);//headconst int N = 2e5 + 5;pdd f[N];double sum[N];int main() {    int n, q, l, r;    double s, a, b, x, y;    scanf("%lf %lf %lf", &s, &a, &b);    scanf("%d", &n);    for (int i = 1; i <= n; i++) scanf("%lf %lf", &f[i].fi, &f[i].se);    sum[0] = 0;    for (int i = 1; i <= n; i++) {        sum[i] = sum[i-1] + f[i].se - f[i].fi;    }    scanf("%d", &q);    while(q--) {        scanf("%lf %lf", &x, &y);        double c1 = (a*y - s*x)/(y - s);        double c2 = (b*y - s*x)/(y - s);        int t = lower_bound(f+1, f+n+1, pdd(c1, 0)) - f;        double ans = 0;        if(t == 1) l = 1;        else {            l = t;            if(c1 < f[t-1].se) ans += f[t-1].se - c1;        }        int tt = lower_bound(f+1, f+n+1, pdd(c2, 0)) - f;        if(tt == 1) r = tt-1;        else {            r = tt-1;            if(c2 < f[tt-1].se) ans -= f[tt-1].se - c2;        }        if(r >= l) ans += sum[r] - sum[l-1];        printf("%.10f\n", ans * (y - s) / y);    }    return 0;}