几何+二分——cf1016E

/*
把所有的阻挡物按x轴排序，对于每个点i，二分去找左右两端点[l,r]，求[l,r]未被阻挡的距离 
*/
#include<bits/stdc++.h>
using namespace std;
#define N 400005
#define ll long long
 
typedef double db;
const db eps=1e-6;
const db pi=acos(-1);
int sign(db k){if (k>eps) return 1; else if (k<-eps) return -1; return 0;}
int cmp(db k1,db k2){return sign(k1-k2);}
int inmid(db k1,db k2,db k3){return sign(k1-k3)*sign(k2-k3)<=0;}// k3 在 [k1,k2] 内 
struct point{
    db x,y;
    point operator + (const point &k1) const{return (point){k1.x+x,k1.y+y};}
    point operator - (const point &k1) const{return (point){x-k1.x,y-k1.y};}
    point operator * (db k1) const{return (point){x*k1,y*k1};}
    point operator / (db k1) const{return (point){x/k1,y/k1};}
    int operator == (const point &k1) const{return cmp(x,k1.x)==0&&cmp(y,k1.y)==0;}
    db abs(){return sqrt(x*x+y*y);}
    db abs2(){return x*x+y*y;}
    db dis(point k1){return ((*this)-k1).abs();}
    point unit(){db w=abs(); return (point){x/w,y/w};}
};
int inmid(point k1,point k2,point k3){return inmid(k1.x,k2.x,k3.x)&&inmid(k1.y,k2.y,k3.y);}
db cross(point k1,point k2){return k1.x*k2.y-k1.y*k2.x;}
db dot(point k1,point k2){return k1.x*k2.x+k1.y*k2.y;}
point getLL(point k1,point k2,point k3,point k4){
    db w1=cross(k1-k3,k4-k3),w2=cross(k4-k3,k2-k3); return (k1*w2+k2*w1)/(w1+w2);
}
struct Seg{
    int l,r;
}s[N];
ll n,sy,a,b,sum[N];//sum[i]表示第i块隔板的长度 
point L,R,k1,k2,k3;
 
int main(){
    cin>>sy>>a>>b>>n;
    L=(point){a*1.0,sy*1.0},R=(point){b*1.0,sy*1.0};
    for(int i=1;i<=n;i++)
        scanf("%d%d",&s[i].l,&s[i].r);
    for(int i=1;i<=n;i++)
        sum[i]=sum[i-1]+s[i].r-s[i].l; 
    int q;cin>>q;
    while(q--){
        scanf("%lf%lf",&k1.x,&k1.y);
        k2=getLL(k1,L,(point){-1,0},(point){1,0});//和x轴左端交点
        k3=getLL(k1,R,(point){-1,0},(point){1,0});//和x轴右端交点
        
        if(k2.x>=s[n].r || k3.x<=s[1].l){//没阻挡物 
            cout<<0<<'\n';continue;
        }
        
        int l=1,r=n,mid,pos1=n+1;
        while(l<=r){//找最靠左的左端点>=k2.x的段 
            mid=l+r>>1;
            if(s[mid].l>=k2.x)
                pos1=mid,r=mid-1;
            else l=mid+1;
        }
        
        l=1,r=n;
        int pos2=0;
        while(l<=r){//找最靠右的右端点<=k3.x的段 
            mid=l+r>>1;
            if(s[mid].r<=k3.x)
                pos2=mid,l=mid+1;
            else r=mid-1; 
        }
        
        db len=0;
        if(pos2>=pos1)len+=sum[pos2]-sum[pos1-1];
        if(pos1==pos2+2){
            len=k3.x-k2.x;//在同一条线段上相交
            printf("%.10lf\n",len/(k3.x-k2.x)*(b-a)); 
            continue;
        }
        if(pos1>1) //前一项的长度 
            len+=max(0.0,s[pos1-1].r-k2.x);
        if(pos2<n) //后一项的长度 
            len+=max(0.0,k3.x-s[pos2+1].l);
        printf("%.10lf\n",len/(k3.x-k2.x)*(b-a));
    }
}