洛谷3707 [SDOI2017] 相关分析 【线段树】

分析：

化简一下就行了，注意一下平方和公式的运用以及精度的误差。

代码：

#include<bits/stdc++.h>
using namespace std;

const int maxn = 105000;

int n,m;

int x[maxn],y[maxn];

struct node{
    int lazx1,lazx2,lazy1,lazy2;
    double multi,sumx,sumy,sqr;
}T[maxn<<2];

long long ump(int l,int r){
    return (1ll*r*(r+1)*(2*r+1)-1ll*(l-1)*l*(2*l-1))/6;
}

void fugai(int now,int tl,int tr,int s,int t){
    T[now].lazx2 = T[now].lazy2 = 0;
    T[now].lazx1 = s; T[now].lazy1 = t;
    T[now].sumx = 1ll*((s+tl)+(s+tr))*(tr-tl+1)/2;
    T[now].sumy = 1ll*((t+tl)+(t+tr))*(tr-tl+1)/2;
    T[now].sqr = 1ll*(tr-tl+1)*s*s+1ll*s*(tl+tr)*(tr-tl+1)+ump(tl,tr);
    T[now].multi=1ll*(tr-tl+1)*s*t+1ll*(s+t)*(tl+tr)*(tr-tl+1)/2+ump(tl,tr);
}

void add(int now,int tl,int tr,int s,int t){
    T[now].lazx2+=s;T[now].lazy2 += t;
    T[now].multi+=1ll*s*T[now].sumy+1ll*t*T[now].sumx+1ll*s*t*(tr-tl+1);
    T[now].sqr +=1ll*s*s*(tr-tl+1)+2ll*s*T[now].sumx;
    T[now].sumx += 1ll*s*(tr-tl+1); T[now].sumy += 1ll*t*(tr-tl+1);
}

void push_up(int now){
    T[now].multi = T[now<<1].multi+T[now<<1|1].multi;
    T[now].sumx = T[now<<1].sumx+T[now<<1|1].sumx;
    T[now].sumy = T[now<<1].sumy+T[now<<1|1].sumy;
    T[now].sqr = T[now<<1].sqr+T[now<<1|1].sqr;
}

void push_down1(int now,int tl,int tr){
    int mid = (tl+tr)/2;
    fugai(now<<1,tl,mid,T[now].lazx1,T[now].lazy1);
    fugai(now<<1|1,mid+1,tr,T[now].lazx1,T[now].lazy1);
    T[now].lazx1 = T[now].lazy1 = 123456;
}

void push_down2(int now,int tl,int tr){
    int mid = (tl+tr)/2;
    add(now<<1,tl,mid,T[now].lazx2,T[now].lazy2);
    add(now<<1|1,mid+1,tr,T[now].lazx2,T[now].lazy2);
    T[now].lazx2 = T[now].lazy2 = 0;
}

void read(){
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)scanf("%d",&x[i]);
    for(int i=1;i<=n;i++) scanf("%d",&y[i]);
}

node merge(node ai,node bi){
    return (node){0,0,0,0,ai.multi+bi.multi,ai.sumx+bi.sumx,ai.sumy+bi.sumy,ai.sqr+bi.sqr};
}

void build_tree(int now,int l,int r){
    if(l == r){
    T[now].lazx1 = T[now].lazy1 = 123456;
    T[now].multi = 1ll*x[l]*y[l];
    T[now].sumx = x[l];T[now].sumy = y[l];
    T[now].sqr = 1ll*x[l]*x[l];
    }else{
    int mid = (l+r)/2;
    build_tree(now<<1,l,mid);
    build_tree(now<<1|1,mid+1,r);
    T[now].lazx1 = T[now].lazy1 = 123456;
    push_up(now);
    }
}

node Query(int now,int tl,int tr,int l,int r){
    if(tl >= l && tr <= r){return T[now];}
    if(tl > r || tr < l){return (node){0,0,0,0,0,0,0,0};}
    if(T[now].lazx1<=maxn||T[now].lazy1<=maxn) push_down1(now,tl,tr);
    if(T[now].lazx2||T[now].lazy2) push_down2(now,tl,tr);
    int mid = (tl+tr)/2;
    node ans = merge(Query(now<<1,tl,mid,l,r),Query(now<<1|1,mid+1,tr,l,r));
    push_up(now);
    return ans;
}

void Modify1(int now,int tl,int tr,int l,int r,int s,int t){
    if(tl >= l && tr <= r){
    add(now,tl,tr,s,t);
    return;
    }
    if(tl > r || tr < l){return;}
    if(T[now].lazx1<=maxn||T[now].lazy1<=maxn) push_down1(now,tl,tr);
    if(T[now].lazx2||T[now].lazy2) push_down2(now,tl,tr);
    int mid = (tl+tr)/2;
    Modify1(now<<1,tl,mid,l,r,s,t);
    Modify1(now<<1|1,mid+1,tr,l,r,s,t);
    push_up(now);
}

void Modify2(int now,int tl,int tr,int l,int r,int s,int t){
    if(tl >= l && tr <= r){
    fugai(now,tl,tr,s,t);
    return;
    }
    if(tl > r || tr < l){return;}
    if(T[now].lazx1<=maxn||T[now].lazy1<=maxn) push_down1(now,tl,tr);
    if(T[now].lazx2||T[now].lazy2) push_down2(now,tl,tr);
    int mid = (tl+tr)/2;
    Modify2(now<<1,tl,mid,l,r,s,t);
    Modify2(now<<1|1,mid+1,tr,l,r,s,t);
    push_up(now);
}

void work(){
    build_tree(1,1,n);
    for(int i=1;i<=m;i++){
    int cas; scanf("%d",&cas);
    if(cas == 1){
        int l,r; scanf("%d%d",&l,&r);
        node forw = Query(1,1,n,l,r);
        double pjx = 1.0*forw.sumx/(r-l+1),pjy = 1.0*forw.sumy/(r-l+1);
        double res=forw.multi+pjx*pjy*(r-l+1)-pjx*forw.sumy-pjy*forw.sumx;
        res /= 1.0*(forw.sqr+pjx*pjx*(r-l+1)-2.0*pjx*forw.sumx);
        printf("%.10lf
",res);
    }else{
        if(cas == 2){
        int l,r,s,t; scanf("%d%d%d%d",&l,&r,&s,&t);
        Modify1(1,1,n,l,r,s,t);
        }else{
        int l,r,s,t; scanf("%d%d%d%d",&l,&r,&s,&t);
        Modify2(1,1,n,l,r,s,t);
        }
    }
    }
}

int main(){
    read();
    work();
    return 0;
}