bzoj
luogu

題意

你需要維護兩個數組\(\{x_i\},\{y_i\}\)，資瓷一下三種操作。
1、給出\(l,r\)，設\(\overline x,\overline y\)分別表示區間\([l,r]\)內\(x_i,y_i\)的平均數，求：
\[\frac{\sum_{i=l}^r(x_i-\overline x)(y_i-\overline y)}{\sum_{i=l}^r(x_i-\overline x)^2}\]
2、給出\(l,r,S,T\)，令區間\([l,r]\)內每個\(x_i+=S,y_i+=T\)
3、給出\(l,r,S,T\)，令區間\([l,r]\)內每個\(x_i=S+i,y_i=i+T\)

sol

把詢問式的括號拆了，利用\(\overline x*(r-l+1)=\sum_{i=l}^rx_i\)化簡一下式子：
\[\frac{\sum x_iy_i+\frac 1{r-l+1}\sum x_i*\sum y_i}{\sum x_i^2+\frac 1{r-l+1}\sum x_i*\sum x_i}\]
所以只要維護區間的\(\sum x,\sum y,\sum xy,\sum x^2\)就行了。
操作2是一個區間加，有：
\(\sum(x+\Delta x)(y+\Delta y)=\sum xy+\sum x*\Delta y+\sum y*\Delta x+(r-l+1)*\Delta x*\Delta y\)

code

#include<cstdio>
#include<algorithm>
using namespace std;
const int N = 1e5+5;
int n,m,tag_c[N<<2];double X[N],Y[N],tag_x[N<<2],tag_y[N<<2
 
];
struct data{double x,y,xy,xx;}t[N<<2];
data operator + (data a,data b){
    return (data){a.x+b.x,a.y+b.y,a.xy+b.xy,a.xx+b.xx};
}
void build(int x,int l,int r){
    if (l==r){
        t[x].x=X[l];t[x].y=Y[l];
        t[x].xy=t[x].x*t[x].y;t[x].xx=t[x].x*t[x].x;
        return;
    }
    int mid=l+r>>1;
    build(x<<1,l,mid);build(x<<1|1,mid+1,r);
    t[x]=t[x<<1]+t[x<<1|1];
}
void cover1(int x,int l,int r,double S,double T){
    double len=r-l+1;
    t[x].xy+=t[x].x*T+t[x].y*S+S*T*len;
    t[x].xx+=2*t[x].x*S+S*S*len;
    t[x].x+=S*len;t[x].y+=T*len;
    tag_x[x]+=S;tag_y[x]+=T;
}
double cal1(int x){return (double)x*(x+1)/2;}
double cal2(int x){return (double)x*(x+1)*(2*x+1)/6;}
void cover2(int x,int l,int r){
    t[x].x=t[x].y=cal1(r)-cal1(l-1);
    t[x].xy=t[x].xx=cal2(r)-cal2(l-1);
    tag_c[x]=1;tag_x[x]=tag_y[x]=0;
}
void pushdown(int x,int l,int r){
    int mid=l+r>>1;
    if (tag_c[x]){
        cover2(x<<1,l,mid);cover2(x<<1|1,mid+1,r);
        tag_c[x]=0;
    }
    if (tag_x[x]||tag_y[x]){
        cover1(x<<1,l,mid,tag_x[x],tag_y[x]);
        cover1(x<<1|1,mid+1,r,tag_x[x],tag_y[x]);
        tag_x[x]=tag_y[x]=0;
    }
}
void modify1(int x,int l,int r,int ql,int qr,double S,double T){
    if (l>=ql&&r<=qr) {cover1(x,l,r,S,T);return;}
    pushdown(x,l,r);int mid=l+r>>1;
    if (ql<=mid) modify1(x<<1,l,mid,ql,qr,S,T);
    if (qr>mid) modify1(x<<1|1,mid+1,r,ql,qr,S,T);
    t[x]=t[x<<1]+t[x<<1|1];
}
void modify2(int x,int l,int r,int ql,int qr){
    if (l>=ql&&r<=qr) {cover2(x,l,r);return;}
    pushdown(x,l,r);int mid=l+r>>1;
    if (ql<=mid) modify2(x<<1,l,mid,ql,qr);
    if (qr>mid) modify2(x<<1|1,mid+1,r,ql,qr);
    t[x]=t[x<<1]+t[x<<1|1];
}
data query(int x,int l,int r,int ql,int qr){
    if (l>=ql&&r<=qr) return t[x];
    pushdown(x,l,r);int mid=l+r>>1;
    if (qr<=mid) return query(x<<1,l,mid,ql,qr);
    if (ql>mid) return query(x<<1|1,mid+1,r,ql,qr);
    return query(x<<1,l,mid,ql,qr)+query(x<<1|1,mid+1,r,ql,qr);
}
int main(){
    scanf("%d%d",&n,&m);
    for (int i=1;i<=n;++i) scanf("%lf",&X[i]);
    for (int i=1;i<=n;++i) scanf("%lf",&Y[i]);
    build(1,1,n);while (m--){
        int opt,l,r;scanf("%d%d%d",&opt,&l,&r);
        if (opt==1){
            data ans=query(1,1,n,l,r);
            double p=ans.xy-ans.x*ans.y/(r-l+1);
            double q=ans.xx-ans.x*ans.x/(r-l+1);
            printf("%.10lf\n",p/q);
        }else{
            double S,T;scanf("%lf%lf",&S,&T);
            if (opt==3) modify2(1,1,n,l,r);
            modify1(1,1,n,l,r,S,T);
        }
    }
    return 0;
}

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