【BZOJ 4821】【SDOI 2017】相关分析

把公式化开来得到：
a=∑ri=l(xiyi−x¯yi−xiy¯+x¯y¯)∑ri=l(x2i−2x¯xi+x¯2)
把分子的第二项拿出来结合∑ri=l就可以得到x¯∑ri=lyi=x¯y¯∗(r−l+1)，第三项同理，分母同理，再合并一下就可以得到：
a=∑xiyi−x¯y¯∗(r−l+1)∑ri=lx2i−x¯2∗(r−l+1)
然后用线段树维护一下∑x、∑y、∑xy、∑x2即可。

2号询问就是个简单的区间加，注意到：
∑(x+Δx)(y+Δy)
=∑(xy+Δxy+Δyx+ΔxΔy)
=∑xy+y∑Δx+x∑Δy+∑ΔxΔy，就可以维护了。
3号询问可以看做是先将第i号元素变为i然后执行一次2号操作。用一下平方和公式即可。
注意：当执行3号操作的覆盖操作时要将所有2号操作的lazy-tag清除掉。

#include<cmath>#include<cstdio>#include<vector>#include<queue>#include<cstring>#include<iomanip>#include<stdlib.h>#include<iostream>#include<algorithm>#define ll long long#define inf 1000000000#define mod 1000000007#define ls x<<1#define rs x<<1|1#define N 100005#define fo(i,a,b) for(i=a;i<=b;i++)#define fd(i,a,b) for(i=a;i>=b;i--)using namespace std;struct TREE{double sumx,sumy,sumxy,sumxx,tag_x,tag_y; int l,r,len,tag_c;} tree[N*4];double xx[N],yy[N];int n,m,i,opt,l,r,dx,dy;void pushup(int x){    tree[x].sumx = tree[ls].sumx + tree[rs].sumx;    tree[x].sumy = tree[ls].sumy + tree[rs].sumy;    tree[x].sumxy = tree[ls].sumxy + tree[rs].sumxy;    tree[x].sumxx = tree[ls].sumxx + tree[rs].sumxx;}void build(int x,int l,int r){    tree[x].len = r - l + 1; tree[x].l = l; tree[x].r = r;    if (l == r)         {            tree[x].sumx = xx[l]; tree[x].sumy = yy[l];            tree[x].sumxy = xx[l] * yy[l];             tree[x].sumxx = xx[l] * xx[l];            return;        }    int mid = (l + r) >> 1;    build(ls,l,mid); build(rs,mid+1,r); pushup(x);    }double calc(int x) {return (double)x*(x+1)*(2*x+1)/6;}void update_a(int x,double dx,double dy){    tree[x].sumxy += tree[x].sumx * dy + tree[x].sumy * dx + tree[x].len * dx * dy;    tree[x].sumxx += tree[x].sumx * dx * 2 + tree[x].len * dx * dx;    tree[x].sumx += dx * tree[x].len;    tree[x].sumy += dy * tree[x].len;    tree[x].tag_x += dx; tree[x].tag_y += dy;}void update_c(int x){    tree[x].sumx = tree[x].sumy = (double)(tree[x].l+tree[x].r)*tree[x].len/2;    tree[x].sumxy = tree[x].sumxx = calc(tree[x].r) - calc(tree[x].l-1);    tree[x].tag_c = 1;    tree[x].tag_x = tree[x].tag_y = 0;}void pushdown(int x){    if (tree[x].tag_c) {update_c(ls); update_c(rs); tree[x].tag_c = 0;}    if (tree[x].tag_x || tree[x].tag_y)        {            update_a(ls,tree[x].tag_x,tree[x].tag_y);            update_a(rs,tree[x].tag_x,tree[x].tag_y);            tree[x].tag_x = tree[x].tag_y = 0;        }}void add(int x,int l,int r,int L,int R,int dx,int dy){    if (L <= l && r <= R) {update_a(x,dx,dy); return;}    pushdown(x);    int mid = (l + r) >> 1;    if (L <= mid) add(ls,l,mid,L,R,dx,dy);    if (mid < R) add(rs,mid+1,r,L,R,dx,dy);    pushup(x);}void change(int x,int l,int r,int L,int R){    if (L <= l && r <= R) {update_c(x); return;}    pushdown(x);    int mid = (l + r) >> 1;    if (L <= mid) change(ls,l,mid,L,R);    if (mid < R) change(rs,mid+1,r,L,R);    pushup(x);}double query(int x,int l,int r,int L,int R,int opt){    if (L <= l && r <= R)         {            if (opt == 1) return tree[x].sumx;            if (opt == 2) return tree[x].sumy;            if (opt == 3) return tree[x].sumxy;            if (opt == 4) return tree[x].sumxx;        }    pushdown(x);    int mid = (l + r) >> 1; double res = 0;    if (L <= mid) res += query(ls,l,mid,L,R,opt);    if (mid < R) res += query(rs,mid+1,r,L,R,opt);    return res;}int main(){    scanf("%d%d",&n,&m);    fo(i,1,n) scanf("%lf",&xx[i]);    fo(i,1,n) scanf("%lf",&yy[i]);    build(1,1,n);    while (m--)        {            scanf("%d",&opt);            scanf("%d%d",&l,&r);            if (opt == 1)                {                    double res1 = query(1,1,n,l,r,1); double res2 = query(1,1,n,l,r,2);                    double res3 = query(1,1,n,l,r,3); double res4 = query(1,1,n,l,r,4);                    double p = res3 - res1 * res2 / (r-l+1);                    double q = res4 - res1 * res1 / (r-l+1);                    printf("%.10lf\n",p/q);                }   else                {                    scanf("%d%d",&dx,&dy);                    if (opt == 3) change(1,1,n,l,r);                    add(1,1,n,l,r,dx,dy);                }               }    return 0;   }