BZOJ4821 SDOI2017相关分析（线段树）

　　纯粹的码农题。维护x的和、y的和、xy的和、x²的和即可。可能会炸long long。

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long double
#define N 100010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
    while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
int n,m,a[N],b[N],L[N<<2],R[N<<2];
ll lazyaddx[N<<2],lazyaddy[N<<2],s[N],t[N];
bool lazyres[N<<2];
struct data
{
    ll sumx,sumy,sumxy,sqr;
    data operator +(const data&a) const
    {
        data c;
        c.sumx=sumx+a.sumx;
        c.sumy=sumy+a.sumy;
        c.sumxy=sumxy+a.sumxy;
        c.sqr=sqr+a.sqr;
        return c;
    }
}tree[N<<2];
void up(int k){tree[k]=tree[k<<1]+tree[k<<1|1];}
void update(int k,ll x,ll y,bool p)
{
    if (p)
    {
        tree[k].sumx=tree[k].sumy=s[R[k]]-s[L[k]-1];
        tree[k].sumxy=tree[k].sqr=t[R[k]]-t[L[k]-1];
        lazyaddx[k]=lazyaddy[k]=0,lazyres[k]=1;
    }
    tree[k].sqr+=x*x*(R[k]-L[k]+1)+2*x*tree[k].sumx;
    tree[k].sumxy+=x*tree[k].sumy;
    tree[k].sumx+=(R[k]-L[k]+1)*x;
    tree[k].sumxy+=y*tree[k].sumx;
    tree[k].sumy+=(R[k]-L[k]+1)*y;
    lazyaddx[k]+=x,lazyaddy[k]+=y;
}
void down(int k)
{
    update(k<<1,lazyaddx[k],lazyaddy[k],lazyres[k]);
    update(k<<1|1,lazyaddx[k],lazyaddy[k],lazyres[k]);
    lazyaddx[k]=lazyaddy[k]=0;lazyres[k]=0;
}
void build(int k,int l,int r)
{
    L[k]=l,R[k]=r;
    if (l==r) {tree[k].sumx=a[l],tree[k].sumy=b[l],tree[k].sumxy=1ll*a[l]*b[l],tree[k].sqr=1ll*a[l]*a[l];return;}
    int mid=l+r>>1;
    build(k<<1,l,mid);
    build(k<<1|1,mid+1,r);
    up(k);
}
data query(int k,int l,int r)
{
    if (L[k]==l&&R[k]==r) return tree[k];
    if (lazyaddx[k]||lazyaddy[k]||lazyres[k]) down(k);
    int mid=L[k]+R[k]>>1;
    if (r<=mid) return query(k<<1,l,r);
    else if (l>mid) return query(k<<1|1,l,r);
    else return query(k<<1,l,mid)+query(k<<1|1,mid+1,r);
}
void res(int k,int l,int r)
{
    if (L[k]==l&&R[k]==r) {update(k,0,0,1);return;}
    if (lazyaddx[k]||lazyaddy[k]||lazyres[k]) down(k);
    int mid=L[k]+R[k]>>1;
    if (r<=mid) res(k<<1,l,r);
    else if (l>mid) res(k<<1|1,l,r);
    else res(k<<1,l,mid),res(k<<1|1,mid+1,r);
    up(k);
}
void add(int k,int l,int r,int x,int y)
{
    if (L[k]==l&&R[k]==r) {update(k,x,y,0);return;}
    if (lazyaddx[k]||lazyaddy[k]||lazyres[k]) down(k);
    int mid=L[k]+R[k]>>1;
    if (r<=mid) add(k<<1,l,r,x,y);
    else if (l>mid) add(k<<1|1,l,r,x,y);
    else add(k<<1,l,mid,x,y),add(k<<1|1,mid+1,r,x,y);
    up(k);
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj4821.in","r",stdin);
    freopen("bzoj4821.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    n=read(),m=read();
    for (int i=1;i<=n;i++) a[i]=read();
    for (int i=1;i<=n;i++) b[i]=read();
    for (int i=1;i<=n;i++) s[i]=s[i-1]+i,t[i]=t[i-1]+1ll*i*i;
    build(1,1,n);
    while (m--)
    {
        int op=read();
        if (op==1)
        {
            int l=read(),r=read();
            data k=query(1,l,r);
            ll x=(ll)k.sumx/(r-l+1),y=(ll)k.sumy/(r-l+1);
            printf("%.6lf\n",(double)((k.sumxy-k.sumx*y-k.sumy*x+x*y*(r-l+1))/(k.sqr-2*k.sumx*x+x*x*(r-l+1))));
        }
        else
        {
            int l=read(),r=read(),x=read(),y=read();
            if (op==3) res(1,l,r);add(1,l,r,x,y);
        }
    }
    return 0;
}