【[HAOI2012]高速公路】

披着期望外衣的数据结构？

非常毒瘤

我们要求得期望其实就是

[frac{sum_{i=l}^{r}sum_{j=i+1}^{r}dis(i,j)}{binom{r-l+1}{2}}]

好像非常难求的样子

我记得慎老师曾经教过我今天的那道线段期望的初赛题，其实这道题和那道初赛题非常的像

老师教给我的思路是每次都将区间拆成两半来考虑，之后就会得到一个无穷等比数列

好像线段树就非常自然的每次将区间拆成了两半考虑

我们用(d(x))表示线段树(x)节点管辖的区间内所有区间的和，显然合并的时候

[d(x)=d(x<<1)+d(x<<1|1)]

这是不跨区间的情况

还需要考虑跨区间的情况

[d(x)+=ls(x<<1)*len(x<<1|1)+rs(x<<1|1)*len(x<<1)]

(ls,rs)表示从左/右开始的所有区间的长度和

剩下的比较简单随便推一下就好了

有点困难的是(pushdown)的时候(d(x))如何维护，由于这里是抄的题解里的柿子，所以这里手推一波

增加量应该对应到每一个区间上去，于是考虑枚举每一种长度的区间有多少个

[sum_{i=1}^{len}i*(len-i+1)*val]

[=sum_{i=1}^{len}i*(len+1)*val-i^2*val]

[=val*((len+1)*sum_{i=1}^{len}i-sum_{i=1}^{len}i^2)]

[=val*(frac{len*(len+1)^2}{2}-frac{(len+1)(2*len+1)len}{6})]

就好了

代码

#include<iostream>
#include<cstring>
#include<cstdio>
#define re register
#define maxn 100005
#define LL long long
#define int long long
LL l[maxn<<2],r[maxn<<2];
LL tag[maxn<<2],d[maxn<<2],ls[maxn<<2],rs[maxn<<2],s[maxn<<2];
struct node
{
    LL len,lc,rc,sum,t;
};
int n,Q;
inline int read()
{
    char c=getchar();
    int x=0,r=1;
    while(c<‘0‘||c>‘9‘) 
    {
        if(c==‘-‘) r=-1;
        c=getchar();
    }
    while(c>=‘0‘&&c<=‘9‘)
        x=(x<<3)+(x<<1)+c-48,c=getchar();
    return x*r;
}
LL gcd(LL a,LL b)
{
    if(!b) return a;
    return gcd(b,a%b);
}
void build(int x,int y,int i)
{
    l[i]=x,r[i]=y;
    if(x==y) return;
    int mid=x+y>>1;
    build(x,mid,i<<1),build(mid+1,y,i<<1|1);
}
inline void pushup(int i)
{
    d[i]=d[i<<1|1]+d[i<<1];
    d[i]+=(r[i<<1]-l[i<<1]+1)*ls[i<<1|1]+rs[i<<1]*(r[i<<1|1]-l[i<<1|1]+1);
    s[i]=s[i<<1|1]+s[i<<1];
    ls[i]=ls[i<<1],rs[i]=rs[i<<1|1];
    ls[i]+=(r[i<<1|1]-l[i<<1|1]+1)*s[i<<1]+ls[i<<1|1];
    rs[i]+=(r[i<<1]-l[i<<1]+1)*s[i<<1|1]+rs[i<<1];
}
inline void pushdown(int i)
{
    if(!tag[i]) return;
    s[i<<1]+=(r[i<<1]-l[i<<1]+1)*tag[i],s[i<<1|1]+=(r[i<<1|1]-l[i<<1|1]+1)*tag[i];
    d[i<<1]+=((r[i<<1]-l[i<<1]+2)*(r[i<<1]-l[i<<1]+1)/2*(r[i<<1]-l[i<<1]+2)-(r[i<<1]-l[i<<1]+1)*(r[i<<1]-l[i<<1]+2)*(2*(r[i<<1]-l[i<<1]+1)+1)/6)*tag[i];
    d[i<<1|1]+=((r[i<<1|1]-l[i<<1|1]+2)*(r[i<<1|1]-l[i<<1|1]+1)/2*(r[i<<1|1]-l[i<<1|1]+2)-(r[i<<1|1]-l[i<<1|1]+1)*(r[i<<1|1]-l[i<<1|1]+2)*(2*(r[i<<1|1]-l[i<<1|1]+1)+1)/6)*tag[i];
    ls[i<<1]+=(r[i<<1]-l[i<<1]+2)*(r[i<<1]-l[i<<1]+1)/2*tag[i];
    ls[i<<1|1]+=(r[i<<1|1]-l[i<<1|1]+2)*(r[i<<1|1]-l[i<<1|1]+1)/2*tag[i];
    rs[i<<1]+=(r[i<<1]-l[i<<1]+2)*(r[i<<1]-l[i<<1]+1)/2*tag[i];
    rs[i<<1|1]+=(r[i<<1|1]-l[i<<1|1]+2)*(r[i<<1|1]-l[i<<1|1]+1)/2*tag[i];
    tag[i<<1]+=tag[i],tag[i<<1|1]+=tag[i];
    tag[i]=0;
} 
void change(int x,int i,int val)
{
    if(x<=l[i]&&y>=r[i])
    {
        s[i]+=(r[i]-l[i]+1)*val;
        d[i]+=((r[i]-l[i]+2)*(r[i]-l[i]+1)/2*(r[i]-l[i]+2)-(r[i]-l[i]+1)*(r[i]-l[i]+2)*(2*(r[i]-l[i]+1)+1)/6)*val;
        ls[i]+=(r[i]-l[i]+1)*(r[i]-l[i]+2)/2*val;
        rs[i]+=(r[i]-l[i]+1)*(r[i]-l[i]+2)/2*val;
        tag[i]+=val;
        return;
    }
    pushdown(i);
    int mid=l[i]+r[i]>>1;
    if(y<=mid) change(x,i<<1,val);
        else if(x>mid) change(x,i<<1|1,val);
            else change(x,val),change(x,val);
    pushup(i);
}
node query(int x,int i)
{
    if(x<=l[i]&&y>=r[i]) return (node){r[i]-l[i]+1,ls[i],rs[i],s[i],d[i]};
    pushdown(i);
    int mid=l[i]+r[i]>>1;
    if(y<=mid) return query(x,i<<1);
    if(x>mid) return query(x,i<<1|1);
    node L=query(x,R=query(x,i<<1|1),now;
    now.sum=L.sum+R.sum;
    now.t=L.t+R.t;
    now.t+=R.len*L.rc+L.len*R.lc;
    now.len=R.len+L.len;
    now.lc=L.lc,now.rc=R.rc;
    now.lc+=R.len*L.sum+R.lc;
    now.rc+=L.len*R.sum+L.rc;
    return now;
}
signed main()
{
    n=read(),Q=read();
    char opt[1];
    build(1,n-1,1);
    int x,z;
    while(Q--)
    {
        scanf("%s",opt);
        if(opt[0]==‘C‘)
        {
            x=read(),y=read(),z=read();
            if(x<y) change(x,y-1,1,z);
        }
        else
        {
            x=read(),y=read();
            node ans=query(x,1);
            LL r=gcd(ans.t,(y-x)*(y-x+1)/2);
            printf("%lld/%lldn",ans.t/r,(y-x)*(y-x+1)/2/r);
        }
    }
    return 0;
}