【[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 y,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,y,i<<1,val);
		else if(x>mid) change(x,y,i<<1|1,val);
			else change(x,y,i<<1,val),change(x,y,i<<1|1,val);
	pushup(i);
}
node query(int x,int y,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,y,i<<1);
	if(x>mid) return query(x,y,i<<1|1);
	node L=query(x,y,i<<1),R=query(x,y,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,y,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,y-1,1);
			LL r=gcd(ans.t,(y-x)*(y-x+1)/2);
			printf("%lld/%lld\n",ans.t/r,(y-x)*(y-x+1)/2/r);
		}
	}
	return 0;
}