标签：rac define etc 如何 lin signed %s 线段树 else
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;
}标签：rac define etc 如何 lin signed %s 线段树 else
原文地址：https://www.cnblogs.com/asuldb/p/10205749.html
