题解Luogu P2221 [HAOI2012]高速公路

Posted 2021-02-11 yzhang-rp-inf

原题传送门

这道题还算简单

我们要求的期望值：

[frac{sum_{i=l}^rsum_{j=l}^rdis[i][j]}{C_{r-l+1}^{2}}]

当然是上下两部分分别求，下面肥肠容易 ，问题在于如何求上面的

我们珂以把上面的换一个形式（枚举每段路会走几次）：

[sum_{i=l}^ra[i]*(r-i+1)*(i-l+1)]

化简一下这个式子：

[(r-l+1-r*l)*sum1+(r+l)*sum2-sum3]

其中(sum1=sum_{i=l}^ra[i])，(sum2=sum_{i=l}^ra[i]*i)，(sum3=sum_{i=l}^ra[i]*i^2)

这样就珂以用线段树维护了

#include <bits/stdc++.h>
#define ll long long 
#define N 100005
#define getchar nc
using namespace std;
inline char nc(){
    static char buf[100000],*p1=buf,*p2=buf;
    return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;
}
inline ll read()
{
    register ll x=0,f=1;register char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9')x=(x<<3)+(x<<1)+ch-'0',ch=getchar();
    return x*f;
}
inline void write(register ll x)
{
    if(!x)putchar('0');if(x<0)x=-x,putchar('-');
    static int sta[25];register int tot=0;
    while(x)sta[tot++]=x%10,x/=10;
    while(tot)putchar(sta[--tot]+48);
}
struct node{
    ll sum[6],tag;
}tr[N<<3];
int n,m;
ll sum1,sum2,sum3;
inline void build(register int x,register int l,register int r)
{
    if(l==r)
    {
        tr[x].sum[4]=1ll*l;
        tr[x].sum[5]=1ll*l*l;
        return;
    }
    int mid=l+r>>1;
    build(x<<1,l,mid);
    build(x<<1|1,mid+1,r);
    tr[x].sum[4]=tr[x<<1].sum[4]+tr[x<<1|1].sum[4];
    tr[x].sum[5]=tr[x<<1].sum[5]+tr[x<<1|1].sum[5];
}
inline void work(register int x,register int l,register int r,register ll val)
{
    tr[x].sum[1]+=1ll*(r-l+1)*val;
    tr[x].sum[2]+=tr[x].sum[4]*val;
    tr[x].sum[3]+=tr[x].sum[5]*val;
    tr[x].tag+=val;
}
inline void pushdown(register int x,register int l,register int r)
{
    int mid=l+r>>1;
    work(x<<1,l,mid,tr[x].tag);
    work(x<<1|1,mid+1,r,tr[x].tag);
    tr[x].tag=0;
}
inline void pushup(register int x)
{
    tr[x].sum[1]=tr[x<<1].sum[1]+tr[x<<1|1].sum[1];
    tr[x].sum[2]=tr[x<<1].sum[2]+tr[x<<1|1].sum[2];
    tr[x].sum[3]=tr[x<<1].sum[3]+tr[x<<1|1].sum[3];
}
inline void change(register int x,register int l,register int r,register int L,register int R,register ll val)
{
    if(L<=l&&r<=R)
    {
        work(x,l,r,val);
        return;
    }
    if(tr[x].tag)
        pushdown(x,l,r);
    int mid=l+r>>1;
    if(L<=mid)
        change(x<<1,l,mid,L,R,val);
    if(R>mid)
        change(x<<1|1,mid+1,r,L,R,val);
    pushup(x);
}
inline void query(register int x,register int l,register int r,register int L,register int R)
{
    if(L<=l&&r<=R)
    {
        sum1+=tr[x].sum[1]; 
        sum2+=tr[x].sum[2]; 
        sum3+=tr[x].sum[3]; 
        return;
    }   
    if(tr[x].tag)
        pushdown(x,l,r);
    int mid=l+r>>1;
    if(L<=mid)
        query(x<<1,l,mid,L,R);
    if(R>mid)
        query(x<<1|1,mid+1,r,L,R);
}
inline ll gcd(register ll a,register ll b)
{
    return !b?a:gcd(b,a%b);
}
int main()
{
    n=read(),m=read();
    build(1,1,n);
    while(m--)
    {
        char ch=getchar();
        while(ch!='C'&&ch!='Q')
            ch=getchar();
        ll l=read(),r=read()-1;
        if(ch=='C')
        {
            ll v=read();
            change(1,1,n,l,r,v);
        }
        else
        {
            sum1=sum2=sum3=0;
            query(1,1,n,l,r);
            ll ansa=(r-l+1-r*l)*sum1+(r+l)*sum2-sum3;
            ll ansb=(r-l+2)*(r-l+1)/2;
            ll g=gcd(ansa,ansb);
            write(ansa/g),putchar('/'),write(ansb/g),puts("");
        }
    }
    return 0;
}