【USACO 2021 US Open, Gold】United Cows of Farmer John & JZOJ7220

题意：对于i上一个同颜色的位置j，ans+=(j+1~i-1的颜色种类)

做法：考场时傻逼了没想到，题做得少。枚举R，统计区间和。由于同种颜色数量不影响贡献R有顺序只用考虑L，每次颜色重复就把原位置-1，当前位置+1。 

#include<bits/stdc++.h>
#define inf 1000000007
#define mid ((l+r)/2)
#define l(i) ((int)i.length())
using namespace std;
int n,m,a[200005],bz[200005],tr[20000005];
long long ans;
long long read(){
    long long x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+ch-48;ch=getchar();}
    return x*f;
}
long long query(int l,int r,int k,int x,int y)
{
    if(x>y)return 0;
    if(l==x&&r==y)
    {
        return tr[k];
    }
    if(y<=mid)
    {
        return query(l,mid,k*2,x,y);
    }
    if(x>mid)
    {
        return query(mid+1,r,k*2+1,x,y);
    }
    return query(l,mid,k*2,x,mid)+query(mid+1,r,k*2+1,mid+1,y);
}
void ins(int l,int r,int k,int x,int y)
{
    if(l==r)
    {
        tr[k]+=y;
        return;
    }
    if(x<=mid)ins(l,mid,k*2,x,y);
    else ins(mid+1,r,k*2+1,x,y);
    tr[k]=tr[k*2]+tr[k*2+1]; 
}
int main()
{
    //freopen(".in","r",stdin);
    //freopen(".out","w",stdout);
    n=read();
    for(int i=1;i<=n;i++)
    {
        a[i]=read();
        ans+=query(1,n,1,bz[a[i]]+1,i-1);
        ins(1,n,1,i,1);
        if(bz[a[i]])ins(1,n,1,bz[a[i]],-1);
        bz[a[i]]=i;
    }
    printf("%lld\n",ans);
    return 0;
}