总时间复杂度不超过O(n√nlogn)

#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define M 50500
#define SQRT_M 250
using namespace std;
int n,m,block,ans;
int a[M],b[M];
int f[SQRT_M][SQRT_M];
//f[i][i]表示第i块内部的逆序对数
//f[i][j](i<j)表示第i块中的元素与第j块中的元素之间的逆序对数
int g[SQRT_M][SQRT_M];
//g[i][j](i<=j)表示第i块中的元素与第i~j块中的元素之间的逆序对数
//第一维暴力第二维树状数组前缀和
//向上修改向下查询
int equals[SQRT_M][M];
int smaller[SQRT_M][M];
/*
equals[i][j]表示前i块之内j的数量 直接暴力
smaller[i][j]表示前i块之内小于等于j的数的数量 向上修改向下查询
第一维暴力第二维树状数组
*/
int l[SQRT_M],r[SQRT_M],belong[M];
int Merge_Sort(int l,int r,int b[]=::b)
{
    static int c[M];
    int i,mid=l+r>>1,re=0;
     
    if(l==r)
        return 0;
    re=Merge_Sort(l,mid,b)+Merge_Sort(mid+1,r,b);
    int l1=l,l2=mid+1;
    for(i=l;i<=r;i++)
    {
        if( b[l1]<=b[l2] && l1<=mid || l2>r )
            c[i]=b[l1++],re+=l2-mid-1;
        else
            c[i]=b[l2++];
    }
    memcpy( b+l , c+l , sizeof(b[0])*(r-l+1) );
    return re;
}
void Update(int c[],int x,int y,int limit)
{
    for(;x<=limit;x+=x&-x)
        c[x]+=y;
}
int Get_Ans(int c[],int x)
{
    int re=0;
    for(;x;x-=x&-x)
        re+=c[x];
    return re;
}
void Modify(int x,int y)
{
    int i,j,_block=belong[x],temp;
    for(i=1;i<_block;i++)
    {
        temp=upper_bound(b+l[i],b+r[i]+1,y)-upper_bound(b+l[i],b+r[i]+1,a[x]);
        Update(g[i],_block,-temp,(n-1)/block+1);
    }
    for(i=_block+1;(i-1)*block+1<=n;i++)
    {
        temp=lower_bound(b+l[i],b+r[i]+1,y)-lower_bound(b+l[i],b+r[i]+1,a[x]);
        Update(g[_block],i,temp,(n-1)/block+1);
    }
    memcpy(b+l[_block],a+l[_block],sizeof(b[0])*(r[_block]-l[_block]+1) );
    b[x]=y;
    temp=Merge_Sort(l[_block],r[_block]);
    //重构b数组
    Update(g[_block],_block,temp-f[_block][_block],(n-1)/block+1);
    f[_block][_block]=temp;
    //修改f[i][i]和g[i][~]
    for(i=_block;(i-1)*block+1<=n;i++)
    {
        equals[i][a[x]]--,equals[i][y]++;
        Update(smaller[i],a[x],-1,n);
        Update(smaller[i],y,1,n);
    }
    //修改euqal和smaller数组
    a[x]=y;
    //修改a数组
}
int Query(int x,int y)
{
    static int c[M];
    int i,top=0,re=0;
    if(belong[y]-belong[x]<=1)
    {
        memcpy(c+x,a+x,sizeof(a[0])*(y-x+1) );
        return Merge_Sort(x,y,c);
    }
    for(i=belong[x]+1;i<belong[y];i++)
        re+=Get_Ans(g[i],belong[y]-1);
    //BB
    for(i=x;i<=r[belong[x]];i++)
    {
        re+=Get_Ans(smaller[belong[y]-1],a[i])-Get_Ans(smaller[belong[x]],a[i])-
            ( equals[belong[y]-1][a[i]] - equals[belong[x]][a[i]] );
        c[++top]=a[i];
    }
    //AB
    for(i=l[belong[y]];i<=y;i++)
    {
        re+=r[belong[y]-1]-r[belong[x]]-
            ( Get_Ans(smaller[belong[y]-1],a[i])-Get_Ans(smaller[belong[x]],a[i]) );
        c[++top]=a[i];
    }
    //BC
    return re+Merge_Sort(1,top,c);
    //AA+AC+CC
}
int main()
{
    int i,j,k;
    int p,x,y;
    cin>>n;
    for(i=1;i<=n;i++)
        scanf("%d",&a[i]);
    block=static_cast<int>(sqrt(n)+1e-7);
 
    for(i=1;i<=n;i++)
    {
        belong[i]=(i-1)/block+1;
        b[i]=a[i];
    }
    for(i=1;(i-1)*block+1<=n;i++)
    {
        l[i]=(i-1)*block+1,r[i]=min(i*block,n);
        f[i][i]=Merge_Sort(l[i],r[i]);
    }
    //预处理块和f[i][i]
 
    for(i=1;(i-1)*block+1<=n;i++)
        for(j=i+1;(j-1)*block+1<=n;j++)
        {
            int p=l[j];
            for(k=l[i];k<=r[i];k++)
            {
                for(;p<=r[j]&&b[p]<b[k];p++);
                f[i][j]+=p-l[j];
            }
        }
    for(i=1;(i-1)*block+1<=n;i++)
        for(j=i;(j-1)*block+1<=n;j++)
            Update(g[i],j,f[i][j],(n-1)/block+1);
    //预处理f数组和g数组
 
    for(i=1;i<=n;i++)
        for(j=belong[i];j<=(n-1)/block+1;j++)
        {
            equals[j][a[i]]++;
            Update(smaller[j],a[i],1,n);
        }
    //预处理equals数组和smaller数组
 
    cin>>m;
    for(i=1;i<=m;i++)
    {
        scanf("%d%d%d",&p,&x,&y);
        x^=ans;y^=ans;
        if(p==1)
            Modify(x,y);
        else
            printf("%d\n", ans=Query(x,y) );
    }
}

BZOJ 3787 Gty的文艺妹子序列 分块+树状数组