BZOJ 2821 作诗

类似上题的做法，f[i][j]表示第i块到第j块出现偶数次的数有多少个，然后调整一下。

复杂度n√nlogn。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#define maxn 100050
using namespace std;
int n,c,m,a[maxn],cnt[maxn],blo[maxn],ret=0,f[2050][2050],l,r,len,st[maxn],lastans=0;
int flag[maxn],base[maxn],times=0;
int l1,r1,l2,r2,s[maxn],s1[maxn],s2[maxn],top=0;
vector <int> v[maxn];
int read()
{
    char ch;int data=0;
    while (ch<'0' || ch>'9') ch=getchar();
    while (ch>='0' && ch<='9')
    {
        data=data*10+ch-'0';
        ch=getchar();
    }
    return data;
}
void build()
{
    ret=1;
    for (int i=1;i<=n;i++)
    {
        blo[i]=ret;v[a[i]].push_back(i);
        if (!(i%len) && i!=n) {st[ret]=i;ret++;}
    }
    if (n%len) st[ret]=n;
    for (int i=1;i<=ret;i++)
    {
        int t1=0,t2=0;memset(cnt,0,sizeof(cnt));
        for (int j=(i-1)*len+1;j<=n;j++)
        {
            cnt[a[j]]++;
            if (cnt[a[j]]==1) t2++;
            else if (cnt[a[j]]&1) t1--,t2++;
            else t1++,t2--;
            if (!(j%len)) f[i][j/len]=t1;
        }
        if (n%len) f[i][ret]=t1;
    }
}
int find_(int l,int r,int x)
{
    int ans1=-1,ans2=-1,left,right;
    left=0;right=v[x].size()-1;
    while (left<=right)
    {
        int mid=(left+right)>>1;
        if (v[x][mid]>=l) {ans1=mid;right=mid-1;}
        else left=mid+1;
    }
    left=0;right=v[x].size()-1;
    while (left<=right)
    {
        int mid=(left+right)>>1;
        if (v[x][mid]<=r) {ans2=mid;left=mid+1;}
        else right=mid-1;
    }
    if ((ans1==-1) || (ans2==-1)) return 0;
    return ans2-ans1+1;
}
int ask(int l,int r)
{
    times++;
    if (blo[l]+1>blo[r]-1)
    {
        int t1=0,t2=0;
        for (int i=l;i<=r;i++)
        {
            if (flag[a[i]]!=times) {flag[a[i]]=times;t2++;base[a[i]]=cnt[a[i]];cnt[a[i]]++;}
            else
            {
                cnt[a[i]]++;
                if ((cnt[a[i]]-base[a[i]])&1) t1--,t2++;
                else t1++,t2--;
            }
        }
        return t1;
    }
    else
    {
        top=0;l1=l;r1=r;l2=st[blo[l]]+1;r2=st[blo[r]-1];int ret=f[blo[l]+1][blo[r]-1];
        for (int i=l;i<=st[blo[l]];i++) s[++top]=a[i];
        for (int i=st[blo[r]-1]+1;i<=r;i++) s[++top]=a[i];
        sort(s+1,s+top+1);top=unique(s+1,s+top+1)-s-1;
        for (int i=1;i<=top;i++)
            s1[i]=find_(l1,r1,s[i]),s2[i]=find_(l2,r2,s[i]);
        for (int i=1;i<=top;i++)
        {
            if (!s2[i])
            {
                if (s1[i]&1) continue;
                ret++;
            }
            else
            {
                if ((s2[i]&1) && (!(s1[i]&1))) ret++;
                if ((!(s2[i]&1)) && (s1[i]&1)) ret--;
            }
        }
        return ret;
    }
}
int main()
{
    n=read();c=read();m=read();len=80;
    for (int i=1;i<=n;i++) a[i]=read();
    build();
    for (int i=1;i<=m;i++)
    {
        l=read();r=read();
        l=(l+lastans)%n+1;r=(r+lastans)%n+1;
        if (l>r) swap(l,r);
        lastans=ask(l,r);printf("%d
",lastans);
    }
    return 0;
}