【bzoj2724】【蒲公英】【分块】

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Description

Input

修正一下

l = (l_0 + x - 1) mod n + 1, r = (r_0 + x - 1) mod n + 1

Output

Sample Input

6 3
1 2 3 2 1 2
1 5
3 6
1 5

Sample Output

1
2
1

HINT

修正下：

n <= 40000, m <= 50000

题解：

区间众数.

对序列分块,t[i][j]表示第i块到第j块的众数.

把序列按权值和位置排序之后就可以二分查找一个数在一段区间内的出现次数.

所以块之外的部分暴力二分查找即可.

代码：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath> #define N 40010#define M 210using namespace std;int a[N],bl[N],n,m,bk,x,y,cnt,ans,h[N],l[N],r[N],sum[N],t[M][M];struct block{int l,r;}st[M];struct xu{int x,p;}q[N];void getbk(){  bk=sqrt(n);  if (n%bk) cnt=n/bk+1;else cnt=n/bk;  for (int i=1;i<=n;i++) bl[i]=(i-1)/bk+1;  for (int i=1;i<=cnt;i++)     st[i].l=(i-1)*bk+1,st[i].r=i*bk;  }int find(int x){  int l=1,r=n;  while(l<=r){    int mid=(l+r)>>1;    if (x<h[mid]) r=mid-1;    else if (x>h[mid]) l=mid+1;    else return mid;  }}bool cmp(xu a,xu b){  if (a.x==b.x) return a.p<b.p;  else return a.x<b.x;}void pre(){  for (int i=1;i<=n;i++)     q[i].x=a[i],q[i].p=i;  sort(q+1,q+n+1,cmp);  for (int i=1;i<=n;i++){    if (!l[q[i].x]) l[q[i].x]=i;    r[q[i].x]=i;  }  int mx(0),v;  for (int i=1;i<=cnt;i++){    memset(sum,0,sizeof(sum));mx=0;    for (int j=st[i].l;j<=n;j++){       sum[a[j]]++;       if (sum[a[j]]>mx||(sum[a[j]]==mx)&&a[j]<v){          mx=sum[a[j]];v=a[j];       }       t[i][bl[j]]=v;    }         }}int up(int k,int x,int y){   int l=x,r=y,ans;   while(l<=r){     int mid=(l+r)>>1;     if (q[mid].p>k) r=mid-1;     else {ans=mid;l=mid+1;}   }   return ans;}int down(int k,int x,int y){  int l=x,r=y,ans;  while (l<=r){    int mid=(l+r)>>1;    if (q[mid].p<k) l=mid+1;    else {ans=mid;r=mid-1;}  }  return ans;}int query(int x,int y){  int mx(0),v;  if (bl[x]==bl[y]){    for (int i=x;i<=y;i++){       int temp=up(y,l[a[i]],r[a[i]])-down(x,l[a[i]],r[a[i]])+1;       if (temp>mx||(temp==mx)&&(a[i]<v)){          v=a[i];mx=temp;       }    }  }  else{    if (bl[x]+1<bl[y]){      v=t[bl[x]+1][bl[y]-1];      mx=up(y,l[v],r[v])-down(x,l[v],r[v])+1;    }    for (int i=x;i<=st[bl[x]].r;i++){       int temp=up(y,l[a[i]],r[a[i]])-down(x,l[a[i]],r[a[i]])+1;       if (temp>mx||(temp==mx)&&(a[i]<v)){          v=a[i];mx=temp;       }    }    for (int i=st[bl[y]].l;i<=y;i++){       int temp=up(y,l[a[i]],r[a[i]])-down(x,l[a[i]],r[a[i]])+1;       if (temp>mx||(temp==mx)&&(a[i]<v)){         v=a[i];mx=temp;       }    }  }  return v;}int main(){  scanf("%d%d",&n,&m);  for (int i=1;i<=n;i++) scanf("%d",&a[i]);  getbk();   for (int i=1;i<=n;i++) h[i]=a[i];  sort(h+1,h+n+1);  for (int i=1;i<=n;i++) a[i]=find(a[i]);  pre();  for (int i=1;i<=m;i++){    scanf("%d%d",&x,&y);    x=(x+ans-1)%n+1;y=(y+ans-1)%n+1;    if (x>y) swap(x,y);    ans=h[query(x,y)];    printf("%d\n",ans);  }}

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