Bzoj 2724 [Violet 6]蒲公英

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：分块＋二分＋离散化

代码：
点击这里

/** @Author: Heristor* @Date:   2016-07-02 11:00:38* @Last Modified by:   Heristor* @Last Modified time: 2016-07-04 16:05:56*/#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <string>#include <cmath>#include <algorithm>#include <queue>#include <map>#include <set>#include <ctime>#include <vector>using namespace std;#define rep(i,l,r) for(i=l;i<=r;i++)#define ser(i,r,l) for(i=r;i>=l;i--)#define INF 40005#define inf 1000000007typedef long long ll;priority_queue<int >QwQ;int n,m,size,tot,cnt=0,Ans=0,tmp=0;int a[INF],b[INF],d[INF],num[INF],sum[INF],top[INF],last[INF],q[INF],l[INF],r[INF];int t[205][205],f[205][205];int read(){    int k=0,f=1;    char ch;    while(ch<'0' || ch>'9'){        if(ch=='-')f=-1;        ch=getchar();    }    while(ch>='0' && ch<='9')k=(k<<1)+(k<<3)+ch-'0',ch=getchar();    return k*f;}int getnum(int x,int y,int z){    return upper_bound(d+top[z],d+last[z]+1,y)-lower_bound(d+top[z],d+last[z]+1,x);}void solve(int x,int y){    int i,j,k;    tmp=Ans=0;    rep(i,x,y)num[a[i]]=0;    rep(i,x,y){        num[a[i]]++;        if(num[a[i]]>Ans){            Ans=num[a[i]];            tmp=a[i];        }        else if(num[a[i]]==Ans){            tmp=min(tmp,a[i]);        }    }}void init(){    int i,j,k,p,x,y,tx,ty;    n=read(),m=read();    rep(i,1,n)a[i]=read(),b[i]=a[i];    sort(b+1,b+n+1);    k=unique(b+1,b+n+1)-b-1;//离散化去重的时候一定要下标减1    rep(i,1,n)a[i]=lower_bound(b+1,b+k+1,a[i])-b;    rep(i,1,n)num[a[i]]++;    rep(i,1,k)sum[i]=sum[i-1]+num[i];    rep(i,1,k)top[i]=sum[i-1]+1,last[i]=sum[i-1];    rep(i,1,n)d[++last[a[i]]]=i;    size=sqrt(n),tot=n/size;    rep(i,1,tot){        l[i]=r[i-1]+1;        r[i]=l[i]+size-1;    }    if(n%size){        tot++;        l[tot]=r[tot-1]+1;        r[tot]=n;    }    rep(i,1,tot){        tmp=Ans=0;        rep(j,1,k)num[j]=0;        rep(j,i,tot){            rep(p,l[j],r[j]){                num[a[p]]++;                if(num[a[p]]>Ans){                    Ans=num[a[p]];                    tmp=a[p];                }                else if(num[a[p]]==Ans){                    tmp=min(tmp,a[p]);                }            }            t[i][j]=Ans;            f[i][j]=tmp;        }    }    Ans=tmp=0;    while(m--){        cnt=0;        x=read(),y=read();        x=(x+tmp-1)%n+1,y=(y+tmp-1)%n+1;        if(x>y)swap(x,y);        tx=x/size;        ty=y/size;        if(x%size)tx++;        if(y%size)ty++;        if(tx==ty || tx+1==ty){            solve(x,y);            tmp=b[tmp];            printf("%d\n", tmp);            continue;        }        rep(i,x,r[tx])num[a[i]]=0;        rep(i,l[ty],y)num[a[i]]=0;        rep(i,x,r[tx]){            if(!num[a[i]])q[++cnt]=a[i];            num[a[i]]++;        }        rep(i,l[ty],y){            if(!num[a[i]])q[++cnt]=a[i];            num[a[i]]++;        }        tx++,ty--;        Ans=t[tx][ty];        tmp=f[tx][ty];        rep(i,1,cnt){            k=getnum(l[tx],r[ty],q[i]);            if(k+num[q[i]]>Ans){                Ans=k+num[q[i]];                tmp=q[i];            }            else if(k+num[q[i]]==Ans){                tmp=min(tmp,q[i]);            }        }        tmp=b[tmp];//离散化后记得变回原来的权值        printf("%d\n", tmp);    }}void work(){    int i,j,k;}int main(){    freopen("Bzoj2724.in","r",stdin);    freopen("Bzoj2724.out","w",stdout);    init();    work();    return 0;}