ACdream 1427 Nice Sequence

主题链接：http://115.28.76.232/problem?

pid=1427

Nice Sequence

Time Limit: 12000/6000MS (Java/Others)Memory Limit: 128000/64000KB (Java/Others)

Problem Description

Let us consider the sequence a1, a2,..., an of
non-negative integer numbers. Denote as ci,j the number of occurrences of the number i among a1,a2,..., aj.
We call the sequence k-nice if for all i1<i2 and
for all j the following condition is satisfied: ci1,j ≥ ci2,j −k.

Given the sequence a1,a2,..., an and
the number k, find its longest prefix that is k-nice.

Input

The first line of the input file contains n and k (1 ≤ n ≤ 200 000, 0 ≤ k ≤ 200 000). The second line contains n integer numbers ranging from 0 to n.

Output

Output the greatest l such that the sequence a1, a2,..., al is
k-nice.

Sample Input

10 1

0 1 1 0 2 2 1 2 2 3

2 0

1 0

Sample Output

8

0

Source

Andrew Stankevich Contest 23

Manager

mathlover

Submit Statistic

用线段树维护 0到A[i]-1间的最小值。用F[A[i]] 统计频率。推断 0 到 A[i]-1范围内的最小值与F[A[i]]-K的大小就可以。

//#pragma comment(linker, "/STACK:36777216")

#include <functional>

#include <algorithm>

#include <iostream>

#include <fstream>

#include <sstream>

#include <iomanip>

#include <numeric>

#include <cstring>

#include <climits>

#include <cassert>

#include <complex>

#include <cstdio>

#include <string>

#include <vector>

#include <bitset>

#include <queue>

#include <stack>

#include <cmath>

#include <ctime>

#include <list>

#include <set>

#include <map>

using namespace std;

#define LL long long

#define MAXN 200100

struct Node{

    int left,right,v;

};Node Tree[MAXN<<2];

void Build(int id,int left,int right){

    Tree[id].v=0;

    Tree[id].left=left;

    Tree[id].right=right;

    if(left==right) return;

    int mid=(left+right)>>1;

    Build(id<<1,left,mid);

    Build(id<<1|1,mid+1,right);

}

void Update(int id,int pos,int add){

    int left=Tree[id].left,right=Tree[id].right;

    if(left==right){

        Tree[id].v+=add;

        return;

    }

    int mid=(left+right)>>1;

    if(mid>=pos)

        Update(id<<1,pos,add);

    else

        Update(id<<1|1,pos,add);

   Tree[id].v=min(Tree[id<<1].v,Tree[id<<1|1].v);

}

int Query(int id,int Qleft,int Qright){

    int left=Tree[id].left,right=Tree[id].right;

    if(left>=Qleft && right<=Qright)

        return Tree[id].v;

    int mid=(left+right)>>1;

    if(mid>=Qright)

        return Query(id<<1,Qleft,Qright);

    else if(mid<Qleft)

        return Query(id<<1|1,Qleft,Qright);

    int r1=Query(id<<1,Qleft,Qright),r2=Query(id<<1|1,Qleft,Qright);

    return min(r1,r2);;

}

int A[MAXN];

int F[MAXN];

int main(){

    int N,K;

    while(~scanf("%d%d",&N,&K)){

        for(int i=1;i<=N;i++)

            scanf("%d",&A[i]);

        Build(1,0,N);

        memset(F,0,sizeof(F));

        int i;

        for(i=1;i<=N;i++){

            F[A[i]]++;

            Update(1,A[i],1);

            if(A[i]==0)    continue;

            int ans=Query(1,0,A[i]-1);

            if(ans<F[A[i]]-K) break;

        }

        printf("%d\n",i-1);

    }

}