acdream 1427 Nice Sequence

Nice Sequence

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

SubmitStatisticNext Problem

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

题解及代码：

      比赛的时候是队友写的这道题，由于查询的时候要进行优化，所以给他加了一个线段树维护区间最小值就AC了。

      今天看了一下这道题，感觉也不难。它的定义有些难懂啊。事实上就是当我们从左向右输入到一个元素x时，要保证其左側元素中1-（x-1）出现的次数大于等于 x的出现次数-k，找出这种最长的前缀。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <set>
#include <map>
#include <queue>
#include <string>
#define maxn 200010
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
using namespace std;
typedef long long  ll;
int a[200005];

struct segment
{
    int l,r;
    int value;
} son[maxn<<2];


void PushUp(int rt)
{
    son[rt].value=min(son[rt<<1].value,son[rt<<1|1].value);
}


void Build(int l,int r,int rt)
{
    son[rt].l=l;
    son[rt].r=r;
    if(l==r)
    {
        son[rt].value=0;
        return;
    }
    int m=(l+r)/2;
    Build(lson);
    Build(rson);
    PushUp(rt);
}


void Update(int p,int rt)
{
    if(son[rt].l==son[rt].r)
    {
        son[rt].value++;
        return;
    }

    int m=(son[rt].l+son[rt].r)/2;
    if(p<=m)
        Update(p,rt<<1);
    else
        Update(p,rt<<1|1);

    PushUp(rt);
}


int  Query(int l,int r,int rt)
{
    if(son[rt].l==l&&son[rt].r==r)
    {
        return son[rt].value;
    }

    int ret=0;
    int m=(son[rt].l+son[rt].r)/2;

    if(r<=m)
        ret=Query(l,r,rt<<1);
    else if(l>m)
        ret=Query(l,r,rt<<1|1);
    else
    {
        ret=Query(lson);
        ret=min(ret,Query(rson));
    }
    return ret;
}



int main()
{
    int n,k,x,ans=0;
    bool flag=false;
    scanf("%d%d",&n,&k);
    Build(1,n+1,1);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&x);
        if(!flag)
        {
            x++;
            a[x]++;
            Update(x,1);
            if(Query(1,x,1)<a[x]-k)
            flag=true;

            if(!flag) ans=i;
        }
    }
    printf("%d
",ans);
    return 0;
}