Nice sequencetime limit: 12000/6000 ms (Java/others) memory limit: 128000/64000 KB (Java/others) submitstatisticnext problemproblem 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 10 1 1 0 2 2 1 2 2 32 01 0

Sample output

80

Sourceandrew stankevich contest 23 managermathlover

Question and code:

This question was written by teammates during the competition. Because optimization was required during the query, a line segment tree was added to him to maintain the minimum value of the interval.

I read this question today, and it is not difficult. The definition of it is a bit difficult, in fact, when we input an element x from left to right, to ensure that its left element 1-(x-1) if the number of occurrences is greater than or equal to X, locate the longest prefix.

#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|1using 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\n",ans); return 0;}

Acdream 1427 nice sequence