标签：des style http color io os ar java for
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.
10 1 0 1 1 0 2 2 1 2 2 3 2 0 1 0
8 0
题解及代码：
比赛的时候是队友写的这道题，因为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，其实就是当我们从左向右输入到一个元素x时，要保证其左侧元素中1-（x-1）出现的次数大于等于 x的出现次数-k，找出这样的最长的前缀。
#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\n",ans);
return 0;
}
标签：des style http color io os ar java for
原文地址：http://blog.csdn.net/knight_kaka/article/details/39852833
