exam9.3

#	用  户  名	公园	计划	抽卡	总分
19	859乔屹	100 03:15:05	40 03:14:01		140 03:15:05

emm

怎么讲

　　T2我把自己优化掉了40分

优化前:

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#define cin(a)	scanf("%d",&a)
using namespace std;
const int maxn=1e5+5;
int dui[maxn],siz,now,last;
int a[maxn];
int m,n,qu;
int ri[maxn];
int len[maxn];
int sum[maxn];
long long ans[maxn];
struct node{
	int l,r,k;
}t[maxn];
__attribute((always_inline))int h233(const node &a,const node &b)
{return a.r==b.r?a.l<b.l:a.r<b.r;}
int main()
{
//	freopen("ans.in","r",stdin);
//	freopen("a.out","w",stdout);
	cin(n),cin(m),cin(qu);
	for(int q=1;q<=n;q++)
		cin>>a[q];
	for(int q=1;q<=n;q++)
	{
		while(siz<m&&now<n)
		{
			++now;
			if(!dui[a[now]])
				++siz;
			++dui[a[now]];
		}
		if(siz>=m)
			ri[q]=now;
		else
		{
			last=q;
			break;
		}
		dui[a[q]]--;
		if(!dui[a[q]])
			--siz;
	}
	/*for(int q=1;q<=last;q++)
		cout<<ri[q]<<" ";
	cout<<endl;*/
	/*for(int q=1;q<=qu;q++)
	{
		cin(t[q].l),cin(t[q].r),t[q].k=q;
	}
	sort(t+1,t+qu+1,h233);*/
	/*for(int q=1;q<=qu;q++)
		cout<<t[q].l<<" "<<t[q].r<<endl;*/
	for(int q=1,x,y;q<=qu;q++)
	{
		cin(x),cin(y);
		long long ans=0;
		for(int w=x;w<=y;w++)
		{
			//cout<<w<<" "<<y<<" "<<(ri[w]+y-2*w)*(y-ri[w]+1)/2<<endl;
			ans+=max(0,(ri[w]+y-2*w)*(y-ri[w]+1)/2);
		}
		cout<<ans<<endl;
	}
	/*
	int tmp=0;
	for(int q=1;q<=qu;q++)
	{
		if(tmp==t[q].r)
			ans[t[q].k]=sum[t[q].l];
		else
		{
			//cout<<t[q].l<<" "<<t[q].r<<endl;
			sum[t[q].r+1]=0;
			for(int w=t[q].r;w>=t[q].l;w--)
				sum[w]=sum[w+1]+max(0,(ri[w]+t[q].r-2*w)*(t[q].r-ri[w]+1)/2);
			tmp=t[q].r;
			ans[t[q].k]=sum[t[q].l];
			for(int w=t[q].l;w<=t[q].r;w++)
				cout<<sum[w]<<" ";
			cout<<endl<<endl;
		}
	}*/
	/*for(int q=1;q<=qu;q++)
		cout<<ans[q]<<endl;*/
}

　显然是$Theta (n+qn)$的

　$nleq10000$,$qleq10000$

　时限2s

　可以过的对吧

　然而我忘了时限是2s

　强行玄学优化:

　

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#define cin(a)	scanf("%d",&a)
using namespace std;
const int maxn=1e5+5;
int dui[maxn],siz,now,last;
int a[maxn];
int m,n,qu;
int ri[maxn];
int len[maxn];
int sum[maxn];
long long ans[maxn];
struct node{
	int l,r,k;
}t[maxn];
__attribute((always_inline))int h233(const node &a,const node &b)
{return a.r==b.r?a.l<b.l:a.r<b.r;}
int main()
{
//	freopen("ans.in","r",stdin);
//	freopen("a.out","w",stdout);
	cin(n),cin(m),cin(qu);
	for(int q=1;q<=n;q++)
		cin>>a[q];
	for(int q=1;q<=n;q++)
	{
		while(siz<m&&now<n)
		{
			++now;
			if(!dui[a[now]])
				++siz;
			++dui[a[now]];
		}
		if(siz>=m)
			ri[q]=now;
		else
		{
			last=q;
			break;
		}
		dui[a[q]]--;
		if(!dui[a[q]])
			--siz;
	}
	/*for(int q=1;q<=last;q++)
		cout<<ri[q]<<" ";
	cout<<endl;*/
	for(int q=1;q<=qu;q++)
	{
		cin(t[q].l),cin(t[q].r),t[q].k=q;
	}
	sort(t+1,t+qu+1,h233);
	/*for(int q=1;q<=qu;q++)
		cout<<t[q].l<<" "<<t[q].r<<endl;*/
	/*for(int q=1,x,y;q<=qu;q++)
	{
		cin(x),cin(y);
		long long ans=0;
		for(int w=x;w<=y;w++)
		{
			//cout<<w<<" "<<y<<" "<<(ri[w]+y-2*w)*(y-ri[w]+1)/2<<endl;
			ans+=max(0,(ri[w]+y-2*w)*(y-ri[w]+1)/2);
		}
		cout<<ans<<endl;
	}*/
	int tmp=0;
	for(int q=1;q<=qu;q++)
	{
		if(tmp==t[q].r)
			ans[t[q].k]=sum[t[q].l];
		else
		{
			//cout<<t[q].l<<" "<<t[q].r<<endl;
			sum[t[q].r+1]=0;
			for(int w=t[q].r;w>=t[q].l;w--)
				sum[w]=sum[w+1]+max(0,(ri[w]+t[q].r-2*w)*(t[q].r-ri[w]+1)/2);
			tmp=t[q].r;
			ans[t[q].k]=sum[t[q].l];
			/*for(int w=t[q].l;w<=t[q].r;w++)
				cout<<sum[w]<<" ";
			cout<<endl<<endl;*/
		}
	}
	for(int q=1;q<=qu;q++)
		cout<<ans[q]<<endl;
}

　于是WA40了

　以此铭记:

在打题之前,我们一定要看清时限

　以上