AtCoder Beginner Contest 194 E - Mex Min

Probably the meaning of the question:

Give one not more than $1.5\times 10^{}$Array of${}^6$$A_1$To $A_i$Length in $m$ , and a$n$ , give each consecutivennThe smallest number among the smallest numbers that do not appear in$n$ numbers.

Ideas:

Don't you just slide the window as soon as you look at this question (there is a detailed explanation here)! ! ! After pasting on the other oj, I changed it and handed it in directly. As a result, I

looked at the question again and noticed that as long as it is the smallest number among the numbers that did not appear. The data is not big, nn before opening an array record$The$ number of occurrences of each of the$n$ numbers and find the smallest number among the numbers that do not appear, then the double pointer points one at the end and the other at the end, each time the$A_{Tail}$The number of occurrences of is reduced by one, $A_{Head}$Add one to the number of occurrences to determine the $A_{Tail}$Whether the number of occurrences is 0, if it is 0, compare it with the previously recorded minimum value. If the current minimum value that has not appeared is smaller than the minimum value that has not appeared before, the minimum value that has not appeared before must be in the queue. in.
Let’s put it this way, let’s take a more intuitive example of my hand

Input

7 3
2 0 1 2 0 1 0

Output

2

So the output answer is 2, which is intuitive enough, and because $A_i$No more than $1.5\times 10^6$ , and it is more than enough to open such a large array.

Code

#include <bits/stdc++.h>
using namespace std;
const int N=2e6;
int num[N],a[N],n,k,ans=9999999;
bool flag;
void work()
{
    
    
	int head=k,tail=1;
	for(int i=tail;i<=head;i++)
	{
    
    
		num[a[i]]++;
	}
	for(int i=0;;i++)
	{
    
    	
		if(num[i]==0)
		{
    
    
			ans=min(ans,i);
			if(ans==0)
			{
    
    
				cout<<0;
				flag=1;
				return;
			}
			break;
		}
	}	
	while(head<n)
	{
    
    
		num[a[tail]]--;
		tail++;
		num[a[head]]++;
		if(num[a[tail]]==0)
		{
    
    
			ans=min(ans,a[tail]);
		}
		head++;
	}
}
int main()
{
    
    
	scanf("%d%d",&n,&k);
	for(int i=1;i<=n;i++)
		scanf("%d",&a[i]);
	work();
	if(flag==0)
		cout<<ans;
	return 0;
}

Then WA had 5 groups of dead and alive and couldn't change it. After playing the game, I sent it to the seniors of the second grade to take a look. It was found that the seniors of the second and third grades used the tree array, the chairman tree and the line segment tree water respectively. Here is the code for the tree array:
I can’t understand it, so I won’t explain it

#include<bits/stdc++.h>
#define N 1600006
#define LL long long 
#define LB long double
using namespace std;

int n,m;
int ans=1e8,a[N],sp=0;
int mp[N],tr[N];

inline int qr()
{
    
    
	char a=0;int w=1,x=0;
	while(a<'0'||a>'9'){
    
    if(a=='-')w=-1;a=getchar();}
	while(a<='9'&&a>='0'){
    
    x=(x<<3)+(x<<1)+(a^48);a=getchar();}
	return x*w;
}

const int K=1500002;

inline void add(int x,int k)
{
    
    
	for(register int i=x;i<=K;i+=i&-i) tr[i]+=k;
}

inline int que(int x)
{
    
    
	int res=0;
	for(register int i=x;i;i-=i&-i) res+=tr[i];
	return res;
}

int jac[123];
inline int ask()
{
    
    
	int pos=1;
	for(register int i=21;i>=0;i--)
		if(pos+jac[i]-1<=K)
		{
    
    
			int op=que(pos+jac[i]-1);
			if(op==pos+jac[i]-1)
				pos+=jac[i];
		}
	return pos;
}

int main()
{
    
    
	n=qr();
	m=qr();
	jac[0]=1;
	for(register int i=1;i<=21;i++) jac[i]=jac[i-1]<<1;
	for(register int i=1;i<=n;i++) a[i]=qr()+1;
	for(register int i=1;i<=n;i++)
	{
    
    
		mp[a[i]]++;
		if(mp[a[i]]==1) add(a[i],1);
		if(i>=m)
		{
    
    
			ans=min(ans,ask());
			mp[a[i-m+1]]--;
			if(!mp[a[i-m+1]]) add(a[i-m+1],-1);
		}
	}
	printf("%d\n",ans-1);
	return 0;
}

Written by this god

Then my code was simplified a bit after I changed it

#include <bits/stdc++.h>
using namespace std;
const int N=2e6;
int num[N],a[N],n,k,ans=9999999;
bool flag;
void work()
{
    
    
	int tou=1,wei=k;
	for(int i=1;i<=k;i++)num[a[i]]++;
	for(int i=0;;i++)
	if(num[i]==0){
    
    ans=i;break;}
	if(ans==0)return;
	while(wei<n){
    
    
		num[a[++wei]]++;
		num[a[tou]]--;
		if(num[a[tou]]==0)
		ans=min(ans,a[tou]);
		tou++;
	}
}
int main()
{
    
    
	scanf("%d%d",&n,&k);
	for(int i=1;i<=n;i++)
		scanf("%d",&a[i]);
	work();
	cout<<ans;
	return 0;
}

In fact, the main reason is that my tail end point moved early, so I should first judge whether $a_{Tail}$Whether the number of occurrences is $0$ then let the tail move.
Well, that's it. This is the easiest E question in AtCOder's ABC, but the mistake in the exam made me miss 500 points. Gan, this reminds me ofthe T1 of theprevious simulation game.Iwas going to write memoization, but the reverse of x and y caused the array to cross the boundary. The maximum result of the legal situation in the problem, I found the maximum process in the legal situation the amount.