AcWing 799. 最长连续不重复子序列

Posted 2022-02-09 MangataTS

题目连接

https://www.acwing.com/problem/content/801/

思路

用一个map或者数组vis标记我们以及存在的元素，然后定义两个指针，如果当前这个元素没被访问过，那么直接让r指针往后移动，然后去一个最大值，然后移动l指针到a[r]这个值相同的位置即可

代码

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define mod 1000000009
#define endl "\\n"
#define PII pair<int,int>

// 高精度乘高精 C = A + B, A >= 0, B >= 0
vector<int> Hadd(vector<int> &A, vector<int> &B)

	if (A.size() < B.size()) return Hadd(B, A);
	
	vector<int> C;
	int t = 0;
	for (int i = 0; i < A.size(); i ++ )
	
		t += A[i];
		if (i < B.size()) t += B[i];
		C.push_back(t % 10);
		t /= 10;
	
	
	if (t) C.push_back(t);
	return C;

// 高精减高精 C = A - B, 满足A >= B, A >= 0, B >= 0
vector<int> Hsub(vector<int> &A, vector<int> &B)

	vector<int> C;
	for (int i = 0, t = 0; i < A.size(); i ++ )
	
		t = A[i] - t;
		if (i < B.size()) t -= B[i];
		C.push_back((t + 10) % 10);
		if (t < 0) t = 1;
		else t = 0;
	
	
	while (C.size() > 1 && C.back() == 0) C.pop_back();
	return C;

// 高精乘第精 C = A * b, A >= 0, b >= 0
vector<int> Hmul(vector<int> &A, int b)

	vector<int> C;
	
	int t = 0;
	for (int i = 0; i < A.size() || t; i ++ )
	
		if (i < A.size()) t += A[i] * b;
		C.push_back(t % 10);
		t /= 10;
	
	
	while (C.size() > 1 && C.back() == 0) C.pop_back();
	
	return C;

// 高精除低精 A / b = C ... r, A >= 0, b > 0
vector<int> Hdiv(vector<int> &A, int b, int &r)

	vector<int> C;
	r = 0;
	for (int i = A.size() - 1; i >= 0; i -- )
	
		r = r * 10 + A[i];
		C.push_back(r / b);
		r %= b;
	
	reverse(C.begin(), C.end());
	while (C.size() > 1 && C.back() == 0) C.pop_back();
	return C;

int dx[4]=0,-1,0,1,dy[4]=-1,0,1,0;

ll ksm(ll a,ll b) 
	ll ans = 1;
	for(;b;b>>=1LL) 
		if(b & 1) ans = ans * a % mod;
		a = a * a % mod;
	
	return ans;


ll lowbit(ll x)return -x & x;

const int N = 2e6+10;
int n,a[N];
map<int,int> vis;

int main()

	cin>>n;
	for(int i = 1;i <= n; ++i) 
		cin>>a[i];
	
	int l = 1,r = 1;
	int ans = 1;
	while(l <= n && r <= n) 
		while(r <= n && vis[a[r]] == 0) 
			vis[a[r++]]++;
		ans = max(ans,r-l);
		while(l <= r && vis[a[r]] != 0) 
			vis[a[l++]]--;
		
	
	cout<<ans<<endl;
	
	return 0;