AcWing 799. 最长连续不重复子序列

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
using namespace std;
#define ll long long
#define mod 1000000009
#define endl "\n"
#define PII pair

// 高精度乘高精 C = A + B, A >= 0, B >= 0
vector Hadd(vector &A, vector &B)
{
	if (A.size() = 0
vector Hsub(vector &A, vector &B)
{
	vector C;
	for (int i = 0, t = 0; i = 0
vector Hmul(vector &A, int b)
{
	vector C;
	
	int t = 0;
	for (int i = 0; i = 0, b > 0
vector Hdiv(vector &A, int b, int &r)
{
	vector 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 vis;

int main()
{
	cin>>n;
	for(int i = 1;i >a[i];
	}
	int l = 1,r = 1;
	int ans = 1;
	while(l