ARC128D

考虑我们直接$dp$。
那么需要快速的求出一段是否可以被消掉只剩两端。
我们可以考虑反过来做的。
我们知道如果全为$abab$型或者$aa$型则无法消掉
那么我们要前缀和，以及遇到$aa$则另起一套，否则我们记录最后一段$abab$的$f$和。

#include <bits/stdc++.h>

#define in read()
#define fi first
#define se second
#define pb push_back
#define rep(i, x, y) for(int i = (x); i <= (y); i++)
#define per(i, x, y) for(int i = (x); i >= (y); i--)

using namespace std;

using pii = pair < int , int >;
using vec = vector < int >;
using veg = vector < pii >;
using ll = long long;

int read() {
    int x = 0; bool f = 0; char ch = getchar(); while(!isdigit(ch)) f |= ch == '-',ch = getchar();
    while(isdigit(ch)) x = x * 10 + (ch ^ 48),ch = getchar(); return f ? -x : x;
}

const int N = 2e5 + 10;
const int mod = 998244353;

int a[N], s[N], n, f[N];

bool check(int l, int r) {
	if(r > l + 1 && a[l] == a[l + 1]) return 1;
	if(r > l + 1 && a[r] == a[r - 1]) return 1;
	return 0;
}

int pos[N];

int main() {
#ifndef ONLINE_JUDGE
    freopen("1.in","r",stdin);
#endif
    n = in; rep(i, 1, n) a[i] = in;
	int l = 1, ts = 0;
	rep(i, 1, n) {
		if(i == 1) f[i] = 1;
		if(a[i] == a[i - 1]) { l = i, ts = 0; f[i] = f[i - 1]; s[i] = (s[i - 1] + f[i]) % mod; continue; }
		f[i] = (f[i] + s[i - 1] - s[l - 1]) % mod;
		if(i - 3 >= l && a[i] == a[i - 2] && a[i - 1] == a[i - 3]) ts = (ts + f[i - 3]) % mod, f[i] = (f[i] + mod - ts) % mod; else ts = 0;
		f[i] = (f[i] + mod) % mod;
		s[i] = (s[i - 1] + f[i]) % mod;
		cerr << i << " " << l << " " << f[i] << endl;
	}
	printf("%d\n", f[n]); return 0;
}