题解

#include <cstdio>
#include <cstring>
#define int long long 
const int mod = 998244353;
const int N = 200010;

int fpow(int a, int b) {
    int res = 1;
    while (b) {
        if (b & 1) res = res * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return res;
}

int fac[2 * N + 10], inv[2 * N + 10];

void init(int n) {
    fac[0] = 1;
    for (int i = 1; i <= n; i++) fac[i] = fac[i - 1] * i % mod;
    inv[n] = fpow(fac[n], mod - 2);
    for (int i = n - 1; i >= 0; i--) inv[i] = inv[i + 1] * (i + 1) % mod;
}

int AA(int a, int b) {
    int res = 1;
    for (int i = a; i >= 1; i -= 2) {
        res = res * i % mod;
    }
    return res;
}

int C(int a, int b) {
    if (a < b) return 0;
    return fac[a] * inv[b] % mod * inv[a - b] % mod;
}

int a[N];

signed main() {
    int n;
    scanf("%lld", &n);
    init(2 * N);
    scanf("%lld", &a[1]);
    int l = 1, r = 1;
    int A = 0, B = 0;
    int ans = 1;
    for (int i = 2; i <= n; i++) {
        scanf("%lld", &a[i]);
        if (a[i] == a[i - 1]) {
            r = i;
        } else {
            if ((r - l) % 2) {
                printf("0");
                return 0;
            }
            if (r - l >= 2) {
                int numa = r - l - 1;
                int numb = (r - l + 1) / 2;
                ans = ans * AA(numa, numb) % mod * C(numb + B, B) % mod;
                A += r - l - 1;
                B += (r - l + 1) / 2;
            }
            l = i, r = i;
        }
    }
    if ((r - l) % 2) {
        printf("0");
        return 0;
    }
    if (r - l >= 2) {
        int numa = r - l - 1;
        int numb = (r - l + 1) / 2;
        ans = ans * AA(numa, numb) % mod * C(numb + B, B) % mod;
    }
    printf("%lld", ans);
    return 0;
}