考虑分类计数，讨论“没有 DD”、“有 DD 无 o”、“有 DDo 无 S”三种情况。
- 没有
DD，枚举有几种大写字母出现过； - 剩下两种情况，考虑设 \(f_{i,0/1}\) 分别表示两种情况的方案数。\(f_{i,0}\) 可以从 \(f_{i-1,0}\) 填大写字母转移，也可以枚举第一个出现两次的字母；\(f_{i,1}\) 可以从 \(f_{i-1,0}\) 和 \(f_{i-1,1}\) 填小写字母转移。
时间复杂度 \(O(52^2n)\)，精细实现可过。
code
// Problem: F - Anti-DDoS
// Contest: AtCoder - パナソニックグループプログラミングコンテスト2023（AtCoder Beginner Contest 301）
// URL: https://atcoder.jp/contests/abc301/tasks/abc301_f
// Memory Limit: 1024 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)
#include <bits/stdc++.h>
#define pb emplace_back
#define fst first
#define scd second
#define mems(a, x) memset((a), (x), sizeof(a))
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef double db;
typedef long double ldb;
typedef pair<ll, ll> pii;
const int maxn = 300100;
const ll mod = 998244353;
inline ll qpow(ll b, ll p) {
ll res = 1;
while (p) {
if (p & 1) {
res = res * b % mod;
}
b = b * b % mod;
p >>= 1;
}
return res;
}
ll n, fac[maxn], ifac[maxn], f[maxn][2], pw[maxn], c[maxn][30], g[maxn][30];
ll h1[30][30], h2[30][30];
bool vis[128];
char s[maxn];
inline ll C(ll n, ll m) {
if (n < m || n < 0 || m < 0) {
return 0;
} else {
return fac[n] * ifac[m] % mod * ifac[n - m] % mod;
}
}
void solve() {
scanf("%s", s + 1);
n = strlen(s + 1);
fac[0] = 1;
for (int i = 1; i <= n + 50; ++i) {
fac[i] = fac[i - 1] * i % mod;
}
ifac[n + 50] = qpow(fac[n + 50], mod - 2);
for (int i = n + 49; ~i; --i) {
ifac[i] = ifac[i + 1] * (i + 1) % mod;
}
pw[0] = 1;
for (int i = 1; i <= n; ++i) {
pw[i] = pw[i - 1] * 26 % mod;
}
for (int i = 0; i <= n + 50; ++i) {
for (int j = 0; j <= min(i, 26); ++j) {
c[i][j] = C(i, j);
g[i][j] = c[i][j] * pw[i - j] % mod;
}
}
for (int i = 0; i <= 26; ++i) {
for (int j = 0; j <= 26; ++j) {
h1[i][j] = c[i][j] * fac[j] % mod;
if (j) {
h2[i][j] = c[i][j - 1] * fac[j - 1] % mod * j % mod;
}
}
}
bool flag = 0;
int cnt = 0, d = 0, T = 0;
for (int i = 1; i <= n; ++i) {
for (int j = 0; j < 52; ++j) {
if (s[i] != '?') {
if (isupper(s[i]) && s[i] != 'A' + j) {
continue;
} else if (islower(s[i]) && s[i] != 'a' + (j - 26)) {
continue;
}
}
if (j < 26 && !flag) {
if (vis['A' + j]) {
int t = 26 - d;
for (int k = 0; k <= min(26, cnt); ++k) {
f[i][0] = (f[i][0] + g[cnt][k] * h1[t][k]) % mod;
}
} else if (cnt) {
int t = 25 - d;
for (int k = 1; k <= min(26, cnt); ++k) {
f[i][0] = (f[i][0] + g[cnt][k] * h2[t][k]) % mod;
}
}
}
if (j < 26) {
f[i][0] = (f[i][0] + f[i - 1][0]) % mod;
}
if (j >= 26) {
f[i][1] = (f[i][1] + f[i - 1][0] + f[i - 1][1]) % mod;
}
}
if (isupper(s[i]) && vis[s[i]]) {
flag = 1;
}
if (isupper(s[i])) {
vis[s[i]] = 1;
++d;
}
cnt += (s[i] == '?');
}
ll ans = (f[n][0] + f[n][1]) % mod;
if (!flag) {
int k = 26 - d;
for (int i = 0; i <= min(k, cnt); ++i) {
ans = (ans + C(cnt, i) * pw[cnt - i] % mod * C(k, i) % mod * fac[i] % mod) % mod;
}
}
printf("%lld\n", ans);
}
int main() {
int T = 1;
// scanf("%d", &T);
while (T--) {
solve();
}
return 0;
}