题意
Sol
首先不难想到一种暴力dp，设(f[i][a][b][c])表示还有(i)轮没打，场上有(a)个1血，(b)个2血，(c)个三血
发现状态数只有(s = 166)个，复杂度为(O(ns))
矩乘优化一下复杂度为(O(s^3 logn T))，还是过不去。
因为每次询问都是独立的，那么可以预处理出(2^i)的转移矩阵，回答询问只需要拿一个行向量去乘log个矩阵
构造矩阵的时候可以加一个列向量表示期望
#include<bits/stdc++.h>
#define LL long long
using namespace std;
const int B = 60, mod = 998244353;
template <typename A, typename B> inline bool chmin(A &a, B b){if(a > b) {a = b; return 1;} return 0;}
template <typename A, typename B> inline bool chmax(A &a, B b){if(a < b) {a = b; return 1;} return 0;}
template <typename A, typename B> inline LL add(A x, B y) {if(x + y < 0) return x + y + mod; return x + y >= mod ? x + y - mod : x + y;}
template <typename A, typename B> inline void add2(A &x, B y) {if(x + y < 0) x = x + y + mod; else x = (x + y >= mod ? x + y - mod : x + y);}
LL mul(int x, int y) {return 1ll * x * y % mod;}
inline LL read() {
char c = getchar(); LL x = 0, f = 1;
while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
return x * f;
}
int fp(int a, int p) {
int base = 1;
while(p) {
if(p & 1) base = mul(base, a);
a = mul(a, a); p >>= 1;
}
return base;
}
int T, M, K;
namespace S3 {
int id[11][11][11], cnt, Lim;
int ans[168];
LL inv[11];
struct Ma {
int m[168][168];
Ma() {
memset(m, 0, sizeof(m));
}
void init() {
for(int i = 0; i <= Lim; i++) m[i][i] = 1;
}
void print() {
for(int i = 1; i <= Lim; i++, puts(""))
for(int j = 1; j <= Lim; j++)
printf("%d ", m[i][j]);
}
Ma operator * (const Ma &rhs) const {
Ma gg = {};
for(int i = 1; i <= Lim; i++)
for(int j = 1; j <= Lim; j++) {
__int128 tmp = 0;
for(int k = 1; k <= Lim; k++)
tmp += mul(m[i][k], rhs.m[k][j]);
tmp %= mod;
gg.m[i][j] = tmp;
}
return gg;
}
}f[B + 1];
void Pre() {
for(int i = 1; i <= K + 1; i++) inv[i] = fp(i, mod - 2);
for(int a = 0; a <= K; a++)
for(int b = 0; a + b <= K; b++)
for(int c = 0; a + b + c <= K; c++)
id[a][b][c] = ++cnt;
for(int a = 0; a <= K; a++)
for(int b = 0; a + b <= K; b++)
for(int c = 0; a + b + c <= K; c++) {
int down = inv[a + b + c + 1], tag = (a + b + c < K), now = id[a][b][c];
if(a) f[0].m[now][id[a - 1][b][c]] = mul(a, down);
if(b) f[0].m[now][id[a + 1][b - 1][c + tag]] = mul(b, down);
if(c) f[0].m[now][id[a][b + 1][c - 1 + tag]] = mul(c, down);
f[0].m[now][now] = down;
f[0].m[now][cnt + 1] = down;
}
f[0].m[cnt + 1][cnt + 1] = 1;
Lim = cnt + 1;
for(int i = 1; i <= B; i++) f[i] = f[i - 1] * f[i - 1];
}
int tmp[168];
void mul(Ma a) {
memset(tmp, 0, sizeof(tmp));
for(int j = 1; j <= Lim; j++)
for(int i = 1; i <= Lim; i++)
add2(tmp[j], 1ll * ans[i] * a.m[i][j] % mod);
memcpy(ans, tmp, sizeof(tmp));
}
void MatrixPow(LL p) {
for(int i = 0; p; p >>= 1, i++)
if(p & 1)
mul(f[i]);
}
void work() {
Pre();
while(T--) {
LL n = read();
memset(ans, 0, sizeof(ans)); ans[id[0][0][1]] = 1;
MatrixPow(n);
cout << ans[cnt + 1] << '
';
}
}
}
namespace S2 {
int id[11][11], cnt, Lim;
int ans[168];
LL inv[11];
struct Ma {
int m[168][168];
Ma() {
memset(m, 0, sizeof(m));
}
void init() {
for(int i = 0; i <= Lim; i++) m[i][i] = 1;
}
void print() {
for(int i = 1; i <= Lim; i++, puts(""))
for(int j = 1; j <= Lim; j++)
printf("%d ", m[i][j]);
}
Ma operator * (const Ma &rhs) const {
Ma gg = {};
for(int i = 1; i <= Lim; i++)
for(int j = 1; j <= Lim; j++) {
__int128 tmp = 0;
for(int k = 1; k <= Lim; k++)
tmp += mul(m[i][k], rhs.m[k][j]);
tmp %= mod;
gg.m[i][j] = tmp;
}
return gg;
}
}f[B + 1];
void Pre() {
for(int i = 1; i <= K + 1; i++) inv[i] = fp(i, mod - 2);
for(int a = 0; a <= K; a++)
for(int b = 0; a + b <= K; b++)
id[a][b] = ++cnt;
for(int a = 0; a <= K; a++)
for(int b = 0; a + b <= K; b++) {
int down = inv[a + b + 1], tag = (a + b < K), now = id[a][b];
if(a) f[0].m[now][id[a - 1][b]] = mul(a, down);
if(b) f[0].m[now][id[a + 1][b - 1 + tag]] = mul(b, down);
f[0].m[now][now] = down;
f[0].m[now][cnt + 1] = down;
}
f[0].m[cnt + 1][cnt + 1] = 1;
Lim = cnt + 1;
for(int i = 1; i <= B; i++) f[i] = f[i - 1] * f[i - 1];
}
int tmp[168];
void mul(Ma a) {
memset(tmp, 0, sizeof(tmp));
for(int j = 1; j <= Lim; j++)
for(int i = 1; i <= Lim; i++)
add2(tmp[j], 1ll * ans[i] * a.m[i][j] % mod);
memcpy(ans, tmp, sizeof(tmp));
}
void MatrixPow(LL p) {
for(int i = 0; p; p >>= 1, i++)
if(p & 1)
mul(f[i]);
}
void work() {
Pre();
while(T--) {
LL n = read();
memset(ans, 0, sizeof(ans)); ans[id[0][1]] = 1;
MatrixPow(n);
cout << ans[cnt + 1] << '
';
}
}
}
namespace S1 {
int N, f[12][9][9][9];
int inv(int a) {
return fp(a, mod - 2);
}
void work() {
N = 11;
for(int i = 1; i <= N; i++) {
for(int a = 0; a <= K; a++) {
for(int b = 0; a + b <= K; b++) {
for(int c = 0; a + b + c <= K; c++) {
int down = a + b + c + 1;
if(a) add2(f[i][a][b][c], mul(mul(a, inv(down)), f[i - 1][a - 1][b][c]));
if(b) {
if(down <= K) add2(f[i][a][b][c], mul(mul(b, inv(down)), f[i - 1][a + 1][b - 1 + (M == 2)][c + (M == 3)]));
else add2(f[i][a][b][c], mul(mul(b, inv(down)), f[i - 1][a + 1][b - 1][c]));
}
if(c) {
if(down <= K) add2(f[i][a][b][c], mul(mul(c, inv(down)), f[i - 1][a][b + 1 + (M == 2)][c - 1 + (M == 3)]));
else add2(f[i][a][b][c], mul(mul(c, inv(down)), f[i - 1][a][b + 1][c - 1]));
}
add2(f[i][a][b][c], mul(inv(down), f[i - 1][a][b][c] + 1));
}
}
}
}
while(T--) {
int n = read();
printf("%d
", f[n][M == 1][M == 2][M == 3]);
}
}
}
int main() {
T = read(); M = read(); K = read();
if(M == 1) S1::work();
else if(M == 2) S2::work();
else S3::work();
return 0;
}