結果

問題 No.2529 Treasure Hunter
ユーザー syndro_6
提出日時 2023-11-03 22:11:09
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 52 ms / 2,000 ms
コード長 4,717 bytes
コンパイル時間 1,988 ms
コンパイル使用メモリ 199,904 KB
最終ジャッジ日時 2025-02-17 18:24:59
ジャッジサーバーID
(参考情報)
judge3 / judge5
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ファイルパターン 結果
sample AC * 1
other AC * 22
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

// (⁠◕⁠ᴗ⁠◕⁠✿⁠)
#include <bits/stdc++.h>
#define rep(i, n) for (ll i = 0; i < (n); i++)
#define srep(i, s, n) for (ll i = s; i < (n); i++)
#define len(x) ((int)(x).size())
#define all(x) (x).begin(), (x).end()
using namespace std;
template<typename T> using vc = vector<T>;
template<typename T> using vv = vc<vc<T>>;
template<typename T> using vvv = vv<vc<T>>;
using vi = vc<int>;using vvi = vv<int>; using vvvi = vv<vi>;
using ll = long long;using vl = vc<ll>;using vvl = vv<ll>; using vvvl = vv<vl>;
using ld = long double; using vld = vc<ld>; using vvld = vc<vld>; using vvvld = vc<vvld>;
using uint = unsigned int;
using ull = unsigned long long;
const ld pi = 3.141592653589793;
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3f;
// const ll mod = 1000000007;
const ll mod = 998244353;
inline bool inside(long long y, long long x, long long H, long long W) {return 0 <= (y) and (y) < (H) and 0 <= (x) and (x) < (W); }
#define debug(var) do{std::cout << #var << " : \n";view(var);}while(0)
template<typename T> void view(T e){cout << e << endl;}
template<typename T> void view(const vc<T>& v){for(const auto& e : v){ cout << e << " "; } cout << endl;}
template<typename T> void view(const vv<T>& vv){ for(const auto& v : vv){ view(v); } }
struct modint {
ll val;
using mint = modint;
modint() {};
modint(ll v) {
if (0 <= v && v < mod) val = v;
else{
v %= mod;
if (v < 0) v += mod;
val = v;
}
}
mint operator++(int){
val++;
if (val == mod) val = 0;
return *this;
}
mint operator--(int){
if (val == 0) val = mod;
val--;
return *this;
}
mint& operator+=(const mint& v){
val += v.val;
if (val > mod) val -= mod;
return *this;
}
friend mint operator+(const mint& u, const mint& v){
return mint(u) += v;
}
mint& operator-=(const mint& v){
val -= v.val;
if (val < 0) val += mod;
return *this;
}
friend mint operator-(const mint& u, const mint& v){
return mint(u) -= v;
}
mint& operator*=(const mint& v){
val *= v.val;
if (val > mod) val %= mod;
return *this;
}
friend mint operator*(const mint& u, const mint& v){
return mint(u) *= v;
}
mint pow(ll n){
mint a = *this, ret = 1;
while (n > 0){
if (n & 1) ret *= a;
a *= a;
n >>= 1;
}
return ret;
}
mint operator/=(const mint& v){
val *= mint(v).pow(mod - 2).val;
if (val > mod) val %= mod;
return *this;
}
friend mint operator/(const mint& u, const mint& v){
return mint(u) /= v;
}
friend bool operator==(const mint& u, const mint& v){
return u.val == v.val;
}
friend bool operator!=(const mint& u, const mint& v){
return u.val != v.val;
}
friend bool operator>(const mint& u, const mint& v){
return u.val > v.val;
}
friend bool operator<(const mint& u, const mint& v){
return u.val < v.val;
}
};
using mint = modint;
struct Binoms {
public :
Binoms(int n){
fact = vc<mint>(1, 1);
inv_fact = vc<mint>(1, 1);
update(n);
};
void update(int n){
if (sz >= n) return;
fact.resize(n + 1, 1);
inv_fact.resize(n + 1, 1);
for (int i = sz; i < n; i++) fact[i + 1] = fact[i] * (i + 1);
inv_fact[n] = 1 / fact[n];
for (int i = n - 1; i >= sz + 1; i--) inv_fact[i] = inv_fact[i + 1] * (i + 1);
sz = n;
return;
}
mint C(int n, int k){
if (sz < n) update(n);
return fact[n] * inv_fact[n - k] * inv_fact[k];
}
mint fac(int n){
if (sz < n) update(n);
return fact[n];
}
private :
int sz = 0;
vc<mint> fact, inv_fact;
};
int main(){
int T; cin >> T;
while (T--){
ll N, M; cin >> N >> M;
vv<mint> dp(M + 1, vc<mint>(3, 0));
vv<mint> to(3, vc<mint>(3, 0));
dp[0][0] = 1;
to[0][0] = 1;
to[0][1] = N;
to[0][2] = (N <= 3 ? 0 : N * (N - 1) / 2 - N);
to[1][0] = 1;
to[1][1] = N - 1;
to[1][2] = (N <= 3 ? 0 : N * (N - 1) / 2 - 2 * N + 3);
to[2][0] = 1;
to[2][1] = N - 2;
to[2][2] = N * (N - 1) / 2 - 3 * N + 7;
rep(i, M) rep(j, 3) rep(k, 3){
dp[i + 1][k] += dp[i][j] * to[j][k];
}
mint ret = dp[M][0] + dp[M][1] + dp[M][2];
cout << ret.val << endl;
}
}
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