結果
問題 | No.1621 Sequence Inversions |
ユーザー |
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提出日時 | 2021-07-22 21:47:59 |
言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 40 ms / 3,000 ms |
コード長 | 4,082 bytes |
コンパイル時間 | 2,490 ms |
コンパイル使用メモリ | 153,768 KB |
最終ジャッジ日時 | 2025-01-23 06:29:46 |
ジャッジサーバーID (参考情報) |
judge4 / judge5 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 26 |
ソースコード
#pragma GCC optimize("O3")#pragma GCC optimize("unroll-loops")#include<iostream>#include<string>#include<cstdio>#include<vector>#include<cmath>#include<algorithm>#include<functional>#include<iomanip>#include<queue>#include<ciso646>#include<random>#include<map>#include<set>#include<bitset>#include<stack>#include<unordered_map>#include<unordered_set>#include<utility>#include<cassert>#include<complex>#include<numeric>#include<array>using namespace std;//#define int long longtypedef long long ll;typedef unsigned long long ul;typedef unsigned int ui;constexpr ll mod = 998244353;const ll INF = mod * mod;typedef pair<int, int>P;#define stop char nyaa;cin>>nyaa;#define rep(i,n) for(int i=0;i<n;i++)#define per(i,n) for(int i=n-1;i>=0;i--)#define Rep(i,sta,n) for(int i=sta;i<n;i++)#define rep1(i,n) for(int i=1;i<=n;i++)#define per1(i,n) for(int i=n;i>=1;i--)#define Rep1(i,sta,n) for(int i=sta;i<=n;i++)#define all(v) (v).begin(),(v).end()typedef pair<ll, ll> LP;typedef long double ld;typedef pair<ld, ld> LDP;const ld eps = 1e-12;const ld pi = acosl(-1.0);ll mod_pow(ll x, ll n, ll m = mod) {if (n < 0) {ll res = mod_pow(x, -n, m);return mod_pow(res, m - 2, m);}if (abs(x) >= m)x %= m;if (x < 0)x += m;ll res = 1;while (n) {if (n & 1)res = res * x % m;x = x * x % m; n >>= 1;}return res;}struct modint {ll n;modint() :n(0) { ; }modint(ll m) :n(m) {if (n >= mod)n %= mod;else if (n < 0)n = (n % mod + mod) % mod;}operator int() { return n; }};bool operator==(modint a, modint b) { return a.n == b.n; }modint operator+=(modint& a, modint b) { a.n += b.n; if (a.n >= mod)a.n -= mod; return a; }modint operator-=(modint& a, modint b) { a.n -= b.n; if (a.n < 0)a.n += mod; return a; }modint operator*=(modint& a, modint b) { a.n = ((ll)a.n * b.n) % mod; return a; }modint operator+(modint a, modint b) { return a += b; }modint operator-(modint a, modint b) { return a -= b; }modint operator*(modint a, modint b) { return a *= b; }modint operator^(modint a, ll n) {if (n == 0)return modint(1);modint res = (a * a) ^ (n / 2);if (n % 2)res = res * a;return res;}ll inv(ll a, ll p) {return (a == 1 ? 1 : (1 - p * inv(p % a, a)) / a + p);}modint operator/(modint a, modint b) { return a * modint(inv(b, mod)); }modint operator/=(modint& a, modint b) { a = a / b; return a; }const int max_n = 1 << 18;modint fact[max_n], factinv[max_n];void init_f() {fact[0] = modint(1);for (int i = 0; i < max_n - 1; i++) {fact[i + 1] = fact[i] * modint(i + 1);}factinv[max_n - 1] = modint(1) / fact[max_n - 1];for (int i = max_n - 2; i >= 0; i--) {factinv[i] = factinv[i + 1] * modint(i + 1);}}modint comb(int a, int b) {if (a < 0 || b < 0 || a < b)return 0;return fact[a] * factinv[b] * factinv[a - b];}modint combP(int a, int b) {if (a < 0 || b < 0 || a < b)return 0;return fact[a] * factinv[a - b];}modint dp[105][10005];void solve() {int n, k; cin >> n >> k;assert(n <= 100);vector<ll> a(n);rep(i, n) {cin >> a[i];assert(a[i] <= 1000000000);}//rep(i, n - 1)assert(a[i] <= a[i + 1]);vector<int> c;sort(all(a));rep(i, n) {int le = i;while (i + 1 < n && a[i] == a[i + 1])i++;c.push_back(i - le + 1);}vector<modint> cdp = { 1 };int sum = c[0];for (int i = 1; i < c.size(); i++) {rep(j, c[i] + 1)rep(k, sum * c[i] + 1)dp[j][k] = 0;dp[0][0] = 1;for (int num = 0; num <= sum; num++) {rep(j, c[i])rep(k, sum * c[i] + 1) {if (dp[j][k] == (modint)0)continue;assert(k + num <= sum * c[i]);dp[j + 1][k + num] += dp[j][k];}}vector<modint> ndp(cdp.size() + sum * c[i]);rep(j, cdp.size())rep(k, sum * c[i] + 1) {modint val = cdp[j] * dp[c[i]][k];ndp[j + k] += val;}swap(cdp, ndp);sum += c[i];}modint ans = 0;if (k < cdp.size())ans = cdp[k];cout << ans << "\n";}signed main() {ios::sync_with_stdio(false);cin.tie(0);//cout << fixed << setprecision(12);//init_f();//init();//expr();//int t; cin >> t; rep(i, t)solve();return 0;}