結果
問題 | No.2243 Coaching Schedule |
ユーザー |
|
提出日時 | 2023-03-10 22:52:49 |
言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
結果 |
TLE
|
実行時間 | - |
コード長 | 5,169 bytes |
コンパイル時間 | 2,093 ms |
コンパイル使用メモリ | 206,492 KB |
最終ジャッジ日時 | 2025-02-11 08:59:50 |
ジャッジサーバーID (参考情報) |
judge3 / judge2 |
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ファイルパターン | 結果 |
---|---|
other | AC * 14 TLE * 23 |
ソースコード
#include "bits/stdc++.h"using namespace std;#define all(x) begin(x),end(x)template<typename A, typename B> ostream& operator<<(ostream &os, const pair<A, B> &p) { return os << '(' << p.first << ", " << p.second << ')'; }template<typename T_container, typename T = typename enable_if<!is_same<T_container, string>::value, typename T_container::value_type>::type>ostream& operator<<(ostream &os, const T_container &v) { string sep; for (const T &x : v) os << sep << x, sep = " "; return os; }#define debug(a) cerr << "(" << #a << ": " << a << ")\n";typedef long long ll;typedef vector<int> vi;typedef vector<vi> vvi;typedef pair<int,int> pi;const int mxN = 1<<19, oo = 1e9;const long long MD = 998244353;template<long long MOD=MD> struct mdint {int d=0;mdint () {d=0;}mdint (long long _d) : d(_d%MOD){if(d<0) d+=MOD;};friend mdint& operator+=(mdint& a, const mdint& o) {a.d+=o.d; if(a.d>=MOD) a.d-=MOD;return a;}friend mdint& operator-=(mdint& a, const mdint& o) {a.d-=o.d; if(a.d<0) a.d+=MOD;return a;}friend mdint& operator*=(mdint& a, const mdint& o) {return a = mdint((ll)a.d*o.d);}mdint operator*(const mdint& o) const {mdint res = *this;res*=o;return res;}mdint operator+(const mdint& o) const {mdint res = *this;res+=o;return res;}mdint operator-(const mdint& o) const {mdint res = *this;res-=o;return res;}mdint operator^(long long b) const {mdint tmp = 1;mdint power = *this;while(b) {if(b&1) {tmp = tmp*power;}power = power*power;b/=2;}return tmp;}friend mdint operator/=(mdint& a, const mdint& o) {a *= (o^(MOD-2));return a;}mdint operator/(const mdint& o) {mdint res = *this;res/=o;return res;}bool operator==(const mdint& o) { return d==o.d;}bool operator!=(const mdint& o) { return d!=o.d;}friend istream& operator>>(istream& c, mdint& a) {return c >> a.d;}friend ostream& operator<<(ostream& c, const mdint& a) {return c << a.d;}};using mint = mdint<MD>;typedef mint cd;void revperm(cd* in, int n) {for(int i=0,j=0;i<n;++i) {if(i<j) swap(in[i],in[j]);for(int k = n >> 1; (j ^= k) < k; k >>= 1);}}cd w[mxN+1]; // stores w^j for each j in [0,n-1]void precomp() {w[0] = 1;int pw = (MD-1)/mxN;w[1] = mint(3)^pw;for(int i= 2;i<=mxN;++i) {w[i] = w[i-1]*w[1];}}void fft(cd* in, int n, bool reverse=false) {int lg = __lg(n);assert(1<<lg == n);int stride = mxN/n;revperm(in,n);for(int s=1;s<=lg;++s) {int pw = 1<<s;int mstride = stride*(n>>s);for(int j=0;j<n;j+=pw) {// do FFT merging on out arraycd* even = in+j, *odd = in+j+pw/2;for(int i=0;i<pw/2;++i) {cd& power = w[reverse?mxN-mstride*i:mstride*i];auto tmp = power*odd[i];odd[i] = even[i] - tmp;even[i] = even[i] + tmp;}}}if(reverse) {mint fac = mint(1)/n;for(int i=0;i<n;++i) in[i]=in[i]*fac;}}vector<cd> polymul(vector<cd>& a, vector<cd>& b) {int n = a.size(), m = b.size(), ptwo = 1;while(ptwo<(n+m)) ptwo*=2;a.resize(ptwo), b.resize(ptwo);fft(a.data(),ptwo);fft(b.data(),ptwo);for(int i=0;i<ptwo;++i)a[i] = a[i]*b[i];fft(a.data(),ptwo,true);a.resize(n+m-1);return a;}const int mxF = 2e6+2;mint fact[mxF], ifact[mxF];cd ncr(int a, int b) {if(b>a or a<0 or b<0) return 0;return fact[a]*ifact[b]*ifact[a-b];}void precomp2() {fact[0]=1;for(int i=1;i<mxF;++i) {fact[i]=fact[i-1]*i;}ifact[mxF-1] = mint(1)/fact[mxF-1];for(int i=mxF-2;i>=0;--i) {ifact[i]=ifact[i+1]*(i+1);}}int main() {precomp();precomp2();// ok need to choose subsets but only of unique// first choose number of days?// then choose how to do it?// you choose a subset of exactly so many// but empty is not allowed.// do it just with empties.// then choose which days to emptyint n,m; cin >> m >> n;map<int,int> cnt;for(int i=0;i<n;++i) {int a; cin >> a;cnt[a]++;}// only sqrt items.// have to subtract those with ncr factors in front.// and one of those is just binomial coefs multiplied.vector<mint> aa(n+1),bb(n+1);for(int i=1;i<=n;++i) {aa[i]=1;for(auto [j,c] : cnt) {if(aa[i]==0) break;aa[i]*=ncr(i,c);}aa[i]*=ifact[i];}mint pw=1;for(int i=0;i<=n;++i) {// /i!bb[i] = ifact[i]*pw;pw*=MD-1;}auto cc = polymul(aa,bb);mint ans=0;for(int i=1;i<=n;++i) {cc[i]*=fact[i];ans+=cc[i];}for(auto [j,c] : cnt) ans*=fact[c];cout << ans << '\n';}