結果
| 問題 | No.2243 Coaching Schedule |
| ユーザー |
 Jeroen Op de Beek
|
| 提出日時 | 2023-03-10 22:52:49 |
| 言語 | C++17 (gcc 12.3.0 + boost 1.83.0) |
| 結果 |
TLE
|
| 実行時間 | - |
| コード長 | 5,169 bytes |
| コンパイル時間 | 2,100 ms |
| コンパイル使用メモリ | 214,400 KB |
| 実行使用メモリ | 26,624 KB |
| 最終ジャッジ日時 | 2024-09-18 04:58:34 |
| 合計ジャッジ時間 | 7,996 ms |
|
ジャッジサーバーID （参考情報） |
judge3 / judge2 |
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テストケース
テストケース表示| 入力 | 結果 | 実行時間 実行使用メモリ |
|---|---|---|
| testcase_00 | AC | 30 ms
26,624 KB |
| testcase_01 | AC | 29 ms
21,068 KB |
| testcase_02 | AC | 29 ms
20,992 KB |
| testcase_03 | AC | 30 ms
20,992 KB |
| testcase_04 | AC | 30 ms
21,120 KB |
| testcase_05 | TLE | - |
| testcase_06 | -- | - |
| testcase_07 | -- | - |
| testcase_08 | -- | - |
| testcase_09 | -- | - |
| testcase_10 | -- | - |
| testcase_11 | -- | - |
| testcase_12 | -- | - |
| testcase_13 | -- | - |
| testcase_14 | -- | - |
| testcase_15 | -- | - |
| testcase_16 | -- | - |
| testcase_17 | -- | - |
| testcase_18 | -- | - |
| testcase_19 | -- | - |
| testcase_20 | -- | - |
| testcase_21 | -- | - |
| testcase_22 | -- | - |
| testcase_23 | -- | - |
| testcase_24 | -- | - |
| testcase_25 | -- | - |
| testcase_26 | -- | - |
| testcase_27 | -- | - |
| testcase_28 | -- | - |
| testcase_29 | -- | - |
| testcase_30 | -- | - |
| testcase_31 | -- | - |
| testcase_32 | -- | - |
| testcase_33 | -- | - |
| testcase_34 | -- | - |
| testcase_35 | -- | - |
| testcase_36 | -- | - |
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 array
cd* 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 empty
int 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';
}