結果
問題 | No.2615 ペアの作り方 |
ユーザー |
|
提出日時 | 2024-01-26 21:29:31 |
言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 50 ms / 2,000 ms |
コード長 | 2,446 bytes |
コンパイル時間 | 2,313 ms |
コンパイル使用メモリ | 204,760 KB |
最終ジャッジ日時 | 2025-02-18 22:57:28 |
ジャッジサーバーID (参考情報) |
judge1 / judge3 |
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ファイルパターン | 結果 |
---|---|
other | AC * 21 |
ソースコード
#include <bits/stdc++.h>using namespace std;long long mod = 998244353;//入力が必ずmod未満の時に使う.struct mint{long long v = 0;mint(){} mint(int a){v = a;} mint(long long a){v = a;}long long val(){return v;}void modu(){v %= mod;}mint repeat2mint(long long a,long long b){mint ret = 1,p = a; int Log = 60;if(b <= 2e9) Log = 30;for(int i=0; i<Log; i++){if(b&(1LL<<i)) ret *= p; p *= p;}return ret;};mint operator+(mint b){return (v+b.v)%mod;}mint operator-(mint b){return (v-b.v+mod)%mod;}mint operator*(mint b){return v*b.v%mod;}mint operator/(mint b){if(b.v == 0) assert(false);return v*(repeat2mint(b.v,mod-2)).v%mod;}void operator+=(mint b){v = (v+b.v)%mod;}void operator-=(mint b){v = (v-b.v+mod)%mod;}void operator*=(mint b){v = v*b.v%mod;}void operator/=(mint b){if(b.v == 0) assert(false);v = v*repeat2mint(b.v,mod-2).v%mod;}void operator++(int){v = (v+1)%mod; return;}void operator--(int){v = (v-1+mod)%mod; return;}bool operator==(mint b){if(v == b.v) return true; else return false;}bool operator!=(mint b){if(v != b.v) return true; else return false;}bool operator>(mint b){if(v > b.v) return true; else return false;}bool operator>=(mint b){if(v >= b.v) return true; else return false;}bool operator<(mint b){if(v < b.v) return true; else return false;}bool operator<=(mint b){if(v <= b.v) return true; else return false;}mint pow(long long x){return repeat2mint(v,x);}mint inv(){return mint(1)/v;}};int main(){ios_base::sync_with_stdio(false);cin.tie(nullptr);int N; cin >> N;vector<long long> X(N),Y(N);for(auto &x : X) cin >> x;for(auto &y : Y) cin >> y;sort(X.begin(),X.end()); sort(Y.rbegin(),Y.rend());vector<mint> fac(N+1,1),inv(N+1,1),facinv(N+1,1);for(int i=1; i<=N; i++) fac.at(i).v = fac.at(i-1).v*i %mod;for(int i=2; i<=N; i++) inv.at(i).v = mod-(mod/i*inv.at(mod%i).v %mod);for(int i=2; i<=N; i++) facinv.at(i).v = facinv.at(i-1).v*inv.at(i).v %mod;auto nCr = [&](int n, int r) -> mint {return fac.at(n)*facinv.at(r)*facinv.at(n-r);};int pos = 0;while(pos != N){if(X.at(pos) < Y.at(pos)) pos++;else break;}mint answer = fac.at(pos)*fac.at(N-pos);cout << answer.v << endl;}