結果
問題 | No.2685 Cell Proliferation (Easy) |
ユーザー |
👑 |
提出日時 | 2024-03-20 23:41:25 |
言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
結果 |
WA
|
実行時間 | - |
コード長 | 2,191 bytes |
コンパイル時間 | 4,466 ms |
コンパイル使用メモリ | 256,100 KB |
最終ジャッジ日時 | 2025-02-20 10:14:47 |
ジャッジサーバーID (参考情報) |
judge2 / judge5 |
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ファイルパターン | 結果 |
---|---|
sample | WA * 3 |
other | WA * 26 |
ソースコード
#include<bits/stdc++.h>#include<atcoder/all>#define rep(i,n) for(int i=0;i<n;i++)using namespace std;using namespace atcoder;typedef long long ll;typedef vector<int> vi;typedef vector<long long> vl;typedef vector<vector<int>> vvi;typedef vector<vector<long long>> vvl;typedef long double ld;typedef pair<int, int> P;ostream& operator<<(ostream& os, const modint& a) {os << a.val(); return os;}template <int m> ostream& operator<<(ostream& os, const static_modint<m>& a) {os << a.val(); return os;}template <int m> ostream& operator<<(ostream& os, const dynamic_modint<m>& a) {os << a.val(); return os;}template<typename T> istream& operator>>(istream& is, vector<T>& v){int n = v.size(); assert(n > 0); rep(i, n) is >> v[i]; return is;}template<typename U, typename T> ostream& operator<<(ostream& os, const pair<U, T>& p){os << p.first << ' ' << p.second; return os;}template<typename T> ostream& operator<<(ostream& os, const vector<T>& v){int n = v.size(); rep(i, n) os << v[i] << (i == n - 1 ? "\n" : " "); returnos;}template<typename T> ostream& operator<<(ostream& os, const vector<vector<T>>& v){int n = v.size(); rep(i, n) os << v[i] << (i == n - 1 ? "\n" : "");return os;}template<typename T> void chmin(T& a, T b){a = min(a, b);}template<typename T> void chmax(T& a, T b){a = max(a, b);}using mint = modint998244353;struct S{mint val1, val2;int s;long long t;S(mint val1, mint val2, int s, long long t) : val1(val1), val2(val2), s(s), t(t) {}};mint q;S op(S a, S b){return S(a.val1 + b.val1 * q.pow(a.s), a.val2 + b.val2 * q.pow(a.t) + b.val1 * q.pow(a.s), a.s + b.s, (long long)a.s * b.s + a.t + b.t);}S e(){return S(0, 0, 0, 0LL);}int main(){int p1, p2, q1, q2, t;cin >> p1 >> p2 >> q1 >> q2 >> t;mint p = (mint)(p1)/ p2;q = (mint)(q1)/ q2;segtree<S, op, e> seg(t + 1);for(int i = t; i >= 0; i--){if(i == t){seg.set(t, S(1, 1, 1, 1));}else{auto [val1, val2, tmp1, tmp2] = seg.prod(i + 1, t + 1);cout << val1 << ' ' << val2 << ' ' << tmp1 << ' ' << tmp2 << "\n";seg.set(i, S(val2 * p, val2 * p, 1, 1));}}auto [val1, val2, tmp1, tmp2] = seg.prod(0, t + 1);cout << val2 << "\n";return 0;}