結果

問題 No.2685 Cell Proliferation (Easy)
ユーザー 👑 binap
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
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ソースコード

diff #
プレゼンテーションモードにする

#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" : " "); return
    os;}
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;
}
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