結果

問題 No.2927 Reverse Polish Equation
ユーザー 👑 binap
提出日時 2024-10-12 15:44:01
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 452 ms / 2,000 ms
コード長 3,347 bytes
コンパイル時間 4,356 ms
コンパイル使用メモリ 255,956 KB
最終ジャッジ日時 2025-02-24 18:09:10
ジャッジサーバーID
（参考情報） 		judge1 / judge4
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 3
other AC * 43
権限があれば一括ダウンロードができます

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;
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 <int m> istream& operator>>(istream& is, static_modint<m>& a) {long long x; is >> x; a = x; return is;}
template <int m> istream& operator>>(istream& is, dynamic_modint<m>& a) {long long x; is >> x; a = x; return is;}
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> ostream& operator<<(ostream& os, const set<T>& se){for(T x : se) os << x << " "; os << "\n"; return os;}
template<typename T> ostream& operator<<(ostream& os, const unordered_set<T>& se){for(T x : se) os << x << " "; os << "\n"; return os;}
template<typename S, auto op, auto e> ostream& operator<<(ostream& os, const atcoder::segtree<S, op, e>& seg){int n = seg.max_right(0, [](S){return 
    true;}); rep(i, n) os << seg.get(i) << (i == n - 1 ? "\n" : " "); return os;}
template<typename S, auto op, auto e, typename F, auto mapping, auto composition, auto id> ostream& operator<<(ostream& os, const atcoder
    ::lazy_segtree<S, op, e, F, mapping, composition, id>& seg){int n = seg.max_right(0, [](S){return true;}); rep(i, n) os << seg.get(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);}
const long long INF = 100100100100100;
int main(){
    int n;
    long long y;
    cin >> n >> y;
    vector<string> t(n);
    cin >> t;
    auto calc = [&](long long x){
        deque<long long> de;
        rep(i, n){
            string s = t[i];
            if(s == "X") de.push_back(x);
            else if(s == "+"){
                auto val1 = de.back(); de.pop_back();
                auto val2 = de.back(); de.pop_back();
                de.push_back(val1 + val2);
            }else if(s == "max"){
                auto val1 = de.back(); de.pop_back();
                auto val2 = de.back(); de.pop_back();
                de.push_back(max(val1, val2));
            }else if(s == "min"){
                auto val1 = de.back(); de.pop_back();
                auto val2 = de.back(); de.pop_back();
                de.push_back(min(val1, val2));
            }else{
                flush(cout);
                long long val = stoll(s);
                flush(cout);
                de.push_back(val);
            }
        }
        long long val = de.back();
        return val;
    };
    long long ng = -1, ok = INF;
    while(ok - ng > 1){
        long long mid = (ok + ng) / 2;
        auto res = calc(mid);
        if(res >= y) ok = mid;
        else ng = mid;
    }
    {
        auto res = calc(ok);
        if(res == y) cout << ok << "\n";
        else cout << "-1\n";
    }
    return 0;
}
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