結果

問題 No.708 （＋ー）の式
ユーザー akusyounin2412
提出日時 2019-12-09 20:57:23
言語 C++11
(gcc 13.3.0)
結果
AC  
実行時間 2 ms / 2,000 ms
コード長 5,045 bytes
コンパイル時間 1,631 ms
コンパイル使用メモリ 129,028 KB
実行使用メモリ 5,376 KB
最終ジャッジ日時 2024-06-23 05:38:12
合計ジャッジ時間 2,355 ms
ジャッジサーバーID
（参考情報） 		judge1 / judge3
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 * 12
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 <iostream>
# include <algorithm>
#include <array>
# include <cassert>
#include <cctype>
#include <climits>
#include <numeric>
# include <vector>
# include <string>
# include <set>
# include <map>
# include <cmath>
# include <iomanip>
# include <functional>
# include <tuple>
# include <utility>
# include <stack>
# include <queue>
# include <list>
# include <bitset>
# include <complex>
# include <chrono>
# include <random>
# include <limits.h>
# include <unordered_map>
# include <unordered_set>
# include <deque>
# include <cstdio>
# include <cstring>
#include <stdio.h>
#include<time.h>
#include <stdlib.h>
#include <cstdint>
#include <cfenv>
#include<fstream>
//#include <bits/stdc++.h>
using namespace std;
using LL = long long;
using ULL = unsigned long long;
long long MOD = 1000000000 + 7;//998244353;
constexpr long long INF = numeric_limits<LL>::max() / 2;
const double PI = acos(-1);
#define fir first
#define sec second
#define thi third
#define debug(x) cerr<<#x<<": "<<x<<'\n'
typedef pair<LL, LL> Pll;
typedef pair<LL, pair<LL, LL>> Ppll;
typedef pair<LL, pair<LL, bitset<100001>>> Pbll;
typedef pair<LL, pair<LL, vector<LL>>> Pvll;
typedef pair<LL, LL> Vec2;
struct Tll { LL first, second, third; };
struct Fll { LL first, second, third, fourd; };
typedef pair<LL, Tll> Ptll;
#define rep(i,rept) for(LL i=0;i<rept;i++)
#define Rrep(i,mf) for(LL i=mf-1;i>=0;i--)
void YN(bool f) {
    if (f)
        cout << "YES" << endl;
    else
        cout << "NO" << endl;
}
void yn(bool f) {
    if (f)
        cout << "Yes" << endl;
    else
        cout << "No" << endl;
}
struct Edge { LL to, cost; };
struct edge { LL from, to, cost; };
vector<vector<int>>g;
vector<edge>edges;
vector<Pll>v;
map<Pll, Pll>ma;
set<LL>st;
LL h, w, n, m, k, t, s, p, q, last, cnt, sum, ans, a[210000], b[210000], dp[2100000];
string str, ss;
bool f;
char c;
int di[4][2] = { { 0,1 },{ 1,0 },{ 0,-1 },{ -1,0 } };
LL plu(LL a,LL b) {
    return a + b;
}
LL minu(LL a, LL b) {
    return a-b;
}
LL multi(LL a, LL b) {
    return a * b;
}
LL _xor(LL a, LL b) {
    return a ^ b;
}
LL _and(LL a, LL b) {
    return a & b;
}
LL _or(LL a, LL b) {
    return a | b;
}
LL how_to(string s) {
    LL ret = 0;
    for (int i = 0; i < s.size(); i++) {
        ret *= 10;
        ret += (s[i]-'0');
    }
    return ret;
}
template<class T>struct operate {
    int p; // 1:(+,-) 2:(*,/)
    char c; // + - * ^ etc...
    function<T(T, T)>f; // +:(a,b)=a+b
    operate() {};
    operate(int _p, char _c, function<T(T, T)>_f) { p = _p, c = _c, f = _f; };
};
template<class T>struct Parser {
    // results
    int root;                       // vals[root] is the answer
    vector<T> vals;                 // value of each node
    vector<int> left, right;        // the index of left-node, right-node
    function<T(string)> how_to_vals;  // evaluation methods
    vector<operate<T>> operators; // all operatorss are single character
    int priority_max = 2;
    // generate nodes
    int newnode(string op, int lp, int rp, T val = T()) {
        bool flag = 0;
        left.push_back(lp); right.push_back(rp);
        for (int i = 0; i < operators.size(); i++)
            if (op.size() == 1 && op[0] == operators[i].c) 
                vals.push_back(operators[i].f(vals[lp], vals[rp])), flag = 1;
        if (flag == 0)
            vals.push_back(val);
        return (int)vals.size() - 1;
    }
    void eval_init(function<T(string)>func) {
        how_to_vals = func;
    }
    void operate_init(operate<T> op) {
        operators.push_back(op);
    }
    // main solver
    T solve(const string &S) {
        int p = 0;
        for (int i = 0; i < operators.size(); i++) priority_max = max(operators[i].p, priority_max);
        root = expr(S, p);
        return vals[root];
    }
    // parser
    int expr(const string &S, int &p, int pri=1) {
        int lp;
        if (pri == priority_max) lp = value(S, p);
        else lp = expr(S, p, pri + 1);
        while (p < (int)S.size()) {
            bool flag = 0;
            for (int i = 0; i < operators.size(); i++) {
                if (S[p] == operators[i].c&&pri == operators[i].p)flag = 1;
            }
            if (!flag)break;
            string op;
            op.push_back(S[p]); ++p;
            if (priority_max == pri) {
                int rp = value(S, p);
                lp = newnode(op, lp, rp);
            }
            else {
                int rp = expr(S, p, pri + 1);
                lp = newnode(op, lp, rp);
            }
        }
        return lp;
    }
    int value(const string &S, int &p) {
        if (S[p] == '(') {
            ++p;                    // skip '('
            int lp = expr(S, p);
            ++p;                    // skip ')'
            return lp;
        }
        else {
            /* each process */
            string c;
            while (p < S.size()) {
                bool flag = 0;
                for (int i = 0; i < operators.size(); i++)
                    if (S[p] == operators[i].c)flag = 1;
                if (S[p] == ')')flag = 1;
                if (S[p] == '(')flag = 1;
                if (flag)break;
                c.push_back(S[p]); ++p;
            };
            return newnode(c, -1, -1, how_to_vals(c));
        }
    }
};
int main() {
    Parser<LL> prs;
    prs.operate_init((operate<LL>(1, '+', plu)));
    prs.operate_init((operate<LL>(1, '-', minu)));
    prs.operate_init((operate<LL>(2, '*', multi)));
    prs.operate_init((operate<LL>(3, '^', _xor)));
    prs.operate_init((operate<LL>(3, '&',_and)));
    prs.operate_init((operate<LL>(3, '|', _or)));
    prs.eval_init(how_to);
    cin >> str;
    prs.solve(str);
    cout << prs.vals[prs.root] << endl;
    return 0;
}
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