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【算法题】CCF CSP第三题练习

2020-02-03 14:54:56  阅读:293  来源: 互联网

标签:string int ++ 算法 num && fom CCF CSP


 

 

 

 

 样例全部没问题,但是只有40分,不知道哪里出问题了:

#include <iostream>
#include <string>
#include <map>
#include <sstream>

using namespace std; 

class Fomular
{
private:
    string s, sr, sp;
    map<string,int> reactant;
    map<string,int> product;
    map<string,int> rElement;
    map<string,int> pElement;

    bool isLowercase(char a)
    {
        if (a >= 'a' && a <= 'z')
            return true;
        return false;
    }

    bool isUppercase(char a)
    {
        if (a >= 'A' && a <= 'Z')
            return true;
        return false;
    }

    bool isDigit(char a)
    {
        if (a >= '0' && a <= '9')
            return true;
        return false;
    }

    void split(string s, decltype(product) &m)
    {
        int a{0}, b{0}, i, j, tt;
        string t;
        for (i = 0; i < s.length(); i++)
        {
            if (s[i] == '+')
            {
                b = i;
                t = s.substr(a, b-a);
                for (j = 0; j < t.length() && isDigit(t[j]); ++j);
                stringstream ss;
                if (j == 0)
                    ss << "1";
                else
                    ss << t.substr(0, j);
                ss >> tt;
                m[t.substr(j, t.length()-j)] = tt;
                a = i+1;
            }
        }
        b = i;
        t = s.substr(a, b-a);
        for (j = 0; j < t.length() && isDigit(t[j]); ++j);
        stringstream ss;
        if (j == 0)
            ss << "1";
        else
            ss << t.substr(0, j);
        ss >> tt;
        m[t.substr(j, t.length()-j)] = tt;
    }

    void elemCnt(string fom, decltype(pElement) &pE, int mul = 1)
    {
        string t;
        int level{0}, num;
        for(int j = 0; j < fom.size(); j++)
        {
            if (isUppercase(fom[j]) && isDigit(fom[j+1]) && j+1 < fom.size())
            {
                t = string{fom[j]};
                int k = j+1;
                num = 0;
                while(isDigit(fom[k]) && k < fom.size())
                {
                    num *= 10;
                     num += fom[k] - '0';
                    k++;
                }
                if (num == 0)
                    num = 1;
                pE[t] += num * mul;
                j = k - 1;
            }
            else if (isUppercase(fom[j]) && isLowercase(fom[j+1]) && j+1 < fom.size())
            {
                t = string{fom[j]};
                t.append(string{fom[j+1]});
                int k = j+2;
                num = 0;
                while(isDigit(fom[k]) && k < fom.size())
                {
                    num *= 10;
                     num += fom[k] - '0';
                    k++;
                }
                if (num == 0)
                    num = 1;
                pE[t] += num * mul;
                j = k - 1;
            }
            else if (isUppercase(fom[j]))
            {
                t = string{fom[j]};
                pE[t] += mul;
            }
            else if (fom[j] == '(')
            {
                int numBack{1}, m, k;
                for (k = j; k < fom.size(); k++)
                {
                    if (fom[k] == '(')
                        level++;
                    else if (fom[k] == ')')
                    {
                        level--;
                        if (level == 0)
                            m = k;
                    }
                    if (level == 0 && isUppercase(fom[k]))
                        break;
                }
                numBack = 0;
                for (int a = m+1; a < k; a++)
                {
                    numBack *= 10;
                     numBack += fom[a] - '0';
                }
                if (numBack == 0)
                    numBack = 1;
                elemCnt(fom.substr(j+1, m-j-1), pE, numBack * mul);
                j = k;
            }
        }
    }

    void elemCount(decltype(product) &p, decltype(pElement) &pE)
    {        
        for(auto i: p)
        {
            string fom{i.first};
            elemCnt(fom, pE, i.second);
        }
    }

public:
    Fomular(string s_)
    {
        int eq;
        s = s_;
        eq = s.find('=');
        sr = s.substr(0,eq);
        sp = s.substr(eq+1,s.length()-eq-1);
        split(sr, reactant);
        split(sp, product);
        elemCount(reactant, rElement);
        elemCount(product,  pElement);
        
        //----print------
        // for (auto i: rElement)
        // {
        //     cout << i.first << "#" << i.second << "  ";
        // }
        // cout << "=  ";
        // for (auto i: pElement)
        // {
        //     cout << i.first << "#" << i.second << "  ";
        // }
        // cout << endl;
        // ----print------
    }
    char getAnswer()
    {
        if (rElement.size() != pElement.size())
            return 'N';
        for (auto i:rElement)
        {
            if(i.second != pElement[i.first])
                return 'N';
        }
        return 'Y';
    }

};

int main()
{
    int n;
    cin >> n;
    string s;
    getline(cin, s);
    for (int i = 0; i < n; ++i)
    {
        getline(cin, s);
        Fomular f{s};
        cout << f.getAnswer() << endl;
    }
    return 0;
}

 

标签:string,int,++,算法,num,&&,fom,CCF,CSP
来源: https://www.cnblogs.com/joeyzhao/p/12255640.html

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