Ggthjy

Inequality

Given n inequalities about X, how many of them can hold at the same time at most?

The inequalities are in the follwing forms:

X < C

X <= C

X = C

X > C

X >= C

input:

First line with an integer n.

Following n lines, each line with an inequality.

Data Limit:

1<=N<=50,0<=C<=1000

output

One line with an integer, denoting the answer.

Sample Input

4

X = 1

X = 2

X = 3

X > 0

Sample Output

2

Here is my code:

#include <iostream>

#include <string>



#include <vector>



#include <map>

#include <set>



using namespace std;









int main() {

    //cout<<"请输入:"<<endl;

    int T;

    cin >> T;

    int count = T;



    vector<float> arrayOne(T, 0);

    vector<int> arrayTwo(T, 0);

    map<string, int> symbols = {

        pair<string, int>(">", 1),

        pair<string, int>(">=", 2),

        pair<string, int>("=", 3),

        pair<string, int>("<=", 4),

        pair<string, int>("<", 5)

    };



    while(T--)

    {

        string sOne;

        cin >> sOne;

        string sThree;

        cin >> sThree;

        arrayTwo[count-T-1] = symbols.find(sThree)->second;

        std::string sTwo;

        std::cin >> sTwo;

        float num = stof(sTwo);

        arrayOne[count-T-1] = num;

    }

    set<float> answerSet;

    for (float numPiece: arrayOne){

        answerSet.insert(numPiece-0.5);

        answerSet.insert(numPiece);

        answerSet.insert(numPiece+0.5);

    }

    int answer = 0;

    for (set<float>::iterator testNum=answerSet.begin(); testNum != answerSet.end(); ++testNum){

        //   cout<<*testNum;

        int passNum = 0;

        int i = 0;

        for (int numTwo: arrayTwo){

            switch (numTwo) {

                case 1:

                {

                    if (*testNum > arrayOne[i]) {

                        passNum += 1;

                    }

                }

                    break;

                case 2:

                {

                    if (*testNum >= arrayOne[i]) {

                        passNum += 1;

                    }

                }

                    break;

                case 3:

                {

                    if (*testNum == arrayOne[i]) {

                        passNum += 1;

                    }

                }

                    break;

                case 4:

                {

                    if (*testNum <= arrayOne[i]) {

                        passNum += 1;

                    }

                }

                    break;

                case 5:

                {

                    if (*testNum < arrayOne[i]) {

                        passNum += 1;

                    }

                }

                    break;

                default:

                    break;

            }

            i++;

        }

        answer = max(answer, passNum);

    }

    cout<<answer<<endl;

    return 0;

}

Very intuitive , just get all the boundary numbers, then find the max passes.

How to make it more efficient?