Allmän lösning

#include <iostream>
#include <vector>
#include <string>

using namespace std;

vector<string> split(string a, char c) {
    vector<string> ekvVector;
    string tmp = "";
    for (int i = 0; i < a.size(); i++) {
        if (a[i] == c) {
            ekvVector.push_back(tmp);
            tmp = "";
        } else {
            tmp += a[i];
        }
    }
    if (tmp != "") ekvVector.push_back(tmp);
    return ekvVector;
}
int skitMetod(string a, int yType) {
    int endResult = 0;
    int bPos = 0;
    int ePos = 0;
    for (int i = 0; i < a.size(); i++) {
        if (a[i] == 'y') {
            bPos = i;
            if (yType == 0) {
                a.erase(bPos, 4);
                break;
            } else if (yType == 1) {
                a.erase(bPos, 5);
                break;
            } else if (yType == 2) {
                a.erase(bPos, 2);
                break;
            }
        }
    }
    if (a != "") {
        endResult = stoi(a);
    } else {
        endResult = 1;
    }
    return endResult;
}

int main() {
    string ekvString = "";
    vector<string> ekvList;
    cout << "Skriv in din ekvation i formen av ayBis+byPrim+cyF" << "\n";
    cin >> ekvString;
    cout << "\n";
    ekvList = split(ekvString, '+');
    double a, b, c;
    string yBis = "yBis";
    cout << ekvList.size() << "\n";
    for (int i = 0; i < ekvList.size(); i++) {
        if (ekvList[i].find(yBis) != string::npos) {
            cout << "yBis found!" << "\n";
            cout << skitMetod(ekvList[i], 0) << "\n";
            a = skitMetod(ekvList[i], 0);
        } else if (ekvList[i].find("yPrim") != string::npos) {
            cout << "yPrim found!" << "\n";
            cout << skitMetod(ekvList[i], 1) << "\n";
            b = skitMetod(ekvList[i], 1);
        } else if (ekvList[i].find("yF") != string::npos) {
            cout << "yF found!" << "\n";
            cout << skitMetod(ekvList[i], 2) << "\n";
            c = skitMetod(ekvList[i], 2);
        }
    }

    b = b / a;
    c = c / a;

    double root1 = ((b / 2)*(-1)) + (sqrt((b / 2)*(b / 2) - c));
    double root2 = ((b / 2)*(-1)) - (sqrt((b / 2)*(b / 2) - c));

    cout << "\n" << "C1e^(" << root1 << "x) + C2e^(" << root2 << "x)\n" << "\n";

#ifdef _DEBUG
        system("pause");
#endif // _DEBUG
}