题解 | 二次方程计算器

#include<iostream>
#include<cmath>
using namespace std;
void f(int& a, int& b, int& c, string s) {
    int flag = 1, tmp = 0;
    for (int i = 0; i < s.size(); i++) {
        if (s[i] >= '0' && s[i] <= '9') {
            tmp = tmp * 10 + (s[i] - '0');
        } else if (s[i] == 'x') {
            if (i == 0)tmp = 1;
            if (s[i - 1] < '0' || s[i - 1] > '9')tmp = 1;
            if (i + 1 < s.size() && s[i + 1] == '^') { //x^2
                tmp *= flag;
                a += tmp;
                tmp = 0;
                flag = 1;
                i += 2;
            } else {
                tmp *= flag;
                b += tmp;
                tmp = 0;
                flag = 1;
            }
        } else {
            tmp *= flag;
            c += tmp;
            tmp = 0;
            flag = 1;
        }
        if (s[i] == '-')flag = -1;
    }
    if (tmp != 0)c += (tmp * flag);
}
int main() {
    string s;
    cin >> s;
    int a1 = 0, b1 = 0, c1 = 0, a2 = 0, b2 = 0, c2 = 0;
    //查x^2
    int pos = s.find("=");
    string s1 = s.substr(0, pos), s2 = s.substr(pos + 1);
    f(a1, b1, c1, s1);
    f(a2, b2, c2, s2);
    int a = a1 - a2, b = b1 - b2, c = c1 - c2;
//  cout<<"a: "<<a<<"  b:  "<<b<<"  c:  "<<c<<endl;
    if (b * b < 4 * a * c)cout << "No Solution" << endl;
    else {
        int k = b * b - 4 * a * c;
        double end = sqrt((double)k);
        double x1 = (0 - (double)b - end) / (2 * (double)a);
        double x2 = (-(double)b + end) / (2 * (double)a);
        if (x1 < x2) {
            printf("%.2f %.2f\n", x1, x2);
        } else {
            printf("%.2f %.2f\n", x2, x1);
        }
    }
}