题解 | #二次方程计算器#

//主要的思路是，遍历二次式字符串，计算常数项、一次项、二次项的系数和，再利用求根公式得出答案
#include <iostream>
#include <string>
#include <cmath>
#include <cstdio>
using namespace std;

int main() {
    string str, str2;
    cin >> str;
    string left = str.substr(0, str.find('=')),
           right = str.substr(str.find('=') + 1);
    for (int i = 0; i < right.length(); i++) { //右侧表达式异号
        if (right[i] == '+') right[i] = '-';
        else if (right[i] == '-') right[i] = '+';
    }
    str2 = left + '-' + right + "!!"; //式子归于左边
    //cout<<str2<<endl;//查看整理后的式子
    int a = 0, b = 0, c = 0;
    for (int i = 0; i < str2.length() - 2; i++) {
        if (str2[i] <= '9' && str2[i] >= '0') {
            if (str2[i + 1] == '+' || str2[i + 1] == '-' || str2[i + 1] == '!') { //常数
                if (i == 0) { //首项的情况
                    c = c + str2[i] - '0';
                } else { //非首项的情况
                    if (str2[i - 1] == '+') c = c + str2[i] - '0';
                    if (str2[i - 1] == '-') c = c - str2[i] + '0';
                }
            } else if (str2[i + 1] == 'x' && str2[i + 2] != '^') { //一次项系数
                if (i == 0) { //首项的情况
                    b = b + str2[i] - '0';
                } else { //非首项的情况
                    if (str2[i - 1] == '+') b = b + str2[i] - '0';
                    if (str2[i - 1] == '-') b = b - str2[i] + '0';
                }
            } else if (str2[i + 1] == 'x' && str2[i + 2] == '^') { //二次项系数
                if (i == 0) { //首项的情况
                    a = a + str2[i] - '0';
                } else { //非首项的情况
                    if (str2[i - 1] == '+') a = a + str2[i] - '0';
                    if (str2[i - 1] == '-') a = a - str2[i] + '0';
                }
            }
        } else if (str2[i] == 'x') {
            if (i == 0) { //首项的情况
                if (str2[i + 1] != '^') { //一次项x
                    b++;
                } else if (str2[i + 1] == '^') {
                    a++;
                }
            } else { //非首项的情况
                if (str2[i - 1] <= '9' && str2[i - 1] >= '0') continue; //常数已判断过
                else if (str2[i - 1] == '+') {
                    if (str2[i + 1] != '^') { //一次项x
                        b++;
                    } else if (str2[i + 1] == '^') { //二次项x
                        a++;
                    }
                } else if (str2[i - 1] == '-') {
                    if (str2[i + 1] != '^') { //一次项x
                        b--;
                    } else if (str2[i + 1] == '^') { //二次项x
                        a--;
                    }
                }
            }
        }
    }
    //cout << "a:" << a << " " << "b:" << b << " " << "c:" << c;//查看各项系数和
    if(pow(b, 2) - 4 * a * c<0){
        printf("No Solution");//无解
        return 0;
    }
    double answer1 = (-b + pow(pow(b, 2) - 4 * a * c, 0.5)) / (2 * a),
           answer2 = (-b - pow(pow(b, 2) - 4 * a * c, 0.5)) / (2 * a);
    printf("%.2f", min(answer1, answer2));
    printf(" ");
    printf("%.2f", max(answer1, answer2));
    return 0;
}