题解 | #计算一元二次方程#3个+0-0要改

+0和-0不一样的情况要改3处，

//
// Created by Administrator on 2023/4/5.
//

#include <cstdio>
#include <cmath>
void equation(float a, float b, float c){
    if(a == 0){//a==0情况
        printf("Not quadratic equation\n");
    }else{//a≠0情况
        float zelta = b*b - 4*a*c;
        float answer, answer1, answer2;
        if(zelta == 0){
            answer = 0.5*(b*(-1)+sqrt(zelta))/a;
            float mp = -b + sqrt(zelta);
            if(mp == 0){//这个改动，+0和-0，不一样
                printf("x1=x2=%.2f\n", mp);
            }else{
                printf("x1=x2=%.2f\n", answer);
            }

        }else if(zelta > 0){
            answer1 = 0.5*(b*(-1)-sqrt(zelta))/a;
            answer2 = 0.5*(b*(-1)+sqrt(zelta))/a;
            printf("x1=%.2f;x2=%.2f\n", answer1, answer2);
        }else{//zelta小于0
            float shibu, xubu;
            shibu = -1*b/(2*a);
            xubu = sqrt(-1*zelta)/(2.0*a);
            if(shibu == -0){//+0和-0不一样
                shibu *= -1;
            }
            if(xubu <= -0){//+0和-0不一样
                xubu *= -1;
            }
            printf("x1=%.2f-%.2fi;x2=%.2f+%.2fi\n",  shibu, xubu, shibu, xubu);
        }
    }

}

int main(){
    float a, b, c;
    while (scanf("%f %f %f", &a, &b, &c) != EOF){
        equation(a, b, c);
    }
    return 0;
}