[编程入门]自定义函数求一元二次方程

解题思路:

注意事项:return 0;//强制退出

else if(z==0)//“==”不是“=”

参考代码:

#include<stdio.h>

#include<math.h>

float x1,x2,a,b,c,z;

void h1(){

x1=( -b+sqrt(z) )/( 2*a );

    x2=( -b-sqrt(z) )/( 2*a );

}

void h2(){

x1=x2=( -b )/( 2*a );

} 

void h3(){

z=sqrt(-z);

    printf("x1=%.3f+%.3fi x2=%.3f-%.3fi\n",-b/(2*a),z/(2*a),-b/(2*a),z/(2*a));

}

int main(){

scanf("%f %f %f",&a,&b,&c);

z=b*b-4*a*c;

if(z>0){

h1();

} 

else if(z==0){

 h2();

} 

else{

h3();

return 0;//强制退出

} 

printf("x1=%.3f x2=%.3f\n",x1,x2);

    return 0;

}