帮忙看看这个程序哪有问题.求一元二次方程解的问题
#include<stdio.h>
#include<math.h>
float x1,x2,p,q,disc;
great_than_zero(float a,float b)
{
x1=(-b+sqrt(disc))/(2*a);
x2=(-b+sqrt(disc))/(2*a);
}
zero(float a,float b)
{
x1=x2=(-b)/(2*a);
}
small_than_zero(float a,float b)
{
p=(-b)/(2*a);
q=sqrt(disc)/(2*a);
}
main()
{
float a,b,c;
printf("please input equation\n");
scanf("%f,%f,%f",&a,&b,&c);
printf("the equation is %5.2fx*x+2*%5.2f*x+%5.2f=0",a,b,c);
disc=b*b-4*a*c;
if (disc>0)
{
great_than_zero(a,b);
printf("x1=%5.2f,x2=%5.2f",x1,x2);
}
else if (disc==0)
{
zero(a,b);
printf("x1=x2=%5.2f",x1);
}
else
{
small_than_zero(a,b);
printf("x1=%5.2f+%5.2f",p,q);
}
}

OK测试通过:
#include<stdio.h>
#include<math.h>
double x1,x2,p,q,disc;
void great_than_zero(float a,float b)
{
x1=(-b+sqrt(disc))/(2*a);
x2=(-b+sqrt(disc))/(2*a);
}
void zero(float a,float b)
{
x1=x2=(-b)/(2*a);
}
void small_than_zero(float a,float b)
{
p=(-b)/(2*a);
q=sqrt(disc)/(2*a);
}
int main(void)
{
float a,b,c;
printf("please input equation\n");
scanf("%f,%f,%f",&a,&b,&c);
printf("the equation is %5.2fx*x+2*%5.2f*x+%5.2f=0\n",a,b,c);
disc=b*b-4*a*c;
if (disc>0)
{
great_than_zero(a,b);
printf("x1=%5.2lf,x2=%5.2lf\n",x1,x2);
}
else if (disc==0)
{
zero(a,b);
printf("x1=x2=%5.2lf\n",x1);
}
else
{
small_than_zero(a,b);
printf("x1=%5.2lf+%5.2lf*i\n",p,q);
printf("x2=%5.2lf-%5.2lf*i\n",p,q);
}
printf("\n");
return 0;
}

恩,非常的感谢 ,
可我为什么输入1,2,4以后的结果是x1=-1.00+-1.#j*i
x2=-1.00--1.#j*i
怎么出来符号#和j呢

哦 你只要把这个函数
void small_than_zero(float a,float b)
{
p=(-b)/(2*a);
q=sqrt(disc)/(2*a);
}
中的disc改为-disc就行了.

#include <stdio.h>
#include <math.h>
#include <stdlib.h>

main()
{
float a, b, c;
float d;
float x1, x2;
int bool;

printf("input a:\n");
scanf("%f", &a);
printf("input b:\n");
scanf("%f", &b);
printf("input c:\n");
scanf("%f", &c);

d = b * b - 4 * a* c;
if(d > 0) bool = 1;
else if(d == 0) bool = 0;
else bool = -1;
bool = bool + 2;
switch(bool){
case 1: break;

case 2:{
x1 = (-b + sqrt(d)) / 2 * a;
x2 = (-b - sqrt(d)) / 2 * a;
printf("%f\t%f\n", x1, x2);
break;
}
case 3:{
x1 = (-b + sqrt(d)) / 2 * a;
x2 = (-b - sqrt(d)) / 2 * a;
printf("%f\t%f\n", x1, x2);
break;
}
}
getch();
}



我以前写的一个，看看哦，希望可以帮你