8.编程实现输入一串字符串:根据输入的a、b、c的值,求ax2+bx+c=0方程的解。

来源：蜘蛛抓取(WebSpider) 时间：2023-04-07 07:48 标签： 编程实现输入一串字符串
搜索数学#include
#include
void main()
{
float a,b,c,d,e,f,g;
void root0(float,float);
void root1(float,float);
void root2(float,float,float);
printf("Please give values to the variale a, b and c respectively:\na=");
scanf("%f",&a);
printf("b=");
scanf("%f",&b);
printf("c=");
scanf("%f",&c);
d=b*b-4*a*c;
e=fabs(b);
f=-0.5*b/a;
g=0.5*sqrt(e)/a;
if(a=0)
root0(b,c);
else
{
if(d>0)
root1(g,f);
else
root2(g,f,d);
}
}
void root0(float x,float y)
{
if(x=0)
printf("The equation is extremely degenerate.\n");
else
printf("The equation is degenerate and has one solution: x=%.3f .\n",-y/x);
}
void root1(float x,float y)
{
float x1;
float x2;
x1=y+x;
x2=y-x;
printf("The equation has two real solutions: x1=%.3f , x2=%.3f.\n",x1,x2);
}
void root2(float x,float y,float z)
{
if(z<0)
printf("The equation has two complex solutions: x1=%.3f+%.3fi , x2=%.3f-%.3fi .\n",y,x,y,x);
else
printf("The equation has two multiple real solutions: x1=x2=%.3f .\n",y);
}
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