- 技术
- 习题7.2 求方程ax2 + bx + c = 0的根，用3个函数分别求当b2-4ac大于零、等于零、和小于零时的方程的根并输出结果。要求从主函数输入a,b,c的值
#include <stdio.h>
#include <math.h>
int main()
{
void fun1(double p
,double q
);
void fun2(double i
,double j
,double k
);
void fun3(double x
,double y
,double z
);
double a
,b
,c
,disc
;
printf("请输入方程式的a,b,c三个值：\n");
scanf("%lf,%lf,%lf",&a
,&b
,&c
);
if(fabs(a
)<=1e-6)
printf("不是二次方程式!\n");
else
{
disc
=b
*b
-4*a
*c
;
if(fabs(disc
)<=1e-6)
fun1(a
,b
);
else if(disc
>1e-6)
fun2(a
,b
,c
);
else
fun3(a
,b
,c
);
}
return 0;
}
void fun1(double p
,double q
)
{
printf("ax2+bx+c=0方程式，当b*b - 4*a*c = 0时：\n");
printf("此方式程有两个相等的实根：%8.4f\n",-q
/(2*p
));
}
void fun2(double i
,double j
,double k
)
{
double x
,y
,z
;
z
=j
*j
-4*i
*k
;
x
=(-j
+sqrt(z
))/(2*i
);
y
=(-j
-sqrt(z
))/(2*i
);
printf("ax2+bx+c=0方程式，当b*b - 4*a*c > 0时：\n");
printf("此方式有两个不相等的实根：%8.4f and %8.4f\n",x
,y
);
}
void fun3(double x
,double y
,double z
)
{
double a
,b
,c
;
a
=y
*y
-4*x
*z
;
b
=-y
/(2*x
);
c
=sqrt(-a
)/(2*x
);
printf("ax2+bx+c=0方程式，当b*b - 4*a*c < 0时：\n");
printf("此方式有两个共轭复根：\t%8.4f+%8.4fi and %8.4f-%8.4fi\n",b
,c
,b
,c
);
}
Processed:
0.014, SQL:
9
