Question

如何用C语言程序解方程?

最好复制过来一个程序段 以供参考

Answer 1

#include "stdio.h" 
#include "math.h" 
/*求一元二次方程ax*x+bx+c=0的解*/ 
main() 
{ 
	float a,b,c,x1,x2,d; 
	printf("请输入a："); 
	scanf("%f",&a); 
	printf("请输入b："); 
	scanf("%f",&b); 
	printf("请输入c："); 
	scanf("%f",&c); 
	d=b*b-4*a*c; 
	if(d < 0) 
		printf("方程没有实数解。\n"); 
	if (d==0) 
	{ 
		x1=(-b)/(2*a); 
		printf("x1=%f\n",x1); 
	} 
	if (d>0) 
	{ 
		x1=(-b+sqrt(d))/(2*a); 
		x2=(-b-sqrt(d))/(2*a); 
		printf("x1=%f,x2=%f\n",x1,x2);} 
}
请输入a：12
请输入b：34
请输入c：4
x1=-0.122985,x2=-2.710348
Press any key to continue

Answer 2

上课时编的，测试过可用。/* Note:Your choice is C IDE */
#include "stdio.h"
#include <math.h>
void main()
{
int a,b,c,d,e;
e=1;
while (e)
{printf("求一个一元二次方程的根");
printf("ax^2+bx+c=0\n");
printf("请输入a=");
scanf("%d",&a);
if(a==0)
{
break;
}printf("请输入b=");
scanf("%d",&b);
printf("请输入c=");
scanf("%d",&c);
d=b*b-4*a*c;
if(d<0)
{
printf("无实数解\n");
}
else if(d==0)
{
printf("只有一个根为：%f\n",-b/(2.0*a));
}
else
{printf("有两个根为：%f%f\n",(-b+sqrt(e))/(2.0*a),(-b-sqrt(e))/(2.0*a));
}}
}

Answer 3

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

double x1,x2,p;

float file1(float a,float b)
{
x1=(-b+sqrt(p))/2*a;
x2=(-b-sqrt(p))/2*a;
return 0;
}

float file2(float a,float b)
{
x1=x2=(-b+sqrt(p))/2*a;
return 0;
}

void main()
{
float a,b,c;
printf("请依次输入方程的三个系数：\n");
scanf("%f%f%f",&a,&b,&c);
p=b*b-4*a*c;
printf("方程是：%.1f*x*x + %.1f*x + %.1f = 0\n",a,b,c);
if(p>0)
{
file1(a,b);
printf("X1=%f\tX2=%f\n",x1,x2);
}
else if(p==0)
{
file2(a,b);
printf("X1=%f\tX2=%f\n",x1,x2);
}
else printf("方程无解");
}

Answer 4

#include "stdio.h"
#include "conio.h"
#include "math.h"int fun(double a,double b,double c,double* result,int* resCount)
{
double p=0;
p=b*b-4*a*c;
if(p>0)
{
result[0]=(-b+sqrt(p))/(2*a);
result[1]=(-b-sqrt(p))/(2*a);
*resCount=2;
return 1;
}
else if(p==0)
{
result[0]=result[1]=-b/(2*a);
*resCount=1;
return 1;
}
else
return 0;
}void main()
{
double a[5]={0};
int iResCount=0;
printf("please Enter the a, b ,c values:");
scanf("%lf%lf%lf",a,a+1,a+2);
if(fun(a[0],a[1],a[2],a+3,&iResCount))
{
if(iResCount==2)
printf("Quadratic equation's Solution set is: %.4lf , %.4lf\n",a[3],a[4]);
else
printf("Quadratic equation's Solution set is: %lf \n",a[3]);
}
else
{
printf("Quadratic equation has no Solution set\n");
}
getchar();
getchar();
}

Answer 5

int a,x,b,c;printf("请输入整数A,X,B,C的值")；sanf("%d%d%d%d",&a,&x,&b,&c);a*x*x+b*x+c=0；printf("x的值为%d",&x); （因为没有经过运行，不知道是否有些问题，但大体上这样就可以了）