第2个回答 推荐于2017-09-11
#include <stdio.h>
#include <conio.h>
#include <math.h>
int main(void)
{
double a,b,c,d,e,f;
printf("请输入: a,b,c:");
scanf("%lf %lf %lf",&a,&b,&c);
d=b*b-4*a*c;
if( 0 == a )
{
if( 0 == b )
{
if( 0 == c )
{
printf("参数都为零,方程无意义\n");
}
else
{
printf("ab都为零,c不为零,方程不成立\n");
}
}
else
{
printf("x=%lf\n",-c/b);
}
}
else if( d>=0 )
{
printf("x1=%lf\n",(-b+sqrt(d))/(2*a));
printf("x2=%lf\n",(-b-sqrt(d))/(2*a));
}
else
{
printf("x1=%lf+%lf\n",-b/(2*a),sqrt(-d)/(2*a));
printf("x2=%lf-%lf\n",-b/(2*a),sqrt(-d)/(2*a));
}
getch();
return 0;
}
第3个回答 2015-04-02
#include <stdio.h>
int main()
{
int a,b,m,c,d,n;
double x=0,y=0;
printf("Enter three integers(a,b,m):");
scanf("%d,%d,%d",&a,&b,&m);
printf("Enter three integers(c,d,n):");
scanf("%d,%d,%d",&c,&d,&n);
printf("%dx + %dy=%d\n",a,b,m);
printf("%dx + %dy=%d\n",c,d,n);
x=(m*d-b*n)/(a*d-b*c);
y=(m*c-a*n)/(b*c-a*d);
printf("The two lines intersect at point:(%.2lf,%.2lf)\n",x,y);
return 0;
}
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