求方程的根,用三个函数分别求当b^2-4ac大于0、等于0、和小于0时的根,并输出结果.从主函数输入a、b、c#include#includefloat p,q;void m(int a,int b,int c,float d){\x05p=(-b+sqrt(d))/(2*a);\x05q=(-b-sqrt(d))/(2*a);\x05pri
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![求方程的根,用三个函数分别求当b^2-4ac大于0、等于0、和小于0时的根,并输出结果.从主函数输入a、b、c#include#includefloat p,q;void m(int a,int b,int c,float d){\x05p=(-b+sqrt(d))/(2*a);\x05q=(-b-sqrt(d))/(2*a);\x05pri](/uploads/image/z/14253890-50-0.jpg?t=%E6%B1%82%E6%96%B9%E7%A8%8B%E7%9A%84%E6%A0%B9%2C%E7%94%A8%E4%B8%89%E4%B8%AA%E5%87%BD%E6%95%B0%E5%88%86%E5%88%AB%E6%B1%82%E5%BD%93b%5E2-4ac%E5%A4%A7%E4%BA%8E0%E3%80%81%E7%AD%89%E4%BA%8E0%E3%80%81%E5%92%8C%E5%B0%8F%E4%BA%8E0%E6%97%B6%E7%9A%84%E6%A0%B9%2C%E5%B9%B6%E8%BE%93%E5%87%BA%E7%BB%93%E6%9E%9C.%E4%BB%8E%E4%B8%BB%E5%87%BD%E6%95%B0%E8%BE%93%E5%85%A5a%E3%80%81b%E3%80%81c%23include%23includefloat+p%2Cq%3Bvoid+m%28int+a%2Cint+b%2Cint+c%2Cfloat+d%29%7B%5Cx05p%3D%28-b%2Bsqrt%28d%29%29%2F%282%2Aa%29%3B%5Cx05q%3D%28-b-sqrt%28d%29%29%2F%282%2Aa%29%3B%5Cx05pri)
求方程的根,用三个函数分别求当b^2-4ac大于0、等于0、和小于0时的根,并输出结果.从主函数输入a、b、c#include#includefloat p,q;void m(int a,int b,int c,float d){\x05p=(-b+sqrt(d))/(2*a);\x05q=(-b-sqrt(d))/(2*a);\x05pri
求方程的根,用三个函数分别求当b^2-4ac大于0、等于0、和小于0时的根,并输出结果.从主函数输入a、b、c
#include
#include
float p,q;
void m(int a,int b,int c,float d)
{
\x05p=(-b+sqrt(d))/(2*a);
\x05q=(-b-sqrt(d))/(2*a);
\x05printf("x1=%.3f x2=%.3f",p,q);
}
void f(int a,int b,int c,float d)
{
\x05p=-b/(2*a);
\x05printf("x1=x2=%.3f",p);
}
void j(int a,int b,int c,float d)
{
\x05p=-b/(2*a);
\x05q=sqrt(-d)/(2*a);
\x05printf("x1=%.3f+%.3fi x2=%.3f+%.3fi",p,q,p,q);\x05
}
int main()
{
\x05int a,b,c;
\x05float d;
\x05scanf("%d%d%d",&a,&b,&c);
\x05d=b*b-4*a*c;
\x05if(d>0)
\x05{
\x05\x05m(a,b,c,d);
\x05}
\x05else if(d=0)
\x05{
\x05\x05f(a,b,c,d);
\x05}
\x05else
\x05{
\x05\x05j(a,b,c,d);
\x05}
\x05return 0;
}
求方程的根,用三个函数分别求当b^2-4ac大于0、等于0、和小于0时的根,并输出结果.从主函数输入a、b、c#include#includefloat p,q;void m(int a,int b,int c,float d){\x05p=(-b+sqrt(d))/(2*a);\x05q=(-b-sqrt(d))/(2*a);\x05pri
#include
#include
float p,q;
void m(int a,int b,int c,float d)
{
p=(-b+sqrt(d))/(2*a);
q=(-b-sqrt(d))/(2*a);
printf("x1=%.3f\n x2=%.3f",p,q);
}
void f(int a,int b,int c,float d)
{
p=-b/(2*a); printf("x1=x2=%.3f\n",p);
}
void j(int a,int b,int c,float d)
{
printf("无实根\n");
}
int main()
{
int a,b,c;
float d;
printf("Input A B C\n");
scanf("%d%d%d",&a,&b,&c);
d=b*b-4.0*a*c;
if(d>0)
{
m(a,b,c,d);
}
else if(d==0)
{
f(a,b,c,d);
}
else
{
j(a,b,c,d);
}
printf("\n");
return 0;
}
我也是略作修改 C++里运行没问题了