3.求解方程ax+bx+c=0的根,要求 (1)画出N-S流程图 （2）写出伪代码 （3）写出相应程序

3.求解方程ax+bx+c=0的根,要求 (1)画出N-S流程图 （2）写出伪代码 （3）写出相应程序

应该是Axx+Bx+C=0的根吧,我这写了个C语言的代码,至于第一步和第二步你还是看着代码自己完成吧,总不能完全依赖别人是吧...
// Solve Equation.cpp :Defines the entry point for the console application.
//
#include
#include
double a,b,c;
void initialize();//初始化工作
void Process();//求解过程
int main(int *argc,int *argv[])
{
int i;
char choice;
do{
i=0;
initialize();
Process();
printf(\t go on?(y/n)\n);
getchar();
scanf(%c,&choice);
if((choice=='y')||(choice=='Y'))
i=1;
}while(i);
return 0;
}
void initialize()
{
printf(-------------axx+bx+c=0----------------\n);
printf(please input the factors:\n);
printf(a=);
scanf(%lf,&a);
printf(b=);
scanf(%lf,&b);
printf(c=);
scanf(%lf,&c);
}
void Process()
{
double dt;//判别式
double x1,x2;
x1=x2=0;
dt=b*b-4*a*c;
if(0==a)
{
if(0==b)
printf(error:'a' and 'b' can`t be both zero!);
else if(0!=c)
printf(\tX=%lf\n-------the quation has only one root\n,-b/c);
else
printf(\tX=0\n-------the quation has only one root\n);
}
else
{
if(dt>=0)
{
dt=sqrt(dt);
x1=(-b+dt)/(2*a);
x2=(b+dt)/(2*a);
printf(\tX1=%lf,x1);
printf(\n\tX2=%lf,x2);
printf(\n-------the quation has two real root.);
}
else
{
dt=sqrt(-dt);
dt=dt/(2*a);
x1=(-b)/(2*a);
x2=b/(2*a);
printf(\tX1=%lf+j%lf,x1,dt);
printf(\n\tX2=%lf-j%lf,x2,dt);
printf(\n-------the quation has two imaginary root\n);
}
}
}