void main ()
{
long double *y=NULL,*x=NULL,a=0.0,R=0.0,R1=0.0,xf=0.0,xd=0.0,xw=0.0;
int ii=0,ij=0,num=0,num1=0,a1=1;
printf(\"估计需要多少塔板：\");
scanf (\"%d\
y=(long double *)malloc(sizeof(long double)*num1);
x=(long double *)malloc(sizeof(long double)*num1);
printf(\"输入相对挥发度α：\");
scanf (\"%lf\
printf (\"输入精馏段回流比R:\");
scanf(\"%lf\
printf (\"输入精馏段回流比R1:\");
scanf(\"%lf\
printf (\"输入进料易挥发组分摩尔分数xf:\");
scanf(\"%lf\
printf (\"输入塔顶易挥发组分摩尔分数xd:\");
scanf(\"%lf\
printf (\"输入塔釜易挥发组分摩尔分数xw:\");
scanf(\"%lf\
printf (\"开始计算理论踏板\\n\");
*y=xd;
for (ii=0;ii<=num1;ii++)
{
*(x+ii)=*(y+ii)/(a-(a-1)*(*(y+ii)));
printf(\"根据相平衡关系式x%d=%lf\\n\
if (*(x+ii)<=xw)
{
printf(\"理论踏板%d\\n\
break;
}
if (*(x+ii)>=xf)
{
*(y+ii+1)=R/(R+1)*(*(x+ii))+xd/(R+1);
printf (\"根据精馏段操作线方程y%d=%lf\\n\
num++;
printf(\"第%d块板\\n\
}
else
{
while(a1)
{
printf(\"\\n第%d块板是最佳进料板\\n\\n\
a1=0;
}
*(y+ii+1)=(R1+1)/R1*(*(x+ii))+xw/R1;
printf (\"根据提馏段操作线方程y%d=%lf\\n\
num++;
printf(\"第%d块板\\n\
}
}
}