Expanding Rods(ZJU_2370二分法)

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//使用二分法解方程,测试数据过了,但提交WA.(已改正确)

地址:http://acm.zju.edu.cn/show_problem.php?pid=2370

#include <stdio.h>
#include <math.h>
int main()
{
double l,c,n,ll,t,min,max,mid,y,r,h;
while(scanf("%lf%lf%lf",&l,&n,&c))
{
  if(l==-1&&n==-1&&c==-1)
   break;
  if(n==0||c==0)
   printf("%.3lf/n",0);
  else
  {
  ll=(1+c*n)*l;
  t=l/ll;
  //printf("%lf/n",t);
  min=0;
  max=3.1415927;

  while(true)
  {
   mid=(min+max)/2;
   y=sin(mid)-t*mid;
   if(y>0.00000000000001)
    min=mid;
   else if(y<0)
    max=mid;
   else break;
  }
  //printf("%lf/n",mid);
  r=(l/2.0)/sin(mid);
  h=r-r*cos(mid);
  printf("%.3lf/n",h);
  }
}
}

//正确版本:

#include <stdio.h>
#include <math.h>
int main()
{
double l,c,n,ll,t,min,max,mid,y,r,h;
while(scanf("%lf %lf %lf",&l,&n,&c)!=-1)
{
if(l==-1&&n==-1&&c==-1)
break;
if(n*c==0)
printf("%.3lf/n",0.0);
else
{
ll=(1+c*n)*l;
t=l/ll;
//printf("%lf/n",t);
min=0;
max=acos(-1)/2;

for(int i=0;i<100;i++)
{
   mid=(min+max)/2;
   y=sin(mid)-t*mid;
   if(y>0)
    min=mid;
   else
    max=mid;
}
//printf("%lf/n",mid);
h=(1-cos(mid))*ll/2/mid;
printf("%.3lf/n",h);
}
}
}