基本思路就是二分法逼近，代码不长
         请各位给些意见

#include <cstdio>
#include <cmath>
//#include <iostream>
using namespace std;
int main()
{
    double l,n,c,t,u,d,r,l1,angle,tmp;
    while(scanf("%lf%lf%lf",&l,&n,&c)&& (l!=-1))
    {
        if(n==0.00) 
        { 
            printf("%.3lf\n",0); 
            continue; 
        }  
        l1 = l*(1+n*c);
        tmp = l/l1;
        u = 3.1415926*6/7;
        d = 0;
        while(fabs(u-d)>0.000000000001)
        {
        //这个函数是单调的，所以用二分法逼近
            angle = (u+d)/2;t = sin(angle)/angle;
            if(fabs(t-tmp)<0.0000000000000001)
               break;
            if(t>tmp)
                d = angle;
            else
                u = angle;
        }
      //  r = l1/2/angle;
    //    tmp = r*(1-cos(angle));
    //    printf("%.3lf\n",l1/angle/2*(1-cos(angle)));
        printf("%.3lf\n",l1/angle/2*(1-cos(angle)));    
    }    
    return 0;
}

• 一般来说，我用的pi不会少过20位小数，能用牛迭就尽量用 (1035) frkstyc 2005-07-06 00:51:02
• Re:用了double，3.1415926,还是WA，无法理解，请指点一下 (1112) winsweet 2008-05-22 21:17:18