x为初始值，y为函数目标值
   fx是函数值，ffx是一阶导


double Newton(double x,double y)
{
double xn1, xn,fx,ffx;
xn1 = x;
do{
    xn = xn1;
    fx = sin(xn)/xn - y;
    ffx = (xn*cos(xn)-sin(xn))/(xn*xn);
    xn1 = xn - (fx/ffx);
}while(abs(fx) > 1.0e-12);
//printf("%.6lf    %.6lf\n",sin(xn)/xn, y);
return xn1;
}
int main()
{
    double l,n,c,l1,angle,tmp,pi;
    while(scanf("%lf%lf%lf",&l,&n,&c)&& (l!=-1))
    {
        if(n==0.00||l<0.001) 
        { 
            printf("%.3lf\n",0); 
            continue; 
        }  
        l1 = l*(1+n*c);
        pi = 3.1415926535897932384626433832795;
        angle = Newton(pi / 2,1/(1+n*c));
        printf("%.3lf\n",l1/angle/2*(1-cos(angle)));    
    }    
    return 0;
} 

• 这个是我的迭代 (289) frkstyc 2005-07-06 01:41:22
  • 没话说了，兄弟太仗义了，谢谢 (309) sunmoonstar_love 2005-07-06 01:48:46
  • 是不是应该对r迭代而不是alpha? (0) HatTricker 2006-08-28 10:57:04
  • 请问那个r = lp * lp * 0.25 / sqrt(lp * lp - l * l);中的r是怎么确定的呢？这个式子不太懂。 (311) miaomiaomiaomiao 2007-08-25 20:30:30