二分法求函数零点

long double l=INT_MIN,r=INT_MAX,mid,eps=1e-6;
while(r-l>eps)
{
  	mid=(l+r)/2;
  	if(f(mid)<0) l=mid;
  	else r=mid;
}
cout<<l<<endl;

三分法求单谷函数最小值

long double l=INT_MIN,r=INT_MAX,lmid,rmid,eps=1e-6;
while(r-l>eps)
{
  	lmid=l+(r-l)/3;
  	rmid=r-(r-l)/3;
  	if(f(lmid)<f(rmid)) r=rmid;
  	else l=lmid;
}
cout<<f(l)<<endl;

求导二分法求单谷函数最小值

inline long double df(long double x)
{
    long double dx=1e-6;
    return (f(x+dx)-f(x-dx))/(dx*2);
}

signed main()
{
    long double l=INT_MIN,r=INT_MAX,mid,eps=1e-6;
    while(r-l>eps)
    {
        mid=(l+r)/2;
        if(df(mid)<0) l=mid;
        else r=mid;
    }
    cout<<f(l)<<endl;
    return 0;
}

优化三分法求单谷函数最小值

long double l=INT_MIN,r=INT_MAX,lmid,rmid,eps=1e-6;
while(r-l>eps)
{
  	lmid=(l+r)/2-eps*0.01;
  	rmid=(l+r)/2+eps*0.01;
  	if(f(lmid)<f(rmid)) r=rmid;
  	else l=lmid;
}
cout<<f(l)<<endl;