三分法

double l = 0,r = 10000;
while(r-l>=0.01){//精度问题
    double m1 = l + (r-l)/3.0,m2 = r - (r-l)/3.0;
    if(f(m1)<f(m2))
        l = m1;
    else
        r = m2;
}

rush!
原文地址:https://www.cnblogs.com/LH2000/p/14631430.html