HJ107 求解立方根

HJ107 求解立方根

#include
#include
using namespace std;
 
class HJ107 {
private:
    static double abs(double x) {
        if (x < 0) {return -x;}
        return x;
    }
public:
    static double getCubeRoot(double c, double err = 1e-5) {
        double x = c;
        while (abs(c-x*x*x) >= err) {
            x = (2*x + c/(x*x)) / 3.0;
        }
        return x;
    }    
};
 
int main() {
    double c = 0.0;
    cin >> c;
    
    // 精确到0.01, 最后1位再四舍五入
    double ans = HJ107::getCubeRoot(c, 1e-2);
    // 保留1位小数
    if (ans > 0) {
        ans = int(10 * (ans + 0.05))/10.0;
    } else {
         ans = int(10 * (ans - 0.05))/10.0;
    }
    
    // 保留n位小数, setprecision参数传n+1
    cout << setprecision(2) << ans << endl;
}

y = x ^ (1/3)

x = y ^ 3;

y^3 - x = 0;

f(y) = y^3 - x;

f'(y) = 3 * (y ^ 2);

根据牛顿迭代法:

Xn1 = Xn - f(Xn) / f'(Xn)

yn1 = y - (y^3 - x) / (3 * y ^ 2) = (2*y + x / y^2) / 3

yn1 为n+1项, y 为n项，得出递推关系式

x换成待开立方需要传入的参数c， y换成x， 迭代后yn1 = y, 得到迭代表达式:

x = (2*x + c /(x * x) )/ 3.0

结束 c == x * x * x,  x^3 -> c 结果无限接近, 得到循环条件:

abs(c - x * x * x) >= err