$\texttt{\large\color{#12a1e8}本\color{#50ce50}蒟蒻\color{#12a1e8}的第\color{#c291e8}七\color{#12a1e8}篇\color{#92a1e8}题\color{#92a9d8}解}$

${\large\color{#92a1e8}手动开3次方}$

$\texttt{\small\color{#e51211}难度：入门}$

last update:2023/4/1

pow()可以求一个数的多次方（通用于c++和python），是常用函数，然后输出的时候取两位即可

${\large\color{#52a1e8}Python：}$

a=int(input())
result=float(pow(a,1/3))
#round(result,2)
print('%.2f'%result)

${\large\color{#52a1e8}C++：}$

#include <bits/stdc++.h>
using namespace std;
int main()
{
 float n,y;
 scanf("%f",&n);
 y=(float)pow (n,1.0/3);
 printf("%.2f\n",y);
 return 0;
}

a=int(input())
b=float(pow(a,1/3))
print('%.2f'%b)

#include #include // 用于比较误差 using namespace std;

// 计算x的三次方根的函数，使用牛顿迭代法 double cbrt(double x, double tolerance) { if (x == 0) { return 0; }

double guess = x / 3; // 初始猜测值
double next_guess;

do {
    next_guess = (2 * guess + x / (guess * guess)) / 3; // 牛顿迭代公式
    if (abs(next_guess - guess) < tolerance) { // 如果误差小于容忍度，则停止迭代
        return next_guess;
    }
    guess = next_guess;
} while (true);

}

int main() { double x; cout << "Enter a number to calculate its cube root: "; cin >> x;

double tolerance = 1e-10; // 设置误差容忍度
double result = cbrt(x, tolerance);

if (result != -1) {

    cout << "The cube root of " << x << " is approximately " << result << endl;
    cout << "Standard library result: " << cbrt(x) << endl; // 使用标准库函数进行比较
}

return 0;

}

a=int(input())
print("%.2f" %(a**(1/3)))

cmath库实现

#include <bits/stdc++.h>
using namespace std;
int main()
{
    double n;
    cin >> n;
    printf("%.2lf",pow(n,1.0/3));
    return 0;
}

二分查找

#include <bits/stdc++.h>
using namespace std;
int main() 
{
    double n;
    cin >> n;
    double l = -10000, r = 10000;
    while (r - l >= 1e-8) {
        double mid = (r + l) / 2;
        if (mid * mid * mid >= n) r = mid;
        else l = mid;
    }
    printf("%.2lf", l);
}

#include<bits/stdc++.h>
using namespace std;
int main()
{
	double n,b;
	cin>>n;
	b=pow(n,1.0/3);
	printf("%.2lf",b);

枚举

#include <iostream>
using namespace std;
int n;
int main(){
    cin>>n;
    for (double i=1;i<=500;i=i+0.0001){
        if((n-i*i*i)<0.0001){
            printf("%.2f",i);
            break;
        }
    }
    return 0;
}

c++