【小数二分】Acwing790.数的三次方根

Acwing790.数的三次方根

题解

做法一：通过二分运算到符合精度要求

#include <iostream>
#include <cstdio>

using namespace std;

#define f 1e-8

int main()
{
    double x;
    cin >> x;
    double bg = -10000, ed = 10000, md, temp;
    while(ed-bg > f){
        md = (bg+ed)/2;
        temp = md*md*md;
        if(temp >= x)
            ed = md;
        else
            bg = md;
    }
    printf("%.6lf\n",md);
    return 0;
}

做法二：进行多次循坏，达到精度要求

#include <iostream>
#include <cstdio>
#include <cmath>

using namespace std;

#define f 1e-6

int main()
{
    double x;
    cin >> x;
    double bg = -10000, ed = 10000, md, temp;
    for(int i = 0; i < 100; ++i){
        md = (bg+ed)/2;
        temp = pow(md,3);
        if(temp >= x)
            ed = md;
        else
            bg = md;
    }
    printf("%.6lf\n",md);
    return 0;
}

标签：md,bg,10000,temp,ed,方根,include,Acwing790,小数
来源： https://www.cnblogs.com/czy-algorithm/p/16243932.html