樂鑫筆試題-----開根號

類似https://blog.csdn.net/SHAOYEZUIZUISHAUI/article/details/106452586

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//注意審題：不用任何數學庫函數，所以我們只用加減乘除
#include <iostream>
using namespace std;
//二分搜索求開n次方
//double epr = 10E-14;
#define epr 10E-14
double Mysqrt(double m, int n)
{
    //double 十進制小數點後十五位精確
    double left = 0;
    double right = m;
    //必須保證第一次while循環可以進去
    double mid = 0;
    double temp=m;
    while (mid - temp>epr ||mid - temp<-epr)
    {
        temp = m;
        mid = left + (right - left) / 2;
        
        //m除以mid n-1次，比如求開方，就除mid一次，再和mid比較
        for (int i = 0; i<n-1; i++)
            temp = temp / mid;
        //二分
        if (mid>temp)
            //因爲是保留小數點12位，所以，right=mid而非二分查找的mid+1
            right = mid;
        else
            left = mid;
    }

    return temp;
}
int main()
{
    double m;
    int n;
    while (cin >> m >> n)
    {
        //輸出13位有效數字，前提是有足夠多的有效數字
        //cout.precision(13);
        //cout << Mysqrt(m, n);
        printf("%.12lf\n",Mysqrt(m,n));
    }
}