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用二分法求数的平方根问题

bisection square-root codeblocks c++

2

Nameless King · 技术社区 · 7 年前

#include<iostream>
#include<cmath>
using namespace std;
double bisection(double errorVal, double userNum){
    double upper=userNum, lower=0;
    double mid=(lower+upper)/2.0;;
    while(mid*mid!=userNum){
        double mid=(lower+upper)/2.0;
        if(mid*mid>userNum){
            upper=mid;
        } else {
            lower=mid;
        }
    }
    return mid;
}

int main(){
    double errorVal=0, userNum=0;
    std::cout<<"Please enter a number (larger than 0) to calculate its square root, and the desired margin of error."<<std::endl;
    std::cin>>userNum>>errorVal;
    bisection(errorVal,userNum);
    std::cout<<"The calculated result is "<<bisection(errorVal,userNum)<<". The error is "<<abs(bisection(errorVal,userNum)-sqrt(userNum))<<"."<<std::endl;
}

我也想知道如何正确实施 errorVal ,以指定允许的误差幅度。谢谢

1 回复 | 直到 7 年前

1

Stefan 7 年前

误差值用于修复执行浮点运算时出现的任何舍入误差。

以下陈述很少是正确的,因此您的循环可能会持续很长时间。

while(mid*mid==userNum)

计算后比较两个浮点的常用方法是

fabs(x1-x2) < e //where, fabs retrieves the absolute value, 
                //x1,2 are the numbers to compare 
                //and e is the epsilon chosen.

因此,固定误差值(通常称为epsilon)也会固定循环。

double bisection(double errorVal, double userNum){
    double upper=userNum, lower=0;
    double mid=(lower+upper)/2.0;

     //error val added
     //** fabs(mid*mid - userNum) < errorVal is true if the numers are "equal"
     //** and you want to run the loop as long as the are NOT "equal" 
     while(!(fabs(mid*mid - userNum) < errorVal)){

        mid=(lower+upper)/2.0;
        if(mid*mid>userNum){
           upper=mid;
        } else {
            lower=mid;
        }
    }
    return mid;
}

请参阅: http://www.cplusplus.com/reference/cmath/fabs/

https://randomascii.wordpress.com/2012/02/25/comparing-floating-point-numbers-2012-edition/