Question

此代码解决了三次函数的根。抛开任何数学概念，我特别想做三次循环（因为它的立方体）。因此，用户将输入根的猜测，v THREE TIMES。

现在问题出现在FIRST DO-WHILE LOOP之后，变量v的值似乎没有超过用户重新输入SECOND和THIRD DO-WHILE LOOP的值。

请帮帮我们！

#include <iostream>
#include <cmath>
#include <iomanip>
using namespace std;

const double R = 0.082054;

double f(double a,double b,double P,double T,double v);        
double f(double a,double b,double P,double T,double v)
{
    double q =P*pow(v,3)-(P*b+R*T)*pow(v,2)+a*v-a*b;
    return q;
}
double fPrime(double a,double b,double P,double T,double v);
double fPrime(double a,double b,double P,double T,double v)
{
    double s =3*P*pow(v,2)-2*(P*b+R*T)*pow(v,1)+a;
    return s;
}

int main()
{
    double a,b,P,T,Tc;
    cin >> a;
    cin >> b;
    cin >> P;
    cin >> T;

    double v,v1=0,fv,fv1,N,e;                                   //FIRST DO-WHILE
    int iteration = 0;
    cout << "\nUse initial guess for molal volume, v: ";
    cin >> v;
    cout << "\nUse maximum number of iterations, N: ";
    cin >> N;
    cout << "\nUse tolerance, e: ";
    cin >> e;
    do
    {
        v = v1;
        fv = f(a,b,P,T,v);
        fv1 = fPrime(a,b,P,T,v);
        if (fv1==0)
        {
           cout << "Math Error";
           break;
        }
        v1 = v-(fv/fv1);
        cout << iteration+1 << setw(12) << v << setw(16) << v1 << setw(16) << abs(v1-v) << endl;
        iteration++;
    }
    while(iteration<N && abs(v1-v)>e);
    cout << "The approximate root/solution is " << v1 << endl;

    iteration = 0;                                              //SECOND DO-WHILE
    cout << "\nUse initial guess for molal volume, v: ";
    cin >> v;
    cout << "\nUse maximum number of iterations, N: ";
    cin >> N;
    cout << "\nUse tolerance, e: ";
    cin >> e;
    do
    {
        v = v1;
        fv = f(a,b,P,T,v);
        fv1 = fPrime(a,b,P,T,v);
        if (fv1==0)
        {
           cout << "Math Error";
           break;
        }
        v1 = v-(fv/fv1);
        cout << iteration+1 << setw(12) << v << setw(16) << v1 << setw(16) << abs(v1-v) << endl;
        iteration++;
    }
    while(iteration<N && abs(v1-v)>e);
    cout << "The approximate root/solution is " << v1 << endl;

    iteration = 0;                                              //THIRD DO-WHILE
    cout << "\nUse initial guess for molal volume, v: ";
    cin >> v;
    cout << "\nUse maximum number of iterations, N: ";
    cin >> N;
    cout << "\nUse tolerance, e: ";
    cin >> e;
    do
    {
        v = v1;
        fv = f(a,b,P,T,v);
        fv1 = fPrime(a,b,P,T,v);
        if (fv1==0)
        {
           cout << "Math Error";
           break;
        }
        v1 = v-(fv/fv1);
        cout << iteration+1 << setw(12) << v << setw(16) << v1 << setw(16) << abs(v1-v) << endl;
        iteration++;
    }
    while(iteration<N && abs(v1-v)>e);
    cout << "The approximate root/solution is " << v1 << endl;

return 0;
}

Answer 1

在我看来你的行'cin＆gt;＆gt; v;'在你的代码中没有意义，因为每次你读v时，你都会立即将v1的值分配给v，其中'v = v1'。所以你最初的猜测就丢失了。此外，由于第二次执行的退出条件与第一次执行的退出条件相同，第二次执行只应执行一次，执行后的v值应等于v1 = v-fv / fv1，所以你永远不会得到一个与第一个值不同的v的新第二个值。

如何使用do-while循环连续使用所涉及变量的新值（由用户重新输入）？

1 个答案: