5 /*description:
6 This program which approximates one root of a polynomial
7 in a closed interval of the real number line and use the
8 bisection method to approximate the root.
9 */
10
11 #include <iostream>
12 #include <string>
13 #include <sstream>
14 #include <vector>
15 #include <iomanip>
16 #include <cmath>
17
18 using namespace std;
19
20 double polynomial(vector<double>,double,double,int);
21
22 int main()
23 {
24 int iterations;
25 cout << "How many iterations? ";
26 cin >> iterations;//gets the number of iterations
27
28 double left_point,right_point,middle;
29 cout << "Enter the endpoints of the interval containing the root: ";
30 cin >> left_point >> right_point;
31 cin.ignore();
32
33 vector<double> coefficients;
34 cout << "Enter the polynomial coefficients, ascending in degree: ";
35 string line;
36 getline(cin,line);// gets an entire line of input
37 istringstream input_string(line); // use input_string like cin
38
39 double num;
40 while (input_string >>num)
41 {
42 coefficients.push_back(num);
43 }
44
45 cout << "Root estimate: " << setprecision(8) //Print at least 8 significant digits
46 << polynomial(coefficients,left_point,right_point,iterations);
47
48 }
49
50 //function
51 double polynomial(vector<double> coefficients,double left_point,double right_point,int iterations)
52 {
53 double middle;
54 for (int n = 0; n < iterations; n++)
55 {
56 middle = (left_point + right_point) / 2; //use the bisection method
57 double sum1 = 0, sum2 = 0, sum3 = 0;
58 for (int i = 0; i < coefficients.size(); i++)
59 {
60 sum1 = sum1 + coefficients[i]*pow(left_point,i);
61 sum2 = sum2 + coefficients[i]*pow(middle,i);
62 sum3 = sum3 + coefficients[i]*pow(right_point,i);
63 }
64
65 if (sum1*sum2 > 0)
66 left_point = middle;
67 else
68 right_point = middle;
69
70 }
71
72 return middle;
73 }
以上是我的一位朋友为她的项目写的节目。但是她试图找到cin.ignore的替代品，因为她的教授禁止使用它。我无法想到，所以我在这里寻求帮助，它不能像其他任何其他库函数一样明确。