成对乘以c ++双向链表的相对元素

Question

我已经获得了以下任务：使用给定的双链实数列表，您必须将列表中的相反元素相乘（首先是最后一个，第二个是最后一个，等等）并将此产品添加到新的清单。即：我们有这个清单：

1.1 2.2 3.3 4.4 5.5 

然后我们打印

1.1 * 5.5 = 6.05; 
2.2 * 4.4 = 9.68; 
3.3 * 3.3 = 10.89;

最后的清单是：

6.05 9.68 10.89 

我想出了以下天真的算法：

#include <iostream>
#include <list>

using namespace std;

int main() {
    double x = 0;   
    double q = 0;   //for the product of the elements
    list <double> user_values, calculated_products;

    //data entry
    while ( cin >> x) {
        user_values.push_back(x);
        if (cin.get() == '\n') break;
    }

    //pairwise multiplication of the opposite elements (х1 * хn; x2 * xn-1; etc.):
    for (auto p = user_values.begin(); p!=(user_values.end()); ++p){
        cout << (*p) << " * " << (*user_values.rbegin()) << " = " ;
        q = (*p) * (*user_values.rbegin());  //result of the multiplication
        cout << q  << "; " << endl;
        calculated_products.push_back(q);  //saving result to the new list
        user_values.pop_back();   //removing the last element of the list, in order to iterate backwards. This is probably the most confusing part.
    }

    //result output:
    cout << "we have such list of products: " << endl;
    for (const auto& t: calculated_products){
        cout << t << " ";
    }
    cout << endl;
    return 0;
}

由于向后迭代列表的元素是有问题的，我只找到了删除列表的最后元素的选项。

所以我想知道是否有人能想出更优雅的算法来做到这一点，或者至少改进上面的算法。

Answer 1

您可以使用rbegin()从后向前迭代：

auto i1 = user_values.begin();
auto i2 = user_values.rbegin();
double bufResult = 0;   //for the product of the elements

for(int i=0; i<user_values.size()/2; i++)
{
    bufResult = (*i1) * (*i2);  //result of the multiplication
    cout << (*i1) << " * " << (*i2) << " = " << bufResult << "; " << endl;
    calculated_products.push_back(bufResult);  //saving result to the new list
    i1++;
    i2++;
}