[英]Pairwise multiplication of the opposite elements of the c++ doubly-linked list
我已經完成了以下任務:使用給定的實數雙向鏈接列表,您必須將列表的相反元素相乘(第一個與最后一個,第二個與最后一個減號,依此類推),然后將此乘積添加到新列表。 即:我們有該列表:
1.1 2.2 3.3 4.4 5.5
然后我們打印
1.1 * 5.5 = 6.05;
2.2 * 4.4 = 9.68;
3.3 * 3.3 = 10.89;
最后的列表是:
6.05 9.68 10.89
我提出了以下朴素算法:
#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;
}
由於向后遍歷列表元素是有問題的,因此我僅找到了刪除列表的最后一個元素的選項。
因此,我想知道是否有人會想出一種更優雅的算法來做到這一點,或者至少完善上面的算法。
您可以使用rbegin()
來回迭代:
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++;
}
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