C ++如何仅对存储在向量中的偶数求和？

Question

我是 C++ 新手，我运行完全不了解如何仅对存储在 C++ 向量中的偶数求和。

任务本身要求用户输入一些随机整数，当输入为 0 时停止，然后 return 偶数的数量以及这些偶数的总和。

据我所知：

#include <algorithm>
#include <functional>
#include <iostream>
#include <vector>
#include <numeric>

using namespace std;

int main()

{
    vector<int> vet;
    int s = 1;
        while (s != 0) {
        std::cin >> s;
        vet.push_back(s);
    }


    int n = count_if(vet.begin(), vet.end(),
        [](int n) { return (n % 2) == 0; });
    cout << n << endl;
    

    //here is the start of my problems and lack of undertanding. Basically bad improv from previous method
    int m = accumulate(vet.begin(), vet.end(), 0,
        [](int m) { for (auto m : vet) {
              return (m % 2) == 0; });
              cout << m << endl;          //would love to see the sum of even values here

        
        
   return 0;
}

Answer 1

传递给std::accumulate的函数取2个值：当前累加值和当前元素值

你应该做的是如果它是偶数就加上它，如果不是偶数就不要改变。

int m = accumulate(vet.begin(), vet.end(), 0,
        [](int cur, int m) {
            if ((m % 2) == 0) {
                return cur + m; // add this element
            } else {
                return cur; // make no change
            }
        });

Answer 2

这是我的解决方案（抱歉，如果不对我写在我的 phone 上）

你不需要向量形式，你只需要直接从输入中检查数字是否可以被 2 整除我的解决方案：（有点难看）

#include <iostream>

using namespace std;

    int main()
    {
        int s {1};
        int sum{};
        int countNum{};
        while (s != 0)
        {
            cin >> s;
            if (s % 2 == 0)
            {
                sum += s;
                countNum++;
            }
        }
        cout << countNum << ' ' << sum;
    }

Answer 3

我真的不知道你想在代码的第二部分做什么，但你可以通过这种方式对偶数求和，我想告诉你另一件事，当你使用命名空间标准时，你不会需要写std::cin 只能直接写cin

#include <iostream>
#include <vector>

using namespace std;

int main()

{
    vector<int> vet;
    int s = 1;
    //Take Input
    while (s != 0) {
        cin >> s;
        vet.push_back(s);
    }
    //count elements
    int elements_count = vet.size(); //vet.size() return the total number of elements of vector

    //store the sum here
    int sum=0;

    //loop on the vector and sum only even numbers
    for(int i=0;i<elements_count;i++){
        if(vet[i] %2 ==0)
            sum += vet[i];//check of the element of index i in the vector is even if it true it will add to sum
    }
    cout << sum;

   return 0;
}

Answer 4

从 c++20 开始，您可以分离出检查偶数的逻辑，以及对这些值求和的逻辑：

auto is_even = [](int i) { return i % 2 == 0; };

auto evens = vet | std::views::filter(is_even);

auto sum = std::accumulate(std::begin(evens), std::end(evens), 0);

这是一个demo。

Answer 5

int sumEven=0;
int v[100];
int n;//number of elements you want to enter in the array
do{cout<<"Enter n";
  cin>>n;}while(n<=0);
//in a normal 1 dimensional array
for(int i=0;i<n;i++)
if(v[i]%2==0)
    sumEven+=v[i];

//in a vector
vector<int> v;
for(vector<int>::iterator it=v.begin();it!=v.end();it++)
if(*it%2==0)
    sumEven+=v[i];

Answer 6

Range-based 循环

A range-based for 循环可以说是 STL 算法的有效替代方案，特别是在算法的运算符为 non-trivial.

的情况下

在 C++14 和 C++17 中

例如在 immediately-executed 可变 lambda 中包装 range-based even-only 累积 for 循环：

#include <iostream>
#include <vector>

int main() {
    // C++17: omit <int> and rely on CTAD.
    const std::vector<int> v{1, 10, 2, 7, 4, 5, 8, 13, 18, 19};
    const auto sum_of_even_values = [sum = 0, &v]() mutable {
        for (auto val : v) { 
            if (val % 2 == 0) { sum += val; }
        }
        return sum;
    }();
    
    std::cout << sum_of_even_values; // 42
}

C++20

从 C++20 开始，您可以在 range-based for 循环以及 the ranges library, allowing you to declare a binary comparator in the initialization statement of the range-based for loop, and subsequently apply it the range-expression of the loop, together with the std::ranges::filter_view 适配器中使用初始化语句：

#include <iostream>
#include <vector>
#include <ranges>

int main() {
    const std::vector v{1, 10, 2, 7, 4, 5, 8, 13, 18, 19};

    const auto sum_of_even_values = [sum = 0, &v]() mutable {
        for (auto is_even = [](int i) { return i % 2 == 0; };
             auto val : v | std::ranges::views::filter(is_even)) { 
            sum += val;
        }
        return sum;
    }();
    
    std::cout << sum_of_even_values; // 42
}

Answer 7

与上面的答案类似，但是如果您还想保留偶数向量，这里有两种方法。

#include <algorithm>
#include <iostream>
#include <numeric>
#include <vector>

int main() {

    std::vector<int> vec = {1,2,3,4,5,6,7,8,9,10};

    // Hold onto what we know is the right answer.
    int known_sum = 2+4+6+8+10;

    // Copy only even values into another vector
    std::vector<int> even_values;
    std::copy_if(vec.begin(), vec.end(),
                std::back_inserter(even_values),
                [](int val){ return val%2==0; });

    // Compute sum from even values vector
    int even_value_sum = std::accumulate(even_values.begin(), even_values.end(), 0);

    // Compute sum from original vector
    int even_value_second = std::accumulate(vec.begin(), vec.end(), 0, 
        [](int current_sum, int new_value) {
            return new_value%2==0 ? current_sum + new_value:current_sum;
        }
    );

    // These should all be the same.
    std::cout << "Sum from only even vector: " << even_value_sum << std::endl;
    std::cout << "Sum from binary op in std accumulate: " << even_value_second << std::endl;
    std::cout << "Known Sum: " << known_sum << std::endl;

}

C ++如何仅对存储在向量中的偶数求和？

C++ How to sum only even values stored in vector?

c++

sum

vector

cycle

Range-based 循环

在 C++14 和 C++17 中

C++20