std::ranges::fill_n

template< class T, std::output_iterator<const T&> O >
constexpr O fill_n( O first, std::iter_difference_t<O> n, const T& value );

template< class O, class T = std::iter_value_t<O> >
requires std::output_iterator<O, const T&>
constexpr O fill_n( O first, std::iter_difference_t<O> n, const T& value );

struct fill_n_fn
{
    template<class O, class T = std::iter_value_t<O>>
    requires std::output_iterator<O, const T&>
    constexpr O operator()(O first, std::iter_difference_t<O> n, const T& value) const
    {
        for (std::iter_difference_t<O> i {}; i != n; ++first, ++i)
            *first = value;
        return first;
    }
};

inline constexpr fill_n_fn fill_n {};

#include <algorithm>
#include <complex>
#include <iostream>
#include <string>
#include <vector>

void println(const auto& v)
{
    for (const auto& elem : v)
        std::cout << ' ' << elem;
    std::cout << '\n';
}

int main()
{
    constexpr auto n{8};
    
    std::vector<std::string> v(n, "▓▓░░");
    println(v);
    
    std::ranges::fill_n(v.begin(), n, "░░▓▓");
    println(v);

    std::vector<std::complex<double>> nums{{1, 3}, {2, 2}, {4, 8}};
    println(nums);
    #ifdef __cpp_lib_algorithm_default_value_type
        std::ranges::fill_n(nums.begin(), 2, {4, 2});
    #else
        std::ranges::fill_n(nums.begin(), 2, std::complex<double>{4, 2});
    #endif
    println(nums);
}

Вывод:

 ▓▓░░ ▓▓░░ ▓▓░░ ▓▓░░ ▓▓░░ ▓▓░░ ▓▓░░ ▓▓░░
 ░░▓▓ ░░▓▓ ░░▓▓ ░░▓▓ ░░▓▓ ░░▓▓ ░░▓▓ ░░▓▓
 (1,3) (2,2) (4,8)
 (4,2) (4,2) (4,8)