如何在 C++ 中将 1 放在零的 martix（或二维数组）中的随机位置[关闭]

Question

我想创建一个 martix 9x9，里面有 10 个，用 C/C++ 随机填充

例如它应该看起来像这样：

array[[ 0, 1, 0, 0, 0, 1, 0, 0, 0, ]
      [ 0, 0, 0, 0, 0, 0, 0, 0, 0, ]
      [ 0, 1, 0, 0, 0, 0, 0, 0, 0, ]
      [ 0, 0, 0, 0, 0, 0, 0, 1, 0, ]
      [ 0, 1, 0, 0, 0, 1, 0, 0, 0, ]
      [ 0, 0, 0, 0, 0, 0, 0, 0, 0, ]
      [ 0, 0, 0, 0, 0, 0, 0, 0, 1, ]
      [ 0, 0, 1, 0, 0, 0, 0, 0, 0, ]
      [ 1, 0, 0, 0, 0, 1, 0, 0, 0, ]]

Answer 1

有不同的方法。您可以用

1

填充前 10 个位置，然后调用

std::shuffle

将它们放在随机位置。您还可以创建一个可用职位列表，然后从该列表中随机选择 10 个。

后者的注释示例：

#include <array>    // array
#include <iostream> // 
#include <numeric>  // iota
#include <random>   // mt19937, uniform_int_distribution

int main() {
    constexpr unsigned side_size = 9;
    // a seeded pseudo number generator
    std::mt19937 prng(std::random_device{}());
    

    // all zeroes:
    std::array<std::array<int, side_size>, side_size> arr_to_fill{};

    // create an array with all available positions number from 0-
    std::array<unsigned, side_size * side_size> positions;
    std::iota(positions.begin(), positions.end(),
              0);  // [0, side_size*side_size)

    unsigned ones = 10;  // the number of 1's we want

    for (unsigned i = 0; i < ones; ++i) {
        unsigned last = positions.size() - 1 - i;  // last unpicked pos
        // distribution of random numbers from 0 to last (inclusive)
        std::uniform_int_distribution dist(0u, last);

        unsigned idx = dist(prng);         // get random index ...
        unsigned pos = positions[idx];     // ... to pick a position
        positions[idx] = positions[last];  // copy last unpicked position

        // put a 1 in the chosen positions place:
        arr_to_fill[pos / side_size][pos % side_size] = 1;
    }
    // print result:
    for (auto& inner : arr_to_fill) {
        for (auto v : inner) std::cout << ' ' << v;
        std::cout << '\n';
    }
}