如何以组合形式 C++ 打印此表？ [关闭]

Question

我需要打印这张表

2
0
4 2
3 3
6 4 2
6 6 6
8 6 4 2
9 9 9 9

我为以下结果编写了这段代码

#include <iostream>
using namespace std;

int main(){
    const int N = 9; 
    for(int i = 0; i <= N; i += 3){   
        for (int j = 0; j <= i; j +=3) {
            cout << i << " ";  
        }       
        cout << endl; 
    }
    cout << "\n";
    for(int i = 2; i <= N; i += 2){    
        for (int j = i; j > 0; j -= 2) {
            cout << j  << " ";    
        }         
        cout << endl; 
    }
    return 0;
}

我的结果：

0
3 3
6 6 6
9 9 9 9

2
4 2
6 4 2
8 6 4 2

所需结果：

2
0
4 2
3 3
6 4 2
6 6 6
8 6 4 2
9 9 9 9

Answer 1

#include <iostream>
using namespace std;

int main(){
    const auto n = 4;
    auto count = 0;
    for (auto i = 2; i <= n * 2; i += 2)
    {
        for (auto j = i; j > 0; j -= 2)
            std::cout << j << " ";
        std::cout << std::endl;
        for (auto j = 0; j < (i == 2 ? i : i + 2); j += 3)
            std::cout << count * 3 << " ";
        ++count;
        std::cout << std::endl;
    }


  return 0;
}

编辑：已更正...

有点接近答案
2
0 ==> 真
4 2
0 3 ==> 应该是 3 3
6 4 2
0 3 6 ==> 应该是 6 6 6
8 6 4 2
0 3 6 ==> 应该是 9 9 9

Answer 2

当您提供重复次数时，这是另一种方法（例如在@Ruks 答案中）：

void printSequence(unsigned int repeats)
{
    int n = 2;
    for(int i = 1; i < repeats; i++)
    {
        n+=2*i;
    }
    //n - number of all numbers in sequence for given number of repeats

    int step = 0;
    int numsPerRow = 1;

    for(int i = 0; i < n; i+=step)
    {
        for(int j = numsPerRow; j > 0; j--)
        {
            std::cout << 2*j << " ";
        }
        std::cout << std::endl;
        for(int j = 0; j < numsPerRow; j++)
        {
            std::cout << step+numsPerRow-1 << " ";
        }
        std::cout << std::endl;

        step+=2;
        numsPerRow++;
    }
}

Answer 3

使用这个：

void print_sequence(unsigned long long const repeat = 4, unsigned long long const i = 0)
{
    for (auto j = (i + 1ull) * 2ull; j > 2ull; j -= 2ull)
        std::cout << j << " ";
    std::cout << (repeat > 0ull && repeat < -1ull ? "2\n" : "");
    for (auto j = 0ull; j < i; j++)
        std::cout << i * 3ull << " ";
    std::cout << (repeat > 0ull && repeat < -1ull ? std::to_string(i * 3ull) + "\n" : "");
    if (repeat < -1ull && i + 1ull < repeat)
        print_sequence(repeat, i + 1ull);
}

编辑：我能想到的最短最强的方法……

然后这样称呼它：

print_sequence();

如果你不想要 4 次：

print_sequence(10)

它会重复你想要的次数......

亲切的问候，

鲁克斯。