h
ho
hou
hous
house

#include <iostream>
#include <string.h>
using namespace std;
int main()
{
   char cuvant[100];
   int i, k;
   
   cin >> cuvant;
   
   for (i = 0; i < strlen(cuvant); i++)
   {
       for (k = 0; k < i; k++)
       {
           if (k == 0)
           {
               cout << cuvant[k] << endl;
           }
           else
           {
               for (k = 1; k <= i; k++)
               {
                   if (k == i) cout << endl;
                   cout << cuvant[k];
               }
           }
       }
   }
}

#include <iostream>
#include <string>
#include <string_view>
int main() {
  std::string s;
  std::cin >> s;
  for (std::string::size_type i = 0, size = s.size(); i != size; ++i)
    std::cout << std::string_view{s.c_str(), i + 1} << '\n';
}

#include <algorithm>
#include <iostream>
#include <iterator>
#include <string>
int main() {
  std::string s;
  std::cin >> s;
  for (auto const& ch : s) {
    std::copy(s.c_str(), (&ch + 1),
              std::ostream_iterator<decltype(ch)>(std::cout));
    std::cout << '\n';
  }
}

尽管如此，我认为对于你的学习进度来说，使用调试器自己找出问题会更好。以下是你的代码存在的问题：
对于 `i=0`（外部循环的第一次迭代），`for(k=0;k<i;k++)` 不会执行，因为 `k<0` 计算结果为 `false`。
而且，在两个嵌套的 `for` 循环中更改运行变量 (`k`) 大多数情况下也表示有问题。
所以你想做的是：你想要创建每个可能的前缀，因此你想要创建长度为 `1` 到 `n` 的 `n` 个字符串。所以你的外部循环的第一个想法是正确的。但你在内部部分过于复杂了。
对于内部部分，你想要打印从索引 `0` 到 `i` 的所有字符。
```cpp
int main() {
    char cuvant[100];
    std::cin >> cuvant;
    // 循环遍历字符串的长度
    for (int i = 0, size = strlen(cuvant); i < size; i++) {
        // 打印从 0 到 i 的所有字符 (k<=0)
        for (int k = 0; k <= i; k++) {
            std::cout << cuvant[k];
        }
        // 打印换行符
        std::cout << std::endl;
    }
}

int main() {
    std::string s;
    std::cin >> s;
    for (std::size_t i = 0, size = s.size(); i < size; i++) {
        std::cout << s.substr(0, i + 1) << std::endl;
    }
}

<details>
<summary>英文:</summary>
Even so, I think it would be better for your learning progress to use a debugger to finger out the problem yourself. Here the problems with your code:
For the `i=0` (the first iteration of your outer loop)  the `for(k=0;k&lt;i;k++)` will not be executed at all, as `k&lt;0` evaluates to `false`.
And having a running variable (`k`) that you change in two `for` loops that are nested, is most of the time also an indication that something is wrong.
So what you want to do: You want to create each possible prefix, so you want to create `n` strings with the length of `1` to `n`. So your first idea with the outer loop is correct. But you overcomplicate the inner part.
For the inner part, you want to print all chars from the index `0` up to `i`.

std::cin >> cuvant;
// loop over the length of the string
for (int i = 0, size = strlen(cuvant); i < size; i++) {
    // print all chars from 0 upto to i (k<=0)
    for (int k = 0; k <= i; k++) {
        std::cout << cuvant[k];
    }
    // print a new line after that
    std::cout << std::endl;
}

But instead of reinventing the wheel I would use the functions the std provides:

</details>
# 答案3
**得分**: 0
以下是翻译好的部分：
对于这个非常简单的字符串后缀任务，您可以使用以下代码：
```cpp
void main()
{
    std::string s = "house";
    std::string s2;
    for(char c : s) 
    {
        s2 += c;
        cout << s2 << endl;
    }
}

#include <iostream>
using namespace std;
int main()
{
   char cuvant[100];
   int i,k;
   cin >> cuvant;
   for(i=0; i<strlen(cuvant); i++)
   {
        for (k = 0; k <= i; ++k)
        {
            cout << cuvant[k];
        }
        cout << endl;
    }
}