Question

在我的程序早期，我让程序一次生成一个向量，9个元素。这9组中的每一组都是相关数据。后来我想一次处理每9组（例如每行打印9个）。

我应该怎么做？我的尝试：

 for (unsigned int i = 0; i < solutions.size()/9; i++)
 {
  cout << i << ": " << solutions[0+ i*9] << " " << solutions[1+ i*9] << " " << solutions[2 + i*9] << " " << solutions[3+ i*9] << " " << solutions[4+ i*9] << " " << solutions[5 + i*9] << " " << solutions[6+ i*9] << " " << solutions[7+ i*9] << " "<< solutions[8 + i*9] << endl;
  cout << endl;
 }

我显然在这里搞砸了。有什么帮助吗？

Answer 1

我建议使用结构。

struct elems {
    elements elems[9];
};
std::ostream& operator<<(std::ostream& out, const elems& elems) {
    for(int i = 0; i < 9; i++) 
        out << elems.elems[i];
    return out;
}
std::vector<elems> solutions;

Answer 2

为什么不使用矢量矢量？

vector< vector <int> > solution;

for (unsigned int i = 0; i < solutions.size(); i++)
{
    for(int j=0; j<9; j++) {
       cout << solution[i][j] << " ";
    }

    cout << endl;
}

Answer 3

简单修复，您只需要一个临时索引变量并在整个循环中递增它，同时每次将主索引增加9。

for(unsigned int i = 0; i < solutions.size(); i += 9) {
   int j = i;
  std::cout << i << ": " << solutions[j++] << " " << solutions[j++] << " " << solutions[j++] << " " << solutions[j++] << " " << solutions[j++] << " " << solutions[j++] << " " << solutions[j++] << " " << solutions[j++] << " "<< solutions[j++] << std::endl << std::endl;
}

如何处理向量中的数据组？

3 个答案: