从文本文件加载不同的数据类型

我是StackExchange和c++的新手，所以如果我没有很好地描述问题，请道歉。我需要一些家庭作业的帮助。我正在寻找一种方法来加载这个文件并存储它的数据。文件有不同的数据类型，所以我到目前为止想到的是使用struct数组来存储所有的值，使用三重嵌套的for循环将数据保存在正确的变量中。

这是。dat文件的样子，用"注释"来帮助描述发生了什么。

100A 2     // model number, 2 different versions of that model
0 0 0      // number of quarters, dimes, and nickels
5          //number of items in the vending machine
1A 1034  5 // Code combonation, ID Number, Quantity
1B 1000 10
1C 1100 10
1D 1123 20
1E 1222  5
0 0 0      // number of quarter, dimes, nickels in 2nd model
7          // number of items in the second version of that model
1A 2180 20
1B 1283 20
1C 3629  5
1D 3649  3
1E 4051 15
1F 4211  1
1G 5318  5
100B 3     // New model, with 3 different versions of itself.
2 10 5     //everything repeats like model 100A
7 

这是我想出来的代码

#include <fstream>
#include <iostream>
#include <iomanip>
#include <string>
using namespace std;
struct VMdata
 {
    ifstream inFile;
    string model[1];
    int version[1];
    int q[5];
    int d[5];
    int n[5];
    int size[5];
    string id[30];
    int code[30];
    int num[30];
    char dummy;
};
int main()
{
    VMdata New;
    cout << fixed << setprecision(2) << showpoint;
    New.inFile.open("machines.dat");

    cout << "Model Data" 
           << endl   << endl;      

    int count1 = 0;
    int count2= 0;
    int count3 = 0;
    for (int i = 0; i < 2; i ++)
    {
        cout << "i :" << count1 << endl;
       New.inFile >> New.model[i] >> New.version[i]; // loads model number and number of versions i times
       count1 = count1 + 1;
       for (int j = 0; j < New.version[count1 -1]; j ++)
       {
           cout <<"j :" << count2 << endl;
       New.inFile >> New.q[count2] >> New.d[count2] >> New.n[count2]  >> New.size[count2]; // loads number of q, d, n, j times
       count2 = count2 + 1;
       for (int k = 0; k < New.size[count2 - 1]; k++)
        {
            cout << "k :" << count3 << endl;
           New.inFile >> New.id[count3] >> New.code[count3] >> New.num[count3]; // loads id number, code number, and total number k times
           count3 = count3 + 1;
        }
       }
    }
    New.inFile.close();
     count1 = 0;
    count2= 0;
    count3 = 0;
    cout << endl;
        for ( int i = 0; i < 2; i ++)
    {
       cout << New.model[i]  << setw(12) << New.version[i] << endl << endl;
        count1 = count1 + 1;
       for (int j = 0; j < New.version[count1 -1]; j ++)
       {
       cout << New.q[count2] << setw(12) << New.d[count2] << setw(12) << New.n[count2] << endl << setw(12) << New.size[count2] << endl;
      count2 = count2 + 1;
       for (int k = 0; k < New.size[count2 - 1]; k++)
        {
           cout << New.id[count3] << setw(12) << New.code[count3] << setw(12) << New.num[count3] << endl << endl;
           count3 = count3 + 1;
        }
       }
    }

      return 0;
}

这是我的测试输出。

100A           2
3          15           0 // should be 0, 0, 0
 5
1A        1034           5
1B        1000          10
1C        1100          10
1D        1123          20
1E        1222           5
0           0           0
  7
1A        2180          20
1B        1283          20
1C        3629           5
1D        3649           3
1E        4051          15
1F        4211           1
1G        5318           5
♥       ☻            3 // the heart and smile should be 100B lol
2          10           5
  7
1A        2180          10
1B        1283          10
1C        3629           5
1D        3649           3
1E        4051          15
1F        4211          10
1G        3026           5
5           6           3
  6
1A        6626           5
1B        6155           5
1C        5982          10
1D        5573           3
1E        5454          10
1F        5336          50
10          10          10
 5
1A        1034           5
1B        1000           5
1C        1100           5
1D        1123           5
1E        1210          12
Press any key to continue . . .

如您所见，第一台机器的25分硬币、1角硬币和5分硬币的数量是错误的，第二台机器的型号名称是错误的。如果有人有任何建议，我将不胜感激。