如何制作这个数字形状？

Question

Write a program to display the below pattern with n rows, where n is in the range between 1 and 100. The variable n should be entered by the user. If the user input is between 1 and 100 then output the pyramid as given below, otherwise prompt the user to enter n again.

Here is the sample output: Enter the number of rows: 6

1

2 3

3 4 5

4 5 6 7

5 6 7 8 9

6 7 8 9 10 11

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this is my code it shows a close answer but not correct.

 int num=1  ,  counter=1;
   cout << "Enter the number of rows: " ;
   cin>>num;

       for(int i=0;i<=num;i++)
       {
           for(int j=0;j<=i;j++)
           {
               cout<<counter<<" ";
               counter++;
           }
           cout<<endl;
       }

Answer 1

int num = 1, counter = 1, temp = 1;
cout << "Enter the number of rows: ";
cin >> num;

for (int i = 0; i < num; i++)
{
    for (int j = 0; j <= i; j++)
    {
        cout << temp << " ";
        temp++;
    }
    counter++;
    temp = counter;
    cout << endl;
}

The variable temp serves as a counter for the rows, meanwhile the variable counter is responsible for the starter numbers in the first column.

Answer 2

you should either set counter in each run of outer loop, right before entering the inner one with something like counter = i+1 , the way you wrote this code value of this variable carries over to the next iteration and keeps going only upwards.

Alternate solution would be to print j instead and work on inner loop, it should then start with i and the first number that wouldn't qualify would be 2*i this way for i equal 2 the sequence would be 2 3 ,

either way you should also rework your outer loop, as right now it starts with 0 and ends with num meaning it goes through num+1 iterations

Answer 3

int rows, i, j = 0, number = 0, counter = 0;
cout << "Enter the number of rows: ";
cin >> rows;

for (i = 1; i <= rows; i++)
{
  while (j != 1 * i)
  {
    if (counter <= rows)
    {
      cout << i+j << " ";
      counter++;
    }
   ++j;
  }
number=counter=j=0;
cout << endl;
}