This task is to find the number of 'For' loops present in the given cpp program.
One example: main()
{
      float a[4];
      float k;
      float c;
      int i;
      for(i=0;i<4;i++)
      {
                      scanf("%f",&a[i]);
      }
      scanf("%f",&c);
      c=2*3.1415926*c/360;
      k=(a[0]+a[1]+a[2]+a[3])/2;
      double s,sq;
      sq=(k-a[0])*(k-a[1])*(k-a[2])*(k-a[3]) - a[0]*a[1]*a[2]*a[3]*(double)cos(c/2)*(double)cos(c/2);
    
      if(sq<0)
      printf("Invalid input");
      else
      {
          s=sqrt(sq);
          printf("%.4f",s);
      }
    
}

Solution is here: 1
Explanation: The number of 'for' loops in the code is given by the number of 'for' string present in the code. Since we can find the exact number of for loops, this is a good example.

Now, solve this: //**********************************************
//**???1090 ????*************************
//**?????? 1000012741*********************
//**???2010?12?7?*************************
//**********************************************
//??change???a???????p???????
int change (int a, int p)      
{
	int i, count = 0;                             //count??????????
	for (i = p; i < a; i++)
	{
		if (a % i == 0 && a / i >= i)             //??????????????
		{
			count++;                              //???1
			int k, t;
			k = (int) sqrt(a / i);                //?k?a/i????
			for (t = 2; t <= k; t++)
			{
				if ((a / i) % t == 0)             //??a???2?k??????
				{
					count += change (a / i, i);   //?a/i??????
					break;
				}   
			}  
				
		}
	}
	return count;                                 //??count?
}
int main()
{
	int n, i, a;                                  //n?????????a?????
	cin >> n;
	for (i = 1; i <= n; i++)
	{
		int total = 0;                            //total?????
		cin >> a;
		total += change (a, 2);                   //??change??
		cout << total + 1 << endl;                //??1?????????????
	}
	return 0;
}
//**********************************************
Solution:
3