If you use a coupon, it is optimal to order the cheapest dish. Thus, it is sufficient to compute

• the price when the coupon is not used (\(P\) yen) and
• the price when the coupon is used, and the cheapest dish is ordered (let this be \(D_{\min}\) yen),

and print the cheapest among them, \(\min\lbrace P, Q + D_{\min}\rbrace\).

To find the price of the cheapest dish \(D_{\min}\), you need to inspect all the dishes \(1, 2, \ldots, N\) in order, and compare their prices. One can “inspect dishes \(1, 2, \ldots, N\) in order” with the loop feature (like the for statement), which is a standard feature of a programming language.

The following is an example of accepted code for this problem in C++ language.

#include <iostream>
using namespace std;

int main(void)
{
  int n, p, q;
  cin >> n >> p >> q;
  
  int d_min = 1000000000, d;
  for(int i = 1; i <= n; i++){
    cin >> d;
    d_min = min(d, d_min);
  }
  cout << min(p, q+d_min) << endl;
  
  return 0;
}