For \(i = 1, 2, \ldots, N-1\) in this order, exchange the currency of country \(i\) for that of country \(i+1\) as many times as possible.

Since exchange can be done a plethora of times, processing them one by one does not finish in the execution time limit. However, if you have \(A_i\) units of the currency of country \(i\), you can exchange the currency of country \(i\) for that of country \((i+1)\) \(\lfloor A_i / S_i \rfloor\) times, so one can perform the exchange \(\lfloor A_i / S_i \rfloor\) at once in order to obtain \(\lfloor A_i / S_i \rfloor \times T_i\) units of the currency of country \((i+1)\).

Therefore, the problem can be solved by receiving \(A = (A_1, A_2, \ldots, A_N)\) as input, adding \(\lfloor A_i / S_i \rfloor \times T_i\) to \(A_{i+1}\) for \(i = 1, 2, \ldots, N-1\) in this order, and then printing the value \(A_N\).

In order to perform some process “for \(i = 1, 2, \ldots, N-1\) in this order,” one can use the loop feature (like a for statement), which is a standard feature in a programming language. Also check out the specification of division in your language to correctly obtain the result of division \(A_i/S_i\) rounded down.

The following is sample code in C++ language.

#include <iostream>
using namespace std;
typedef long long ll;

ll n;
ll a[200001], s[200001], t[200001];

int main(void)
{
  cin >> n;
  for(int i = 1; i <= n; i++) cin >> a[i];
  for(int i = 1; i <= n-1; i++) cin >> s[i] >> t[i];
  
  for(int i = 1; i <= n-1; i++) a[i+1] += a[i]/s[i] * t[i];
  cout << a[n] << endl;
  
  return 0;
}