Solution 1: Try all permutations

The problem can be solved by trying all permutations of \(A\).

Sample code (C++):

#include<bits/stdc++.h>

using namespace std;

int main(){
  vector<int> a(3);
  cin >> a[0] >> a[1] >> a[2];
  if(a[0]*a[1]==a[2]){cout << "Yes\n"; return 0;}
  if(a[0]*a[2]==a[1]){cout << "Yes\n"; return 0;}
  if(a[1]*a[0]==a[2]){cout << "Yes\n"; return 0;}
  if(a[1]*a[2]==a[0]){cout << "Yes\n"; return 0;}
  if(a[2]*a[0]==a[1]){cout << "Yes\n"; return 0;}
  if(a[2]*a[1]==a[0]){cout << "Yes\n"; return 0;}
  cout << "No\n";
  return 0;
}

In case of C++, one can use a feature called next_permutation to try all permutations.

Sample code (C++):

#include<bits/stdc++.h>

using namespace std;

int main(){
  vector<int> a(3);
  cin >> a[0] >> a[1] >> a[2];
  sort(a.begin(),a.end());
  do{
    if(a[0]*a[1]==a[2]){cout << "Yes\n"; return 0;}
  }while(next_permutation(a.begin(),a.end()));
  cout << "No\n";
  return 0;
}

Solution 2: Utilize commutative law

For two integers \(A\) and \(B\) we have \(A \times B = B \times A\), so whether a[0]*a[1]==a[2] and whether a[1]*a[0]==a[2] are equivalent.
This helps us reduce the conditional branches to three.

Sample code (C++):

#include<bits/stdc++.h>

using namespace std;

int main(){
  vector<int> a(3);
  cin >> a[0] >> a[1] >> a[2];
  if(a[0]*a[1]==a[2]){cout << "Yes\n"; return 0;}
  if(a[1]*a[2]==a[0]){cout << "Yes\n"; return 0;}
  if(a[2]*a[0]==a[1]){cout << "Yes\n"; return 0;}
  cout << "No\n";
  return 0;
}

Solution 3: Utilize the ordering

For integers \(A,B,C \ge 1\), if \(A \times B = C\), then we have \(A \le C\) and \(B \le C\).
Thus, one can show that if the answer is Yes, then \(A_1 \times A_2 = A_3\) holds after sorting \(A\) in ascending order.
Thus, one can sort \(A\) and check only once.

Sample code (C++):

#include<bits/stdc++.h>

using namespace std;

int main(){
  vector<int> a(3);
  cin >> a[0] >> a[1] >> a[2];
  sort(a.begin(),a.end());
  if(a[0]*a[1]==a[2]){cout << "Yes\n"; return 0;}
  cout << "No\n";
  return 0;
}