Naive approach

\(A _ 1,A _ 2,\ldots,A _ N\) are all equal if and only if any two elements in \(A\) are equal.

Thus, it is sufficient to read \(A\), inspect every pairs of elements, and determine if they have the same value.

The following is sample code.

#include <iostream>
#include <vector>

using namespace std;

int main() {
    int N;
    cin >> N;

    vector<int> A(N);
    for(int i = 0; i < N; ++i) // Read A
        cin >> A[i];

    for(int i = 0; i < N; ++i)
        for(int j = 0; j < N; ++j)
            if(A[i] != A[j]){ // If there is a pair of different values,
                cout << "No" << endl; // print No and
                return 0; // exit
            }

    // If no pair has different values,
    cout << "Yes" << endl; // print Yes and
    return 0; // exit
}

#include <iostream>
#include <vector>

using namespace std;

int main() {
    int N;
    cin >> N;
    
    vector<int> A(N);
    for(auto&& a : A) // Read A
        cin >> a;
    
    for(auto a : A)
        for(auto b : A)
            if(a != b){ // If there is a pair of different values,
                cout << "No" << endl; // print No and
                return 0; // exit
            }
    
    // If no pair has different values,
    cout << "Yes" << endl; // print Yes and
    return 0; // exit
}

N = int(input())
A = list(map(int, input().split())) # Read A

for a in A:
    for b in A:
        if a != b: # If there is a pair of different values,
            print('No') # print No and
            break # escape from the loop
    else: # If no pair has different values,
        continue # continue the loop
    break
else: # If no pair appeared to be different,
    print('Yes') # print Yes

A faster way

\(A _ 1,A _ 2,\ldots,A _ N\) are all equal if and only if \(A _ 1\) equals all of \(A _ 2,A _ 3,\ldots,A _ N\).

Thus, it is sufficient to read \(N\) and \(A _ 1\) first, and determine if each of the remaining elements equals \(A _ 1\). Here, it is convenient to print No and terminate the program once a different value is found, and print Yes after all the elements have been scanned.

The following is sample code.

#include <iostream>

using namespace std;

int main() {
    int N, A1; // Read the initial element of A beforehand
    cin >> N >> A1;

    for (int i = 1; i < N; ++i) { // For each of the remaining elements of A,
        int A;
        cin >> A; // read it,
        if (A != A1) { // and if they are not equal, print No and exit
            cout << "No" << endl;
            return 0;
        }
    }
    // If the execution reaches here after scanning all the elements, print Yes
    cout << "Yes" << endl;
    return 0;
}

N = int(input())
A1, *A = map(int, input().split()) # Split A into the first and the remaining values

for a in A:
    if A1 != a: # Print No if any of the element is different from the first
        print('No')
        break
else: # If `break` not occurred, print Yes
    print('Yes')