Solution 1

There are \(6\) permutations of \((A_1,A_2,A_3)\). By checking for each permutation if it forms an arithmetic sequence, the problem can be solved.

Sample Code (C++)

#include<bits/stdc++.h>
using namespace std;

int main(){
    int A1,A2,A3; cin >> A1 >> A2 >> A3;
    string ans = "No";
    //(A1,A2,A3)
    if(A3-A2 == A2-A1) ans = "Yes";
    //(A1,A3,A2)
    if(A2-A3 == A3-A1) ans = "Yes";
    //(A2,A1,A3)
    if(A3-A1 == A1-A2) ans = "Yes";
    //(A2,A3,A1)
    if(A1-A3 == A3-A2) ans = "Yes";
    //(A3,A1,A2)
    if(A2-A1 == A1-A3) ans = "Yes";
    //(A3,A2,A1)
    if(A1-A2 == A2-A3) ans = "Yes";
    cout << ans << endl;
}

Also, by the fact that when \((a,b,c)\) is an arithmetic sequence then \((c,b,a)\) is also an arithmetic sequence, one can reduce the number of conditional branches to three.

Sample Code (Python)

A1,A2,A3 = map(int,input().split())
ans = "No"
#(A1,A2,A3)
if A3-A2 == A2-A1:
    ans = "Yes"
#(A1,A3,A2)
if A2-A3 == A3-A1:
    ans = "Yes"
#(A2,A1,A3)
if A3-A1 == A1-A2:
    ans = "Yes"
print(ans)

Solution 2

An arithmetic sequence must be monotonically increasing or decreasing, because the sign (positive, negative, or zero) of \((b-a)\) and \((c-b)\) must be the same. Therefore, you can easily get AC by first receiving \((A_1,A_2,A_3)\) as a sequence and sort in the increasing or decreasing order, and then check if it forms an arithmetic sequence.

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());
    string ans = "No";
    if(A[2]-A[1] == A[1]-A[0]) ans = "Yes";
    cout << ans << endl;
}

Sample Code (Python)

A = list(map(int,input().split()))
A.sort()
ans = "No"
if A[2]-A[1] == A[1]-A[0]:
  ans = "Yes"
print(ans)