A C++ program of adding numbers without adding the carry

Trans bought a calculator at creatnx's store. Unfortunately, it is fake. It has many bugs. One of them is adding two numbers without carrying. Example expression: 12 + 9 will have result 11 in his calculator. Given an expression in the form a + b, please output result from that calculator. Constraints: 1 ≤ T ≤ 100 1 ≤ a, b ≤ 109

My code pass the test cases where 1 ≤ a, b ≤ 9 But failed for the original contraints. Could not find the case where my code giving wrong answer.

My Code

#include <iostream>
#include<string>
#define ll long long int
using namespace std;

int main()
{
  ll t;
  cin>>t;
  while(t--){
    string a, b;
    cin>>a>>b;
    ll max_last = max(a.size(), b.size());
    ll min_last = min(a.size(), b.size());
    ll result[max_last];
    ll i=max_last-1;
    for(ll  j= min_last-1; j>=0; i--, j--){
        ll sum = a[i]-'0' + b[j]-'0';
        result[i] = sum ;
    }
    while(i>=0){
        result[i] = a[i]-'0';
        i--;
    }

    ll sum = 0;
    for(ll i=0; i<max_last; i++){a
        sum = sum*10 + result[i]%10;
    }
    cout<<sum<<"\n";

 }
 return 0;
}

Tester's Code

#include <bits/stdc++.h>

using namespace std;

signed main() {
 int t;
 cin >> t;
 assert(1 <= t && t <= 100);
 while(t--) {
    int a, b;
    cin >> a >> b;
    assert(1 <= a && a <= 1'000'000'000);
    assert(1 <= b && b <= 1'000'000'000);
    vector<int> A, B;
    while(a) {
        A.push_back(a % 10);
        a /= 10;
    }
    while(b) {
        B.push_back(b % 10);
        b /= 10;
    }
    while(A.size() < B.size()) A.push_back(0);
    while(B.size() < A.size()) B.push_back(0);
    for(int i = 0; i < A.size(); i++) {
        A[i] += B[i];
    }
    int ans = 0;
    reverse(begin(A), end(A));
    for(auto it: A) {
        ans = ans * 10 + it % 10;
    }
    cout << ans << endl;
  }
 }