ADAGAME.cpp

//
//  main.cpp
//  practice
//
//  Created by Mahmud on 11/06/17.
//  Copyright © 2017 Mahmud. All rights reserved.
//


/*
 O(T * MAX(N) * MAX(STEPS)) solution using dynamic programming
 Just follow the statement:
 If it is Ada's turn, maximize current answer by next iteration;
 Else minimize current answer by next iteration;
 And at the end, if the result is bigger than original value Ada wins, else she loses.
 */

#include <iostream>
#include <cstring>
#include <cmath>

using namespace std;

const int MAX = 10005;

int T;
int turns;
int digits[4];
string number;
string dp[101][10][10][10][10];

string solve(int step, string number) {
    if (step > turns) {
        return number;
    }
    for (int i = 0; i < 4; i ++) {
        digits[i] = number[i] - '0';
    }
    string &result = dp[step][digits[0]][digits[1]][digits[2]][digits[3]];
    if (result != "-1") {
        return result;
    }
    if (step & 1) {
        result = "0000";
    }
    else {
        result = "9999";
    }
    for (int i = 0; i < 4; i ++) {
        string to = number;
        to[i] = (to[i] - '0' + 1) % 10 + '0';
        if (step & 1) {
            result = max(result, solve(step + 1, to));
        }
        else {
            result = min(result, solve(step + 1, to));
        }
    }
    return result;
}

int main() {
    cin >> T;
    while (T --) {
        for (int i = 1; i < 101; i ++) { // turns
            for (int j = 0; j < 10; j ++) { // first digit
                for (int k = 0; k < 10; k ++) { // second digit
                    for (int q = 0; q < 10; q ++) { // third digit
                        for (int l = 0; l < 10; l ++) { // forth digit
                            dp[i][j][k][q][l] = "-1";
                        }
                    }
                }
            }
        }
        cin >> number >> turns;
        if (solve(1, number) > number) {
            cout << "Ada" << endl;
        }
        else {
            cout << "Vinit" << endl;
        }
    }
    
    return 0;
}