Takahashi repeats the same move at the period of \((A+C)\) seconds. Specifically, for a non-negative integer \(k\), Takahashi’s move between \(k\) and \(k+1\) seconds passed since he started will be:

• walks \(B\) meters, if the remainder of \(k\) divided by \(A+C\) is less than \(A\); or
• stops moving otherwise.

Similarly, Aoki’s move will be:

• walks \(E\) meters, if the remainder of \(k\) divided by \(D+F\) is less than \(D\); or
• stops moving otherwise.

By simulating the moves for each \(k = 0, 1, \dots, X - 1\), we can compute how many meters the two advanced. Therefore, the problem has been solved in a total time complexity of \(O(X)\).

Sample code (C++):

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

int main() {
    int a, b, c, d, e, f, x;
    cin >> a >> b >> c >> d >> e >> f >> x;
    int takahashi = 0, aoki = 0;
    for (int k = 0; k < x; ++k) {
        if (k % (a + c) < a) {
            takahashi += b;
        }
        if (k % (d + f) < d) {
            aoki += e;
        }
    }
    if (takahashi > aoki) {
        cout << "Takahashi\n";
    } else if (takahashi < aoki) {
        cout << "Aoki\n";
    } else {
        cout << "Draw\n";
    }
    return 0;
}

This problem can be solved in a faster way (in a total time complexity of \(O(1)\).

Let’s consider every \((A+C)\)-second chunk of Takahashi’s move as a single period. Takahashi advances \(A \times B\) in every period. In the first \(k\) seconds (\(0 \leq k \leq A + C\)) of a period, he advances \(\min(A, k) \times B\) meters.

Suppose that Takahashi completes exactly \(q\) periods during the first \(X\) seconds since he starts, then \(q =\left\lfloor \frac{X}{A + C} \right\rfloor\). The remaining \(X - (A + C) \times q\) seconds belongs to the same period, so the sum of distance Takahashi advances in the first \(X\) seconds is \(A \times B \times q + \min(A, X - (A + C) \times q) \times B\) meters.

The same applies for Aoki, so the distances that the two advance can be calculated in a constant time.

Sample code (Python):

def solve(a, b, c, x):
  q = x // (a + c)
  r = x % (a + c);
  return (q * a + min(a, r)) * b;

a, b, c, d, e, f, x = map(int, input().split())

takahashi = solve(a, b, c, x)
aoki = solve(d, e, f, x)

if takahashi > aoki:
  print("Takahashi")
elif takahashi < aoki:
  print("Aoki")
else:
  print("Draw")