TLX - TROC17A. Firework Festival

You can practice the question here.

Given two arrays A and B, each of size N. Determine whether the sum of A[i]^B[i] for all 1 <= i <= N, is odd or even. Display the output 0 if the sum is even, or 1 if the sum is odd.

Input Format

N
A[1] A[2] ... A[N]
B[1] B[2] ... B[N]

Sample Input

5
2 4 6 8 2
1 2 3 4 5

Sample Output

0

Constraints

1 <= N <= 100
1 <= A[i], B[i] <= 100

Usually for Problem A, most of the solutions are brute-force. However, if we take the edge case 100, A[0] ^ B[0] + A[1] ^ B[1] + ... + A[99] ^ B[99] where A[i] and B[i] are 100, the overall result would occur overflow. We can notice that this problem is all about parity. The exponent B[i] of A[i] does not change the parity of A[i] ^ B[i]. Hence, we can conclude that

even ^ odd  -> even
even ^ even -> even
odd ^ even -> odd
odd ^ even -> odd

and we know that

even + even -> even
odd + odd -> even
even + odd -> odd

Therefore, we can simply sum all the values and check if it is even or odd.

int main()
{
    int n; cin >> n;
    vi a(n), b(n);
    READ(a);
    READ(b);
    int sum = 0;
    REP(i, n) sum += a[i];
    OUT((sum & 1));
    return 0;
}