Answer to Question #294213 in C++ for Ran

Question

Consider a square matrix A of size nxn and an integer x which is an element of A. find the row number R and column number C of X in A, and calculate the sum R and C. If the sum is even, find the sum of the digits of all even numbers in the matrix, and if the sum is odd then find the sum of digits of all odd numbers in matrix

Answer 1

#include <iostream> // standart C++ library for input and output stream

using namespace std; // including namespace std

int calculate_sum_digits(int x) { // function that returns sum of digits of number x
    int answer = 0;
    while (x >= 0) {
        answer += x % 10; // add the left digit of x
        x /= 10; // delete the left digit of x
    }
    return answer;
}

int main() {
    int n, x;
    cin >> n >> x;
    int A[n][n];
    int even_sum = 0, odd_sum = 0;
    int R = -1, C = -1;
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            cin >> A[i][j];
            if (A[i][j] % 2 == 0)
                even_sum += calculate_sum_digits(A[i][j]); // add the sum of digits if A[i][j] is even
            else
                odd_sum += calculate_sum_digits(A[i][j]); // same but if A[i][j] is odd
            if (A[i][j] == x) {
                R = i;
                C = j; // found element x
            }
        }
    }
    if (R == -1 && C == -1) {
        cout << "No element x in matrix A, exiting program..."; // there is no element x in A
        exit(0);
    }
    if ((R + C) % 2 == 0) { // if R + C is even
        cout << even_sum; // print sum of digits of even numbers in A
    } else {
        cout << odd_sum; // print sum of digits of odd numbers in A
    }
}