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+#include <iostream>
+
+/* Тази задача може да се реши без матрици.
+ * Все пак идеята е да упражним точно тях.
+ */
+
+int main() {
+ const unsigned int SIZE = 5;
+ int mat[SIZE][SIZE];
+
+ for (unsigned int i = 0; i < SIZE; i++)
+ for (unsigned int j = 0; j < SIZE; j++)
+ std::cin >> mat[i][j];
+
+ unsigned int countZeroes = 0;
+ for (unsigned int i = 0; i < SIZE; i++)
+ for (unsigned int j = 0; j < SIZE; j++)
+ countZeroes += mat[i][j] == 0;
+
+ /* Бройката нули е повече от бройката на всички елементи които не са нули.
+ * Алтернативно, може и да се пази втори брояч за ненулеви елементи.
+ */
+ if (countZeroes > SIZE * SIZE - countZeroes)
+ std::cout << "Sparse" << std::endl;
+ else
+ std::cout << "Not sparse" << std::endl;
+
+ return 0;
+}