<span>Если в явном виде:
</span>
¬A&¬C = ¬A&(B v <span>¬B)&¬C = </span>¬A&B&¬C v ¬A&¬B&¬C
C&B = (A v ¬A)&B&C = A&B&C v <span>¬A&B&C
</span>¬A&B = ¬A&B&(C v <span>¬C)</span> = ¬A&B&C v ¬A&B&<span>¬C
Т.е. исходное выражение =</span>
<span>¬A&B&¬C v ¬A&¬B&<span>¬C v </span></span><span>A&B&C v <span>¬A&B&C v </span></span><span><span>¬A&B&C v ¬A&B&<span>¬C
</span></span>Если убрать повторы, получится</span>
A&B&C v <span>¬A&B&C v </span><span>¬A&B&¬C v </span><span>¬A&¬B&<span>¬C =
</span></span>= (A & ¬A)&B&C v <span>¬A&(B v </span><span>¬B)&</span><span>¬C =</span>
= B&C v <span>¬A&</span><span>¬C</span>
#include <iostream>
using namespace std;
int main() {
int x, n;
int res = 0;
cout << "Введите два числа: ";
cin >> x >> n;
for (int i = 1; i <= x; i++)
res += n;
cout << endl << x << " * " << n << " = " << res << endl;
return 0;
}