Another small problem before I go to sleep tonight: There is an array A[N] of N numbers. You have to compose an array Output[N] such that Output[i] will be equal to multiplication of all the elements of A[N] except A[i]. For example Output[0] will be multiplication of A[1] to A[N-1] and Output[1] will be multiplication of A[0] and from A[2] to A[N-1]. Solve it without division operator and in O(n). ...
Someone says this is an old lame trick. I think it鈥檚 simple and clever use of XOR. How/Why does it work? It鈥檚 built around the properties of the XOR ^ operator, who has the following properties: A ^ B = B ^ A (commutative) A ^ 0 = A A ^ 1 = ~A A ^ A = 0 So, you can see how it get鈥檚 applied here: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 #include <stdio .h> int main(void) { unsigned int a, b; // ... populate somehow "a" and "b"... printf("a = %d - b = %d\n", a, b); a ^= b; // store in "a" the value of "a XOR b" b ^= a; // store in "b" the value of "a XOR b XOR b" = "a XOR 0" = "a" a ^= b; // store in "a" the velue of "a XOR b XOR a" = "b XOR 0" = "b" printf("a = %d - b = %d\n", a, b); } Neat.