Added a program to check if two integers are of opposite signs.

README.md

@@ -70,6 +70,7 @@ Include contains single header implementation of data structures and some algori
 | Given a number x and two positions (from right side) in binary representation of x, write a function that swaps n right bits at given two positions and returns the result. It is also given that the two sets of bits do not overlap.|[swapSetOfBits.cpp](bit_manipulation/swapSetOfBits.cpp)|
 | Add two numbers without using any arithmetic operators | [addition_without_operators.cpp](bit_manipulation/addition_without_operators.cpp)|
 |Louise and Richard play a game. They have a counter set to N. Louise gets the first turn and the turns alternate thereafter. In the game, they perform the following operations: <ul><li>If N is not a power of 2, reduce the counter by the largest power of 2 less than N.</li></ul><ul><li>If N is a power of 2, reduce the counter by half of N.</li></ul> The resultant value is the new N which is again used for subsequent operations.The game ends when the counter reduces to 1, i.e., N == 1, and the last person to make a valid move wins. <ul><li> Given N, your task is to find the winner of the game. If they set counter to 1, Richard wins, because its Louise' turn and she cannot make a move.</li></ul><ul><li>Input Format : -The first line contains an integer T, the number of testcases. T lines follow. Each line contains N, the initial number set in the counter.</ul></li> |[counter_game.cpp](bit_manipulation/counter_game.cpp)|
+|Determine if two integers are of opposite signs|[check_opposite_signs.cpp](bit_manipulation/check_opposite_signs.cpp)|
 
 
 ### Cracking the coding interview problems

bit_manipulation/check_opposite_signs.cpp

@@ -0,0 +1,35 @@
+/*
+ * Given two integers, using bit manipulations determine if they are of opposite signs.
+ * Most Significant Bit (MSB) of number represents sign of the number. If it is 1, it represents
+ * a negative value, if it is 0, it represents a positive value.
+ * If MSB of two numbers are different, their XOR would be 1, otherwise it would be 0.
+ * Thus, result of XOR of two numbers will have MSB 1 if they are of opposite signs,
+ * 0 other wise, in other words, XOR of two numbers would be negative (MSB 1) if they are of
+ * opposite signs.
+ * Source : http://graphics.stanford.edu/~seander/bithacks.html
+ */
+
+#include <iostream>
+
+ bool are_of_different_signs(int a, int b)
+ {
+     return ((a ^ b) < 0);
+ }
+
+ int main()
+ {
+     int a, b;
+     std::cout << "Enter number 1: ";
+     std::cin >> a;
+     std::cout << "Enter number 2:";
+     std::cin >> b;
+     if (are_of_different_signs(a, b))
+     {
+         std::cout << a << " and " << b << " are of different signs" << std::endl;
+     }
+     else
+     {
+         std::cout << a << " and " << b << " are of same signs" << std::endl;
+     }
+     return 0;
+ }