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Merge pull request #234 from mahek5/newcode

Added a Bitwise operation program

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Competitive Coding/Bitwise Operations/Largest_power_of_2
- Largest_power_2.cpp
- readme.md

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	`1`	`+#include <iostream>`
	`2`	`+using namespace std;`
	`3`	`+`
	`4`	`+int main() {`
	`5`	`+ long N;`
	`6`	`+ cin>>N;`
	`7`	`+ largest_power(N);}`
	`8`	`+void largest_power(long N)`
	`9`	`+ {`
	`10`	`+ //changing all right side bits to 1.`
	`11`	`+ N = N\| (N>>1);`
	`12`	`+ N = N\| (N>>2);`
	`13`	`+ N = N\| (N>>4);`
	`14`	`+ N = N\| (N>>8);`
	`15`	`+`
	`16`	`+`
	`17`	`+ //as now the number is 2 * x-1, where x is required answer, so adding 1 and dividing it by`
	`18`	`+ 2.`
	`19`	`+ printf((N+1)>>1;`
	`20`	`+`
	`21`	`+ }`

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	`1`	`+# Finding the largest power of 2 (most significant bit in binary form), which is less than or`
	`2`	`+equal to the given number N.`
	`3`	`+`
	`4`	`+Let’s say binary form of a N is {1111}base2 which is equal to 15.`
	`5`	`+15 = 2^4-1, where 4 is the number of bits in N.`
	`6`	`+`
	`7`	`+This property can be used to find the largest power of 2 less than or equal to N. How?`
	`8`	`+If we somehow, change all the bits which are at right side of the most significant bit of N to 1, then the number will become x + (x-1) = 2 * x -1 , where x is the required answer.`
	`9`	`+Example:`
	`10`	`+Let’s say N = 21 = {10101}, here most significant bit is the 4th one. (counting from 0th digit) and so the answer should be 16.`
	`11`	`+So lets change all the right side bits of the most significant bit to 1. Now the number changes to`
	`12`	`+{11111} = 31 = 2 * 16 -1 = Y (let’s say).`
	`13`	`+Now the required answer is (Y+1)>>1 or (Y+1)/2.`

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