Maximum AND

Problem
https://www.hackerearth.com/practice/basic-programming/bit-manipulation/basics-of-bit-manipulation/practice-problems/algorithm/maximum-and/

Solution

	#include <iostream>
	#include <cstdlib>
	using namespace std;
	int main(int argc, char** argv) {
	int T;
	long int A,B;
	long int max=0;
	cin>>T;
	for(int i=0;i<T;i++)//input test cases
	{
	cin>>A>>B;//input A,B
	for(int j=A;j<B;j++)// loop all pairs between A,B
	{
	for(int k=j+1;k<=B;k++)
	{
	if((k&j)>max){max=(k&j);}
	}
	}
	cout<<max<<endl;
	max=0;//reset max for next test case
	}
	return 0;
	}

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Toggle bits given range

Problem 
http://practice.geeksforgeeks.org/problems/toggle-bits-given-range/0

Solution

	#include <iostream>
	#include <cstdlib>
	using namespace std;
	int main(int argc, char** argv) {
	int t;
	int N,L,R;
	cin>>t;// test cases
	for(int i=0;i<t;i++)
	{
	cin>>N>>L>>R;
	for(int i=L-1;i<R;i++)
	{
	N^=1<<i;
	}
	cout<<N<<endl;
	}
	return 0;
	}

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Monk and his Friend

Problem 

https://www.hackerearth.com/practice/basic-programming/bit-manipulation/basics-of-bit-manipulation/practice-problems/algorithm/monk-and-his-friend/

Solution

	#include <iostream>
	#include <cstdlib>
	#include <string>
	#include <math.h>
	using namespace std;
	int main(int argc, char** argv) {
	long int N,P;
	long int XOR;
	int counter=0;
	int t;
	cin>>t;// test cases
	for(int i=0;i<t;i++)
	{
	cin>>N>>P;
	XOR=N^P;
	for(int i=0;i<=31;i++)
	{
	if(XOR&(1<<i)){counter++;}
	}
	cout<<counter<<endl;
	counter=0;
	}
	return 0;
	}

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Monk and the box of cookies

Problem

https://www.hackerearth.com/practice/basic-programming/bit-manipulation/basics-of-bit-manipulation/practice-problems/algorithm/monk-and-the-box-of-cookies/

Solution

	#include <iostream>
	#include <cstdlib>
	using namespace std;
	int main(int argc, char** argv) {
	int t,N,b;
	int p=0;
	cin>>t;
	int record[32];//record for every position
	for(int i=0;i<t;i++)
	{
	for(int x=0;x<31;x++){record[x]=0;}//intialize records
	cin>>N;
	for(int j=0;j<N;j++)
	{
	cin>>b;
	for(int k=0;k<31;k++)
	{
	if(b&(1<<k)){record[k]++;}//if position k has set bit , increment its record
	}
	}
	int max=record[0];
	for(int x=0;x<31;x++)
	{
	if(max<record[x]){max=record[x];p=x;}//find most occurring position has set bit
	}
	cout<<p<<endl;
	}
	return 0;
	}

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The Great Kian

Problem
https://www.hackerearth.com/practice/basic-programming/implementation/basics-of-implementation/practice-problems/algorithm/the-great-kian/

Solution

	#include <iostream>
	#include <cstdlib>
	#include <string>
	#include <math.h>
	using namespace std;
	int main(int argc, char** argv) {
	long long int n;
	cin>>n;
	long long int a[n];
	long long int s1=0;
	long long int s2=0;
	long long int s3=0;
	for(int i=0;i<n;i++)
	{
	cin>>a[i];
	}
	for(int j=0;j<n;j+=3)
	{
	s1+=a[j];
	}
	cout<<s1<<" ";
	for(int j=1;j<n;j+=3)
	{
	s2+=a[j];
	}
	cout<<s2<<" ";
	for(int j=2;j<n;j+=3)
	{
	s3+=a[j];
	}
	cout<<s3;
	return 0;
	}

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Little Jhool and psychic powers

Problem

https://www.hackerearth.com/practice/basic-programming/implementation/basics-of-implementation/practice-problems/algorithm/psychic-powers/

Solution

	#include <iostream>
	#include <cstdlib>
	#include <string>
	#include <math.h>
	using namespace std;
	int main(int argc, char** argv) {
	string s;
	cin>>s;
	int i=0;
	int sum=0;
	int flag=0;
	while(s[i]!='\0')// loop all elements
	{
	for(int j=i;j<=i+5;j++)// check every 6 consecutive numbers
	{
	sum+=s[j];
	}
	if(sum==288||sum==294)// if you found 6 consecutive ones or zeros (sum is ascii , because input is string)
	{
	flag=1;break;
	}
	i++;// move to next element
	sum=0;//reset sum
	}
	if(flag){cout<<"Sorry, sorry!"<<endl;}
	else if(flag==0){cout<<"Good luck!";}
	return 0;
	}

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Count Digits

Problem 

https://www.hackerearth.com/practice/basic-programming/implementation/basics-of-implementation/practice-problems/algorithm/count-digits-1/

Solution

	#include <iostream>
	#include <cstdlib>
	#include <string>
	using namespace std;
	int main(int argc, char** argv) {
	string s;
	cin>>s;
	int i=0;
	int c[10]={0,0,0,0,0,0,0,0,0};
	while(s[i]!='\0')
	{
	switch(s[i])
	{
	case 48:c[0]++;i++;break;
	case 49:c[1]++;i++;break;
	case 50:c[2]++;i++;break;
	case 51:c[3]++;i++;break;
	case 52:c[4]++;i++;break;
	case 53:c[5]++;i++;break;
	case 54:c[6]++;i++;break;
	case 55:c[7]++;i++;break;
	case 56:c[8]++;i++;break;
	case 57:c[9]++;i++;break;
	}
	}
	// print count of each digit
	for(int j=0;j<=9;j++)
	{
	cout<<j<<" "<<c[j]<<endl;
	} return 0;
	}

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