Hadoop Setup Cool Links:
Found some cool links to setup hadoop on Virtualbox.

Found that cloning the machine before setting up will save us a good amount of effort just is case some steps are not followed properly.

http://cs.smith.edu/dftwiki/index.php/Setup_Virtual_Hadoop_Cluster_under_Ubuntu_with_VirtualBox


sudo -s
editor/nano/vi /etc/hostname
editor/nano/vi /etc/hosts
shutdown -ry now

Although everything after Download hadoop from hadoop directory should be from:
http://codesfusion.blogspot.com/2013/10/setup-hadoop-2x-220-on-ubuntu.html 

Actually this one is  better than the second link:
https://www.digitalocean.com/community/articles/how-to-install-hadoop-on-ubuntu-13-10 

Project Euler Problem 25 Solution

Important concepts. Binet's Formula for finding the nth fibonacci number
Useful link:http://www.geekality.net/2009/11/06/project-euler-problem-25/

Code is just one line: (int) Math.Ceiling((1000 + Math.Log10(5) / 2 - 1) / Math.Log10(1.61803))

1.61803 is the approximate equivalent of golden ratio ((1+sqrt(5))2)

Project Euler Problem 24 Solution

#include "stdafx.h"
#include "stdio.h"
#include "conio.h"
#include "iostream"
#include "fstream"
#include "string"
#include "list"
#include "vector"
#include "map"
#include <algorithm>//Sameer, to find the elements
using namespace std;

int factorial(int k)
{
    int fact = 1;
    for(int counter = k; counter>=1; counter--)
    {
        fact = fact * counter;
    }
    return fact;

}


int main()
{ //for permuation of 1,2,3
    vector<int> firstperm;//={0,1,2,3,4,5,6,7,8,9};
    //IMP:The first permutation we get here is actually the 0th permutation in reality. So, in the final answer, we will actually get 1000,001th permutation. We will need to go back which will be easy because its lexicographic i.e in order
    vector<int> millionthoneloc;
    int quo=0;   
    int millsum=1000000,rem=1;
    for(int i=0;i<=9;i++)
    {
        firstperm.push_back(i);
    }
   
    for(int i=9;i>=0;i--)
    {
      
           
             div_t divresult;
             divresult = div (millsum,factorial(i));           
             rem=divresult.rem;
             millsum=rem;           
             quo=divresult.quot;
       
           
             cout<<endl<<"millsum: "<<millsum;
             cout<<" quo: "<<quo;
             cout<<" rem: "<<rem;

         {
             millionthoneloc.push_back(firstperm[quo]);
             firstperm.erase(firstperm.begin()+quo);

           
         }

    }
   
   

    cout<<endl<<"1,000,001th permutation is: ";
    int i1=millionthoneloc.size();
    for(int i2=0;i2<i1;i2++)
    {
        cout<<millionthoneloc[i2];
       
    }

    return 0;
}

//Following links were helpful:
//http://blog.mycila.com/2009/05/project-euler-problem-24.html (Spoiler, this has answer)
//http://en.wikipedia.org/wiki/Factoradic
//http://www.eggwall.com/2012/01/project-euler-in-r-problem-24.html
//Spoiler alert:
//Answer adjustment explained here. Answer too is mentioned here:
// So, at 1000,001th -> 2783915604. (since 2nd last digit is 0, this number is first in the range for 6 i.e after this we will have various 6 combinations i.e after 604 we have 640....)
//So, now lets see what this says about 5 in this permutation. There are many combinations before this for 5. So, lets adjust what 1,000,000 lexicographic permutation will be.
//4 precedes 6. And the permutation before this one will be the maximum permutation for 4. Hence, it should be 460.
//All in all, the whole number should be 2783915460

//It says that this is the fifth permutation for 5 out of 6 namely:
//5046
//5064
//5406
//5460
//5604
//5640

Project Euler Problem 23 Solution

//Project Euler Problem 23 Solution
//Language Used:C++

#include "stdafx.h"
#include "stdio.h"
#include "conio.h"
#include "iostream"
#include "fstream"
#include "string"
#include "list"
#include "vector"
#include "map"

using namespace std;



int SumOfDivisors(int i)
{
int sum = 0;
for (int j=1;j<i;j++)//Just proper Divisors
{
if(i%j ==0)
{
sum += j;
}
}
return sum;
}
int main()
{

long long int NonAbundantSum = 0, totalSum=0;
vector <int> Abundant;//using push_back will actually help us create a dynamic vector and hence no need to define vector size.
vector <int> Numbers;
int abundantcounter=0, nonabundantcounter=1;

vector <int> SumNo;
string printword ="";

for(int i=1; i<=/*28123*/20161; i++)
{
    Numbers.push_back(i);
    totalSum += i;// find sum of all the numbers from 1 to 28123 and then subract all the values in SumNo

    if (i<SumOfDivisors(i))//abundant if its sum exceeds the no itself
        {
            Abundant.push_back(i);   
           
        }
}
//Sameer: I modified a piece of code here where I was actually storing the Sum of abundant numbers and then trying to do the subtraction. This was lengthy as I realized that even after running my code for 3 minutes, there was no answer. So, then I referred to one of the articles(mentioned below) and stored all the numbers from 1 to 28122 inside a vector which was then used in mapping locations of sums of abundant numbers.
//Also, according to wolfram every number after 20161 can be expressed as abundant number. So, we can cut
for (int j=0; j<Abundant.size();j++)
{
    // Compare abundant[j] to every other element AND ITSELF in the vector.
    for (int p=0;p<Abundant.size();p++)
    {
        bool found = false;
        int temp = Abundant[j]+Abundant[p]-1;//-1 is because eg j=p=25, we need to check 50. however Numbers vector begins at 1, Numbers[0]=1. So, Numbers[49]=50
        if(temp<=/*28123-1*/20161-1) //for -1 Same logic applies here
        {
           
            totalSum -= Numbers[temp];
            Numbers[temp] = 0;
   
        }
    }
}
NonAbundantSum = totalSum;//The remaining value is the non abundant sum (sum of numbers that cannot be expressed as the sum of two abundant numbers
cout<<endl<<NonAbundantSum;


return 0;
}

/*Following links were helpful:
//http://www.mathblog.dk/project-euler-23-find-positive-integers-not-sum-of-abundant-numbers/
//http://mathworld.wolfram.com/AbundantNumber.html
//http://siralansdailyramble.blogspot.com/2009/04/project-euler-23.html
*/

Project Euler Solved Solutions

Project Euler Problem 13
Posted on January 17, 2012 by Sameer
Problem 13
#include “stdio.h”
#include “conio.h”
#include “iostream”
#include “fstream”
#include “string”
#include “list”
using namespace std;
void print(list<__int8> *integer)   
{
int size=integer->size();
char *buffer= new char[size+1];
*(buffer+size)=’\0′;
int j=size-1;
for (list<__int8>::iterator i=integer->begin();i!=integer->end();i++, j–)
{
*(buffer+j)=*i+’0′;
}
cout<<buffer;
delete buffer;
}
void main ()
{
string STRING;
size_t   chars_read = 0;
const int SIZE=51;
char buffer[SIZE]=”\0″;    //initiating all values of  the arrqay to 0;
list<__int8> *integer = new list<__int8>();//Sameer,dont want to use pointers but list wont accept anything wihtout pointer
ifstream infile (“num.txt”);
//the number of numbers to add is 100
for (int i=0;i<100;i++)
{
if (infile.eof())
{
break;
}
else
{
//read 51 because the return character will be the ‘\0′
infile.read(buffer,SIZE);
if(integer->empty())
{
for (int i=SIZE-1;i>=0;i–)// add the digits into the list in reverse order.
{
integer->push_back(buffer[i]-’0′);// deducting the character value by ’0′ char value gives us the integer.
}
}
else
{
/*__int8 is used to reduce memory consumption.
loop through the integer array and buffer. add the numbers.*/
int sum =0;
list<__int8>::iterator it=integer->begin();
for (int j=SIZE-1;j>=0;j–)
{//I think this variable is worngfully called sum. It should be called partial sum since it adds 2 single digit numbers (max can be 9+9 = 18) and the rest it shifts over to the next number
sum += (buffer[j] -’0′) + *it;
if(sum>9)
*it++ = sum -10; //18-10 = 8 equals remainder or 19-10 = 9 (if there is a remainder from previous sum value i.e sum/=10 equals 1
else
*it++ = sum;
sum/=10; //so sum will either be 1 or 0
}
while(it!=integer->end() && sum>0)
{
sum+=*it;
if(sum>9)
*it++ = sum -10;
else
*it++ = sum;
sum/=10;
}
if (sum>0 && it==integer->end())
{
integer->push_back(sum);
}
}
}
}
print(integer);
delete integer;
}
 
Following link was helpful:
http://mafahir.wordpress.com/2010/12/04/adding-50-digit-numbers-cc-implementation/

Project Euler Problem 22
Posted on March 14, 2012 by Sameer
Project Euler Problem 22
Language: C++
#include "stdafx.h"
#include "stdio.h"
#include "conio.h"
#include "iostream"
#include "fstream"
#include "string"
#include "list"
#include "vector"
#include "map"
#include "sstream"
#include "iomanip"//for using setw
using namespace std;

map <long int, string> CharNumberValue;
vector <string> line(103);
int number[103][103];

int main ()
{
vector <vector <string> > data;
vector <string> NamesUnsorted;
vector <char[300]> CharUnsorted;
long long int tempNameValue=0;
//vector <string> NamesSorted;
string NamesSorted;
vector <int> NameValue;
ifstream infile( "names.txt" );

while (infile)
{
string s;
if (!getline( infile, s )) break;

istringstream ss( s );

while (ss)
{
string s;
if (!getline( ss, s, ',' )) break;
NamesUnsorted.push_back( s );
}

// data.push_back( NamesUnsorted );//meaningless

}
//selection sort or insertion sort: Since its a large list, I will go with insertion sort.
for (int iCv=1;iCv<NamesUnsorted.size();iCv++)
{
//the new value to be inserted into a temporary location
NamesSorted = NamesUnsorted[iCv];
// k is the index of the number to the left of the iCv.
int k;

for (k = iCv-1; k >= 0 && NamesUnsorted[k] > NamesSorted; k--)
{
NamesUnsorted[k+1] = NamesUnsorted[k];
}
NamesUnsorted[k+1] = NamesSorted;

}

//for(int iCv=0;iCv<NamesUnsorted.size();iCv++) cout<<NamesUnsorted[iCv]<<" ";
//cout<<endl;

//after the sort, multiply the position to the char value
//string to char conversion
for(int iCv=0;iCv<NamesUnsorted.size();iCv++)
{
char *a=new char[NamesUnsorted[iCv].size()+1];
a[NamesUnsorted[iCv].size()]=0;
memcpy(a,NamesUnsorted[iCv].c_str(),NamesUnsorted[iCv].size());
int i=0, temp=0;

while(a[i]!='\0')
{
//take another var and multiply the i
if(a[i] != '"')
{
temp += int(a[i]) ;
temp -=64; //For ascii TO alphabetical places
}

i++;
}

temp *= (iCv+1); //Mul the location to value-----array starts at 0 so +1 i.e val of iCv + 1
tempNameValue += temp;
cout<<endl<<" Count Value of "<<NamesUnsorted[iCv]<<" is: "<< temp<<".Total Value is: "<< tempNameValue;
}

if (!infile.eof())
{
cerr << "Fooey!\n";
}

return 0;
}
Following link was helpful:
http://www.daniweb.com/software-development/cpp/threads/64778/1780074#post1780074

Project Euler Problem 21
Posted on February 22, 2012 by Sameer
Project Euler Problem 21
Language: C++
#include "stdafx.h"
#include "stdio.h"
#include "conio.h"
#include "iostream"
#include "fstream"
#include "string"
#include "list"
#include "vector"
#include "map"
using namespace std;
map<int, int> DivisorSumMap ;

int SumOfDivisors(int i)
{
int sum = 0;
for (int j=1;j<i;j++)//Just proper Divisors
{

if(i%j ==0)
{
sum += j;
}
}
return sum;
}
int main()
{

long long int AmicableSum = 0;
string printword ="";

for(int i=1; i< 10000; i++)
{

DivisorSumMap[i] = SumOfDivisors(i);

}
for (int i=2;i< 10000; i++)
{
int xval1,yval1,xval2,yval2;

map< int, int>:: iterator it1 = DivisorSumMap.find(i);
xval1=it1->first;
yval1=it1->second;
//if some values in the map are greater than the 10000, just take the value less than 10000 into consideration
if(yval1>10000)
{
yval2=SumOfDivisors(yval1);
}
else
{
map<int, int>:: iterator it2 = DivisorSumMap.find(yval1);
xval2=it2->first;
yval2=it2->second;
}

if(yval2==xval1)
{

AmicableSum +=xval1;
if(xval1 != xval2)
{
AmicableSum +=xval2;
}
else//if they are perfect
{
AmicableSum -=xval2;
}
//To avoid re adding
DivisorSumMap[xval1]=92;
DivisorSumMap[xval2]=91;
cout<<xval1 << " and " << xval2 << "are amicable.Amicable Sum was "<<AmicableSum<<endl;

}

}

return 0;
}


Project Euler Problem 20
Posted on January 26, 2012 by Sameer
Project Euler Problem 20
Language Used:C++ (Not external library like GMP or additional functions)
#include “stdio.h”
#include “conio.h”
#include “iostream”
#include “fstream”
#include “string”
#include “list”
#include “vector”
#include “map”
//#include “math.h”
void main()
{
//////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
long k;
int count1=0;
cout<<”Enter a number whose factorial needs to be calculated:”;
cin>>k;
//////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
/////////////////code for numbers which are less than 33 and greater than 3///////////////////////////////////////////////////////////
if (k<=33)
{
unsigned double long fact=1;
fact=1;
for(int b=k;b>=1;b–)
{
fact=fact*b;
}
cout<<”The factorial of “<<k<<” is “<<fact<<endl;
}
/////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
////////////////code for numbers which are greater than 33 ///////////////////////////////////////////////////////////
else
{
//Sameer, Increased the size of the array from 10000 to 100000
int numArr[100000];
int total,rem=0,count;        //rem use to save remainder of division(Carry Number).
register int i;             //register will enhance performance by minimizing access time
//int i;
for(i=0;i<100000;i++)
numArr[i]=0;            //set all array on NULL.
numArr[100000]=1; //start from end of array.
for(count=2;count<=k;count++)
{
while(i>0)
{
total=numArr[i]*count+rem;  //rem will add the carry number to the next array digit that is multiplied to the count
rem=0;
if(total>9)
{
numArr[i]=total%10;
rem=total/10;
}
else
{
numArr[i]=total;   //numArr[i] is being accessed in this for loop because we have made i as register which means the memory is allocated
}
i–;
}
rem=0;
total=0;
i=100000;
}
cout<<”The factorial of “<<k<<” is “;
for(i=0;i<100000;i++)
{
if(numArr[i]!=0 || count==1)
{
cout<<numArr[i];
count=1;
count1 +=numArr[i];
}
}
cout<<endl;
}
cout<< endl<<count1;
}
 
This link was helpful:
http://www.daniweb.com/software-development/cpp/code/250350

Project Euler Problem 67 and 18
Posted on January 19, 2012 by Sameer
The following is the solution to two similar problems: Problem 18 and Problem 67
Language:C++
#include “stdafx.h”
#include “stdio.h”
#include “conio.h”
#include “iostream”
#include “fstream”
#include “string”
#include “list”
#include “vector”
#include “map”
using namespace std;
map CharNumberValue;
vector line(103);
int number[103][103];
int getMax(int a, int b) {
int c = a – b;
int k = (c >> 31) & 0×1;
int max = a – k * c;
return max;
}
void main ()
{
string STRING;
char col[5000];
size_t chars_read = 0;
int i=0,j=0, partialsum[15];
int depth;
ifstream infile;
infile.open (“Prob67.txt”);//change for Prob18
while( infile ) {
if( infile )
{
string Tline;
std::getline(infile,Tline);
line[i] = Tline;
//following 3 lines convert the string to char
char *a=new char[line[i].size()+1];
a[line[i].size()]=0;
memcpy(a,line[i].c_str(),line[i].size());
//following 4 lines seperates the values after whitespace
char *p = strtok(a, ” “);
while (p)
{
number[i][j]=atoi(p);//convert p to int
//cout<
}
Following links were helpful:
http://blog.vi-kan.net/2010/project-euler-problem-18-67/
http://stackoverflow.com/questions/5792721/problem-with-project-euler-problem-18

Project Euler Problem 17
Posted on January 17, 2012 by Sameer
Problem 17
 
#include “stdio.h”
#include “conio.h”
#include “iostream”
#include “fstream”
#include “string”
#include “list”
#include “vector”
#include “map”
using namespace std;
map<long int, string> CharNumberValue ;
int main()
{
CharNumberValue[0] =”zero”;
CharNumberValue[1] =”one”;
CharNumberValue[2] =”two”;
CharNumberValue[3] =”three”;
CharNumberValue[4] =”four”;
CharNumberValue[5] =”five”;
CharNumberValue[6] =”six”;
CharNumberValue[7] =”seven”;
CharNumberValue[8] =”eight”;
CharNumberValue[9] =”nine”;
CharNumberValue[10] =”ten”;
CharNumberValue[11] =”eleven”;
CharNumberValue[12] =”twelve”;
CharNumberValue[13] =”thirteen”;
CharNumberValue[14] =”fourteen”;
CharNumberValue[15] =”fifteen”;
CharNumberValue[16] =”sixteen”;
CharNumberValue[17] =”seventeen”;
CharNumberValue[18] =”eighteen”;
CharNumberValue[19] =”nineteen”;
CharNumberValue[20] =”twenty”;
CharNumberValue[30] =”thirty”;
CharNumberValue[40] =”forty”;
CharNumberValue[50] =”fifty”;
CharNumberValue[60] =”sixty”;
CharNumberValue[70] =”seventy”;
CharNumberValue[80] =”eighty”;
CharNumberValue[90] =”ninety”;
CharNumberValue[100] =”hundred”;//Sameer,remember 100 is one hundred
CharNumberValue[1000] =”thousand”;//Sameer,remember 1000 is one thousand
long long int count = 0;
string printword =”";
for(int i=1; i<= 1000; i++)
{
if(i<=100)
{
int ten = (i/10)*10;
int unit = i%10;
map<long int, string>:: iterator it = CharNumberValue.find(i);
map<long int, string>:: iterator it1 = CharNumberValue.find(unit);
map<long int, string>:: iterator it2 = CharNumberValue.find(ten);
if(i<10)
{
count  += it1->second.length();
printword = it1->second;
cout<<printword<<endl;
}
if(i>=10 && i<=20)
{
count  += it->second.length();//These are unique
printword = it->second;
cout<<printword<<endl;
}
if(i>20 && i<100)
{
count  += it2->second.length();
printword = it2->second ;
if ((i != 30)&&(i != 40)&&(i != 50)&&(i != 60)&&(i != 70)&&(i != 80)&&(i!= 90)&&(i!= 100))
{
count  += it1->second.length();
printword = it2->second + ” ” + it1->second ;
}
cout<<printword<<endl;
}
if(i==100)
{
count += 10;//Add one hundred
}
}
if(i>100 && i<1000)
{
int hun = i/100;
int ten=i%100;
int mten = (ten/10)*10;//modified ten
int unit = ten%10;
map<long int, string>:: iterator it = CharNumberValue.find(unit);
map<long int, string>:: iterator it1 = CharNumberValue.find(mten);
map<long int, string>:: iterator it2 = CharNumberValue.find(ten);
map<long int, string>:: iterator it3 = CharNumberValue.find(hun);
int counttemp = CharNumberValue[100].length();
count  += it3->second.length() + CharNumberValue[100].length();
printword = it3->second + ” ” + CharNumberValue[100];
if((i != 200)&&(i != 300)&&(i != 400)&&(i != 500)&&(i != 600)&&(i != 700)&&(i != 800)&&(i != 900))
{
if( ten<=20)
{
count  += 3/*for and */+ it2->second.length()   ;//These are unique
printword = it3->second + ” ” + CharNumberValue[100]  + ” and “  + it2->second ;
cout<<printword<<endl;
}
if(ten>20 && ten<=99)
{
count  += 3/*for and */+ it1->second.length();
printword = it3->second + ” ” + CharNumberValue[100]  + ” and “  + it1->second;
if ((ten != 30)&&(ten != 40)&&(ten != 50)&&(ten != 60)&&(ten != 70)&&(ten != 80)&&(ten!= 90)&&(ten!= 100))
{
count  += it->second.length();
printword = it3->second + ” ” + CharNumberValue[100]  + ” and “  + it1->second +” “+ it->second ;
}
cout<<printword<<endl;
}
}
else
{
cout<<printword<<endl;
}
}
}
count += 11;//for one thousand
cout<< count;
return 0;
}
Follwoing links were helpful:
http://blog.functionalfun.net/2008/08/project-euler-problem-17-converting.html


Project Euler Problem 13
Posted on January 17, 2012 by Sameer
Problem 13
#include “stdio.h”
#include “conio.h”
#include “iostream”
#include “fstream”
#include “string”
#include “list”
using namespace std;
void print(list<__int8> *integer)
{
int size=integer->size();
char *buffer= new char[size+1];
*(buffer+size)=’\0′;
int j=size-1;
for (list<__int8>::iterator i=integer->begin();i!=integer->end();i++, j–)
{
*(buffer+j)=*i+’0′;
}
cout<<buffer;
delete buffer;
}
void main ()
{
string STRING;
size_t   chars_read = 0;
const int SIZE=51;
char buffer[SIZE]=”\0″;    //initiating all values of  the arrqay to 0;
list<__int8> *integer = new list<__int8>();//Sameer,dont want to use pointers but list wont accept anything wihtout pointer
ifstream infile (“num.txt”);
//the number of numbers to add is 100
for (int i=0;i<100;i++)
{
if (infile.eof())
{
break;
}
else
{
//read 51 because the return character will be the ‘\0′
infile.read(buffer,SIZE);
if(integer->empty())
{
for (int i=SIZE-1;i>=0;i–)// add the digits into the list in reverse order.
{
integer->push_back(buffer[i]-’0′);// deducting the character value by ’0′ char value gives us the integer.
}
}
else
{
/*__int8 is used to reduce memory consumption.
loop through the integer array and buffer. add the numbers.*/
int sum =0;
list<__int8>::iterator it=integer->begin();
for (int j=SIZE-1;j>=0;j–)
{//I think this variable is worngfully called sum. It should be called partial sum since it adds 2 single digit numbers (max can be 9+9 = 18) and the rest it shifts over to the next number
sum += (buffer[j] -’0′) + *it;
if(sum>9)
*it++ = sum -10; //18-10 = 8 equals remainder or 19-10 = 9 (if there is a remainder from previous sum value i.e sum/=10 equals 1
else
*it++ = sum;
sum/=10; //so sum will either be 1 or 0
}
while(it!=integer->end() && sum>0)
{
sum+=*it;
if(sum>9)
*it++ = sum -10;
else
*it++ = sum;
sum/=10;
}
if (sum>0 && it==integer->end())
{
integer->push_back(sum);
}
}
}
}
print(integer);
delete integer;
}
 
Following link was helpful:
http://mafahir.wordpress.com/2010/12/04/adding-50-digit-numbers-cc-implementation/

Project Euler Problem 8
Posted on January 17, 2012 by Sameer
Problem 8
#include “stdio.h”
#include “conio.h”
void main()
{
int just5=0;
int tempbigi=0,tempbiga=0,tempbigb=0,tempbigc=0,tempbigd=0;
int bigbigi=1,bigbiga=1,bigbigb=1,bigbigc=1,bigbigd=1;
char a1[1001]=”7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450″;
//find product of consecutive 5 numbers
int i=0,a=1,b=2,c=3,d=4,count=0;
//i
while(a1[i] != ‘\0′)
{
just5 = (i+4);
count =0;
tempbigi = 1;
for(int j=i; j<=just5;j++)//j:1,2
{
count++;
int f=a1[j]-’0′;    //7
tempbigi *= f;     //21
}
bigbigi=(tempbigi>bigbigi)?tempbigi:bigbigi;
i += 5;
}
printf(“1st(i)=%d”, bigbigi);
//a
i=1;
while(a1[a] != ‘\0′)
{
just5 = (a+4);
count =0;
tempbiga = 1;
for(int j=i; j<=just5;j++)//j:1,2
{
count++;
int f=a1[j]-’0′;    //7
tempbiga *= f;     //21
}
bigbiga=(tempbiga>bigbiga)?tempbiga:bigbiga;
a += 5;
}
printf(“\n2nd(a) %d”, bigbiga);
//b
i=2;
while(a1[b] != ‘\0′)
{
just5 = (b+4);
count =0;
tempbigb = 1;
for(int j=i; j<=just5;j++)//j:1,2
{
count++;
int f=a1[j]-’0′;    //7
tempbigb *= f;     //21
if(tempbigb == 1666980)
{
printf(“hi”);
}
}
bigbigb=(tempbigb>bigbigb)?tempbigb:bigbigb;
b += 5;
}
printf(“\n3rd(b) %d”, bigbigb);
//c
i=3;
while(a1[c] != ‘\0′)
{
just5 = (c+4);
count =0;
tempbigc = 1;
for(int j=i; j<=just5;j++)//j:1,2
{
count++;
int f=a1[j]-’0′;    //7
tempbigc *= f;     //21
}
bigbigc=(tempbigc>bigbigc)?tempbigc:bigbigc;
c += 5;
}
printf(“\n4th(c) %d”, bigbigc);
//d
i=4;
while(a1[d] != ‘\0′)
{
just5 = (d+4);
count =0;
tempbigd = 1;
for(int j=i; j<=just5;j++)//j:1,2
{
count++;
int f=a1[j]-’0′;    //7
tempbigd *= f;     //21
}
bigbigd=(tempbigd>bigbigd)?tempbigd:bigbigd;
d += 5;
}
printf(“\n5th(d) %d”, bigbigd);
getch();
}

Problem 11
#include “stdio.h”
#include “conio.h”
int mul(int one, int two, int three, int four)
{
int product = one*two*three*four;
return product;
}
void main()
{
int grid [20][20]={
{8, 2, 22, 97, 38, 15, 0, 40, 0, 75, 4, 5, 7, 78, 52, 12, 50, 77, 91, 8},
{49, 49, 99, 40, 17, 81, 18, 57, 60, 87, 17, 40, 98, 43, 69, 48, 4, 56, 62, 0},
{81, 49, 31, 73, 55, 79, 14, 29, 93, 71, 40, 67, 53, 88, 30, 3, 49, 13, 36, 65},
{52, 70, 95, 23, 4, 60, 11, 42, 69, 24, 68, 56, 1, 32, 56, 71, 37, 2, 36, 91},
{22, 31, 16, 71, 51, 67, 63, 89, 41, 92, 36, 54, 22, 40, 40, 28, 66, 33, 13, 80},
{24, 47, 32, 60, 99, 3, 45, 2, 44, 75, 33, 53, 78, 36, 84, 20, 35, 17, 12, 50},
{32, 98, 81, 28, 64, 23, 67, 10, 26, 38, 40, 67, 59, 54, 70, 66, 18, 38, 64, 70},
{67, 26, 20, 68, 2, 62, 12, 20, 95, 63, 94, 39, 63, 8, 40, 91, 66, 49, 94, 21},
{24, 55, 58, 5, 66, 73, 99, 26, 97, 17, 78, 78, 96, 83, 14, 88, 34, 89, 63, 72},
{21, 36, 23, 9, 75, 0, 76, 44, 20, 45, 35, 14, 0, 61, 33, 97, 34, 31, 33, 95},
{78, 17, 53, 28, 22, 75, 31, 67, 15, 94, 3, 80, 4, 62, 16, 14, 9, 53, 56, 92},
{16, 39, 5, 42, 96, 35, 31, 47, 55, 58, 88, 24, 0, 17, 54, 24, 36, 29, 85, 57},
{86, 56, 0, 48, 35, 71, 89, 7, 5, 44, 44, 37, 44, 60, 21, 58, 51, 54, 17, 58},
{19, 80, 81, 68, 5, 94, 47, 69, 28, 73, 92, 13, 86, 52, 17, 77, 4, 89, 55, 40},
{4, 52, 8, 83, 97, 35, 99, 16, 7, 97, 57, 32, 16, 26, 26, 79, 33, 27, 98, 66},
{88, 36, 68, 87, 57, 62, 20, 72, 3, 46, 33, 67, 46, 55, 12, 32, 63, 93, 53, 69},
{4, 42, 16, 73, 38, 25, 39, 11, 24, 94, 72, 18, 8, 46, 29, 32, 40, 62, 76, 36},
{20, 69, 36, 41, 72, 30, 23, 88, 34, 62, 99, 69, 82, 67, 59, 85, 74, 4, 36, 16},
{20, 73, 35, 29, 78, 31, 90, 1, 74, 31, 49, 71, 48, 86, 81, 16, 23, 57, 5, 54},
{1, 70, 54, 71, 83, 51, 54, 69, 16, 92, 33, 48, 61, 43, 52, 1, 89, 19, 67, 48}
};
int max=0,product=0;
//row
for (int i=0; i<20; i++)
{
for(int j=0;j<17;j++)
{
printf(“\n%d %d %d %d”,grid[i][j],grid[i][j+1],grid[i][j+2],grid[i][j+3] );
product = mul(grid[i][j],grid[i][j+1],grid[i][j+2],grid[i][j+3]);
max=(max>product)?max:product;
printf(“\n Maxproduct here is %d”,max);
}
}
//col
for (int i=0; i<17; i++)
{
for(int j=0;j<20;j++)
{
printf(“\n%d %d %d %d”,grid[i][j],grid[i+1][j],grid[i+2][j],grid[i+3][j] );
//row
product = mul(grid[i][j],grid[i+1][j],grid[i+2][j],grid[i+3][j]);
max=(max>product)?max:product;
printf(“\n Maxproduct here is %d”,max);
}
}
//right diagonal
for (int i=0; i<17; i++)
{
for(int j=0;j<17;j++)
{
printf(“\n%d %d %d %d”,grid[i][j],grid[i+1][j+1],grid[i+2][j+2],grid[i+3][j+3] );
//row
product = mul(grid[i][j],grid[i+1][j+1],grid[i+2][j+2],grid[i+3][j+3]);
max=(max>product)?max:product;
printf(“\n Maxproduct here is %d”,max);
}
}
//left diagonal
for (int i=0; i<17; i++)
{
for(int j=0;j<17;j++)
{
printf(“\n%d %d %d %d”,grid[i][j+3],grid[i+1][j+2],grid[i+2][j+1],grid[i+3][j] );
product = mul(grid[i][j+3],grid[i+1][j+2],grid[i+2][j+1],grid[i+3][j]);
max=(max>product)?max:product;
printf(“\n Maxproduct here is %d”,max);
}
}
printf(“\n Final Max is %d”,max);
getch();
}
Following link was helpful:
http://johnnycoder.com/blog/page/2/


Problem 19
Posted on January 24, 2012 by Sameer
Problem 19
Language:C++
using namespace std;
bool isLeapYear(int year) {
bool retval = false;
if (year % 100 == 0) {//this saves comparing every number by 400, just the one with 100
if (year % 400 == 0) {
retval = true;
}
} else if (year % 4 == 0) {
retval = true;
}
return retval;
}
void main()
{
int months[13] =  {0,31,28,31,30,31,30,31,31,30,31,30,31};
int i4=1;
int countsun=0;
string tmonth;
vector <string> month;
tmonth = “”;
month.push_back(tmonth);
tmonth = “jan”;
month.push_back(tmonth);
tmonth = “feb”;
month.push_back(tmonth);
tmonth = “mar”;
month.push_back(tmonth);
tmonth = “apr”;
month.push_back(tmonth);
tmonth = “may”;
month.push_back(tmonth);
tmonth = “jun”;
month.push_back(tmonth);
tmonth = “jul”;
month.push_back(tmonth);
tmonth = “aug”;
month.push_back(tmonth);
tmonth = “sep”;
month.push_back(tmonth);
tmonth = “oct”;
month.push_back(tmonth);
tmonth = “nov”;
month.push_back(tmonth);
tmonth = “dec”;
month.push_back(tmonth);
string tweek;
vector <string> week;
tweek = “”;
week.push_back(tweek);
tweek = “mon”;//since first jan of 1900 was Mon
week.push_back(tweek);
tweek = “tue”;
week.push_back(tweek);
tweek = “wed”;
week.push_back(tweek);
tweek = “thu”;
week.push_back(tweek);
tweek = “fri”;
week.push_back(tweek);
tweek = “sat”;
week.push_back(tweek);
tweek = “sun”;
week.push_back(tweek);
//make 4 loops one inside another for years,months and days
//year (its between 1900 and 2000 not including these years)
for(int i1=1901;i1<2000;i1++)
{
bool leap = isLeapYear(i1);//check if leapyear or not
for (int i2=1;i2<=12;i2++)
{
int monthdays=months[i2];//based on the year and month selected, get the number of days in a month.
if( leap == true && i2 == 2)
{
monthdays = 29;
}
for (int i3=1;i3<=monthdays;i3++)
{
if(i4<=7)
{
cout<<endl<<”Today is “<<week[i4]<<” “<<i3<<” “<<month[i2]<< ” “<<i1;
if(week[i4]==”sun” && i3 == 1)
{countsun++;}
i4++;
}
else
{
i4=1;
cout<<endl<<”Today is “<<week[i4]<<” “<<i3<<” “<<month[i2]<< ” “<<i1;
if(week[i4]==”sun” && i3 == 1)
{countsun++;}
i4++;
}
}
}
}
cout<<endl<<”Total Sundays: “<<countsun;
}

Project Euler Problem 16
Posted on January 19, 2012 by Sameer
Problem 16
Language:C++
#include “stdafx.h”
#include “stdio.h”
#include “conio.h”
#include “iostream”
#include “fstream”
#include “string”
#include “list”
#include “vector”
#include “map”
#include “string”
using namespace std;
using namespace System;
using namespace System::Collections::Generic;
size_t   int_read = 0;
const int SIZE=10;//need to 2 raised to 1000-30, 1000-30 times
int buffer[SIZE];    //initiating all values of  the array to 0;
list<__int8> *integer = new list<__int8>();
//2 raised to 30 is the maximum value that a long integer can hold.
void print(list<__int8> *integer)
{
int size=integer->size();
char *buffer= new char[size+1];
*(buffer+size)=’\0′;
int j=size-1;
for (list<__int8>::iterator i=integer->begin();i!=integer->end();i++, j–)
{
*(buffer+j)=*i+’0′;
}
cout<<buffer;
delete buffer;
}
int  main ()
{
for (int i = 0; i < 1000-30; i++)
{
buffer[0]=1;
buffer[1]=0;
buffer[2]=7;
buffer[3]=3;
buffer[4]=7;
buffer[5]=4;
buffer[6]=1;
buffer[7]=8;
buffer[8]=2;
buffer[9]=4;
if(integer->empty())
{
for (int i=SIZE-1;i>=0;i–)// add the digits into the list in reverse order.
{
integer->push_back(buffer[i]);// deducting the character value by ’0′ char value gives us the integer.
}
}
else
{
int sum=0;
list<__int8>::iterator it=integer->begin();
for (int j=SIZE-1;j>=0;j–)
{// Sum calculates partial sum. It adds 2 single digit numbers (max can be 9+9 = 18) and the rest it shifts over to the next number
sum += (buffer[j] ) + *it;
if(sum>9)
*it++ = sum -10; //18-10 = 8 equalls remainder or 19-10 = 9 (if there is a remainder from previous sum value i.e sum/=10 equals 1
else
*it++ = sum;
sum/=10; //so sum will either be 1 or 0
}
while(it!=integer->end() && sum>0)
{
sum+=*it;
if(sum>9)
*it++ = sum -10;
else
*it++ = sum;
sum/=10;
}
if (sum>0 && it==integer->end())
{
integer->push_back(sum);
}
}
}
print(integer);
delete integer;
return 0;
}
Project Euler Problem 14
Posted on January 19, 2012 by Sameer
Problem 14
Language: C++
#include “stdafx.h”
#include “stdio.h”
#include “conio.h”
#include “iostream”
#include “fstream”
#include “string”
#include “list”
#include “vector”
using namespace std;
//traverse the array for max term
long int max_array(vector<long int> a ,long int num_elements)
{
long int i, max=-32000;
for (i=0; i<num_elements; i++)
{
if (a[i]>max)
{
max=a[i];
}
}
return(max);
}
//recursive function to calculate and count based on Collatz_conjecture
long int Collatz(  long int c1)
{
long long int b1 = c1;
long int counter = 1;
while (b1 != 1) {
++counter;
b1 = (b1 % 2 == 0) ? b1/2 : 3*b1+1;
}
return counter;
}
int  main()
{
long int t1,max=0,major=0,minor=0,majora=0;
//vector<long int> totalcounter (1000000);
for (long  int i=13;i<1000001;i++)
{
minor=Collatz(i);
major = (major > minor ) ? major : minor;
majora =
//cout << “The Counter for no ” << i <<” is ” << totalcounter[i] << ‘\n’;
cout <<”i:”<<i <<”    “<< major <<’\n’;
}
// max = max_array(totalcounter, 1000000);
//cout << “The max is ” << max << ‘\n’;
cout << “The max is ” << major << ‘\n’;
return 0;
}
Project Euler Problem 12
Posted on January 19, 2012 by Sameer
Problem 12
Language:C
#include “stdafx.h”
#include “stdio.h”
#include “conio.h”
#include “math.h”
long int divisor_count(long int  x)
{
long int count =0 ;
// typecast x to long double for calculating sq root and then typecast it again to increment the counter.
for (long int i=2; i< int(sqrt(long double(x))); i++)
{
if ( x % i  == 0)
{
if (i != int(sqrt(long double(x))))
{
count += 2;
}
else count += 1;
}
}
return count;
}
void main()
{
long int i = 1, answer;
while (i>=1)
{
long int num = long int (0.5 * i * (i+1));
if ( divisor_count(num) >= 501)
{
answer = num;
break;
}
i++;
}
printf(“%li”,answer);
getch();
}
Project Euler Problem 11
Posted on January 17, 2012 by Sameer
Problem 11
#include “stdio.h”
#include “conio.h”
int mul(int one, int two, int three, int four)
{
int product = one*two*three*four;
return product;
}
void main()
{
int grid [20][20]={
{8, 2, 22, 97, 38, 15, 0, 40, 0, 75, 4, 5, 7, 78, 52, 12, 50, 77, 91, 8},
{49, 49, 99, 40, 17, 81, 18, 57, 60, 87, 17, 40, 98, 43, 69, 48, 4, 56, 62, 0},
{81, 49, 31, 73, 55, 79, 14, 29, 93, 71, 40, 67, 53, 88, 30, 3, 49, 13, 36, 65},
{52, 70, 95, 23, 4, 60, 11, 42, 69, 24, 68, 56, 1, 32, 56, 71, 37, 2, 36, 91},
{22, 31, 16, 71, 51, 67, 63, 89, 41, 92, 36, 54, 22, 40, 40, 28, 66, 33, 13, 80},
{24, 47, 32, 60, 99, 3, 45, 2, 44, 75, 33, 53, 78, 36, 84, 20, 35, 17, 12, 50},
{32, 98, 81, 28, 64, 23, 67, 10, 26, 38, 40, 67, 59, 54, 70, 66, 18, 38, 64, 70},
{67, 26, 20, 68, 2, 62, 12, 20, 95, 63, 94, 39, 63, 8, 40, 91, 66, 49, 94, 21},
{24, 55, 58, 5, 66, 73, 99, 26, 97, 17, 78, 78, 96, 83, 14, 88, 34, 89, 63, 72},
{21, 36, 23, 9, 75, 0, 76, 44, 20, 45, 35, 14, 0, 61, 33, 97, 34, 31, 33, 95},
{78, 17, 53, 28, 22, 75, 31, 67, 15, 94, 3, 80, 4, 62, 16, 14, 9, 53, 56, 92},
{16, 39, 5, 42, 96, 35, 31, 47, 55, 58, 88, 24, 0, 17, 54, 24, 36, 29, 85, 57},
{86, 56, 0, 48, 35, 71, 89, 7, 5, 44, 44, 37, 44, 60, 21, 58, 51, 54, 17, 58},
{19, 80, 81, 68, 5, 94, 47, 69, 28, 73, 92, 13, 86, 52, 17, 77, 4, 89, 55, 40},
{4, 52, 8, 83, 97, 35, 99, 16, 7, 97, 57, 32, 16, 26, 26, 79, 33, 27, 98, 66},
{88, 36, 68, 87, 57, 62, 20, 72, 3, 46, 33, 67, 46, 55, 12, 32, 63, 93, 53, 69},
{4, 42, 16, 73, 38, 25, 39, 11, 24, 94, 72, 18, 8, 46, 29, 32, 40, 62, 76, 36},
{20, 69, 36, 41, 72, 30, 23, 88, 34, 62, 99, 69, 82, 67, 59, 85, 74, 4, 36, 16},
{20, 73, 35, 29, 78, 31, 90, 1, 74, 31, 49, 71, 48, 86, 81, 16, 23, 57, 5, 54},
{1, 70, 54, 71, 83, 51, 54, 69, 16, 92, 33, 48, 61, 43, 52, 1, 89, 19, 67, 48}
};
int max=0,product=0;
//row
for (int i=0; i<20; i++)
{
for(int j=0;j<17;j++)
{
printf(“\n%d %d %d %d”,grid[i][j],grid[i][j+1],grid[i][j+2],grid[i][j+3] );
product = mul(grid[i][j],grid[i][j+1],grid[i][j+2],grid[i][j+3]);
max=(max>product)?max:product;
printf(“\n Maxproduct here is %d”,max);
}
}
//col
for (int i=0; i<17; i++)
{
for(int j=0;j<20;j++)
{
printf(“\n%d %d %d %d”,grid[i][j],grid[i+1][j],grid[i+2][j],grid[i+3][j] );
//row
product = mul(grid[i][j],grid[i+1][j],grid[i+2][j],grid[i+3][j]);
max=(max>product)?max:product;
printf(“\n Maxproduct here is %d”,max);
}
}
//right diagonal
for (int i=0; i<17; i++)
{
for(int j=0;j<17;j++)
{
printf(“\n%d %d %d %d”,grid[i][j],grid[i+1][j+1],grid[i+2][j+2],grid[i+3][j+3] );
//row
product = mul(grid[i][j],grid[i+1][j+1],grid[i+2][j+2],grid[i+3][j+3]);
max=(max>product)?max:product;
printf(“\n Maxproduct here is %d”,max);
}
}
//left diagonal
for (int i=0; i<17; i++)
{
for(int j=0;j<17;j++)
{
printf(“\n%d %d %d %d”,grid[i][j+3],grid[i+1][j+2],grid[i+2][j+1],grid[i+3][j] );
product = mul(grid[i][j+3],grid[i+1][j+2],grid[i+2][j+1],grid[i+3][j]);
max=(max>product)?max:product;
printf(“\n Maxproduct here is %d”,max);
}
}
printf(“\n Final Max is %d”,max);
getch();
}
Following link was helpful:
http://johnnycoder.com/blog/page/2/