Problem 8: Sum of primes

Q:Find the sum of all the primes below two million.

Sol:

The logic used to solve the problem was the same as in problem 7 of finding the nth prime.
Code for the problem is also very similar,the small difference being finding the sum.

#include<cstdio>
#include<cmath>
using namespace std;

int main()
{
    int n;
    scanf("%d",&n);
    int arr[500000];
    int i=2;
    arr[0]=2;
    arr[1]=3;
    int k=1,j;
    long long sum=5;
    int nextprime,flag=0,c=1;
    nextprime=5;
    while(nextprime<n)
    {
               flag=0;
               for(j=0;((j < i)&&(arr[j]<=sqrt((double)nextprime)));j++)
               {
                               if(nextprime%arr[j] == 0)
                               {
                                              flag=1;
                                              break;
                               }
               }
               if(flag==0)
               {
                          arr[i]=nextprime;
                          if (arr[i]< n)
                            sum=sum+(long long)(arr[i]);
                          i++;
               }
               if(c==-1)
                        k++;
               nextprime=6*k +c;
               c=-c;
    }
    printf("%lld\n",sum);

}

The program takes 2-3 seconds to execute for the value of n being two million.



Please post your comments and doubts.
Suggestions for better algorithms are always welcome :)

Problem 8: 1000 digit number

Q. Find the greatest product of five consecutive digits in the 1000-digit number.

73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450

Sol:

The method adopted to solve the problem was creating a window of 5 digits, finding the product of numbers in it and then sliding the window over the whole of the number. Each time we advance one digit, a new digit is included in the window which is multiplied to earlier product and an old digit is discarded by which the product is divided. Thus we get a new product each time we slide the window. Finding maximum of the products is the answer. 

This algorithm is simple, but trouble arises when there is a zero in between the number at some place. For those cases. in which zeroes appear, the 5 digit combinations are discarded i.e. they are not allowed to interfere with the window product. Thus awkward values are avoided.

The coding part is 

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main()

{

    int n,i,p=1,max,j;

    FILE *fp=fopen("input.txt","r");

    fscanf(fp,"%d\n",&n);

    char *a=(char*)malloc(n * sizeof(char));

    a[0]='\0';

    char temp[100];

    while(fscanf(fp,"%s",temp)!=EOF)

    {

                    strcat(a,temp);

    }

    int x=0;

    for(i=0;i<5;i++)

    {

                    if((a[x+i]-'0')==0)

                    {

                                 p=1;

                                 x=x+i+1;

                                 i=-1;

                    }

                    else

                        p=p * (a[x+i]-'0');

    }

    max=p;

    int st=x+i;

    for(i=st;i<n;i++)

    {

                    if((a[i]-'0')==0)

                    {

                               p=1;

                               x=i+1;

                               for(j=0;(j<5)&&((x+j) < n);j++)

                               {

                                   if((a[x+j]-'0')==0)

                       {

                                 p=1;

                                 x=x+j+1;

                                 j=-1;

                       }

                       else

                         p=p * (a[x+j]-'0');

                               

                               }

                               i=x+j-1;                

                    }

                    else

                    {

                               p=p/(a[i-5]-'0');

                               p=p*(a[i]-'0');

                    }

                    if(max

                             max=p;

    }

    printf("\n%d\n",max);

}

In the program the input is taken from file input.txt, n denotes the number of digits in the number, the character array 'a' is used to store the number. p denotes the product in the current window and max denotes the maximum product obtained yet.

As it can be seen, when zero occurs, the new product is obtained by finding a nearest window which doesn't have zero.

Please post your comments and doubts...
Suggestion for better algorithms are always welcome.....

Problem 7: Nth prime

Q. What is the 10001st prime number?
Sol:
I got the idea to solve this problem from the algorithm "Sieve of Eratosthenes"  which finds the prime numbers upto a number. It just stores the number in an array selects one element of the array at a time in ascending order and eliminates the numbers in the array which come out to be multiple of the selected.
If we start constructing an array starting with array values 2 and 3(the first two prime numbers) and then continue to incorporate the numbers in the array which are not multiple of previous values in array, we will always get a prime number in the array. Moreover it can be observed that any prime number can be expressed as (6x+1) or (6x-1). So, there is no need to check other numbers for divisiblity by numbers in array.


Then, the algorithm is that we start from 2 and 3 as initial values in array, increment a variable so that it has always the value of the form 6x+1 or 6x-1 and check if not divisible by elements already present in the array. If not divisible by any one of them, the number is entered into the array.


Now, comes the fun part........ coding


#include< cstdio>
using namespace std;

int main()
{
    int n;
    scanf("%d",&n);
    int arr[100000];
    int i=2;
    arr[0]=2;
    arr[1]=3;
    int k=1,j;
    int nextprime,flag=0,c=1;
    nextprime=5;
    while(i < n)
    {
               flag=0;
               for(j=0;((j < i)&&(arr[j]<=sqrt((double)nextprime)));j++)
               {
                               if(nextprime%arr[j] == 0)
                               {
                                              flag=1;
                                              break;
                               }
               }
               if(flag==0)
               {
                          arr[i]=nextprime;
                          i++;
               }
               if(c==-1)
                        k++;
               nextprime=6*k +c;
               c=-c;
    }
    printf("%d\n",arr[n-1]);
}


In the program, a variable k is is used to find the next element (which is 6k+1 or 6k-1) to be considered for position in the array.The element is checked for divisiblity and entered into array if not divisible by any of the numbers in the array....




Please post your comments and doubts...
Suggestion for better algorithms are always welcome.....

Problem 6: Sum of squares or Square of sum !!!!

Q. Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.


Sol: 
I think problem is also one of the easiest problems here. 
The problem can be easily solved through basic knowledge of Arithmetic-Geometric Progression....


We know that sum of squares of numbers from 1 to n is n(n+1)(2n+1)/6 and sum of the numbers from 1 to n is n(n+1)/2.
So, the result can be formulated into an expression


Difference= (n(n+1)/2) ^ 2  - n(n+1)(2n+1)/6 = n(n+1)( 3(n^2) - n -2 ) / 12


Now all that remains is calculation.......  :D


Please post your comments...
Suggestion for better algorithms are always welcome.....

Problem 5:Least evenly divisible number

Q. 2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.What is the smallest number that is evenly divisible by all of the numbers from 1 to 20 ?

Sol:

It can be easily seen that the question is asking the LCM of the numbers.The problem can be solved without even resorting to algorithms.Just compute the prime factorization of numbers and multiply the highest power of each prime.

20 = 2^2 * 5
19 = 19
18 = 2 * 3^2
17 = 17
16 = 2^4
15 = 3 * 5
14 = 2 * 7
13 = 13
11 = 11 

Since all the lesser numbers have been included, they can be safely ignored..

And

lcm= 2^4 * 3^2 * 5 * 7 * 11 * 13 * 17 * 19

Now, the problem seems so simple, but what if we were to calculate the result for natural numbers from 1 to an arbitrary number(say ' n ').

It can be observed that 

LCM(a,b,c,d)=LCM(b,c,LCM(a,d))

To implement it, we can run a loop from 2 to n-1 and at each step find the lcm of loop variable and the previous stage lcm.The lcm can be easily found by applying euclid's algorithm of finding GCD as

 LCM(a,b)=(a*b)/GCD(a,b).

Now the coding part:

int main()

{

scanf("%d",&lastnumber);

long n=20;

 for(i=2;i < 20; i++)

 {

                  n=(n*i)/gcd(n,i);

 }

 printf(" %ld \n", n);

return 0;

}

The euclid's algo can be implemented as:

int gcd(int a,int b)

{

    if(a == 0)

       return b;

    while(b != 0)

        if (a > b)

           a = a - b;

        else

           b = b - a;

    return a;

}

Thats all it is to the problem :D

Please post your comments and doubts...
Suggestion for better algorithms are always welcome.....

Problem 4: Largest palindrome number

Q.Find the largest palindrome number made from the product of two 3-digit numbers.


Sol.
The problem can be solved easily by brute force i.e. checking all the products of 3-digit numbers to be palindrome and then determining the largest among them.
The code for palindrome checking is easy, we can simply reverse the number and check the equality.
int ispalindrome(int n)
{
        int temp=n,rem,sum;
        while(n>0)
        {
              rem=n%10;
              n=n/10;
              sum=sum*10 + rem;
         }
         if(sum==temp)
              return 1;
         else
              return 0;
}


int main()
{
      int big=0;
      for(i=999;i>=100;i--)
          for(i=999;i>=100;i--)
              if(ispalindrome(i*j)&&(big<(i*j)))
                         big=(i*j);
}
As you may have noticed, it takes too much time. So, I used some of my observations regarding the problem to reduce the time.


1. A palindrome number with even number of digits is always divisible by 11. So, at least  one of the numbers must be divisible by 11.Thus the inner loop can be decreased to be decremented by 11 each time.
2.There are many redundant multiplications like 999*998 and then 998*999.That must be reduced.For that we can find the nearest number to 'i' which is divisible by 11 and is less than 'i'.
It can be done easily as the number would be : i-(i%11)
3. The last digit of the palindrome number can't be zero as the first isn't. So, all the numbers which are multiple of 10 can be eliminated from the consideration.


The improved code can be:
int main()
{
         int big=0,i,j;
         for(i=999;i>=101;i--)
         {
                   if(i%10==0)
                          continue;
                   x=i-(i%11);
                   for(j=x;j>=121;j-=11)
                   {
                            n=i*j;
                            if(ispalindrome(n)&& big < n)
                                   big=n;
                   }
          }
          printf("%d",big);
}


Please post your comments and doubts...
Suggestion for better algorithms are always welcome.....
This is me signing off for today....

Problem 3: Largest Prime Factor

Q. What is the largest prime factor of the number 600851475143 ?


Sol:
It can be seen that the largest prime factor of a number is equal to larger of its largest prime factor of its largest factor and the smallest factor.
i.e. largest prime factor(n)=max(smallest factor(n), largest prime factor(largest factor(n)))
or,largest prime factor(n)=max(smallest factor(n), largest prime factor(n/smallestfactor(n)))


using the relation, either the recursive or the iterative approach can be applied to solve the problem..


Now, it can also be seen that while finding the smallest factor, we only need to divide the number concerned by only one even number '2' and rest can be odd numbers. So, all the other even numbers can be eliminated from the process, thus optimizing our program. 
Since, a factor 'i' can divide 'n' multiple times, we can also put a subroutine to divide n by i multiple times until the new number comes out to be not divisible by i, further optimizing the problem...


Now the last and the easiest part , Coding :



long long int solution(long long n)
{

     int i=2;
     long long x;
     if(n%2==0)
     {
               x=n/i;
               while(x%i==0)
                            x=x/i;
               x=solution(x);
               return((i>x)?i:x);
     }
     else
     {
         i=3;    
         while(i < sqrt ( ( double ) n ) )
         {
                                 if(n%i==0)
                                 {
                                           x=n/i;
                                           while(x%i==0)
                                                        x=x/i;
                                           x=solution(x);
                                           return((i>x)?i:x);
                                 }
                                 i=i+2;
         }
         return n;
     }
}



As explained above, the code applies recursive approach to solve the problem and returns the largest prime factor for 'n'..

Please post your comments and doubts...

Better algorithms are always welcome..