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.....