Project Euler #8 - C

Question:

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

Solution:

There are several ways to go about this, but I decided to take an approach that I believe minimizes the number of multiplications required to find the correct solution. First, store this number as one long one-dimensional array. Then, rather than multiplying out every single set of consecutive integers, I decided to look at 6 numbers at a time. For example, lets start at the beginning. 

731671

It is easy to see here that if we were to multiply out the first 5 numbers (73167) and then compare it to the second 5 numbers (31671), that four of the numbers are used twice. Since those 4 numbers are the same in both cases and since we are searching for the highest product of 5 consecutive numbers, we need not multiply out the case which will lead to a lesser solution. Simply put, is the number to the left of the 4 or to the right of the 4 higher (in this case, 7 or 1)? Once that has been decided, multiply out the product and check whether that product is higher than the current highest product. We continue this process until we reach the end of the array. However, with this approach instead of incrementing to the very next number we can make jumps of two! This means that not only does this approach minimize the number of multiplications, but it also cuts the number or required iterations in half!

And now, the code...

#include <stdio.h>

#include <stdlib.h>

int main (void)

{

char * numString = "7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450";

int numArray [1000] = {0};

int i;

int high_index = 0;

int high_sum = 0;

int result = 1;

/* Turn numString into an array for easier referencing */

for (i = 0; i < 1000; i++)

numArray[i] = *(numString + i) - '0';

for(i = 0; i < 995; i+=2)

{

int j = 0;

int sum = 0;

int temp_high_index = i; 

int left  = numArray[i];

int right = numArray[i+5];

for ( j = 0; j < 6; j++)

{

if (numArray[i+j] == 0)

{

/* 

  if a zero is encountered, skip all numbers until the zero has passed. By adding j-1 to i, we skip to the index just BEFORE the 0. However, when the loop turns over the i+=2 will skip just beyond the 0.

*/

i += j-1; 

continue;

}

}

if (left >= right) 

{

for (j = 0; j < 5; j++)

sum += numArray[i+j];

}

else

{

temp_high_index++; 

for (j = 0; j < 5; j++)

sum += numArray[i+j+1];

}

if (sum >= high_sum) 

{

high_sum = sum; 

high_index = temp_high_index; 

}

}

/* get product */

for (i = high_index; i < high_index + 5; i++)

result *= numArray[i];

printf("Highest product is %d\n", result);

return 1;

}

Compile and watch the solution appear before your eyes!