[Project Euler] Problem 11 in C

Problem: Project Euler Problem 11

This problem was very similar to problem 8 of Project Euler.
I used the same concept proposed in optimizing the code (click this)

Optimizations
Skipping product upon seeing certain "low" digit values.
        Upon inspection of product of randomly chosen 4 numbers, which in my case was (78*78*96*83) resulting in a product of 48477312, we can arrive at a conclusion of ignoring any product containing numbers less than or equal to 49.
        This is because the highest possible product with only 3 digits is 99 * 99 * 99 = 970299. So, simply dividing 48477312/970299 = 49.96... So, even in the best case scenario with the number 49, we cannot exceed the arbitrarily found product. Therefore, we don't even need to compute for products with number less than or equal to 49.

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

int main()
{
    int arr[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 i, j, k, tmp;
    //1 = left-down, 2 = down, 3 = right-down, 4 = right
    long long int prod1, prod2, prod3, prod4, greatestProd = 0;
    for (i = 0; i < 20; i++){
        for (j = 0; j < 20; j++){
            prod1 = 1, prod2 = 1, prod3 = 1, prod4 = 1;
            for (k = 0; k < 4; k++){
                //left-down
                tmp = arr[i+k][j-k];
                if (tmp >= 50 && j >= 3 && i <= 16)
                    prod1 *= tmp;
                else
                    prod1 = 0;

                //down
                tmp = arr[i+k][j];
                if (tmp >= 50 && i <= 16)
                    prod2 *= tmp;
                else
                    prod2 = 0;

                //right-down
                tmp = arr[i+k][j+k];
                if (tmp >= 50 && i <= 16 && j <= 16)
                    prod3 *= tmp;
                else
                    prod3 = 0;

                //right
                tmp = arr[i][j+k];
                if (tmp >= 50 && j <= 16)
                    prod4 *= tmp;
                else
                    prod4 = 0;
            }

            if (prod1 > greatestProd)
                greatestProd = prod1;
            if (prod2 > greatestProd)
                greatestProd = prod2;
            if (prod3 > greatestProd)
                greatestProd = prod3;
            if (prod4 > greatestProd)
                greatestProd = prod4;
        }
    }
    printf("%lld\n", greatestProd);

    return 0;
}

Best Execution Time: 0.006s
Answer:

70600674