PROJECT EULER SOLUTION # 18

Solution to problem number 18 of Project Euler.

Question # 18

By starting at the top of the triangle below and moving to adjacent numbers on the row below, the maximum total from top to bottom is 23.

3
7 4
2 4 6
8 5 9 3

That is, 3 + 7 + 4 + 9 = 23.
Find the maximum total from top to bottom of the triangle below:

75
95 64
17 47 82
18 35 87 10
20 04 82 47 65
19 01 23 75 03 34
88 02 77 73 07 63 67
99 65 04 28 06 16 70 92
41 41 26 56 83 40 80 70 33
41 48 72 33 47 32 37 16 94 29
53 71 44 65 25 43 91 52 97 51 14
70 11 33 28 77 73 17 78 39 68 17 57
91 71 52 38 17 14 91 43 58 50 27 29 48
63 66 04 68 89 53 67 30 73 16 69 87 40 31
04 62 98 27 23 09 70 98 73 93 38 53 60 04 23

NOTE: As there are only 16384 routes, it is possible to solve this problem by trying every route. However, Problem 67, is the same challenge with a triangle containing one-hundred rows; it cannot be solved by brute force, and requires a clever method! ;o)

Solution # 18

/***********************************************************************************************************/

#include<stdio.h>

#include<conio.h>

#include<stdlib.h>

#include<time.h>

int get_sum(int *);

int main()

{

       FILE *f;

       int arr[15][15],max1[15],max2[15],i=0,j=0,maximum=0;

      

       f=fopen("D:/d.txt","r");

       i=0;

       for(i=0;i<15;i++)

              for(j=0;j<=i;j++)

              {

                     fscanf(f,"%d",&arr[i][j]);

              }

       for(i=0;i<15;i++)

       {

              for(j=0;j<=i;j++)

              {

                     if(i==0)

                     {

                           max1[j]=arr[i][j];

                           continue;

                     }

                     else if((j-1)<0)

                     {

                           max2[j]=arr[i][j]+max1[j];

                     }

                     else if(j==i)

                           max2[j]=arr[i][j]+max1[j-1];

                     else

                           max2[j]=arr[i][j]+((max1[j-1]>max1[j])?max1[j-1]:max1[j]);          

              }

              if(i==0)

                     continue;

                    

              for(j=0;j<=i;j++)

                     max1[j]=max2[j];

       }

      

       for(i=0;i<15;i++)

              if(maximum<max2[i])

                     maximum=max2[i];

       printf("Answer = %d\n\n",maximum);

      

       printf("\nEXECUTION TIME = %f\n",clock()/(float)CLK_TCK);

       system("pause");

}

/***********************************************************************************************************/