Solution to Problem 18 on ProjectEuler.net


The problem:

—–
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)
—–

My Solution

#include <stdio.h>
#include <math.h>

int main(){
  int mat [15][15];
  int n,i,j,sum,sumfinal,sumparcial,x,y,max;

  for (i=0;i<15;i++)
    for (j=0;j<15;j++)
      if (j<=i){
        scanf("%d",&x);
        mat[i][j]=x;
      }
      else
        mat[i][j]=0;

  int best [15][15];
  best[0][0]=mat[0][0];

  for (i=1;i<15;i++){
    best[i][0] = mat[i][0] + best[i-1][0];
    best[i][i] = mat[i][i] + best[i-1][i-1];
    for (j=1;j<i;j++){
      best[i][j] = mat[i][j] + fmax(best[i-1][j-1],best[i-1][j]);
    }
  }

  max = 0;
  for (j=0;j<15;j++)
    if (best[14][j] > max)
      max = best[14][j];

  printf("%dn",max);
  
  return 0;
}

Leave a Reply

Your email address will not be published. Required fields are marked *