prb018.cpp

/*


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)

*/

#include<iostream>
#include<cstdio>

using namespace std;
int main()
{
	int i,j,parent=0,child=0,p[120]={75,95,64,17,47,82,18,35,87,10,20,4,82,47,65,19,1,23,75,3,34,88,2,77,73,7,63,67,99,65,4,28,6,16,70,92,41,41,26,56,83,40,80,70, 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, 4, 68, 89, 53, 67, 30, 73, 16, 69, 87, 40, 31,4, 62, 98, 27, 23, 9, 70, 98, 73, 93, 38, 53, 60,4, 23};

	for(i=14;i>0;i--)
	{
		child=((i*(i+1))/2);
		parent=(((i-1)*(i))/2);
		for(j=1;j<=i;j++)
		{
			//cout<<"child= "<<child<<"p[child]= "<<p[child]<<"  "<<p[child+1];
			if(p[child]>p[child+1])
				p[parent]=p[parent]+p[child];
			else
				p[parent]=p[parent]+p[child+1];

			parent++;
			child++;
			//getchar();
		}
	}
	cout<<"\n\nSum = "<<p[0];
	getchar();
	return 0;
}