Sums in a Triangle

Sums in a Triangle

Let's consider a triangle of numbers in which a number appears in the first line, two numbers appear in the second line, three in the third line, etc. Develop a program which will compute the largest of the sums of numbers that appear on the paths starting from the top towards the base, so that:

• on each path the next number is located on the row below, more precisely either directly below or below and one place to the right;
• The number of rows is strictly positive, but less than 100
• All numbers are positive integers between O and 99.

Input

In the first line integer n - the number of test cases (equal to about 1000). Then n test cases follow. Each test case starts with the number of lines which is followed by their content.

Output

For each test case write the determined value in a separate line.

Example

Input:
2
3
1
2 1
1 2 3
4
1
1 2
4 1 2
2 3 1 1

Output:
5
9

 Solution:

#include<stdlib.h>
#include<stdio.h>
inline int input()
{
    int a=0;
    char c;
    c=getchar_unlocked();
    while(c<33)
    {
        c=getchar_unlocked();
    }
    while(c>=33)
    {
        a=(a<<3)+(a<<1)+(c-'0');
        c=getchar_unlocked();
    }
    return a;
};

int main()
{
    int t,n,max,i,j;
    int s[100][100],a[100][100];
    t=input();
    while(t--)
    {
        n=input();
        a[0][0]=input();
        s[0][0]=a[0][0];
        max=s[0][0];

        for(i=1;i<n;i++)
         for(j=0;j<=i;j++)
         {
             a[i][j]=input();
             if(j==0)
              s[i][j]=s[i-1][j]+a[i][j];
            else if(i==j)
              s[i][j]=s[i-1][j-1]+a[i][j];
             else
             {
                 if(s[i-1][j]>s[i-1][j-1])
                    {
                        s[i][j]=s[i-1][j]+a[i][j];

                    }

                else
                    {
                        s[i][j]=s[i-1][j-1]+a[i][j];


                     }


             }
             if(max<s[i][j])
                max=s[i][j];
          }
          printf("%d\n",max);
   }
   return 0;
}