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I have a N*N upper triangular matrix with property such that, all its diagonal elements are a1,a2,a3,...,aN. I want that a[i][j] (for all j>i) should be

(a[i][j-1] + a[i+1][j]) / 2.

I have many test cases, and I have to apply this property every time to calculate the answer. What is the most optimal way to do this, so that for all test cases the overall running time is less? Test cases: Inputs are N and a1,a2,...,aN.

To calculate the answer, I need to do:

a[0][0] + a[0][2] + ... + a[0][n-1] + a[2][n-1] + a[4][n-1] + ... + a[n-1][n-1].

My solution (which keeps getting timed out):

#include<stdio.h>
double a[2000][2000];
int main(){
int test;
scanf("%d",&test);
//int arr[2000];
while(test--){
    int n,i,j;
    //scanf("%d",&n);
    scanf("%d",&n);
    for(i=0;i<n;i++){
         int num;
         scanf("%d",&num);
         if(n!=1)
            a[i][i] = num*0.5;
         else
            a[i][i] = num;
     }
    for(j=1;j<n;j++){
         int k=j;
         for(i=0;i<n-j;i++,k++){
             if(i==0 && k==n-1)
                 a[i][k] = (a[i+1][k]+a[i][k-1]);
             else
                 a[i][k] = (a[i+1][k]+a[i][k-1])*0.5;
         }
     }
     float sum=0.0;
     for(i=0;i<n;i+=2){
         if( i != n-1 )
         sum+=a[0][i]+a[n-1-i][n-1];
         else
         sum+=a[0][i];
     }
    printf("%.3f\n",sum);
}
getch();
}

Please provide some hints how to optimize the above code.

share|improve this question
Strange since I have asked the same thing on stack overflow and they said that this question best fits here! – jigsawmnc Oct 6 '12 at 14:45

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