Understanding the execution time of a memory access code with varying strides - performance

Question

I was trying to understand the execution time of the following code for different steps but some figures are really confusing to me. This is a simple code for memory reads with increasing steps.
Here is my code
#include <iostream>
#include <algorithm>
#include <ctime>
#include <stdio.h>
using namespace std;
int main(){
int a=1024;
int numLoop=a;
int temp =0;
int * arr;
arr = (int*) malloc (100000000 * sizeof(int));
for(int array=a; array<(a+1); array+=1) {
for(int step=1; step<(numLoop+1); step*=2) {
clock_t begin = clock();
for(int v=1; v<(step+1); v++) {
for(int i=0; i<array; i+=step) {
temp = (temp+arr[i]) % 33721;
}
}
clock_t end = clock();
double elapsed_secs = double(end - begin) / CLOCKS_PER_SEC;
printf("%0.6lf %d %d\n",elapsed_secs,step,array);
}
}
return 0;
}
I got the following results after execution
0.000073 1 1024
0.000069 2 1024
0.000078 4 1024
0.000099 8 1024
0.000140 16 1024
0.000220 32 1024
0.000432 64 1024
0.000682 128 1024
0.001354 256 1024
0.002693 512 1024
0.005178 1024 1024
Architecture details:
Freescale p4080ds QorIQ processor
Numer of active core:1 (configured)
L1 cache : 32kB
L2 cache : 128kB
L3 cache : 2MB
(More about the architecture is here)
Here the most disturbing thing for me is the last line of the result. When step=1024 then it is accessing arr[0] 1024 times which should be in the cache after first access. Then why does it take the highest execution time( 0.005178) of all?