//use dynamic programming approach
//it is O(n) time and O(1) space
#include<stdio.h>
#define N 5
void main()
{
int a[N]={1,2,3,4,5},i;
int prod1[N];
int p=1;
for(i=0;i<N;++i)
{
prod1[i]=p;
p*=a[i];
}
int prod2[N];
p=1;
for(i=N-1;i>=0;--i)
{
prod2[i]=p;
p*=a[i];
}
int products[N];
for(i=0;i<N;++i)
{
products[i]=prod1[i]*prod2[i];
printf("%d ",products[i]);
}
}
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