@Ankur: then what is the 'best' solution for this? :)
@Balaji: i tried implementing but I dont know which case it fails?? getting
WA now!!
Here is the code:
#include<stdio.h>
int main()
{
long n,gcd=1;
scanf("%d",&n);
long long a[n],b[n],cnt=0,sum=0;
long long min=999999999;
scanf("%lld",&a[0]);
for(long i=1;i<n;i++)
{
scanf("%lld",&a[i]);
b[i-1]=a[i]-a[0];
if(min>b[i-1])
min=b[i-1];
}
for(int k=min;k>0;k--)
{
cnt=0;
for(int i=0;i<n-1;i++)
{
if(b[i]%k==0)
cnt++;
}
if(cnt==n-1)
{
gcd=k;
break;
}
}
sum=((a[n-1]-a[0])/gcd)-n+1;
printf("%lld\n",sum);
return 0;
}
On Sat, Mar 12, 2011 at 2:38 PM, Satyam Kapoor <satyamkapoo...@gmail.com>wrote:
>
> this is gud but not the best soln.
>
> --
> Satyam Kapoor
> B.Tech 2nd year
> Computer Science And Engineering
> M.N.N.I.T ALLAHABAD
>
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