sorry, @satyam: then what is the 'best' solution for this? :)
On Sat, Mar 12, 2011 at 7:06 PM, Akshata Sharma
<[email protected]>wrote:
> @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
> <[email protected]>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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>
>
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