#define MOD 1000000007
....


long long int num[n];

for( int k2=1;k2<n;k2++)

{

for( int k3=0;k3<k2;k3++)

{

if(seq[k3]<seq[k2])

num[k2]=(num[k2]+num[k3])%MOD;
// i don't know why nead %MOD
}

}



.....


could you tell me does c++ iso has the long long int type? why
here need the %MOD.

--
[ See http://www.gotw.ca/resources/clcm.htm for info about ]
[ comp.lang.c++.moderated. First time posters: Do this! ]

puwei wrote:

> #define MOD 1000000007

>
> num[k2]=(num[k2]+num[k3])%MOD;
> // i don't know why nead %MOD

> could you tell me does c++ iso has the long long int type?

Afair long long int is being added to standard c++ in c++0x

http://www.open-std.org/jtc1/sc22/wg...2005/n1811.pdf

> why here need the %MOD.

Well, it's probably needed by the algorithm.


--
[ See http://www.gotw.ca/resources/clcm.htm for info about ]
[ comp.lang.c++.moderated. First time posters: Do this! ]

On 30 Lip, 20:29, puwei <puwei...@gmail.com> wrote:
> #define MOD 1000000007
> ...
>
> long long int num[n];
>
> for( int k2=1;k2<n;k2++)
>
> {
>
> for( int k3=0;k3<k2;k3++)
>
> {
>
> if(seq[k3]<seq[k2])
>
> num[k2]=(num[k2]+num[k3])%MOD;
> // i don't know why nead %MOD
> }
>
> }
>
> ....
>
> could you tell me does c++ iso has the long long int type? why
> here need the %MOD.

To my knowledge long long int is not part of c++ iso.

I do not think that %MOD has something to do with long long type.
If the question was not about %MOD w.r.t. long long type, then this
code sample does not hold the answer.

Cheers
Marcin ¦widerski


--
[ See http://www.gotw.ca/resources/clcm.htm for info about ]
[ comp.lang.c++.moderated. First time posters: Do this! ]