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Re:思路

Posted by team162 at 2012-05-14 12:07:30 on Problem 3928
In Reply To:思路 Posted by:yuanchuanshun at 2010-08-27 20:41:47
> __int64 ans=0,low,up,tmp;
>        for(i=1;i<=n;i++)
>        {          
>            modify(b[i]);
>            if(i==1||i==n) continue;
>            tmp=Getsum(b[i]); 
>            low=tmp-1;   //前面比它小的数的个数; 
>            up= (n-b[i]-i+tmp);  //后面比它大的数的个数; 
>            ans+= low*up;
>            //////
>            up=i-tmp;   //前面比它大的数的个数; 
>            low=b[i]-tmp; //后面比它小的数的个数; 
>            ans+= low*up;
>        }

用离散化一下么?

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