POJ 2299 tree array + discretization

After reading the questions before the class, I thought about the structure in the class. After the class is over 10 minutes, I just handed in A... Okay... This question is so weak!

Question:

Returns the number of reverse orders in a series ....

You can use coordinates for discretization. Then... update the tree array from a large number... sum the query time, and obtain the large number before the number.

Water burst... 1Y

#include<iostream>#include<cstdio>#include<algorithm>using namespace std;struct node{    int val,pos;}a[555555];int b[555555],c[555555];bool cmp( node a,node b ){ return a.val>b.val; }int lowbit( int x ){ return x&-x; }int n;int sum( int x ){ int rec=0; while( x ) {    rec+=c[x];    x-=lowbit(x);  }  return rec;}void add( int x,int val ){ while( x<=n ) {    c[x]+=val;    x+=lowbit(x);  }}int main(){ while( scanf("%d",&n)!=EOF ) {    memset( c,0,sizeof(c) );    if( n==0 )break;    for( int i=1;i<=n;i++ )    {   scanf( "%d",&a[i].val );   a[i].pos=i;   }   sort( a+1,a+n+1,cmp );   for( int i=1;i<=n;i++ )   b[a[i].pos]=i;   __int64 ans=0;   for( int i=1;i<=n;i++ )   {   ans+=sum(b[i]-1);   add(b[i],1);      }      printf( "%I64d\n",ans );  } return 0;}