poj 2299 树状数组

就是求每个数之前比他大的数的个数，然后因为数太大了，先离散化一下

注意一组测试数据

3

1

1

1

答案是0

还有结果是longlong

AC代码如下：

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>using namespace std;const int MAX_N = 5e5;struct Node{    __int64 n;    int id;};Node node[MAX_N+5];int num[MAX_N+5];int sum[MAX_N+5];int N;bool cmp( Node a, Node b ){    if( a.n > b.n ){        return false;    }else{        return true;    }}inline int lowbit( int x ){    return x & -x;}int getsum( int x ){    int ans = 0;    while( x > 0 ){        ans += sum[x];        x -= lowbit( x );    }    return ans;}void updata( int x, int val ){    while( x <= N ){        sum[x] += val;        x += lowbit( x );    }}int main(){    while( scanf( "%d", &N ) && N ){        for( int i = 1; i <= N; i++ ){            scanf( "%I64d", &node[i].n );            node[i].id = i;        }        sort( node + 1, node + 1 + N, cmp );        num[node[1].id] = 1;        for( int i = 2; i <= N; i++ ){            if( node[i].n == node[i-1].n ){                num[node[i].id] = num[node[i-1].id];            }else{                num[node[i].id] = num[node[i-1].id] + 1;            }        }        memset( sum, 0, sizeof( sum ) );        __int64 ans = 0;        for( int i = 1; i <= N; i++ ){            updata( num[i], 1 );            ans += i - getsum( num[i] );        }        printf( "%I64d\n", ans );    }    return 0;}/*3111答案是0*/