hdu-2838 Cow Sorting---逆序对的花费

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=2838

题目大意：

就是求将之前的排列变成一个递增的排列，每交换两个数的代价为两个数的和，求变成递增的排列所需的最小代价为多少。

其实就是求出逆序对的花费（每对逆序对的花费是这个逆序对的和）

解题思路：

用两个树状数组维护即可，一个维护逆序对的数目，一个维护逆序对的值的和

注意：全部用long long

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<map>
#include<set>
#include<cmath>
#include<algorithm>
#include<vector>
#include<sstream>
#define lowbit(i) (i&(-i))
using namespace std;
const int maxn = 1e6 + 10;
typedef long long ll;
ll tree_sum[maxn], tree_num[maxn], n;
void add(ll x, ll d, ll tree[])
{
    while(x <= n)tree[x] += d, x += lowbit(x);
}
ll sum(ll x, ll tree[])
{
    ll ans = 0;
    while(x)
        ans += tree[x], x -= lowbit(x);
    return ans;
}
int main()
{
    while(scanf("%lld", &n) != EOF)
    {
        ll x, ans = 0;
        memset(tree_num, 0, sizeof(tree_num));
        memset(tree_sum, 0, sizeof(tree_sum));
        for(int i = 1; i <= n; i++)
        {
            scanf("%lld", &x);
            add(x, 1, tree_num);
            add(x, x, tree_sum);
            ans += (sum(n, tree_num) - sum(x, tree_num)) * x + sum(n, tree_sum) - sum(x, tree_sum);
        }
        cout<<ans<<endl;
    }
    return 0;
}