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 + ;
 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 = ;
     while(x)
         ans += tree[x], x -= lowbit(x);
     return ans;
 }
 int main()
 {
     while(scanf("%lld", &n) != EOF)
     {
         ll x, ans = ;
         memset(tree_num, , sizeof(tree_num));
         memset(tree_sum, , sizeof(tree_sum));
         for(int i = ; i <= n; i++)
         {
             scanf("%lld", &x);
             add(x, , 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 ;
 }