hdu 2838 Cow Sorting

题意：

给出一个数组，要求把他们排成升序，两个数字交换的代价是x + y，求代价的总和。

思路：

简单的逆序对变形，树状数组维护小于的cnt和sum即可。

代码：

 #include <stdio.h>
 #include <string.h>
 #include <algorithm>
 using namespace std;
 typedef long long ll;
 const int N = 1e5 + ;
 const int M = ;
 int a[N];
 int c[N];
 ll sum[N];
 int lowbit(int x)
 {
     return x&(-x);
 }
 void addcnt(int x)
 {
     for (int i = x;i <= M;i += lowbit(i)) c[i]++;
 }
 int getcnt(int x)
 {
     int ans = ;
     for (int i = x;i > ;i -= lowbit(i)) ans += c[i];
     return ans;
 }
 void addsum(int x,int y)
 {
     for (int i = x;i <= M;i += lowbit(i)) sum[i] += y;
 }
 ll getsum(int x)
 {
     ll ans = ;
     for (int i = x;i > ;i -= lowbit(i)) ans += sum[i];
     return ans;
 }
 int main()
 {
     int n;
     while (scanf("%d",&n) != EOF)
     {
         memset(c,,sizeof(c));
         memset(sum,,sizeof(sum));
         for (int i = ;i < n;i++)
         {
             scanf("%d",&a[i]);
         }
         ll tmp = ;
         ll ans = ;
         for (int i = ;i < n;i++)
         {
             int cnt = getcnt(a[i]);
             cnt = i - cnt;
             ll s = getsum(a[i]);
             s = tmp - s;
             ans += s + 1LL * cnt * a[i];
             addcnt(a[i]);
             addsum(a[i],a[i]);
             tmp += a[i];
         }
         printf("%lld\n",ans);
     }
     return ;
 }