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 + 10;
const int M = 100000;
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 = 0;
    for (int i = x;i > 0;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 = 0;
    for (int i = x;i > 0;i -= lowbit(i)) ans += sum[i];
    return ans;
}
int main()
{
    int n;
    while (scanf("%d",&n) != EOF)
    {
        memset(c,0,sizeof(c));
        memset(sum,0,sizeof(sum));
        for (int i = 0;i < n;i++)
        {
            scanf("%d",&a[i]);
        }
        ll tmp = 0;
        ll ans = 0;
        for (int i = 0;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
",ans);
    }
    return 0;
}