题意：首先每次只能交换相邻的两头牛，并且最后要求升序排列，所以最后整个序列的逆序是0，每次交换只可以消除1个逆序。（令a[i]的逆序是从1到i-1比它大的数的个数。）
思路：对于某个数，要把它变成有序的，那么很容易可以推算出公式就是它自身的逆序数乘自身的值再加上它的逆序数的和，自己算算看看。
吐槽：一开始没想清楚树状数组下标的问题，后来没注意会爆int搞了好久...
#include <cstdio>
#include <queue>
#include <cstring>
#include <iostream>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <map>
#include <string>
#include <set>
#include <ctime>
#include <cmath>
#include <cctype>
using namespace std;
#define maxn 100000+100
#define LL long long
int cas=1,T;
int a[maxn];
LL L[maxn];
LL c[maxn];
int lowbit(int x)
{
return x&(-x);
}
LL sum(int i)
{
LL ans = 0;
while (i)
{
ans +=c[i];
i-=lowbit(i);
}
return ans;
}
void add(int i,int d)
{
while (i<maxn)
{
c[i]+=d;
i+=lowbit(i);
}
}
int main()
{
int n;
while (scanf("%d",&n)!=EOF && n)
{
memset(L,0,sizeof(L));
memset(c,0,sizeof(c));
for (int i = 1;i<=n;i++)
{
scanf("%d",&a[i]);
L[i]=sum(maxn)-sum(a[i]);
add(a[i],1);
// printf("%d ",L[i]);
}
memset(c,0,sizeof(c));
LL ans = 0;
for (int i = 1;i<=n;i++)
{
LL temp = a[i]*L[i];
ans+= temp+sum(a[i]);
add(a[i],a[i]);
}
printf("%lld\n",ans);
}
//freopen("in","r",stdin);
//scanf("%d",&T);
//printf("time=%.3lf",(double)clock()/CLOCKS_PER_SEC);
return 0;
}
Description
Sherlock's N (1 ≤ N ≤ 100,000) cows are lined up to be milked in the evening. Each cow has a unique "grumpiness" level in the range 1...100,000. Since grumpy cows are more likely to damage Sherlock's milking equipment, Sherlock would like to reorder the cows in line so they are lined up in increasing order of grumpiness. During this process, the places of any two cows (necessarily adjacent) can be interchanged. Since grumpy cows are harder to move, it takes Sherlock a total of X + Y units of time to exchange two cows whose grumpiness levels are X and Y.
Please help Sherlock calculate the minimal time required to reorder the cows.
Input
Line 1: A single integer: N
Lines 2..N + 1: Each line contains a single integer: line i + 1 describes the grumpiness of cow i.
Output
Line 1: A single line with the minimal time required to reorder the cows in increasing order of grumpiness.
Sample Input
3 2 3 1
Sample Output
7
Hint
Input Details Three cows are standing in line with respective grumpiness levels 2, 3, and 1. Output Details 2 3 1 : Initial order. 2 1 3 : After interchanging cows with grumpiness 3 and 1 (time=1+3=4). 1 2 3 : After interchanging cows with grumpiness 1 and 2 (time=2+1=3).