学步园

大意：将无序的序列通过两两交换最少需要几次才能有序。

思路：可以证明即数列的逆序对数，于是归并排序求解。

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <string>
using namespace std;

const int MAXN = 500010;

int n;

typedef long long LL;

LL A[MAXN], T[MAXN];

void merge_sort(int x, int y, LL &ans)
{
	if(y-x > 1)
	{
		int m = x + (y-x)/2;
		int p = x, q = m, i = x;
		merge_sort(x, m, ans);
		merge_sort(m, y, ans);
		while(p < m || q < y)
		{
			if(q >= y || (p < m && A[p] < A[q])) T[i++] = A[p++];
			else
			{
				T[i++] = A[q++];
				ans += (m - p);
			}
		}
		for(i = x; i < y; i++) A[i] = T[i];
	}
}

int read_case()
{
	scanf("%d", &n);
	if(!n) return 0;
	for(int i = 0; i < n; i++) scanf("%lld", &A[i]);
	return 1;
}

void solve()
{
	LL ans = 0;
	merge_sort(0, n, ans);
	printf("%lld\n", ans);
}

int main()
{
	while(read_case())
	{
		solve();
	}
	return 0;
}