题意：输入1-n的排列，每次可以交换两个整数。用最少的次数把排列变成1-n的环状序列。

思路：枚举+贪心。依次枚举环上所有的点，正序一遍，倒序一遍，然后贪心求的所需的最小步数（贪心策略是1与1号交换，2与2号交换……）。

code：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <sstream>
#include <string>
#include <vector>
#include <list>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <bitset>using namespace std;typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;const int INF=0x3fffffff;
const int inf=-INF;
const int N=1000000;
const int M=505;
const int mod=1000000007;
const double pi=acos(-1.0);#define cls(x,c) memset(x,c,sizeof(x))
#define cpy(x,a) memcpy(x,a,sizeof(a))
#define fr(i,s,n) for (int i=s;i<=n;i++)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lrt  rt<<1
#define rrt  rt<<1|1
#define middle int m=(r+l)>>1
#define lowbit(x) (x&-x)
#define pii pair<int,int>
#define mk make_pair
#define IN freopen("in.txt","r",stdin);
#define OUT freopen("out.txt","w",stdout);int n,vis[M],v[M*2];
int sol(int *p)
{cls(vis,0);int ct=0,k;fr (i,1,n){if (vis[i]) continue;ct++;k=i;while (!vis[k]) vis[k]=1,k=p[k];}return n-ct;
}
int  main()
{while (~scanf("%d",&n)&&n){fr (i,1,n) scanf("%d",&v[i]);int ans=INF;fr(i,1,2){fr (j,1,n) v[j+n]=v[j];fr(j,1,n) ans=min(ans,sol(v+j));reverse(v+1,v+1+n);}printf("%d\n",ans);}
}