题意：给定一个1-n的排列，用不超过2*n2的操作把他变成升序。每次操作只有两种，一种是交换前两个元素，另外一种是把最后一个元素放到最后一位。

思路：贪心。用双端队列来保存数据，每次当v[0]>v[1]&&v[0]!=n时用交换前两个的策略把当前排好，否则就放后面等待被排。

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=180005;
const int M=2005;
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);deque<int>v;
int s[N],len;int check()
{fr (i,1,v.size()-1)if (v[i]<v[i-1]) return 0;return 1;
}
int main()
{int T,n,x;//scanf("%d",&T);while (~scanf("%d",&n)&&n){v.clear();len=0;fr (i,1,n) scanf("%d",&x),v.push_back(x);while (!check()){if (v[0]>v[1]&&v[0]!=n){swap(v[0],v[1]);s[len++]=1;//s='1'+s;//cout<<"bug1"<<endl;}else{int t=v.back();v.pop_back();//cout<<t<<endl;v.push_front(t);s[len++]=2;//s='2'+s;}}for (int i=len-1;i>=0;i--) printf("%d",s[i]);puts("");//cout<<s<<endl;}
}