BNUOJ-26474 Bread Sorting 逆序对

　　题目链接：http://www.bnuoj.com/bnuoj/problem_show.php?pid=26474

　　题意：给一个数列，可以对三个数操作：把最后一个数放到第一个，前两个数后移一位。问最后能否到达相应的目标序列。。

　　先考虑三个数A B C，变换后两种情况B C A和C A B，可以证得（列举3个数的大小情况，枚举证），这三个序列变换后的逆序对个数的奇偶性是相同的，而且只有这3个序列相同，所以A B C只能到达与之奇偶相同的序列，而且是全部能到达。那么多个数的情况也是一样的，就是多个3元组的扩展。因此如果变换后的逆序奇偶性相同，那么有解，否则无解。。

//STATUS:C++_AC_868MS_2220KB
#include <functional>
#include <algorithm>
#include <iostream>
//#include <ext/rope>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <cstring>
#include <cassert>
#include <cstdio>
#include <string>
#include <vector>
#include <bitset>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <list>
#include <set>
//#include <map>
using namespace std;
//#pragma comment(linker,"/STACK:102400000,102400000")
//using namespace __gnu_cxx;
//define
#define pii pair<int,int>
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define PI acos(-1.0)
//typedef
typedef long long LL;
typedef unsigned long long ULL;
//const
const int N=100010;
const int INF=0x3f3f3f3f;
const int MOD=1e9+7,STA=8000010;
//const LL LNF=1LL<<60;
const double EPS=1e-8;
const double OO=1e15;
const int dx[4]={-1,0,1,0};
const int dy[4]={0,1,0,-1};
const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
//Daily Use ...
inline int sign(double x){return (x>EPS)-(x<-EPS);}
template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
template<class T> inline T Min(T a,T b){return a<b?a:b;}
template<class T> inline T Max(T a,T b){return a>b?a:b;}
template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
//End

int num[N],temp[N];
int n;
int ans;

void sort(int l,int r)
{
    if(l==r)return;
    int i,j,k,mid=(l+r)>>1;
    sort(l,mid);
    sort(mid+1,r);
    for(i=k=l,j=mid+1;i<=mid && j<=r;){
        if(num[i]<num[j]){
            ans=(ans+j-mid-1)&1;
            temp[k++]=num[i++];
        }
        else temp[k++]=num[j++];
    }
    while(i<=mid){
        ans=(ans+j-mid-1)&1;
        temp[k++]=num[i++];
    }
    while(j<=r)
        temp[k++]=num[j++];
    for(i=l;i<=r;i++)
        num[i]=temp[i];
}

int main()
{
 //   freopen("in.txt","r",stdin);
    int i,j,ok;
    while(~scanf("%d",&n))
    {
        ok=ans=0;
        for(i=0;i<n;i++)
            scanf("%d",&num[i]);
        sort(0,n-1);
        ok=ans;
        ans=0;
        for(i=0;i<n;i++)
            scanf("%d",&num[i]);
        sort(0,n-1);
        ok^=ans;

        printf("%s
",ok?"Impossible":"Possible");
    }
    return 0;
}