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。。

  1 //STATUS:C++_AC_868MS_2220KB
  2 #include <functional>
  3 #include <algorithm>
  4 #include <iostream>
  5 //#include <ext/rope>
  6 #include <fstream>
  7 #include <sstream>
  8 #include <iomanip>
  9 #include <numeric>
 10 #include <cstring>
 11 #include <cassert>
 12 #include <cstdio>
 13 #include <string>
 14 #include <vector>
 15 #include <bitset>
 16 #include <queue>
 17 #include <stack>
 18 #include <cmath>
 19 #include <ctime>
 20 #include <list>
 21 #include <set>
 22 //#include <map>
 23 using namespace std;
 24 //#pragma comment(linker,"/STACK:102400000,102400000")
 25 //using namespace __gnu_cxx;
 26 //define
 27 #define pii pair<int,int>
 28 #define mem(a,b) memset(a,b,sizeof(a))
 29 #define lson l,mid,rt<<1
 30 #define rson mid+1,r,rt<<1|1
 31 #define PI acos(-1.0)
 32 //typedef
 33 typedef long long LL;
 34 typedef unsigned long long ULL;
 35 //const
 36 const int N=100010;
 37 const int INF=0x3f3f3f3f;
 38 const int MOD=1e9+7,STA=8000010;
 39 //const LL LNF=1LL<<60;
 40 const double EPS=1e-8;
 41 const double OO=1e15;
 42 const int dx[4]={-1,0,1,0};
 43 const int dy[4]={0,1,0,-1};
 44 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
 45 //Daily Use ...
 46 inline int sign(double x){return (x>EPS)-(x<-EPS);}
 47 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
 48 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
 49 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
 50 template<class T> inline T Min(T a,T b){return a<b?a:b;}
 51 template<class T> inline T Max(T a,T b){return a>b?a:b;}
 52 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
 53 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
 54 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
 55 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
 56 //End
 57 
 58 int num[N],temp[N];
 59 int n;
 60 int ans;
 61 
 62 void sort(int l,int r)
 63 {
 64     if(l==r)return;
 65     int i,j,k,mid=(l+r)>>1;
 66     sort(l,mid);
 67     sort(mid+1,r);
 68     for(i=k=l,j=mid+1;i<=mid && j<=r;){
 69         if(num[i]<num[j]){
 70             ans=(ans+j-mid-1)&1;
 71             temp[k++]=num[i++];
 72         }
 73         else temp[k++]=num[j++];
 74     }
 75     while(i<=mid){
 76         ans=(ans+j-mid-1)&1;
 77         temp[k++]=num[i++];
 78     }
 79     while(j<=r)
 80         temp[k++]=num[j++];
 81     for(i=l;i<=r;i++)
 82         num[i]=temp[i];
 83 }
 84 
 85 int main()
 86 {
 87  //   freopen("in.txt","r",stdin);
 88     int i,j,ok;
 89     while(~scanf("%d",&n))
 90     {
 91         ok=ans=0;
 92         for(i=0;i<n;i++)
 93             scanf("%d",&num[i]);
 94         sort(0,n-1);
 95         ok=ans;
 96         ans=0;
 97         for(i=0;i<n;i++)
 98             scanf("%d",&num[i]);
 99         sort(0,n-1);
100         ok^=ans;
101 
102         printf("%s
",ok?"Impossible":"Possible");
103     }
104     return 0;
105 }