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\n",ok?"Impossible":"Possible");

103     }

104     return 0;

105 }