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 }
