【hash】Three friends ＢＺＯＪ3916: [Baltic2014]friends

 

[Source]: bzoj3916

Reference [blog]

ＢＺＯＪ3916: [Baltic2014]friends

[Hash and hash table] Three Friends

【Baltic2014】【BZOJ3916】friends

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 【answer】

First, the entire hash string, and then divided into three cases, namely, the first half, the middle, second half, trying to remove the three sections of the letters to see whether it together.

If you can, you also need to consider whether it is unique.

The only means: a string S, if there are selected two different string S can constitute the original string.

Code or reference written by someone else.

 

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 using namespace std;
 5 
 6 const int N = 2e6+10;
 7 typedef unsigned long long ull ;
 8 const ull base = 13313;
 9 
10 ull Pow[N],Hash[N];
11 ull Lstr , Rstr , S;
12 char str[N],Ans[N];
13 int n,Mid,pos=-1;
14 
15 //POSSIBLE = Not Odd or the NOT found S A
 16  // the NOT UNIQUE TWO S = Least exist
 . 17  // Else of The answer S =
 18 is  
. 19  // cutting string return hash value 
20 is ULL Cut_str ( int L, int R & lt) {
 21 is      ULL = the Hash ANS [R & lt] - the Hash [L- . 1 ] * Pow [L-R & lt + . 1 ];
 22 is      return ANS;
 23 is  }
 24  
25  BOOL Check ( int I) {
 26 is      // if they are not labeled, the tag returns true
 27      // If labeled, while the string and the string is not the same temporary false if 
28      IF(POS == - . 1 || S == Lstr) {
 29          POS = I;
 30          S = Lstr;
 31 is          return  to true ;
 32      } the else {
 33 is          return  to false ;
 34 is      }
 35  }
 36  int main ()
 37 [  {
 38 is      Scanf ( " % D% S " , & n-, + STR . 1 );
 39      Mid = n->> . 1 ;
 40  
41 is      // case where an even number not constituting 
42 is      IF(N-% 2 == 0 )     return the puts ( " the NOT POSSIBLE " ), 0 ;
 43 is  
44 is      Pow [ 0 ] = . 1 ;
 45      for ( int I = . 1 ; I <= n-; I ++ ) {
 46 is          Pow [I] = POW [I- . 1 ] * Base ;
 47          the Hash [I] = the Hash [I- . 1 ] * Base + (ULL) (STR [I] - ' A ' + . 1 );
 48          // all strings that contain only uppercase English letter 
49     }
50 
51     // delete one character in [1,Mid]
52     for(int i=1 ; i<=Mid ; i++ ){
53         Lstr = Cut_str(1,i-1) * Pow[Mid+1-i] + Cut_str(i+1,Mid+1);
54         Rstr = Cut_str(Mid+2,n);
55         if( Lstr == Rstr ) if( !check(i) ) return puts("NOT UNIQUE"),0;
56     }
57 
58     // delete Mid + 1
59     Lstr = Hash[Mid] ;
60     Rstr = Hash[n] - Hash[Mid+1] * Pow[Mid];
61     if( Lstr == Rstr ) if( !check(Mid+1) ) return puts("NOT UNIQUE"),0;
62 
63     // delete [Mid+2,n]
64 
65     for(int i=Mid+2 ; i<=n; i++ ){
66         Lstr = Hash[Mid];
67         Rstr = Cut_str(Mid+1,i-1) * Pow[n-i] + Cut_str(i+1,n);
68         if( Lstr == Rstr ) if( !check(i) ) return puts("NOT UNIQUE"),0;
69     }
70 
71     if( pos == -1 ){
72         return puts("NOT POSSIBLE"),0;
73     }
74 
75     int len = n/2,cnt = 0 ;
76     for(int i=1; i<=n && len ; i++ ){
77         if( pos == i ) continue;
78         Ans[cnt++] = str[i] ;
79         len -- ;
80     }
81     Ans[cnt] = '\0';
82     printf("%s\n",Ans);
83     return 0 ;
84 }
85 
86 /*
87 
88 7
89 ABXCABC
90 
91 */

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