hdu 3374

Posted 2020-09-07

Problem Description

Give you a string with length N, you can generate N strings by left shifts. For example let consider the string “SKYLONG”, we can generate seven strings:
String Rank 
SKYLONG 1
KYLONGS 2
YLONGSK 3
LONGSKY 4
ONGSKYL 5
NGSKYLO 6
GSKYLON 7
and lexicographically first of them is GSKYLON, lexicographically last is YLONGSK, both of them appear only once.
  Your task is easy, calculate the lexicographically fisrt string’s Rank (if there are multiple answers, choose the smallest one), its times, lexicographically last string’s Rank (if there are multiple answers, choose the smallest one), and its times also.

 

 

Input

  Each line contains one line the string S with length N (N <= 1000000) formed by lower case letters.

 

 

Output

Output four integers separated by one space, lexicographically fisrt string’s Rank (if there are multiple answers, choose the smallest one), the string’s times in the N generated strings, lexicographically last string’s Rank (if there are multiple answers, choose the smallest one), and its times also.

 

 

Sample Input

abcder aaaaaa ababab

 

 

Sample Output

1 1 6 1 1 6 1 6 1 3 2 3

 

 

Author

WhereIsHeroFrom

 

 

Source

HDOJ Monthly Contest – 2010.04.04

 利用nxt数组很好推出循环次数，重点是最小最大表示法。

 1 #include <iostream>
 2 #include <cstdlib>
 3 #include <cstring>
 4 #include <cstdio>
 5 const int N = 1000000 + 11;
 6 using namespace std;
 7 char ss[N],s[N<<1];
 8 int cs,n,a[N<<1],nxt[N];
 9 
10 void pre()
11 {
12     nxt[1] = 0; int k = 0;
13     for(int i = 2; i <= n; ++i)
14     {
15         while(k > 0 && a[i] != a[k+1]) k = nxt[k];
16         if(a[i] == a[k+1]) ++k;
17         nxt[i] = k;
18     }
19 }
20 
21 void Init()
22 {
23     n = strlen(ss);
24     strcpy(s,ss);
25     strcat(s,ss);
26     for(int i = 0; i < ( n << 1 ); ++i) a[i+1] = s[i]; 
27     pre();
28 }
29 
30 int minx()
31 {
32     
33     int i = 1,j = 2,k = 0;
34     while(i <= n && j <= n  && k < n)
35     {
36         if(a[i+k] == a[j+k] ) ++k;
37         else if(a[i+k] < a[j+k]) j = j + k + 1, k = 0;
38         else i = i + k + 1, k = 0;
39         if( i == j ) ++j;
40     }
41     return i;
42 }
43  
44 void Solve()
45 {
46     cs = n - nxt[n]; 
47     if(n % cs == 0) cs = n/cs; else cs = 1; n = n/cs;
48     printf("%d %d ",minx(),cs);
49     for(int i = 1;i <= ( n << 1 ); ++i) a[i] = 0-a[i];
50     printf("%d %d",minx(),cs);
51 }
52 
53 
54 int main()
55 {
56     while(~scanf("%s",ss))
57     {
58         Init();
59         Solve();
60         puts("");
61     }
62     return 0;
63 }