地址：http://acm.hdu.edu.cn/showproblem.php?pid=3374
题目：
String Problem
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3029 Accepted Submission(s): 1230
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.
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
Recommend
lcy
思路：最大最小表示法
1 #include <bits/stdc++.h>
2
3 using namespace std;
4
5 #define MP make_pair
6 #define PB push_back
7 typedef long long LL;
8 typedef pair<int,int> PII;
9 const double eps=1e-8;
10 const double pi=acos(-1.0);
11 const int K=2e6+7;
12 const int mod=1e9+7;
13
14 int nt[K];
15 char sa[K];
16 void kmp_next(char *T,int *next)
17 {
18 next[0]=0;
19 for(int i=1,j=0,len=strlen(T);i<len;i++)
20 {
21 while(j&&T[i]!=T[j]) j=next[j-1];
22 if(T[i]==T[j]) j++;
23 next[i]=j;
24 }
25 }
26 int kmp(char *S,char *T,int *next)
27 {
28 int ans=0;
29 int ls=strlen(S),lt=strlen(T);
30 kmp_next(T,next);
31 for(int i=0,j=0;i<ls;i++)
32 {
33 while(j&&S[i]!=T[j]) j=next[j-1];
34 if(S[i]==T[j]) j++;
35 if(j==lt) ans++;
36 }
37 return ans;
38 }
39 //ff为真表示最小，为假表示最大
40 int mx_mi_express(char *S,bool ff,int len)
41 {
42 int i=0,j=1,k;
43 while(i<len&&j<len)
44 {
45 k=0;
46 while(k<len&&S[i+k]==S[j+k]) k++;
47 if(k==len) return i<=j?i:j;
48 if((ff&&S[i+k]>S[j+k]) || (!ff&&S[i+k]<S[j+k]))
49 {
50 if(i+k+1>j) i=i+k+1;
51 else i=j+1;
52 }
53 else if((ff&&S[i+k]<S[j+k]) || (!ff&&S[i+k]>S[j+k]))
54 {
55 if(j+k+1>i) j=j+k+1;
56 else j=i+1;
57 }
58 }
59 return i<=j?i:j;
60 }
61 int main(void)
62 {
63 int t,n;cin>>t;
64 while(scanf("%s",sa)==1)
65 {
66 kmp_next(sa,nt);
67 int len=strlen(sa);
68 int num=len-nt[len-1];
69 if(num!=len&&len%num==0)num=len/num;
70 else num=1;
71 for(int i=0;i<len;i++)
72 sa[i+len]=sa[i];
73 printf("%d %d %d %d
",1+mx_mi_express(sa,1,len),num,1+mx_mi_express(sa,0,len),num);
74 }
75 return 0;
76 }