HDU 3374 String Proble

String Problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3183    Accepted Submission(s): 1295

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

求循环节，并求字典序最大最小出现的次数

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>
#include <set>
#include <vector>
#include <queue>
using namespace std;
typedef long long ll;
char a[1000010];
int p[1000010];
void getnext()
{
  int i=0,j=-1;
  p[0]=-1;
  while(a[i]!='')
  {
    if(j==-1 || a[i]==a[j])
      p[++i]=++j;
    else j=p[j];
  }
}
int check(bool flag)
{
  int i=0,j=1,k=0;
  while(i<strlen(a) && j<strlen(a) && k<strlen(a))
  {
    int t=a[(j+k)%strlen(a)]-a[(i+k)%strlen(a)];
    if(t==0) k++;
    else
    {
      if(flag)
      {
        if(t>0) j+=k+1;
        else i+=k+1;
      }
      else
      {
        if(t<0) j+=k+1;
        else i+=k+1;
      }
      if(i==j) j++;
      k=0;
    }
  }
  return min(i,j);
}
int main()
{
  while(scanf("%s",a)!=EOF)
  {
    int minn=check(true);
    int maxn=check(false);
    getnext();
    int ans=(strlen(a)%(strlen(a)-p[strlen(a)]))==0?strlen(a)/(strlen(a)-p[strlen(a)]):1;
    printf("%d %d %d %d
",minn+1,ans,maxn+1,ans);
  }
  return 0;
}