hdu 3374 String Problem(kmp+最小表示法)

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

题意：已知字符串 s 现在要求出求 每个字符串的最小循环表示，最大循环表示（输出序号最小的）以及该字符串在其中的次数。

思路：最小表示法 输出序号 next[]数组输出 该字符串出现的次数(其实就是字符串的循环节)

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<string>
#include<vector>
#include<stack>
#include<bitset>
#include<cstdlib>
#include<cmath>
#include<set>
#include<list>
#include<deque>
#include<map>
#include<queue>
#define ll long long int
using namespace std;
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
int moth[]={,,,,,,,,,,,,};
int dir[][]={, ,, ,-, ,,-};
int dirs[][]={, ,, ,-, ,,-, -,- ,-, ,,- ,,};
const int inf=0x3f3f3f3f;
const ll mod=1e9+;
int nextt[+];
void getnext(string s){
    nextt[]=;
    int len=s.length();
    for(int i=,j=;i<=len;i++){
        while(j>&&s[i-]!=s[j]) j=nextt[j];
        if(s[i-]==s[j]) j++;
        nextt[i]=j;
    }
}
int get_min(string s){
    int len=s.length();
    int i=; int j=; int k=;
    while(i<len&&j<len&&k<len){
        int t=s[(i+k)%len]-s[(j+k)%len];
        if(!t) k++;
        else{
            if(t>) i+=(k+); //s[i]>s[j] i推进
            else j+=(k+);
            if(i==j) j++;
            k=;
        }
    }
    return i>j?j:i;
}
int getmax(string s){
    int len=s.length();
    int i=; int j=; int k=;
    while(i<len&&j<len&&k<len){
        int t=s[(i+k)%len]-s[(j+k)%len];
        if(!t) k++;
        else{
            if(t>) j+=(k+);
            else i+=(k+);
            if(i==j) j++;
            k=;
        }
    }
    return i>j?j:i;
}
int main(){
    ios::sync_with_stdio(false);
    string s;
    while(cin>>s){
        getnext(s);
        int len=s.length();
        int ans=(len%(len-nextt[len])==)?(len/(len-nextt[len])):; //有循环节ans为循环节个数 不然为1
        cout<<get_min(s)+<<" "<<ans<<" "<<getmax(s)+<<" "<<ans<<endl;
    }
    return ;
}