HDU - 3374:String Problem (最小表示法模板题)

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.OutputOutput 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

#include<bits/stdc++.h>
using namespace std;
const int maxn=1000010;
char a[maxn]; int Next[maxn],num,L;
void KMP()
{
    for(int i=2,j=0;i<=L;i++){
        while(j&&a[i]!=a[j+1]) j=Next[j];
        if(a[i]==a[j+1]) j++;
        Next[i]=j;
    }
    num=L%(L-Next[L])?1:L/(L-Next[L]);
}
int minexpress()
{
    int i=1,j=2,k=0;
    while(i<=L&&j<=L&&k<L){
        int ti=(i+k>L?i+k-L:i+k);
        int tj=j+k>L?j+k-L:j+k;
        if(a[ti]==a[tj]) k++;
        else {
            if(a[ti]<a[tj]) j+=k+1;
            else i+=k+1;
            if(i==j) j++;
            k=0;
        }
    }
    return min(i,j);
}
int maxexpress()
{
    int i=1,j=2,k=0;
    while(i<=L&&j<=L&&k<L){
        int ti=i+k>L?i+k-L:i+k;
        int tj=j+k>L?j+k-L:j+k;
        if(a[ti]==a[tj]) k++;
        else {
            if(a[ti]>a[tj]) j+=k+1;
            else i+=k+1;
            if(i==j) j++;
            k=0;
        }
    }
    return min(i,j);
}
int main()
{
    int T,Min,Max;
    while(~scanf("%s",a+1)){
        L=strlen(a+1);
        KMP();
        Min=minexpress();
        Max=maxexpress();
        printf("%d %d %d %d\n",Min,num,Max,num);
    }
    return 0;
}