hdu 3374 String Problem （最大最小表示法）

http://acm.hdu.edu.cn/showproblem.php?pid=3374

SKYLONG 1
KYLONGS 2
YLONGSK 3
LONGSKY 4
ONGSKYL 5
NGSKYLO 6
GSKYLON 7

题意：输出的是最小字典序的编号，最小字典序个数，最大字典序编号，最大字典序个数

求有几个这样的字符串直接Kmp求得next数组就ok了，

然后用最小表示法求出字典序最小的，用最大表示法求出字典序最大的

最大最小表示法详解

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>

using namespace std;

#define N 1100000
#define met(a, b) memset(a, b, sizeof(a))

const double PI = acos(-1.0);

typedef long long LL;

int Next
;
char str
, s
;

int Min_Repre ()///最小表示法
{
int i = 0, j = 1, k = 0;
met (str, 0);

int len = strlen (s);

strcpy (str, s);
strcat (str, s);

while (i<len && j<len && k<len)
{
int t = str[ (i+k)%len ] - str[ (j+k)%len ];

if (!t) k++;
else
{
if (t>0) i = i+k+1;
else j = j+k+1;

if (i==j) j++;
k = 0;
}
}
return min (i, j);
}

int Max_Repre ()///最大表示法
{
int i = 0, j = 1, k = 0;
met (str, 0);

int len = strlen (s);
strcpy (str, s);
strcat (str, s);

while (i<len && j<len && k<len)
{
int t = str[ (i+k)%len ] - str[ (j+k)%len ];

if (!t) k++;

else
{
if (t<0) i = i+k+1;
else j = j+k+1;

if (i==j) j++;
k = 0;
}
}
return min (i, j);
}

int Get_Next ()
{
int i = 0, j = -1;
Next[i] = j;

int len = strlen (s);

while (i<len)
{
if (j==-1 || s[i] == s[j])
Next[++i] = ++j;
else j = Next[j];
}

int l = len - Next[len];
if (len % l) return 1;
return len/l;
}

int main ()
{
while (scanf ("%s", s) != EOF)
{
met (Next, 0);

int ans = Get_Next ();
int a1 = Min_Repre ();
int a2 = Max_Repre ();

printf ("%d %d %d %d\n", a1+1, ans, a2+1, ans);
}
return 0;
}