HDU 3374 String Problem

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

最小表示法求出位置，然后kmp求循环节。

表示之前完全不知道有最小表示法这样的东西，推荐看看百度文库里的ppt

差不多可以理解，最大表示法其实就是比较的时候反一下。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn = 1000001;
int n, nt[maxn], N;
char s[maxn];

void getnext()
{
  nt[0] = -1;
  for (int i = 0, j; s[i]; i++)
  {
    j = nt[i];
    while (j >= 0 && s[i] != s[j]) j = nt[j];
    nt[i + 1] = j + 1;
  }
}

int cmp(int x, char a, char b)
{
  if (a > b) return x - 0;
  if (a < b) return 1 - x;
  return 2;
}

int present(int z)
{
  int i, j, k, x, y;
  for (i = 0, j = 1, k = 0; k < n;)
  {
    if (i == j) { j++; k = 0; }
    x = (i + k) % n;  y = (j + k) % n;
    switch (cmp(z, s[x], s[y]))
    {
      case 0:i += k + 1; k = 0; break;
      case 1:j += k + 1; k = 0; break;
      case 2:k++; break;
    }
  }
  return min(i, j);
}

int main()
{
  while (~scanf("%s", s))
  {
    n = strlen(s);
    getnext();
    N = n / (n - nt[n]);
    printf("%d %d %d %d\n", present(0) + 1, N, present(1) + 1, N);
  }
  return 0;
}