HDOJ 3374 String Problem

String Problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1990    Accepted Submission(s): 872

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

abcderaaaaaaababab

 

Sample Output

1 1 6 11 6 1 61 3 2 3

 

Author

WhereIsHeroFrom

 

Source

HDOJ Monthly Contest – 2010.04.04

 

#include <cstdio>#include <cstring>const int N = 1000010;char str[N];int next[N];int len;void get_next() {int i = 0, j = -1;next[i] = j;while (str[i]) {if (j == -1 || str[i] == str[j])next[++i] = ++j;elsej = next[j];}}int get_min(bool flag) {int i, j, k;i = k = 0, j = 1;while (i < len && j < len && k < len) {int t = str[(i + k) % len] - str[(j + k) % len];if (t == 0)++k;else {if (flag) {if (t < 0) i += k + 1;else j += k + 1;}else {if (t < 0) j += k + 1;else i += k + 1;}if (i == j)++j;k = 0;}}return i < j ? i + 1 : j + 1;}int main() {while (~scanf("%s", str)) {len = strlen(str);int x = get_min(0);int y = get_min(1);get_next();int t = len - next[len];if (len % t == 0)t = len / t;elset = 1;printf("%d %d %d %d\n", x, t, y, t);}return 0;}