HDU 3374 String Problem

String Problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1233    Accepted Submission(s): 550

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

 解题方法：先用最小表示法和最大表示法求出其开始位置，然后用KMP算法求出循环次数。

#include <stdio.h>
#include <iostream>
#include <string.h>
using namespace std;

char str[1000005];  
int _next[1000005];  

int Getmin(char str[])
{
    int nLen = strlen(str);
    int i = 0, j = 1, k;
    while(i < nLen && j < nLen)
    {
        for (k = 0; k < nLen; k++)
        {
            if (str[(i + k) % nLen] != str[(j + k) % nLen])
            {
                break;
            }
        }
        if (str[(i + k) % nLen] > str[(j + k) % nLen])
        {
            i += k + 1;
        }
        else
        {
            j += k + 1;
        }
        if (i == j)
        {
            j++;
        }
    }
    return min(i, j);
}

int Getmax(char str[])
{
    int nLen = strlen(str);
    int i = 0, j = 1, k;
    while(i < nLen && j < nLen)
    {
        for (k = 0; k < nLen; k++)
        {
            if (str[(i + k) % nLen] != str[(j + k) % nLen])
            {
                break;
            }
        }
        if (str[(i + k) % nLen] < str[(j + k) % nLen])
        {
            i += k + 1;
        }
        else
        {
            j += k + 1;
        }
        if (j ==i)
        {
            j++;
        }
    }
    return min(i, j);
}

void Getnext(char str[])
{
    int j = 0, k = -1;
    int nLen = strlen(str);
    _next[0] = -1;
    while(j < nLen)
    {
        if (k == -1 || str[k] == str[j])
        {
            j++;
            k++;
            _next[j] = k;
        }
        else
        {
            k = _next[k];
        }
    }
}

int main()
{
    while(scanf("%s", str) != EOF)
    {
        int n = Getmin(str);
        int n1 = Getmax(str);
        Getnext(str);
        int nLen = strlen(str);
        int nCount = 0;
        if (nLen % (nLen - _next[nLen]) == 0)
        {
            nCount = nLen / (nLen - _next[nLen]);
        }
        else
        {
            nCount = 1;
        }
        printf("%d %d %d %d\n", n + 1, nCount, n1 + 1, nCount);
    }
    return 0;
}

 

 

Author

WhereIsHeroFrom