String Problem

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2045 Accepted Submission(s): 891

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

输出：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。

刚开始一直没明白是怎么回事，终于明白了吧，用暴力写了，明明知道时间超限，但我还是交了一次，果断的ＴＬＥ，后来搜了搜，在看了许久后终于明白是怎么一回事，原来是用了两个指针，　用两个指针来动态比较，然后的然后就能求出最小字典串和最大字典串了，另外个数的问题其实只需要ＦｉｎｄＮｅｘｔ就可以求出来了，这个刚开始的时候是一直没想到，我只是想暴力求出来，然后发现自己忽略了这么好用的。

代码：

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

using namespace std;

const int maxn = 2000007;

int Next[maxn], sum;
char s1[maxn], s[maxn];

void FindNext(char S[])
{
    int i=0, j=-1;
    Next[0] = -1;

    int Slen = strlen(S);

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

int GetMin(char s[], int N)
{
    int i=0, j=1;

    while(i<N && j<N)
    {
        int k=0;
        while(s[i+k]==s[j+k] && k<N) k++;
        if(k==N)
            break;

        if(s[i+k]<s[j+k])
        {
            if(j+k>i)
                j += k+1;
            else
                j = i+1;
        }
        else
        {
            if(i+k>j)
                i += k+1;
            else
                i = j+1;
        }
    }

    return min(i, j);
}

int GetMax(char s[], int N)
{
    int i=0, j=1;

    while(i<N && j<N)
    {
        int k=0;
        while(s[i+k]==s[j+k] && k<N) k++;
        if(k==N)
            break;

        if(s[i+k]>s[j+k])
        {
            if(j+k>i)
                j += k+1;
            else
                j = i+1;
        }
        else
        {
            if(i+k>j)
                i += k+1;
            else
                i = j+1;
        }
    }

    return min(i, j);
}


int main()
{
    while(scanf("%s", s1)!=EOF)
    {
        int N = strlen(s1);

        strcpy(s, s1);
        strcat(s, s1);

        FindNext(s);
        int circle = N - Next[N], time=1;

        if(N%circle==0)
            time = N / circle;

        printf("%d %d %d %d
", GetMin(s, N)+1, time, GetMax(s, N)+1, time);
    }

    return 0;
}

View Code

勿忘初心

（KMP 最大表示最小表示）String Problem -- hdu-- 3374

String Problem

（KMP 最大表示最小表示）String Problem　-- hdu-- 3374