hdu 3374 String Problem-优快云博客

本文链接：https://blog.youkuaiyun.com/zhang__liuchen/article/details/78571220

本文介绍了一种快速寻找字符串中最小值和最大值的方法，并通过实例演示如何计算一个字符串经过左移产生的N个字符串中的字典序最前与最后的字符串排名及其出现次数。

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String Problem

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3637 Accepted Submission(s): 1500

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

在我看来，找循环节，现在很简单了，这个题的难点应该是快速找字符串中的最小最大值，这部分要求快，要不然就会超时。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N=1e6;
char str[N];
int Next[N];
void getNext(char *s,int len)
{
    Next[0]=-1;
    int i=0,j=-1;
    while(i<len)
    {
        if(j==-1||s[i]==s[j])
        {
            i++;
            j++;
            Next[i]=j;
        }
        else
            j=Next[j];
    }
}
int mi_ma(char *s,int len,bool flag)
{
    int i=0,j=1,k=0;
    while(i<len&&j<len&&k<len)
    {
        int t=s[(j+k)%len]-s[(i+k)%len];
        if(t==0)
            k++;
        else
        {
            if(flag)
            {
                if(t>0)
                    j+=k+1;
                else
                    i+=k+1;
            }
            else
            {
                if(t>0)
                    i+=k+1;
                else
                    j+=k+1;
            }
            if(i==j)
                j++;
            k=0;
        }
    }
    return min(i,j);
}
int main()
{
    while(~scanf("%s",str))
    {
        int len=strlen(str);
        // for(int i=0;i<=len;i++)getNext(str,)
        //     cout<<Next[i]<<" ";
        // cout<<endl;
        int s=mi_ma(str,len,true);
        int e=mi_ma(str,len,false);
        getNext(str,len);
        int w=len/(len-Next[len]);//循环节
        printf("%d %d %d %d\n",s+1,w,e+1,w);
    }
    return 0;
}

快速找字符串中的最小最大值

参考博客

http://blog.youkuaiyun.com/acm_cxlove/article/details/7909087