Sort String

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我是靠谱客的博主虚幻书本，最近开发中收集的这篇文章主要介绍Sort String，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

链接：https://www.nowcoder.com/acm/contest/141/E

题目描述

Eddy likes to play with string which is a sequence of characters. One day, Eddy has played with a string S for a long time and wonders how could make it more enjoyable. Eddy comes up with following procedure:

1. For each i in [0,|S|-1], let Si be the substring of S starting from i-th character to the end followed by the substring of first i characters of S. Index of string starts from 0.
2. Group up all the Si. Si and Sj will be the same group if and only if Si=Sj.
3. For each group, let Lj be the list of index i in non-decreasing order of Si in this group.
4. Sort all the Lj by lexicographical order.

Eddy can't find any efficient way to compute the final result. As one of his best friend, you come to help him compute the answer!

输入描述:

Input contains only one line consisting of a string S.

1≤ |S|≤ 106
S only contains lowercase English letters(i.e. ).

输出描述:

First, output one line containing an integer K indicating the number of lists.
For each following K lines, output each list in lexicographical order.
For each list, output its length followed by the indexes in it separated by a single space.

示例1

输入

复制

abab

输出

复制

2
2 0 2
2 1 3

示例2

输入

复制

deadbeef

输出

复制

题意：

将第一个字母移到最后，看有几种排序

分析：

用kmp求出循环节的长度len，然后判断n是否整除len，

代码：

#include<bits/stdc++.h>
using namespace std;

const int maxn =1e6+10;
int next1[maxn];     /// 串的next1数组 （看这个串的相似度）
char s1[maxn];

void kmp_pre(char x[],int m,int next1[])
{
    int i,j;
    j=next1[0]=-1;
    i=0;
    while(i<m){
        while(-1!=j&&x[i]!=x[j])j=next1[j];
        next1[++i]=++j;
    }
}
int main()
{
        scanf("%s",s1);
        int len=strlen(s1);
        kmp_pre(s1,len,next1);
        int temp=len-next1[len];
        if(len%temp!=0)
        {
        temp=len;
        printf("%dn",temp);
        for(int i=0;i<temp;i++)
           {
             printf("%d %d",len/temp,i);
             for(int j=1;j<len/temp;j++)
                printf(" %d",i+j*temp);
             printf("n");
           }
        }
        else{
             printf("%dn",temp);
        for(int i=0;i<temp;i++)
           {
             printf("%d %d",len/temp,i);
             for(int j=1;j<len/temp;j++)
                printf(" %d",i+j*temp);
             printf("n");
           }
        }
    return 0;
}