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NC17244. E、Sort String

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:

abab

输出:

2
2 0 2
2 1 3

示例2

输入:

deadbeef

输出:

8
1 0
1 1
1 2
1 3
1 4
1 5
1 6
1 7

原站题解

上次编辑到这里,代码来自缓存 点击恢复默认模板

C++14(g++5.4) 解法, 执行用时: 226ms, 内存消耗: 9996K, 提交时间: 2018-07-27 10:06:56

#include<bits/stdc++.h>
char s[1000001];
int main(){
    scanf("%s",s);
    int len=strlen(s),x,i,j;

    for(x=1;x<=len/2;x++)
        if(len%x==0){
            bool ok=1;
            for(i=0;i<len&&ok;i++)
                if(s[i]!=s[i%x])ok=0;
            if(ok)break;
        }

    if(x>len/2)x=len;
    printf("%d\n",x);
    for(i=0;i<x;i++){
        printf("%d",len/x);
        for(j=0;j<len/x;j++)
            printf(" %d",i+j*x);
        puts("");
    }
    return 0;
}

C++11(clang++ 3.9) 解法, 执行用时: 260ms, 内存消耗: 15924K, 提交时间: 2020-02-28 14:09:17

#include<bits/stdc++.h>
const int maxn=1e6+10,mo=1e9+7;
char s[maxn];
int f[maxn],n,g;
int main()
{
	scanf("%s",s+1);
	n=strlen(s+1);
	f[0]=-1;
	for(int i=1,j=-1;i<=n;f[i++]=++j)
	while(j>=0&&s[j+1]!=s[i]) j=f[j];
	g=(n%(n-f[n])==0)?n-f[n]:n;
	printf("%d\n",g);
	for(int i=0;i<g;i++)
	{
		printf("%d ",n/g);
		for(int j=i;j<n;j+=g) printf("%d ",j);
		puts("");
	}
}

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