前缀

OKR-Periods of Words

题目描述
A string is a finite sequence of lower-case (non-capital) letters of the English alphabet. Particularly, it may be an empty sequence, i.e. a sequence of 0 letters. By A=BC we denotes that A is a string obtained by concatenation (joining by writing one immediately after another, i.e. without any space, etc.) of the strings B and C (in this order). A string P is a prefix of the string !, if there is a string B, that A=PB. In other words, prefixes of A are the initial fragments of A. In addition, if P!=A and P is not an empty string, we say, that P is a proper prefix of A.

A string Q is a period of A, if Q is a proper prefix of A and A is a prefix (not necessarily a proper one) of the string QQ. For example, the strings abab and ababab are both periods of the string abababa. The maximum period of a string A is the longest of its periods or the empty string, if A doesn’t have any period. For example, the maximum period of ababab is abab. The maximum period of abc is the empty string.

Task Write a programme that:

reads from the standard input the string’s length and the string itself,calculates the sum of lengths of maximum periods of all its prefixes,writes the result to the standard output.

对于一个仅含小写字母的字符串

规定字符串

例如 ab 是 abab 的一个周期，因为 ab 是 abab 的 proper 前缀，且 abab 是 ab+ab 的前缀。

求给定字符串所有前缀的最大周期长度之和。

输入格式
In the first line of the standard input there is one integer kk (1\le k\le 1\ 000\ 0001≤k≤1 000 000) - the length of the string. In the following line a sequence of exactly kk lower-case letters of the English alphabet is written - the string.

输出格式
In the first and only line of the standard output your programme should write an integer - the sum of lengths of maximum periods of all prefixes of the string given in the input.

输入输出样例
输入 #1复制

1	`8babababa`

输出 #1复制

题意：

对于给定串S的每个前缀i(记为ai)ai可以等于A
求ai的满足条件的Q
求所有的这些“前缀的前缀”Q的长度和L

Q满足：Q是ai的前缀，Q!=ai，Q+Q能够覆盖ai,且ai是Q+Q的前缀

这里的代码是学习的，原代码地址：https://www.luogu.com.cn/blog/dedicatus545/solution-p3435

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define ll long long
using namespace std;
char a[1000010];
int n,f[1000010];
int main(){
    scanf("%d",&n);//字符串长度
    scanf("%s",a); //给定的字符串
    int i,j;
    ll cnt=0;//记录最终答案
    f[0]=f[1]=0;
    j=0;
    for(i=1;i<n;i++)
    {//求解next
        while(j&&(a[i]!=a[j])) j=f[j];
        j+=(a[i]==a[j]);
        f[i+1]=j;
    }
    /*
    for(int i=0;i<n;i++)
    {
        cout<<f[i]<<" ";
    }
    cout<<endl;
    */
    for(i=1;i<=n;i++)
    {
          j=i;
          //cout<<i<<": "<<"fail["<<i<<"]="<<f[i]<<" ";
          while(f[j]) j=f[j];
          //cout<<"j="<<j<<";";
          if(f[i]!=0) f[i]=j;//记忆化
          //cout<<"fail["<<i<<"]="<<f[i]<<";";
          //cout<<i-j<<endl;

          cnt+=i-j;

    }
    //cout<<endl;
    printf("%lld",cnt);
}