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 Q, 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.
In the first line of the standard input there is one integer $k$ ($1\le k\le 1\ 000\ 000$) - the length of the string. In the following line a sequence of exactly $k$ lower-case letters of the English alphabet is written - the string.
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.
8
babababa24对于一个仅含小写字母的字符串 $a$，$p$ 为 $a$ 的前缀且 $p\ne a$，那么我们称 $p$ 为 $a$ 的 $proper$ 前缀。
规定字符串 Q 表示 a 的周期，当且仅当 Q 是 a 的 proper 前缀且 a 是 Q+Q 的前缀。若这样的字符串不存在，则 a 的周期为空串。
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。
首先介绍一个叫$next$数组的东西
因为当匹配失败时，就会利用$next$数组往前跳，这样就能节省匹配次数
$next$简称$ne$.
对于一个字符串$s$ ，对应的 $ne_i$ 定义为 $s$ 长度为 $i$ 的前缀的最长公共前后缀的长度。
例如，给定一个 $s =$ abacabad.
| $i$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
|---|---|---|---|---|---|---|---|---|
| $s_i$ | a | b | a | c | a | b | a | d |
| $ne_i$ | 0 | 0 | 1 | 0 | 1 | 2 | 3 | 0 |
这段代码可以求出$ne$数组：
cpp
ne[1]=0;
for(int i=2,j=0;i<=n;i++){
while(j&&s[i]!=s[j+1])j=ne[j];
if(s[i]==s[j+1])j++;
ne[i]=j;
}
整个过程就是，如果能匹配，就一直往前进（j++）;否则，不断地利用之前的信息往回跳（j=ne[j]），直到可以匹配为止。
设 $Q_{max}$ 为字符串的最大周期。那么：
| $i$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
|---|---|---|---|---|---|---|---|---|
| $Q_{max}$ | a | b | a | b | ||||
| $s$ | a | b | a | b | a | b | ||
| $Q_{max}Q_{max}$ | a | b | a | b | a | b | a | b |
| 所以，对于一个字符串 $s$ , 它的 $|Q_{max}|$ 就是$|s|-$它最短公共前后缀的长度。 |
最短公共前后缀的长度不好求，但是我们发现可以通过不断j=ne[j]求得。
于是就有了这样一段代码，一趟循环求得本题答案：
cpp
for(int i=1;i<=n;i++){
now=i;
while(ne[now])now=ne[now];
if(ne[i]!=0)ne[i]=now;
ans+=i-now;
}
#include<bits/stdc++.h>
using namespace std;
const int N=1000086;
int n,ne[N],now;
char s[N];
long long ans;
int main(){
cin>>n>>s+1;
ne[1]=0;
for(int i=2,j=0;i<=n;i++){
while(j&&s[i]!=s[j+1])j=ne[j];
if(s[i]==s[j+1])j++;
ne[i]=j;
}
for(int i=1;i<=n;i++){
now=i;
while(ne[now])now=ne[now];
if(ne[i]!=0)ne[i]=now;
ans+=i-now;
}
cout<<ans;
return 0;
}P3435 [POI 2006] OKR-Periods of Words 题解
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