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題目描述

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 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.

一個串是有限個小寫字符的序列,特別的,一個空序列也可以是一個串. 一個串P是串A的前綴, 當且僅當存在串B, 使得 A = PB. 如果 P A 并且 P 不是一個空串,那么我們說 P 是A的一個proper前綴. 定義Q 是A的周期, 當且僅當Q是A的一個proper 前綴并且A是QQ的前綴(不一定要是proper前綴). 比如串 abab 和 ababab 都是串abababa的周期. 串A的最大周期就是它最長的一個周期或者是一個空串(當A沒有周期的時候), 比如說, ababab的最大周期是abab. 串abc的最大周期是空串. 給出一個串,求出它所有前綴的最大周期長度之和.。

輸入輸出格式

輸入格式：

?

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.

?

輸出格式：

?

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.

?

輸入輸出樣例

輸入樣例#1：?
8 babababa 輸出樣例#1：?
24 1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 const int maxn=1000000+5; 5 6 char a[maxn]; 7 int net[maxn]; 8 int n; 9 ll ans; 10 11 inline void fnet() 12 { 13 net[1]=0; 14 for(int i=2;i<=n;i++) 15 { 16 int now=i-1; 17 while(true) 18 { 19 now=net[now]; 20 if(a[now+1]==a[i]) 21 { 22 net[i]=now+1; 23 break; 24 } 25 if(now==0) 26 break; 27 } 28 } 29 return ; 30 } 31 32 33 int main() 34 { 35 scanf("%d",&n); 36 scanf("%s",a+1); 37 fnet(); 38 for(int i=1;i<=n;i++) 39 { 40 int now=i; 41 while(net[now]!=0) 42 { 43 now=net[now]; 44 } 45 if(net[i]!=0) 46 { 47 net[i]=now; 48 } 49 ans+=i-now; 50 } 51 printf("%lld\n",ans); 52 return 0; 53 }

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轉載于:https://www.cnblogs.com/Hammer-cwz-77/p/8546242.html

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