Cyclic Nacklace HDU - 3746

Cyclic Nacklace HDU - 3746
CC always becomes very depressed at the end of this month, he has checked his credit card yesterday, without any surprise, there are only 99.9 yuan left. he is too distressed and thinking about how to tide over the last days. Being inspired by the entrepreneurial spirit of “HDU CakeMan”, he wants to sell some little things to make money. Of course, this is not an easy task.

As Christmas is around the corner, Boys are busy in choosing christmas presents to send to their girlfriends. It is believed that chain bracelet is a good choice. However, Things are not always so simple, as is known to everyone, girl’s fond of the colorful decoration to make bracelet appears vivid and lively, meanwhile they want to display their mature side as college students. after CC understands the girls demands, he intends to sell the chain bracelet called CharmBracelet. The CharmBracelet is made up with colorful pearls to show girls’ lively, and the most important thing is that it must be connected by a cyclic chain which means the color of pearls are cyclic connected from the left to right. And the cyclic count must be more than one. If you connect the leftmost pearl and the rightmost pearl of such chain, you can make a CharmBracelet. Just like the pictrue below, this CharmBracelet’s cycle is 9 and its cyclic count is 2:

Now CC has brought in some ordinary bracelet chains, he wants to buy minimum number of pearls to make CharmBracelets so that he can save more money. but when remaking the bracelet, he can only add color pearls to the left end and right end of the chain, that is to say, adding to the middle is forbidden.
CC is satisfied with his ideas and ask you for help.
Input
The first line of the input is a single integer T ( 0 < T <= 100 ) which means the number of test cases.
Each test case contains only one line describe the original ordinary chain to be remade. Each character in the string stands for one pearl and there are 26 kinds of pearls being described by ‘a’ ~‘z’ characters. The length of the string Len: ( 3 <= Len <= 100000 ).
Output
For each case, you are required to output the minimum count of pearls added to make a CharmBracelet.
Sample Input
3
aaa
abca
abcde
Sample Output
0
2
5

题目大意:

一个字符串,在串后面补全几个字符,使他的全部字符都是最少循环2次以上,求补几个字符
比如
aaa+0->aaa
abca+2->abcabc
abcde+5->abcdeabcde

解:

首先这里用到一个kmp算法的最长相等前后缀算法next[]

然后很重要的一点:补全后字符串循环体L为strlen(s)-next[n]

这里有三种可能性:
1,字符串里已经有两次以上循环,而且都已补全
这种就不用补全了,判断方式:
相等前后缀肯定不为零,
且总长n%L==0

2,字符串里已经有两次以上循环,后续循环未补全
这种要把单次循环的补全,
由于已经有两次以上的循环,那么前后缀肯定要比单次循环要长(懒得证明)
那么 前后缀%单次循环 就是循环剩下来的:
比如 ababa 前后缀是aba 单次循环是 ab 剩下来的就是a
那么我们要补全的就是 单词循环-剩下来的: b
ababa + b = ababab

3, 字符串里没有两次以上循环
这种要补全两次
由于没有两次以上的循环,那么前后缀肯定要比单次循环要短(同上)
那么我们就输出 单词循环-前后缀: l-next[n]
这种可以和第二种合并

代码

#include<iostream>
#include<stack>
#include<queue>
#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<cmath>
#define ll long long
#define oo 1000000007
#define MAXN 1000005
using namespace std;    
char s[MAXN],c[MAXN];
long long next[MAXN];
int main()
{ 
	int n,m,i,j,T;
	ll ans,num;
	cin>>T;
	while (T--)
	{
		ans=0;
		scanf("%s",&s);
		n=strlen(s);
		i=0,j=-1;
		next[0]=-1;
		while(i<n){
			if(j==-1||s[i]==s[j]){
				i++;j++;
				next[i]=j;
			}else{
				j=next[j];
			}
		}
		int l=n-next[n];
		if(next[n]!=0&&n%l==0)
			cout<<0<<endl;
		else
			cout<<l-next[n]%l<<endl;
	}
	return 0;
}