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.
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 ).
For each case, you are required to output the minimum
count of pearls added to make a CharmBracelet.
0
2
5
题目大意：
在字符串的左边或者右边添加最少的字符，使得字符串拥有循环节；
解题思路：
根据直觉，不难发现好像和Next[len]有一定关系，一开始我看了样例，单纯的以为答案就是len-2*Next[len]，十分天真
；
后来才发现，len-Next[len]就是最小循环节：
求出循环节之后，再做一点简单的数学计算就可以得到答案了。
（有一点是，如果循环节的组成是C[1…k]+C[1…m]，那是不是就要在左边加字符，或者要做一些特殊处理呢？显然不是，至于为什么，例如abcdabcdab和badcbadcba）
AC代码：
1 #include<cstdio>
2 #include<cstring>
3 #define MAX 100000+5
4 using namespace std;
5 int Next[MAX];
6 char pat[MAX];
7 int len,cycle;
8 void getnext()
9 {
10 int i=0, j=-1;
11 len=strlen(pat);
12 Next[0]=-1;
13 while(i<len)
14 {
15 if(j == -1 || pat[i] == pat[j]) Next[++i]=++j;
16 else j=Next[j];
17 }
18 }
19 int main()
20 {
21 int t;
22 scanf("%d",&t);
23 while(t--)
24 {
25 scanf("%s",pat);
26 getnext();
27 cycle=len-Next[len];
28 if(len%cycle==0)
29 {
30 if(len/cycle==1) printf("%d\n",cycle);
31 else printf("0\n");
32 }
33 else printf("%d\n",cycle-(len-len/cycle*cycle));
34 }
35 }
