HDU 3746 Cyclic Nacklace （非常有意义的题目）

Cyclic Nacklace

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5475 Accepted Submission(s): 2472

Problem Description
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

题目的意思很简单，就是CC喜欢循环的东西。现在给你一个字符串， 问你需要再加多少个字符串才可以补齐一个循环。

是不是很抽象？ 这里我们就要用到next数组了。

next 数组有人叫前缀数组，我觉得是 前后缀匹配数组

他可以吧前面的字符和后面的字符进行匹配，这样大大节省了时间。
看代码吧。

#include <stdio.h>#include <string.h>#include <algorithm>#include <iostream>#include <queue>using namespace std;int nexts[1000010];char s[1000010];void get_nexts(char s[],int nexts[]){    int i=0,k=-1;    nexts[0]=-1;    int l=strlen(s);    while(i<l)    {        if(k==-1||s[i]==s[k])                                                                                  i++,k++,nexts[i]=k;   //这里要这么写，不写成  nexts[i]=next[k]        else            k=nexts[k];    }}int main(){    int t;    scanf("%d",&t);    getchar();    while(t--)    {        scanf("%s",s);        getchar();        int len=strlen(s);        get_nexts(s,nexts);        int min_repeat=len-nexts[len];//next[len]为这个字符串整体的前后缀字符相匹配的情况。   那么len-next[len]就表明了是整个字符串最小循环节长为多少。        if(len!=min_repeat&&len%min_repeat==0)   //防止一个字符的情况，或者说已经循环完毕了。            printf("0\n");        else            printf("%d\n",min_repeat-len%min_repeat); //len%min_repeat 表明在最小循环节中 多了 多少个字符可以构成一个循环    }    return 0;}