L - Cyclic Nacklace

野性酷女 2022-03-22 15:26 156阅读 0赞

### 题目描述： ###

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  
题意：构建一个完美的字符串。即：给出一个字符串，求在其尾部需要加几个字符使其构成循环节 。

##### 分析： #####

本题很容易想到只要找出给出字符串的最小循环节就可以了。  
那么关键是怎么找这个循环节。一开始，我依旧想到穷举的办法去了。但是很明显不对。 我们应该通过KMP的next数组来找到最小的循环节。  
例： abaabbab， l为字符串长度，由KMP中Next数组求得Next\[l\]=2，ans=l-Next\[l\]就是该串的最小循环节。看似是用最小循环节减去Next\[l\]就得需要的字符数了。  
参考博客：[https://blog.csdn.net/xiaoxiede\_wo/article/details/78423817][https_blog.csdn.net_xiaoxiede_wo_article_details_78423817]

##### 代码： #####

#include"stdio.h"
    #include"string.h"
    //得到next数组
    void Get_next(char text[],int next[])
    {
        int i,j,len;
        len=strlen(text);
        next[0]=-1;
        i=0;
        j=-1;
        while(i<len)
        {
            if(j==-1||text[i]==text[j])
            {
                i++;
                j++;
    
                next[i]=j;
            }
            else
                j=next[j];
        }
    
    }
    int main()
    {
        int N;
        char text[100001];
        int ans,len;
        int next[100001];
        while(~scanf("%d",&N))
        {
            while(N--)
            {
                scanf("%s",text);
                len=strlen(text);
                Get_next(text,next);
                ans=len-next[len];//求出最小循环节
    //如果最小循环节不等于len且又len%ans==0，就说明原字符串已是完美字符串了
                if(ans!=len&&len%ans==0)
                    printf("%d\n",0);
                else
                //否则，我们先看一下最小循环节外的字符能够成几个最小循环节
                //并且多了几个字符，用ans减去多的字符，可求出需要添加几个字符
              //next[len]就是最小循环节外的字符。因为ans=len-next[len]。
                    printf("%d\n",ans-next[len]%ans);
            }
        }
    }

[https_blog.csdn.net_xiaoxiede_wo_article_details_78423817]: https://blog.csdn.net/xiaoxiede_wo/article/details/78423817