hdoj 3746 Cyclic Nacklace 【kmp（len%(len-p[len])公式的灵活运用）】

Cyclic Nacklace

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4450    Accepted Submission(s): 2025

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

3aaaabcaabcde

 

Sample Output

025

 

Author

possessor WC

 

Source

HDU 3rd “Vegetable-Birds Cup” Programming Open Contest

 

思路：

 

          先用getp()函数来求p数组，然后用len%(len-p[len])==0判断是否是循环字符串，但是一定要记住在p[len]!=0的前提下，否则如果没有循环（例如：abc）也满足，如果是循环字符串，直接输出0！否则计算出来最小循环子节min=len-p[len],然后用min-p[len]%min就是所求的结果！

 

代码：

 

#include <stdio.h>#include <string.h>char a[100005];int p[100005];int len;void getp(){int i=0,j=-1;p[0]=-1;while(i<len){if(j==-1||a[i]==a[j]){i++;j++;p[i]=j;}elsej=p[j];}}int main(){int T;scanf("%d",&T);while(T--){scanf("%s",a);len=strlen(a);getp();if(p[len]&&len%(len-p[len])==0)//如果是两次及两次以上的循环p[len]!=0！ printf("0\n");//但是是一次的循环的话，它就会是0！ else//len%(len-p[len])==0说明这个一定是循环字符串！ {int min=len-p[len];//最小循环节 printf("%d\n",min-p[len]%min);//用最小循环节减去重复的字符串的长度！ }//因为有可能是两个以上的时候出现了不能循环的情况，所以要用p[len]%min! }//例如：ababa;min=2，p[len]=3,但是p[len]%min=1，min-p[len]%min=1是所需要的答案！ return 0;}