HDU 3476 Cyclic Nacklice

Cyclic Nacklace

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4037    Accepted Submission(s): 1818

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

这是一道KMP next数组应用的题。题意是给一个字符串，判断它的循环节。如果刚好循环完输出0，否则就输出还缺几个。比如abcabcabc，输出0。abcdeab，还差cde这3个字母，输出3。

题目传送门：HDU3476 Cyclic Nacklice

思路是对给出的字符串求next数组。设输入的字符串长度为m，那么循环节的长度len其实就是m-next[m]。如果字符串长度m刚好能整除循环节的长度len，那么这个字符串就是刚好循环了。值得注意的是，如果输入abcde，那么循环节就是3，我们需要再补一个abcde，也就是输出5。但是这样输入的话，m同样能整除len。这里我们就要加一个判断。如果m%len==0 同时，循环节长度len<字符串长度m，这时候我们就可以输出0。否则，则使用通式。下面给出AC代码：

#include <cstdio>#include <iostream>#include <cstring>using namespace std;#define N 100005char a[N];int nextt[N];int m;void getNext(){    int i,j;    i=0;    j=-1;    nextt[i]=j;    while(i<m)    {        if(j==-1 || a[i]==a[j])        {            i++;            j++;            nextt[i]=j;        }        else            j=nextt[j];    }    return ;}int main(){    int T,len,ans;    scanf("%d",&T);    getchar();    while(T--)    {        gets(a);        m=strlen(a);        memset(nextt,0,sizeof(nextt));        getNext();        len=m-nextt[m];        if(m%len==0 && m!=len)  //类似aaa或ababab的情况。即已经完美循环，循环节不等于总长度，不需要添加            ans=0;        else            ans=len-nextt[m]%len;        printf("%d\n",ans);    }    return 0;}