Cyclic Nacklace(KMP + 最小循环串)

Cyclic Nacklace
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10103 Accepted Submission(s): 4327

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

//昨天刚刚做了一个循环节的题目 今天又做了一道
这道题就是让我们在串的左边或者右边加一些珠子让他形成一个循环串
所以这里需要找到一个最小循环串 其长度就是len - nxt[len]
如果len%(len - nxt[len])==0 并且其len/(len - nxt[len])!=1则说明其本身已经是一个循环串了 不需要再加珠子 直接输出0
否则的话 就需要加 len - nxt[len] - (len - (len/(len - nxt[len])*(len - nxt[len])))个珠子 意思就是用总串长减掉循环的次数乘以循环节的长度 就是除了循环节剩下的珠子的个数 再用循环节的长度 减去剩下珠子的长度就是其最小添加的长度

#include<iostream>#include<string>#include<string.h>#include<stdio.h>using namespace std;int lenp,nxt[100007];char p[100007];void getNext(){    lenp = strlen(p);    int i =0,j =-1;    nxt[0] = -1;    while(i<lenp)    {        if(j==-1||p[i]==p[j])            nxt[++i] = ++j;        else            j = nxt[j];    }}int main(){    int ncase;    scanf("%d",&ncase);    while(ncase--)    {        scanf("%s",p);        getNext();        int ans = lenp - nxt[lenp];       // cout<<ans<<endl;        if(ans!=lenp && lenp % ans==0)            printf("%d\n",0);        else            printf("%d\n", ans - (lenp - (lenp /ans)*ans));    }    return 0;}