HDU 3746 Cyclic Nacklace（KMP）

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

 【题解】 这是一道字符串CMP题，也是刚开始接触这个方面，代码可能不是很完美，谅解~~~~

  题意就是给你个字符串，现在要求这个串最少补多少个字符可以构成一个循环次数为n的串，n为>=2的任意整数。

  如果刚好就是个循环串，就输出0，（需要添加0个字符）

  否则，就求出next[len]，循环节长度就为len-next[len]。

 了解CMP请戳：blog.csdn.net/qq_38538733/article/details/75267945

 【代码】

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>using namespace std;int nex[100005];//不知道为什么我把数组命写成next，一提交就是CE，，所以就改成nex了，char s[100005];void getnext(int n){    int i=0,j=-1;    nex[0]=-1;    while(i<n)    {        if(j == -1||s[i] == s[j])        {            i++;            j++;            nex[i]=j;        }        else            j=nex[j];    }}int main(){    int t;    scanf("%d",&t);    getchar();    while(t--)    {        memset(nex,0,sizeof(nex));        scanf("%s",s);        int len=strlen(s);        getnext(len);        int ans=len-nex[len];        if(ans!=len && len%ans == 0)            printf("0\n");        else            printf("%d\n",ans-nex[len]%ans);    }    return 0;}