HDU 3746 KMP算法

http://acm.hdu.edu.cn/showproblem.php?pid=3746

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

题目大意： 给一个长度大于3的字符串， 你可以在其左边或右边补上任意个字符， 问使该字符串的循环节个数大于1的最少要补的字符数。

思路：依旧要用到KMP的next数组， 下面的博客有对KMP相关的介绍：

https://blog.youkuaiyun.com/xiji333/article/details/88614354

首先要明白在字符串左边补或者右边补都是一样的，我们以在右边补为例，设字符串长度为len， 当len%(len-next[len])==0时，该字符串已经完全循环； 否则需要补上(len-next[len])-len%(len-next[len])个字符。特判next[len]=0的情况， 此时若想满足题意， 则需要在原串右侧补上一个完全一样的字符串。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#define INF 0x3f3f3f3f
using namespace std;
typedef long long ll;

int Next[100005];
char s[100005];

void getnext()
{
    int len=strlen(s);
    int j=0,k=-1;
    Next[0]=-1;
    while(j<len)
    {
        if(k==-1||s[j]==s[k])
        {
            ++j,++k;
            Next[j]=k;
        }
        else
            k=Next[k];
    }
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%s",s);
        getnext();
        int len=strlen(s);
        int temp=len-Next[len];
        if(len%temp==0&&temp!=len)
            printf("%d\n",0);
        else
            printf("%d\n",temp-len%temp);
    }
    return 0;
}