HUD：3746-Cyclic Nacklace（補齊循環節）

Cyclic Nacklace

Time Limit: 2000/1000 MS (Java/Others)
Memory Limit: 32768/32768 K (Java/Others)

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

---

解題心得:

1. 題目說了一大堆，其實就是問你要在這個字符串的後面添加多少個字符纔可以讓字符中的每個循環節都完整。
2. 先找出循環節，而後檢查須要補多少個字符讓每個循環節都完整，若是沒有循環節，那麼就直接在這個字符串後面添加一個相同的字符串。

---

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
#include<vector>
#include<stack>
#include<string>
using namespace std;
const int maxn = 1e5+100;
char s[maxn];
int Next[maxn];

int cal_next(){
    int k = -1;
    Next[0] = -1;
    int len = strlen(s);
    for(int i=1;i<len;i++) {
        while (k > -1 && s[i] != s[k + 1])
            k = Next[k];
        if (s[i] == s[k + 1])
            k++;
        Next[i] = k;
    }
    int cir = len-1-Next[len-1];//循環節的長度
    if(Next[len-1] == -1)//若是整個個字符串都沒循環節就直接添加一個相同的字符串
        return len;
    if(len%cir == 0)//已是完整的循環節
        return 0;
    return cir-len%cir;//須要補齊的字符個數
}

int main(){
    int t;
    scanf("%d",&t);
    while (t--){
        scanf("%s",s);
        int ans = cal_next();
        printf("%d\n",ans);
    }
    return 0;
}