HDU 3746 Cyclic Nacklace (next数组)

Cyclic Nacklace

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

Total Submission(s): 3200    Accepted Submission(s): 1452

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

 

Author

possessor WC

 

Source

HDU 3rd “Vegetable-Birds Cup” Programming Open Contest

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

题目大意：给一个字符串，问在后面最少添加多少字符能让其具有大于1的周期

题目分析：利用kmp的next数组的性质，我们先得到next数组，字符串长度为len，如果next[len] == 0说明这个字符串没有相等的前后缀，因此直接加上一个串长才能得到一个为2的周期，否则令tmp = len - next[len]，这里tmp求得的就是一个周期的串长，如果len % tmp == 0说明总串长是一个周期串长的整数倍，则原串必已具有周期，例如abcabc对于这个串tmp ＝ len - next[len] = 6 - 3 = 3，len % tmp == 0，显然我们能看出来abc就是其一个周期，就不用再加任何字符了，若len % tmp != 0，则我们直接用tmp - len % tmp就是要添加的数量，相当于加上第二个周期内不足一个周期串长的长度

#include <cstdio>#include <cstring>int const MAX = 1e5 + 5;int next[MAX];char s[MAX];void get_next(){    int i = 0, j = -1;    next[0] = -1;    while(s[i] != '\0')    {        if(j == -1 || s[i] == s[j])        {            i++;            j++;            next[i] = j;        }        else            j = next[j];    }}int main(){    int T;    scanf("%d", &T);    while(T--)    {        scanf("%s", s);        int len = strlen(s);        get_next();        if(next[len] == 0)            printf("%d\n", len);        else        {            int tmp = len - next[len];            if(len % tmp == 0)                printf("0\n");            else                printf("%d\n", tmp - len % tmp);        }    }}