HDU 3746 Cyclic Nacklace kmp求字符串的循环节长度

Cyclic Nacklace ，题目链接， click here。

Cyclic Nacklace
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6023 Accepted Submission(s): 2690

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

Author

possessor WC

Source

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

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lcy | We have carefully selected several similar problems for you: 3336 1358 1711 2222 3068

    题目意思：求一个字符串的循环节的长度。    思路：求一个next数组。next[len]代表的是最大匹配数目,那么len-next[len]就是最大匹配数目前的那段,即 循环节;    如果len%(len-next[len])==0，那么代表这个字符串有len/（len-next[len])个周期；    如果len%(len-next[len])!=0,循环节长度-(len%循环节长度)就是该字符串的循环节;    否则，该字符串就是单独的一个字符，循环节的长度就是0。

之前我用过的是一next为next数组的名字，但是一直CE，然后就用了nexts，就过了。好坑… …

#include<bits/stdc++.h>using namespace std;const int maxn = 100005;int nexts[maxn],len;char str[maxn];void getn()//得到next数组；{    int k = -1,i=0;    nexts[0]=  -1;    while(i<len)    {        if(str[k]==str[i]||k==-1)        {            i++;            k++;            nexts[i]=str[i]!=str[k]?k:nexts[k];//代码优化下；        }        else            k=nexts[k];    }}int main(){    int t;    scanf("%d",&t);    while(t--)    {        scanf("%s",str);        len=strlen(str);        getn();        int minn=len-nexts[len];//该字符串的循环节长度;        if(!nexts[len])//如果next[len]==0，证明循环节为0；            printf("%d\n",len);        else if(len%minn)// 如果周期不是一个整数；            printf("%d\n",minn-len%minn);        else            printf("0\n");    }    return 0;}