什麽 tid ges 關系 字符串 cstring ice 解題思路 lin

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

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 題目大意： 在字符串的左邊或者右邊添加最少的字符，使得字符串擁有循環節； 解題思路： 根據直覺，不難發現好像和Next[len]有一定關系，一開始我看了樣例，單純的以為答案就是len-2*Next[len]，十分天真； 後來才發現，len-Next[len]就是最小循環節： 　　 求出循環節之後，再做一點簡單的數學計算就可以得到答案了。 （有一點是，如果循環節的組成是C[1…k]+C[1…m]，那是不是就要在左邊加字符，或者要做一些特殊處理呢？顯然不是，至於為什麽，例如abcdabcdab和badcbadcba） AC代碼：

 1 #include<cstdio>
 2 #include<cstring>
 3 #define MAX 100000+5
 4 using namespace std;
 5 int Next[MAX];
 6 char pat[MAX];
 7 int
 
 len,cycle;
 8 void getnext()
 9 {
10     int i=0, j=-1;
11     len=strlen(pat);
12     Next[0]=-1;
13     while(i<len)
14     {
15         if(j == -1 || pat[i] == pat[j]) Next[++i]=++j;
16         else j=Next[j];
17     }
18 }
19 int main()
20 {
21     int t;
22     scanf("%d",&t);
23     while(t--)
24     {
25         scanf("%s",pat);
26         getnext();
27         cycle=len-Next[len];
28         if(len%cycle==0)
29         {
30             if(len/cycle==1) printf("%d\n",cycle);
31             else printf("0\n");
32         }
33         else printf("%d\n",cycle-(len-len/cycle*cycle));
34     }
35 }

HDU 3746 - Cyclic Nacklace - [KMP求最小循環節]