HDU 3746 Cyclic Nacklace(KMP)

Cyclic Nacklace

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1729    Accepted Submission(s): 755


Problem DescriptionCC 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. 

InputThe 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
). 

OutputFor each case, you are required to output the minimum count of pearls added to make a CharmBracelet. 

Sample Input

3aaaabcaabcde

 

Sample Output

025

 

Authorpossessor WC 

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

Recommendlcy 題意：給定一個字串，要求在兩端添加字母使之成為一個二迴圈字串，要求求出最少添加的字母個數這個題目應該不難想到用KMP去做，這樣只要求最後一個字母的next值就可以了！求完next值後關鍵是這段代碼POS=n-next[n];
if(POS==n)//相當於next[n]==0的情況
printf("%d\n",n);
else if(n%POS==0) printf("0\n");//已經迴圈，POS一定小於n/2這樣才有餘數為0的可能
else
printf("%d\n",POS-n%POS);/*前面next[n]個在迴圈裡面，那麼POS個一定有next[n]在迴圈裡面，減去！*/

#include <iostream>#include <string.h>#include <stdio.h>using namespace std;char str[200001];int next[200001];int n;int get_next(){int i=0,j=-1;next[0]=-1;while(str[i]){if(j==-1 || str[i]==str[j]){i++;j++;next[i]=j;}elsej=next[j];}return 0;}int main(){int t,i,j;int MAX,POS;scanf("%d",&t);while(t--){MAX=-2;POS=0;scanf("%s",str);n=strlen(str);if(n==1){printf("1\n");continue;}get_next();POS=n-next[n];if(POS==n)printf("%d\n",n);else if(n%POS==0) printf("0\n");elseprintf("%d\n",POS-n%POS);}return 0;}