Cyclic Nacklace
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2229 Accepted Submission(s): 986
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:
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 求最少在结尾补几个字符才能使得此串形成循环。 要用非优化的kmp来处理。
//140MS 736K #include#include char pattern[100007]; int next[100007]; void pre(int len) { int i = 0, j = -1; next[0] = -1; while(i != len) { if(j == -1 || pattern[i] == pattern[j]) next[++i] = ++j; else j = next[j]; } } int main() { int t; scanf("%d",&t); while(t--) { scanf("%s",pattern); int m=strlen(pattern); pre(m); int n=m-next[m];//代表循环节的长度 if(n!=m&&m%n==0)printf("0\n");//如果可以多次循环 else printf("%d\n",n-next[m]%n);//取余的作用：abcab，去掉abc } return 0; }