本文主要是介绍hdu 3746 Cyclic Nacklace(KMP 最短循环节),希望对大家解决编程问题提供一定的参考价值,需要的开发者们随着小编来一起学习吧!
1、http://acm.hdu.edu.cn/showproblem.php?pid=3746
2、题目大意:
给定一个字符串T, 在T后面添加x个字符串(让x最小),使得新字符串由前缀字串至少循环两次构成的。
例如,
abca, 只需要再添加2个字母bc, 形成abcabc,就变成了由abc循环两次构成的。
对于长度为len的字符串,假设已经够造完了next数组,那么len-next[len]就是这个字符串的最小循环节。
如果正好len%(len-next[len])==0就说明正好组成完成的循环。
否则,说明还需要再添加几个字母才能补全。
需要补的个数是循环个数len-next[len]-f[len]%(len-next[len]).
f[len]%(len-next[len])表示在最后一个循环节中已经构造了这么多个数。
3、题目:
Cyclic Nacklace
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1598 Accepted Submission(s): 690
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.
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 ).
3 aaa abca abcde
0 2 5
#include<stdio.h>
#include<string.h>
char str[100005];
int f[100005];
void getFail(char *p, int *f)
{int m=strlen(p);f[0]=f[1]=0;for(int i=1; i<m; ++i){int j=f[i];while(j && p[i]!=p[j])j=f[j];f[i+1] = p[i]==p[j]?1+j:0;}
}
int main()
{int t;scanf("%d",&t);while(t--){//memset(str,0,sizeof(str));scanf("%s",str);getFail(str,f);int len=strlen(str);if(f[len]&&len%(len-f[len])==0)printf("0\n");else{printf("%d\n",(len-f[len])-f[len]%(len-f[len]));}}return 0;
}
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