Cyclic Nacklace(kmp求循环节)

本文介绍了一种名为CyclicNacklace的算法,用于计算如何最少地添加彩色珠子将普通项链转换成拥有至少两个循环节的Charm Bracelet。挑战在于只能在链的两端添加,核心思路是利用KMP求循环节。通过实例解析,解决如何在有限预算下制作满足女孩需求的时尚手链。

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Cyclic Nacklace(kmp求循环节)

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

题意: 给一个字符串,问至少需要增加几个字符使修改后的字符串含有至少两个循环节。

思路: 设字符串的长度为l,则x=l-ne[l]为循环节的长度,补充多少个,就用x-l%x即可

AC代码:

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
const int N=1e5+100;
char a[N];
int ne[N],t,l;
void getnext()
{
	ne[0]=-1;
	int i=0,j=-1;
	while(i<l)
	{
		if(j==-1||a[i]==a[j])
		{
			i++;
			j++;
			ne[i]=j;
		}
		else
		j=ne[j];
	}
}
int main()
{
	scanf("%d",&t);
	while(t--)
	{
		scanf("%s",a);
		l=strlen(a);
		memset(ne,0,sizeof(ne));
		getnext();
		int x=l-ne[l];
		int ci=l/x;
		if(l%x==0&&ci>1)
		printf("0\n");
		else
		printf("%d\n",x-l%x);
	}
	return 0;
}
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