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
给你一个字符串让你求最少添加几个字符可以使他变成一个有子串的字符串比如给你abcde你需要变成abcdeabcde才能使他有子串也就是加5个字符。
思路
通对对next数组的理解可以知道里面存储的是最大前缀后缀公共子串的长度那么我们要求需要加多少个字符可以使他变成有循环串，首先先找到他的最小循环串是多长len=（字符串长度-next【字符串长】） next数组所存储的最后一位是最后一个匹配的位置不信你可以动笔在纸上写一下，现在找到了最小循环节我们来判断他本身是不是循环串首先他的循环节长度不能等于自身串长度，并且自己串长度要是循环节长度的倍数len！=自身串长&&自身串长%len==0这样我们直接输出0，如果不是我们就用len-最后出现不循环的长度（最后出现不循环的长度=next【字符串长】%len）输出即可
代码
#include <iostream>
#include <cstdio>
#include <fstream>
#include <algorithm>
#include <cmath>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <cstring>
#include <map>
#include <stack>
#include <set>
#include<algorithm>
#include<stdio.h>
#define N 10000+5
#define inf 0x3f3f3f3f
using namespace std;
typedef long long ll;
int n, m;
int nex[1000005];
char s1[1000005], s2[1000005];
int len2;
void getnex()
{
nex[0] = { -1 };
int k = -1;
int j = 0;
while (j < len2)
{
if (k == -1 || s2[j] == s2[k])
{
++k;
++j;
nex[j] = k;
}
else
{
k = nex[k];
}
}
}
int main()
{
int T;
scanf("%d", &T);
while (T--)
{
memset(nex, -1, sizeof(nex));
scanf("%s", s2);
len2 = strlen(s2);
getnex();//获取next数组
int len = len2 - nex[len2];//最小循环节
if (len != len2 &&len2%len==0)//是循环串
{
puts("0"); continue;
}
else
{
printf("%d\n", len - (nex[len2]%len));//最少加几个字符
}
}
return 0;
}
来源：https://blog.csdn.net/qq_43619680/article/details/98850693