题目链接：https://vjudge.net/problem/HDU-3746
Cyclic Nacklace
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10985 Accepted Submission(s): 4692
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 ).
aaa
abca
abcde
2
5
题解：
求出最小循环节，然后求出最少需要补多少个字符使得其长度能被循环节整除。
注意：当循环节为自身时，需要再补全一份自己。因为题目要求了至少有两个循环节。
代码如下：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <sstream>
#include <algorithm>
using namespace std;
typedef long long LL;
const double eps = 1e-;
const int INF = 2e9;
const LL LNF = 9e18;
const int MOD = 1e9+;
const int MAXN = 1e6+; char x[MAXN];
int Next[MAXN]; void get_next(char x[], int m)
{
int i, j;
j = Next[] = -;
i = ;
while(i<m)
{
while(j!=- && x[i]!=x[j]) j = Next[j];
Next[++i] = ++j;
}
} int main()
{
int T;
scanf("%d", &T);
while(T--)
{
scanf("%s", x);
int len = strlen(x);
get_next(x, len);
int r = len-Next[len]; //找到最小循环节
if(len!=r && len%r==) //如果最小循环节不等于自己，且除得尽，则不需要补全
printf("0\n");
else //否则，补全使其循环
printf("%d\n", r-len%r);
}
}