HDU3746 Cyclic Nacklace(KMP,最小循环节)

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:

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

3aaaabcaabcde

Sample Output

025

思路

先说题意:给你一个字符串，要求将字符串的全部字符最少循环2次需要添加的字符数。
kmp找循环节问题，关于证明，请看：KMP模板，最小循环节

定理是：
如果对于next数组中的 i， 符合
i % ( i - next[i] ) == 0 && next[i] != 0 ,
则说明字符串循环，而且
循环节长度为: i - next[i]
循环次数为: i / ( i - next[i] )
否则的话不循环

代码

#include<cstdio>#include<cstring>#include<string>#include<set>#include<iostream>#include<stack>#include<queue>#include<vector>#include<algorithm>#define mem(a,b) memset(a,b,sizeof(a))#define inf 0x3f3f3f3f#define mod 10000007#define debug() puts("what the fuck!!!")#define N 100000+20#define ll longlongusing namespace std;char s[N];int nxt[N];void get_next(int len){    int j=0,k=-1;    nxt[0]=-1;    while(j<len)        if(k==-1||s[j]==s[k])            nxt[++j]=++k;        else            k=nxt[k];}int main(){    int t;    scanf("%d",&t);    while(t--)    {        scanf("%s",s);        int len=strlen(s);        get_next(len);        int n=len-nxt[len];        if(n!=len&&len%n==0)            puts("0");        else            printf("%d\n",n-nxt[len]%n);    }    return 0;}