HDU-3746-Cyclic Nacklace【KMP求循环节】【经典题】

HDU-3746-Cyclic Nacklace

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
3
aaa
abca
abcde

Sample Output
0
2
5

题目链接：HDU-3746

题目大意：给出一个字符串，可以在这个字符串后面添加字母，将这个字符串变成大于或等于两个的循环，问最少要添加几个字母

题目思路：很经典的一道KMP求循环节的题目，但今天我还是第一次接触

关键代码： 结论： 最小循环节 c = len-next[len - 1]

void getnext()
{
    int i,j;
    Next[0] = 0;
    for(i = 1,j = 0; s[i]; i++)
    {
        while(j > 0 && s[i] != s[j]) j = Next[j-1];
        if(s[i] == s[j]) j++;
        Next[i]=j;
    }
    len = i;
}

以下是代码：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <algorithm>
#include <string>
#include <set>
#include <functional>
#include <numeric>
#include <sstream>
#include <stack>
#include <map>
#include <queue>
#include<iomanip>

using namespace std;
const int N=100005;
char s[N];
int Next[N],len;
void getnext()
{
    int i,j;
    Next[0] = 0;
    for(i = 1,j = 0; s[i]; i++)
    {
        while(j > 0 && s[i] != s[j]) j = Next[j-1];
        if(s[i] == s[j]) j++;
        Next[i]=j;
    }
    len = i;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%s",s);
        getnext();
        int c = len - Next[len - 1];  //最小循环节
        if (len % c == 0)  //正好
        {
            if (len / c > 1) printf("0\n");  //存在2个及以上的循环
            else printf("%d\n",len);
        }
        else
        {
            int ans = c - (len - (len / c * c));  
            /* len / c * c， 完整的循环字母个数 len - len / c * c，多余的字母个数 c - 多余的，即还需要的字母个数 */
            printf("%d\n",ans);
        }
    }
    return 0;
}