HDU3746（KMP求循环节）

Cyclic Nacklace

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4735    Accepted Submission(s): 2165

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

 

Author

possessor WC

 

Source

HDU 3rd “Vegetable-Birds Cup” Programming Open Contest

 

题意：求字符串最少再加上几个字符就可以构成至少由一个循环节构成的字符串

思路：循环节就是cir = len - next[len], 如果cir == len那么则需要添加len个字符，如果cir != len && len % cir == 0则需要添加0个字符，否则需要添加cir - len % cir个字符

/*
ID: LinKArftc
PROG: 3746.cpp
LANG: C++
*/

#include <map>
#include <set>
#include <cmath>
#include <stack>
#include <queue>
#include <vector>
#include <cstdio>
#include <string>
#include <utility>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define eps 1e-8
#define randin srand((unsigned int)time(NULL))
#define input freopen("input.txt","r",stdin)
#define debug(s) cout << "s = " << s << endl;
#define outstars cout << "*************" << endl;
const double PI = acos(-1.0);
const double e = exp(1.0);
const int inf = 0x3f3f3f3f;
const int INF = 0x7fffffff;
typedef long long ll;

const int maxn = 100010;

char str[maxn];
int Next[maxn];

void getNext(char *p) {
    int len = strlen(p);
    int i = 0, j = -1;
    Next[i] = j;
    while (i < len) {
        if (j == -1 || p[i] == p[j]) {
            i ++; j ++;
            Next[i] = j;
        } else j = Next[j];
    }
}

int main() {
    int T;
    scanf("%d", &T);
    while (T --) {
        scanf("%s", str);
        getNext(str);
        int len = strlen(str);
        int cir = len - Next[len];
        if (cir == len) printf("%d
", len);
        else if (len % cir == 0) printf("0
");
        else printf("%d
", cir - len % cir);
    }

    return 0;
}