hdu 3746 KMP next數組的應用

Cyclic Nacklace

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4375    Accepted Submission(s): 1988

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

 

#include <iostream>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <algorithm>
#include <cstdio>
#include <string>
#include <cstring>
#include <cmath>
#include <ctime>
using namespace std;

#pragma comment(linker,"/STACK:102400000,102400000") /// kuo zhan
#define clr(s,x) memset(s,x,sizeof(s))
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lowbit(x) (x&(-x))
#define PB push_back
#define For(i,a,b) for(int i=a;i<b;i++)
#define FOR(i,a,b) for(int i=a;i<=b;i++)

typedef long long               LL;
typedef unsigned int            uint;
typedef unsigned long long      ULL;
typedef vector<int>             vint;
typedef vector<string>          vstring;

void RI (int& x){
    x = 0;
    char c = getchar ();
    while (c == ' '||c == '\n')    c = getchar ();
    bool flag = 1;
    if (c == '-'){
        flag = 0;
        c = getchar ();
    }
    while (c >= '0' && c <= '9'){
        x = x * 10 + c - '0';
        c = getchar ();
    }
    if (!flag)    x = -x;
}
void RII (int& x, int& y){RI (x), RI (y);}
void RIII (int& x, int& y, int& z){RI (x), RI (y), RI (z);}


const double PI  = acos(-1.0);
const int maxn   = 1e6 + 100;
const int maxm   = 1e5 + 100;
const LL mod     = 1e9 + 7;
const double eps = 1e-9;
const int INF    = 0x7fffffff;

/************************************END DEFINE*********************************************/
char s[maxn],t[maxm];
int nxt[maxm];
int T;
void get_next(){
    int len = strlen(t);
    nxt[0] = 0;
    nxt[1] = 0;
    For(i,1,len){
        int j = nxt[i];
        while(j && t[i] != t[j])j = nxt[j];
        nxt[i+1] = t[i] == t[j] ? j+1 : 0;
    }
}

int kmp(){
    get_next();
    int lens = strlen(s), lent = strlen(t);
    int j = 0,cnt = 0;
    For(i,0,lens){
        while(j && t[j] != s[i])j = nxt[j];
        if(t[j] == s[i]) j++;
        if(j == lent){
            cnt ++;
            j = nxt[j];
        }
    }
    return cnt;
}

int main()
{
    RI(T);
    while(T--){
        scanf("%s",t);
        get_next();
        int x = strlen(t) - nxt[strlen(t)];
        if(x != strlen(t) && strlen(t)%x == 0){
            printf("0\n");
        }
        else{
            int ans = x - (strlen(t) % x);
            printf("%d\n",ans);
        }
    }
    return 0;
}