Power Strings

Description

Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "abcdef". If we think of concatenation as multiplication, exponentiation by a non-negative integer is defined in the normal way: a^0 = "" (the empty string) and a^(n+1) = a*(a^n).

Input

Each test case is a line of input representing s, a string of printable characters. The length of s will be at least 1 and will not exceed 1 million characters. A line containing a period follows the last test case.

Output

For each s you should print the largest n such that s = a^n for some string a.

Sample Input

abcd
aaaa
ababab
.

Sample Output

1
4
3

Hint

This problem has huge input, use scanf instead of cin to avoid time limit exceed.

題意：問你字串最多有多少個迴圈節

解題思路：next陣列真是巧妙呀！！通過next陣列，很容易計算出迴圈節的長度（N-nxt[N]），然後通過這個長度就能計算出這道題啦！

#include<iostream>
#include<deque>
#include<memory.h>
#include<stdio.h>
#include<map>
#include<string.h>
#include<algorithm>
#include<vector>
#include<math.h>
#include<stack>
#include<queue>
#include<set>
using namespace std;
typedef long long int ll;

void getNext(char a[],int len,int nxt[]){
    nxt[0]=nxt[1]=0;
    for(int i=1;i<len;i++){
        int j=nxt[i];
        while(j&&a[i]!=a[j])
            j=nxt[j];
        nxt[i+1]=(a[i]==a[j])?j+1:0;
    }
}

int KMP(char a[],char b[],int N,int M,int nxt[]){
    int j=0;
    int ans=0;
    for(int i=0;i<N;i++){
        while(j&&a[i]!=b[j])
            j=nxt[j];
        if(a[i]==b[j])
            j++;
        if(j==M){
            ans++;
        }
    }
    return ans;
}

char a[1000050];//原始串
char b[1000050];//模式串
int nxt[1000050];
int N,M;

int main(){



    while(~scanf("%s",a)){
        N=strlen(a);
        if(a[0]=='.'&&N==1)
            break;

        getNext(a,N,nxt);


        if(N%(N-nxt[N])==0)
            printf("%d\n",N/(N-nxt[N]));
        else
            printf("%d\n",1);


    }



    return 0;
}
 

另外還有後綴陣列解法！速度慢很多，但是也是體現了一個很重要的字尾陣列的思想。詳見程式碼理解。另外，倍增超時。

#include <iostream>
#include <deque>
#include <stdio.h>
#include <map>
#include <string.h>
#include <algorithm>
#include <vector>
#include <math.h>
#include <stack>
#include <queue>
#include <set>
using namespace std;

typedef long long int ll;

const int MAXN = 3000005; //用DC3要開三倍大小

int wa[MAXN], wb[MAXN], wv[MAXN], JS[MAXN]; //計算SA用的輔助陣列

int rk[MAXN], height[MAXN], SA[MAXN]; //三個常用陣列

/***字尾陣列倍增解法***/
int cmp(int *r, int a, int b, int l)
{
    return r[a] == r[b] && r[a + l] == r[b + l];
}

void DA(int *r, int *SA, int n, int m)
{
    int i, j, p, *x = wa, *y = wb, *t;
    for (i = 0; i < m; i++)
        JS[i] = 0;
    for (i = 0; i < n; i++)
        JS[x[i] = r[i]]++;
    for (i = 1; i < m; i++)
        JS[i] += JS[i - 1];
    for (i = n - 1; i >= 0; i--)
        SA[--JS[x[i]]] = i;
    for (j = 1, p = 1; p < n; j *= 2, m = p)
    {
        for (p = 0, i = n - j; i < n; i++)
            y[p++] = i;
        for (i = 0; i < n; i++)
            if (SA[i] >= j)
                y[p++] = SA[i] - j;
        for (i = 0; i < n; i++)
            wv[i] = x[y[i]];
        for (i = 0; i < m; i++)
            JS[i] = 0;
        for (i = 0; i < n; i++)
            JS[wv[i]]++;
        for (i = 1; i < m; i++)
            JS[i] += JS[i - 1];
        for (i = n - 1; i >= 0; i--)
            SA[--JS[wv[i]]] = y[i];
        for (t = x, x = y, y = t, p = 1, x[SA[0]] = 0, i = 1; i < n; i++)
            x[SA[i]] = cmp(y, SA[i - 1], SA[i], j) ? p - 1 : p++;
    }
    return;
}
/*******************/

/***字尾陣列DC3解法***/
#define F(x) ((x) / 3 + ((x) % 3 == 1 ? 0 : tb))
#define G(x) ((x) < tb ? (x)*3 + 1 : ((x)-tb) * 3 + 2)
int c0(int *r, int a, int b)
{
    return r[a] == r[b] && r[a + 1] == r[b + 1] && r[a + 2] == r[b + 2];
}
int c12(int k, int *r, int a, int b)
{
    if (k == 2)
        return r[a] < r[b] || r[a] == r[b] && c12(1, r, a + 1, b + 1);
    else
        return r[a] < r[b] || r[a] == r[b] && wv[a + 1] < wv[b + 1];
}
void sort(int *r, int *a, int *b, int n, int m)
{
    int i;
    for (i = 0; i < n; i++)
        wv[i] = r[a[i]];
    for (i = 0; i < m; i++)
        JS[i] = 0;
    for (i = 0; i < n; i++)
        JS[wv[i]]++;
    for (i = 1; i < m; i++)
        JS[i] += JS[i - 1];
    for (i = n - 1; i >= 0; i--)
        b[--JS[wv[i]]] = a[i];
    return;
}
void DC3(int *r, int *SA, int n, int m)
{
    int i, j, *rn = r + n, *SAn = SA + n, ta = 0, tb = (n + 1) / 3, tbc = 0, p;
    r[n] = r[n + 1] = 0;
    for (i = 0; i < n; i++)
        if (i % 3 != 0)
            wa[tbc++] = i;
    sort(r + 2, wa, wb, tbc, m);
    sort(r + 1, wb, wa, tbc, m);
    sort(r, wa, wb, tbc, m);
    for (p = 1, rn[F(wb[0])] = 0, i = 1; i < tbc; i++)
        rn[F(wb[i])] = c0(r, wb[i - 1], wb[i]) ? p - 1 : p++;
    if (p < tbc)
        DC3(rn, SAn, tbc, p);
    else
        for (i = 0; i < tbc; i++)
            SAn[rn[i]] = i;
    for (i = 0; i < tbc; i++)
        if (SAn[i] < tb)
            wb[ta++] = SAn[i] * 3;
    if (n % 3 == 1)
        wb[ta++] = n - 1;
    sort(r, wb, wa, ta, m);
    for (i = 0; i < tbc; i++)
        wv[wb[i] = G(SAn[i])] = i;
    for (i = 0, j = 0, p = 0; i < ta && j < tbc; p++)
        SA[p] = c12(wb[j] % 3, r, wa[i], wb[j]) ? wa[i++] : wb[j++];
    for (; i < ta; p++)
        SA[p] = wa[i++];
    for (; j < tbc; p++)
        SA[p] = wb[j++];
    return;
}
/***********************/

//計算rank和height陣列
void calheight(int *r, int *SA, int n)
{
    //  memset(height,0,sizeof(height));
    //  memset(rk,0,sizeof(rk));
    int i, j, k = 0;
    for (i = 1; i <= n; i++)
        rk[SA[i]] = i;
    for (i = 0; i < n; height[rk[i++]] = k)
        for (k ? k-- : 0, j = SA[rk[i] - 1]; r[i + k] == r[j + k]; k++)
            ;
}

int N;         //字串長度
int str[MAXN]; //主字串
char cc[MAXN];

int main()
{

   
    while (~scanf("%s", cc))
    {
        if(cc[0]=='.')
            break;
        N = strlen(cc);
        for (int i = 0; i < N; i++)
            str[i] = int(cc[i]);

        str[N] = 0; //必須要末尾補0！！！！！

        DC3(str, SA, N + 1, 300); //N是沒有補0的大小，算SA時要把末尾0計算進去，所以要N+1
        calheight(str, SA, N);   //計算height時不用末尾0

        int flag = 0;
        for (int i = 1; i <= N/2+1; i++)
        {
            //三個條件
            if (N % i == 0 && rk[0] == rk[i] + 1 && height[rk[0]] == N - i)
            {
                printf("%d\n", N / i);
                flag = 1;
                break;
            }
        }
        if (!flag)
            printf("1\n");

        // for (int i = 0; i <= N; i++)
        //     cout << SA[i] << " ";
        // cout << endl;

        // for (int i = 0; i <= N; i++)
        //     cout << rk[i] << " ";
        // cout << endl;

        // for (int i = 0; i <= N; i++)
        //     cout << height[i] << " ";
        // cout << endl;
    }

    return 0;
}