POJ2406-Power Strings-KMP循环节/哈希循环节

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.

 

题意：

求每个字符串最多是由多少个相同的子字符串重复连接而成的。

如：ababab 则最多有3个 ab 连接而成。

 

思路：两种方法求循环节的模板题。

 

哈希：

 

#include<stdio.h>
#include<string.h>
typedef unsigned long long ull;
const int N=1e6+20;
char a[N];
ull p[N],sum[N],x=131;

void init()
{
    p[0]=1;
    for(int i=1; i<N; i++)
        p[i]=p[i-1]*x;
}

int main()
{
    init();
    while(~scanf("%s",a+1))
    {
        if(a[1]=='.')
            break;
        int len=strlen(a+1);
        sum[0]=0;
        for(int i=1;i<=len;i++)//主串哈希值
            sum[i]=sum[i-1]*x+a[i];
        for(int i=1; i<=len; i++)
        {
            if(len%i!=0)
                continue;//说明不存在该循环节
            int flag=0;
            for(int j=i; j<=len; j+=i)
            {   //ababab 
                //i=2时 -> j=2、4、6
                if((sum[j]-sum[j-i]*p[i])!=sum[i])
                //(sum[2]-sum[2-2]*p[2])!=sum[2]
                //(sum[4]-sum[4-2]*p[2])!=sum[2]
                //(sum[6]-sum[6-2]*p[2])!=sum[2]
                {
                    flag=1;
                    break;
                }
            }
            if(flag==0)
            {
                printf("%d\n",len/i);
                break;
            }
        }
    }
    return 0;
}

 

 

 

 

KMP：

 思想：

比如abcabcabcabc ，nextt[len]=9,len=12，所以len-next[len]=3肯定是len=12的因数，并且此时len-next[len]=3也肯定为最短循环节的长度，所以len/（len-next[len]）=12/（12-9）=12/3=4，即循环节的个数，也就是出现过几次abc。

如果不能整除说明不存在循环节，直接输出1即可。

int w=len-nextt[len];
if(len%w==0)
    printf("%d\n",len/w);
else
    printf("1\n");

 

详细代码如下：

#include<stdio.h>
#include<string.h>
const int N=1e6+20;

int nextt[N],len;
char a[N];

void getnext()
{
    int i=0,j=-1;
    nextt[0]=-1;
    while(i<len)
    {
        if(j==-1||a[i]==a[j])
        {
            nextt[++i]=++j;
        }
        else
            j=nextt[j];
    }
}

int main()
{
    while(~scanf("%s",a))
    {
        if(a[0]=='.')
            break;
        len=strlen(a);
        getnext();
        len=strlen(a);
        int w=len-nextt[len];
        if(len%w==0)
            printf("%d\n",len/w);
        else
            printf("1\n");
    }
    return 0;
}