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+;
 char a[N];
 ull p[N],sum[N],x=;

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

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

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==)
    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+;

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

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

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