POJ 2406 Power Strings

 

Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 44595		Accepted: 18629

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

题目大意： 给定一个字符串，问最多是多少个相同子串不重叠连接构成。

解题思路：KMP的next数组应用。这里主要是如何判断是否有这样的子串，和子串的个数。

　　　　　　若为abababa，显然除其本身外，没有子串满足条件。而分析其next数组，next[7] = 5，next[5] = 3，next[3] = 1，

　　　　　　即str[2..7]可由ba子串连接构成，那怎么否定这样的情况呢？很简单，若该子串满足条件，

　　　　　　则len%sublen必为0。sunlen可由上面的分析得到为len-next[len]。

            因为子串是首尾相接，len/sublen即为substr的个数。

            若L%(L-next[L])==0,n = L/(L-next[L]),else n = 1

　　　　　　对KMP不太熟悉的可以看这里:从头到尾理解KMP算法

 

 

AC代码：

#include <stdio.h>
#include <string.h>
char str[1000010];
int next[1000010];
void getnext()  //获取next数组
{
    int i = 0,j = -1;
    next[0] = -1;
    int len = strlen(str);
    while (i < len)
    {
        if (j == -1 || str[i] == str[j])
        {
            i ++;
            j ++;
            next[i] = j;
        }
        else
            j = next[j];
    }
}
int main()
{
   while (~scanf("%s",str))
   {
       if (strcmp(str,".")==0)  //如果str数组为“.”则结束
            break;
       int len = strlen(str);
       getnext();
       if (len%(len-next[len])==0)
            printf("%d\n",len/(len-next[len]));
       else
            printf("1\n");
   }
   return 0;
}