POJ 2406 Power Strings

Posted 2020-08-08 风中的簌雨

 

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：

 1 #include <stdio.h>
 2 #include <string.h>
 3 char str[1000010];
 4 int next[1000010];
 5 void getnext()  //获取next数组
 6 {
 7     int i = 0,j = -1;
 8     next[0] = -1;
 9     int len = strlen(str);
10     while (i < len)
11     {
12         if (j == -1 || str[i] == str[j])
13         {
14             i ++;
15             j ++;
16             next[i] = j;
17         }
18         else
19             j = next[j];
20     }
21 }
22 int main()
23 {
24    while (~scanf("%s",str))
25    {
26        if (strcmp(str,".")==0)  //如果str数组为“.”则结束
27             break;
28        int len = strlen(str);
29        getnext();
30        if (len%(len-next[len])==0)
31             printf("%d\\n",len/(len-next[len]));
32        else
33             printf("1\\n");
34    }
35    return 0;
36 }

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