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POJ 2406 Power Strings
题目连接：http://poj.org/problem?id=2406
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
abcdaaaaababab.
Sample Output
143
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
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 else19 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 else33 printf("1\n");34 }35 return 0;36 }
POJ 2406 Power Strings
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