poj 2406 Power Strings【kmp】

电玩女神 2022-08-21 02:17 26阅读 0赞

Power Strings

**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

题意：给以字符串【长度小于壹佰万】，字符串一定为某一长度的前缀重复多次组成的，求这个字符串是由该长度的前缀重复几次组成的？

解析：设前缀重复n次，则字符串长度len%n==0；根据next数组的意义，n = len - next\[len\];则出现次数为k = len / (len - next\[len\]);

Accept代码【C++】 【110MS】

#include <cstdio>
    #include <cstring>
    
    using namespace std;
    
    char str[1000001];
    int p[1000001];
    
    void get_p(int len) {
    	int i = 0, j = -1;
    	p[0] = -1;
    	while(i != len) {
    		if(j == -1 || str[i] == str[j])
    			p[++i] = ++j;
    		else
    			j = p[j];
    	}
    }
    
    int main() {
    	while(scanf("%s", str), str[0] != '.') {
    		int len = strlen(str);
    		get_p(len);
    		int a = 0;
    		if(len % (len - p[len]) == 0)
    			printf("%d\n", len / (len - p[len]));
    		else
    			printf("1\n");
    	}
    	return 0;
    }