poj 2406 Power Strings【kmp】-蒲公英云

poj 2406 Power Strings【kmp】

电玩女神 2022-08-21 02:17 128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;
}