poj 2406 Power Strings【KMP】

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

Hint

This problem has huge input, use scanf instead of cin to avoid time limit exceed.

题意：

给你一段字符串，问你完全重复（没有多余字符）某一字串最多的次数。

思路：

根据最大匹配长度，关键看最后一位匹配的最大长度。前缀和为后缀的最大匹配长度有没重合部分，那就最大为2，一旦中间有其他字符结果就为一，有重合部分时，那就中间就没多余的字符干扰了。

给几组测试数据：

aaabaaab => 2

abcdabc => 1

abcabcab => 1

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#define max_n 1000010
using namespace std;
char str1[max_n], str2[max_n];
int f[max_n];

void getnext() {
int len = strlen(str2);
int i = 0, j = -1;
f[0] = -1;
while(i < len) {
if(j == -1 || str2[j] == str2[i])
f[++i] = ++j;
else
j = f[j];
}
//  for(int i = 1; i <= len; i++) {
//      printf("---%d---\n", f[i]);
//  }
}

void solve() {
int len = strlen(str2);
int res = len - f[len];
if(len % res == 0) printf("%d\n", len / res);
else printf("1\n");
}

int main() {
while(scanf("%s", str2) && str2[0] != '.') {
getnext();
solve();
}
return 0;
}