xtu字符串 B. Power Strings

B. Power Strings

Time Limit: 3000ms

Memory Limit: 65536KB

64-bit integer IO format: %lld      Java class name: Main

 

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.

 

解题：求字符串的循环节长度。利用KMP的适配数组。如果字符长度可以被（字符长度-fail[字符长度])整除，循环节这是这个商，否则循环节长度为1，即就是这个字符本身。

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <vector>
#include <climits>
#include <algorithm>
#include <cmath>
#define LL long long
#define INF 0x3f3f3f
using namespace std;
const int maxn = 1000100;
char str[maxn];
int fail[maxn];
void getFail(int &len) {
    int i,j;
    len = strlen(str);
    fail[0] = fail[1];
    for(i = 1; i < len; i++) {
        j = fail[i];
        while(j && str[j] != str[i]) j = fail[j];
        fail[i+1] = str[j] == str[i] ? j+1:0;
    }
}
int main() {
    int len;
    while(gets(str) && str[0] != '.') {
        getFail(len);
        if(len%(len-fail[len])) puts("1");
        else printf("%d
",len/(len-fail[len]));
    }
    return 0;
}