POJ 2406 Power Strings

题目链接

http://poj.org/problem?id=2406

题目

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

重新定义字符串乘法：两个字符串相乘等价于两个字符串首尾相连。现给定一个字符串s，将s用字符串a表示，求最大的n使得a^n=s。

题解

根据最小循环节的定义，loop=len-pre[len]。其中pre为kmp算法中的前缀数组（在有些书中叫next数组）。
将字符串s看作由最小循环节组成的话，如果len恰好能整除loop，则ans=len/loop；否则，ans=len。

代码

#include <cstdio>#include <cstring>using namespace std;const int maxn=1e6+100;char t[maxn];int pre[maxn],len,loop;void init(){    memset(pre,0,sizeof(pre));}void getpre(){    memset(pre,0,sizeof(pre));    int j=0;    for(int i=2;i<=len;i++)    {        while(j>0 && t[j+1]!=t[i]) j=pre[j];        if(t[j+1]==t[i]) j++;        pre[i]=j;    }}int main(){    while(scanf("%s",t+1))    {        init();        if(strcmp(t+1,".")==0)            break;        len=strlen(t+1);        getpre();        loop=len-pre[len];        printf("%d\n",len%loop==0?len/loop:1);    }    return 0;}