POJ 2406 Power Strings 数据结构+KMP

 Power Strings

Time Limit:3000MS    Memory Limit:65536KB    64bit IO Format:%lld & %llu

SubmitStatus

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

abcdaaaaababab.

Sample Output

143

Hint

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

解题思路：
一看就知道考的是kmp的失配函数/next数组的含义
看最后一个字符的f[i] 如果n%(n-f[n])！=0那就是不循环的
否则循环节就是n/(n-f[n])
面对这种题，只要自己对失配函数和next理解到一定程度，基本不用太多思考

#include<cstdio>#include<cstring>#include<iostream>using namespace std;const int maxn = 1000005 ;char str[maxn] ;int f[maxn] ;void getfail(){    f[0] = 0 ;    f[1] = 0 ;    int n = strlen(str);    for(int i=1;i<n;i++){        int j = f[i] ;        while(j&&str[i]!=str[j])j = f[j] ;        f[i+1] = (str[i]==str[j]?j+1:0) ;    }    return ;}int main(){    while(~scanf("%s",str)){        if(str[0]=='.')break ;        getfail();//        for(int i=0;i<=strlen(str);i++){//            printf("%d ",f[i]);//        }printf("\n");        int n = strlen(str) ;        int t = n-f[n];        if(n%t==0){            printf("%d\n",n/t);        }else{            printf("1\n");        }    }    return 0;}