Power Strings

最新推荐文章于 2023-06-22 23:47:57 发布

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本文介绍了一种用于解决字符串幂次分解问题的高效算法。该算法利用KMP算法中的next数组来快速找到输入字符串的最大重复周期，从而确定该字符串是否可以表示为某个基串的多次重复，并计算出重复次数。

Power Strings

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.

next数组求循环周期

code:

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int MAXN = 1010000;
char s[MAXN];
int Next[MAXN];
void getNext(){
    int i = -1,j = 0;
    int lens = strlen(s);
    Next[0] = -1;
    while(j < lens){
       if(i == -1 || s[i] == s[j]){
          i++,j++;
          Next[j] = i;
       }
       else
          i = Next[i];
    }
    return;
}
int main(){
    while(~scanf("%s",s)&&s[0]!='.'){//这里有个小坑，题目没有说明但是样例中有个.没有输出结果说明代表结束
        getNext();
        int len = strlen(s);
        int tmp = len-Next[len];
        if(len%tmp==0)//一定要先判断整除，只有整除才说明是循环的，否则都是1
            printf("%d\n",len/tmp);
        else
            printf("1\n");
    }
    return 0;
}