poj2406 Power Strings

最新推荐文章于 2022-07-21 20:55:14 发布

UMR小豪

最新推荐文章于 2022-07-21 20:55:14 发布

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分类专栏： kmp 文章标签： kmp poj

本文链接：https://blog.youkuaiyun.com/qq_33362864/article/details/52810144

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本文介绍了一个字符串处理问题的解决方法，即寻找输入字符串的最大循环节。通过定义字符串的乘法和幂运算，利用next数组特性高效求解。给出具体代码实现，并提示使用scanf避免超时。

Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 44918		Accepted: 18762

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

Hint

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

题意：找最大循环节

思路：直接借用next数组的性质求解

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int MAXN=1000000+10;
char s[MAXN];
int next[MAXN];
int l;
void get_next()
{
    int i=0,j=-1;
    next[0]=-1;
    while(i<l)
    {
        if(j==-1||s[i]==s[j])next[++i]=++j;
        else j=next[j];
    }
}
int main()
{
    while(~scanf("%s",s)&&s[0]!='.')
    {
         l=strlen(s);
        get_next();
        int  t=l-next[l];
        if(l%t==0)printf("%d\n",l/t);
        else puts("1");
    }
    return 0;
}