poj2406 Power Strings(KMP)

最新推荐文章于 2019-10-18 13:40:44 发布

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最新推荐文章于 2019-10-18 13:40:44 发布

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分类专栏： KMP

本文链接：https://blog.youkuaiyun.com/zhanghaoxian1/article/details/73740706

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本文介绍了一个字符串处理问题——PowerStrings，旨在找出一个给定字符串能由多少个相同连续子串组成。通过KMP算法的next数组特性来解决该问题，找到最长重复模式。

Power Strings

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.

题意：给定一个字符串，让你求出他最多由几个相同的连续子串连接而成。

分析：在这里我们假设这个字符串的长度是len，那么如果len可以被len-next[len]整除的话，我们就可以说len-next[len]就是那个最短子串的长度。

代码

#include <cstdio>
#include <string>
using namespace std;

int l,p[1000000];
char a[1000000];

int main()
{

    while (scanf("%s",a),a[0]!='.')
    {
        l=strlen(a);
        p[0]=-1;
        int j=-1;
        for (int i=1;i<l;i++)
        {
            while ((j>-1)&&(a[j+1]!=a[i])) j=p[j];
            if (a[j+1]==a[i]) j++;
            p[i]=j;
        }
        int ans=1;
        if (l%(l-p[l-1]-1)==0) ans=l/(l-p[l-1]-1);
        printf("%d\n",ans);
    }
}