POJ 2406 Power Strings（kmp）

最新推荐文章于 2019-06-14 11:10:14 发布

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最新推荐文章于 2019-06-14 11:10:14 发布

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分类专栏： KMP 文章标签： kmp

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本文针对PowerStrings问题，介绍了一种高效求解字符串循环节的方法。通过分析字符串的next数组，快速定位循环节长度及重复次数。文章提供了一个C++实现示例。

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Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 47749		Accepted: 19903

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.

Source

Waterloo local 2002.07.01

题意：给出一个字符串，找出其循环节并判断其循环了几次。

本题考查了对next数组的理解若有不懂请戳这里

不难发现，如果字符串长度len可以整除next[ len ]的话，循环节长度就等于 len - next[ len ] ，否则循环节长度就等于1。

#include <stdio.h>
#include <string.h>
using namespace std;

const int N = 1e6 + 10;
char a[N];
int nex[N];

void get_nex(int l){
    int i = 0, j = -1;
    nex[0] = -1;
    while(i < l){
        if(j == -1 || a[i] == a[j]){
            i++, j++;
            nex[i] = j;
        }
        else j = nex[j];
    }
}

int main(){
    while(scanf("%s", &a), a[0] != '.'){
        int l = strlen(a);
        get_nex(l);
        if(!(l%(l-nex[l]))) printf("%d\n", l/(l-nex[l]));
        else printf("1\n");
    }
}