POJ 1961 （KMP）

最新推荐文章于 2021-09-01 21:48:51 发布

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最新推荐文章于 2021-09-01 21:48:51 发布

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本文链接：https://blog.youkuaiyun.com/Lcc_2016/article/details/51946128

本文介绍了一种使用KMP算法解决字符串周期性问题的方法，通过构建特定的next数组来高效找出所有有效周期的终点及其对应的周期数量。

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Description

For each prefix of a given string S with N characters (each character has an ASCII code between 97 and 126, inclusive), we want to know whether the prefix is a periodic string. That is, for each i (2 <= i <= N) we want to know the largest K > 1 (if there is one) such that the prefix of S with length i can be written as A^K ,that is A concatenated K times, for some string A. Of course, we also want to know the period K.

Input

The input consists of several test cases. Each test case consists of two lines. The first one contains N (2 <= N <= 1 000 000) – the size of the string S.The second line contains the string S. The input file ends with a line, having the
number zero on it.

Output

For each test case, output "Test case #" and the consecutive test case number on a single line; then, for each prefix with length i that has a period K > 1, output the prefix size i and the period K separated by a single space; the prefix sizes must be in increasing order. Print a blank line after each test case.

Sample Input

3
aaa
12
aabaabaabaab
0

Sample Output

Test case #1
2 2
3 3

Test case #2
2 2
6 2
9 3
12 4

大致题意：

给你一个长度为n的字符串，输出字符串循环节个数大于1的循环节的终点和个数。

用KMP做就行，话说KMP的next前缀数组很不好懂啊。

具体代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
using namespace std;

const int Max=1e6+100;
char str[Max];
int _next[Max];

void Next(int len)
{
    int i=-1,j=0;
    _next[0]=-1;
    while(j<len)
    {
        if(i==-1||str[i]==str[j])
        {
            ++i,++j;
            _next[j]=i;
        }
        else
            i=_next[i];
    }
    return;
}

int main()
{
    int n,num,count1=1;
    while(~scanf("%d",&n)&&n)
    {
        scanf("%s",str);
        Next(n);
        printf("Test case #%d\n",count1++);
        for(int i=2; i<=n; i++)
        {
            if(i%(i-_next[i])==0&&i/(i-_next[i])>1)
                printf("%d %d\n",i,i/(i-_next[i]));
        }
        printf("\n");
    }
    return 0;
}