poj 1961 Period 【kmp】

最新推荐文章于 2019-10-01 23:15:42 发布

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最新推荐文章于 2019-10-01 23:15:42 发布

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分类专栏： KMP 文章标签： KMP

本文链接：https://blog.youkuaiyun.com/LOOKQAQ/article/details/81945276

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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。具体任务是对于给定的字符串，找出所有前缀的周期K及对应的前缀长度i，并提供了一个完整的C++实现示例。

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Time limit

3000 ms

Memory limit

30000 kB

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

题意：给你一个字符串，计算周期为K，前缀为 i 这两个量（具有周期且周期至少为1），要求i至少从2开始。kmp裸题。

#include<iostream>
#include<cstdio>
#include<cstring>
const int maxn = 1e8+10;
using namespace std;
char s[maxn];
int next[maxn];
void get_next()  //求next[]数组
{
    int i=0,j=-1;
    next[0]=-1;
    int m=strlen(s);
    while(i<m)
    {
        if(j==-1 || s[i]==s[j])
        {
            i++,j++;
            next[i]=j;
        }
        else
            j=next[j];
    }
}
int main()
{
    int n;
    int p=1;
    while(~scanf("%d",&n) && n!=0)
    {
        scanf("%s",s);
        printf("Test case #%d\n",p++);
        get_next();
        int q = strlen(s);
        for(int i=2;i<=q;i++)
        {
            if(i%(i-next[i]) == 0 && next[i] != 0)  //&&后面的不能省
                printf("%d %d\n",i,i/(i-next[i]));
        }
        printf("\n");
    }
    return 0;
}