【KMP】 poj1961 Period

字符串周期性前缀算法

最新推荐文章于 2019-10-01 23:15:42 发布

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本文介绍了一种算法，用于解决字符串前缀周期性问题。该算法通过KMP算法求解字符串的循环节长度，进而找出所有具有周期性的前缀及其周期。文中提供了一个完整的C++实现示例。

Period

http://poj.org/problem?id=1961

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

题目：给定字符串S，求其前n位重复的次数。比如aabaabaabaab，前2位是aa，a重复了2次，前6位是aabaab，aab重复了2次，前9位是aabaabaab，aab重复了3次，前12位是aabaabaabaab，aab重复了4次。

所以输出的结果就是

2 2
6 2
9 3
12 4

题解：通过KMP求循环节长度，然后枚举每个点判断。

#include<cstdio>
#include<cstring>
using namespace std;
char s[1000005];
int next[1000005];
void get_next()
{
    int i=0,j=-1;
    next[0]=-1;
    for(;s[i];)
        if(j==-1||s[i]==s[j])
        {
            ++i;
            ++j;
            next[i]=j;
        }
        else
           j=next[j];
}
int main()
{
    int t,n;
    for(int cas=1;scanf("%d",&n),n;++cas)
    {
        scanf("%s",s);
        get_next();
        printf("Test case #%d\n",cas);
        for(int i=1;i<=n;++i)
            if(i%(i-next[i])==0)
               if(i/(i-next[i])>1)
                  printf("%d %d\n",i,i/(i-next[i]));
        printf("\n");
    }
    return 0;
}