HDU 1358 Period 【KMP】（子串循环次数）

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本文深入探讨了如何识别给定字符串中具有循环结构的子串，并提供了求解最大循环结及其循环次数的算法实现。通过实例演示，帮助理解循环子串的概念及其在字符串分析中的应用。

Period

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3315 Accepted Submission(s): 1660

Problem 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 file 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

/*
题解：求含有循环结的子串的位置及循环结的循环次数。
最大循环结：cir = len - next[len]
*/

#include<cstdio>
#include<cstring>
#define N 1000002
int next[N]; 
void getnext(char *p)
{
    int i=0,len=strlen(p);
    int j=next[0]=-1;
    while(i<len)
    {
        if(j==-1||p[i]==p[j])
        {
            i++,j++;
            next[i]=j;
        }
        else j=next[j];
    }
}
int main()
{
    int n,cas=0;
    char p[N];
    while(scanf("%d",&n)&&n)
    {
        scanf("%s",p);
        getnext(p);
        printf("Test case #%d\n",++cas);
        for(int i=0; i<n; i++)
        {
            int cir=i+1-next[i+1];//最大循环结 
            //printf("--%d\n",cir);
            if(i+1!=cir&&(i+1)%cir==0) printf("%d %d\n",i+1,(i+1)/cir);
        }
        printf("\n");
    }
    return 0;
}