HDU1358 Period（KMP中求字符串的前缀是否为周期串，若是，求循环节长度和循环次数）

最新推荐文章于 2021-10-04 20:37:54 发布

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本文介绍了一种用于检测字符串前缀是否为周期串的算法，并详细解释了如何通过构造next数组来判断循环节长度和循环次数。该算法适用于处理大规模字符串数据，通过实例演示了其在实际应用中的有效性。

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

题意：

求字符串的前缀是否为周期串，若是，打印循环节长度和循环次数。

思路：

当next数组满足i%(i-next[i])==0 && next[i]!=0。说明当前字符串中前缀为周期串，且循环节长度为i-next[i] ,循环次数为i/(i-next[i]);

代码：

#include <iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
using namespace std;
const int MAXN=1000000+100;
char P[MAXN];
int f[MAXN];
int m;
void getFail(char *P,int *f)
{
    f[0]=f[1]=0;
    for(int i=1;i<m;i++)
    {
        int j=f[i];
        while(j && P[i]!=P[j]) j=f[j];
        f[i+1]=(P[i]==P[j])?j+1:0;
    }
}
int main()
{
    int Case=1;
    int n;
    while(scanf("%d",&n)&&n)
    {
        scanf("%s",P);
        m=strlen(P);
        getFail(P,f);
        printf("Test case #%d\n",Case++);
        for(int i=1;i<=m;i++)
            if(f[i]&&i%(i-f[i])==0)
                printf("%d %d\n",i,i/(i-f[i]));
        printf("\n");
    }
    return 0;
}