HDU1358：Period KMP跳转数组

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#acm #算法 #hdu #kmp

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本文针对给定字符串，探讨如何高效地找出所有前缀的最长周期，并通过代码实现这一算法。利用KMP算法原理，对跳转数组进行特殊处理，以达到快速解决问题的目的。

Period

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8691 Accepted Submission(s): 4149

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

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题意：给定长度的字符串，找出其循环子串，输出当前位置长度

思路：运用定理即可，对初始化跳转数组的函数加一条判断语句，注意长度大于1

AC代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=1000000+50;
int f [maxn];
int ls,la;
char s[maxn];
void getfill(char* s)
{
    memset(f,0,sizeof(f));
    for(int i=1;i<ls+1;i++)
    {
        int j=f[i];
        while(j && s[i]!=s[j])
            j=f[j];
        f[i+1]=(s[i]==s[j])?j+1:0;
        if(i%(i-f[i])==0){
           int index=i;
           int len=i/(i-f[i]);
           if(len>1) printf("%d %d\n",index,len);
        }
    }
}

int main()
{
    int n;
    //char s[maxn];
    int cnt=0;
    while(~scanf("%d",&n),n){
        scanf("%s",s);
        ls=strlen(s);
        printf("Test case #%d\n",++cnt);
        getfill(s);
        printf("\n");
    }
}