poj 1961(next数组的循环节)

最新推荐文章于 2020-03-08 21:48:27 发布

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本文介绍了一种用于检测字符串前缀是否具有周期性的算法，并通过一个具体的编程实例展示了如何使用next数组来解决这一问题。该算法能够高效地找出每个前缀的最大周期。

Period

Time Limit: 3000MS		Memory Limit: 30000K
Total Submissions: 13318		Accepted: 6263

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

Source

Southeastern Europe 2004

搞清楚next数组的意义就行了，不懂自己打出next就知道了。

AC代码：

#include<iostream>
#include<cstring>
#include<stdio.h>
using namespace std;
int n;
int next[1000005];
void getnext(char *s){
    int i,j;
    i=0; j=-1;
    next[0]=-1;
    while(s[i]){
        if(j==-1 || s[i]==s[j]){
            i++;
            j++;
            next[i]=j;
        }
        else
            j=next[j];
    }
}
int main(){
    int cas=1;
    char str[1000005];
    while(scanf("%d",&n)!=EOF && n){
        scanf("%s",str);
        getnext(str);
        printf("Test case #%d\n",cas++);
        for(int i=2;i<=n;i++){
            if(next[i]>0 && 0==i%(i-next[i]))
                printf("%d %d\n",i,i/(i-next[i]));
        }
        printf("\n");
    }
    return 0;
}