E - Period

最新推荐文章于 2021-11-26 16:01:51 发布

转载最新推荐文章于 2021-11-26 16:01:51 发布 · 150 阅读

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原文链接：http://www.cnblogs.com/joeylee97/p/6674581.html

本文介绍了一种用于检测字符串前缀是否为周期性序列的算法，通过KMP算法预处理得到Next数组，进而判断每个前缀是否存在周期性重复，并确定其周期长度。

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.

InputThe input file consists of several test cases. Each test case consists of two lines. The first one contains N (2 <= N <= 1 000 000) �C 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.
OutputFor 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


结论：j%(j-Nex[t])==0 说明该前缀是一个周期为j/(j-Next[j])的周期子序列
证明： j - Next[j] 是子串在失配时候的右移长度，说明在j位置失配的时候子串转移到next[j]处继续匹配，说明j之前的部分和next[j]之前的部分是相同的，也就是说j-Next[j]
是一个周期

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<vector>
#include<string>
#include<cstring>
using namespace std;
#define MAXN 1000001
typedef long long LL;
/*
*/
char s[MAXN];
int Next[MAXN];
void kmp_pre(int m)
{
    int j,k;
    j = 0;
    k = -1;
    Next[0] = -1;
    while(j<m)
    {
        if(k==-1||s[j]==s[k])
            Next[++j] = ++k;
        else
            k = Next[k];
    }
}
int main()
{
    int m;
    int cas=1;
    while(scanf("%d",&m),m)
    {
        scanf("%s",s);
        kmp_pre(m);
        printf("Test case #%d\n",cas++);
        for(int i=1;i<=m;i++)
            if((i)%(i-Next[i])==0&&i/(i-Next[i])>1)
                printf("%d %d\n",i,(i)/(i-Next[i]));
        cout<<endl;
    }
}

转载于:https://www.cnblogs.com/joeylee97/p/6674581.html