LA 3026 Period KMP算法求循環節

題目鏈接

https://vjudge.net/problem/UVALive-3026

題目

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 AK, 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 ≤ 1000000) 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

題意

給定一個長度爲n的字符串S，求它每個前綴的最短循環節。換句話說，對於每個i，求一個最大的整數K，使得S的前i個字符組成的前綴是某個字符重複K（K>1)次得到。輸出所有存在K的i和對應的K。

解題

設pre[i]表示S[i…len]的最長後綴S[i…i+pre[i]-1]與前綴[1…pre[i]]恰好相等。
上面的pre數組我叫它前綴數值，也有說法叫next數值。都差不多。
求pre的過程顯然可以通過遞推去解決。
j=pre[i-1].
pre[i]=j+1,當S[i]==S[j+1].
否則不停的j=pre[j]，找到使得S[j+1]==S[i]的那個j或者j=0.

長度爲N的字符串的最小循環節就是：N - pre[N]。

AC代碼

#include <bits/stdc++.h>
using namespace std;

const int maxn=1e6+7;
int pre[maxn],len;
char str[maxn];

void getPre()
{
    memset(pre,0,sizeof(pre));
    int j=0;
    for(int i=2;i<=len;i++)
    {
        while(j>0 && str[j+1]!=str[i]) j=pre[j];
        if(str[j+1]==str[i]) j++;
        pre[i]=j;
    }
}
int main()
{
    int kase=0;
    while(~scanf("%d",&len),len)
    {
        scanf("%s",str+1);
        getPre();
        printf("Test case #%d\n",++kase);
        //for(int i=1;i<=len;i++)
           // printf("%d %d\n",i,pre[i]);
        for(int i=2;i<=len;i++)
        {
           // int loop=i-pre[i];
            if(i%(i-pre[i])==0 && pre[i]) printf("%d %d\n",i,i/(i-pre[i]));
        }
        putchar(10);
    }
    return 0;
}