poj1961 & hdu1358 Period【KMP】

Period

Time Limit: 3000MS		Memory Limit: 30000K
Total Submissions: 20436		Accepted: 9961

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

题意：

 求一个字符串的所有前缀的最短循环节。

思路：

首先，一个字符串要是可以由他的子串循环而成的话，那么这个字符串的长度一定是子串长度len的倍数。并且一定有S[len+1 ~ i] = S[1 ~ i- len]

而KMP求出的nxt数组，表示的就是对于每一个i，S[i - nxt[i] + 1 ~ i] = S[1 ~ nxt[i]]

因此，当i - nxt[i]能整除i时，S[1 ~ i - nxt[i]]就是S[1 ~ i]的最小循环元。

#include <iostream>
#include <set>
#include <cmath>
#include <stdio.h>
#include <cstring>
#include <algorithm>
#include <map>
using namespace std;
typedef long long LL;
#define inf 0x7f7f7f7f

int n;
const int maxn = 1e6 + 5;
int nxt[maxn];
char s[maxn];

void getnxt()
{
    nxt[1] = 0;
    for(int i = 2, j = 0; i <= n; i++){
        while(j > 0 && s[i] != s[j + 1]){
            j = nxt[j];
        }
        if(s[i] == s[j + 1])j++;
        nxt[i] = j;
    }
}

int main()
{
    int cas = 1;
    while(scanf("%d", &n) != EOF && n){
        scanf("%s", s + 1);
        getnxt();
        printf("Test case #%d
", cas++);
        for(int i = 1; i <= n; i++){
            if(i % (i - nxt[i]) == 0 && i / (i - nxt[i]) > 1){
                printf("%d %d
", i, i / (i - nxt[i]));
            }
        }
        printf("
");
    }
    return 0;
}