[POJ 1961] Period【KMP】

  • 2018-02-20
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Problem:

Time Limit: 3000MS Memory Limit: 30000K

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 AK ,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

Solution:

这道题和[POJ 2406] Power Strings【KMP】相似,都是将一个字符串表示成某个子串重复若干次的形式。

只是这道题需要求出所有重复超过 1 次前缀串和对应子串长度。

所以遍历字符串,相同方法判断即可。

Code: O(TN) [5568K, 485MS]

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
using namespace std;

int T = 0, N, nxt[1000005];
char S[1000005];

int main(){
	while(scanf("%d", &N) != EOF && N){
		scanf("%s", S + 1);
		nxt[1] = 0;
		for(register int i = 2, j = 0; i <= N; i++){
			while(j && S[j + 1] != S[i]) j = nxt[j];
			if(S[j + 1] == S[i]) j++;
			nxt[i] = j;
		}
		printf("Test case #%d\n", ++T);
		for(register int i = 2; i <= N; i++)
			if(nxt[i] && i % (i - nxt[i]) == 0)
				printf("%d %d\n", i, i / (i - nxt[i]));
		puts("");
	}
	return 0;
} 

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OPEN AT 2017.12.10

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