POJ 1961 Period

最新推荐文章于 2019-08-31 11:05:30 发布

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Period

Time Limit: 3000MS		Memory Limit: 30000K
Total Submissions: 18558		Accepted: 9019

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

题目链接：http://poj.org/problem?id=1961

题意：输入一个字符串str，求其中循环节出现的位置和出现的次数。

解题思路：最近在学KMP，所以就找了一些KMP的题目，这个就是其中之一啦，根据KMP中的失配函数（fail数组）去寻找循环节，我们知道失配函数fail[i]是指str[0..k]==str[i-k...i]的最大k，如果前缀str[0...i]是一个循环节，那么将它重复k次（k>=1）是str的一个子串，那么它的充要条件就是，i不等于i-fail[i]且i可以整除i-fail[i]，循环节出现的次数就是i/(i-fail[i])。

AC代码：

#include<cstdio>
#include<algorithm>
using namespace std;
const int MAXN = 1e6 + 10;
char str[MAXN];
int fail[MAXN];
int n;
void get_fail()//失配函数 
{
	fail[0]=0;
	fail[1]=0;
	for(int i=1;i<n;i++)
	{
		int j=fail[i];
		while(j && str[i]!=str[j]) j=fail[j];
		fail[i+1]=str[i]==str[j]?j+1:0;
	}
}
void solve()
{
	get_fail();//失配函数，得到失配数组fail 
	for(int i=1;i<=n;i++)
	{
		int l=i-fail[i];//循环长度 
		if(i!=l && i%l==0)//当该循环节不是第一次出现 
		{
			printf("%d %d\n",i,i/l);//输出字符串的下标和出现的次数 
		}
	}
	printf("\n");
}
int main(void)
{
	int kase=0;//测试组数 
	scanf("%d",&n);//字符串的长度 
	while(n)
	{
		getchar();//吃掉每次输入n的换行符 
		gets(str);
		printf("Test case #%d\n",++kase);
		solve();//处理函数 
		scanf("%d",&n);
	}
	return 0;
}