[LA3026]Period

试题描述

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.

输入

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.

输出

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.

输入示例

3
aaa
12
aabaabaabaab
0

输出示例

Test case #1
2 2
3 3Test case #2
2 2
6 2
9 3
12 4

数据规模及约定

见“输入”

题解

KMP 裸题，对于位置 i，它指向的失配的位置为 f[i+1]，那么当 f[i+1] > 1 且 (i - f[i+1] + 1) | i 时答案为 i / (i - f[i+1] + 1). 我 KMP 从 1 开始做的所以前面的式子可能会奇怪一些。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <stack>
#include <vector>
#include <queue>
#include <cstring>
#include <string>
#include <map>
#include <set>
using namespace std;int read() {int x = 0, f = 1; char c = getchar();while(!isdigit(c)){ if(c == '-') f = -1; c = getchar(); }while(isdigit(c)){ x = x * 10 + c - '0'; c = getchar(); }return x * f;
}#define maxn 1000010
int n, f[maxn];
char S[maxn];int main() {n = read();int kase = 0; // bool fl = 1;while(n) {scanf("%s", S + 1);
//      if(!fl) putchar('\n'); fl = 0;printf("Test case #%d\n", ++kase);f[1] = f[2] = 1;for(int i = 2; i <= n; i++) {int u = f[i];while(u > 1 && S[u] != S[i]) u = f[u];f[i+1] = S[u] == S[i] ? u + 1 : u;if(f[i+1] > 1 && i % (i + 1 - f[i+1]) == 0) printf("%d %d\n", i, i / (i + 1 - f[i+1]));}putchar('\n');n = read();}return 0;
}

转载于:https://www.cnblogs.com/xiao-ju-ruo-xjr/p/5753884.html

[LA3026]Period相关推荐

LA3026 Period
题意大概是,给你一个字符串,如果该字符串的某一个前缀Si是由若干个相同的字符串循环构成,就输出i和最短循环节长度利用KMP算法里面失配函数可以解决这个问题,失配函数f(i)可以定义为字符串的前缀Si ...
时间周期 java_周期和持续时间 / Period and Duration
# 周期和持续时间 / Period and Duration 当您编写代码来指定一段时间时,请使用最符合您需要的类或方法: Duration 类, Period 类或 ChronoUnit.betw ...
java between_Java8 Period.between方法坑及注意事项
在使用Java8 新特性中关于Period.between的方法时需注意该方法获取日期的区间问题. @Test public void test1(){ LocalDate from = LocalD ...
HDU 5908 Abelian Period 暴力
Abelian Period 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5908 Description Let S be a number st ...
Java 8 – Period and Duration examples
Few examples to show you how to use Java 8 Duration, Period and ChronoUnit objects to find out the d ...
Back Stub Period
http://www.investopedia.com/terms/p/portfolio.asp#axzz1lnDvrq00 The last interim period in the life ...
JDK8中的新时间API:Duration Period和ChronoUnit介绍
文章目录简介 Duration Period ChronoUnit 简介在JDK8中,引入了三个非常有用的时间相关的API:Duration,Period和ChronoUnit. 他们都是用来对时 ...
Java8中使用以下类计算日期时间差：Period、Duration、ChronoUnit
一.简述在Java8中,我们可以使用以下类来计算日期时间差异: 1.Period 2.Duration 3.ChronoUnit 二.Period类主要是Period类方法getYears(),g ...
【POJ - 1961】Period（KMP，循环节问题）
题干: For each prefix of a given string S with N characters (each character has an ASCII code between ...

[LA3026]Period

[LA3026]Period相关推荐

最新文章

热门文章