杰拉斯的博客

不容易系列之(3)—— LELE的RPG难题

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15209 Accepted Submission(s): 6025

Description

人称“AC女之杀手”的超级偶像LELE最近忽然玩起了深沉，这可急坏了众多“Cole”（LELE的粉丝,即"可乐"）,经过多方打探，某资深Cole终于知道了原因，原来，LELE最近研究起了著名的RPG难题:

有排成一行的ｎ个方格，用红(Red)、粉(Pink)、绿(Green)三色涂每个格子，每格涂一色，要求任何相邻的方格不能同色，且首尾两格也不同色．求全部的满足要求的涂法.

以上就是著名的RPG难题.

如果你是Cole,我想你一定会想尽办法帮助LELE解决这个问题的;如果不是,看在众多漂亮的痛不欲生的Cole女的面子上,你也不会袖手旁观吧?

Input

输入数据包含多个测试实例,每个测试实例占一行,由一个整数N组成，(0<n<=50)。

Output

对于每个测试实例，请输出全部的满足要求的涂法，每个实例的输出占一行。

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Children’s Queue

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5660 Accepted Submission(s): 1755

Description

There are many students in PHT School. One day, the headmaster whose name is PigHeader wanted all students stand in a line. He prescribed that girl can not be in single. In other words, either no girl in the queue or more than one girl stands side by side. The case n=4 (n is the number of children) is like
FFFF, FFFM, MFFF, FFMM, MFFM, MMFF, MMMM
Here F stands for a girl and M stands for a boy. The total number of queue satisfied the headmaster’s needs is 7. Can you make a program to find the total number of queue with n children?

Input

There are multiple cases in this problem and ended by the EOF. In each case, there is only one integer n means the number of children (1<=n<=1000)

Output

For each test case, there is only one integer means the number of queue satisfied the headmaster’s needs.

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Number Sequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 53991 Accepted Submission(s): 12146

Description

A number sequence is defined as follows:

f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7.

Given A, B, and n, you are to calculate the value of f(n).

Input

The input consists of multiple test cases. Each test case contains 3 integers A, B and n on a single line (1 <= A, B <= 1000, 1 <= n <= 100,000,000). Three zeros signal the end of input and this test case is not to be processed.

Output

For each test case, print the value of f(n) on a single line.

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在竞赛中，遇到大数据时，往往读文件成了程序运行速度的瓶颈，需要更快的读取方式。相信几乎所有的C++学习者都在cin机器缓慢的速度上栽过跟头，于是从此以后发誓不用cin读数据。还有人说Pascal的read语句的速度是C/C++中scanf比不上的，C++选手只能干着急。难道C++真的低Pascal一等吗？答案是不言而喻的。一个进阶的方法是把数据一下子读进来，然后再转化字符串，这种方法传说中很不错，但具体如何从没试过，因此今天就索性把能想到的所有的读数据的方式都测试了一边，结果是惊人的。

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骨牌铺方格

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14699 Accepted Submission(s): 7085

Description

在2×n的一个长方形方格中,用一个1× 2的骨牌铺满方格,输入n ,输出铺放方案的总数.
例如n=3时,为2× 3方格，骨牌的铺放方案有三种,如下图：

Input

输入数据由多行组成，每行包含一个整数n,表示该测试实例的长方形方格的规格是2×n (0<n<=50)。

Output

对于每个测试实例，请输出铺放方案的总数，每个实例的输出占一行。

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