杰拉斯的博客

本文没有咬文嚼字的地方，只是一个技巧的分享，十分简单，简单到流泪。

网页视觉层面主要是由形式（或叫布局）、色彩、图片和文字信息组成，设计师通常对形式感关注的比较多，因为视觉冲击力、设计差异性或创新大多都仰赖形式呈现，而色彩主要影响整体观感、设计品质以及受众情绪，很多时候我们设计了一个不错的形式却未能做出这个稿子应有的品质，会不会太可惜？

形式需要思考创造，图片素材需要学习处理，文字需要梳理编排，但网页色彩是不是一定需要有天生的色感、丰富的理论和多年经验沉淀才能运用自如？

当然不！

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]实验项目：ACM程序设计基础(5)
实验目的：掌握C++程序设计基础。
实验要求：使用VC++6.0实现实验要求。
实验内容：

1.编写一个函数实现如下功能：
输入：7
输出：

   1   8  14  19  23  26  28
       2   9  15  20  24  27
           3  10  16  21  25
               4  11  17  22
                   5  12  18
                       6  13
                           7

输入：5
输出：

   1   6  10  13  15
       2   7  11  14
           3   8  12
               4   9
                   5

(提示：使用setw（int n）函数对齐，该函数在iomanip.h中，动态二维数组的程序如下：

int **a = new int*[n];	//n行
for(int i = 0; i < n; ++i)
	a[i] = new int[m];	//m列

2.由1..9这九个数字组成九位数（无重复数字）能被11整除，求最大、最小者。
3.附加题：
给定n个矩阵A1A2…An， 其中Ai与Ai+1是可乘的。考察这n个矩阵的连乘积A1A2..An，如何确定计算矩阵连乘积的计算次序，使得依此次序计算矩阵连乘积需要的数乘次数最少。
例如A1=30×35、A2=35×15、A3=15×5、A4=5×10、A5=10×20、A6=20×25
最小乘数为15125。

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最优矩阵连乘积

Accepted: 10 Total Submit: 18
Time Limit: 1000ms Memony Limit: 32768KB

Description

在科学计算中经常要计算矩阵的乘积。矩阵A和B可乘的条件是矩阵A的列数等于矩阵B的行数。若A是一个p×q的矩阵，B是一个q×r的矩阵，则其乘积C=AB是一个p×r的矩阵。其标准计算公式为：

由该公式知计算C=AB总共需要pqr次的数乘。

为了说明在计算矩阵连乘积时加括号方式对整个计算量的影响，我们来看一个计算3个矩阵{A1，A2，A3}的连乘积的例子。设这3个矩阵的维数分别为10×100，100×5和5×50。若按第一种加括号方式((A1A2)A3)来计算，总共需要10×100×5+10×5×50=7500次的数乘。若按第二种加括号方式(A1(A2A3))来计算，则需要的数乘次数为100×5×50+10×100×50=75000。第二种加括号方式的计算量是第一种加括号方式的计算量的10倍。由此可见，在计算矩阵连乘积时，加括号方式，即计算次序对计算量有很大影响。

于是，人们自然会提出矩阵连乘积的最优计算次序问题，即对于给定的相继n个矩阵{A1,A2,…,An}(其中Ai的维数为pi-1×pi ，i=1,2,…,n)，如何确定计算矩阵连乘积A1A2…An的一个计算次序(完全加括号方式)，使得依此次序计算矩阵连乘积需要的数乘次数最少。

Input

有若干种案例，每种两行，第一行是一个非负整数n表示矩阵的个数，n=0表示结束。接着有n行，每行两个正整数，表示矩阵的维数。

Ouput
对应输出最小的乘法次数。

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实验项目：ACM程序设计基础(4)
实验目的：掌握C++程序设计基础。
实验要求：使用VC++6.0实现实验要求。
实验内容：

1.设有n个活动的集合E={1,2,…n}，其中每个活动都要求使用同一资源，如演讲会场等，而在同一时间内只有一个活动能使用这一资源。每个活动i都有一个要求使用该资源的起始时间Si和一个结束时间Fi，且Si<Fi，求出最多可以安排多少个活动使用该资源，并给出一个安排方案，如：

最多安排的资源个数为4，安排方案为：11 2 8 1
Sample Input
11 12 14 5 7 0 6 3 5 8 12 5 9 2 13 8 11 3 8 6 10 1 4
Sample Output
11 2 8 1 (4)
2.用KMP算法实现实验，输入两个只包含小写字母的字符串，判断第二个字符串是否是第一个字符串的子串，是则输出第二字符串在第一个字符串的起始位置，不是则输出NO。例如：
输入：abedsadfdseg
dsa
输出：4
3. Crashing Balloon问题，见Crashing Balloon。

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Crashing Balloon

Time Limit: 2 Seconds Memory Limit: 65536 KB

Description

On every June 1st, the Children's Day, there will be a game named "crashing balloon" on TV. The rule is very simple. On the ground there are 100 labeled balloons, with the numbers 1 to 100. After the referee shouts "Let's go!" the two players, who each starts with a score of "1", race to crash the balloons by their feet and, at the same time, multiply their scores by the numbers written on the balloons they crash. After a minute, the little audiences are allowed to take the remaining balloons away, and each contestant reports his\her score, the product of the numbers on the balloons he\she's crashed. The unofficial winner is the player who announced the highest score.

Inevitably, though, disputes arise, and so the official winner is not determined until the disputes are resolved. The player who claims the lower score is entitled to challenge his\her opponent's score. The player with the lower score is presumed to have told the truth, because if he\she were to lie about his\her score, he\she would surely come up with a bigger better lie. The challenge is upheld if the player with the higher score has a score that cannot be achieved with balloons not crashed by the challenging player. So, if the challenge is successful, the player claiming the lower score wins.

So, for example, if one player claims 343 points and the other claims 49, then clearly the first player is lying; the only way to score 343 is by crashing balloons labeled 7 and 49, and the only way to score 49 is by crashing a balloon labeled 49. Since each of two scores requires crashing the balloon labeled 49, the one claiming 343 points is presumed to be lying.

On the other hand, if one player claims 162 points and the other claims 81, it is possible for both to be telling the truth (e.g. one crashes balloons 2, 3 and 27, while the other crashes balloon 81), so the challenge would not be upheld.

By the way, if the challenger made a mistake on calculating his/her score, then the challenge would not be upheld. For example, if one player claims 10001 points and the other claims 10003, then clearly none of them are telling the truth. In this case, the challenge would not be upheld.

Unfortunately, anyone who is willing to referee a game of crashing balloon is likely to get over-excited in the hot atmosphere that he\she could not reasonably be expected to perform the intricate calculations that refereeing requires. Hence the need for you, sober programmer, to provide a software solution.

Input

Pairs of unequal, positive numbers, with each pair on a single line, that are claimed scores from a game of crashing balloon.

Output

Numbers, one to a line, that are the winning scores, assuming that the player with the lower score always challenges the outcome.

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