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发表于 更新于 分类于 阅读次数： Valine：
本文字数： 3.6k 阅读时长 ≈ 3 分钟

题目链接：POJ 1860

Description

Several currency exchange points are working in our city. Let us suppose that each point specializes in two particular currencies and performs exchange operations only with these currencies. There can be several points specializing in the same pair of currencies. Each point has its own exchange rates, exchange rate of A to B is the quantity of B you get for 1A. Also each exchange point has some commission, the sum you have to pay for your exchange operation. Commission is always collected in source currency.

For example, if you want to exchange 100 US Dollars into Russian Rubles at the exchange point, where the exchange rate is 29.75, and the commission is 0.39 you will get (100 - 0.39) * 29.75 = 2963.3975RUR.

You surely know that there are N different currencies you can deal with in our city. Let us assign unique integer number from 1 to N to each currency. Then each exchange point can be described with 6 numbers: integer A and B - numbers of currencies it exchanges, and real RAB, CAB, RBA and CBA - exchange rates and commissions when exchanging A to B and B to A respectively.

Nick has some money in currency S and wonders if he can somehow, after some exchange operations, increase his capital. Of course, he wants to have his money in currency S in the end. Help him to answer this difficult question. Nick must always have non-negative sum of money while making his operations.

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发表于 更新于 分类于 阅读次数： Valine：
本文字数： 2.3k 阅读时长 ≈ 2 分钟

题目链接：HDU 5183

Problem Description

When given an array $(a_0,a_1,a_2,⋯a_{n−1})$ and an integer $K$, you are expected to judge whether there is a pair $(i,j)(0≤i≤j<n)$ which makes that $NP−sum(i,j)$ equals to $K$ true. Here $NP−sum(i,j)=a_i−a_{i+1}+a_{i+2}+⋯+(−1)^{j−i}a_j$

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发表于 更新于 分类于 阅读次数： Valine：
本文字数： 837 阅读时长 ≈ 1 分钟

写程序时想要用 cin 对输入的合法性检查，于是学习了一下 cin.fail() 函数，顺便学习了类似的一些函数。

cin 对象包含了一个描述流状态的数据成员。流状态有 3 个标志位：eofbit，badbit 和 failbit。

当 cin 操作到达文件末尾时，eofbit 置 1。

当流被破坏时，badbit 置 1。例如：试图读取不可访问的文件、写入写保护的磁盘、写入的设备剩余空间不足等。

cin 操作未能读取到预期的字符时 failbit 会置 1。例如：要输入到一个整型变量中，输入的却是字符时 failbit 置 1。I/O 失败时 failbit 也可能置 1。

如果 3 个状态位都为 0 表示一切顺利。

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发表于 更新于 分类于 阅读次数： Valine：
本文字数： 3.4k 阅读时长 ≈ 3 分钟

题目链接：POJ 2287

Description

Here is a famous story in Chinese history.

That was about 2300 years ago. General Tian Ji was a high official in the country Qi. He likes to play horse racing with the king and others.

Both of Tian and the king have three horses in different classes, namely, regular, plus, and super. The rule is to have three rounds in a match; each of the horses must be used in one round. The winner of a single round takes two hundred silver dollars from the loser.

Being the most powerful man in the country, the king has so nice horses that in each class his horse is better than Tian’s. As a result, each time the king takes six hundred silver dollars from Tian.

Tian Ji was not happy about that, until he met Sun Bin, one of the most famous generals in Chinese history. Using a little trick due to Sun, Tian Ji brought home two hundred silver dollars and such a grace in the next match.

It was a rather simple trick. Using his regular class horse race against the super class from the king, they will certainly lose that round. But then his plus beat the king’s regular, and his super beat the king’s plus. What a simple trick. And how do you think of Tian Ji, the high ranked official in China?

Were Tian Ji lives in nowadays, he will certainly laugh at himself. Even more, were he sitting in the ACM contest right now, he may discover that the horse racing problem can be simply viewed as finding the maximum matching in a bipartite graph. Draw Tian’s horses on one side, and the king’s horses on the other. Whenever one of Tian’s horses can beat one from the king, we draw an edge between them, meaning we wish to establish this pair. Then, the problem of winning as many rounds as possible is just to find the maximum matching in this graph. If there are ties, the problem becomes more complicated, he needs to assign weights 0, 1, or -1 to all the possible edges, and find a maximum weighted perfect matching…

However, the horse racing problem is a very special case of bipartite matching. The graph is decided by the speed of the horses — a vertex of higher speed always beat a vertex of lower speed. In this case, the weighted bipartite matching algorithm is a too advanced tool to deal with the problem.

In this problem, you are asked to write a program to solve this special case of matching problem.

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发表于 更新于 分类于 阅读次数： Valine：
本文字数： 1.4k 阅读时长 ≈ 1 分钟

题目链接：HDU 1028

Problem Description

“Well, it seems the first problem is too easy. I will let you know how foolish you are later.” feng5166 says.

“The second problem is, given an positive integer N, we define an equation like this:

N=a[1]+a[2]+a[3]+…+a[m];

a[i]>0,1<=m<=N;

My question is how many different equations you can find for a given N.

For example, assume N is 4, we can find:

4 = 4;

4 = 3 + 1;

4 = 2 + 2;

4 = 2 + 1 + 1;

4 = 1 + 1 + 1 + 1;

so the result is 5 when N is 4. Note that “4 = 3 + 1” and “4 = 1 + 3” is the same in this problem. Now, you do it!”

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发表于 更新于 分类于 阅读次数： Valine：
本文字数： 1.2k 阅读时长 ≈ 1 分钟

题目链接：HDU 2087

Problem Description

一块花布条，里面有些图案，另有一块直接可用的小饰条，里面也有一些图案。对于给定的花布条和小饰条，计算一下能从花布条中尽可能剪出几块小饰条来呢？

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发表于 更新于 分类于 阅读次数： Valine：
本文字数： 2.3k 阅读时长 ≈ 2 分钟

题目链接：HDU 2899

Problem Description

Now, here is a fuction:

F(x) = 6x^7+8x^6+7x^3+5x^2-yx (0 <= x <=100)

Can you find the minimum value when x is between 0 and 100.

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