第三十次codeforces竞技结束 #296 Div 2

Problems # Name A Playing with Paper standard input/output 2 s, 256 MB x2429 B Error Correct System standard input/output 2 s, 256 MB x953 C Glass Carving standard input/output 2 s, 256 MB x584 第三十次……好想到紫啊好想紫55555，两个1A，14

Problems

 

 

#	Name
A	Playing with Paper standard input/output 2 s, 256 MB		x2429
B	Error Correct System standard input/output 2 s, 256 MB		x953
C	Glass Carving standard input/output 2 s, 256 MB		x584

第三十次……好想到紫啊好想紫55555，两个1A，14min，然后我看Standing的时候你们造么！ 28名啊！！！

然后被C骗住了，我以为是zkw线段树……居然是考察数据结构set的问题……

然后……Failed System Test…… 原因？

A. Playing with Paper

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

One day Vasya was sitting on a not so interesting Maths lesson and making an origami from a rectangular a mm ?×? bmm sheet of paper (a?>?b). Usually the first step in making an origami is making a square piece of paper from the rectangular sheet by folding the sheet along the bisector of the right angle, and cutting the excess part.

After making a paper ship from the square piece, Vasya looked on the remaining (a?-?b) mm ?×? b mm strip of paper. He got the idea to use this strip of paper in the same way to make an origami, and then use the remainder (if it exists) and so on. At the moment when he is left with a square piece of paper, he will make the last ship from it and stop.

Can you determine how many ships Vasya will make during the lesson?

Input

The first line of the input contains two integers a, b (1?≤?b?a?≤?1012) — the sizes of the original sheet of paper.

Output

Print a single integer — the number of ships that Vasya will make.

Sample test(s)

input

2 1

output

2

input

10 7

output

6

input

1000000000000 1

output

1000000000000

Note

Pictures to the first and second sample test.

说一张长方形的纸~，每次都以较短边为边长裁一个正方形下来折个纸船，问长宽已知的长方形纸能折多少个纸船。

我想说……做题的时候有这个图么！！！

每次如果裁剪的效果没有达到a的剩余部分

其实看到这个图大家应该就立马懂了，每次不要一裁裁一个正方形，咱们要裁一串，就是 a/b 个。

Code：

#include <cmath> 
#include <cctype>
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a) b;
}

int main()
{
	long long a=0,b=0,c=0,t=0;
	cin>>a>>b;
	while(a!=b)
	{
		if(a%b)
		{
			c+=(a/b);
			a=a%b;
			t=a;
			a=b;
			b=t;
		}
		else
		{
			c+=(a/b);
			a=b;
		}
	}
	cout<br>

<p>
</p>
<p>
</p>
<p>
</p>
<p>
</p>
<p>
</p>
<p>
</p>
<h2 id="B-Error-Correct-System">B. Error Correct System</h2>
<p>
</p>
<p>
time limit per test</p>
2 seconds
<p>
</p>
<p>
memory limit per test</p>
256 megabytes
<p>
</p>
<p>
input</p>
standard input
<p>
</p>
<p>
output</p>
standard output

<p>
</p>
<p>
Ford Prefect got a job as a web developer for a small company that makes towels. His current work task is to create a search engine for the website of the company. During the development process, he needs to write a subroutine for comparing strings <span><em>S</em></span> and <span><em>T</em></span> of
 equal length to be "similar". After a brief search on the Internet, he learned about the<span>Hamming distance</span> between two strings <span><em>S</em></span> and <span><em>T</em></span> of
 the same length, which is defined as the number of positions in which <span><em>S</em></span> and <span><em>T</em></span> have
 different characters. For example, the Hamming distance between words "<span>permanent</span>" and "<span>pergament</span>"
 is two, as these words differ in the fourth and sixth letters.</p>
<p>
Moreover, as he was searching for information, he also noticed that modern search engines have powerful mechanisms to correct errors in the request to improve the quality of search. Ford doesn't know much about human beings, so he assumed that the most common
 mistake in a request is swapping two arbitrary letters of the string (not necessarily adjacent). Now he wants to write a function that determines which two letters should be swapped in string<span><em>S</em></span>,
 so that the Hamming distance between a new string <span><em>S</em></span> and string <span><em>T</em></span> would
 be as small as possible, or otherwise, determine that such a replacement cannot reduce the distance between the strings.</p>
<p>
Help him do this!</p>

<p>
</p>
<p>
Input</p>
<p>
The first line contains integer <span><em>n</em></span> (<span>1?≤?<em>n</em>?≤?200?000</span>)
 — the length of strings <span><em>S</em></span> and <span><em>T</em></span>.</p>
<p>
The second line contains string <span><em>S</em></span>.</p>
<p>
The third line contains string <span><em>T</em></span>.</p>
<p>
Each of the lines only contains <span>lowercase</span> Latin letters.</p>

<p>
</p>
<p>
Output</p>
<p>
In the first line, print number <span><em>x</em></span> — the minimum possible Hamming distance between strings <span><em>S</em></span> and <span><em>T</em></span> if
 you swap at most one pair of letters in <span><em>S</em></span>.</p>
<p>
In the second line, either print the indexes <span><em>i</em></span> and <span><em>j</em></span> (<span>1?≤?<em>i</em>,?<em>j</em>?≤?<em>n</em></span>, <span><em>i</em>?≠?<em>j</em></span>),
 if reaching the minimum possible distance is possible by swapping letters on positions <span><em>i</em></span> and <span><em>j</em></span>,
 or print "<span>-1 -1</span>", if it is not necessary to swap characters.</p>
<p>
If there are multiple possible answers, print any of them.</p>

<p>
</p>
<p>
Sample test(s)</p>
<p>
</p>
<p>
</p>
<p>
input</p>
<pre class="brush:php;toolbar:false">9
pergament
permanent

output

1
4 6

input

6
wookie
cookie

output

1
-1 -1

input

4
petr
egor

output

2
1 2

input

6
double
bundle

output

2
4 1

Note

In the second test it is acceptable to print i?=?2, j?=?3.

所谓海明距离，就是字符串s和字符串t字符不同的位置个数，比如acm和acg，有一个字符不同，所以是1.

题目问，如果最多允许把s中的两个不同位置的字符调换位置，那么调换后（也可以不调换）海明距离最小是多少。

假设原先海明距离是k，不调换肯定依然是k，调换的话如果要比k小只有2种情况：

1）[s]A [t]B 和某处的[s]B [t]A 调换位置，那么有两个不同处被解决了，即海明距离小了2个，最小为k-2.

2）[s]A [t]B 和某处的[s]B [t]C 调换位置，或

[s]A [t]B 和某处的[s]C [t]A 调换位置， 那么有一个不同处被解决了，即海明距离小了2个，最小为k-1.

那么，该怎么写呢?

由于字符的不同处可多可少，感觉如果全都消耗时间空间比较浪费所以我使用的是map，有的就放进来，读的过程中还可以同时把海明距离数出来，这是O(n)。

用map找由于其本质是红黑树，所以是在O(lgn)的时间内寻找到，综合来看最坏情况也是nlgn，可行。

Code：

#include <map>
#include <cmath> 
#include <cctype>
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef pair<char> pcc;
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a) b;
}

map <pcc> mpi;
map <char> mci,mcii;

int main()
{
	int flag=0,x=-1,y=-1;
	int n=0,cnt=0;	cin>>n;
	string s,t;	cin>>s>>t;
	mpi.clear(); mci.clear(); mcii.clear();
	for(int i=0;i<n if cnt mpi flag="2,x=mpi[make_pair(t[i],s[i])],y=i+1;" mci forget a half god... mcii cout return><br>
<br>


<p>
</p>
<h2 id="C-Glass-Carving">C. Glass Carving</h2>
<p>
</p>
<p>
time limit per test</p>
2 seconds
<p>
</p>
<p>
memory limit per test</p>
256 megabytes
<p>
</p>
<p>
input</p>
standard input
<p>
</p>
<p>
output</p>
standard output

<p>
</p>
<p>
Leonid wants to become a glass carver (the person who creates beautiful artworks by cutting the glass). He already has a rectangular <span><em>w</em></span> mm <span>?×?</span> <span><em>h</em></span> mm
 sheet of glass, a diamond glass cutter and lots of enthusiasm. What he lacks is understanding of what to carve and how.</p>
<p>
In order not to waste time, he decided to practice the technique of carving. To do this, he makes vertical and horizontal cuts through the entire sheet. This process results in making smaller rectangular fragments of glass. Leonid does not move the newly made
 glass fragments. In particular, a cut divides each fragment of glass that it goes through into smaller fragments.</p>
<p>
After each cut Leonid tries to determine what area the largest of the currently available glass fragments has. Since there appear more and more fragments, this question takes him more and more time and distracts him from the fascinating process.</p>
<p>
Leonid offers to divide the labor — he will cut glass, and you will calculate the area of the maximum fragment after each cut. Do you agree?</p>

<p>
</p>
<p>
Input</p>
<p>
The first line contains three integers <span><em>w</em>,?<em>h</em>,?<em>n</em></span> (<span>2?≤?<em>w</em>,?<em>h</em>?≤?200?000</span>, <span>1?≤?<em>n</em>?≤?200?000</span>).</p>
<p>
Next <span><em>n</em></span> lines contain the descriptions of the cuts. Each description has the form <span><em>H y</em></span> or <span><em>V x</em></span>.
 In the first case Leonid makes the horizontal cut at the distance <span><em>y</em></span> millimeters (<span>1?≤?<em>y</em>?≤?<em>h</em>?-?1</span>)
 from the lower edge of the original sheet of glass. In the second case Leonid makes a vertical cut at distance <span><em>x</em></span> (<span>1?≤?<em>x</em>?≤?<em>w</em>?-?1</span>)
 millimeters from the left edge of the original sheet of glass. It is guaranteed that Leonid won't make two identical cuts.</p>

<p>
</p>
<p>
Output</p>
<p>
After each cut print on a single line the area of the maximum available glass fragment in mm<span><span>2</span></span>.</p>

<p>
</p>
<p>
Sample test(s)</p>
<p>
</p>
<p>
</p>
<p>
input</p>
<pre class="brush:php;toolbar:false">4 3 4
H 2
V 2
V 3
V 1

output

8
4
4
2

input

7 6 5
H 4
V 3
V 5
H 2
V 1

output

28
16
12
6
4

Note

Picture for the first sample test:

Picture for the second sample test:  

有一块大大的玻璃~~~

每次横着或者竖着砍一刀，砍完不要拿走，下一道也得砍到这条线所在的所有部件。

每次砍完之后输出当前碎片中最大面积的部件的面积。

首先要知道的是，千万别一个一个部件的比较面积大小哦，因为每刀都是垂直或者水平的，所以每部分的面积都等于所对应的长边线段和短边线段的积。

即：我们要求的其实只是长边上的最长线段和短边上的最长线段，输出它们的积。

嘛，也顺带用两个例程说明下这三个函数的使用方法吧~ 毕竟set感觉并不是那么太常见（灵活使用的大大们请无视这句^_^）

// erasing from multiset
#include <iostream>
#include <set>

int main ()
{
  std::multiset<int> mymultiset;
  std::multiset<int>::iterator it;

  // insert some values:
  mymultiset.insert (40);                            // 40
  for (int i=1; i<br>
<pre class="brush:php;toolbar:false">// set::find
#include <iostream>
#include <set>

int main ()
{
  std::set<int> myset;
  std::set<int>::iterator it;

  // set some initial values:
  for (int i=1; i<br>
<pre class="brush:php;toolbar:false">// set::insert (C++98)
#include <iostream>
#include <set>

int main ()
{
  std::set<int> myset;
  std::set<int>::iterator it;
  std::pair<:set>::iterator,bool> ret;

  // set some initial values:
  for (int i=1; i<br>
说完了函数的使用方法，嘛，就来看看这题的解题Code吧
<h3 id="Code">Code：</h3>
<pre class="brush:php;toolbar:false">#include <set>
#include <cmath> 
#include <cctype>
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
#define i insert

char ch;
int n,w,h,t;
set<int> x,y,*s;
set<int>::iterator l,r;
multiset<int> lx,ly,*ls;

int main()
{
	cin>>w>>h>>n; x.i(0);x.i(w);y.i(0);y.i(h);lx.i(w);ly.i(h);
    for (int j=1; j>ch>>t;
        if (ch=='H') 
		{s=&y; ls=&ly;} 
		else 
		{s=&x; ls=&lx;}
		s->i(t); l=r=s->find(t); l--; r++; 
		ls->erase(ls->find(*r-*l));
		ls->i(t-*l);
		ls->i(*r-t);
		cout<br>
<br>

<p><br>
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<p><br>
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<p><br>
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