本文提供CodeForces比赛527题目的解析与代码实现，涵盖纸张折叠、错误纠正系统、玻璃雕刻及派系问题等算法挑战。
A题：http://codeforces.com/contest/527/problem/A
One day Vasya was sitting on a not so interesting Maths lesson and making an origami from a rectangular a mm × b mm 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?
The first line of the input contains two integers a, b (1 ≤ b < a ≤ 1012) — the sizes of the original sheet of paper.
Print a single integer — the number of ships that Vasya will make.
2 1
2
10 7
6
1000000000000 1
1000000000000
Pictures to the first and second sample test.
代码：
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<string>
#include<iostream>
#include<sstream>
#include<set>
#include<map>
#include<queue>
#include<bitset>
#include<vector>
#include<fstream>
using namespace std;
typedef long long ll;
const int M = 10;
int m[30][30];
ll deal(ll a,ll b)
{
if(a<b)
swap(a,b);
if(b==0)
return 0;
if(a==b)
return 1;
return a/b+deal(b,a%b);
}
int main() {
ll a,b;
while(cin>>a>>b)
{
cout<<deal(a,b)<<endl;
}
return 0;
}
代码：
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<string>
#include<iostream>
#include<sstream>
#include<set>
#include<map>
#include<queue>
#include<bitset>
#include<vector>
#include<fstream>
using namespace std;
const int M = 10;
int m[30][30];
int main() {
int n;
while(cin>>n)
{
memset(m,0,sizeof(m));
string s,t;
cin>>s>>t;
int ret = 0;
for(int i=1;i<=s.length();i++)
{
char c1 = s[i-1];
char c2 = t[i-1];
m[c1-'a'][c2-'a'] = i;
if(c1==c2)
ret++;
}
int ret1 = -1;
int ret2 = -1;
for(char c1 = 'a';c1<='z';c1++)
{
for(char c2='a';c2<='z';c2++)
{
if(c1!=c2&&m[c1-'a'][c2-'a']>0&&m[c2-'a'][c1-'a']>0)
{
ret1 = m[c1-'a'][c2-'a'];
ret2 = m[c2-'a'][c1-'a'];
}
}
}
if(ret1>0)
{
cout<<n-(ret+2)<<endl;
cout<<ret1<<" "<<ret2<<endl;
continue;
}
for(char c1 = 'a';c1<='z';c1++)
{
for(char c2='a';c2<='z';c2++)
{
if(c1!=c2&&m[c1-'a'][c2-'a']>0)
{
for(char c3='a';c3<='z';c3++)
{
if(m[c2-'a'][c3-'a']>0&&c2!=c3)
{
ret1 = m[c1-'a'][c2-'a'];
ret2 = m[c2-'a'][c3-'a'];
}
}
}
}
}
if(ret1>0)
{
cout<<n-(ret+1)<<endl;
cout<<ret1<<" "<<ret2<<endl;
continue;
}
cout<<n-ret<<endl;
cout<<ret1<<" "<<ret2<<endl;
}
return 0;
}C题： http://codeforces.com/contest/527/problem/C
Leonid wants to become a glass carver (the person who creates beautiful artworks by cutting the glass). He already has a rectangular wmm × h mm sheet of glass, a diamond glass cutter and lots of enthusiasm. What he lacks is understanding of what to carve and how.
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.
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.
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?
The first line contains three integers w, h, n (2 ≤ w, h ≤ 200 000, 1 ≤ n ≤ 200 000).
Next n lines contain the descriptions of the cuts. Each description has the form H y or V x. In the first case Leonid makes the horizontal cut at the distance y millimeters (1 ≤ y ≤ h - 1) from the lower edge of the original sheet of glass. In the second case Leonid makes a vertical cut at distance x (1 ≤ x ≤ w - 1) millimeters from the left edge of the original sheet of glass. It is guaranteed that Leonid won't make two identical cuts.
After each cut print on a single line the area of the maximum available glass fragment in mm2.
4 3 4 H 2 V 2 V 3 V 1
8 4 4 2
7 6 5 H 4 V 3 V 5 H 2 V 1
28 16 12 6 4
Picture for the first sample test:
代码：
#include <vector>
#include <list>
#include <map>
#include <set>
#include <queue>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
using namespace std;
typedef long long ll;
set<int> H, W;
multiset<int> MH, MW;
int main()
{
int w, h, n;
cin >> w >> h >> n;
W.insert(0);
W.insert(w);
H.insert(0);
H.insert(h);
MH.insert(h);
MW.insert(w);
while (n--)
{
char x;
int a, b, c;
cin >> x >> c;
if (x != 'V')
swap(W, H), swap(MW, MH);
W.insert(c);
a = *(--W.find(c));
b = *(++W.find(c));
MW.erase(MW.find(b - a));
MW.insert(b - c);
MW.insert(c - a);
if(x!='V')
swap(W, H), swap(MW, MH);
cout << ((ll)(*(--MW.end())))**(--MH.end()) << endl;
}
}D题： http://codeforces.com/contest/527/problem/D
The clique problem is one of the most well-known NP-complete problems. Under some simplification it can be formulated as follows. Consider an undirected graph G. It is required to find a subset of vertices C of the maximum size such that any two of them are connected by an edge in graph G. Sounds simple, doesn't it? Nobody yet knows an algorithm that finds a solution to this problem in polynomial time of the size of the graph. However, as with many other NP-complete problems, the clique problem is easier if you consider a specific type of a graph.
Consider n distinct points on a line. Let the i-th point have the coordinate xi and weight wi. Let's form graph G, whose vertices are these points and edges connect exactly the pairs of points (i, j), such that the distance between them is not less than the sum of their weights, or more formally: |xi - xj| ≥ wi + wj.
Find the size of the maximum clique in such graph.
The first line contains the integer n (1 ≤ n ≤ 200 000) — the number of points.
Each of the next n lines contains two numbers xi, wi (0 ≤ xi ≤ 109, 1 ≤ wi ≤ 109) — the coordinate and the weight of a point. All xi are different.
Print a single number — the number of vertexes in the maximum clique of the given graph.
4 2 3 3 1 6 1 0 2
3
If you happen to know how to solve this problem without using the specific properties of the graph formulated in the problem statement, then you are able to get a prize of one million dollars!
The picture for the sample test.
代码：
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<string>
#include<iostream>
#include<sstream>
#include<set>
#include<map>
#include<queue>
#include<bitset>
#include<vector>
#include<fstream>
using namespace std;
typedef long long ll;
const int M = 10;
int m[30][30];
int main() {
int n;
while(cin>>n)
{
vector<pair<ll,ll> >all;
for(int i=0;i<n;i++)
{
ll x,w;
cin>>x>>w;
all.push_back(make_pair(x-w,x+w));
}
sort(all.begin(),all.end());
ll last = -9999999999999;
ll ans = 0;
for(int i=0;i<n;i++)
{
pair<ll,ll> p = all[i];
if(last<=p.first)
last = p.second,ans++;
else if(last>p.second)
last = p.second;
}
cout<<ans<<endl;
}
return 0;
}
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