2008 ACM-ICPC SEERC

http://codeforces.com/gym/101462

 因为某些原因，这场比赛只进行了4小时，我队只过了6题：A,B,E,F,H,I

A题水题，最长上升子序列

#include <bits/stdc++.h>
using namespace std;
#define ll long long
int n;
int a[100005];
int b[100005];

int search(int num,int low,int high)
{
	int mid;
	while (low<=high)
	{
		mid=(low+high)/2;
		if (num>b[mid]) low=mid+1;
		else high=mid-1;
	}
	return low;
}

int dp(int n)
{
	int len,pos;
	b[1]=a[1];
	len=1;
	for (int i=2;i<=n;i++)
	{
		if (a[i]>b[len])
		{
			len=len+1;
			b[len]=a[i];
		}
		else
		{
			pos=search(a[i],1,len);
			b[pos]=a[i];
		}
	}
	return len;
}	

int main()
{
	freopen("A.IN","r",stdin);
	while (~scanf("%d",&n))
	{
		for (int i=1;i<=n;i++)
		{
			scanf("%d",&a[i]);
		}
		printf("%d\n",dp(n));
	}
	return 0;
}

B题是预处理+莫比乌斯反演，莫比乌斯反演用nlogn的容斥原理替换也可以，一开始想到先用每个数去判，看有多少个数是这个数的倍数，然后容斥原理，容斥用莫比乌斯反演去做可以更快，O(n^2)，但是感觉这样还是会超时，后来发现可以预处理，即对于一个数可以先求出这个数的所有因子，sqrt(n)求即可。所以时间复杂度为O(n^(3/2))

#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define maxn 10000
int pri[maxn+5];
int mu[maxn+5];
int p[10005];
int n;

void getMu()
{
	mu[1]=1;
	for (int i=2;i<=maxn;i++)
	{
		if (pri[i]==0)
		{
			pri[++pri[0]]=i;
			mu[i]=-1;
		}
		for (int j=1;j<=pri[0] && pri[j]<=maxn/i;j++)
		{
			pri[pri[j]*i]=1;
			if (i%pri[j]) mu[i*pri[j]]-=mu[i];
			else
			{
				mu[i*pri[j]]=0;
				break;
			}
		}
	}
}

int main()
{
	freopen("B.IN","r",stdin);
	getMu();
	while (~scanf("%d",&n))
	{
		memset(p,0,sizeof(p));
		for (int i=1;i<=n;i++)
		{
			int x;
			scanf("%d",&x);
			for (int j=1;j*j<=x;j++)
			{
				if (x%j==0)
				{
					if (j*j==x) p[j]++;
					else
					{
						p[j]++;
						p[x/j]++;
					}
				}
			}
		}
		ll ans=0;
		for (int i=1;i<=maxn;i++)
		{
			ans+=(ll)mu[i]*p[i]*(p[i]-1)*(p[i]-2)*(p[i]-3)/24;
		}
		printf("%I64d\n",ans);
	}
	return 0;
}

C题：是一道很裸的2-SAT的模板题

#include <bits/stdc++.h>
using namespace std;
#define ll long long
const int maxn=2005;
const int maxm=2000010;
struct Edge
{
	int to,next;
}edge[maxm];
int head[maxn],tot;
void init()
{
	tot=0;
	memset(head,-1,sizeof(head));
}
void addedge(int u,int v)
{
	edge[tot].to=v;
	edge[tot].next=head[u];
	head[u]=tot++;
}
bool vis[maxn];
int s[maxn],top;
bool dfs(int u)
{
	if (vis[u^1]) return false;
	if (vis[u]) return true;
	vis[u]=true;
	s[top++]=u;
	for (int i=head[u];i!=-1;i=edge[i].next)
		if (!dfs(edge[i].to))
			return false;
	return true;
}
bool Twosat(int n)
{
	memset(vis,false,sizeof(vis));
	for (int i=0;i<n;i+=2)
	{
		if (vis[i] || vis[i^1]) continue;
		top=0;
		if (!dfs(i))
		{
			while (top) vis[s[--top]]=false;
			if (!dfs(i^1)) return false;
		}
	}
	return true;
}
int n,m;

int main()
{
	freopen("C.IN","r",stdin);
	while (~scanf("%d%d",&n,&m))
	{
		init();
		for (int i=1;i<=m;i++)
		{
			int x,y;
			scanf("%d%d",&x,&y);
			if (x>0) x=(x<<1)-2;
			else x=((-x)<<1)-1;
			if (y>0) y=(y<<1)-2;
			else y=((-y)<<1)-1;
			addedge(x,y^1);
			addedge(y,x^1);
		}
		if (Twosat(2*n)) printf("1\n");
		else printf("0\n");
	}
	return 0;
}

E题水题，凸多边形求面积，直接抄模板即可

#include <bits/stdc++.h>
using namespace std;
#define ll long long
const double eps=1e-8;
struct point
{
	double x,y;
	point(){}
	point (double _x,double _y)
	{
		x=_x;
		y=_y;
	}
	double operator ^(const point &b) const
	{
		return x*b.y-y*b.x;
	}
}p[1000005];
int n;

double CalcArea(point p[],int n)
{
	double res=0;
	for (int i=0;i<n;i++)
	{
		res+=(p[i]^p[(i+1)%n])/2;
	}
	return fabs(res);
}

int main()
{
	freopen("E.IN","r",stdin);
	while (~scanf("%d",&n) && n!=0)
	{
		for (int i=0;i<n;i++)
		{
			scanf("%lf%lf",&p[i].x,&p[i].y);
		}
		double area;
		if (n<3) area=0;
		else area=CalcArea(p,n);
		printf("%.0f\n",area);
	}
	return 0;
}

F题并查集，一开始不会把一个元素独立出来，没办法，写了一个O(n)的居然过了

#include <bits/stdc++.h>
using namespace std;
#define ll long long
int n;
int f[10005];
int p[10005];

void init()
{
	for (int i=1;i<=n;i++)
	{
		f[i]=i;
	}
}

int find_father(int x)
{
	if (f[x]==x) return x;
	else return f[x]=find_father(f[x]);
}

void Union(int x,int y)
{
	f[find_father(y)]=find_father(x);
}

void erase(int x)
{
	if (f[x]==x)
	{
		int flag=-1;
		for (int i=1;i<=n;i++)
		{
			if (f[i]==x)
			{
				p[i]=1;
				flag=i;
			}
			else p[i]=0;
		}
		if (flag!=-1)
		{
			for (int i=1;i<=n;i++)
			{
				if (p[i]==1) f[i]=flag;
			}
		}
		f[x]=x;
	}
	else
	{
		int y=find_father(x);
		for (int i=1;i<=n;i++)
		{
			if (f[i]==x) f[i]=y;
		}
		f[x]=x;
	}
}

int main()
{
	freopen("F.IN","r",stdin);
	while (~scanf("%d",&n))
	{
		int n1=0,n2=0;
		init();
		while (1)
		{
			char s[10];
			scanf("%s",s);
			if (s[0]=='c')
			{
				int x,y;
				scanf("%d%d",&x,&y);
				if (find_father(x)!=find_father(y)) Union(x,y);
			}
			else if (s[0]=='q')
			{
				int x,y;
				scanf("%d%d",&x,&y);
				if (find_father(x)==find_father(y)) n1++;
				else n2++;
			}
			else if (s[0]=='d')
			{
				int x;
				scanf("%d",&x);
				erase(x);
			}
			else if (s[0]=='e')
			{
				break;
			}
		}
		printf("%d , %d\n",n1,n2);
	}
	return 0;
}

H题队友写的

#include <bits/stdc++.h>
using namespace std;
#define ll long long
int eular(int n)
{
	int ans=n;
	for(int i=2;i*i<=n;i++)
	{
		if(n%i==0)
		{
			ans-=ans/i;
			while(n%i==0)
				n/=i;
		}
	}
	if(n>1)ans-=ans/n;
	return ans;
}
int main()
{
	int T,x;
	freopen("H.IN","r",stdin);
	while(scanf("%d",&T)!=EOF)
	{
		long long a=1;
		while(T--)
		{
			scanf("%d",&x);
			a=a*(long long)eular(x);
			a=a%1000000007;
		}
		printf("%lld\n",a);
	}
	return 0;
}

I题队友过的

#include <bits/stdc++.h>
using namespace std;
#define ll long long
int n,c;
int a[100005];
int b[100005];
int ans = 0;

int main()
{
	freopen("I.IN","r",stdin);
	while (~scanf("%d%d",&n,&c))
	{
		if(n == 0 || c == 0) ans = 0;
		else {
			if(c % (2*n) == 0)
			{
				ans = c / (2*n);
			}
			else{
				int t1 = c / (2*n);
				int t2 = t1 + 1;
				if(t2*(c-t2*n) > t1*(c-t1*n))
				{
					ans = t2;	
				}
				else ans = t1;
			}
		}
		printf("%d\n",ans);
	}
	return 0;
}