Codeforces Round #488 Div. 1

　　A：枚举每个点判断是否同时在两个正方形中即可。

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
	int x=0,f=1;char c=getchar();
	while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
	while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
	return x*f;
}
struct data
{
	int x,y;
	void get(){x=read();y=read();}
}a[4],b[4];
bool isin1(int x,int y)
{
	int LEFT=1000,RIGHT=-1000,UP=-1000,DOWN=1000;
	for (int i=0;i<4;i++)
	LEFT=min(LEFT,a[i].x),
	RIGHT=max(RIGHT,a[i].x),
	UP=max(UP,a[i].y),
	DOWN=min(DOWN,a[i].y);
	if (LEFT<=x&&RIGHT>=x&&UP>=y&&DOWN<=y) return 1;
	else return 0;
}
bool isin2(int x,int y)
{
	int LEFT=1000,RIGHT=-1000,UP=-1000,DOWN=1000;
	for (int i=0;i<4;i++)
	LEFT=min(LEFT,b[i].x),
	RIGHT=max(RIGHT,b[i].x),
	UP=max(UP,b[i].y),
	DOWN=min(DOWN,b[i].y);
	int mid1=(LEFT+RIGHT)/2,mid2=(UP+DOWN)/2;
	if (abs(x-mid1)+abs(y-mid2)<=mid1-LEFT) return 1;
	else return 0;
}
void win(){cout<<"Yes";exit(0);}
signed main()
{
#ifndef ONLINE_JUDGE
	freopen("a.in","r",stdin);
	freopen("a.out","w",stdout);
#endif
	for (int i=0;i<4;i++) a[i].get();
	for (int i=0;i<4;i++) b[i].get();
	for (int i=-100;i<=100;i++)
		for (int j=-100;j<=100;j++)
		if (isin1(i,j)&&isin2(i,j)) win();
	cout<<"No";
	return 0;
	//NOTICE LONG LONG!!!!!
}

　　B：莫名其妙的题。首先要知道这俩人只想知道他们的共同数字是什么，而不是对方的数字对是什么。先判断啥都不知道能否确定该数字，只要考虑每一对仅有一个数字相同的数对，如果其共同数字都相同，即可确定该数为答案。否则仍考虑每一对可能的数对，如果某一种情况下有一方无法确定共同数字，即输出-1。最后输出0即可。

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 100
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
	int x=0,f=1;char c=getchar();
	while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
	while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
	return x*f;
}
int n,m;
struct data{int x,y;
}a[N],b[N];
int calc(data a,data b)
{
	return (a.x==b.x)+(a.x==b.y)+(a.y==b.x)+(a.y==b.y);
}
int same(data a,data b)
{
	if (a.x==b.x||a.x==b.y) return a.x;
	else return a.y;
}
signed main()
{
#ifndef ONLINE_JUDGE
	freopen("a.in","r",stdin);
	freopen("a.out","w",stdout);
#endif
	n=read(),m=read();
	for (int i=1;i<=n;i++)
	{
		a[i].x=read(),a[i].y=read();
		if (a[i].x>a[i].y) swap(a[i].x,a[i].y);
	}
	for (int i=1;i<=m;i++)
	{
		b[i].x=read(),b[i].y=read();
		if (b[i].x>b[i].y) swap(b[i].x,b[i].y);
	}
	int ans=0,uuu;
	for (int i=1;i<=n;i++)
		for (int j=1;j<=m;j++)
		if (calc(a[i],b[j])==1) ans++,uuu=same(a[i],b[j]);
	bool flag=0;
	for (int i=1;i<=n;i++)
		for (int j=1;j<=m;j++)
		if (calc(a[i],b[j])==1&&same(a[i],b[j])!=uuu) flag=1;
	if (!flag) {cout<<uuu;return 0;}
	ans=0;
	for (int i=1;i<=n;i++)
		for (int j=1;j<=m;j++)
		if (calc(a[i],b[j])==1)
		{
			for (int x=1;x<=n;x++)
			if (calc(a[x],b[j])==1&&same(a[x],b[j])!=same(a[i],b[j])) {cout<<-1;return 0;}
			for (int x=1;x<=m;x++)
			if (calc(a[i],b[x])==1&&same(a[i],b[x])!=same(a[i],b[j])) {cout<<-1;return 0;}
		}
	cout<<0;
	return 0;
	//NOTICE LONG LONG!!!!!
}

　　C：显然答案位置至少在一对飞船的交点上，这个数量级是O(nm)的。预处理每个位置能消灭哪些飞船，bitset存储，然后暴力枚举每一对位置，bitset上or一下count一下就完了。

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<bitset>
using namespace std;
#define ll long long
#define N 122
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
	int x=0,f=1;char c=getchar();
	while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
	while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
	return x*f;
}
int n,m,a[N],b[N],ans;
bitset<N> f[N>>1][N>>1];
signed main()
{
#ifndef ONLINE_JUDGE
	freopen("a.in","r",stdin);
	freopen("a.out","w",stdout);
#endif
	n=read(),m=read();
	for (int i=1;i<=n;i++) a[i]=read();
	for (int i=1;i<=m;i++) b[i]=read();
	for (int i=1;i<=n;i++)
		for (int j=1;j<=m;j++)
		{
			for (int x=1;x<=n;x++)
				for (int y=1;y<=m;y++)
				if (a[i]-a[x]==b[y]-b[j]) f[i][j][x]=1,f[i][j][y+n]=1;
		}
	for (int i=1;i<=n;i++)
		for (int j=1;j<=m;j++)
			for (int x=1;x<=n;x++)
				for (int y=1;y<=m;y++)
				ans=max(ans,(int)(f[i][j]|f[x][y]).count());
	cout<<ans;			
	return 0;
	//NOTICE LONG LONG!!!!!
}

　　D：显然二分答案。相当于每将一个功率=x的任务放在第一批，就为功率<x的任务提供了一个垃圾桶。于是按功率从大到小排序，设f[i][j]为前i个任务剩余j个垃圾桶的最小值。唯一的问题是功率相同的任务间不能提供垃圾桶，状态再加一维处理即可。

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 55
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
	int x=0,f=1;char c=getchar();
	while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
	while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
	return x*f;
}
int n;
ll f[N][N][N];
struct data
{
	ll x,y;
	bool operator <(const data&a) const
	{
		return x>a.x||x==a.x&&y<a.y;
	}
}a[N];
bool check(ll k)
{
	memset(f,42,sizeof(f));
	f[0][0][0]=0;
	for (int i=1;i<=n;i++)
	{
		int t=i;
		while (t<n&&a[t+1].x==a[i].x) t++;
		for (int j=i;j<=t;j++)
		{
			for (int x=0;x<=n;x++)
				for (int y=0;y<=n;y++)
				{
					f[j][x][y]=f[j-1][x+1][y];
					if (y) f[j][x][y]=min(f[j][x][y],f[j-1][x][y-1]+a[j].x-k*a[j].y);
				}
		}
		i=t;
		for (int j=n;j>=0;j--)
			for (int x=0;x<=j;x++)
			f[i][j][0]=min(f[i][j][0],f[i][j-x][x]);
	}
	for (int i=0;i<=n;i++) if (f[n][i][0]<=0) return 1;
	return 0;
}
signed main()
{
#ifndef ONLINE_JUDGE
	freopen("a.in","r",stdin);
	freopen("a.out","w",stdout);
#endif
	n=read();
	for (int i=1;i<=n;i++) a[i].x=1000ll*read();
	for (int i=1;i<=n;i++) a[i].y=read();
	sort(a+1,a+n+1);
	ll l=0,r=10000000000000000ll,ans;
	while (l<=r)
	{
		ll mid=l+r>>1;
		if (check(mid)) ans=mid,r=mid-1;
		else l=mid+1;
	}
	cout<<ans;
	return 0;
	//NOTICE LONG LONG!!!!!
}

　　E：显然每个数有用的信息仅仅是其和x的大小，根据这个设为01后，即相当于求和为k的区间个数。又可以转换为前缀和之差为k的区间个数。裸的FFT。

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 1100010
#define double long double
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
	int x=0,f=1;char c=getchar();
	while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
	while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
	return x*f;
}
int n,m,f[N],r[N];
ll ans[N];
const double PI=3.14159265358979324;
struct complex
{
    double x,y;
    complex operator +(const complex&a) const
    {
        return (complex){x+a.x,y+a.y};
    }
    complex operator -(const complex&a) const
    {
        return (complex){x-a.x,y-a.y};
    }
    complex operator *(const complex&a) const
    {
        return (complex){x*a.x-y*a.y,x*a.y+y*a.x};
    }
}a[N],b[N];
void DFT(int n,complex *a,int p)
{
	for (int i=0;i<n;i++) r[i]=(r[i>>1]>>1)|(i&1)*(n>>1);
    for (int i=0;i<n;i++) if (i<r[i]) swap(a[i],a[r[i]]);
    for (int i=2;i<=n;i<<=1)
    {
        complex wn=(complex){cos(2*PI/i),p*sin(2*PI/i)};
        for (int j=0;j<n;j+=i)
        {
            complex w=(complex){1,0};
            for (int k=j;k<j+(i>>1);k++,w=w*wn)
            {
                complex x=a[k],y=w*a[k+(i>>1)];
                a[k]=x+y,a[k+(i>>1)]=x-y;
            }
        }
    }
}
void mul(int n,complex *a,complex *b)
{
    DFT(n,a,1),DFT(n,b,1);
    for (int i=0;i<n;i++) a[i]=a[i]*b[i];
    DFT(n,a,-1);
    for (int i=0;i<n;i++) a[i].x=a[i].x/n;
}
signed main()
{
#ifndef ONLINE_JUDGE
	freopen("a.in","r",stdin);
	freopen("a.out","w",stdout);
#endif
	n=read(),m=read();
	for (int i=1;i<=n;i++) f[i]=(read()<m);
	for (int i=1;i<=n;i++) f[i]+=f[i-1];
	for (int i=0;i<=n;i++) a[f[i]].x++,b[n-f[i]].x++;
	int t=1;while (t<=(n<<1)) t<<=1;
	mul(t,a,b);
	for (int i=0;i<=n;i++) ans[i]=(ll)(a[i+n].x+0.5);
	ans[0]-=n+1;ans[0]/=2;
	for (int i=0;i<=n;i++) printf("%I64d ",ans[i]);
	return 0;
	//NOTICE LONG LONG!!!!!
}

　　居然还有我能在场上做五题的div1