Codeforces Round #440 Div. 1

　　A：显然应该尽量拆成4。如果是奇数，先拆一个9出来即可。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 100010
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 q,n;
signed main()
{
#ifndef ONLINE_JUDGE
	freopen("a.in","r",stdin);
	freopen("a.out","w",stdout);
#endif
	q=read();
	while (q--)
	{
		n=read();int ans=0;
		if (n<4) {cout<<-1<<endl;continue;}
		if (n&1)
		{
			if (n<9) {cout<<-1<<endl;continue;}
			n-=9;ans++;
			if (n==2) {cout<<-1<<endl;continue;}
		}
		ans+=n/4;
		printf("%d\n",ans);
	}
	return 0;
	//NOTICE LONG LONG!!!!!
}

　　B：注意到由异或的消去性，事实上可以通过2n次询问得到所有n²种询问的结果。然后若第一个数确定，整个排列就确定了，暴力枚举即可。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 10010
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,a[N],b[N],c[N],d[N];
signed main()
{
	n=read();
	int u;
	for (int i=0;i<n;i++)
	{
		cout<<'?'<<' '<<0<<' '<<i<<endl;
		cin>>c[i];
		if (c[i]==0) u=i;
	}
	for (int i=0;i<n;i++)
	{
		cout<<'?'<<' '<<i<<' '<<u<<endl;
		cin>>a[i];
	}
	cout<<"!"<<endl;int ans=0;
	for (int i=0;i<n;i++)
	{
		for (int j=0;j<n;j++) d[j]=i^c[j];
		bool flag=0;
		for (int j=0;j<n;j++)
		{
			b[j]=i^a[j];
			if (b[j]==u&&i!=j)  {flag=1;break;}
		}
		for (int j=0;j<n;j++) if (b[d[j]]!=j) {flag=1;break;}
		if (!flag)
		{
			sort(b,b+n);
			for (int j=0;j<n;j++) if (b[j]!=j) {flag=1;break;}
			if (!flag) ans++;
		}
	}
	cout<<ans<<endl;
	for (int i=0;i<n;i++)
	{
		for (int j=0;j<n;j++) d[j]=i^c[j];
		bool flag=0;
		for (int j=0;j<n;j++)
		{
			b[j]=i^a[j];
			if (b[j]==u&&i!=j)  {flag=1;break;}
		}
		for (int j=0;j<n;j++) if (b[d[j]]!=j) {flag=1;break;}
		if (!flag)
		{
			for (int j=0;j<n;j++) cout<<(i^a[j])<<' ';return 0;
		}
	}
	return 0;
	//NOTICE LONG LONG!!!!!
}

　　C：同行同列相邻点连边，对每个连通块分别考虑。如果连通块构成一棵树，显然只要不选择所有直线，每种方案都能被构造出来。如果有环，考虑环上所有直线都可以被选中，而增加一个点和其相连至多会增加一条未选中直线，令该点选择该直线即可，所以所有直线都能同时被选中，所以方案都能被构造出来。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<map>
#include<cassert>
using namespace std;
#define ll long long
#define N 100010
#define P 1000000007
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,ans=1,p[N],X[N],Y[N],t,u,v;
map<int,int> row,line;
bool flag[N];
struct data2{int to,nxt;
}edge[N<<4];
struct data
{
	int i,x,y;
	bool operator <(const data&a) const
	{
		return x<a.x;
	}
}a[N];
bool cmp1(const data&a,const data&b)
{
	return a.x<b.x;
}
void addedge(int x,int y)
{
	t++;edge[t].to=y,edge[t].nxt=p[x],p[x]=t;
}
int ksm(int a,int k)
{
	int s=1;
	for (;k;k>>=1,a=1ll*a*a%P) if (k&1) s=1ll*s*a%P;
	return s;
}
void dfs(int k,int from)
{
	flag[k]=1;
	if (row.find(X[k])==row.end()) row[X[k]]=1,u++;
	if (line.find(Y[k])==line.end()) line[Y[k]]=1,u++;
	for (int i=p[k];i;i=edge[i].nxt)
	if (edge[i].to!=from)
		if (flag[edge[i].to]) v=0;
		else dfs(edge[i].to,k);
}
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++) X[i]=a[i].x=read(),Y[i]=a[i].y=read(),a[i].i=i;
	sort(a+1,a+n+1);
	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++) addedge(a[j].i,a[j+1].i),addedge(a[j+1].i,a[j].i);
		i=t;
	}
	for (int i=1;i<=n;i++) swap(a[i].x,a[i].y);
	sort(a+1,a+n+1);
	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++) addedge(a[j].i,a[j+1].i),addedge(a[j+1].i,a[j].i);
		i=t;
	}
	for (int i=1;i<=n;i++)
	if (!flag[i])
	{
		row.clear();line.clear();
		u=0;v=1;dfs(i,i);
		ans=1ll*ans*(ksm(2,u)-v)%P;
	}
	cout<<ans;
	return 0;
	//NOTICE LONG LONG!!!!!
}

　　D：若两数互质，考虑其最小质因子，若其乘积<=n则显然两数距离为2，否则注意到可以通过2*质因子来过渡，所以只要两个数都不是>=n/2的质数或1距离就为3，否则为0。大力容斥大讨论即可。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 10000010
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,prime[N],phi[N],p[N],v[N],cnt;
ll ans;
bool flag[N],f[N];
signed main()
{
#ifndef ONLINE_JUDGE
	freopen("d.in","r",stdin);
	freopen("d.out","w",stdout);
#endif
	n=read();
	flag[1]=1;phi[1]=1;
	for (int i=2;i<=n;i++)
	{
		if (!flag[i]) prime[++cnt]=i,phi[i]=i-1,p[i]=i,v[i]=-1;
		for (int j=1;j<=cnt&&prime[j]*i<=n;j++)
		{
			flag[prime[j]*i]=1;
			p[prime[j]*i]=prime[j];
			if (i%prime[j]==0) {phi[prime[j]*i]=phi[i]*prime[j];break;}
			phi[prime[j]*i]=phi[i]*(prime[j]-1);
		}
	}
	ll tot=0;
	for (int i=2;i<=n;i++) tot+=phi[i],v[p[i]]++;
	ans+=1ll*n*(n-1)/2-tot; //dis=1
	int u=1;for (int i=1;i<=cnt;i++) if (prime[i]>n/2) u++;
	tot-=1ll*u*(n-1);tot+=1ll*u*(u-1)/2; //dis=0 tot=dis2+dis3
	ans+=tot*3;
	/*for (int i=2;i<=n;i++)
		for (int j=2;j<i;j++)
		if (gcd(i,j)==1&&1ll*p[i]*p[j]<=n) ans--;
	cout<<ans;return 0;*/
	for (int i=1;i<=cnt;i++)
	if (prime[i]<=n/2)
	{
		for (int j=1;j<i;j++)
		if (prime[i]*prime[j]<=n) ans--;
		else break;
		for (int j=1;j<=cnt;j++)
		if (prime[j]<=n/prime[i]) ans-=v[prime[j]];
		else break;
		for (int j=prime[i]*2;j<=n;j+=prime[i])
		if (p[j]<=n/prime[i]) ans++;
	}
	else break;
	u=0;
	for (int i=2;i<=n;i++)
	if (flag[i])
	{
		ans-=i-2;
		ans+=i-1-phi[i];
		ans+=u;
		int v=i;while (v>1) ans-=!f[p[v]],f[p[v]]=1,v/=p[v];
		v=i;while (v>1) f[p[v]]=0,v/=p[v];
	}
	else u++;
	cout<<ans;
	return 0;
	//NOTICE LONG LONG!!!!!
}