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
",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!!!!!
}