A：考虑每对最大值最小值的贡献即可。
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 300010
#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,a[N],p[N],x,ans;
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]=read();
sort(a+1,a+n+1);
p[0]=1;for (int i=1;i<=n;i++) p[i]=2ll*p[i-1]%P;
for (int i=2;i<=n;i++)
{
ans=(ans+1ll*a[i]*(p[i-1]-1))%P;
x=(2ll*x+a[i-1])%P;
ans=(ans-x+P)%P;
}
cout<<ans;
return 0;
//NOTICE LONG LONG!!!!!
}
B：考虑每次都询问x=i y=i+1。这样得到的是最近的被标记点是在左边还是右边。这样二分一下可以确定一个标记点的位置。同样二分找第二个点，由于要使其不和第一个点重复，对第一个点的两边分别考虑，二分过程中需要保证第一个点不对询问造成影响，讨论一下一些边界即可。
#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 n,k;
char s[10];
bool ask(int x,int y)
{
cout<<1<<' '<<x<<' '<<y<<endl;
scanf("%s",s);
return s[0]=='T';
}
bool check(int x)
{
if (x<=0||x>n) return 0;
if (x==1)
{
return ask(1,2);
}
if (x==n)
{
return ask(n,n-1);
}
if (ask(x,x-1)&&ask(x,x+1)) return 1;else return 0;
}
signed main()
{
n=read(),k=read();
int x=0;
if (ask(n,n-1)) x=n;
else
{
int l=1,r=n-1;
while (l<r)
{
int mid=l+r>>1;
if (ask(mid,mid+1))
{
r=mid;
}
else
{
l=mid+1;
}
}
x=l;
}
int l=1,r=x-1;
while (l<r)
{
int mid=l+r>>1;
if (ask(mid,mid+1)) r=mid;
else l=mid+1;
}
if (check(l)&&x!=l) cout<<2<<' '<<l<<' '<<x<<endl;
else
{
int l=x+1,r=n;
while (l<r)
{
int mid=l+r>>1;
if (ask(mid+1,mid)) l=mid+1;
else r=mid;
}
if (check(x)&&x!=l) cout<<2<<' '<<x<<' '<<l<<endl;
else
{
if (x>1) if (ask(x-1,x)) {cout<<2<<' '<<x-1<<' '<<x<<endl;return 0;}
if (x<n) if (ask(x+1,x)) {cout<<2<<' '<<x<<' '<<x+1<<endl;return 0;}
}
}
return 0;
//NOTICE LONG LONG!!!!!
}
C：打表可以发现，对于2k*2k的矩阵，将其分成左上左下右上右下四部分，其中左上和右下相同，左下和右上相同，且左下是左上+2k-1复制而来。于是直接记搜即可。
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<map>
using namespace std;
#define ll long long
#define N 100010
#define P 1000000007
#define mp(x,y) make_pair((x),(y))
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;
}
typedef pair<int,int> pii;
struct data
{
int x,y,k;
bool operator <(const data&a) const
{
return x<a.x||x==a.x&&y<a.y||x==a.x&&y==a.y&&k<a.k;
}
bool operator ==(const data&a) const
{
return x==a.x&&y==a.y&&k==a.k;
}
};
map<data,pii> f;
pii operator +(pii a,pii b)
{
return mp((a.first+b.first)%P,(a.second+b.second)%P);
}
pii calc(int x,int y,int k)
{
if (x==0||y==0||k<=0) return mp(0,0);
if (x==1&&y==1) return mp(1,1);
k=min(k,x+y-1);
if (x>y) swap(x,y);
data qwq=(data){x,y,k};
if (f.find(qwq)!=f.end()) return f[qwq];
int u=1;
while ((u<<1)<y) u<<=1;
pii ans;
if (x<=u)
{
ans=calc(x,u,k);
pii tmp=calc(x,y-u,k-u);
ans=ans+tmp;ans.first=(ans.first+1ll*tmp.second*u)%P;
}
else
{
ans=calc(u,u,k);
ans=ans+calc(x-u,y-u,k);
pii tmp=calc(x-u,u,k-u);
ans=ans+tmp;ans.first=(ans.first+1ll*tmp.second*u)%P;
tmp=calc(u,y-u,k-u);
ans=ans+tmp;ans.first=(ans.first+1ll*tmp.second*u)%P;
}
return f[qwq]=ans;
}
int bruteforce(int x1,int y1,int x2,int y2,int k)
{
int ans=0;
for (int i=x1;i<=x2;i++)
for (int j=y1;j<=y2;j++)
if ((i-1^j-1)+1<=k) ans=(ans+(i-1^j-1)+1)%P;
return ans;
}
signed main()
{
int q=read();
while (q--)
{
int x1=read(),y1=read(),x2=read(),y2=read(),k=read();
int ans=0;
ans=(ans+calc(x2,y2,k).first)%P;
ans=(ans-calc(x1-1,y2,k).first+P)%P;
ans=(ans-calc(x2,y1-1,k).first+P)%P;
ans=(ans+calc(x1-1,y1-1,k).first)%P;
printf("%d
",ans);
//printf("%d
",bruteforce(x1,y1,x2,y2,k));
}
return 0;
//NOTICE LONG LONG!!!!!
}
D：搬上LIS的单调队列做法，设f[j]为前i位长度为j的单增序列的最后一位的最小值。考虑转移，显然f[j]=min(f[j],max(f[j-1]+1,l[i])) (f[j-1]<r[i])。注意到显然有f[j]<f[j+1]，所以不考虑l[i]限制的话，上述转移不需要取min，符合条件就可以直接赋值。那么二分找到l[i]和r[i]对应的边界，就相当于将区间平移并+1，然后区间内对l[i]取max。可以用splay维护。
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define lson tree[k].ch[0]
#define rson tree[k].ch[1]
#define N 300010
#define inf 1000000001
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],root,cnt;
struct data{int ch[2],fa,x,lazymax,lazyadd,s;
}tree[N<<1];
void up(int k){tree[k].s=tree[lson].s+tree[rson].s+1;}
void access(int k){while (k) up(k),k=tree[k].fa;}
int newnode(int v)
{
int k=++cnt;
tree[k].s=1;
tree[k].x=v;
return k;
}
void build(int &k,int l,int r)
{
if (l>r) return;
int mid=l+(r-l)/2;
if (mid<=0) k=newnode(mid);
else k=newnode(inf+mid);
build(lson,l,mid-1);
build(rson,mid+1,r);
tree[lson].fa=tree[rson].fa=k;
up(k);
}
int whichson(int k){return tree[tree[k].fa].ch[1]==k;}
void update(int k,int lazymax,int lazyadd)
{
tree[k].x+=lazyadd;
tree[k].x=max(tree[k].x,lazymax);
tree[k].lazyadd+=lazyadd;
tree[k].lazymax+=lazyadd;
tree[k].lazymax=max(tree[k].lazymax,lazymax);
}
void down(int k)
{
update(lson,tree[k].lazymax,tree[k].lazyadd);
update(rson,tree[k].lazymax,tree[k].lazyadd);
tree[k].lazymax=tree[k].lazyadd=0;
}
void push(int k){if (k!=root) push(tree[k].fa);down(k);}
void move(int k)
{
int fa=tree[k].fa,gf=tree[fa].fa,p=whichson(k);
if (fa!=root) tree[gf].ch[whichson(fa)]=k;tree[k].fa=gf;
tree[fa].ch[p]=tree[k].ch[!p];tree[tree[k].ch[!p]].fa=fa;
tree[k].ch[!p]=fa,tree[fa].fa=k;
up(fa),up(k);
}
void splay(int k,int rt)
{
push(k);
while (tree[k].fa!=rt)
{
int fa=tree[k].fa;
if (tree[fa].fa!=rt)
if (whichson(fa)^whichson(k)) move(k);
else move(fa);
move(k);
}
if (rt==0) root=k;
}
int find(int k,int x)
{
if (tree[lson].s-1==x) return k;
if (tree[lson].s-1>x) return find(lson,x);
else return find(rson,x-tree[lson].s-1);
}
int findsuf(int k,int x)
{
if (k==0) return -2;
down(k);
if (tree[k].x<=x) return findsuf(rson,x);
else
{
int t=findsuf(lson,x);
if (t==-2) return k;
else return t;
}
}
int findpre(int k,int x)
{
if (k==0) return -2;
down(k);
if (tree[k].x>=x) return findpre(lson,x);
else
{
int t=findpre(rson,x);
if (t==-2) return k;
else return t;
}
}
int getrank(int k)
{
splay(k,0);
return tree[lson].s-1;
}
int split(int l,int r)
{
int p=find(root,l-1),q=find(root,r+1);
splay(p,0);splay(q,p);
return tree[q].ch[0];
}
void del(int x)
{
int k=split(x,x);
tree[tree[k].fa].ch[0]=0;
access(tree[k].fa);
}
void ins(int x,int v)
{
int k=split(x,x);
lson=newnode(v);
tree[lson].fa=k;
access(lson);
}
int query(int x)
{
return tree[split(x,x)].x;
}
void add(int l,int r)
{
int k=split(l,r);
update(k,0,1);
}
void getmax(int l,int r,int x)
{
if (l>r) return;
int k=split(l,r);
update(k,x,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]=read(),b[i]=read();
build(root,-1,n+1);
for (int i=1;i<=n;i++)
{
int l=getrank(findsuf(root,a[i]));
int r=getrank(findpre(root,b[i]))+1;
if (l>r) continue;
del(r);ins(l-1,query(l-1));add(l,r);getmax(l,r,a[i]);
}
for (int i=n;i>=1;i--) if (query(i)<inf) {cout<<i;break;}
return 0;
//NOTICE LONG LONG!!!!!
}
E：众所周知φ(nm)=φ(n)*φ(m)*gcd(n,m)/φ(gcd(n,m))，于是莫比乌斯反演一下，最后大约要求f(D)=Σdis(u,v)*φ(au)*φ(av) (D|au,av)。考虑点分，可以转化为对每个点u求g(D,u)=2deepu*φ(au)*φ(av) (D|au,av)，每个点对所有因子加一下贡献即可。
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
#define ll long long
#define N 200010
#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,a[N],p[N],t;
int prime[N],phi[N],mobius[N],inv[N],cnt;
int deep[N],size[N],val[N],f[N],ans;
bool flag[N];
vector<int> factor[N];
struct data{int to,nxt;
}edge[N<<1];
void addedge(int x,int y){t++;edge[t].to=y,edge[t].nxt=p[x],p[x]=t;}
inline void inc(int &x,int y){x+=y;if (x>=P) x-=P;}
void make(int k,int from)
{
size[k]=1;
for (int i=p[k];i;i=edge[i].nxt)
if (!flag[edge[i].to]&&edge[i].to!=from)
{
deep[edge[i].to]=deep[k]+1;
make(edge[i].to,k);
size[k]+=size[edge[i].to];
}
}
int findroot(int k,int from,int s)
{
int mx=0;
for (int i=p[k];i;i=edge[i].nxt)
if (!flag[edge[i].to]&&edge[i].to!=from&&size[edge[i].to]>size[mx]) mx=edge[i].to;
if ((size[mx]<<1)>s) return findroot(mx,k,s);
else return k;
}
void work(int k,int from,int op)
{
for (int i=0;i<factor[a[k]].size();i++)
if (op==1) inc(val[factor[a[k]][i]],phi[a[k]]);
else inc(val[factor[a[k]][i]],P-phi[a[k]]);
for (int i=p[k];i;i=edge[i].nxt)
if (!flag[edge[i].to]&&edge[i].to!=from) work(edge[i].to,k,op);
}
void update(int k,int from)
{
for (int i=0;i<factor[a[k]].size();i++)
inc(f[factor[a[k]][i]],2ll*deep[k]*phi[a[k]]%P*val[factor[a[k]][i]]%P);
for (int i=p[k];i;i=edge[i].nxt)
if (!flag[edge[i].to]&&edge[i].to!=from) update(edge[i].to,k);
}
void solve(int k)
{
make(k,k);
k=findroot(k,k,size[k]);
flag[k]=1;deep[k]=0;make(k,k);
work(k,k,1);
for (int i=p[k];i;i=edge[i].nxt)
if (!flag[edge[i].to])
{
work(edge[i].to,edge[i].to,-1);
update(edge[i].to,edge[i].to);
work(edge[i].to,edge[i].to,1);
}
work(k,k,-1);
for (int i=p[k];i;i=edge[i].nxt)
if (!flag[edge[i].to]) solve(edge[i].to);
}
//f(D)=��2deepu*��(au)*��(av) (D|au,D|av)
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]=read();
for (int i=1;i<n;i++)
{
int x=read(),y=read();
addedge(x,y),addedge(y,x);
}
flag[1]=1;mobius[1]=phi[1]=1;
for (int i=2;i<=n;i++)
{
if (!flag[i]) prime[++cnt]=i,mobius[i]=-1,phi[i]=i-1;
for (int j=1;j<=cnt&&prime[j]*i<=n;j++)
{
flag[prime[j]*i]=1;
if (i%prime[j]==0) {phi[prime[j]*i]=phi[i]*prime[j];break;}
mobius[prime[j]*i]=-mobius[i];
phi[prime[j]*i]=phi[i]*(prime[j]-1);
}
}
inv[1]=1;for (int i=2;i<=n;i++) inv[i]=P-1ll*(P/i)*inv[P%i]%P;
memset(flag,0,sizeof(flag));
for (int i=1;i<=n;i++)
for (int j=i;j<=n;j+=i)
factor[j].push_back(i);
solve(1);
for (int i=1;i<=n;i++)
for (int j=i;j<=n;j+=i)
ans=(ans+1ll*f[j]*(P+mobius[j/i])%P*i%P*inv[phi[i]])%P;
cout<<1ll*ans*inv[n]%P*inv[n-1]%P;
return 0;
//NOTICE LONG LONG!!!!!
}