突然发现上一场edu忘记写了（
A：签到。
#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],ans;
bool flag[N];
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=a[i];
ans++;
while (i<x)
{
i++;
x=max(x,a[i]);
}
}
cout<<ans;
return 0;
//NOTICE LONG LONG!!!!!
}
B：显然仅当最左为>或最右为<时才能消去整个字符串。于是找到最左的>和最右的<取min即可。
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 110
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 T,n;
char s[N];
signed main()
{
#ifndef ONLINE_JUDGE
freopen("a.in","r",stdin);
freopen("a.out","w",stdout);
#endif
T=read();
while (T--)
{
n=read();
scanf("%s",s+1);
int ans=n-1;
int last=0;
for (int i=n;i>=1;i--) if (s[i]=='<') {last=i;break;}
ans=min(ans,n-last);
int first=n;
for (int i=1;i<=n;i++) if (s[i]=='>') {first=i;break;}
ans=min(ans,first-1);
cout<<ans<<endl;
}
return 0;
//NOTICE LONG LONG!!!!!
}
C：枚举beauty最小值，则显然应该选择beauty不小于该数的前k大。堆维护。
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<queue>
#include<vector>
using namespace std;
#define ll long long
#define N 300010
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;
bool operator <(const data&a) const
{
return y<a.y;
}
}a[N];
priority_queue<int,vector<int>,greater<int> > q;
signed main()
{
#ifndef ONLINE_JUDGE
freopen("c.in","r",stdin);
freopen("c.out","w",stdout);
#endif
n=read(),m=read();
for (int i=1;i<=n;i++) a[i].x=read(),a[i].y=read();
sort(a+1,a+n+1);
ll tot=0,ans=0;
for (int i=n;i>=1;i--)
{
q.push(a[i].x);tot+=a[i].x;
if (n-i+1>m) tot-=q.top(),q.pop();
ans=max(ans,tot*a[i].y);
}
cout<<ans;
return 0;
//NOTICE LONG LONG!!!!!
}
D：大胆猜想不用证明。即让1与所有相邻点配对。
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<queue>
#include<vector>
using namespace std;
#define ll long long
#define N 510
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,f[N][N];
signed main()
{
#ifndef ONLINE_JUDGE
freopen("c.in","r",stdin);
freopen("c.out","w",stdout);
#endif
n=read();
int ans=0;
for (int i=2;i<n;i++) ans+=i*(i+1);
cout<<ans;
return 0;
//NOTICE LONG LONG!!!!!
}
E：不存在奇回文串的充要条件是不存在长度为3的回文串。黑白染色（也就是奇偶分开考虑），则相当于相邻数不同。已确定的数将序列分割成若干区间，显然这些区间之间无关。然后随便dp一发。
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define int long long
#define N 200010
#define P 998244353
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],c[N],f[N],g[N],ans;
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;
}
int calc(int *a,int n)
{
int ans=1;
bool flag=0;
for (int i=1;i<=n;i++) if (a[i]!=-1) {flag=1;break;}
if (!flag)
{
ans=m;for (int i=2;i<=n;i++) ans=1ll*ans*(m-1)%P;return ans;
}
int l=1,r=n;
if (a[1]==-1)
{
while (a[l+1]==-1) l++;
for (int i=1;i<=l;i++) ans=1ll*ans*(m-1)%P;
l++;
}
if (a[n]==-1)
{
while (a[r-1]==-1) r--;
for (int i=n;i>=r;i--) ans=1ll*ans*(m-1)%P;
r--;
}
for (int i=l;i<=r;i++)
{
if (i==r) break;
int t=i+1;
while (a[t]==-1) t++;
if (a[i]==a[t]) ans=1ll*ans*f[t-i]%P;
else ans=1ll*ans*g[t-i]%P;
i=t-1;
}
return ans;
}
signed main()
{
#ifndef ONLINE_JUDGE
freopen("c.in","r",stdin);
freopen("c.out","w",stdout);
#endif
n=read(),m=read();
for (int i=1;i<=n;i++) a[i]=read();
f[0]=1;for (int i=1;i<=n;i++) f[i]=(ksm(m-1,i-1)+P-f[i-1])%P;
g[0]=0;for (int i=1;i<=n;i++) g[i]=(ksm(m-1,i-1)+P-g[i-1])%P;
int u=0,v=0;
for (int i=1;i<=n;i++)
if (i&1) b[++u]=a[i];
else c[++v]=a[i];
ans=1ll*calc(b,u)*calc(c,v)%P;
cout<<ans;
return 0;
//NOTICE LONG LONG!!!!!
}
F：https://www.cnblogs.com/Gloid/p/9392532.html 用上这一场的D的结论（woc怎么看着这么羞耻），然后相当于动态图连通性问题，码了个线段树分治+带撤销并查集，得卡卡常。
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<map>
#include<vector>
using namespace std;
#define ll long long
#define int long long
#define N 300010
#define mp(x,y) make_pair((x),(y))
#define left(x) (x)
#define right(x) ((x)+n)
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;
int n,t,fa[N<<1],lsize[N<<1],rsize[N<<1],L[N<<2],R[N<<2],lazy[N<<2];
ll ans,Q[N];
map<int,int> id[N];
vector<pii> edge[N<<2];
struct data{int x,y,z,l,r;
}a[N];
vector<data> stk[N<<2];
void build(int k,int l,int r)
{
L[k]=l,R[k]=r;
if (l==r) return;
int mid=l+r>>1;
build(k<<1,l,mid);
build(k<<1|1,mid+1,r);
}
void add(int k,int l,int r,int x,int y)
{
if (L[k]==l&&R[k]==r) {edge[k].push_back(mp(x,y));return;}
int mid=L[k]+R[k]>>1;
if (r<=mid) add(k<<1,l,r,x,y);
else if (l>mid) add(k<<1|1,l,r,x,y);
else add(k<<1,l,mid,x,y),add(k<<1|1,mid+1,r,x,y);
}
int find(int x){return fa[x]==x?x:find(fa[x]);}
ll calc(int x){return 1ll*lsize[x]*rsize[x];}
void solve(int k)
{
lazy[k]=ans;
for (int i=0;i<edge[k].size();i++)
{
int x=edge[k][i].first,y=edge[k][i].second;
int p=find(left(x)),q=find(right(y));
if (p!=q)
{
ans-=calc(p);ans-=calc(q);
if (lsize[p]+rsize[p]<rsize[p]+rsize[q]) swap(p,q);
stk[k].push_back((data){q,p,fa[q],lsize[p],rsize[p]});
fa[q]=p;lsize[p]+=lsize[q];rsize[p]+=rsize[q];
ans+=calc(p);
}
}
if (L[k]==R[k]) Q[L[k]]=ans;
if (L[k]<R[k]) solve(k<<1),solve(k<<1|1);
ans=lazy[k];
for (int i=((int)stk[k].size())-1;i>=0;i--)
{
fa[stk[k][i].x]=stk[k][i].z;
lsize[stk[k][i].y]=stk[k][i].l;
rsize[stk[k][i].y]=stk[k][i].r;
}
}
signed main()
{
#ifndef ONLINE_JUDGE
freopen("f.in","r",stdin);
freopen("f.out","w",stdout);
#endif
n=read();
for (int i=1;i<=n;i++) fa[left(i)]=left(i),fa[right(i)]=right(i),lsize[left(i)]=1,rsize[right(i)]=1;
for (int i=1;i<=n;i++)
{
int x=read(),y=read();
if (id[x][y]==0)
{
id[x][y]=++t;
a[t].x=x,a[t].y=y,a[t].l=i;
}
else
{
a[id[x][y]].r=i-1;
id[x][y]=0;
}
}
for (int i=1;i<=t;i++) if (a[i].r==0) a[i].r=n;
build(1,1,n);
for (int i=1;i<=t;i++) add(1,a[i].l,a[i].r,a[i].x,a[i].y);
solve(1);
for (int i=1;i<=n;i++) printf("%I64d ",Q[i]);
return 0;
//NOTICE LONG LONG!!!!!
}
G：没补。
小小号的第一场。因为E一直wa非常自闭。result：rank 98 rating +214