【CSP-S 2019模拟】题解

T1：

$O(n^2)$枚举后用逆元算第三个即可
用$hash table$存即可
脑残写了个奇奇妙妙的$n^2log$$wa3$个$T3$个

#include<bits/stdc++.h>
using namespace std;
#define cs const
#define re register
#define pb push_back
#define pii pair<int,int>
#define fi first
#define se second
#define bg begin
#define ll long long
cs int RLEN=1<<20|1;
inline char gc(){
	static char ibuf[RLEN],*ib,*ob;
	(ib==ob)&&(ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
	return (ib==ob)?EOF:*ib++;
}
inline int read(){
	char ch=gc();
	int res=0;bool f=1;
	while(!isdigit(ch))f^=ch=='-',ch=gc();
	while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
	return f?res:-res;
}
template<class tp>inline void chemx(tp &a,tp b){a<b?a=b:0;}
template<class tp>inline void chemn(tp &a,tp b){a>b?a=b:0;}
int mod;
inline int add(int a,int b){a+=b-mod;return a+(a>>31&mod);}
inline void Add(int &a,int b){a+=b-mod,a+=a>>31&mod;}
inline int dec(int a,int b){a-=b;return a+(a>>31&mod);}
inline void Dec(int &a,int b){a-=b,a+=a>>31&mod;}
inline int mul(int a,int b){static ll r;r=1ll*a*b;return r>=mod?r%mod:r;}
inline void Mul(int &a,int b){static ll r;r=1ll*a*b;a=r>=mod?r%mod:r;}
inline int ksm(int a,int b,int res=1){for(;b;b>>=1,Mul(a,a))(b&1)&&(Mul(res,a),1);return res;}
inline int Inv(int x){return ksm(x,mod-2);}
struct Map{
	static cs int Mod=2718343;
	int key[Mod],val[Mod];
	Map(){memset(key,-1,sizeof(key));}
	inline int nxt(int p){
		return p+1==Mod?0:p+1;
	}
	inline void insert(int k){
		int p=k%Mod;
		for(;~key[p]&&key[p]!=k;p=nxt(p));
		key[p]=k,val[p]++;
	}
	inline int find(int k){
		int p=k%Mod;
		for(;~key[p]&&key[p]!=k;p=nxt(p));
		if(key[p]!=k)return 0;
		return val[p];
	}
}mp,mp2;
cs int N=2505;
int n,a[N],inv[N],cnt,ans;
int main(){
	#ifdef Stargazer
	freopen("lx.in","r",stdin);
	#endif
	n=read(),mod=read();
	for(int i=1;i<=n;i++){
		a[i]=read();
	}
	sort(a+1,a+n+1);
	for(int i=1;i<=n;i++){
		mp2.insert(a[i]);
	}
	cnt=unique(a+1,a+n+1)-a-1;
	mp.insert(a[1]%mod);
	for(int i=1;i<=cnt;i++)inv[i]=Inv(a[i]%mod);
	for(int i=2;i<=cnt;i++){
		for(int j=i+1;j<=cnt;j++){
			int iv=mul(inv[i],inv[j]);
			ans+=mp.find(iv);
		}
		mp.insert(a[i]%mod);
	}
	for(int i=1;i<=cnt;i++)if(mp2.find(a[i])>1){
		int iv=mul(inv[i],inv[i]);
		ans+=mp.find(iv);
		if(mul(mul(a[i],a[i]),a[i])==1)ans--;
	}
	for(int i=1;i<=cnt;i++){
		if(mp2.find(a[i])>=3&&mul(mul(a[i],a[i]),a[i])==1)ans++;
	}
	cout<<ans;return 0;
}

T2：

发现大概就是一个有限制的二维偏序
从大往小排序后
$f[i][j]$表示前$i$个放了$j$个的最长长度
枚举上一个更新即可
发现第二维没有用处
可以把前面的答案离线加进去线段树维护$max$即可

#include<bits/stdc++.h>
using namespace std;
#define cs const
#define re register
#define pb push_back
#define pii pair<int,int>
#define fi first
#define se second
#define bg begin
#define ll long long
cs int RLEN=1<<20|1;
inline char gc(){
	static char ibuf[RLEN],*ib,*ob;
	(ib==ob)&&(ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
	return (ib==ob)?EOF:*ib++;
}
inline int read(){
	char ch=gc();
	int res=0;bool f=1;
	while(!isdigit(ch))f^=ch=='-',ch=gc();
	while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
	return f?res:-res;
}
template<class tp>inline void chemx(tp &a,tp b){a<b?a=b:0;}
template<class tp>inline void chemn(tp &a,tp b){a>b?a=b:0;}
cs int N=100005;
int n,m,t[N];
struct node{
	int w,v;
	friend inline bool operator <(cs node &a,cs node &b){
		return a.w==b.w?a.v>b.v:a.w>b.w;
	}
}p[N];
namespace Seg{
	int mx[N<<2];
	#define lc (u<<1)
	#define rc ((u<<1)|1)
	#define mid ((l+r)>>1)
	void clear(int u,int l,int r){
		mx[u]=0;if(l==r)return;
		clear(lc,l,mid),clear(rc,mid+1,r);
	}
	inline void pushup(int u){
		mx[u]=max(mx[lc],mx[rc]);
	}
	void update(int u,int l,int r,int p,int k){
		if(l==r){chemx(mx[u],k);return;}
		if(p<=mid)update(lc,l,mid,p,k);
		else update(rc,mid+1,r,p,k);
		pushup(u);
	}
	int query(int u,int l,int r,int st,int des){
		if(st<=l&&r<=des)return mx[u];
		int res=0;
		if(st<=mid)chemx(res,query(lc,l,mid,st,des));
		if(mid<des)chemx(res,query(rc,mid+1,r,st,des));
		return res;
	}
	#undef lc
	#undef rc
	#undef mid
}
vector<int>upd[N];
int f[N],b[N],cnt;
inline void solve(){
	n=read();
	for(int i=1;i<=n;i++)p[i].w=read(),p[i].v=read(),b[i]=p[i].v;
	sort(p+1,p+n+1);
	sort(b+1,b+n+1),cnt=unique(b+1,b+n+1)-b-1;
	for(int i=1;i<=n;i++)p[i].v=lower_bound(b+1,b+cnt+1,p[i].v)-b;
	m=read();
	for(int i=1;i<=m;i++)t[i]=read();
	sort(t+1,t+m+1);
	reverse(t+1,t+m+1);
	for(int pos=0,i=1;i<=n;i++){
		while(pos<m&&p[i].w<=t[pos+1]){
			for(int j=0;j<upd[pos].size();j++)
			Seg::update(1,1,cnt,p[upd[pos][j]].v,f[upd[pos][j]]);
			pos++;
		}
		if(p[i].w<=t[pos]){
			f[i]=Seg::query(1,1,cnt,p[i].v,cnt)+1;
			if(f[i]<pos)Seg::update(1,1,cnt,p[i].v,f[i]);
			else upd[f[i]+1].pb(i);
		}
	}
	int res=0;
	for(int i=1;i<=n;i++)chemx(res,f[i]);
	for(int i=1;i<=n+1;i++)upd[i].clear(),f[i]=0;
	Seg::clear(1,1,n);
	cout<<res<<'
';
}
int main(){
	int T=read();
	while(T--)solve();
	return 0;
}

T3：

想到了$dp$，但是在算重的地方脑残了，只需要强制当前新加进去的子树的根的编号最小即可

其实顺着考试想的算重可以枚举子树个数$k$
最后只需要除以$k!$实际上也是最终答案

然后就可以推出$zy$的$Exp$做法了

#include<bits/stdc++.h>
using namespace std;
#define cs const
#define re register
#define pb push_back
#define pii pair<int,int>
#define fi first
#define se second
#define bg begin
#define ll long long
cs int RLEN=1<<20|1;
inline char gc(){
	static char ibuf[RLEN],*ib,*ob;
	(ib==ob)&&(ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
	return (ib==ob)?EOF:*ib++;
}
inline int read(){
	char ch=gc();
	int res=0;bool f=1;
	while(!isdigit(ch))f^=ch=='-',ch=gc();
	while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
	return f?res:-res;
}
template<class tp>inline void chemx(tp &a,tp b){a<b?a=b:0;}
template<class tp>inline void chemn(tp &a,tp b){a>b?a=b:0;}
cs int mod=998244353;
inline int add(int a,int b){a+=b-mod;return a+(a>>31&mod);}
inline void Add(int &a,int b){a+=b-mod,a+=a>>31&mod;}
inline int dec(int a,int b){a-=b;return a+(a>>31&mod);}
inline void Dec(int &a,int b){a-=b,a+=a>>31&mod;}
inline int mul(int a,int b){static ll r;r=1ll*a*b;return r>=mod?r%mod:r;}
inline void Mul(int &a,int b){static ll r;r=1ll*a*b;a=r>=mod?r%mod:r;}
inline int ksm(int a,int b,int res=1){for(;b;b>>=1,Mul(a,a))(b&1)&&(Mul(res,a),1);return res;}
inline int Inv(int x){return ksm(x,mod-2);}
cs int N=505;
int fac[N],ifac[N];
int f[N][N];
int wr[N],k,n,L,R;
inline void init(){
	fac[0]=ifac[0]=1;
	for(int i=1;i<N;i++)fac[i]=mul(fac[i-1],i);
	ifac[N-1]=Inv(fac[N-1]);
	for(int i=N-2;i;i--)ifac[i]=mul(ifac[i+1],i+1);
}
inline int C(int n,int m){return n<m?0:mul(fac[n],mul(ifac[m],ifac[n-m]));}
int main(){
	#ifdef Stargazer
	freopen("lx.cpp","r",stdin);
	#endif
	init();
	n=read(),k=read();
	for(int i=1;i<=k;i++)wr[read()]=1;
	L=read(),R=read();
	for(int i=1;i<=n;i++)f[i][1]=1;
	for(int j=2;j<=n;j++){
		for(int i=1;i<=n;i++)
		for(int k=1;k<=i-1;k++)
		Add(f[j][i],mul(f[j][i-k],mul(f[j-1][k],C(i-2,k-1))));
		for(int i=1;i<=n;i++)if(wr[i])f[j][i]=0;
	}
	for(int i=L;i<=R;i++){
		cout<<dec(f[i][n],f[i-1][n])<<" ";
	}
}

总结：

代码能力稀撇
脑子还蠢
情况总是考虑不全

这样下去可能真的$noip$就退役了