【2020省选模拟】题解

T1：

考场写了个
递归归并的假的复杂度的做法
我是个傻逼
结果还过了90分
把递归然后归并看哪个放前面换成用堆维护即可

#include<bits/stdc++.h>
using namespace std;
#define cs const
#define re register
#define pb push_back
#define pii pair<int,int>
#define ll long long
#define fi first
#define se second
#define bg begin
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<typename tp>inline void chemx(tp &a,tp b){a<b?a=b:0;}
template<typename tp>inline void chemn(tp &a,tp b){a>b?a=b:0;}
cs int N=200005;
int v[N],a1[N],a2[N],n;
#define lc (u<<1)
#define rc ((u<<1)|1)
#define mid ((l+r)>>1)
struct Seg{
	pii tr[N<<2];
	inline pii merge(cs pii &a,cs pii &b){return a.fi<b.fi?a:b;}
	void build(int u,int l,int r,int *a){
		if(l==r){tr[u].se=l;tr[u].fi=a[l];return;}
		build(lc,l,mid,a),build(rc,mid+1,r,a);
		tr[u]=merge(tr[lc],tr[rc]);
	}
	pii query(int u,int l,int r,int st,int des){
		if(st<=l&&r<=des)return tr[u];
		if(des<=mid)return query(lc,l,mid,st,des);
		if(mid<st)return query(rc,mid+1,r,st,des);
		return merge(query(lc,l,mid,st,des),query(rc,mid+1,r,st,des));
	}
}Seg[2];
#undef lc
#undef rc
#undef mid
struct node{
	int l,r,vl,vr;
	node(int a=0,int b=0,int c=0,int d=0):l(a),r(b),vl(c),vr(d){}
	friend inline bool operator <(cs node &a,cs node &b){
		return v[a.vl]>v[b.vl];
	}
};
priority_queue<node>q;
inline node get(int l,int r){
	pii p=Seg[l&1].query(1,1,n,l,r),q=Seg[(l&1)^1].query(1,1,n,p.se+1,r);
	return node(l,r,p.se,q.se);
}
int main(){
	#ifdef Stargazer
	freopen("lx.in","r",stdin);
//	freopen("my.out","w",stdout);
	#endif
	n=read();
	for(int i=1;i<=n;i++){
		v[i]=read();
		if(i&1)a1[i]=v[i],a2[i]=1e9;
		else a2[i]=v[i],a1[i]=1e9;
	}
	Seg[1].build(1,1,n,a1),Seg[0].build(1,1,n,a2);
	q.push(get(1,n));
	while(q.size()){
		node now=q.top();q.pop();
		cout<<v[now.vl]<<" "<<v[now.vr]<<" ";
		if(now.l<now.vl)q.push(get(now.l,now.vl-1));
		if(now.vl<now.vr-1)q.push(get(now.vl+1,now.vr-1));
		if(now.vr<now.r)q.push(get(now.vr+1,now.r));
	}
	return 0;
}

T2：

普及组数位 $dp$

嫌麻烦写了个 $log^2n$ 的做法
但好像 $log$ 的还要好写些？
我是个智障

#include<bits/stdc++.h>
using namespace std;
#define cs const
#define re register
#define pb push_back
#define pii pair<int,int>
#define ll long long
#define fi first
#define se second
#define bg begin
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;
}
inline ll readll(){
    char ch=gc();
    ll 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<typename tp>inline void chemx(tp &a,tp b){a<b?a=b:0;}
template<typename tp>inline void chemn(tp &a,tp b){a>b?a=b:0;}
ll pw[20];
inline int getin(char c){
	if(isdigit(c))return c-'0';
	else return 10+c-'A';
}
inline void write(int x){
	if(x<10)cout<<x;
	else putchar('A'+x-10);
}
namespace solve1{
int dig[20],cnt;
inline void calc1(ll pos,int res){
	cnt=0;
	while(pos)dig[++cnt]=pos%16,pos/=16;
	reverse(dig+1,dig+cnt+1);
	write(dig[res]);puts("");
}
inline void solve(ll n){
	for(int len=1;;len++){
		if(n/len<=pw[len]-pw[len-1]){
			return calc1(pw[len-1]+(n-1)/len,(n-1)%len+1);
		}
		n-=(pw[len]-pw[len-1])*len;
	}
}
}
namespace solve2{
ll f[20][2][2],s[20];
int whi;
int dig[20],cnt;
ll dfs(int pos,int lim,int zero){
	if(!pos)return 0;
	if(f[pos][lim][zero])return f[pos][lim][zero];
	ll ret=0;
	for(int L=(lim)?dig[pos]:15,i=0;i<=L;i++){
		ret+=dfs(pos-1,lim&(i==dig[pos]),zero&(i==0));
		if(whi==0&&i==0&&!zero){
			if(lim&&i==dig[pos])
			ret+=s[pos-1]+1;
			else ret+=pw[pos-1];
		}
		if(whi!=0&&i==whi){
			if(lim&&i==dig[pos])
			ret+=s[pos-1]+1;
			else ret+=pw[pos-1];
		}
	//	cout<<pos<<" "<<i<<" "<<ret<<'\n';
	}
	return f[pos][lim][zero]=ret;
}
inline void calc2(ll n,int res){
	cnt=0;ll pos=n;int anc=0;
	while(pos)dig[++cnt]=pos%16,pos/=16;
	reverse(dig+1,dig+cnt+1);
	for(int i=1;i<=res;i++)if(dig[i]==whi)anc++;
//	cout<<anc<<" "<<n-1<<'\n';
	cnt=0,pos=n-1;
	while(pos)dig[++cnt]=pos%16,pos/=16;
	for(int i=1;i<=cnt;i++)s[i]=s[i-1]+dig[i]*pw[i-1];
	cout<<anc+dfs(cnt,1,1)<<'\n';
}
inline void solve(ll n){
	char ch=gc();
	while(isspace(ch))ch=gc();
	whi=getin(ch);
	memset(f,0,sizeof(f));
	for(int len=1;;len++){
	if(n/len<=pw[len]-pw[len-1]){
		return calc2(pw[len-1]+(n-1)/len,(n-1)%len+1);
	}
	n-=(pw[len]-pw[len-1])*len;
}	
}
}
inline void solve(){
	int op=read();
	if(op==1)solve1::solve(readll());
	else solve2::solve(readll());
}
int main(){
	#ifdef Stargazer
	freopen("lx.in","r",stdin);
//	freopen("my.out","w",stdout);
	#endif
	pw[0]=1;
	for(int i=1;pw[i-1]<=1e18;i++)pw[i]=pw[i-1]*16;
	int T=read();
	while(T--)solve();
}

T3：

我是个菜逼
完全做不来

设 $f[i][j]$ 表示有 $i$ 行 $j$ 列且每行都有黑色的方案数
考虑一列一列枚举所有前面列都没有且这列有的行，插入到原来的 $i$ 行中
$f[i][j]*(1+i+{i\choose 2})->f[i][j+1]$
$f[i][j]*{i+k+2\choose k+2}->f[i+k][j+1]$
第二个的系数是因为存在原来 $i$ 行在第 $j$ 列有的情况
所以可以考虑加入 $k+2$ 行，其中有两个代表原来的 $i$ 行中的
再多设 $2$ 个可以放的位置来处理不存在原来 $i$ 行做开头/结尾的情况

然后就可以了

#include<bits/stdc++.h>
using namespace std;
#define cs const
#define re register
#define pb push_back
#define pii pair<int,int>
#define ll long long
#define fi first
#define se second
#define bg begin
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<typename tp>inline void chemx(tp &a,tp b){a<b?a=b:0;}
template<typename tp>inline void chemn(tp &a,tp b){a>b?a=b:0;}
cs int mod=998244353,G=3;
inline int add(int a,int b){return (a+=b)>=mod?(a-mod):a;}
inline int dec(int a,int b){a-=b;return 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 Add(int &a,int b){(a+=b)>=mod?(a-=mod):0;}
inline void Dec(int &a,int b){a-=b,a+=a>>31&mod;}
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=32005;
int rev[N];
inline void init_rev(int lim){
	for(int i=0;i<lim;i++)rev[i]=(rev[i>>1]>>1)|((i&1)*(lim>>1));
}
inline void ntt(int *f,int lim,int kd){
	for(int i=0;i<lim;i++)if(i>rev[i])swap(f[i],f[rev[i]]);
	for(int a0,a1,mid=1;mid<lim;mid<<=1){
		int wn=ksm(G,(mod-1)/(mid<<1));
		for(int i=0;i<lim;i+=mid<<1)
		for(int j=0,w=1;j<mid;j++,Mul(w,wn))
		a0=f[i+j],a1=mul(f[i+j+mid],w),f[i+j]=add(a0,a1),f[i+j+mid]=dec(a0,a1);
	}
	if(kd==-1){
		reverse(f+1,f+lim);
		for(int i=0,iv=Inv(lim);i<lim;i++)Mul(f[i],iv);
	}
}
int fac[N],ifac[N];
inline void init_inv(cs int len=N-5){
	fac[0]=ifac[0]=1;
	for(int i=1;i<=len;i++)fac[i]=mul(fac[i-1],i);
	ifac[len]=Inv(fac[len]);
	for(int i=len-1;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 lim,n,m,f[N],a[N],g[N];
int main(){
	#ifdef Stargazer
	freopen("lx.in","r",stdin);
	#endif
	n=read(),m=read();
	lim=1;init_inv();
	while(lim<=n*2)lim<<=1;
	init_rev(lim);
	for(int i=1;i<=n;i++)g[i]=ifac[i+2];
	ntt(g,lim,1);
	a[0]=1;
	for(int i=1;i<=m;i++){
		for(int j=0;j<=n;j++)f[j]=mul(a[j],ifac[j]);
		ntt(f,lim,1);
		for(int j=0;j<lim;j++)Mul(f[j],g[j]);
		ntt(f,lim,-1);
		for(int j=0;j<=n;j++)a[j]=add(mul(f[j],fac[j+2]),mul(a[j],(1+j+C(j,2))%mod));
		for(int j=0;j<lim;j++)f[j]=0;
	}
	int ret=0;
	for(int i=0;i<=n;i++)Add(ret,mul(C(n,i),a[i]));
	cout<<ret<<'\n';
}

posted @ 2020-02-02 17:03 Stargazer_cykoi 阅读(215) 评论(0) 编辑收藏举报

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【2020省选模拟】题解

T1：

T2：

T3：

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