【2020省选模拟】题解

胡策题

感觉出了一道送分的

瑠璃色の物語

我的送分题
$Solution$

#include
using namespace std;
#define cs const
#define re register
#define pb push_back
#define pii pair
#define ll long long
#define y1 shinkle
#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++;
}
#define gc getchar
inline int read(){
    char ch=gc();
    int res=0;bool f=1;
    while(!isdigit(ch))f^=ch=='-',ch=gc();
    while(isdigit(ch))res=(res*10)+(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*10)+(ch^48),ch=gc();
    return f?res:-res;
}
inline char readchar(){
	char ch=gc();
	while(isspace(ch))ch=gc();
	return ch;
}
inline int readstring(char *s){
	int top=0;char ch=gc();
	while(isspace(ch))ch=gc();
	while(!isspace(ch)&&ch!=EOF)s[++top]=ch,ch=gc();
	s[top+1]='\0';return top;
}
template<typename tp>inline void chemx(tp &a,tp b){a=max(a,b);}
template<typename tp>inline void chemn(tp &a,tp b){a=min(a,b);}
cs int mod=998244353;
inline int add(int a,int b){return (a+b)>=mod?(a+b-mod):(a+b);}
inline int dec(int a,int b){return (a<b)?(a-b+mod):(a-b);}
inline int mul(int a,int b){static ll r;r=(ll)a*b;return (r>=mod)?(r%mod):r;}
inline void Add(int &a,int b){a=(a+b)>=mod?(a+b-mod):(a+b);}
inline void Dec(int &a,int b){a=(a<b)?(a-b+mod):(a-b);}
inline void Mul(int &a,int b){static ll r;r=(ll)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);}
inline int fix(ll x){x%=mod;return (x<0)?x+mod:x;}
typedef vector<int> poly;
namespace Poly{
	cs int C=19,M=(1<<C)|5;
	int *w[C+1],rev[M],iv[M];
	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 init_w(){
		for(int i=1;i<=C;i++)w[i]=new int[(1<<(i-1))|1];
		int wn=ksm(3,(mod-1)/(1<<C));w[C][0]=1;
		for(int i=1,l=1<<(C-1);i<l;i++)w[C][i]=mul(w[C][i-1],wn);
		for(int i=C-1;i;i--)
		for(int j=0,l=1<<(i-1);j<l;j++)w[i][j]=w[i+1][j<<1];
		iv[0]=iv[1]=1;
		for(int i=2;i<M;i++)iv[i]=mul(mod-mod/i,iv[mod%i]);
	}
	inline void dft(int *f,int lim){
		for(int i=0;i<lim;i++)if(i>rev[i])swap(f[i],f[rev[i]]);
		for(int mid=1,l=1,a0,a1;mid<lim;mid<<=1,l++)
		for(int i=0;i<lim;i+=mid<<1)
		for(int j=0;j<mid;j++)
		a0=f[i+j],a1=mul(w[l][j],f[i+j+mid]),f[i+j]=add(a0,a1),f[i+j+mid]=dec(a0,a1);
	}
	inline void ntt(poly &f,int lim,int kd){
		dft(&f[0],lim);
		if(kd==-1){
			reverse(f.bg()+1,f.bg()+lim);
			for(int i=0;i<lim;i++)Mul(f[i],iv[lim]);
		}
	}
	inline poly operator +(poly a,poly b){
		if(a.size()<b.size())a.resize(b.size());
		for(int i=0;i<b.size();i++)Add(a[i],b[i]);
		return a;
	}
	inline poly operator -(poly a,poly b){
		if(a.size()<b.size())a.resize(b.size());
		for(int i=0;i<b.size();i++)Dec(a[i],b[i]);
		return a;
	}
	inline poly operator *(poly a,int b){
		for(int i=0;i<a.size();i++)Mul(a[i],b);return a;
	}
	inline poly operator *(poly a,poly b){
		if(!a.size()||!b.size())return poly(0);
		int deg=a.size()+b.size()-1;
		if(a.size()<=32||b.size()<=32){
			poly c(deg,0);
			for(int i=0;i<a.size();i++)
			for(int j=0;j<b.size();j++)
			Add(c[i+j],mul(a[i],b[j]));
			return c;
		}int lim=1;while(lim<deg)lim<<=1;
		init_rev(lim);
		a.resize(lim),ntt(a,lim,1);
		b.resize(lim),ntt(b,lim,1);
		for(int i=0;i<lim;i++)Mul(a[i],b[i]);
		ntt(a,lim,-1),a.resize(deg);
		return a;
	}
	inline poly Inv(poly a,int deg){
		poly b(1,::Inv(a[0])),c;
		for(int lim=4;lim<(deg<<2);lim<<=1){
			c.resize(lim>>1);init_rev(lim);
			for(int i=0;i<(lim>>1);i++)c[i]=(i<a.size()?a[i]:0);
			c.resize(lim),ntt(c,lim,1);
			b.resize(lim),ntt(b,lim,1);
			for(int i=0;i<lim;i++)Mul(b[i],dec(2,mul(b[i],c[i])));
			ntt(b,lim,-1),b.resize(lim>>1);
		}b.resize(deg);return b;
	}
	inline poly deriv(poly a){
		for(int i=0;i+1<a.size();i++)a[i]=mul(a[i+1],i+1);
		a.pop_back();return a;
	}
	inline poly integ(poly a){
		a.pb(0);
		for(int i=a.size()-1;i;i--)a[i]=mul(a[i-1],iv[i]);
		a[0]=0;return a;
	}
	inline poly Ln(poly a,int deg){
		a=integ(Inv(a,deg)*deriv(a)),a.resize(deg);return a;
	}
	inline poly Exp(poly a,int deg){
		poly b(1,1),c;
		for(int lim=2;lim<(deg<<1);lim<<=1){
			c=Ln(b,lim);
			for(int i=0;i<lim;i++)c[i]=dec(i<a.size()?a[i]:0,c[i]);
			Add(c[0],1);
			b=b*c,b.resize(lim);
		}b.resize(deg);return b;
	}
	inline poly ksm(poly a,int b,int deg){
		return Exp(Ln(a,deg)*b,deg);
	}
	inline poly Mulx(poly a){
		a.pb(0);
		for(int i=a.size()-1;i;i--)a[i]=a[i-1];
		a[0]=0;return a;
	}
	inline poly Divx(poly a){
		for(int i=0;i+1<a.size();i++)a[i]=a[i+1];
		a.pop_back();return a;
	}
}
using namespace Poly;
cs int N=200005;
int n,m;
int c[N],v[N],a[N];
vector<int> tp[N];
int g[N];
poly build(int l,int r){
	if(l==r){
		poly a(2);
		a[0]=1,a[1]=dec(0,g[l]);
		return a;
	}int mid=(l+r)>>1;
	return build(l,mid)*build(mid+1,r);
}
inline poly get(int ln,int lm){
	poly f=build(1,ln),g(ln+1);
	for(int i=0;i<=ln;i++)g[i]=mul(f[i],ln-i);
	g=g*Inv(f,lm),g.resize(lm);return g;
	//for(int i=0;i
	//return Inv(now,lm);
}
poly build(){
	poly f(n+1);
	for(int i=1;i<=n;i++)if(tp[i].size()){
		int lt=n/i;
		for(int j=0;j<tp[i].size();j++)g[j+1]=tp[i][j];
		vector<int> vl=get(tp[i].size(),lt+1);
		for(int j=1;j<=lt;j++){
			Add(f[i*j],mul(iv[j],vl[j]));
		}
	}
	f=Exp(f,n+1);
	return f;
}
char rd[100],bt[100];
int main(){
	init_w();
	m=read(),n=read();
	for(int i=1;i<=m;i++)c[i]=read(),assert(c[i]<=n&&c[i]!=0);
	for(int i=1;i<=m;i++)v[i]=read(),assert(v[i]>0&&v[i]<mod);
	for(int i=1;i<=m;i++)tp[c[i]].pb(v[i]);
	poly f=build();
	poly g=Mulx(Inv(f,n));
	poly now=Inv(poly(1,1)-g,n+1);
	now.resize(n+1);
	now=now*ksm(f,n,n+1);
	cout<<now[n]<<'\n';
	return 0;
}

Naiive

$fsy$的神仙题

考虑串建成$Sa$或者后缀树上的两点的贡献
只需要按照$lcp$合并，这样需要的是找到在原数列中位置的前驱后继
询问就变成矩形取$max$,单点询问，但发现矩形比较特殊可以做

强制在线写个主席树即可

#include
using namespace std;
#define cs const
#define pb push_back
#define pii pair
#define fi first
#define se second
#define ll long long
#define bg begin
namespace IO{
	
cs int RLEN=1<<20|1;
char ibuf[RLEN],*ib,*ob;
inline char gc(){
	(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=0;
	while(!isdigit(ch))f=(ch=='-'),ch=gc();
	while(isdigit(ch))res=(res*10)+(ch^48),ch=gc();
	return f?-res:res;
}
inline int readstring(char *s){
	int top=0;char ch=gc();
	while(isspace(ch))ch=gc();
	while(!isspace(ch)&&ch!=EOF)s[++top]=ch,ch=gc();
	s[top+1]='\0';return top;
}

}
using IO::read;
using IO::readstring;
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=100005;

namespace Seg{
	cs int N=::N*400;
	#define mid ((l+r)>>1)
	int lc[N],rc[N],tot,mx[N];
	inline int copy(int r){
		int u=++tot;lc[u]=lc[r],rc[u]=rc[r],mx[u]=mx[r];return u;
	}
	void update(int &u,int l,int r,int p,int k){
		u=copy(u);chemx(mx[u],k);
		if(l==r)return;
		(p<=mid)?update(lc[u],l,mid,p,k):update(rc[u],mid+1,r,p,k);
	}
	int query(int &u,int l,int r,int st){
		if(!u)return 0;
		if(st<=l)return mx[u];
		if(mid<st)return query(rc[u],mid+1,r,st);
		return max(mx[rc[u]],query(lc[u],l,mid,st));
	}
	#undef mid
}
int rt[N];
vector<pii> upd[N];
namespace Sa{
	int n,m,sa[N],sa2[N],rk[N],a[N],buc[N],ht[N];
	set<int>st[N];
	typedef set<int>::iterator It;
	inline void Sort(){
		for(int i=1;i<=n;i++)buc[rk[sa2[i]]]++;
		for(int i=1;i<=m;i++)buc[i]+=buc[i-1];
		for(int i=n;i>=1;i--)sa[buc[rk[sa2[i]]]--]=sa2[i];
		for(int i=0;i<=m;i++)buc[i]=0;
	}
	int id[N],fa[N];
	inline int find(int x){return fa[x]==x?x:fa[x]=find(fa[x]);}
	void Ins(int l,int r,int vl){
		if(l>r)swap(l,r);
		upd[r].pb(pii(l,vl));
	}
	inline void build(char *s,int _n){
		n=_n;m=26;
		for(int i=1;i<=n;i++)a[i]=s[i]-'a'+1;
		for(int i=1;i<=n;i++)rk[i]=a[i],sa2[i]=i;
		Sort();
		for(int i=1,pos=0;i<=n&&pos<n;i<<=1){
			pos=0;
			for(int j=n-i+1;j<=n;j++)sa2[++pos]=j;
			for(int j=1;j<=n;j++)if(sa[j]>i)sa2[++pos]=sa[j]-i;
			Sort(),swap(rk,sa2);
			rk[sa[1]]=1,pos=1;
			for(int j=2;j<=n;j++)
			rk[sa[j]]=(sa2[sa[j]]==sa2[sa[j-1]]&&sa2[sa[j]+i]==sa2[sa[j-1]+i])?pos:++pos;
			m=pos;
		}
		for(int i=1,k=0,j;i<=n;ht[rk[i++]]=k)
		for(j=sa[rk[i]-1],k?k--:0;a[i+k]==a[j+k];k++);
		for(int i=1;i<n;i++)id[i]=i+1;
		for(int i=1;i<=n;i++)fa[i]=i,st[i].insert(sa[i]);
		sort(id+1,id+n,[](cs int &i,cs int &j){return ht[i]>ht[j];});
		for(int i=1;i<n;i++){
			int u=id[i],v=u-1,h=ht[u];u=find(u),v=find(v);
			if(st[u].size()>st[v].size())swap(u,v);fa[u]=v;
			for(int x:st[u]){
				It it=st[v].lower_bound(x);
				if(it!=st[v].end())Ins(x,*it,h);
				if(it!=st[v].bg())Ins(x,*--it,h);
			}
			for(int x:st[u])st[v].insert(x);st[u].clear();
		}
	}
	
}
int n,m;
char str[N];
int last;
int main(){
	#ifdef Stargazer
	freopen("lx.in","r",stdin);
	freopen("my.out","w",stdout);
	#endif
	n=read(),m=read();
	readstring(str);
	Sa::build(str,n);
	for(int i=0;i<=n;i++){
		if(i)rt[i]=rt[i-1];
		for(int j=0;j<upd[i].size();j++){
			pii x=upd[i][j];
			Seg::update(rt[i],1,n,x.fi,x.se);
		}
	}
	for(int t=1;t<=m;t++){
		int l=(read()+last-1)%n+1,r=(read()+last-1)%n+1;
		if(l>r)swap(l,r);	
		cout<<(last=Seg::query(rt[r],1,n,l))<<'\n';
	}return 0;
}

gcd卷积

$ldx$的神仙题

直接调和级数容斥做是$O(55nlogmod)$
如果用质因数次数高维前缀和容斥并预处理幂次是$O(55nloglogn+55modlog)$的

#include
using namespace std;
#define cs const
#define pb push_back
#define pii pair
#define fi first
#define se second
#define ll long long
#define bg begin
namespace IO{
	
cs int RLEN=1<<20|1;
char ibuf[RLEN],*ib,*ob;
inline char gc(){
	(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=0;
	while(!isdigit(ch))f=(ch=='-'),ch=gc();
	while(isdigit(ch))res=(res*10)+(ch^48),ch=gc();
	return f?-res:res;
}
inline int readstring(char *s){
	int top=0;char ch=gc();
	while(isspace(ch))ch=gc();
	while(!isspace(ch)&&ch!=EOF)s[++top]=ch,ch=gc();
	s[top+1]='\0';return top;
}

}
using IO::read;
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=23333;
inline int add(int a,int b){return (a+b>=mod)?(a+b-mod):(a+b);}
inline int dec(int a,int b){return (a<b)?(a-b+mod):(a-b);}
inline int mul(int a,int b){static ll r;r=(ll)a*b;return (r>=mod)?(r%mod):r;}
inline void Add(int &a,int b){a=(a+b>=mod)?(a+b-mod):(a+b);}
inline void Dec(int &a,int b){a=(a<b)?(a-b+mod):(a-b);}
inline void Mul(int &a,int b){static ll r;r=(ll)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))if(b&1)Mul(res,a);return res;}
inline int Inv(int x){return ksm(x,mod-2);}
cs int N=100005;
int n,a[N],s[N],pr[N],tot,pw[mod],lm[N];
bool vis[N];
inline void init_sieve(){
	for(int i=2;i<=n;i++){
		if(!vis[i])pr[++tot]=i,lm[tot]=n/i;
		for(int j=1;j<=tot&&i*pr[j]<=n;j++){
			vis[i*pr[j]]=1;
			if(i%pr[j]==0)break;
		}
	}
}
inline void fgt(int *f){
	for(int i=1;i<=tot;i++)
	for(int j=lm[i];j;j--)Add(f[j],f[j*pr[i]]);
}
inline void ifgt(int *f){
	for(int i=1;i<=tot;i++)
	for(int j=1;j<=lm[i];j++)Dec(f[j],f[j*pr[i]]);
}
int main(){
	#ifdef Stargazer
	freopen("lx.in","r",stdin);
	freopen("my.out","w",stdout);
	#endif
	n=read();init_sieve();
	for(int i=1;i<=n;i++)s[i]=1;
	for(int i=1;i<=55;i++){
		for(int j=1;j<=n;j++)a[j]=read();
		fgt(a);
		int t=read();
		for(int j=0;j<mod;j++)pw[j]=ksm(j,t);
		for(int j=1;j<=n;j++)assert(a[j]<mod),Mul(s[j],pw[a[j]]);
	}
	ifgt(s);
	for(int i=1;i<=n;i++)cout<<s[i]<<" ";puts("");
	return 0;
}

构造题

$zxy$的神仙题
做不来QwQ
$solution$