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  • 【CSP2019模拟】题解

    T1:

    本场送分题
    建个图求最长路即可
    由于T2T2花了太久并没有来的及写

    被踩爆了

    #include<bits/stdc++.h>
    using namespace std;
    const int RLEN=1<<20|1;
    inline char gc(){
        static char ibuf[RLEN],*ib,*ob;
        (ob==ib)&&(ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
        return (ob==ib)?EOF:*ib++;
    }
    #define gc getchar
    inline int read(){
        char ch=gc();
        int res=0,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;
    }
    #define ll long long
    #define re register
    #define pii pair<int,int>
    #define pic pair<int,char>
    #define fi first
    #define se second
    #define pb push_back
    #define cs const
    #define bg begin
    #define poly vector<int>
    #define chemx(a,b) ((a)<(b)?(a)=(b):0)
    #define chemn(a,b) ((a)>(b)?(a)=(b):0)
    cs int N=1e5,M=31;
    int n,m,tot,in[N];
    vector<int> e[N],E[N];
    inline void addedge(int u,int v){
    	e[u].pb(v);
    }
    inline void add(int u,int v){
    	E[u].pb(v),in[v]++;
    }
    int pos[M][M][M][M];
    struct node{
    	int a,b,c,d;
    	node(int _1=0,int _2=0,int _3=0,int _4=0):a(_1),b(_2),c(_3),d(_4){}
    	friend inline bool operator <(cs node &a,cs node &b){
    		return a.a<=b.a&&a.b<=b.b&&a.c<=b.c&&a.d<=b.d;
    	}
    	friend inline node operator +(cs node &a,cs node &b){
    		return node(a.a+b.a,a.b+b.b,a.c+b.c,a.d+b.d);
    	}
    	friend inline node operator -(cs node &a,cs node &b){
    		return node(a.a-b.a,a.b-b.b,a.c-b.c,a.d-b.d);
    	}
    }s[55],t[55],idx[N];
    char ss[N];
    ll val[N],w[N],mxv[N];
    __int128 tval;
    int dfn[N],low[N],tim,vis[N],bel[N],belnum;
    stack<int> stk;
    inline ll calc(int i,int j,int k,int p){
    	__int128 res=tval;
    	for(int x=1;x<=i;x++)res/=x;
    	for(int x=1;x<=j;x++)res/=x;
    	for(int x=1;x<=k;x++)res/=x;
    	for(int x=1;x<=p;x++)res/=x;
    	return res;
    }
    void dfs(int u){
    	dfn[u]=low[u]=++tim;
    	stk.push(u),vis[u]=1;
    	for(int &v:e[u]){
    		if(!dfn[v])dfs(v),chemn(low[u],low[v]);
    		else if(vis[v])chemn(low[u],dfn[v]);
    	}
    	if(low[u]==dfn[u]){
    		belnum++;
    		int tmp;
    		do{
    			tmp=stk.top();
    			stk.pop();
    			vis[tmp]=0;
    			bel[tmp]=belnum;
    			w[belnum]+=val[tmp];
    		}while(tmp!=u);
    	}
    }
    int pp[4];
    int main(){
    	n=read(),m=read();
    	for(int i=1;i<=m;i++){
    		scanf("%s",ss+1);
    		memset(pp,0,sizeof(pp));
    		for(int j=1,l=strlen(ss+1);j<=l;j++)
    		pp[ss[j]-'A']++;
    		s[i]=node(pp[0],pp[1],pp[2],pp[3]);
    		memset(pp,0,sizeof(pp));
    		scanf("%s",ss+1);
    		for(int j=1,l=strlen(ss+1);j<=l;j++)
    		pp[ss[j]-'A']++;
    		t[i]=node(pp[0],pp[1],pp[2],pp[3]);
    	}
    	tval=1;
    	for(int i=2;i<=n;i++)tval*=i;
    	for(int i=0;i<=n;i++)
    	for(int j=0;i+j<=n;j++)
    	for(int k=0;i+j+k<=n;k++)
    	pos[i][j][k][n-i-j-k]=++tot,val[tot]=calc(i,j,k,n-i-j-k),idx[tot]=node(i,j,k,n-i-j-k);
    	for(int i=1;i<=tot;i++){
    		node to;for(int j=1;j<=m;j++)
    		if(s[j]<idx[i]){
    			to=idx[i]-s[j]+t[j];
    			addedge(i,pos[to.a][to.b][to.c][to.d]);
    		}
    	}
    	for(int i=1;i<=tot;i++)if(!dfn[i])dfs(i);
    	for(int i=1;i<=tot;i++){
    		for(int &v:e[i]){
    			if(bel[v]!=bel[i])
    			add(bel[i],bel[v]);
    		}
    	}
    	queue<int> q;
    	for(int i=1;i<=belnum;i++)
    	if(!in[i])q.push(i);
    	ll ans=0;
    	while(!q.empty()){
    		int u=q.front();q.pop();
    		chemx(ans,mxv[u]+w[u]);
    		for(int &v:E[u]){
    			in[v]--,chemx(mxv[v],mxv[u]+w[u]);
    			if(!in[v])q.push(v);
    		}
    	}
    	cout<<ans;
    }
    

    T2:

    以前考过的一道CFCF原题
    不过当时并没有写就是了

    结果为了保险写了双取模哈希光荣TT了,白丢1515

    #include<bits/stdc++.h>
    using namespace std;
    const int RLEN=1<<20|1;
    inline char gc(){
        static char ibuf[RLEN],*ib,*ob;
        (ob==ib)&&(ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
        return (ob==ib)?EOF:*ib++;
    }
    #define gc getchar
    inline int read(){
        char ch=gc();
        int res=0,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;
    }
    #define ll long long
    #define re register
    #define pii pair<int,int>
    #define pic pair<int,char>
    #define fi first
    #define se second
    #define pb push_back
    #define cs const
    #define bg begin
    #define poly vector<int>
    #define chemx(a,b) ((a)<(b)?(a)=(b):0)
    #define chemn(a,b) ((a)>(b)?(a)=(b):0)
    cs int mod=1e9+7;
    inline int add(int a,int b){return (a+=b)>=mod?a-mod:a;}
    inline void Add(int &a,int b){(a+=b)>=mod?a-=mod:0;}
    inline int dec(int a,int b){return (a-=b)<0?a+mod:a;}
    inline void Dec(int &a,int b){(a-=b)<0?a+=mod:0;}
    inline int mul(int a,int b){return 1ll*a*b%mod;}
    inline void Mul(int &a,int b){a=1ll*a*b%mod;}
    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=2005;
    int n,m;
    char a[2][N],s[N];
    int f[2][N][N];
    int pw1[N],pw2[N];
    cs int bas1=1331,mod1=1920643713,bas2=233,mod2=771939933;
    inline void init(){
    	pw1[0]=pw2[0]=1;
    	for(int i=1;i<N;i++)pw1[i]=1ll*pw1[i-1]*bas1%mod1,pw2[i]=1ll*pw2[i-1]*bas2%mod2;
    }
    struct Hash{
    	int has1[N],has2[N];
    	inline void init(char *s){
    		for(int i=1,len=strlen(s+1);i<=len;i++)
    		has1[i]=(1ll*has1[i-1]*bas1+s[i]-'a'+1)%mod1,has2[i]=(1ll*has2[i-1]*bas2+s[i]-'a'+1)%mod2;
    	}
    	inline pii query(int l,int r){
    		return pii(((has1[r]-1ll*has1[l-1]*pw1[r-l+1])%mod1+mod1)%mod1,((has2[r]-1ll*has2[l-1]*pw2[r-l+1])%mod2+mod2)%mod2);
    	}
    }S,pre[2],suf[2];
    inline bool operator ==(cs pii &a,cs pii &b){
    	return a.fi==b.fi&&a.se==b.se;
    }
    inline int calc(char *s,int dir){
    	memset(f,0,sizeof(f));
    	int res=0;
    	for(int i=0;i<=n;i++){
    		for(int k=2;k*2<=m&&k<=i;k++)if(!dir||k*2!=m){
    			if(suf[0].query(n-i+1,n-i+k)==S.query(1,k)&&pre[1].query(i-k+1,i)==S.query(k+1,2*k))Add(f[1][i][2*k],1);
    			if(suf[1].query(n-i+1,n-i+k)==S.query(1,k)&&pre[0].query(i-k+1,i)==S.query(k+1,2*k))Add(f[0][i][2*k],1);
    		}
    		f[0][i][0]=1,f[1][i][0]=1;
    		for(int k=2;2*k<=m&&i+k<=n;k++)if(!dir||k*2!=m){
    			if(pre[0].query(i+1,i+k)==S.query(m-2*k+1,m-k)&&suf[1].query(n-i-k+1,n-i)==S.query(m-k+1,m))Add(res,f[0][i][m-2*k]);
    			if(pre[1].query(i+1,i+k)==S.query(m-2*k+1,m-k)&&suf[0].query(n-i-k+1,n-i)==S.query(m-k+1,m))Add(res,f[1][i][m-2*k]);
    		}
    		for(int j=0;j<m;j++){
    			if(a[0][i+1]==s[j+1])Add(f[0][i+1][j+1],f[0][i][j]);
    			if(a[1][i+1]==s[j+1])Add(f[1][i+1][j+1],f[1][i][j]);
    			if(j+2<=m){
    				if(a[1][i+1]==s[j+1]&&a[0][i+1]==s[j+2])Add(f[0][i+1][j+2],f[1][i][j]);
    				if(a[0][i+1]==s[j+1]&&a[1][i+1]==s[j+2])Add(f[1][i+1][j+2],f[0][i][j]);
    			}
    		}
    		Add(res,f[0][i][m]),Add(res,f[1][i][m]);
    	}
    	return res;
    }
    int ans;
    int main(){
    	init();
    	scanf("%s",a[0]+1);
    	scanf("%s",a[1]+1);
    	n=strlen(a[0]+1);
    	pre[0].init(a[0]);
    	reverse(a[0]+1,a[0]+n+1);
    	suf[0].init(a[0]);
    	reverse(a[0]+1,a[0]+n+1);
    	pre[1].init(a[1]);
    	reverse(a[1]+1,a[1]+n+1);
    	suf[1].init(a[1]);
    	reverse(a[1]+1,a[1]+n+1);
    	scanf("%s",s+1);
    	m=strlen(s+1);
    	S.init(s);
    	Add(ans,calc(s,0));
    	if(m==1){cout<<ans;return 0;}
    	reverse(s+1,s+m+1);
    	S.init(s);
    	Add(ans,calc(s,1));
    	if(m==2){
    		for(int i=1;i<=n;i++){
    			if(a[0][i]==s[1]&&a[1][i]==s[2])Dec(ans,1);
    			if(a[1][i]==s[1]&&a[0][i]==s[2])Dec(ans,1);
    		}
    	}
    	cout<<ans<<'
    ';
    }
    

    T3:

    卷一个idkid_k
    就是相当于要求i=1nimsum_{i=1}^ni^m这样一个东西
    先快速插值后多点求值完杜教筛即可

    本机跑了30s30sojoj上只跑了4s4s
    什么鬼

    #include<bits/stdc++.h>
    using namespace std;
    const int RLEN=1<<20|1;
    inline char gc(){
        static char ibuf[RLEN],*ib,*ob;
        (ob==ib)&&(ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
        return (ob==ib)?EOF:*ib++;
    }
    #define gc getchar
    inline int read(){
        char ch=gc();
        int res=0,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;
    }
    #define ll long long
    #define re register
    #define pii pair<int,int>
    #define pic pair<int,char>
    #define fi first
    #define se second
    #define pb push_back
    #define cs const
    #define bg begin
    #define poly vector<int>
    #define chemx(a,b) ((a)<(b)?(a)=(b):0)
    #define chemn(a,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 void Add(int &a,int b){(a+=b)>=mod?a-=mod:0;}
    inline int dec(int a,int b){return (a-=b)<0?a+mod:a;}
    inline void Dec(int &a,int b){(a-=b)<0?a+=mod:0;}
    inline int mul(int a,int b){return 1ll*a*b%mod;}
    inline void Mul(int &a,int b){a=1ll*a*b%mod;}
    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 C=19;
    poly w[C+1];
    int rev[(1<<20)|5];
    inline void init_w(){
    	for(int i=1;i<=C;i++)w[i].resize(1<<(i-1));
    	w[C][0]=1;
    	int wn=ksm(G,(mod-1)/(1<<C));
    	for(int i=1;i<1<<(C-1);i++)w[C][i]=mul(w[C][i-1],wn);
    	for(int i=C-1;i;i--)
    	for(int j=0;j<1<<(i-1);j++)
    	w[i][j]=w[i+1][j<<1];
    }
    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(poly &f,int lim,int kd){
    	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);
    	if(kd==-1){
    		reverse(f.bg()+1,f.bg()+lim);
    		for(int i=0,iv=Inv(lim);i<lim;i++)Mul(f[i],iv);
    	}
    }
    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,poly b){
    	int deg=a.size()+b.size()-1,lim=1;
    	if(deg<=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;
    	}
    	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=a,c.resize(lim>>1);
    		init_rev(lim);
    		b.resize(lim),ntt(b,lim,1);
    		c.resize(lim),ntt(c,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 operator /(poly a,poly b){
    	int deg=a.size()-b.size()+1;
    	reverse(a.bg(),a.end());
    	reverse(b.bg(),b.end());
    	a=a*Inv(b,deg),a.resize(deg);
    	reverse(a.bg(),a.end());
    	return a;
    }
    inline poly operator %(poly a,poly b){
    	if(a.size()<b.size())return a;
    	a=a-(a/b)*b;
    	a.resize(b.size()-1);return a;
    }
    inline poly integ(poly a){
    	for(int i=0;i<(int)a.size()-1;i++)a[i]=mul(a[i+1],i+1);
    	a.pop_back();return a;
    }
    cs int N=1000005;
    namespace fastcalc{
    	poly f[N<<2];
    	#define lc (u<<1)
    	#define rc ((u<<1)|1)
    	#define mid ((l+r)>>1)
    	void build(int u,int l,int r,int *v){
    		if(l==r){f[u].clear(),f[u].pb(dec(0,v[l]%mod)),f[u].pb(1);return;}
    		build(lc,l,mid,v),build(rc,mid+1,r,v);
    		f[u]=f[lc]*f[rc];
    	}
    	void calc(int u,int l,int r,poly g,int *v){
    		if(l==r){v[l]=g[0];return;}
    		calc(lc,l,mid,g%f[lc],v),calc(rc,mid+1,r,g%f[rc],v);
    	}
    	inline void calc(poly coef,int n,int *v){
    		build(1,1,n,v);
    		calc(1,1,n,coef,v);
    	}
    	#undef lc
    	#undef rc
    	#undef mid
    }
    namespace interpolation{
    	poly f[N<<2];
    	int val[N];
    	#define lc (u<<1)
    	#define rc ((u<<1)|1)
    	#define mid ((l+r)>>1)
    	void build(int u,int l,int r){
    		if(l==r){f[u].clear(),f[u].pb(dec(0,l)),f[u].pb(1);return;}
    		build(lc,l,mid),build(rc,mid+1,r);
    		f[u]=f[lc]*f[rc];
    	}
    	poly calc(int u,int l,int r,int *v){
    		if(l==r){return poly(1,v[l]);}
    		return calc(lc,l,mid,v)*f[rc]+f[lc]*calc(rc,mid+1,r,v);
    	}
    	inline poly calc(int n,int *v){
    		build(1,1,n);
    		for(int i=1;i<=n;i++)val[i]=i;
    		fastcalc::calc(integ(f[1]),n,val);
    		for(int i=1;i<=n;i++)val[i]=mul(Inv(val[i]),v[i]);
    		return calc(1,1,n,val);
    	}
    	#undef lc
    	#undef rc
    	#undef mid
    }
    int n,m,tot;
    int pr[N],s1[N],s2[N],g1[N],g2[N],mu[N];
    int all[N],num,idx[N];
    bitset<N> vis;
    inline int get(int x){
    	return x<=N-5?g1[x]:g2[n/x];
    }
    int du(int x){
    	if(x<=N-5)return s1[x];
    	if(vis[n/x])return s2[n/x];
    	vis[n/x]=1;
    	int res=1;
    	for(int i=2,nxt;i<=x;i=nxt+1){
    		nxt=x/(x/i);
    		Dec(res,mul(dec(get(nxt),get(i-1)),du(x/i)));
    	}
    	return s2[n/x]=res;
    }
    inline void init(){
    	cs int len=N-5;
    	s1[1]=g1[1]=mu[1]=1;
    	for(int i=2;i<=len;i++){
    		if(!vis[i])pr[++tot]=i,mu[i]=mod-1,g1[i]=ksm(i,m);
    		s1[i]=mul(g1[i],mu[i]);
    		for(int j=1;i*pr[j]<=len&&j<=tot;j++){
    			vis[i*pr[j]]=1,g1[i*pr[j]]=mul(g1[i],g1[pr[j]]);
    			if(i%pr[j]==0)break;
    			mu[i*pr[j]]=mod-mu[i];
    		}
    	}
    	for(int i=2;i<=len;i++)Add(s1[i],s1[i-1]),Add(g1[i],g1[i-1]);
    	vis.reset();
    }
    int main(){
    	init_w();
    	n=read(),m=read();
    	init();
    	if(n>N-5){
    		for(int i=1,nxt;i<=n;i=nxt+1){
    			nxt=n/(n/i);
    			if(nxt>N-5)all[++num]=nxt,idx[num]=nxt;
    		}
    		poly res=interpolation::calc(m+2,g1);
    		fastcalc::calc(res,num,all);
    		for(int i=1;i<=num;i++)g2[n/idx[i]]=all[i];
    	}
    	cout<<du(n);
    }
    
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    dev_queue_xmit 发生了什么?skb还会在哪里缓存
    内核blackhole
    网卡多ip 再看arp; arp队列也会缓存skb
    tcp发送缓冲区中的数据都是由产生数据的进程给推送到ip层还是有定时任务触发?
    socket有没有同步写一说(怎么判定数据一定达到了对端?还得用户态)
    貌似要看看时钟了
    读写JSON作配置文件
  • 原文地址:https://www.cnblogs.com/stargazer-cyk/p/12328413.html
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