【Codeforces #1316】CodeCraft-20 (Div. 2) 简要题解

A

傻逼题

#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;
}
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();
	return top;
}
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=1e9+7;
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){if(a==0&&b==0)return 0;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(int x){return (x<0)?x+mod:x;}
int main(){
	#ifdef Stargazer
	freopen("lx.in","r",stdin);
	#endif
	int T=read();
	while(T--){
		int n=read(),mx=read(),v=0;
		for(int i=1;i<=n;i++)v+=read();
		cout<<min(v,mx)<<'\n';
	}
}

B

枚举 $k$对几种情况讨论一下最后的字符串长啥样

#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;
}
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();
	return top;
}
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=1e9+7;
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){if(a==0&&b==0)return 0;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(int x){return (x<0)?x+mod:x;}
cs int N=100005;
char s[N],tp[N],ans[N];int n,mx;
int main(){
	#ifdef Stargazer
	freopen("lx.in","r",stdin);
	#endif
	int T=read();
	while(T--){
		n=read();mx=0;
		readstring(tp);
		for(int i=1;i<=n;i++)ans[i]='z';
		for(int i=n+1;i<=n+n;i++)tp[i]=tp[i-n];
		mx=1;
		for(int k=0;k<n;k++){int fg=1;
			for(int i=k+1,p=1;p<=n;p++,i++)s[p]=(i<=n)?(tp[i]):(((n&1)^(k&1))?tp[n-i+k+1]:tp[i]);
			for(int j=1;j<=n;j++)if(ans[j]!=s[j]){fg=0;
				if(ans[j]<s[j])break;
				if(ans[j]>s[j]){
					mx=k+1;
					for(int i=1;i<=n;i++)ans[i]=s[i];
					break;
				}
			}
			if(fg==1)chemn(mx,k+1);
		}
		for(int i=1;i<=n;i++)putchar(ans[i]);puts("");
		cout<<mx<<'\n';
	}
}

C

脑残莽了个 $mtt$
直接找到两个多项式中第一个不为 $0$的位置即可
所以那个 $gcd$性质有啥用？？

#include<bits/stdc++.h>
using namespace std;
#define cs const
#define pb push_back
#define pii pair<int,int>
#define fi first
#define se second
#define ll long long
#define re register
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;
}
cs int N=3000005;
int mod;
namespace Poly{
struct plx{
	double x,y;
	plx(double _x=0,double _y=0):x(_x),y(_y){}
	friend inline plx operator +(cs plx &a,cs plx &b){
		return plx(a.x+b.x,a.y+b.y);
	}
	friend inline plx operator -(cs plx &a,cs plx &b){
		return plx(a.x-b.x,a.y-b.y);
	}
	friend inline plx operator *(cs plx &a,cs plx &b){
		return plx(a.x*b.x-a.y*b.y,a.x*b.y+a.y*b.x);
	}
	inline plx conj()cs{return plx(x,-y);}
};
#define poly vector<plx>
cs int C=21,M=(1<<15)-1;
cs double pi=acos(-1);
plx *w[C+1];
int rev[(1<<C)|5];
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 plx[(1<<(i-1))|1];
	w[C][0]=plx(1,0);
	for(int i=1;i<(1<<(C-1));i++){
		w[C][i]=plx(cos(pi*i/(1<<(C-1))),sin(pi*i/(1<<(C-1))));
	}
	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 fft(plx *f,int lim,int kd){
	for(int i=0;i<lim;i++)if(i>rev[i])swap(f[i],f[rev[i]]);
	plx a0,a1;
	for(int mid=1,l=1;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=f[i+j+mid]*w[l][j],f[i+j]=a0+a1,f[i+j+mid]=a0-a1;
	if(kd==-1){
		reverse(f+1,f+lim);
		for(int i=0;i<lim;i++)f[i].x/=lim,f[i].y/=lim;
	}
}
plx a[(1<<C)|5],b[(1<<C)|5],c[(1<<C)|5],d[(1<<C)|5],da,db,dc,dd;
inline void mul(int *A,int *B,int lim,int *ret){
	for(int i=0;i<lim;i++)a[i]=plx(A[i]&M,A[i]>>15),b[i]=plx(B[i]&M,B[i]>>15);
	init_rev(lim);
	fft(a,lim,1),fft(b,lim,1);
	for(int i=0;i<lim;i++){
		int j=(lim-i)&(lim-1);
		da=(a[i]+a[j].conj())*plx(0.5,0);
		db=(a[j].conj()-a[i])*plx(0,0.5);
		dc=(b[i]+b[j].conj())*plx(0.5,0);
		dd=(b[j].conj()-b[i])*plx(0,0.5);
		c[i]=(da*dc)+((da*dd)*plx(0,1));
		d[i]=(db*dd)+((db*dc)*plx(0,1));
	}
	fft(c,lim,-1),fft(d,lim,-1);
	for(int i=0;i<lim;i++){
		ll da=(ll)(d[i].x+0.5)%mod,db=(ll)(d[i].y+0.5)%mod,dc=(ll)(c[i].y+0.5)%mod,dd=(ll)(c[i].x+0.5)%mod;
		ret[i]=((da<<30)+((db+dc)<<15)+dd)%mod;
	}
}
}
int n,m,a[N],b[N],lim,ans[N];
int main(){
	#ifdef Stargazer
	freopen("lx.in","r",stdin);
	#endif
	Poly::init_w();
	n=read(),m=read(),mod=read();
	for(int i=0;i<n;i++)a[i]=(read()%mod+mod)%mod;
	for(int i=0;i<m;i++)b[i]=(read()%mod+mod)%mod;
	lim=1;
	while(lim<(n+m))lim<<=1;
	Poly::mul(a,b,lim,ans);
	for(int i=0;i<n+m-1;i++){
		if(ans[i]){
			cout<<i<<'\n';return 0;
		}
	}
}

D

由于 $-1$之间可以直接两两互相走就合法
于是模拟即可

#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;
}
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;
}
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();
	return top;
}
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=1e9+7;
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){if(a==0&&b==0)return 0;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(int x){return (x<0)?x+mod:x;}
cs int N=1005;
pii op[N][N];
int st[N][N];
int n;
cs int tx[5]={0,-1,0,1,0};
cs int ty[5]={0,0,-1,0,1};
queue<pii> q;
inline int ni(int k){
	if(k==1)return 3;
	if(k==2)return 4;
	if(k==3)return 1;
	if(k==4)return 2;
}
inline void write(int x){
	if(x==1)putchar('U');
	if(x==2)putchar('L');
	if(x==3)putchar('D');
	if(x==4)putchar('R');
	if(x==5)putchar('X');
}
inline bool link(int i,int j,int c,int d,int v1,int v2){
	if(op[c][d].fi==-1){
		st[i][j]=v1;
		if(!st[c][d])st[c][d]=v2;
		return true;
	}return false;
}
void dfs(int i,int j,int v){
	if(st[i][j])return;
	st[i][j]=v;
	for(int k=1;k<=4;k++){
		int px=i+tx[k],py=j+ty[k];
		if(1<=px&&px<=n&&1<=py&&py<=n&&op[px][py]==op[i][j])
		dfs(px,py,ni(k));
	}
}
int main(){
	#ifdef Stargazer
	freopen("lx.in","r",stdin);
	#endif
	n=read();
	for(int i=1;i<=n;i++)
	for(int j=1;j<=n;j++)op[i][j].fi=read(),op[i][j].se=read();
	for(int i=1;i<=n;i++)
	for(int j=1;j<=n;j++){
		if(op[i][j].fi==-1){
			int now=!!st[i][j];
			for(int k=1;k<=4&&!now;k++){
				int px=i+tx[k],py=j+ty[k];
				now=link(i,j,px,py,k,ni(k));
			}
			if(!now){cout<<"INVALID\n";return 0;}
		}
		else if(op[i][j].fi==i&&op[i][j].se==j)dfs(i,j,5);
	}
	for(int i=1;i<=n;i++)
	for(int j=1;j<=n;j++)if(!st[i][j]){cout<<"INVALID\n";return 0;}
	cout<<"VALID\n";
	for(int i=1;i<=n;i++,puts(""))
	for(int j=1;j<=n;j++)write(st[i][j]);
}

E

傻逼状压 $dp$

#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;
}
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();
	return top;
}
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,M=(1<<7)+1;
int n,p,k,a[N],v[N][7],id[N];
ll f[N][M];
int cnt[M];
int main(){
	#ifdef Stargazer
	freopen("lx.in","r",stdin);
	#endif
	n=read(),p=read(),k=read();
	int lim=1<<p;
	for(int i=1;i<=n;i++)a[i]=read();
	for(int i=1;i<=n;i++){id[i]=i;
		for(int j=0;j<p;j++)v[i][j]=read();
	}
	sort(id+1,id+n+1,[](cs int&x,cs int&y){return a[x]>a[y];});
	memset(f,128,sizeof(f));
	f[0][0]=0;
	for(int i=1;i<=lim;i++)cnt[i]=cnt[i>>1]+(i&1);
	for(int i=0;i<n;i++){
		int x=id[i+1];
		for(int s=0;s<lim;s++){
			ll now=f[i][s];
			for(int j=0;j<p;j++)if(!(s&(1<<j))){
				chemx(f[i+1][s|(1<<j)],now+v[x][j]);
			}
			if(i+1-cnt[s]<=k)
			chemx(f[i+1][s],now+a[x]);
			else chemx(f[i+1][s],now);
		}
	}
	ll res=0;
	for(int s=0;s<lim;s++)chemx(res,f[n][s]);
	cout<<res<<'\n';
}

F

傻逼题
考虑任意一对点的贡献
然后发现可以直接线段树或者平衡树维护了

#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;
}
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;
}
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();
	return top;
}
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=1e9+7;
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){if(a==0&&b==0)return 0;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(int x){return (x<0)?x+mod:x;}
cs int N=300005;
int pw[N],ip[N];
pii p[N<<1];
int pos[N<<1],upd[N],n,q;
namespace Seg{
	cs int N=::N<<3;
	int ls[N],rs[N],s[N],siz[N];
	#define lc (u<<1)
	#define rc ((u<<1)|1)
	#define mid ((l+r)>>1)
	inline void pushup(int u){
		siz[u]=siz[lc]+siz[rc];
		s[u]=add(add(s[lc],s[rc]),mul(mul(ls[lc],rs[rc]),ip[siz[lc]]));
		ls[u]=add(ls[lc],mul(ls[rc],pw[siz[lc]]));
		rs[u]=add(rs[lc],mul(rs[rc],ip[siz[lc]]));
	}
	void build(int u,int l,int r){
		if(l==r){
			if(p[l].se<=n)
				s[u]=0,ls[u]=p[l].fi,rs[u]=mul(p[l].fi,ip[1]),siz[u]=1;
			return;
		}
		build(lc,l,mid),build(rc,mid+1,r);
		pushup(u);
	}
	void update(int u,int l,int r,int ps,int kd){
		if(l==r){
			if(kd==1)s[u]=0,ls[u]=p[l].fi,rs[u]=mul(p[l].fi,ip[1]),siz[u]=1;
			else s[u]=ls[u]=rs[u]=siz[u]=0;
			return;
		}
		if(ps<=mid)update(lc,l,mid,ps,kd);
		else update(rc,mid+1,r,ps,kd);
		pushup(u);
	}
	#undef lc
	#undef rc
	#undef mid
}
int main(){
	#ifdef Stargazer
	freopen("lx.in","r",stdin);
	#endif
	pw[0]=ip[0]=1;
	for(int i=1;i<N;i++)pw[i]=mul(pw[i-1],2);
	for(int i=1,iv=Inv(2);i<N;i++)ip[i]=mul(ip[i-1],iv);
	n=read();
	for(int i=1;i<=n;i++){
		int x=read();
		p[i]=pii(x,i);
	}
	q=read();
	for(int i=1;i<=q;i++){
		int t=read(),x=read();
		upd[i]=t;
		p[i+n]=pii(x,i+n);
	}
	sort(p+1,p+n+q+1);
	for(int i=1;i<=n+q;i++)pos[p[i].se]=i;
	Seg::build(1,1,n+q);
	cout<<Seg::s[1]<<'\n';
	for(int i=1;i<=q;i++){
		Seg::update(1,1,n+q,pos[upd[i]],0);
		pos[upd[i]]=pos[n+i];
		Seg::update(1,1,n+q,pos[upd[i]],1);
		cout<<Seg::s[1]<<'\n';
	}
}