21.7.24 test

\(NOIP\) 模拟赛

考差了。

T1签到题。注意存在字符串长度为0，不能直接模。\(100\rightarrow0\)

代码：

#include<bits/stdc++.h>
using namespace std;
int k1,k2,k3,len;
string s,t;
int s1[85],s2[85],s3[85];
char c1[85],c2[85],c3[85];
signed main(){
	freopen("a.in","r",stdin);
	freopen("a.in","w",stdout);
	cin>>k1>>k2>>k3>>s;
	t=s;len=s.length();
	for(int i=0;i<len;i++){
		if(s[i]>='a'&&s[i]<='i')s1[++s1[0]]=i,c1[s1[0]]=s[i];
		else if(s[i]>='j'&&s[i]<='r')s2[++s2[0]]=i,c2[s2[0]]=s[i];
		else s3[++s3[0]]=i,c3[s3[0]]=s[i];
	}
	k1%=s1[0],k2%=s2[0],k3%=s3[0];
	for(int i=1;i<=s1[0];i++){
		int v=i+k1;if(v>s1[0])v%=s1[0];
		t[s1[v]]=c1[i];
	}
	for(int i=1;i<=s2[0];i++){
		int v=i+k2;if(v>s2[0])v%=s2[0];
		t[s2[v]]=c2[i];
	}
	for(int i=1;i<=s3[0];i++){
		int v=i+k3;if(v>s3[0])v%=s3[0];
		t[s3[v]]=c3[i];
	}
	cout<<t;
	return 0;
}

---

T2

可以看出是平面最近点对，之前看到过但没做。。。\(100\rightarrow40\)

经典\(O(n\log n)\) 分治做法。

代码：

#include<bits/stdc++.h>
using namespace std;
#define int long long
#define in read()
inline int read(){
	int p=0,f=1;
	char c=getchar();
	while(c>'9'||c<'0'){if(c=='-')f=-1;c=getchar();}
	while(c>='0'&&c<='9'){p=p*10+c-'0';c=getchar();}
	return p*f;
}
const int N=100005;
struct node{
	int x,y;
}t[N],A[N],B[N];
int n;
inline bool cmp1(node a,node b){return a.x<b.x;}
inline bool cmp2(node a,node b){return a.y<b.y;}
inline int dis(node a,node b){
	int tx=a.x-b.x,ty=a.y-b.y;
	return tx*tx+ty*ty;
}
inline int binl(int key,int bn){
	int l=1,r=bn,mid;
	while(l<r){
		mid=(l+r)>>1;
		if(A[mid].y>key)r=mid;
		else l=mid+1;
	}
	return l;
}
inline int binr(int key,int bn){
	int l=1,r=bn,mid;
	while(l<r){
		mid=(l+r+1)>>1;
		if(A[mid].y<key)l=mid;
		else r=mid-1;
	}
	return l;
}
inline int solve(int l,int r){
	if(l==r)return 0x7fffffffffffffff;
	int mid=(l+r)>>1;
	int d1=solve(l,mid),d2=solve(mid+1,r),d=min(d1,d2),an=0,bn=0;
	for(int i=mid;t[mid].x-t[i].x<d&&i>=l;i--)
		A[++an]=t[i];
	for(int i=mid+1;t[i].x-t[mid].x<d&&i<=r;i++)
		B[++bn]=t[i];
	sort(B+1,B+1+bn,cmp2);
	sort(A+1,A+1+an,cmp2);
	int tl=binl(B[1].y-d,an),tr=binr(B[1].y+d,an);
	for(int i=1;i<=bn;i++){
		while(A[tl].y<=B[i].y-sqrt(d)&&tl<an)tl++;
		while(A[tr+1].y<B[i].y+sqrt(d)&&tr<bn)tr++;
		for(int j=tl;j<=tr;j++)
			d=min(d,dis(B[i],A[j]));
	}
	return d;
}
signed main(){
	n=in;
	for(int i=1;i<=n;i++){
		t[i].y=t[i-1].y+in;
		t[i].x=i;
	}
	sort(t+1,t+1+n,cmp1);
	printf("%lld",solve(1,n));
	return 0;
}

---

T3

需要对两个函数进行分析。正解链剖线段树。但暴力dfs加小优化也能A。我有想过优化但认为其能被卡，所以\(100\rightarrow40\)

代码：

#include<bits/stdc++.h>
using namespace std;
#define in read()
inline int read(){
	int p=0,f=1;
	char c=getchar();
	while(c>'9'||c<'0'){if(c=='-')f=-1;c=getchar();}
	while(c>='0'&&c<='9'){p=p*10+c-'0';c=getchar();}
	return p*f;
}
const int N=200005;
struct edge{
	int v,next;
}e[N*2];
int head[N],en;
void insert(int u,int v){
	e[++en].v=v;
	e[en].next=head[u];
	head[u]=en;
}
int n,q,a[N],fa[N],f1[N],f2[N];
void dfs(int u){
	int t1=f1[u],t2=f2[u];
	if(f1[fa[u]]+1<a[u])f1[u]=a[u],f2[u]=1;
	else if(f1[fa[u]]+1>a[u])f1[u]=f1[fa[u]]+1,f2[u]=f2[fa[u]];
	else f1[u]=f1[fa[u]]+1,f2[u]=f2[fa[u]]+1;
	if(t1==f1[u]&&t2==f2[u])return ;
	for(int i=head[u];i;i=e[i].next){
		int v=e[i].v;
		if(v==fa[u])continue;
		fa[v]=u;
		dfs(v);
	}
}
signed main(){
	//freopen("c.in","r",stdin);
	//freopen("c.out","w",stdout);
	n=in;
	for(int i=1;i<=n;i++)
		a[i]=in;
	for(int i=1;i<n;i++){
		int u=in,v=in;
		insert(u,v);
		insert(v,u);
	}
	f1[1]=a[1],f2[1]=1;
	for(int i=head[1];i;i=e[i].next){
		int v=e[i].v;
		fa[v]=1;
		dfs(v);
	}
	q=in;
	for(int i=1;i<=q;i++){
		int opt=in;
		if(opt==0){
			int u=in,val=in;
			a[u]=val;
			dfs(u);
		}
		if(opt==1){
			int u=in;
			cout<<f1[u]<<" "<<f2[u]<<'\n';
		}
	}
	return 0;
}

---

T4

在用权值线段树求LIS时做些修改即可，但我写挂了。\(100\rightarrow0\)

代码：

#include<bits/stdc++.h>
using namespace std;
const int N=50005;
const int mod=1000000007;
int n;
int a[N];
int qn,q[N];
int bin(int key){
	int l=1,r=qn;
	int mid;
	while(l<r){
		mid=(l+r)>>1;
		if(q[mid]>key)r=mid-1;
		else if(q[mid]<key)l=mid+1;
		else{l=mid;break;}
	}
	return l;
}

int len[N],fro[N];
//seg_tree----------------
#define m(x) tree[x].m
#define sum(x) tree[x].sum
struct node{
	int sum,m;
}tree[4*N];
void built(int l,int r,int p){
	m(p)=sum(p)=0;
	if(l==r)return ;
	int mid=(l+r)>>1;
	built(l,mid,p<<1);
	built(mid+1,r,p<<1|1);
	return ;
}
struct temp{int maxn,many;};
temp query(int l,int r,int p,int ql,int qr){
	temp ans;
	ans.many=ans.maxn=0;
	if(ql>qr)return ans;
	if(l>=ql&&r<=qr){
		ans.maxn=m(p);
		ans.many=sum(p);
		return ans;
	}
	int mid=(l+r)>>1;
	if(ql<=mid)ans=query(l,mid,p<<1,ql,qr);
	if(qr>mid){
		if(!ans.maxn)ans=query(mid+1,r,p<<1|1,ql,qr);
		else{
			temp ans1=query(mid+1,r,p<<1|1,ql,qr);
			if(ans.maxn==ans1.maxn) ans.many+=ans1.many;
			else if(ans.maxn<ans1.maxn)ans.maxn=ans1.maxn,ans.many=ans1.many;
		}
	}
	return ans;
}
void update(int l,int r,int p,int u,int d){
	if(l==r){
		if(d==m(p))sum(p)++;
		if(d>m(p)){m(p)=d;sum(p)=1;}
		return ;
	}
	int mid=(l+r)>>1;
	if(u<=mid)update(l,mid,p<<1,u,d);
	else update(mid+1,r,p<<1|1,u,d);
	if(m(p<<1)==m(p<<1|1)){
		m(p)=m(p<<1);
		sum(p)=sum(p<<1)+sum(p<<1|1);
	}
	else if(m(p<<1)>m(p<<1|1)){
		m(p)=m(p<<1);
		sum(p)=sum(p<<1);
	}
	else{
		m(p)=m(p<<1|1);
		sum(p)=sum(p<<1|1);
	}
	return ;
}
//------------------------
signed main(){
	freopen("d.in","r",stdin);
	freopen("d.out","w",stdout);
	cin>>n;
	for(int i=1;i<=n;i++){
		cin>>a[i];
		q[i]=a[i];
	}
	sort(q+1,q+1+n);
	qn=unique(q+1,q+1+n)-(q+1);
	for(int i=1;i<=n;i++)
		a[i]=bin(a[i]);
	built(1,qn,1);
	int maxlen=0,tans=0;
	for(int i=1;i<=n;i++){
		temp t=query(1,qn,1,1,a[i]-1);
		len[i]=1+t.maxn;
		fro[i]=t.many;

		update(1,qn,1,a[i],len[i]);
		if(len[i]>maxlen){
			maxlen=len[i];
			tans=fro[i]%mod;
		}
		else if(len[i]==maxlen)
			tans=(tans+fro[i])%mod;
	}
	//for(int i=1;i<=n;i++)
	//	cout<<len[i]<<" ";
	//cout<<endl;
	//for(int i=1;i<=n;i++)
	//	cout<<fro[i]<<" ";
	//cout<<maxlen<<" "<<tans;
	cout<<tans;
	return 0;
}

---

考的不理想.