CSP-S2019题解

D1T1

判断每一位是否超过一半，如果超了就把后面的反过来

注意不要把k+1

#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#define fo(a,b,c) for (a=b; a<=c; a++)
#define fd(a,b,c) for (a=b; a>=c; a--)
using namespace std;

unsigned long long p[64];
unsigned long long m;
int n,i,j,k,l;

int main()
{
	freopen("code.in","r",stdin);
	freopen("code.out","w",stdout);
	
	p[0]=1;
	fo(i,1,63)
	p[i]=p[i-1]*2;
	
	cin>>n>>m;
	
	fd(i,n-1,0)
	if (m<p[i])
	printf("0");
	else
	printf("1"),m=p[i]-(m-p[i])-1;
	printf("
");
	
	fclose(stdin);
	fclose(stdout);
	
	return 0;
}

D1T2

主席树碾过

找最后一个前缀和相同的

#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#define fo(a,b,c) for (a=b; a<=c; a++)
#define fd(a,b,c) for (a=b; a>=c; a--)
#define min(a,b) (a<b?a:b)
#define max(a,b) (a>b?a:b)
using namespace std;

int tr[10500001][5]; //sonl sonr max maxid sum
int b[500001];
int fa[500001];
long long ans[500001];
int d[500001];
int N,n,i,j,k,l,len,Find,Find2;
long long Ans;
char ch;

void New(int t,int x)
{
	++len;
	tr[len][0]=tr[tr[t][x]][0];
	tr[len][1]=tr[tr[t][x]][1];
	tr[len][2]=tr[tr[t][x]][2];
	tr[len][3]=tr[tr[t][x]][3];
	tr[len][4]=tr[tr[t][x]][4];
	
	tr[t][x]=len;
}

void change(int t,int l,int r,int x,int s)
{
	int mid=(l+r)/2;
	
	if (d[s]>tr[t][2])
	{
		tr[t][2]=d[s];
		tr[t][3]=s;
	}
	
	if (l==r)
	{
		++tr[t][4];
		return;
	}
	
	if (x<=mid)
	{
		New(t,0);
		change(tr[t][0],l,mid,x,s);
	}
	else
	{
		New(t,1);
		change(tr[t][1],mid+1,r,x,s);
	}
}

void find1(int t,int l,int r,int x,int y)
{
	if (x>y) return;
	
	int mid=(l+r)/2;
	
	if (x<=l && r<=y)
	{
		if (tr[t][2]>Find)
		Find=tr[t][2],Find2=tr[t][3];
		
		return;
	}
	
	if (x<=mid && tr[t][0])
	find1(tr[t][0],l,mid,x,y);
	if (mid<y && tr[t][1])
	find1(tr[t][1],mid+1,r,x,y);
}

int find2(int t1,int t2,int l,int r,int x)
{
	int mid=(l+r)/2;
	
	if (l==r)
	return tr[t2][4]-tr[t1][4];
	
	if (x<=mid)
	return find2(tr[t1][0],tr[t2][0],l,mid,x);
	else
	return find2(tr[t1][1],tr[t2][1],mid+1,r,x);
}

int main()
{
	freopen("brackets.in","r",stdin);
	freopen("brackets.out","w",stdout);
	
	scanf("%d",&n);N=n+n+1;
	fo(i,1,n)
	{
		ch=getchar();
		while (ch!='(' && ch!=')')
		ch=getchar();
		
		if (ch=='(')
		b[i]=1;
		else
		b[i]=-1;
	}
	fo(i,2,n)
	scanf("%d",&fa[i]);
	
	d[1]=1;
	len=n;
	
	change(1,1,N,b[1]+(n+1),1);
	fo(i,2,n)
	{
		tr[i][0]=tr[fa[i]][0];
		tr[i][1]=tr[fa[i]][1];
		tr[i][2]=tr[fa[i]][2];
		tr[i][3]=tr[fa[i]][3];
		tr[i][4]=tr[fa[i]][4];
		
		d[i]=d[fa[i]]+1;
		b[i]+=b[fa[i]];
		
		Find=0;
		Find2=0;
		
		find1(i,1,N,1,b[i]-1+(n+1));
		
		ans[i]=find2(Find2,i,1,N,b[i]+(n+1));
		if (b[i]==0 && !Find2)
		++ans[i];
		
		change(i,1,N,b[i]+(n+1),i);
	}
	fo(i,1,n)
	{
		ans[i]+=ans[fa[i]];
		Ans^=ans[i]*i;
	}
	
	printf("%lld
",Ans);
	
	fclose(stdin);
	fclose(stdout);
	
	return 0;
}

D1T3

枚举每个数最终到哪个点上，因为直接判边与边的关系不好搞，所以用链表维护每个点相连的边之间的关系

关系有三种：第一条，两条边相邻，最后一条

再建两个点表示头和尾，枚举时判断是否合法即可

因为剩下无限制的边随便放必定合法

#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#define fo(a,b,c) for (a=b; a<=c; a++)
#define fd(a,b,c) for (a=b; a>=c; a--)
#define min(a,b) (a<b?a:b)
using namespace std;

int a[4002][2];
int ls[2001];
int b[2001];
int pre[2001][2002];
int nxt[2001][2002];
int ans[2001];
int d[2001];
int d1[2001];
int d2[2001];
int T,n,i,j,k,l,len,mn,tot;
bool bz;

void New(int x,int y)
{
	++len;
	a[len][0]=y;
	a[len][1]=ls[x];
	ls[x]=len;
}

void dfs(int Fa,int Ls,int t)
{
	int id,i,j,k,l;
	
	if (t==2 && Fa==0)
	n=n;
	
	if (Ls && nxt[t][Ls]==Ls && pre[t][n+1]==n+1 && (pre[t][Ls]!=0 || d[t]==1))
	mn=min(mn,t);
	
	for (i=ls[t]; i; i=a[i][1])
	if (a[i][0]!=Fa)
	{
		if (t==3)
		n=n;
		
		id=i/2;
		
		if (!Ls && pre[t][id]==id && nxt[t][0]==0 && (nxt[t][id]!=n+1 || d[t]==1) || Ls && pre[t][Ls]!=id && nxt[t][Ls]==Ls && pre[t][id]==id && (pre[t][Ls]!=0 || nxt[t][id]!=n+1 || d[t]==1))
		dfs(t,id,a[i][0]);
	}
}

void Dfs(int Fa,int t)
{
	int i;
	
	if (t==mn)
	{
		bz=1;
		return;
	}
	
	for (i=ls[t]; i; i=a[i][1])
	if (a[i][0]!=Fa)
	{
		++tot;
		d1[tot]=a[i][0];
		d2[tot]=i/2;
		
		Dfs(t,a[i][0]);
		
		if (bz)
		return;
		
		--tot;
	}
}

int main()
{
	freopen("tree.in","r",stdin);
	freopen("tree.out","w",stdout);
	
	scanf("%d",&T);
	for (;T;--T)
	{
		memset(ls,0,sizeof(ls));
		memset(d,0,sizeof(d));
		len=1;
		
		scanf("%d",&n);
		fo(i,1,n)
		scanf("%d",&b[i]),++d[i];
		fo(i,2,n)
		{
			scanf("%d%d",&j,&k);
			
			New(j,k);
			New(k,j);
			
			++d[j];
			++d[k];
		}
		
		if (n==1)
		{
			printf("1
");
			continue;
		}
		
		fo(i,1,n)
		{
			fo(j,0,n+1)
			pre[i][j]=nxt[i][j]=j;
		}
		
		fo(i,1,n)
		{
			bz=tot=0;
			mn=n+1;
			
			dfs(0,0,b[i]);
			Dfs(0,b[i]);
			
			nxt[b[i]][0]=d2[1];
			pre[b[i]][nxt[b[i]][d2[1]]]=0;
			
			if (nxt[b[i]][d2[1]]!=d2[1])
			pre[b[i]][d2[1]]=nxt[b[i]][d2[1]]=-1;
			
			--d[b[i]];
			
			fo(j,1,tot-1)
			{
				k=pre[d1[j]][d2[j]];
				l=nxt[d1[j]][d2[j+1]];
				
				nxt[d1[j]][k]=l;
				pre[d1[j]][l]=k;
				
				if (pre[d1[j]][d2[j]]!=d2[j])
				nxt[d1[j]][d2[j]]=pre[d1[j]][d2[j]]=-1;
				if (nxt[d1[j]][d2[j+1]]!=d2[j+1])
				nxt[d1[j]][d2[j+1]]=pre[d1[j]][d2[j+1]]=-1;
				
				--d[d1[j]];
			}
			
			pre[d1[tot]][n+1]=d2[tot];
			nxt[d1[tot]][pre[d1[tot]][d2[tot]]]=n+1;
			
			if (pre[d1[tot]][d2[tot]]!=d2[tot])
			pre[d1[tot]][d2[tot]]=nxt[d1[tot]][d2[tot]]=-1;
			
			--d[d1[tot]];
			
			ans[i]=mn;
		}
		
		fo(i,1,n)
		printf("%d ",ans[i]);
		printf("
");
	}
}

D2T1

n^3m显然，把总数和钦定的数做差即可续走一个n

#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#define fo(a,b,c) for (a=b; a<=c; a++)
#define fd(a,b,c) for (a=b; a>=c; a--)
#define add(a,b) a=((a)+(b))%998244353
#define mod 998244353
#define Mod 998244351
using namespace std;

int a[101][2001];
long long sum[101];
long long f[101][201];
int n,m,i,j,k,l;
long long ans;

int main()
{
	freopen("meal.in","r",stdin);
	freopen("meal.out","w",stdout);
	
	ans=1;
	scanf("%d%d",&n,&m);
	fo(i,1,n)
	{
		fo(j,1,m)
		{
			scanf("%d",&a[i][j]);
			add(sum[i],a[i][j]);
		}
		ans=ans*(sum[i]+1)%mod;
	}
	--ans;
	
	fo(l,1,m)
	{
		memset(f,0,sizeof(f));
		f[0][100]=1;
		
		fo(i,0,n-1)
		{
			fo(j,-i,i)
			if (f[i][j+100])
			{
				add(f[i+1][j+1+100],f[i][j+100]*a[i+1][l]);
				add(f[i+1][j-1+100],f[i][j+100]*(sum[i+1]-a[i+1][l]));
				add(f[i+1][j+100],f[i][j+100]);
			}
		}
		
		fo(j,1,n)
		add(ans,-f[n][j+100]);
	}
	
	printf("%lld
",(ans+mod)%mod);
}

D2T2

找规律，每次从后选最靠右的合法段

处理出每个前缀的最小末段和，单调栈优化

证明见uoj

#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#define fo(a,b,c) for (register int a=b; a<=c; a++)
#define fd(a,b,c) for (register int a=b; a>=c; a--)
using namespace std;

__int128 ans,s;
long long a[40000001];
long long f[40000001];
int d[40000001];
int P[100001];
int L[100001];
int R[100001];
int n,type,i,j,k,l,h,t;
long long x,y,z,m;

void Printf(__int128 t)
{
	if (t)
	{
		Printf(t/10);
		printf("%d",(int)(t%10));
	}
}

int main()
{
	freopen("partition.in","r",stdin);
	freopen("partition.out","w",stdout);
	
	scanf("%d%d",&n,&type);
	if (!type)
	{
		fo(i,1,n)
		scanf("%lld",&a[i]),a[i]+=a[i-1];
	}
	else
	{
		scanf("%lld%lld%lld%lld%lld%lld",&x,&y,&z,&a[1],&a[2],&m);
		fo(i,1,m)
		scanf("%d%d%d",&P[i],&L[i],&R[i]);
		
		fo(i,3,n)
		a[i]=(x*a[i-1]+y*a[i-2]+z)%1073741824;
		
		fo(i,1,m)
		{
			fo(j,P[i-1]+1,P[i])
			a[j]=a[j]%(R[i]-L[i]+1)+L[i];
		}
		
		fo(i,1,n)
		a[i]+=a[i-1];
	}
	
	h=t=1;
	d[1]=0;
	fo(i,1,n)
	{
		while (h<t && f[d[h+1]]<=a[i])
		++h;
		
		f[i]=a[i]-a[d[h]]+a[i];
		while (h<=t && f[d[t]]>=f[i])
		--t;
		
		d[++t]=i;
	}
	
	l=n;
	fd(i,n,1)
	if (f[i-1]<=a[l])
	{
		s=a[l]-a[i-1];
		ans+=s*s;
		
		l=i-1;
	}
	
	Printf(ans);
	printf("
");
}

D2T3

重心性质：

1.重心一定在重链上

如果不在那么一定可以向重链方向移动

2.一个点是重心当前仅当该点的重儿子size<=n/2

若不满足则可以向重儿子方向移动，移动后必更优

---

考虑计算断掉每条边的贡献，向下直接倍增找到最后一个总size-当前size<=总size/2的点，这个点&其父亲可能是重心

向上的就边做边维护倍增数组，回溯时修改断边深度浅的点

#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#define fo(a,b,c) for (a=b; a<=c; a++)
#define fd(a,b,c) for (a=b; a>=c; a--)
#define max(a,b) (a>b?a:b)
using namespace std;

int a[600001][2];
int ls[300001];
int fa[300001];
int nx[300001][19];
int Nx[300001][19];
int size[300001];
int T,n,i,j,k,l,len;
long long ans;

void New(int x,int y)
{
	++len;
	a[len][0]=y;
	a[len][1]=ls[x];
	ls[x]=len;
}

void dfs1(int Fa,int t)
{
	int i,mx1=0,mx2=0;
	
	fa[t]=Fa;
	
	size[t]=1;
	for (i=ls[t]; i; i=a[i][1])
	if (a[i][0]!=Fa)
	{
		dfs1(t,a[i][0]);
		size[t]+=size[a[i][0]];
		
		if (size[a[i][0]]>mx1)
		mx1=size[a[i][0]],mx2=a[i][0];
	}
	
	nx[t][0]=mx2;
	fo(i,1,18)
	nx[t][i]=nx[nx[t][i-1]][i-1];
}

void dfs2(int Fa,int t)
{
	int i,j,mx1=0,mx2=0,Mx1=0,Mx2=0,Size;
	
	for (i=ls[t]; i; i=a[i][1])
	{
		if (size[a[i][0]]>mx1)
		{
			Mx1=mx1,Mx2=mx2;
			mx1=size[a[i][0]],mx2=a[i][0];
		}
		else
		if (size[a[i][0]]>Mx1)
		Mx1=size[a[i][0]],Mx2=a[i][0];
	}
	
	for (i=ls[t]; i; i=a[i][1])
	if (a[i][0]!=Fa)
	{
		Size=size[t];
		fa[t]=a[i][0];
		size[t]=n-size[a[i][0]];
		
		if (a[i][0]==mx2)
		{
			if (Mx1>n-Size)
			Nx[t][0]=Mx2;
			else
			Nx[t][0]=Fa;
		}
		else
		{
			if (mx1>n-Size)
			Nx[t][0]=mx2;
			else
			Nx[t][0]=Fa;
		}
		
		fo(j,1,18)
		Nx[t][j]=Nx[Nx[t][j-1]][j-1];
		
		k=t;
		fd(j,18,0)
		if (Nx[k][j] && size[t]-size[Nx[k][j]]<=size[t]/2)
		k=Nx[k][j];
		
		ans+=k;
		if (k!=t && !(size[t]&1) && size[t]-size[k]==size[t]/2)
		ans+=fa[k];
		
		dfs2(t,a[i][0]);
		
		fa[t]=Fa;
		size[t]=Size;
	}
	
	fo(j,0,18)
	Nx[t][j]=nx[t][j];
}

int main()
{
	freopen("centroid.in","r",stdin);
	freopen("centroid.out","w",stdout);
	
	scanf("%d",&T);
	for (;T;--T)
	{
		memset(ls,0,sizeof(ls));
		ans=len=0;
		
		scanf("%d",&n);
		fo(i,2,n)
		{
			scanf("%d%d",&j,&k);
			
			New(j,k);
			New(k,j);
		}
		
		dfs1(0,1);
		
		fo(i,1,n)
		{
			fo(j,0,18)
			Nx[i][j]=nx[i][j];
			
			if (i>1)
			{
				k=i;
				fd(j,18,0)
				if (nx[k][j] && size[i]-size[nx[k][j]]<=size[i]/2)
				k=nx[k][j];
				
				ans+=k;
				if (k!=i && !(size[i]&1) && size[i]-size[k]==size[i]/2)
				ans+=fa[k];
			}
		}
		
		dfs2(0,1);
		
		printf("%lld
",ans);
	}
}