Codeforces Global Round 1 自闭记

其他 2019-02-08 14:08:04 阅读次数: 0

　　A：签到。

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long

char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
	int x=0,f=1;char c=getchar();
	while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
	while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
	return x*f;
}
int b,k,a[100010];
signed main()
{
/*#ifndef ONLINE_JUDGE
	freopen("a.in","r",stdin);
	freopen("a.out","w",stdout);
	const char LL[]="%I64d\n";
#endif*/
	b=read(),k=read();b&=1;
	for (int i=1;i<=k;i++) a[i]=read();
	if (b==0)
	{
		if (a[k]&1) cout<<"odd";
		else cout<<"even";
	}
	else
	{
		int s=0;
		for (int i=1;i<=k;i++) s+=a[i];
		if (s&1) cout<<"odd";
		else cout<<"even";
	}
	return 0;
	//NOTICE LONG LONG!!!!!
}

　　B：显然先连接距离较小的点。

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 100010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
	int x=0,f=1;char c=getchar();
	while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
	while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
	return x*f;
}
int n,m,k,a[N],b[N],ans;
signed main()
{
/*#ifndef ONLINE_JUDGE
	freopen("a.in","r",stdin);
	freopen("a.out","w",stdout);
	const char LL[]="%I64d\n";
#endif*/
	n=read(),m=read(),k=read();
	for (int i=1;i<=n;i++) a[i]=read();
	ans=a[n]-a[1]+1;
	for (int i=2;i<=n;i++) b[i]=a[i]-a[i-1];
	sort(b+2,b+n+1);reverse(b+2,b+n+1);
	for (int i=2;i<=k;i++) ans-=b[i]-1;
	cout<<ans;
	return 0;
	//NOTICE LONG LONG!!!!!
}

　　C：当a!=2k-1时，令k为满足2k-1>a的最小正整数，显然可以令b=2k-1^a使答案成为2k-1，并且显然不可能更优。a=2k-1时直接暴力打表。

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 100010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
	int x=0,f=1;char c=getchar();
	while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
	while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
	return x*f;
}
int q,ans[1<<25];
signed main()
{
/*#ifndef ONLINE_JUDGE
	freopen("a.in","r",stdin);
	freopen("a.out","w",stdout);
	const char LL[]="%I64d\n";
#endif*/
	q=read();
	for (int i=1;i<=22;i++) 
	{
		for (int j=1;j<(1<<i)-1;j++)
		ans[(1<<i)-1]=max(ans[(1<<i)-1],gcd((1<<i)-1&j,(1<<i)-1^j));
	}
	ans[(1<<23)-1]=178481;
	ans[(1<<24)-1]=5592405;
	ans[(1<<25)-1]=1082401;
	while (q--)
	{
		int x=read();
		if (ans[x]>0) printf("%d\n",ans[x]);
		else
		{
			while (x!=(x&-x)) x^=x&-x;
			printf("%d\n",(x<<1)-1);
		}
	}
	return 0;
	//NOTICE LONG LONG!!!!!
}

　　E：https://www.cnblogs.com/Gloid/p/10060025.html 作为一个做过所谓原题的选手看了1h这个题感到十分自闭。事实上第一眼就想到了这个原题，发现并不一样之后就去梦游了。毕竟当时也不是用前缀和的做法做的，而是自己想了半天搞了个做法还觉得挺nb，反应不过来也挺正常。但现在感觉我这做法好像也能类似的搬过来？自闭了啊？

　　比较差分集合是否相同即可，因为每次操作相当于交换差分数组中相邻两数。注意特判首尾。

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 100010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
	int x=0,f=1;char c=getchar();
	while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
	while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
	return x*f;
}
int n,q[N<<4],head,tail,cnt;
ll a[N],b[N];
bool flag[N];
signed main()
{
/*#ifndef ONLINE_JUDGE
	freopen("a.in","r",stdin);
	freopen("a.out","w",stdout);
	const char LL[]="%I64d\n";
#endif*/
	n=read();
	for (int i=1;i<=n;i++) a[i]=read();
	for (int i=1;i<=n;i++) b[i]=read();
	if (a[1]!=b[1]||a[n]!=b[n]) {cout<<"No";return 0;}
	for (int i=n;i>=1;i--) a[i]-=a[i-1];
	for (int i=n;i>=1;i--) b[i]-=b[i-1];
	sort(a+1,a+n+1);
	sort(b+1,b+n+1);
	for (int i=1;i<=n;i++) if (a[i]!=b[i]) {cout<<"No";return 0;}
	cout<<"Yes";
	return 0;
	//NOTICE LONG LONG!!!!!
}

　　D：注意到一定存在最优方案使得同种顺子最多出两次。然后就是普及组dp了。我这种弱智怎么可能注意的到啊。

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<vector> 
using namespace std;
#define ll long long
#define N 1000010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
	int x=0,f=1;char c=getchar();
	while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
	while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
	return x*f;
}
int n,m,u,a[N],f[N][3][3];
signed main()
{
#ifndef ONLINE_JUDGE
	freopen("a.in","r",stdin);
	freopen("a.out","w",stdout);
	const char LL[]="%I64d\n";
#endif
	n=read(),m=read();
	for (int i=1;i<=n;i++) a[read()]++;
	memset(f,200,sizeof(f));
	f[0][0][0]=0;
	for (int i=1;i<=m+2;i++)
		for (int j=0;j<=2;j++)
			for (int k=0;k<=2;k++)
				for (int x=0;x<=2;x++)
				if (k<=a[i]&&j+k<=a[i-1]&&j+k+x<=a[i-2])
				f[i][j][k]=max(f[i][j][k],f[i-1][x][j]+x+(a[i-2]-j-k-x)/3);
	cout<<f[m+2][0][0];
	return 0;
	//NOTICE LONG LONG!!!!!
}

　　F：对dfs序建线段树维护当前点到每个点的距离（非叶子设为inf），询问按dfs序离线，然后整棵树dfs一遍并维护距离即可，这个维护仅仅是线段树上的区间加和区间减。我觉得比D简单多了！算了云选手没资格说话。

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
#define ll long long
#define N 500010
#define inf 100000000000000000ll
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
	int x=0,f=1;char c=getchar();
	while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
	while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
	return x*f;
}
int n,m,fa[N],p[N],len[N],dfn[N],id[N],L[N<<2],R[N<<2],size[N],t,cnt,cur;
ll deep[N],ans[N],tree[N<<2],lazy[N<<2];
struct data
{
	int to,len;
	bool operator <(const data&a) const
	{
		return to<a.to;
	}
};
vector<data> edge[N];
struct data2
{
	int x,l,r,i;
	bool operator <(const data2&a) const
	{
		return dfn[x]<dfn[a.x];
	}
}q[N];
void addedge(int x,int y,int z){edge[x].push_back((data){y,z});}
void dfs(int k)
{
	dfn[k]=++cnt;id[cnt]=k;size[k]=1;
	for (int i=0;i<edge[k].size();i++)
	{
		deep[edge[k][i].to]=deep[k]+edge[k][i].len;
		dfs(edge[k][i].to);
		size[k]+=size[edge[k][i].to];
	}
	if (edge[k].size()) deep[k]=inf;
}
void up(int k){tree[k]=min(tree[k<<1],tree[k<<1|1]);}
void build(int k,int l,int r)
{
	L[k]=l,R[k]=r;
	if (l==r)
	{
		tree[k]=deep[id[l]];
		return;
	}
	int mid=l+r>>1;
	build(k<<1,l,mid);
	build(k<<1|1,mid+1,r);
	up(k);
}
void update(int k,ll x){tree[k]+=x;lazy[k]+=x;}
void down(int k){update(k<<1,lazy[k]),update(k<<1|1,lazy[k]),lazy[k]=0;}
void add(int k,int l,int r,int x)
{
	if (l>r) return;
	if (L[k]==l&&R[k]==r) {update(k,x);return;}
	if (lazy[k]) down(k);
	int mid=L[k]+R[k]>>1;
	if (r<=mid) add(k<<1,l,r,x);
	else if (l>mid) add(k<<1|1,l,r,x);
	else add(k<<1,l,mid,x),add(k<<1|1,mid+1,r,x);
	up(k);
}
ll query(int k,int l,int r)
{
	if (L[k]==l&&R[k]==r) return tree[k];
	if (lazy[k]) down(k);
	int mid=L[k]+R[k]>>1;
	if (r<=mid) return query(k<<1,l,r);
	else if (l>mid) return query(k<<1|1,l,r);
	else return min(query(k<<1,l,mid),query(k<<1|1,mid+1,r));
}
void work(int k)
{
	while (k==q[cur+1].x) cur++,ans[q[cur].i]=query(1,q[cur].l,q[cur].r);
	for (int i=0;i<edge[k].size();i++)
	{
		int to=edge[k][i].to;
		add(1,dfn[to],dfn[to]+size[to]-1,-edge[k][i].len);
		add(1,1,dfn[to]-1,edge[k][i].len);
		add(1,dfn[to]+size[to],n,edge[k][i].len);
		work(to);
		add(1,dfn[to],dfn[to]+size[to]-1,edge[k][i].len);
		add(1,1,dfn[to]-1,-edge[k][i].len);
		add(1,dfn[to]+size[to],n,-edge[k][i].len);
	}
}
signed main()
{
#ifndef ONLINE_JUDGE
	freopen("a.in","r",stdin);
	freopen("a.out","w",stdout);
#endif
	n=read(),m=read();
	for (int i=2;i<=n;i++)
	{
		fa[i]=read();len[i]=read();
		addedge(fa[i],i,len[i]);
	}
	for (int i=1;i<=n;i++) sort(edge[i].begin(),edge[i].end());
	dfs(1);
	for (int i=1;i<=m;i++) q[i].x=read(),q[i].l=read(),q[i].r=read(),q[i].i=i;
	sort(q+1,q+m+1);
	build(1,1,n);
	work(1);
	for (int i=1;i<=m;i++) printf("%I64d\n",ans[i]);
	return 0;
	//NOTICE LONG LONG!!!!!
}

　　div1+div2的场从来没进过前400，自闭了。

　　result：rank 420 rating +22

猜你喜欢

转载自www.cnblogs.com/Gloid/p/10356065.html
今日推荐
TIOBE 5 月榜单：Fortran “复活”进入 Top 10
GCC 14.1 发布
面壁智能发布 Eurux-8x22B 开源大模型 —— 堪称「理科状元」
开源日报 | 谷歌扶持鸿蒙上位；开源Rabbit R1；Docker加持的安卓手机；微软的焦虑和野心；海尔电器把开放平台关了
中国码农的“35岁魔咒”
蘭雅 CorelDRAW 插件 2024.5.1 国际劳动节版，免费下载
Arc Browser for Windows 1.0 正式 GA
90后程序员开发视频搬运软件、不到一年获利超 700 万，结局很刑！
周排行
Java自定义时间格式
同步整形电路
在开发中最最最常用的字符串的属性大集合
Linux 查看端口占用并杀掉
Java基础四：ArrayList
多线程之死锁就是这么简单
mysql 基础命令集
awk 命令详解
Centos6.3编译安装nginx+php步骤
OCR （Optical Character Recognition，光学字符识别）
每日归档
更多
2024-05-08(42)
2024-05-07(14)
2024-05-06(40)
2024-05-05(0)
2024-05-04(7)
2024-05-03(19)
2024-05-02(0)
2024-05-01(4)
2024-04-30(1)
2024-04-29(40)

代码天地 © 2018 codetd.com

境外服务器   最新电视剧2022