Codeforces Round #494 (Div 3) (A~E)

目录

• Codeforces 1003
  • A.Polycarp's Pockets
  • B.Binary String Constructing
  • C.Intense Heat
  • D.Coins and Queries
  • E.Tree Constructing(贪心)
  • F.Abbreviation

---

Codeforces 1003

比赛链接

这么做是不是不太好啊233

A.Polycarp's Pockets

#include <cstdio>
#include <cctype>
#include <algorithm>
#define gc() getchar()
const int N=500;
int n,A[N],tm[N];
inline int read()
{
	int now=0,f=1;register char c=gc();
	for(;!isdigit(c);c=='-'&&(f=-1),c=gc());
	for(;isdigit(c);now=now*10+c-'0',c=gc());
	return now*f;
}
int main()
{
	n=read();
	for(int i=1; i<=n; ++i) ++tm[read()];
	int res=0;
	for(int i=1; i<=100; ++i) res=std::max(res,tm[i]);
	printf("%d",res);
	return 0;
}

B.Binary String Constructing

#include <cstdio>
#include <cctype>
#include <algorithm>
#define gc() getchar()
const int N=10000;
int A,B,X;
char s[N];
inline int read()
{
	int now=0,f=1;register char c=gc();
	for(;!isdigit(c);c=='-'&&(f=-1),c=gc());
	for(;isdigit(c);now=now*10+c-'0',c=gc());
	return now*f;
}
int main()
{
	A=read(),B=read(),X=read();//A:0 B:1
	int n=A+B;
//	if(X==1)
//	{
//		for(int i=1; i<=A; ++i) putchar('0');
//		for(int i=A+1; i<=n; ++i) putchar('1');
//		return 0;
//	}
	const char f=A>=B?'1':'0', s=A>=B?'0':'1';
	if(A>B) std::swap(A,B);
	for(int i=1; i<=X; ++i)
		putchar((i&1)?s:f), --((i&1)?B:A);
	if(X&1)
	{
		for(int i=X+1; i<=X+B; ++i) putchar(s);
		for(int i=X+B+1; i<=n; ++i) putchar(f);
	}
	else
	{
		for(int i=X+1; i<=X+A; ++i) putchar(f);
		for(int i=X+A+1; i<=n; ++i) putchar(s);
	}
	return 0;
}

C.Intense Heat

#include <cstdio>
#include <cctype>
#include <algorithm>
#define gc() getchar()
#define eps (1e-8)
const int N=5007;
int n,K,A[N];
inline int read()
{
	int now=0,f=1;register char c=gc();
	for(;!isdigit(c);c=='-'&&(f=-1),c=gc());
	for(;isdigit(c);now=now*10+c-'0',c=gc());
	return now*f;
}
int main()
{
	n=read(),K=read();
	for(int i=1; i<=n; ++i) A[i]=read();
	double ans=0;
	for(int i=1; i<=n; ++i)//O(n^2) 这种暴力我最会写了233
	{
		int sum=0;
		for(int j=i; j<=n; ++j)
		{
			sum+=A[j];
			if(j-i+1>=K) ans=std::max(ans,1.0*sum/(j-i+1));
		}
	}
	printf("%.10lf",ans);
	return 0;
}

D.Coins and Queries

#include <cmath>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <algorithm>
#define gc() getchar()
const int N=2e5+5, INF=0x7fffffff;
int n,Q,A[N],tot[999],num[999];
inline int read()
{
	int now=0;register char c=gc();
	for(;!isdigit(c);c=gc());
	for(;isdigit(c);now=now*10+c-'0',c=gc());
	return now;
}
inline int Count(int x){
	for(int i=0; i<31; ++i) if(x&(1<<i)) return i;
}
int Query()
{//写的好像麻烦了 当时非常zz
	memcpy(num,tot,sizeof tot);
	int B=read(),ans=0;
	for(int i=30; ~i; --i)
	{
		if(!(B&(1<<i))) continue;
		if(num[i]) ++ans;
		else
		{
			int need=1<<i;
			for(int j=i-1; ~j&&need; --j)
			{
				if(!num[j]) continue;
				int v=std::min(num[j],need>>j);//need/(1<<j);//not ceil!
				ans+=v, need-=v*(1<<j), num[j]-=v;
			}
			if(need) return -1;
		}
	}
	return ans;
}
int main()
{
	n=read(), Q=read();
	for(int i=1; i<=n; ++i) ++tot[Count(read())];
	while(Q--) printf("%d\n",Query());
	return 0;
}

---

比赛结束后

E.Tree Constructing(贪心)

贪心策略显然 但可能要注意特殊情况(比如D=N-1,K=1)。

WA了四次结果结束后发现脑抽写错一行 改了就A了啊mmp

#include <cstdio>
#include <cctype>
#include <algorithm>
#define gc() getchar()
const int N=4e5+7,M=N<<1;
int n,D,K,now,dgr[N],Enum,H[N],nxt[M],to[M];
inline int read()
{
	int now=0;register char c=gc();
	for(;!isdigit(c);c=gc());
	for(;isdigit(c);now=now*10+c-'0',c=gc());
	return now;
}
inline void AddEdge(int u,int v)
{
	++dgr[u], to[++Enum]=v, nxt[Enum]=H[u], H[u]=Enum;
	++dgr[v], to[++Enum]=u, nxt[Enum]=H[v], H[v]=Enum;
}
void Link(int x,int dep)
{
	if(!dep) return;
	for(int i=K-dgr[x]; i>0&&now<n; --i) AddEdge(x,++now), Link(now,dep-1);
}
bool Solve()
{
	if(D>=n) return 0;
	for(int i=1; i<=D; ++i) AddEdge(i,i+1);
	now=D+1;
	for(int i=2; i<=D&&now<n; ++i) Link(i,D-std::max(i-1,D+1-i));
	for(int i=1; i<=n; ++i) if(dgr[i]>K) return 0;
//	DFS1(1,1,0), Max=0, DFS1(V,V,0);
//	if(Max!=D) return 0;
	return now>=n;
}
void DFS_for_Ans(int x,int f)
{
	for(int i=H[x]; i; i=nxt[i])
		if(to[i]!=f) printf("%d %d\n",x,to[i]), DFS_for_Ans(to[i],x);
}
int main()
{
	n=read(), D=read(), K=read();
	if(Solve()) puts("YES"), DFS_for_Ans(1,1);
	else puts("NO");
	return 0;
}

F.Abbreviation

待补