AtCoder Beginner Contest 382 题解

A - Daily Cookie

	int n,k;
	cin>>n>>k;
	string p;
	cin>>p;
	int c=0;
	Rep(i,p.length()) c+=p[i]=='@';
	cout<<n-c+min(k,c);

B - Daily Cookie 2

	int n,k;
	cin>>n>>k;
	string p;
	cin>>p;
	RepD(i,n-1) {
		if(p[i]=='@' && k) --k,p[i]='.';
	}
	cout<<p;

C - Kaiten Sushi

pair<int,int> a[(int)2e5+10];
int main()
{
//	freopen("A.in","r",stdin);
//	freopen(".out","w",stdout);
	
	int n,m;
	cin>>n>>m;
	int l=0;
	For(i,n) {
		int p=read();
		if(l>0 && -a[l].fi<=p) continue;
		a[++l]={-p,i};
	}
	For(i,m) {
		int p=read();
		int id=upper_bound(a+1,a+1+l,mp(-p,0))-a;
		if(id>l) puts("-1");else
		cout<<a[id].se<<endl;
//		cout<<lower_bound(a+1,a+1+l,mp(p,0))-a<<endl;
	}	
	return 0;
}

D - Keep Distance

int m,a[1010]={-9},n;
int x=0;
void dfs(int l) {
	if(l>n) {
		++x;
		return;
	}
	Fork(i,a[l-1]+10,m) {
		if(m-i>=10*(n-l)){
			a[l]=i;dfs(l+1);
		}else break;
	}
}
void dfs2(int l) {
	if(l>n) {
		PRi(a,n)
		return;
	}
	Fork(i,a[l-1]+10,m) {
		if(m-i>=10*(n-l)){
			a[l]=i;dfs2(l+1);
		}else break;
	}
}
int main()
{
//	freopen("A.in","r",stdin);
//	freopen(".out","w",stdout);
	
	cin>>n>>m;
	dfs(1);cout<<x<<endl;
	dfs2(1);

	return 0;
}

E - Expansion Packs

int n,x;
double g[5010][5010]={},p[5010];
double f[5010];
int main()
{
//	freopen("e.in","r",stdin);
//	freopen(".out","w",stdout);
	cin>>n>>x;
	For(i,n) cin>>p[i];
	For(i,n) p[i]/=100.;
	g[0][0]=1;
//	PRi(p,n)
	For(i,n) {
		Rep(j,i+1)g[i][j]=g[i-1][j]*(1-p[i])+g[i-1][j-1]*p[i];
	}
//	Rep(j,n+1) cout<<g[n][j]<<endl;
	f[0]=0;
	For(i,x) {
		f[i]=1;
		For(j,n) if(i-j>0)f[i]+=f[i-j]*g[n][j];
		f[i]/=(1.-g[n][0]);
	}
	cout<<setprecision(10)<<f[x]<<endl;
	return 0;
}

G - Tile Distance 3

构造题

如图所示，已知K，对每个K*K的矩形横纵切K-1刀。按上面的样子拼。

2个矩形相邻当且仅当有有长度为正的公共边。

求任意2个点（保证不在格点上）所在的矩形间，至少经过几个相邻的矩形。

考虑通过旋转平移把其中一个点固定到(0,0)这个方块中。

平移和旋转，将其中一个点放到(i=0,j=0)的矩形中。

在通过翻转x=k/2,y=k/2把另一个点放到第一象限或xy正半轴。

然后分类讨论。

假设另一个点在y正半轴。

向右移动的过程中，直接移动， ( i , j , k ) − > ( i , j , k ′ ) (i,j,k)->(i,j,k') (i,j,k)−>(i,j,k′)只需要2步，可以优化。向上同理。

#include<bits/stdc++.h> 
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,0x3f,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define MEMx(a,b) memset(a,b,sizeof(a));
#define INF (0x3f3f3f3f)
#define F (1000000007)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int> 
#define pi pair<int,int>
#define SI(a) ((a).size())
#define Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;
#define PRi2D(a,n,m) For(i,n) { \
						For(j,m-1) cout<<a[i][j]<<' ';\
						cout<<a[i][m]<<endl; \
						} 
#pragma comment(linker, "/STACK:102400000,102400000")
#define ALL(x) (x).begin(),(x).end()
#define gmax(a,b) a=max(a,b);
#define gmin(a,b) a=min(a,b);
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return ((a-b)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
inline int read()
{
	int x=0,f=1; char ch=getchar();
	while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
	while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
	return x*f;
} 
bool iseven(ll x) {return x%2==0;}
ll k,sx,sy,tx,ty;
tuple<int,ll,ll,ll> get(ll x,ll y) {
	ll px=(x>=0)?(x/k):((x+1)/k-1),py=(y>=0)?(y/k):((y+1)/k-1),j;
	int d=0; //0 - 1 |
	if(iseven(px)==iseven(py)) d=0,j=(y%k+k)%k;else d=1,j=(x%k+k)%k;
	return {d,px,py,j};
}
pair<ll,ll> rotate90(ll x,ll y) {
	return {y,-x-1};
}

int main()
{
//	freopen("g.in","r",stdin);
//	freopen(".out","w",stdout);
	int T;
	
	k=3;
	
	cin>>T;
	while(T--) {
		cin>>k>>sx>>sy>>tx>>ty;
		auto [d1,i1,j1,k1]=get(sx,sy);
		if(d1) {
			tie(sx,sy)=rotate90(sx,sy);
			tie(tx,ty)=rotate90(tx,ty);
		}
		ll newsx=(sx%k+k)%k,newsy=(sy%k+k)%k;
		ll mvx=newsx-sx,mvy=newsy-sy;
		sx+=mvx;
		sy+=mvy;
		tx+=mvx;
		ty+=mvy;
		tie(d1,i1,j1,k1)=get(sx,sy);
		auto [d2,i2,j2,k2]=get(tx,ty);
		assert(d1==0 && i1==0 && j1==0);
		if(i2<0) { //flip x==k/2
			i2=-i2;
			if(d2%2==1) k2=k-1-k2;
		}
		if(j2<0) { //flip y==k/2
			j2=-j2;
			k1=k-1-k1;
			if(d2%2==0) k2=k-1-k2;
		}
		if(i2==0 && j2==0) {
			if(k1==k2) puts("0");
			else if(k1==k2+1 || k1==k2-1) puts("1");
			else puts("2");
			continue;
		}	
		
		ll ans=0;	
		ll dis=(i2==j2)?(i2-1) : (min(i2,j2));
		ans+=dis*2;
		i2-=dis,j2-=dis;
		
		if(i2==1 && j2==1) {
			ans+=min(k-1-k1+2,2+min(2ll,k2));
			cout<<ans<<endl;
			continue;
		}
		if(i2>0) {
			ans+=i2/2*(min(3ll,k)+1)+(i2%2)*(min(k2+1,3ll));
			cout<<ans<<endl;
			continue;
		}
		if(j2>0) {
			ll minO1O2=min(k-k1-1,2ll)+1;
			if(j2%2==1) cout<<ans+minO1O2+j2/2*min(4ll,k+1)<<endl;
			else {
				cout<<ans+minO1O2+min(3ll,k2+1)+(j2>=4)*(j2/2-1)*min(4ll,k+1)<<endl;
			}
		}
		
	}
	return 0;
}