CTU Open Contest 2019部分题解

eg：出题人是真的喜欢喝啤酒- -只出了七题，还有三题没怎么看。
比赛传送门

A-Beer Barrels

题意：长度为k的数字由a,b组成，问其中字符c出现的次数。
题解：分类讨论。若$a=b=c,ans=k$，若$a\neq b$且$a=c$或$b=c$，那么答案就为$\sum_{i=0}^kC_k^i(k-i)$,若$a\neq c$且$b\neq c$答案为0.

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=1e5+10;
const int mod=1e9+7;
const int INF=0x3f3f3f3f;
ll read(){
    ll f=1,x=0;char ch;
    do{ch=getchar();if(ch=='-')f=-1;}while(ch<'0'||ch>'9');
    do{x=x*10+ch-'0';ch=getchar();}while(ch>='0'&&ch<='9');
    return f*x;
}
int a,b,k,c;
ll quick(ll a,ll b){
	ll res=1;
	while(b){
		if(b&1)res=res*a%mod;
		a=a*a%mod;
		b>>=1;
	}
	return res;
}
ll solve(){
	if(a!=c && b!=c)return 0;
	if(k==0)return 0;
	if(a==c && b==c && a==b)return 1ll*k;
	if(a==c || b==c){
		return 1ll*k*quick(2,k-1)%mod;
	}
}
int main(){
	scanf("%d%d%d%d",&a,&b,&k,&c); 
	printf("%lld\n",solve());
	return 0;
}

B-Beer Bill

题意：阅读理解题，算出数字乘上竖线的和，若没有数字则按照42来算，向上取整到10.
题解：模拟即可。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=1e5+10;
const int mod=1e9+7;
const int INF=0x3f3f3f3f;
ll read(){
    ll f=1,x=0;char ch;
    do{ch=getchar();if(ch=='-')f=-1;}while(ch<'0'||ch>'9');
    do{x=x*10+ch-'0';ch=getchar();}while(ch>='0'&&ch<='9');
    return f*x;
}
char s[maxn];
int ans=0;
int main(){
//	freopen("1.txt","r",stdin);
	while(~scanf("%s",&s)){
		int len=strlen(s),cnt=0,num=0,pos=0;
		for(int i=len-1;i>=0;i--){
			if(s[i]=='|')cnt++;
			if(s[i]==','){
				pos=i;
				break;
			}
		}
		for(int i=0;i<pos;i++){
			num=num*10+s[i]-'0';
		}
		if(!num)num=42;
		ans=ans+num*cnt;
	}
	while(ans%10!=0)ans++;
	printf("%d,-\n",ans);
	return 0;
}

C-Beer Coasters

题意：求一个圆和矩形的面积交。
题解：计算几何模板题，实现的方法似乎是三角剖分之类的，具体博客很多。

#include<bits/stdc++.h>
using namespace std;  
#define INF 0x3f3f3f3f  
#define eps 1e-17  
#define pi acos(-1.0)  
typedef long long ll;  
int dcmp(double x){  
    if(fabs(x)<eps)return 0;  
    return x>0?1:-1;  
}  
struct Point{  
    double x,y;  
    Point(double _x=0,double _y=0){  
        x=_x;y=_y;  
    }  
};  
Point operator + (const Point &a,const Point &b){  
    return Point(a.x+b.x,a.y+b.y);  
}  
Point operator - (const Point &a,const Point &b){  
    return Point(a.x-b.x,a.y-b.y);  
}  
Point operator * (const Point &a,const double &p){  
    return Point(a.x*p,a.y*p);  
}  
Point operator / (const Point &a,const double &p){  
    return Point(a.x/p,a.y/p);  
}  
bool operator < (const Point &a,const Point &b){  
    return a.x<b.x||(dcmp(a.x-b.x)==0&&a.y<b.y);  
}  
bool operator == (const Point &a,const Point &b){  
    return dcmp(a.x-b.x)==0&&dcmp(a.y-b.y)==0;  
}  
double Dot(Point  a,Point b){  
    return a.x*b.x+a.y*b.y;  
}  
double Length(Point a){  
    return sqrt(Dot(a,a));  
}  
double Angle(Point a,Point b){  
    return acos(Dot(a,b)/Length(a)/Length(b));  
}  
double angle(Point a){  
    return atan2(a.y,a.x);  
}  
double Cross(Point a,Point b){  
    return a.x*b.y-a.y*b.x;  
}  
Point vecunit(Point a){  
    return a/Length(a);  
}  
Point Normal(Point a){  
    return Point(-a.y,a.x)/Length(a);  
}  
Point Rotate(Point a,double rad){  
    return Point(a.x*cos(rad)-a.y*sin(rad),a.x*sin(rad)+a.y*cos(rad));  
}  
double Area2(Point a,Point b,Point c){  
    return Length(Cross(b-a,c-a));  
}  
bool OnSegment(Point p,Point a1,Point a2){    
    return dcmp(Cross(a1-p,a2-p))==0&&dcmp(Dot(a1-p,a2-p))<=0;    
}    
struct Line{  
    Point p,v;  
    double ang;  
    Line(){};  
    Line(Point p,Point v):p(p),v(v){  
        ang=atan2(v.y,v.x);  
    }  
    bool operator < (const Line &L) const {  
        return ang<L.ang;  
    }  
    Point point(double d){  
        return p+(v*d);  
    }  
};  
bool OnLeft(const Line &L,const Point &p){  
    return Cross(L.v,p-L.p)>=0;  
}  
Point GetLineIntersection(Point p,Point v,Point q,Point w){  
    Point u=p-q;  
    double t=Cross(w,u)/Cross(v,w);  
    return p+v*t;  
}  
Point GetLineIntersection(Line a,Line b){  
    return GetLineIntersection(a.p,a.v,b.p,b.v);  
}  
double PolyArea(vector<Point> p){  
    int n=p.size();  
    double ans=0;  
    for(int i=1;i<n-1;i++)  
        ans+=Cross(p[i]-p[0],p[i+1]-p[0]);  
    return fabs(ans)/2;  
}  
struct Circle{      
    Point c;      
    double r;      
    Circle(){}      
    Circle(Point c, double r):c(c), r(r){}      
    Point point(double a) {//根据圆心角求点座标          
        return Point(c.x+cos(a)*r, c.y+sin(a)*r);      
    }      
};     
    
bool InCircle(Point x,Circle c){    
    return dcmp(c.r-Length(c.c-x))>=0;    
}    
bool OnCircle(Point x,Circle c){    
    return dcmp(c.r-Length(c.c-x))==0;    
}    
int getSegCircleIntersection(Line L,Circle C,Point *sol){    
    Point nor=Normal(L.v);    
    Line p1=Line(C.c,nor);    
    Point ip=GetLineIntersection(p1,L);    
    double dis=Length(ip-C.c);    
    if(dcmp(dis-C.r)>0)return 0;    
    Point dxy=vecunit(L.v)*sqrt(C.r*C.r-dis*dis);    
    int ret=0;    
    sol[ret]=ip+dxy;    
    if(OnSegment(sol[ret],L.p,L.point(1)))ret++;    
    sol[ret]=ip-dxy;    
    if(OnSegment(sol[ret],L.p,L.point(1)))ret++;    
    return ret;    
}    
double SegCircleArea(Circle C,Point a,Point b){    
    double a1=angle(a-C.c);    
    double a2=angle(b-C.c);    
    double da=fabs(a1-a2);    
    if(da>pi)da=pi*2-da;    
    return dcmp(Cross(b-C.c,a-C.c))*da*C.r*C.r/2.0;    
}    
double PolyCircleArea(Circle C,Point *p,int n){    
    double ret=0;    
    Point sol[2];    
    p[n]=p[0];    
    for(int i=0;i<n;i++){    
        double t1,t2;    
        int cnt=getSegCircleIntersection(Line(p[i],p[i+1]-p[i]),C,sol);  //判断线段与圆有几个交点，  
        if(cnt==0){  //0个交点，判断线段在多边形内部还是外部。  
            if(!InCircle(p[i],C)||!InCircle(p[i+1],C))ret+=SegCircleArea(C,p[i],p[i+1]); //外部直接计算圆弧面积   
            else ret+=Cross(p[i+1]-C.c,p[i]-C.c)/2;  //内部计算三角形面积。  
        }    
        if(cnt==1){    
            if(InCircle(p[i],C)&&(!InCircle(p[i+1],C)||OnCircle(p[i+1],C)))ret+=Cross(sol[0]-C.c,p[i]-C.c)/2,ret+=SegCircleArea(C,sol[0],p[i+1]);//,cout<<"jj-1"<<endl;  
            else ret+=SegCircleArea(C,p[i],sol[0]),ret+=Cross(p[i+1]-C.c,sol[0]-C.c)/2;//,cout<<"jj-2"<<endl;  
        }    
        if(cnt==2){
            if((p[i]<p[i+1])^(sol[0]<sol[1]))swap(sol[0],sol[1]);    
            ret+=SegCircleArea(C,p[i],sol[0]);    
            ret+=Cross(sol[1]-C.c,sol[0]-C.c)/2;    
            ret+=SegCircleArea(C,sol[1],p[i+1]);    
        }       
    }    
    return fabs(ret);    
}
Point p[5];  
int main(){     
	double R,x1,y1,x2,y2,x3,y3;   
    while(cin>>x1>>y1>>R>>x2>>y2>>x3>>y3){    
        Circle C=Circle(Point(x1,y1),R);    
        if(x2>x3)swap(x2,x3);  
        if(y2>y3)swap(y2,y3);  
        p[0]=Point(x2,y2);    
		p[2]=Point(x3,y3);    
        p[1]=Point(x3,y2);    
        p[3]=Point(x2,y3);    
        double ans=PolyCircleArea(C,p,4);  
        if(ans < -eps) ans = -ans;
		printf("%.4lf\n",ans);         
    }  
     return 0;    
}  

F-Beer Marathon

题意：给出n个数字，可以通过加法或减法调整数字，每次代价就是加或减的数字，问如何使得代价最小令整个数组为公差k的等差数列。
题解：先将数组按从小到大的顺序排序，令$b_i=a_i-i*k$，这个bi即为将ai移到最接近的ik的代价，再将b数组从小到大排序，选择中间的那个元素作为基准，左右依次调整。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=1e6+10;
const int mod=1e9+7;
const int INF=0x3f3f3f3f;
ll read(){
    ll f=1,x=0;char ch;
    do{ch=getchar();if(ch=='-')f=-1;}while(ch<'0'||ch>'9');
    do{x=x*10+ch-'0';ch=getchar();}while(ch>='0'&&ch<='9');
    return f*x;
}
int n,k,a[maxn];
ll ans;
struct arr{
	ll val;
	int rnk;
}b[maxn];
bool cmp(arr a,arr b){
	return a.val<b.val;
}
int main(){
	scanf("%d%d",&n,&k);
	for(int i=1;i<=n;i++)scanf("%d",&a[i]);
	sort(a+1,a+1+n);
	for(int i=1;i<=n;i++)
		b[i].val=1ll*a[i]-1ll*i*k,b[i].rnk=a[i];
	sort(b+1,b+1+n,cmp);
	int pos;	
	for(int i=1;i<=n;i++)		
		if(a[i]==b[n/2+1].rnk){
			pos=i;break;
		}
	int i=pos-1,j=pos+1;
	ll t=1ll*a[pos];
	while(i>0){			
		ans=ans+abs(1ll*(t-k)-1ll*a[i]);
		t=t-1ll*k;
		i--;
	}
	t=a[pos];
	while(j<n+1){
		ans=ans+abs(1ll*(t+k)-1ll*a[j]);
		t=t+1ll*k;
		j++;
	}

	printf("%lld\n",ans);
	return 0;
}

G-Beer Mugs

题意：长度为n的字符串找出最长连续子串，使得最多只有一个字母出现过奇数次。
题解：因为字母只有a～t，$2^{20}$是显然可以接受的，考虑状态压缩。对于字符串，我们开一个a数组表示其前缀异或和，并用一个2e6的数组记录所有可能的状态。之后我们仅需反向向前寻找，看这个状态是否之前就出现过，比较一下取max就好。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=3e5+10;
const int maxm=2e6+10;
const int mod=1e9+7;
const int INF=0x3f3f3f3f;
ll read(){
    ll f=1,x=0;char ch;
    do{ch=getchar();if(ch=='-')f=-1;}while(ch<'0'||ch>'9');
    do{x=x*10+ch-'0';ch=getchar();}while(ch>='0'&&ch<='9');
    return f*x;
}
int n,ans;
char s[maxn];
int f[22];
int a[maxn],vis[maxm],pos[maxm];
int main(){
	scanf("%d",&n);
	scanf("%s",s+1);
	f[0]=1;vis[0]=1;pos[0]=0;
	for(int i=1;i<=19;i++)f[i]=f[i-1]*2;
	for(int i=1;i<=n;i++){
		a[i]=a[i-1]^(f[s[i]-'a']);
		if(!vis[a[i]])vis[a[i]]=1,pos[a[i]]=i;
	}
	for(int i=n;i>=1;i--){
		for(int j=0;j<=20;j++)
			if(vis[a[i]^f[j]] && pos[a[i]^f[j]]<i)
				ans=max(ans,i-pos[a[i]^f[j]]);
	}
	
	printf("%d\n",ans);
	return 0;
}

H-Screamers in the Storm

题意：给出一个m*n的棋盘，有羊和狼两种棋子，狼每次都向右移动一格若超过边界则回到最左边，羊每次向下移动一格若超过边界则回到最上面。如果狼和羊相遇，狼会吃羊，并在这个格子上留下一个尸体。羊会吃草。每一个格子每过三轮就会长草，羊如果五回合吃不到草就会死，狼如果十回合吃不到羊就会死，有尸体的格子不会长草，问T轮游戏之后棋盘的状态。S表示羊，W表示狼，#表示草，*表示尸体，.表示正常格子。
题解：大模拟。

//赛后写的，参考了巨神的代码
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=1e5+10;
const int mod=1e9+7;
const int INF=0x3f3f3f3f;
ll read(){
    ll f=1,x=0;char ch;
    do{ch=getchar();if(ch=='-')f=-1;}while(ch<'0'||ch>'9');
    do{x=x*10+ch-'0';ch=getchar();}while(ch>='0'&&ch<='9');
    return f*x;
}
struct Sheep{
	int x,y;
	int flag;
};
struct Wolf{
	int x,y;
	int flag;
};
vector<Sheep>s;
vector<Wolf>w;
char a[22][22];
int f[22][22];
int n,m,T;
void move(){
	for(int i=0;i<s.size();i++){
		s[i].x++;
		if(s[i].x>=n)s[i].x=0;
		s[i].flag++;
	}
	for(int i=0;i<w.size();i++){
		w[i].y++;
		if(w[i].y>=m)w[i].y=0;
		w[i].flag++;
	}
	vector<pair<int,int> > v;
	for(int i=0;i<s.size();i++){
        for(int j=0;j<w.size();j++){
            if(s[i].x==w[j].x && s[i].y==w[j].y){
                v.push_back({s[i].x,s[i].y});
                w[j].flag=0;
            }
        }
    }
    vector<Sheep> z;
    for(int i=0;i<s.size();i++){
    	int x=s[i].x,y=s[i].y,flag=0;
    	for(int j=0;j<v.size();j++)
            if(v[j].first==x && v[j].second==y) flag = 1;
        if(!flag) z.push_back(s[i]);
        else f[x][y]=-INF;
	}
	s=z;
	vector<Wolf> q;
    for(int i=0;i<w.size();i++){
        if(w[i].flag<10)q.push_back(w[i]); 
        else f[w[i].x][w[i].y]=-INF;
    }
    w=q;
    z.resize(0);
    for(int i=0;i<s.size();i++){
        int x=s[i].x,y=s[i].y;
        if(f[x][y]==3){
            s[i].flag=0;
            f[x][y]=-1;
        }
        if(s[i].flag<5) z.push_back(s[i]);
        else f[x][y]=-INF;
    }
    s=z;
    
    for(int i=0;i<n;i++)
        for(int j=0;j<m;j++){
            if(f[i][j]==-INF)continue;
			f[i][j]=min(3,f[i][j]+1);
		}
}
int main(){
	scanf("%d%d%d",&T,&n,&m);
	for(int i=0;i<n;i++)scanf("%s",&a[i]);
	for(int i=0;i<n;i++)
		for(int j=0;j<m;j++)
			if(a[i][j]=='S'){
				Sheep s1;
				s1.x=i,s1.y=j,s1.flag=0;
				s.push_back(s1); 
			}else if(a[i][j]=='W'){
				Wolf w1;
				w1.x=i,w1.y=j,w1.flag=0;
				w.push_back(w1);
			}
	while(T--)move();
	for(int i=0;i<s.size();i++)f[s[i].x][s[i].y]=4;
	for(int i=0;i<w.size();i++)f[w[i].x][w[i].y]=5;
	for(int i=0;i<n;i++){
		for(int j=0;j<m;j++){
			if(f[i][j]==4)printf("S");
			else if(f[i][j]==5)printf("W");
			else if(f[i][j]==-INF)printf("*");
			else if(f[i][j]==3)printf("#");
			else printf(".");
		}
		printf("\n");
	}
	return 0;
}

J-Beer Vision

题意：给出n个点，他们可能是由n个点中的部分点彼此平移构成的，问可能由多少个不同的向量平移所形成这n个点。
题解：以$(x_1,y_1)$为基准，每个点到其的距离就为$(d_x,d_y)$判断这个向量是否合法即可。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=1e3+10;
const int mod=1e9+7;
const int INF=0x3f3f3f3f;
ll read(){
    ll f=1,x=0;char ch;
    do{ch=getchar();if(ch=='-')f=-1;}while(ch<'0'||ch>'9');
    do{x=x*10+ch-'0';ch=getchar();}while(ch>='0'&&ch<='9');
    return f*x;
}
int n,ans;
int x[maxn],y[maxn];
unordered_map<int,unordered_map<int,int> > mmap,vis;
bool check(int dx,int dy){
	for(int i=1;i<=n;i++){
		int flag1=1,flag2=1;
		if(mmap[x[i]+dx][y[i]+dy])flag1=0;
		if(mmap[x[i]-dx][y[i]-dy])flag2=0;
		if(flag1 && flag2)return false;
	}
	return true;
}
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++){
		scanf("%d%d",&x[i],&y[i]);
		mmap[x[i]][y[i]]=1;
	}
	for(int i=2;i<=n;i++){
		int dx=x[1]-x[i];
		int dy=y[1]-y[i];
		if(vis[dx][dy]==0){
			if(check(dx,dy))ans++;
			vis[dx][dy]=1;
		}
		if(vis[-dx][-dy]==0){
			if(check(-dx,-dy))ans++;
			vis[-dx][-dy]=1;
		}
	}
	printf("%d\n",ans);
	return 0;
}