精华内容
下载资源
问答
  • 从原点开始,按顺序弹向下目标点,找到线段与对应墙的交点作为反射点,判断是否合法 Code #include using namespace std ; #define rep(i, a, b) for(int i = (a); i (b); i++) #define red(i...

    Problem

    Description

    Janine recently went to her local game store and bought “Hole in One”, a new mini-golf game for her computer. As indicated by the name, the objective of the game is to shoot a ball into a hole using just one shot. The game also borrows elements from brick breaker style games: in the playing field, several walls are placed that will be destroyed upon being hit by the ball. The score of a successful shot depends on the number of destroyed walls, so Janine wonders: what is the maximum number of walls that can be hit while performing a “Hole in One”?
    For the purposes of this problem you can think of the playing field as a cartesian plane with the initial position of the ball at the origin. The walls are non-intersecting axis-parallel line segments in this plane (i.e., parallel to either the x axis or the y axis). The diameter of the ball is negligible so it is represented as a single point.
    这里写图片描述
    Figure H.1: Illustration of the first sample input: The ball first bounces off two walls at points 1 and 2. When it passes point 3 the wall has already vanished.

    Whenever the ball hits a wall, two things happen:

    • The direction of the ball changes in the usual way: the angle of incidence equals the angle of reflection.
    • The wall that the ball touched is destroyed. Following common video game logic, no rubble of the wall remains; it will be as though it vanished.

    The behaviour of the ball is also affected by the power of the shot. In particular, an optimal shot may need to first roll over the hole, then hit some more walls, and only later drop into the hole.

    Input

    The input consists of:

    • one line with one integer n (0 ≤ n ≤ 8), the number of walls;
    • one line with two integers x and y, the coordinates of the hole;
    • n lines each with four integers x1,y1,x2 and y2 (either x1=x2 , or y1=y2 , but not both), representing a wall with end points (x1,y1) and (x2,y2).

    The hole is not at the origin and not on a wall. The walls do not touch or intersect each other.
    No wall lies completely on the x axis or the y axis. All coordinates in the input are integers with
    absolute value at most 1 000.

    Output

    If there is no way to shoot the ball such that it reaches the hole, print “impossible”. Otherwise, print the maximum number of walls that can be destroyed in a single “Hole in One” shot.

    Sample Input

    Sample Input 1
    3
    4 2
    1 1 1 2
    2 1 2 2
    3 1 3 2
    Sample Input 2
    1
    2 0
    1 -1 1 1
    Sample Input 3
    2
    -2 4
    2 4 2 2
    0 6 -2 6

    Sample Output

    Sample Output 1
    2
    Sample Output 2
    Impossible
    Sample Output 3
    2

    Source

    NWERC 2015 Problem H

    Solution

    这里写图片描述

    1. 枚举所有可能的撞击墙壁的方案,包括撞击哪些墙,和撞击的顺序
    2. 从目标点T开始,计算小球每一次弹射的目标点(也就是按逆序对每堵墙取轴对称点)
    3. 从原点开始,按顺序弹向下一目标点,找到线段与对应墙的交点作为反射点,判断是否合法

    Code

    #include <bits/stdc++.h>
    using namespace std;
    #define rep(i, a, b) for(int i = (a); i <= (b); i++)
    #define red(i, a, b) for(int i = (a); i >= (b); i--)
    #define ll long long
    #define y1 yy1
    #define abs Abs
    #define Vector Point
    
    inline int read() {
        int x = 0, f = 1; char c = getchar();
        while(!isdigit(c)) { if (c == '-') f = -1; c = getchar(); }
        while(isdigit(c)) { x = x * 10 + c - '0'; c = getchar(); }
        return x * f;
    }
    template<typename tn> void cmax(tn& a, tn b) { if (b > a) a = b; }
    template<typename tn> void cmin(tn& a, tn b) { if (b < a) a = b; }
    template<typename tn> tn abs(tn x) { return x < 0 ? -x : x; }
    
    const int N = 20;
    const double eps = 1e-8;
    struct LINE{
        double x1, y1, x2, y2; int tag;
    }w[N];
    struct Point{
        double x, y;
        Point(double x = 0, double y = 0):x(x),y(y) { }
    }p[N];
    int n, flag = 0, line[N], l[N], vis[N], used[N];
    double xs, ys, xt, yt;
    
    Vector operator + (Vector A, Vector B) { return Vector(A.x + B.x, A.y + B.y); }
    Vector operator - (Point A, Point B) { return Vector(A.x - B.x, A.y - B.y); }
    int dcmp(double x) {
        if (fabs(x) < eps) return 0;
        else return x < 0 ? -1 : 1;
    }
    bool operator == (const Point& A, const Point& B) {
        return dcmp(A.x - B.x) == 0 && dcmp(A.y - B.y) == 0;
    }
    double dot(Vector A, Vector B) { return A.x * B.x + A.y * B.y; }
    double cross(Vector A, Vector B) { return A.x * B.y - A.y * B.x; }
    Point Get_Inter(Point p, Vector v, LINE w) {
        double xt, yt;
        if (w.tag) {
            yt = w.y1;
            xt = (yt - p.y) * v.x / v.y + p.x;
        }else {
            xt = w.x1;
            yt = (xt - p.x) * v.y / v.x + p.y;
        }
        return Point(xt, yt);
    }
    bool On_Wall(Point p, LINE w) {
        if (w.tag) return (p.x >= w.x1 && p.x <= w.x2);
        else return (p.y >= w.y1 && p.y <= w.y2);
    }
    bool On_Seg(Point p, Point a1, Point a2) { return dcmp(dot(a1 - p, a2 - p)) <= 0; }
    void solve(int m) {
        memset(used, 0, sizeof(used));
        double X = xt, Y = yt;
        rep(i, 1, m) l[i] = line[i]; l[m + 1] = 0;
        red(i, m, 1) {
            if (w[l[i]].tag) Y = 2 * w[l[i]].y1 - Y;
            else X = 2 * w[l[i]].x1 - X;
            p[i] = Point(X, Y);
        }
        p[++m] = Point(xt, yt);
        Point now = Point(0, 0);
        rep(i, 1, m) {
            Vector dir = p[i] - now;
            Point rev_p;
            if (i != m) {
                if (w[l[i]].tag == 1 && dcmp(dir.y) == 0) return;
                if (w[l[i]].tag == 0 && dcmp(dir.x) == 0) return;
                rev_p = Get_Inter(now, dir, w[l[i]]);
                if (!On_Seg(rev_p, now, p[i])) return;
                if (!On_Wall(rev_p, w[l[i]])) return;
            }else rev_p = p[m];
            rep(j, 1, n) {
                if (used[j] || j == l[i]) continue;
                if (w[j].tag == 1 && dcmp(dir.y) == 0) continue;
                if (w[j].tag == 0 && dcmp(dir.x) == 0) continue;
                Point tmp = Get_Inter(now, dir, w[j]);
                if (!On_Wall(tmp, w[j])) continue;
                if (On_Seg(tmp, now, rev_p)) return; 
            }
            now = rev_p; used[l[i]] = 1;
        }
        flag = 1;
        return;
    }
    void dfs1(int lim, int dep) {
        if (dep > lim) { solve(lim); return; }
        rep(i, 1, n) {
            if (vis[i]) continue;
            line[dep] = i; vis[i] = 1;
            dfs1(lim, dep + 1);
            if (flag) return;
            vis[i] = 0;
        }
    }
    
    int main() {
        freopen("H.in", "r", stdin);
        freopen("H.out", "w", stdout);
        n = read();
        if (n == 0) { printf("0\n"); return 0; }
        xs = ys = 0;
        xt = (double)read(); yt = (double)read();
        rep(i, 1, n) {
            int x1 = read(), y1 = read(), x2 = read(), y2 = read();
            if (x1 > x2) swap(x1, x2);
            if (y1 > y2) swap(y1, y2);
            w[i].x1 = (double)x1; w[i].y1 = (double)y1;
            w[i].x2 = (double)x2; w[i].y2 = (double)y2;
            w[i].tag = dcmp(x1 - x2) != 0;
        }
        red(num, n, 0) {
            memset(vis, 0, sizeof(vis));
            dfs1(num, 1);
            if (flag) { printf("%d\n", num); break; }
        }
        if (!flag) printf("impossible\n");
        return 0;
    }
    

    尾声

    判断合法状态真的很蛋疼。。炸到只有三十分,后来发现是没有判断南辕北辙的情况,也就是这样
    || × –> ×
    是不是很像一个表情。。

    End.

    展开全文
  • [Nwerc2015]Hole in One一杆进洞 Description Janine recently went to her local game store and bought “Hole in One”, a new mini-golf game for her computer. As indicated by the name, the objective...

    [Nwerc2015]Hole in One一杆进洞

    Description
    Janine recently went to her local game store and bought “Hole in One”, a new mini-golf game for her computer. As indicated by the name, the objective of the game is to shoot a ball into a hole using just one shot. The game also borrows elements from brick breaker style games: in the playing field, several walls are placed that will be destroyed upon being hit by the ball. The score of a successful shot depends on the number of destroyed walls, so Janine wonders: what is the maximum number of walls that can be hit while performing a “Hole in One”?
    For the purposes of this problem you can think of the playing field as a cartesian plane with the initial position of the ball at the origin. The walls are non-intersecting axis-parallel line segments in this plane (i.e., parallel to either the x axis or the y axis). The diameter of the ball is negligible so it is represented as a single point.
    Figure 1: Illustration of the first sample input: The ball first bounces off two walls at points 1 and 2. When it passes point 3 the wall has already vanished.
    Whenever the ball hits a wall, two things happen:
    The direction of the ball changes in the usual way: the angle of incidence equals the angle of reflection.
    The wall that the ball touched is destroyed. Following common video game logic, no rubble of the wall remains; it will be as though it vanished.
    The behaviour of the ball is also affected by the power of the shot. In particular, an optimal shot may need to first roll over the hole, then hit some more walls, and only later drop into the hole.
    The input consists of:
    one line with one integer n (0≤n≤8), the number of walls;
    one line with two integers x and y, the coordinates of the hole;
    nn lines each with four integers x1, y1, x2, and y2 (either x1=x2, or y1=y2, but not both), representing a wall with end points (x1,y1) and (x2,y2).
    The hole is not at the origin and not on a wall. The walls do not touch or intersect each other. No wall lies completely on the x axis or the y axis. All coordinates in the input are integers with absolute value at most 1000.
    If there is no way to shoot the ball such that it reaches the hole, print “impossible”. Otherwise, print the maximum number of walls that can be destroyed in a single “Hole in One” shot.
    珍妮最近去当地的游戏商店买了“一杆进洞”,一个新的迷你高尔夫电脑游戏。正如名称所示,这个游戏的目标是,将球只用一击击进洞里。游戏还借鉴了打砖块风格的游戏元素:在场上,放置了一些墙,它们被球击中就会被摧毁。一个成功的击球的得分取决于破坏墙的数量,所以珍妮不禁要问:在一次“一杆进洞”中能够摧毁的最多的墙数量是多少?
    对于这个问题的意向,你可以认为,比赛场地为笛卡尔坐标平面,球的最初位置在原点。墙壁为在该平面中互不相交且与坐标轴平行的线段(即平行于x轴或y轴)。球的直径非常小以至于把它当成一个点。
    图1:第一个样例输入的插图:球首先经过1,2号点,并在3号点时穿过了被摧毁的墙。
    每当球碰到一堵墙,会发生:
    球的方向以通常的方式改变:入射角等于反射角。
    被球触碰的墙被破坏。 游戏逻辑规定,没有墙的废墟存在,认为它消失了。
    球的行为也受到击球的力影响。特别的是,最佳的击球可能需要将鼠标放在洞上,然后摧毁一些墙壁,并且只是延迟入洞。
    Input
    一行一个整数n(0≤n≤8),墙壁的数量。
    一行两个整数x和y,洞的坐标。
    n行,每行各有四个整数x1,y1,x2与y2(x1 = x2或y1 = y2,但不会满足两者),表示墙壁的两个端点。
    洞不会在原点也不会在墙上。墙壁不会互相接触或交叉。没有墙会完全在x轴或y轴上。输入的所有的坐标都是整数且绝对值最大1000。
    Output
    如果没有办法一击入洞,输出“impossible”。否则,输出在一次“一杆进洞”中能够摧毁的最多的墙的数量。
    Sample Input
    Sample Input 1
    3
    4 2
    1 1 1 2
    2 1 2 2
    3 1 3 2
    Sample Input 2
    1
    2 0
    1 -1 1 1
    Sample Input 3
    2
    -2 4
    2 4 2 2
    0 6 -2 6
    Sample Output
    Sample Output 1
    2
    Sample Output 2
    Impossible
    Sample Output 3
    2

    Solution
    首先从大到小枚举答案,再答案!枚举方案,通过镜面对称终点得到路线,然后平方验证路线
    注意细节

    Code

    #include <bits/stdc++.h>
    using namespace std;
    #define MS(_) memset(_,0,sizeof(_))
    template<typename T> inline void read(T &x){
        x=0; T f=1; char ch=getchar();
        while (!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar(); }
        while (isdigit(ch))  {x=x*10+ch-'0'; ch=getchar(); }
        x*=f;
    }
    
    const int INF=0x7fffffff;
    const int MaxN=100;
    const double eps=0;
    struct Vec{
        double x,y;
        Vec(){}
        Vec(double _x,double _y):x(_x),y(_y){}
    }p1[MaxN],p2[MaxN],T,p[MaxN];
    inline Vec operator + (const Vec &a,const Vec &b){return Vec(a.x+b.x,a.y+b.y);}
    inline Vec operator - (const Vec &a,const Vec &b){return Vec(a.x-b.x,a.y-b.y);}
    template<typename T>inline Vec operator * (const Vec &a,T b){return Vec(a.x*b,a.y*b);}
    template<typename T>inline Vec operator * (T &a,const Vec &b){return Vec(a*b.x,a*b.y);}
    template<typename T>inline Vec operator / (const Vec &a,T b){return Vec(a.x/b,a.y/b);}
    inline double cross(const Vec &a,const Vec &b){return a.x*b.y-b.x*a.y;}
    inline double dot(const Vec &a,const Vec &b){return a.x*b.x+a.y*b.y;}
    typedef Vec Poi;
    int n,N,type[MaxN],used[MaxN],vis[MaxN],lst[MaxN],ans;
    
    inline Poi calnxt(const Vec &a,int no){return type[no]?Vec(a.x,p1[no].y*2-a.y):Vec(p1[no].x*2-a.x,a.y);}
    inline Vec caldot(Poi a,Poi b,Poi c,Poi d){
        if (a.x==b.x){
            if (c.x==d.x) return Vec(-INF,-INF);
            if (c.y==d.y) {swap(a,c);swap(b,d);}
        }    
        double k=(b.y-a.y)/(b.x-a.x),dt=b.y-k*b.x;
        if (c.x==d.x) return Vec(c.x,k*c.x+dt);
        else return Vec((c.y-dt)/k,c.y);
    }
    inline bool shot(){
        Poi now=Vec(0.,0.); MS(used);
        for (int i=N;i;i--){
            Poi nxt=p[i],wall1=p1[lst[i]],wall2=p2[lst[i]];
            Poi point=caldot(now,nxt,wall1,wall2);
            if(dot(point-wall1,point-wall2)>0||dot(point-now,point-nxt)>0) return false;
            used[lst[i]]=1;
            for (int j=1;j<=n;j++) if (!used[j]){
                Poi point2=caldot(now,point,p1[j],p2[j]);
                if(dot(point2-p1[j],point2-p2[j])<=0&&dot(point2-now,point2-point)<=0) return false;
            }
            now=point;
        }
        Poi nxt=p[0];
        for (int j=1;j<=n;j++) if (!used[j]){
            Poi point2=caldot(now,nxt,p1[j],p2[j]);
            if(dot(point2-p1[j],point2-p2[j])<=0&&dot(point2-now,point2-nxt)<=0) return false;
        }
        return true;
    }
    
    inline void check(){
        p[0]=T;
        for (int i=1;i<=N;i++)p[i]=calnxt(p[i-1],lst[i]);
        if (shot()) ans=N;  
    }
    void dfs(int dep){
        if (~ans) return;
        if (dep>N) {check();return;}
        for(int i=1;i<=n;i++){
            if (!vis[i]){vis[lst[dep]=i]=1; dfs(dep+1); vis[i]=0;}
            if (~ans) return;
        }   
    }
    int main(){
        read(n);
        read(T.x);read(T.y);
        for(int i=1;i<=n;i++){read(p1[i].x);read(p1[i].y);read(p2[i].x);read(p2[i].y);}
        for(int i=1;i<=n;i++) 
            if (p1[i].x==p2[i].x){type[i]=0; if (p1[i].y>p2[i].y) swap(p1[i],p2[i]);}
            else{type[i]=1; if (p1[i].x>p2[i].x) swap(p1[i],p2[i]);}
    
        ans=-1;
        for(int i=n;i>=0;i--){
            N=i; dfs(1); 
            if (~ans) break;
        }
        if (!~ans) puts("impossible"); else printf("%d\n",ans);
        return 0;
    }
    展开全文
  • 这是一个微信环境下的高乐夫球场预订小程序,包括了球场介绍、新闻动态、订场服务、一杆进洞、发现和用户中心等模块,可为同类预订服务小程序开发提供设计思路参考,界面效果如截图所示,不过因本地微信环境问题,...
  • 快乐广播稿4篇.doc

    2021-01-15 21:25:17
    快乐广播稿4篇 *目录快乐广播稿我读书我快乐校园广播稿... 第三个洞开始,长老打出超完美的成绩,几乎都是一杆进洞。 长老兴奋莫名,到打第七个洞时,天使又跑去找上帝:上帝呀,你不是要惩罚长老吗?为何还不见有...
  • 最近许多科技公司都发布了新品,但都是通过线上发布,而4月24日佳明则是发布...高尔夫讲求的是力度和方向,一杆进洞固然是最精彩的进球,而APPROACH Z82内置全球41000+球道图,GPS定位即时的球道果岭图,旗杆锁定功...

    最近许多科技公司都发布了新品,但都是通过线上发布,而4月24日佳明则是发布了一款型号叫APPROACH Z82的高尔夫GPS激光测距仪,通过这款产品,可以让用户看到前方球洞和自己的距离,并且可以测出风速和风向,从而调整球杆和击球的方式,看着也是很实用呀。

    97a311fa60359b94b69a3ef1eb2e17b7.png

    高尔夫讲求的是力度和方向,一杆进洞固然是最精彩的进球,而APPROACH Z82内置全球41000+球道图,GPS定位即时的球道果岭图,旗杆锁定功能,高精度测距,坡道补偿距离以及风速和风向检测,戴上APPROACH Z82去打高尔夫球就如开挂一般呀。

    4b3166eafc18a793ac73242b4d8469a3.png

    我们在打高尔夫的时候,往往天气是很难预测的,所以每次的风向和风速都深深影响着我们的判断,而APPROACH Z82则是为用户提供风速和风向的数据,用户则是通过数据来调整自己的挥杆方向以及力度,从而达到更易进球的效果,就算不能一杆进洞,也能更加准确的让球去到球洞旁。

    81374192e1ea62fad0dba4067d5a9d48.png

    很难想象,APPROACH Z82还内置了全球大部分的高尔夫球场图,一眼望去,就能看到前方的球道视图、果岭视图以及障碍物,这样在挥杆的时候就能通过调整来躲过障碍物了,大大提高进球的效率。

    171ed8b56f399c4b5ae70f58925b67b4.png

    而当用户瞄准旗杆之后,画面中就会闪现黄色的旗杆图标,并且会发出震动提醒,这样就大大的帮助用户稳爪旗杆了,准确率提升很多。

    7c4e84b1ea55821f07cac6795b3b848d.png

    碰到坡道怎么办,APPROACH Z82带有坡度补偿功能,开启后可以更加精准的计算球与球洞的坡度补偿距离,因为只计算直线距离,球就会因为坡度产生距离差,降低进球的几率,通过坡度补偿就能让用户更加精准的入球了。

    d8cc51cbaf91f64acc357c8d302e40d5.png

    而APPROACH Z82也是带有手机App的,可以同步电子记分卡、统计数据分析、分享竞赛排行,并且可以通过APP来查找自己不知道放在哪的APPROACH Z82设备,因为APPROACH Z82是内置GPS的,而通过手机的Garmin Golf App定位到设备的位置,让用户在广阔的高尔夫场轻松找回丢失的APPROACH Z82。

    ae26e7f6f57f68e310d4b577ac161728.png

    其实APPROACH Z82是APPROACH Z80的升级款,新增的功能特别多,开启GPS的情况下,续航也能去到15小时,而全黑的设计看上去更加百搭而且更酷,对于喜爱高尔夫运动的人士而言,佳明这款APPROACH Z82可以提高准确率,而对于新手而言,则是作用没那么大,比较适合有一定球龄的用户。

    展开全文
  • 于是,巴菲特的球友们决定和他打一个赌,如果他不能一杆进洞,只需付10美元;而如果他做到了,可以获得20000美元,但巴菲特拒绝了,他说:“如果你不学会在小事上约束自己,你在大事情上也不会受到内心

    巴菲特曾经说过,他的投资行为没有超过常人的能力范围,他之所以成功,与他强大的自制力是分不开的,他从来不参与自己不能控制的事情。
        有一次,巴菲特参加了一场户外高尔夫运动,在三天内一干进洞的成绩为零。于是,巴菲特的球友们决定和他打一个赌,如果他不能一杆进洞,只需付10美元;而如果他做到了,可以获得20000美元,但巴菲特拒绝了,他说:“如果你不学会在小事上约束自己,你在大事情上也不会受到内心的约束。”
        在40多年的投资生涯中巴菲特正是因为蔑视和回避了众多的市场诱惑,他才躲过了20世纪60年代的“电子风潮”,20世纪80年代的“生物概念”。20世纪末的“网络闹剧”。
         对个人而言,投资是一种自由度很大的行为,没人监督、管理、限制你的操作,很多投资行为靠自己的决策和智慧。但现实中想巴菲特一样能够自制的投资者不多。
        要培养自己的自制力,就必须在平时投资时多思考,只有多方面想国家的宏观面、股市所处的区域,才能更清晰地认识、理解这个市场。看大势赚大钱,只有站的高,才能望得远。要具备一种全局性的战略投资眼光。
       换个角度,从大多数投资失败者来说,他们的不冷静主要是:
        一、根源------缺乏自制力,投身股市,自制力是极为主要的,很多投资者出现失误不是应为他们不明白投资的原理,而是在于明白道理确做不到。
       二、从众心理,-------这点在中国人的个性中有着强烈的体现,正所谓人气价值的影响远远超出了实际价值。但中国人最信奉人气价值。

      三、不能拒绝市场诱惑,--------  在股市中存在着很多诱惑,而人们往往存在着侥幸心理,认为自己一定能成功,高估了自己的能力。但是,要知道,在股市中存在着很多自己无法控制的风险,风险出现,你的金钱就变成了众多埋葬品之一了。


    展开全文
  • 圣诞送美服,球技大练兵活动19日推出以来,受到了新老玩家的一致追捧,接收截图的信箱每天都被玩家的信件塞得满满的,一个个漂亮的一杆进洞、老鹰球、长距离推杆进洞,玩家们在游戏中的精彩表现连工作人员看到都艳羡...
  • 有人给孤独下一个定义:贵为九球之王,一杆进洞,却四下无人,这便是孤独。不少年轻人就把自己过成为了这样。 出门逛街的时间,一个人逛逛停停。也想过要叫伴侣一路进去,找遍了列表所有人的名单,最初想了想仍是算...
  • 重点在理解题意,最大的牌是顺子+同花。 扑克牌型 ...下面列出从大到小的扑克牌面....它是所有德州扑克中的王牌,即使您...好比打高尔夫球一杆进洞一样。它是由T(10)到Ace的清一色同花组成。  131 - The Psyc
  • 扑克牌型 下面列出从大到小的扑克牌面. 这是所有的扑克通用规则。...好比打高尔夫球一杆进洞一样。它是由T(10)到Ace的清一色同花组成。  131 - The Psychic Poker Player" title="Uv
  • 扑克牌

    2012-03-04 15:07:21
    扑克牌型 下面列出从大到小的扑克牌面. 这是所有的扑克通用规则。...好比打高尔夫球一杆进洞一样。它是由T(10)到Ace的清一色同花组成。 131 - The Psychic Poker Player" alt="Uva 131 - T
  • 第四天伍兹发飙,最后四个拿下三个博蒂(低于标准杆一杆),以六十六杆作收。四天总杆数两百八十一杆,名列第三。 职业高尔夫球赛进行四天,以总杆决定最后名次。常常可以看到选手前后一天杆数
  • 这就和动态学习率一样了,如果你每一杆都用同样的力气打,如果是轻轻打还好,会越来越接近洞口,运气好最终也能进洞。如果每一杆力气一样并且是很大的力气,那你可能是在和自己过不去了。下面贴上最直观的小实验和...
  • 想想看,没有一杆完成的高尔夫比赛,你需要一打过去, 你每打出一杆的目的就是离球洞越近越好,直到把它打。 目的是驱动我们潜能的关键,是主导一切的力量, 它可以影响我们的行为,激励我们制造达到目的的...
  • UVALIVE6886 Golf Bot (FFT)

    2018-07-16 18:48:00
     然后对于m个询问 问能否在两杆内把球打进洞 题解:平方一下就好 注意一下x0的系数为1表示打一杆  才发现数组应该开MAXN * 4 之前写的题数据有点不严谨了 #include <stdio.h> #include <algorithm...
  • 有些球友有疑问,在果岭上离洞口二尺,球推出去后在距洞穴尺的地方头碰到球第二次这罚不罚,还有依现有规定沙坑,救球连击罚不罚?比较好奇这两尺推杆后来有没有?已经这么惨了,还要罚吗?尽管高尔夫规则...
  • 绝妙对联

    千次阅读 2006-10-31 18:05:00
    绝妙对联 (1)苏小妹绝对解兄愁 一人给苏东坡出一上联: 架上丝瓜酷如吊 东坡想了半天... 男出上联:一杆枪,两只弹,三十八年没有参过战; 女对下联:一个,两扇门,二十五载从未过人。 横批:月明松(藏月日...
  • 电磁阀定义及原理

    2020-12-23 11:23:48
    通过控制阀体的运动来堵塞或泄漏不同的排油孔,而且油孔始终是打开的,液压油将进入不同的排油管,然后通过油压将油推向活塞,活塞依次移动活塞,活塞驱动机构。因此,机械运动是控制电磁铁的电流。 1、直动式...

空空如也

空空如也

1 2
收藏数 21
精华内容 8
关键字:

一杆进洞