CodeForces 47E. Cannon（离线暴力+数学）

E. Cannon

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Bertown is under siege! The attackers have blocked all the ways out and their cannon is bombarding the city. Fortunately, Berland intelligence managed to intercept the enemies' shooting plan. Let's introduce the Cartesian system of coordinates, the origin of which coincides with the cannon's position, the Ox axis is directed rightwards in the city's direction, the Oy axis is directed upwards (to the sky). The cannon will make n more shots. The cannon balls' initial speeds are the same in all the shots and are equal to V, so that every shot is characterized by only one number alphai which represents the angle at which the cannon fires. Due to the cannon's technical peculiarities this angle does not exceed 45 angles (π / 4). We disregard the cannon sizes and consider the firing made from the point (0, 0).

The balls fly according to the known physical laws of a body thrown towards the horizon at an angle:

vx(t) = V·cos(alpha)vy(t) = V·sin(alpha)  –  g·tx(t) = V·cos(alpha)·ty(t) = V·sin(alpha)·t  –  g·t2 / 2

Think of the acceleration of gravity g as equal to 9.8.

Bertown defends m walls. The i-th wall is represented as a vertical segment (xi, 0) - (xi, yi). When a ball hits a wall, it gets stuck in it and doesn't fly on. If a ball doesn't hit any wall it falls on the ground (y = 0) and stops. If the ball exactly hits the point (xi, yi), it is considered stuck.

Your task is to find for each ball the coordinates of the point where it will be located in the end.

Input

The first line contains integers n and V (1 ≤ n ≤ 104, 1 ≤ V ≤ 1000) which represent the number of shots and the initial speed of every ball. The second line contains n space-separated real numbers alphai (0 < alphai < π / 4) which represent the angles in radians at which the cannon will fire. The third line contains integer m (1 ≤ m ≤ 105) which represents the number of walls. Then follow m lines, each containing two real numbers xi and yi (1 ≤ xi ≤ 1000, 0 ≤ yi ≤ 1000) which represent the wall’s coordinates. All the real numbers have no more than 4 decimal digits. The walls may partially overlap or even coincide.

Output

Print n lines containing two real numbers each — calculate for every ball the coordinates of its landing point. Your answer should have the relative or absolute error less than 10 - 4.

Examples

input

Copy

2 10
0.7853
0.3
3
5.0 5.0
4.0 2.4
6.0 1.9

output

Copy

5.000000000 2.549499369
4.000000000 0.378324889

input

Copy

2 10
0.7853
0.3
2
4.0 2.4
6.0 1.9

output

Copy

10.204081436 0.000000000
4.000000000 0.378324889

题意：

   有一门大炮，坐标在(0,0)(0,0)，和mm堵墙，现在大炮要射nn发炮弹，每发炮弹的初始速度v是一样的，射击角度为α(0<α<π/4)，假设射击后经过时间t，重力加速度g=9.8，则有：

   x​(t)=v∗cos(α)

   y​(t)=v∗sin(α)−g∗t

   x(t)=vx​(t)∗t

   y(t)=v∗sin(α)∗t−g∗t2/2

   给定m堵墙墙顶坐标(xi​,yi​)，墙垂直于xx坐标轴，炮弹如果打到墙上，就会卡住；如果掉到地上，也不会再滚动。

   求这n发炮弹最终的位置

                                                      ----translate by 守望、copy from 洛谷

  题解：显然如果速度相同，角度在45度以内，那么角度越大的射的越高越远，所以如果矮的能越过的墙高的也能越过，把问题离线下来按照角度排序，模拟每个球会怎么走就可以了，显然每堵墙只会被访问一遍，均摊复杂度O(1)，总复杂度O(n)

代码如下：

  

#include<set>
#include<map>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

struct bomb
{
    int id;
    double a,ansx,ansy;
} b[];

struct wall
{
    double x,y;
} w[];

int n,m;
double v;

int cmp1(wall x,wall y)
{
    if(x.x==y.x) return x.y<y.y;
    return x.x<y.x;
}

int cmp2(bomb x,bomb y)
{
    return x.a<y.a;
}

int cmp3(bomb x,bomb y)
{
    return x.id<y.id;
}

int main()
{
    double g=9.8;
    scanf("%d %lf",&n,&v);
    for(int i=; i<=n; i++)
    {
        scanf("%lf",&b[i].a);
        b[i].id=i;
    }
    scanf("%d",&m);
    for(int i=; i<=m; i++)
    {
        scanf("%lf%lf",&w[i].x,&w[i].y);
    }
    sort(w+,w+m+,cmp1);
    sort(b+,b+n+,cmp2);
    int r=;
    for(;; r++)
    {
        if(r>m)
        {
            r=m+;
            break;
        }
        double t=w[r].x/(cos(b[].a)*v);
        if(v*sin(b[].a)*t-g*t*t/<=w[r].y)
        {
            break;
        }
    }
    for(int i=; i<=n; i++)
    {
        while()
        {
            if(r>m)
            {
                r=m+;
                break;
            }
            double t=w[r].x/(cos(b[i].a)*v);
            if(v*sin(b[i].a)*t-g*t*t/>=w[r].y) r++;
            else break;
        }
        double t=w[r].x/(cos(b[i].a)*v);
        if(r<=m)
        {
            b[i].ansy=(v*sin(b[i].a)*t)-(g*t*t/);
            b[i].ansx=w[r].x;
        }
        else
        {
            b[i].ansy=;
            b[i].ansx=(v*sin(b[i].a)/g)*v*cos(b[i].a)*;
        }
    }
    sort(b+,b+n+,cmp3);
    for(int i=; i<=n; i++)
    {
        if(b[i].ansy<)
        {
            b[i].ansy=;
            b[i].ansx=(v*sin(b[i].a)/g)*v*cos(b[i].a)*;
        }
        printf("%.9lf %.9lf\n",b[i].ansx,b[i].ansy);
    }
}