HDU 4445 Crazy Tank --枚举

题意: n个物体从高H处以相同角度抛下,有各自的初速度,下面[L1,R1]是敌方坦克的范围,[L2,R2]是友方坦克,问从某个角度抛出,在没有一个炮弹碰到友方坦克的情况下,最多的碰到敌方坦克的炮弹数。

解法: 枚举角度,将pi/2分成1000份,然后枚举,通过方程 v*sin(theta)*t - 1/2*g*t^2 = -H 解出t,然后 x = v*cos(theta)*t算出水平距离,直接统计即可。

代码:

#include <iostream>

#include <cstdio>

#include <cstring>

#include <cstdlib>

#include <cmath>

#include <algorithm>

#define pi acos(-1.0)

#define eps 1e-8

using namespace std;

#define N 207

#define g 9.8



double V[N];

double H;



int sgn(double x)

{

    if(x > eps) return 1;

    if(x < -eps) return -1;

    return 0;

}



double calc(double theta,double v)

{

    double up = v*sin(theta) + sqrt(v*v*sin(theta)*sin(theta)+2.0*g*H);

    double down = g;

    return v*cos(theta)*up/down;

}



int main()

{

    double L1,R1,L2,R2;

    int n,i,j;

    while(scanf("%d",&n)!=EOF && n)

    {

        scanf("%lf%lf%lf%lf%lf",&H,&L1,&R1,&L2,&R2);

        if(sgn(L1-L2) == 0 && sgn(R1-R2) == 0) { puts("0"); continue; }

        for(i=1;i<=n;i++) scanf("%lf",&V[i]);

        double delta = pi*0.001;

        int Maxi = 0;

        for(i=0;i<=1000;i++)

        {

            double theta = delta*i - pi/2.0;

            int cnt = 0;

            for(j=1;j<=n;j++)

            {

                double x = calc(theta,V[j]);

                if(sgn(x-L2) >= 0 && sgn(x-R2) <= 0)

                {

                    cnt = 0;

                    break;

                }

                if(sgn(x-L1) >= 0 && sgn(R1-x) >= 0)

                    cnt++;

            }

            Maxi = max(Maxi,cnt);

        }

        cout<<Maxi<<endl;

    }

    return 0;

}
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