The Descent

In this puzzle, the aim is to destroy the moutains below your ship, which is moving space invader style, before colliding with them. The idea is to perform a linear search and shoot the highest moutain each time an ammo is available.

C

#include <stdlib.h>
#include <stdio.h>

int main(int argc, char** argv)
{
    int oldY = -1;
    int target = -1;
    int x, y, max, h;

    for (;;) {
        scanf("%d%d", &x, &y);
        
        /* One new ammo is available each time the height change 
           The highest moutain is the new target */
        if (y != oldY) {
            oldY = y;
            max = 0;
            for (int i = 0; i < 8; i++) {
                scanf("%d", &h);
                if (h > max) {
                    target = i;
                    max = h;
                }
            }
        }
        // If height hasn't change, flushes stdin
        else
            for (int i = 0; i < 8; i++)
                scanf("%d", &h);
        
        printf(x == target ? "FIRE\n" : "HOLD\n");
    }
    return EXIT_SUCCESS;
}

Java

import java.util.*;
import java.math.*;

class Player {
    public static void main(String args[]) {
        Scanner in = new Scanner(System.in);

        int N = in.nextInt();
    
        int[] surfaceX = new int[N];
        int[] surfaceY = new int[N];
        for (int i = 0; i < N; i++) {
            surfaceX[i] = in.nextInt();
            surfaceY[i] = in.nextInt();
        }
        
        while (true) {
            int X = in.nextInt();
            int Y = in.nextInt();
            int dX = in.nextInt();
            int dY = in.nextInt();
            in.nextInt(); in.nextInt(); in.nextInt();

            // Finds landing altitude by Looking for 2 consecutive points w/ same Y
            int groundY = -1;
            for (int i = 0; (i < N && groundY == -1); i++)
                if (surfaceX[i] <= X && X <= surfaceX[i+1])
                    groundY = surfaceY[i];

            /* Finds out if it is safe to start braking only next turn, by
               simulating it and analyzing vertical speed at landing
               If it is safe, waits, else starts braking*/
            double vdY = dY - 8.555;
            double vY = Y - 36.665 + 5*dY;
            long t = Math.round((-40-vdY)/0.289);
            if (vY + t*(vdY+0.289*(1+t)/2) > groundY)
                System.out.println("0 0");
            else
                System.out.println("0 4");
        }
    }
}

Python 3

import math

N = int(input())
surfaceX, surfaceY = [], []
for i in range(N):
    landX, landY = [int(j) for j in input().split()]
    surfaceX.append(landX)
    surfaceY.append(landY)

while 1:
    X, Y, dX, dY, fuel, rotate, power = [int(i) for i in input().split()]
    
    # Finds landing altitude by Looking for 2 consecutive points w/ same Y
    groundY, i = -1, 0
    while (i < N) and (groundY == -1):
        if (surfaceX[i] <= X) and (X <= surfaceX[i+1]):
            groundY = surfaceY[i]
        i+=1
        
    # Finds out if it is safe to start braking only next turn, by
    # simulating it and analyzing vertical speed at landing
    # If it is safe, waits, else starts braking
    vdY , vY = dY-8.555, Y-36.665+5*dY;
    t = math.ceil((-40-vdY)/0.289);
    if (vY + t*(vdY+0.289*(1+t)/2) > groundY):
        print(0, 0)
    else:
        print(0, 4)