output_description: Print the maximum possible that a train can cover in the run.  
  Output is considered correct if its absolute difference from the judge's
  output is at most 10^{-3}.
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      N
      t_1 t_2 t_3 … t_N
      v_1 v_2 v_3 … v_N
attempt: num = int(input())
  in1 = input().split(" ")
  in2 = input().split(" ")
  times = [0]
  ct = 0
  for a in range(num):
      in1[a] = int(in1[a])
      in2[a] = int(in2[a])
  for a in range(num):
      ct += in1[a]
      times.append(ct)
  vlimit = []
  for a in range(1, num):
      vlimit.append(min(in2[a - 1], in2[a]))
  vlimit.append(in2[num - 1])
  print(in1, in2)
  print(times, vlimit)
  
  bairitu = 300
  
  last = times[len(times) - 1]
  limit = [0 for a in range(last * bairitu + 1)]
  
  for a in range(last * bairitu):
      limit[a] = round(min(a / bairitu, last - a / bairitu), 2)
  for a in range(len(vlimit)):
      start = times[a] * bairitu
      end = times[a + 1] * bairitu
      for b in range(start, end):
          limit[b] = min(limit[b], in2[a])
  # print(str(limit))
  
  for a in range(len(vlimit)):
      for b in range(0, times[a + 1] * bairitu):
          limit[b] = min(limit[b], times[a + 1] - b / bairitu + vlimit[a])
      for b in range(times[a + 1] * bairitu + 1, last * bairitu + 1):
          limit[b] = min(limit[b], b / bairitu - times[a + 1] + vlimit[a])
  
  ans = 0
  for a in range(1, len(limit)):
      ans += (limit[a] + limit[a - 1]) / bairitu / 2
  print(ans)
  
problem_description: Statement
  
  In the year 2168, AtCoder Inc., which is much larger than now, is starting a
  limited express train service called _AtCoder Express_.
  
  In the plan developed by the president Takahashi, the trains will run as
  follows:
  
    * A train will run for (t_1 + t_2 + t_3 + ... + t_N) seconds.
    * In the first t_1 seconds, a train must run at a speed of at most v_1 m/s (meters per second). Similarly, in the subsequent t_2 seconds, a train must run at a speed of at most v_2 m/s, and so on.
  
  According to the specifications of the trains, the acceleration of a train
  must be always within ±1m/s^2. Additionally, a train must stop at the
  beginning and the end of the run.
  
  Find the maximum possible distance that a train can cover in the run.
samples: [{"input": "1\n    100\n    30", "output": "2100.000000000000000\n    \n\n![ ](https://img.atcoder.jp/abc076/69c1f4241b608bc36f1f08dd4184d3f0.png)\n\nThe maximum distance is achieved when a train runs as follows:\n\n  * In the first 30 seconds, it accelerates at a rate of 1m/s^2, covering 450 meters.\n  * In the subsequent 40 seconds, it maintains the velocity of 30m/s, covering 1200 meters.\n  * In the last 30 seconds, it decelerates at the acceleration of -1m/s^2, covering 450 meters.\n\nThe total distance covered is 450 \\+ 1200 \\+ 450 = 2100 meters.\n\n* * *"}, {"input": "2\n    60 50\n    34 38", "output": "2632.000000000000000\n    \n\n![ ](https://img.atcoder.jp/abc076/a3e07ea723f50df04461165bc2cc8890.png)\n\nThe maximum distance is achieved when a train runs as follows:\n\n  * In the first 34 seconds, it accelerates at a rate of 1m/s^2, covering 578 meters.\n  * In the subsequent 26 seconds, it maintains the velocity of 34m/s, covering 884 meters.\n  * In the subsequent 4 seconds, it accelerates at a rate of 1m/s^2, covering 144 meters.\n  * In the subsequent 8 seconds, it maintains the velocity of 38m/s, covering 304 meters.\n  * In the last 38 seconds, it decelerates at the acceleration of -1m/s^2, covering 722 meters.\n\nThe total distance covered is 578 \\+ 884 \\+ 144 \\+ 304 \\+ 722 = 2632 meters.\n\n* * *"}, {"input": "3\n    12 14 2\n    6 2 7", "output": "76.000000000000000\n    \n\n![ ](https://img.atcoder.jp/abc076/77f821f590cb19d8e449303a102422dc.png)\n\nThe maximum distance is achieved when a train runs as follows:\n\n  * In the first 6 seconds, it accelerates at a rate of 1m/s^2, covering 18 meters.\n  * In the subsequent 2 seconds, it maintains the velocity of 6m/s, covering 12 meters.\n  * In the subsequent 4 seconds, it decelerates at the acceleration of -1m/s^2, covering 16 meters.\n  * In the subsequent 14 seconds, it maintains the velocity of 2m/s, covering 28 meters.\n  * In the last 2 seconds, it decelerates at the acceleration of -1m/s^2, covering 2 meters.\n\nThe total distance covered is 18 \\+ 12 \\+ 16 \\+ 28 \\+ 2 = 76 meters.\n\n* * *"}, {"input": "1\n    9\n    10", "output": "20.250000000000000000\n    \n\n![ ](https://img.atcoder.jp/abc076/ebde8cbeb649ae7fd338180c0562ae0b.png)\n\nThe maximum distance is achieved when a train runs as follows:\n\n  * In the first 4.5 seconds, it accelerates at a rate of 1m/s^2, covering 10.125 meters.\n  * In the last 4.5 seconds, it decelerates at the acceleration of -1m/s^2, covering 10.125 meters.\n\nThe total distance covered is 10.125 \\+ 10.125 = 20.25 meters.\n\n* * *"}, {"input": "10\n    64 55 27 35 76 119 7 18 49 100\n    29 19 31 39 27 48 41 87 55 70", "output": "20291.000000000000"}]