AtCoder Beginner Contest 076
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Problem Statement
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In the year 2168, AtCoder Inc., which is much larger than now, is starting a limited express train service called AtCoder Express.
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In the plan developed by the president Takahashi, the trains will run as follows:
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- 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.
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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.
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Find the maximum possible distance that a train can cover in the run.
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Constraints
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$1 \leq N \leq 100$$1 \leq t_i \leq 200$$1 \leq v_i \leq 100$- All input values are integers.
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Input
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Input is given from Standard Input in the following format:
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N
t_1 t_2 t_3 … t_N
v_1 v_2 v_3 … v_N翻译中…
Output
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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 $10^{-3}$.
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样例
Input #1
1
100
30
Output #1
2100.000000000000000
The maximum distance is achieved when a train runs as follows:
- In the first $30$ seconds, it accelerates at a rate of $1m/s^2$, covering $450$ meters.
- In the subsequent $40$ seconds, it maintains the velocity of $30m/s$, covering $1200$ meters.
- In the last $30$ seconds, it decelerates at the acceleration of $-1m/s^2$, covering $450$ meters.
The total distance covered is $450$ + $1200$ + $450$ = $2100$ meters.
Input #2
2
60 50
34 38
Output #2
2632.000000000000000
The maximum distance is achieved when a train runs as follows:
- In the first $34$ seconds, it accelerates at a rate of $1m/s^2$, covering $578$ meters.
- In the subsequent $26$ seconds, it maintains the velocity of $34m/s$, covering $884$ meters.
- In the subsequent $4$ seconds, it accelerates at a rate of $1m/s^2$, covering $144$ meters.
- In the subsequent $8$ seconds, it maintains the velocity of $38m/s$, covering $304$ meters.
- In the last $38$ seconds, it decelerates at the acceleration of $-1m/s^2$, covering $722$ meters.
The total distance covered is $578$ + $884$ + $144$ + $304$ + $722$ = $2632$ meters.
Input #3
3
12 14 2
6 2 7
Output #3
76.000000000000000
The maximum distance is achieved when a train runs as follows:
- In the first $6$ seconds, it accelerates at a rate of $1m/s^2$, covering $18$ meters.
- In the subsequent $2$ seconds, it maintains the velocity of $6m/s$, covering $12$ meters.
- In the subsequent $4$ seconds, it decelerates at the acceleration of $-1m/s^2$, covering $16$ meters.
- In the subsequent $14$ seconds, it maintains the velocity of $2m/s$, covering $28$ meters.
- In the last $2$ seconds, it decelerates at the acceleration of $-1m/s^2$, covering $2$ meters.
The total distance covered is $18$ + $12$ + $16$ + $28$ + $2$ = $76$ meters.
Input #4
1
9
10
Output #4
20.250000000000000000
The maximum distance is achieved when a train runs as follows:
- In the first $4.5$ seconds, it accelerates at a rate of $1m/s^2$, covering $10.125$ meters.
- In the last $4.5$ seconds, it decelerates at the acceleration of $-1m/s^2$, covering $10.125$ meters.
The total distance covered is $10.125$ + $10.125$ = $20.25$ meters.
Input #5
10
64 55 27 35 76 119 7 18 49 100
29 19 31 39 27 48 41 87 55 70
Output #5
20291.000000000000
题目信息
题面与样例版权归 AtCoder 及出题人所有；中文翻译由 AI 生成（CI 管线），如有出入请以原文为准。