{"problem":{"name":"AtCoder Express","description":{"content":"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 r","description_type":"Markdown"},"platform":"AtCoder","limit":{"time_limit":2000,"memory_limit":262144},"difficulty":"None","is_remote":true,"is_sync":true,"sync_url":null,"sign":"abc076_d"},"statements":[{"statement_type":"Markdown","content":"In the year 2168, AtCoder Inc., which is much larger than now, is starting a limited express train service called _AtCoder Express_.\nIn the plan developed by the president Takahashi, the trains will run as follows:\n\n*   A train will run for $(t_1 + t_2 + t_3 + ... + t_N)$ seconds.\n*   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.\n\nAccording 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.\nFind the maximum possible distance that a train can cover in the run.\n\n## Constraints\n\n*   $1 \\leq N \\leq 100$\n*   $1 \\leq t_i \\leq 200$\n*   $1 \\leq v_i \\leq 100$\n*   All input values are integers.\n\n## Input\n\nInput is given from Standard Input in the following format:\n\n$N$\n$t_1$ $t_2$ $t_3$ … $t_N$\n$v_1$ $v_2$ $v_3$ … $v_N$\n\n[samples]","is_translate":false,"language":"English"}],"meta":{"iden":"abc076_d","tags":[],"sample_group":[["1\n100\n30","2100.000000000000000\n\n![image](https://img.atcoder.jp/abc076/69c1f4241b608bc36f1f08dd4184d3f0.png)\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."],["2\n60 50\n34 38","2632.000000000000000\n\n![image](https://img.atcoder.jp/abc076/a3e07ea723f50df04461165bc2cc8890.png)\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."],["3\n12 14 2\n6 2 7","76.000000000000000\n\n![image](https://img.atcoder.jp/abc076/77f821f590cb19d8e449303a102422dc.png)\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."],["1\n9\n10","20.250000000000000000\n\n![image](https://img.atcoder.jp/abc076/ebde8cbeb649ae7fd338180c0562ae0b.png)\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."],["10\n64 55 27 35 76 119 7 18 49 100\n29 19 31 39 27 48 41 87 55 70","20291.000000000000"]],"created_at":"2026-03-03 11:01:14"}}