AtCoder Express(数学+二分)

D - AtCoder Express

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Time limit : 2sec / Memory limit : 256MB

Score : 400 points

Problem 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 (t1+t2+t3+…+tN) seconds.
• In the first t1 seconds, a train must run at a speed of at most v1 m/s (meters per second). Similarly, in the subsequent t2 seconds, a train must run at a speed of at mostv2 m/s, and so on.

According to the specifications of the trains, the acceleration of a train must be always within ±1m⁄s2. 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.

Constraints

• 1≤N≤100
• 1≤ti≤200
• 1≤vi≤100
• All input values are integers.

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Input

Input is given from Standard Input in the following format:

N
t1 t2 t3 … tN
v1 v2 v3 … vN

Output

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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Sample Input 1

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1
100
30

Sample Output 1

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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⁄s2, 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⁄s2, covering 450 meters.

The total distance covered is 450 + 1200 + 450 = 2100 meters.

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Sample Input 2

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2
60 50
34 38

Sample Output 2

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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⁄s2, 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⁄s2, 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⁄s2, covering 722 meters.

The total distance covered is 578 + 884 + 144 + 304 + 722 = 2632 meters.

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Sample Input 3

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3
12 14 2
6 2 7

Sample Output 3

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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⁄s2, 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⁄s2, 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⁄s2, covering 2 meters.

The total distance covered is 18 + 12 + 16 + 28 + 2 = 76 meters.

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Sample Input 4

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1
9
10

Sample Output 4

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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⁄s2, covering 10.125 meters.
• In the last 4.5 seconds, it decelerates at the acceleration of −1m⁄s2, covering 10.125 meters.

The total distance covered is 10.125 + 10.125 = 20.25 meters.

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Sample Input 5

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10
64 55 27 35 76 119 7 18 49 100
29 19 31 39 27 48 41 87 55 70

Sample Output 5

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20291.000000000000

//题意：读了老半天，就是说一列火车在 n 段路上行驶，给出行驶的时间，和最大速度限制，起点终点速度要为 0 ，问最大可行驶的路程是多少?
//应该是头次做这种题吧， 所以，做的比较慢，先逆序扫一遍，可得到在每段路上行驶，为了到终点为 0 速度，限制速度是多少。
然后正序累加，具体操作是，二分（加速时间+匀速时间)的最大值，然后就可以轻松解决了

 # include <bits/stdc++.h>
 using namespace std;
 # define eps 1e-
 # define INF 1e20
 # define pi  acos(-1.0)
 # define MX
 struct Node
 {
     double t,v;
     double ev; //从终点的最大速度
 }seg[MX];

 int n;
 double spd, ans;

 void gogo(int x)
 {
     double l=, r=seg[x].t;
     double res = ;
     while (l<r-eps) // <
     {
         double mid = (l+r)/;
         if (mid + spd > seg[x].v-eps)
         {
             if ((seg[x].v - seg[x+].ev) + mid < seg[x].t+eps)
             {
                 res = mid;
                 l = mid;
             }
             else r = mid;
         }
         else
         {
             if (mid + spd - (seg[x].t-mid) - seg[x+].ev < -eps) //<=
             {
                 res = mid;
                 l = mid;
             }
             else r = mid;
         }
     }

     double jia, yun, jian;
     if (spd + res < seg[x].v)
     {
         jia = res;
         yun = ;
         jian = seg[x].t - jia;
     }
     else
     {
         jia = seg[x].v - spd;
         yun = res - jia;
         jian = seg[x].t - jia - yun;
     }
     ans += 1.0/2.0*jia*jia + spd * jia;
     ans += yun * seg[x].v;

     if (res+spd < seg[x].v)
     {
         ans += -1.0/2.0*jian*jian +  (spd+jia) * jian;
         spd = spd + jia - jian;
     }
     else
     {
         ans += -1.0/2.0*jian*jian +  seg[x].v * jian;
         spd = seg[x].v - jian;
     }

 }

 int main()
 {
     scanf("%d",&n);
     for (int i=;i<=n;i++)
         scanf("%lf",&seg[i].t);
     for (int i=;i<=n;i++)
         scanf("%lf",&seg[i].v);
     spd=0.0;
     for (int i=n;i>=;i--)
     {
         if (spd+seg[i].t < seg[i].v+eps) //<=
             spd += seg[i].t;
         else
             spd = seg[i].v;
         seg[i].ev = spd;
     }
     ans = ;
     spd = ;
     seg[n+] = (Node){0.0, 0.0, 0.0};

     for (int i=;i<=n;i++)
     {
         gogo(i);
     }
     printf("%.5f\n",ans);
     return ;
 }