C. Discrete Acceleration
题解：二分。二分枚举距离，分别求出从前往后的时间与从后往前的时间，让距离近似相等。
#include <iostream>
#include <cstdio>
typedef long long ll;
using namespace std;
const int N = 1e5+10;double a[N];int L;
int n;double get1(double x){double pre = 0,t = 0,speed = 1.0;for(int i = 1;i <= n;i++){if(a[i] <= x){t += (a[i]-pre)/speed;pre = a[i];}else break;speed += 1.0;}t += (x-pre)/speed;return t;
}double get2(double x){double pre = L,t = 0,speed = 1.0;for(int i = n;i >= 1;i--){if(a[i] >= x){t += (pre-a[i])/speed;pre = a[i];}else break;speed += 1.0;}t += (pre-x)/speed;return t;
}void solve(){scanf("%d%d",&n,&L);for(int i = 1;i <= n;i++){scanf("%lf",a+i);}double l = 0,r = L;while(r-l > 1e-6){double mid = (l+r)/2.0;//cout<<l<<" "<<r<<" "<<mid<<endl;double t1 = get1(mid);double t2 = get2(mid);if(t1 < t2){l = mid;}else {r = mid;}}printf("%.10lf\n",get1(l));
}
int main(){int t;scanf("%d",&t);while(t--){solve();}return 0;
}