【HDOJ6581】Vacation（模拟）

题意：有标号从0到n的n+1辆车，每辆车离终点的距离为s[i]，车长为l[i]，速度为v[i]，规定不能超车，问何时所有车都能过终点线

n<=1e5，保证

思路：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned int uint;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<ll,ll> Pll;
typedef vector<int> VI;
typedef vector<PII> VII;
typedef pair<ll,int>P;
#define N  2100000
#define M  151000
#define fi first
#define se second
#define MP make_pair
#define pi acos(-1)
#define mem(a,b) memset(a,b,sizeof(a))
#define rep(i,a,b) for(int i=(int)a;i<=(int)b;i++)
#define per(i,a,b) for(int i=(int)a;i>=(int)b;i--)
#define lowbit(x) x&(-x)
#define Rand (rand()*(1<<16)+rand())
#define id(x) ((x)<=B?(x):m-n/(x)+1)
#define ls p<<1
#define rs p<<1|1

const ll MOD=1e9+7,inv2=(MOD+1)/2;
      double eps=1e-6;
      ll INF=1e18;
      ll inf=5e13;
      int dx[4]={-1,1,0,0};
      int dy[4]={0,0,-1,1};

int l[N],s[N],v[N];

int read()
{
   int v=0,f=1;
   char c=getchar();
   while(c<48||57<c) {if(c=='-') f=-1; c=getchar();}
   while(48<=c&&c<=57) v=(v<<3)+v+v+c-48,c=getchar();
   return v*f;
}


int main()
{
    //freopen("1.in","r",stdin);
    //freopen("1.out","w",stdout);
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        rep(i,1,n+1) l[i]=read();
        rep(i,1,n+1) s[i]=read();
        rep(i,1,n+1) v[i]=read();
        double sum=0,ans=1.0*s[1]/v[1];
        rep(i,2,n+1)
        {
            sum+=l[i];
            ans=max(ans,1.0*(sum+s[i])/v[i]);
        }
        printf("%.10f
",ans);
    }



    return 0;
}