题干：
On vacations n pupils decided to go on excursion and gather all together. They need to overcome the path with the length l meters. Each of the pupils will go with the speed equal to v1. To get to the excursion quickly, it was decided to rent a bus, which has seats for k people (it means that it can't fit more than k people at the same time) and the speed equal to v2. In order to avoid seasick, each of the pupils want to get into the bus no more than once.
Determine the minimum time required for all n pupils to reach the place of excursion. Consider that the embarkation and disembarkation of passengers, as well as the reversal of the bus, take place immediately and this time can be neglected.
Input
The first line of the input contains five positive integers n, l, v1, v2 and k (1 ≤ n ≤ 10 000, 1 ≤ l ≤ 109, 1 ≤ v1 < v2 ≤ 109, 1 ≤ k ≤ n) — the number of pupils, the distance from meeting to the place of excursion, the speed of each pupil, the speed of bus and the number of seats in the bus.
Output
Print the real number — the minimum time in which all pupils can reach the place of excursion. Your answer will be considered correct if its absolute or relative error won't exceed 10 - 6.
Examples
Input
5 10 1 2 5
Output
5.0000000000
Input
3 6 1 2 1
Output
4.7142857143
Note
In the first sample we should immediately put all five pupils to the bus. The speed of the bus equals 2 and the distance is equal to 10, so the pupils will reach the place of excursion in time 10 / 2 = 5
题目大意：
n个人，一段路程为L，走路的速度为v1，公交车的速度v2（公交车只有一辆，且v1<v2），车中最多坐k人，每人最多做一次车，求所有人到终点的最小时间。 (1 ≤ n ≤ 10 000, 1 ≤ l ≤ 109, 1 ≤ v1 < v2 ≤ 109, 1 ≤ k ≤ n)
解题报告：
因为最佳答案不一定是把所有人载到终点再回去载下一波人，且注意到时间符合单调性，可以二分。首先求出一共需要用多少波公交车（因为v1<v2，所以充分利用公交车肯定可以使得时间最短），然后维护当前所有人所在的位置，然后求出载的这一波人在当前二分的可用时间下可以最短用多久的公交车，就让他用这么久的公交车，然后就让公交车返程来载下一波人。注意统计时间的时候，公交车是不需要返程的，所以不需要加上这一部分时间。
AC代码：
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define FF first
#define SS second
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
typedef pair<int,int> PII;
const int MAX = 2e5 + 5;
const double eps = 1e-8;
ll n,k;
double L,v1,v2;
bool ok(double time) {ll all = n/k + (n%k>0);double pos=0,t=0;for(int i = 1; i<=all; i++) {double tmpt = (L-pos-(time-t)*v1)/(v2-v1);if(i == all) {t += tmpt;break;}double chepos = tmpt * v2;//车本次走的的最远长度 double huit = (chepos-tmpt*v1)/(v1+v2); pos = pos + (tmpt+huit) * v1;t += tmpt+huit;}return t <= time;
}int main()
{cin>>n>>L>>v1>>v2>>k; double l = 0,r = L,mid,ans=r;int cnt=0;while(cnt++<500) {mid=(l+r)/2;if(ok(mid)) r = mid,ans=mid;else l = mid;} printf("%.8f\n",ans);return 0 ;
}
