Chat: 本來想隨便看看題寫一下， 發現自己現在理解能力跌倒谷底orz。 看了好久纔看懂這個題是個什麼意思。。。
Description：
Natasha is going to fly on a rocket to Mars and return to Earth. Also, on the way to Mars, she will land on n−2n−2 intermediate planets. Formally: we number all the planets from 11 to nn. 11 is Earth, nn is Mars. Natasha will make exactly nn flights: 1→2→…n→11→2→…n→1.
Flight from xx to yy consists of two phases: take-off from planet xx and landing to planet yy. This way, the overall itinerary of the trip will be: the 11-st planet →→ take-off from the 11-st planet →→ landing to the 22-nd planet →→ 22-nd planet →→ take-off from the 22-nd planet →→ …… →→ landing to the nn-th planet →→ the nn-th planet →→ take-off from the nn-th planet →→ landing to the 11-st planet →→ the 11-st planet.
The mass of the rocket together with all the useful cargo (but without fuel) is mm tons. However, Natasha does not know how much fuel to load into the rocket. Unfortunately, fuel can only be loaded on Earth, so if the rocket runs out of fuel on some other planet, Natasha will not be able to return home. Fuel is needed to take-off from each planet and to land to each planet. It is known that 11 ton of fuel can lift off aiai tons of rocket from the ii-th planet or to land bibi tons of rocket onto the ii-th planet.
For example, if the weight of rocket is 99 tons, weight of fuel is 33 tons and take-off coefficient is 88 (ai=8ai=8), then 1.51.5 tons of fuel will be burnt (since 1.5⋅8=9+31.5⋅8=9+3). The new weight of fuel after take-off will be 1.51.5tons.
Please note, that it is allowed to burn non-integral amount of fuel during take-off or landing, and the amount of initial fuel can be non-integral as well.
Help Natasha to calculate the minimum mass of fuel to load into the rocket. Note, that the rocket must spend fuel to carry both useful cargo and the fuel itself. However, it doesn't need to carry the fuel which has already been burnt. Assume, that the rocket takes off and lands instantly.
Input
The first line contains a single integer nn (2≤n≤10002≤n≤1000) — number of planets.
The second line contains the only integer mm (1≤m≤10001≤m≤1000) — weight of the payload.
The third line contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤10001≤ai≤1000), where aiai is the number of tons, which can be lifted off by one ton of fuel.
The fourth line contains nn integers b1,b2,…,bnb1,b2,…,bn (1≤bi≤10001≤bi≤1000), where bibi is the number of tons, which can be landed by one ton of fuel.
It is guaranteed, that if Natasha can make a flight, then it takes no more than 109109 tons of fuel.
Output
If Natasha can fly to Mars through (n−2)(n−2) planets and return to Earth, print the minimum mass of fuel (in tons) that Natasha should take. Otherwise, print a single number −1−1.
It is guaranteed, that if Natasha can make a flight, then it takes no more than 109109 tons of fuel.
The answer will be considered correct if its absolute or relative error doesn't exceed 10−610−6. Formally, let your answer be pp, and the jury's answer be qq. Your answer is considered correct if |p−q|max(1,|q|)≤10−6|p−q|max(1,|q|)≤10−6.
input
2
12
11 8
7 5
output
10.0000000000
input
3
1
1 4 1
2 5 3
output
-1
input
6
2
4 6 3 3 5 6
2 6 3 6 5 3
output
85.4800000000
Note
Let's consider the first example.
Initially, the mass of a rocket with fuel is 2222 tons.
- At take-off from Earth one ton of fuel can lift off 1111 tons of cargo, so to lift off 2222 tons you need to burn 22 tons of fuel. Remaining weight of the rocket with fuel is 2020 tons.
- During landing on Mars, one ton of fuel can land 55 tons of cargo, so for landing 2020 tons you will need to burn 44 tons of fuel. There will be 1616 tons of the rocket with fuel remaining.
- While taking off from Mars, one ton of fuel can raise 88 tons of cargo, so to lift off 1616 tons you will need to burn 22 tons of fuel. There will be 1414 tons of rocket with fuel after that.
- During landing on Earth, one ton of fuel can land 77 tons of cargo, so for landing 1414 tons you will need to burn 22 tons of fuel. Remaining weight is 1212 tons, that is, a rocket without any fuel.
In the second case, the rocket will not be able even to take off from Earth.
題目大意：
有n個星球， 問從第一個星球出發到所有的星球起飛降落再回到第一個星球所需要的最小耗燃料量。（因爲題上寫只能在第一個星球加燃料， 意思就是讓你求一開始要帶多少燃料）。
解題思路：
一翻譯完大致思路差不多都出來了， 因爲要計算最小的所以肯定回到第一個星球如果油量剩餘的不多不少是最好的。 這樣我們可以就倒着推出一開始的燃料量， 題上又給出了計算的方法， 那麼我們直接拿來用就可以了。
代碼：
#include<bits/stdc++.h>
using namespace std;
//tools:
//ios::sync_with_stdio(false);
//freopen("input.txt", "r", stdin);
typedef long long ll;
typedef unsigned long long ull;
const int dir[9][2] = { 0, 1, 0, -1, 1, 0, -1, 0, 1, 1, 1, -1, -1, 1, -1, -1, 0, 0 };
const ll ll_inf = 0x7fffffff;
const int inf = 0x3f3f3f;
const int mod = (int)9901;
const int Max = (int)1e3 + 51;
int a[Max], b[Max];
int main() {
int n;
double m;
int flag = 0;
scanf("%d %lf", &n, &m);
for (int i = 0; i < n; ++i) {
scanf("%d", &a[i]);
if (a[i] <= 1) {
flag = 1;
}
}
for (int i = 0; i < n; ++i) {
scanf("%d", &b[i]);
if (b[i] <= 1) {
flag = 1;
}
}
if (flag) {
printf("-1\n");
return 0;
}
double sum = m, fuel = 0, ans = 0;
for (int i = 0; i < n; ++i) {
fuel = sum / (a[i] - 1);
sum += fuel;
ans += fuel;
fuel = (sum / (b[i] - 1));
sum += fuel;
ans += fuel;
}
printf("%.10f\n", ans);
return 0;
}