C. Fly
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
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 aiaitons 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 1.51.5 tons.
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.
题意：火箭原本重为m，带有k重量的燃料，从一星球降落飞，需要消耗（m+k）/ai重量的燃料，再从一星球起飞需要消耗（m+k'）/bi，现在问，至少要多少燃料，才能从1星球飞到n星球再飞回来。
思路：题意已经说明k最大为1e9，所以通过二分枚举k是否满足条件即可算出最小的k。
#include<bits/stdc++.h>
#define lson l,m
#define rson m+1,r
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = int(1e5) + 100;
const int maxm=100;
const int BN = 30;
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const int mod = (int)1e9 + 7;
const double EPS = 1e-12;
inline int DB(double x) {
if (x<-EPS) return -1;
if (x>EPS) return 1;
return 0;
}
int a[maxn],b[maxn];
bool solve(int n,int m,double x) {
for(int i=1; i<=n; i++) {
double tmp=(x+m)/a[i];
if(DB(x-tmp)<0) {
return false;
}
x-=tmp;
tmp=(x+m)/b[i];
if(DB(x-tmp)<0) {
return false;
}
x-=tmp;
}
return true;
}
double search_bit(int n,int m) {
double l=0,r=1e9;
double mid;
for(int i=1; i<=200; i++) {
mid=(l+r)*0.5;
if(solve(n,m,mid))
r=mid;
else
l=mid;
}
if(DB(l-1e9)!=0)
return l;
else
return -1;
}
int main() {
int n,m;
while(~scanf("%d %d",&n,&m)) {
for(int i=1; i<=n; i++) scanf("%d",&a[i]);
for(int i=1; i<=n; i++) scanf("%d",&b[i]);
double ans=search_bit(n,m);
if(DB(ans+1)==0)
printf("-1\n");
else
printf("%.12lf\n",ans);
}
return 0;
}