Codeforces Round #499 (Div. 2) C Fly题解

题目

http://codeforces.com/contest/1011/problem/C

Natasha is going to fly on a rocket to Mars and return to Earth. Also, on the way to Mars, she will land on $\mathrm{n-2n-2 intermediate\; planets.\; Formally:\; we\; number\; all\; the\; planets\; from 11 to nn. 11 is\; Earth, nn is\; Mars.\; Natasha\; will\; make\; exactly nn flights: 1\to 2\to \dots n\to 11\to 2\to \dots n\to 1.}$

Flight from $\mathrm{xx to yy consists\; of\; two\; phases:\; take-off\; from\; planet xx and\; landing\; to\; planet yy.\; This\; way,\; the\; overall\; itinerary\; of\; the\; trip\; will\; be:\; the 11-st\; planet \to \to take-off\; from\; the 11-st\; planet \to \to landing\; to\; the 22-nd\; planet \to \to 22-nd\; planet \to \to take-off\; from\; the 22-nd\; planet \to \to \dots \dots \to \to landing\; to\; the nn-th\; planet \to \to the nn-th\; planet \to \to take-off\; from\; the nn-th\; planet \to \to landing\; to\; the 11-st\; planet \to \to the 11-st\; planet.}$

The mass of the rocket together with all the useful cargo (but without fuel) is $\mathrm{mm tons.\; However,\; Natasha\; does\; not\; know\; how\; much\; fuel\; to\; load\; into\; the\; rocket.\; Unfortunately,\; fuel\; can\; only\; be\; loaded\; on\; Earth,\; so\; if\; the\; rocket\; runs\; out\; of\; fuel\; on\; some\; other\; planet,\; Natasha\; will\; not\; be\; able\; to\; return\; home.\; Fuel\; is\; needed\; to\; take-off\; from\; each\; planet\; and\; to\; land\; to\; each\; planet.\; It\; is\; known\; that 11 ton\; of\; fuel\; can\; lift\; off aiaitons\; of\; rocket\; from\; the ii-th\; planet\; or\; to\; land bibi tons\; of\; rocket\; onto\; the ii-th\; planet.}$

For example, if the weight of rocket is $\mathrm{99 tons,\; weight\; of\; fuel\; is 33 tons\; and\; take-off\; coefficient\; is 88 (ai=8ai=8),\; then 1.51.5 tons\; of\; fuel\; will\; be\; burnt\; (since 1.5\cdot 8=9+31.5\cdot 8=9+3).\; The\; new\; weight\; of\; fuel\; after\; take-off\; will\; be 1.51.5 tons.}$

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 $\mathrm{nn (2\le n\le 10002\le n\le 1000) —\; number\; of\; planets.}$

The second line contains the only integer $\mathrm{mm (1\le m\le 10001\le m\le 1000) —\; weight\; of\; the\; payload.}$

The third line contains $\mathrm{nn integers a1,a2,\dots ,ana1,a2,\dots ,an (1\le ai\le 10001\le ai\le 1000),\; where aiai is\; the\; number\; of\; tons,\; which\; can\; be\; lifted\; off\; by\; one\; ton\; of\; fuel.}$

The fourth line contains $\mathrm{nn integers b1,b2,\dots ,bnb1,b2,\dots ,bn (1\le bi\le 10001\le bi\le 1000),\; where bibi is\; the\; number\; of\; tons,\; which\; can\; be\; landed\; by\; one\; ton\; of\; fuel.}$

It is guaranteed, that if Natasha can make a flight, then it takes no more than ${\mathrm{109109 tons\; of\; fuel.}}^{}$

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 ${\mathrm{109109 tons\; of\; fuel.}}^{}$

The answer will be considered correct if its absolute or relative error doesn't exceed ${\mathrm{10-610-6.\; Formally,\; let\; your\; answer\; be pp,\; and\; the\; jury\text{'}s\; answer\; be qq.\; Your\; answer\; is\; considered\; correct\; if |p-q|max(1,|q|)\le 10-6|p-q|max(1,|q|)\le 10-6.}}^{}$

Examples

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
题意：
有n个星球，从1飞到n，路径为1->2->3->……->n->1,数组a中a[i]表示从第i个星球起飞一吨燃料可以支持多少重量起飞(飞船重量m和燃料重量总和为总重量)，

数组b中b[i]表示降落到第i个星球一吨燃料可以支持多少重量降落(飞船重量m和燃料重量总和为总重量)。
思路：
反向思考，设当前操作需要燃料x吨，除当前燃料所需外飞船总重量为y,当前1吨燃料可以支持的重量为t，则（x+y）/t=x;
所以x=y/(t-1);再进行前一步操作时将y加上x，由此式子也可看出t必须大于1，否则无法完成。
代码：

#include<bits/stdc++.h>
#define fi first
#define se second
#define INF 0x3f3f3f3f
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define pqueue priority_queue
#define NEW(a,b) memset(a,b,sizeof(a))
const double pi=4.0*atan(1.0);
const double e=exp(1.0);
const int maxn=3e6+8;
typedef long long LL;
typedef unsigned long long ULL;
//typedef pair<LL,LL> P;
const LL mod=1e9+7;
using namespace std;
int a[1100];
int b[1100];
int c[2200];
int main(){
    int n,m;
    scanf("%d%d",&n,&m);
    for(int i=0;i<n;i++){
        scanf("%d",&a[i]);
    }
    for(int i=0;i<n;i++){
        scanf("%d",&b[i]);
    }
    int cnt=0;
    for(int i=0;i<n;i++){
        c[cnt++]=a[i];
        if(i!=n-1){
            c[cnt++]=b[i+1];
        }
        else{
            c[cnt++]=b[0];
        }
    }//记录操作顺序.
    double g=double(m);
    double k;
　　 //反向操作.
    for(int i=cnt-1;i>=0;i--){
        if(c[i]<=1){
            printf("-1
");
            return 0;
        }
        k=g/(double(c[i])-1.0);
        g+=k;
    }
    double ans=g-(double)m;
    printf("%.10lf
",ans);
}