Codeforces Round #499 (Div. 2) C. Fly(数学+思维模拟)

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 $\mathrm{n-2n-2 intermediate\; planets.\; Formally:\; we\; number\; all\; the\; planets\; from 11 to nn. 11 is\; Earth, nn is\; Mars.\; Natasha\; will\; make\; exactly nn flights: 1\to 2\to \dots n\to 11\to 2\to \dots n\to 1.}$

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 $\mathrm{mm tons.\; However,\; Natasha\; does\; not\; know\; how\; much\; fuel\; to\; load\; into\; the\; rocket.\; Unfortunately,\; fuel\; can\; only\; be\; loaded\; on\; Earth,\; so\; if\; the\; rocket\; runs\; out\; of\; fuel\; on\; some\; other\; planet,\; Natasha\; will\; not\; be\; able\; to\; return\; home.\; Fuel\; is\; needed\; to\; take-off\; from\; each\; planet\; and\; to\; land\; to\; each\; planet.\; It\; is\; known\; that 11 ton\; of\; fuel\; can\; lift\; off aiai tons\; of\; rocket\; from\; the ii-th\; planet\; or\; to\; land bibi tons\; of\; rocket\; onto\; the ii-th\; planet.}$

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

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

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

${\mathrm{Output}}^{}$

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 ${\mathrm{10-610-6.\; Formally,\; let\; your\; answer\; be pp,\; and\; the\; jury\text{'}s\; answer\; be qq.\; Your\; answer\; is\; considered\; correct\; if |p-q|max(1,|q|)\le 10-6|p-q|max(1,|q|)\le 10-6.}}^{}$

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 $\mathrm{2222 tons.}$

• At take-off from Earth one ton of fuel can lift off $\mathrm{1111 tons\; of\; cargo,\; so\; to\; lift\; off 2222 tons\; you\; need\; to\; burn 22 tons\; of\; fuel.\; Remaining\; weight\; of\; the\; rocket\; with\; fuel\; is 2020 tons.}$
• During landing on Mars, one ton of fuel can land $\mathrm{55 tons\; of\; cargo,\; so\; for\; landing 2020 tons\; you\; will\; need\; to\; burn 44 tons\; of\; fuel.\; There\; will\; be 1616 tons\; of\; the\; rocket\; with\; fuel\; remaining.}$
• While taking off from Mars, one ton of fuel can raise $\mathrm{88 tons\; of\; cargo,\; so\; to\; lift\; off 1616 tons\; you\; will\; need\; to\; burn 22 tons\; of\; fuel.\; There\; will\; be 1414 tons\; of\; rocket\; with\; fuel\; after\; that.}$
• During landing on Earth, one ton of fuel can land $\mathrm{77 tons\; of\; cargo,\; so\; for\; landing 1414 tons\; you\; will\; need\; to\; burn 22 tons\; of\; fuel.\; Remaining\; weight\; is 1212 tons,\; that\; is,\; a\; rocket\; without\; any\; fuel.}$

In the second case, the rocket will not be able even to take off from Earth.

 

 

题意：

给你飞机初始重量，飞机要在所有机场起飞降落。一开始会携带一定重量的汽油，然后每次飞的过程会消耗掉相应汽油，而每个机场给的值是1吨汽油能让他飞多少。

求一开始最少携带多少汽油。

解题思路：

拿第一组样例来说

11 8
7 5

欲最小，回到起点时候肯定就耗光了汽油，而最后一个待过的机场是 7 机场，因为2*7=12+2,2吨汽油能让他飞14，恰好是飞机重量+耗油重量，依次回推2*8=14+2,16+4=5*4,20+2=11*2

得 x*(a[i]或者b[i])=s+x  ,s为下一站油和机身的和，x和具体某个机场的耗油量。

说是数学，其实就是证明下就能发现，无论a[i] b[i]用哪个顺序，最后耗油最小量都一样

 

再注意点细节，就ok了

 

#include <bits/stdc++.h>
using namespace std;
#define ll            long long
#define FastRead        ios_base::sync_with_stdio(0);cin.tie(0)

const int maxn=1e5+20;
const int mod=1e9+7;

int n,a[1005],b[1005];
double s;
int main()
{
    cin>>n>>s;
    int f=1;
    for(int i=1;i<=n;i++) {cin>>a[i];if(a[i]<=1) f=0;}
    for(int i=1;i<=n;i++) {cin>>b[i];if(b[i]<=1) f=0;}
    if(f==0) { puts("-1");return 0; }
    double sum=0,x;
    for(int i=1;i<=n;i++)
    {
        x=s/(a[i]-1),s+=x,sum+=x;
        x=s/(b[i]-1),s+=x,sum+=x;
    }
    printf("%.10f\n",sum);
    return 0;
}