Three-level Laser（cf Round#472 C）

最新推荐文章于 2018-11-04 07:47:47 发布

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#高精度除法 #二分

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本文探讨了在特定约束条件下，如何通过选择合适的原子状态来最大化三能级激光的能量转换效率。介绍了问题背景、解决思路及算法实现细节，并提供了一个高精度的解决方案。

An atom of element X can exist in n distinct states with energies E₁ < E₂ < ... < E_n. Arkady wants to build a laser on this element, using a three-level scheme. Here is a simplified description of the scheme.

Three distinct states i, j and k are selected, where i < j < k. After that the following process happens:

initially the atom is in the state i,
we spend E_k - E_i energy to put the atom in the state k,
the atom emits a photon with useful energy E_k - E_j and changes its state to the state j,
the atom spontaneously changes its state to the state i, losing energy E_j - E_i,
the process repeats from step 1.

Let's define the energy conversion efficiency as $\text{[math]}$ , i. e. the ration between the useful energy of the photon and spent energy.

Due to some limitations, Arkady can only choose such three states that E_k - E_i ≤ U.

Help Arkady to find such the maximum possible energy conversion efficiency within the above constraints.

Input

The first line contains two integers n and U (3 ≤ n ≤ 10⁵, 1 ≤ U ≤ 10⁹) — the number of states and the maximum possible difference between E_k and E_i.

The second line contains a sequence of integers E₁, E₂, ..., E_n (1 ≤ E₁ < E₂... < E_n ≤ 10⁹). It is guaranteed that all E_i are given in increasing order.

Output

If it is not possible to choose three states that satisfy all constraints, print -1.

Otherwise, print one real number η — the maximum possible energy conversion efficiency. Your answer is considered correct its absolute or relative error does not exceed 10^- 9.

Formally, let your answer be a, and the jury's answer be b. Your answer is considered correct if $\text{[math]}$ .

Examples

Input

Copy

4 4
1 3 5 7

Output

0.5

Input

Copy

10 8
10 13 15 16 17 19 20 22 24 25

Output

0.875

Input

Copy

3 1
2 5 10

Output

-1

Note

In the first example choose states 1, 2 and 3, so that the energy conversion efficiency becomes equal to $\text{[math]}$ .

In the second example choose states 4, 5 and 9, so that the energy conversion efficiency becomes equal to $\text{[math]}$ .

题目链接：传送门

题意：

给定n个严格递增的数（E[1],E[2],...,E[n]），从中任意找三个数，设下标分别为i,j,k，要求E[i]、E[j]、E[k]相互之间的差值都小于U，且i<j<k，问(E[k]-E[j])/(E[k]-E[i])的最大值是多少，若不存在这样的i,j,k，则输出-1。

思路：

设 res = (E[k]-E[j])/(E[k]-E[i]) = 1 - (E[j]-E[i])/(E[k]-E[i]);

可以看出E[k]越大，res越大 ; E[j]越小，res越大。因此，要在可选范围内，让E[k]的值尽可能大，E[j]尽可能小。

因为 E[k]<E[i]+U 且 E[j]>E[i]，所以，可以枚举 i 从1 -> n，找到E[ ]中小于E[i]+U的最大值，即为E[k]，E[j]则为E[i+1]。找到遍历 i 的过程中res的最大值即可。

当然，若在[ E[i] , E[i]+U ]的范围内没有3个数，则对于该 i ,res不存在。

优化：用二分查找E[ ]中小于E[i]+U的最大值，总复杂度为O( nlogn ) ;

坑点：题目对答案的精度有要求，要在1e-9之内，而double只能到1e-6，所以要自己模拟一下除法运算，高精度输出。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
#include <vector>
#include <algorithm>
using namespace std;

const int MAX = 1e5 + 100;

int n;
long long U;
long long e[MAX];

//二分查找E[ ]中小于E[i]+U的最大值的下标
long long bin_search(long long x)
{
	int l = 0;
	int r = n;
	while (r - l >= 1)
	{
		int i = (l + r) >> 1;
		if (e[i] == x)
			return i;
		else if (e[i]<x)
			l = i + 1;
		else
			r = i;
	}
	if (e[l] > x)
	{
		for (int i = l; i >= 0; i--)
		{
			if (e[i] <= x)
				return i;
		}
	}
	else if (e[l] == x)
		return l;
	else
	{
		for (int i = l; i < n; i++)
		{
			if (e[i] > x)
				return i - 1;
		}
	}
}

//模拟除法运算
void div(long long a,long long b,long long *num)
{
    for(int i=0;i<20;i++)
    {
        num[i]=a/b;
        a=a%b;
        a*=10;
    }
}

int main()
{
	scanf("%d%lld", &n, &U);
	for (int i = 0; i < n; i++)
		scanf("%lld", &e[i]);
	long  long ans[100];
	memset(ans,0,sizeof(ans));
	for (int i = 0; i < n; i++)
	{
		int num;
		if (e[i] + U >= e[n - 1])
			num = n - 1;
		else
			num = bin_search(e[i] + U);
		if (num - i + 1 >= 3)
		{
			//ans = max(ans, (long double)(e[num] - e[i + 1]) / (e[num] - e[i]));
            long long tmp[100];
			div(e[num] - e[i + 1],e[num] - e[i],tmp);
			for(int j=0;j<20;j++)
			{
                if(tmp[j]>ans[j])
                {
                    for(int k=0;k<20;k++)
                        ans[k]=tmp[k];
                    break;
                }
                else if(tmp[j]==ans[j])
                    continue;
                else
                    break;
			}
		}
	}
    //若ans的值为0,则输出-1
    int flag=0;
    for(int i=0;i<20;i++)
    {
        if(ans[i]!=0)
        {
            flag=1;
            break;
        }
    }
    if(flag==0)
    {
        printf("-1\n");
    }
	else
    {
        printf("%lld.",ans[0]);
        for(int i=1;i<20;i++)
            printf("%lld",ans[i]);
        printf("\n");
    }
	return 0;
}