Codeforces Beta Round #62 ——B. Energy exchange

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#二分 #codeforces

文章探讨了如何通过合理分配能源来解决行星面临的能源危机，利用数学算法帮助角色Petya找到每个能源储存装置能容纳的最大等量能源。详细介绍了输入输出格式、数据范围和解决方案，包括数组排序、二分查找和能量转移计算。对于数组元素相同的特殊情况，文章进行了额外的判断以确保正确解题。

It is well known that the planet suffers from the energy crisis. Little Petya doesn't like that and wants to save the world. For this purpose he needs every accumulator to contain the same amount of energy. Initially every accumulator has some amount of energy: the i-th accumulator has a_i units of energy. Energy can be transferred from one accumulator to the other. Every timex units of energy are transferred (x is not necessarily an integer)k percent of it is lost. That is, if x units were transferred from one accumulator to the other, amount of energy in the first one decreased byx units and in other increased by $\text{[math]}$ units.

Your task is to help Petya find what maximum equal amount of energy can be stored in each accumulator after the transfers.

Input

First line of the input contains two integers n andk (1 ≤ n ≤ 10000, 0 ≤ k ≤ 99) — number of accumulators and the percent of energy that is lost during transfers.

Next line contains n integers a₁, a₂, ... , a_n — amounts of energy in the first, second, ..,n-th accumulator respectively (0 ≤ a_i ≤ 1000, 1 ≤ i ≤ n).

Output

Output maximum possible amount of energy that can remain in each of accumulators after the transfers of energy.

The absolute or relative error in the answer should not exceed 10^- 6.

Sample test(s)

Input

3 50
4 2 1

Output

2.000000000

Input

2 90
1 11

Output

1.909090909

Codeforces (c) Copyright 2010-2014 Mike Mirzayanov

二分求最大，我的做法是先把数组从大到小排列，在二分的时候判断，如果当前的能量已经大于最大值，那我们可以从中拿出多余的那部分，积累起来；否则，分配积累的能量给它，我们假设它需要t能量，那么真正需要的能量就是t*100/(100-k)；，假如处理到某个i时，积累的能量不够使它达到我们要的最大值，那么就说明这么最大值太大了，要小一点，如果处理完所有的，说明最大值还可以再大，如此下去就能找到答案了，但是对于所有数都相同的情况，一开始我没注意,WA了，之后加了个判断就过了

#include<map>    
#include<set>    
#include<list>    
#include<stack>    
#include<queue>    
#include<vector>    
#include<cmath>    
#include<cstdio>    
#include<cstring>    
#include<iostream>    
#include<algorithm>    
    
using namespace std; 

double ener[10010];

int cmp(double a, double b)
{
	return a > b;
}

int main()
{
	int n, k;
	while(~scanf("%d%d", &n ,&k))
	{
		bool flag = false;
		for(int i = 0; i < n; i++)
		{
			scanf("%lf", &ener[i]);
			if(i >= 1)
			{
				if(ener[i] != ener[i - 1])
				{
					flag = true;
				}
			}
		}
		if(n == 1 || !flag )
		{
			printf("%.9f\n", ener[0]);
			continue;
		}
		sort(ener, ener + n,cmp);
		double l = ener[n - 1];
		double r = ener[0];
		double mid;
		double ans;
		while( r - l > 1e-9 )
		{
			bool flag = true;
			double cur_ener = 0;
			mid = (l + r) / 2;
			for(int i = 0; i < n; i++)
			{
				if(ener[i] - mid >= 1e-9)
					cur_ener += (ener[i] - mid);
				else
				{
					double dis = 1.0 * (mid - ener[i]) * 100 / (100 - k);
					if(dis - cur_ener > 1e-9)
					{
						flag = false;
						break;
					}
					else
						cur_ener -= dis;
				}
			}
			if(flag)
			{
				ans = mid;
				l = mid;
			}
			else
				r = mid;
		}
		printf("%.9f\n", ans);
	}
	return 0;
}