Algorithm

B. Energy exchange
time limit per test
2 seconds
memory limit per test
256 megabytes
Problem Link- https://codeforces.com/problemset/problem/68/B
input
standard input
output
standard output

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 ai units of energy. Energy can be transferred from one accumulator to the other. Every time x 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 by x units and in other increased by  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 and k (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, ... , an — amounts of energy in the first, second, .., n-th accumulator respectively (0 ≤ ai ≤ 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.

Examples
input
Copy
3 50
4 2 1
output
Copy
2.000000000
input
Copy
2 90
1 11
output
Copy
1.909090909



 

Code Examples

#1 Code Example with C++ Programming

Code - C++ Programming

#include <cstdio>
#include <vector>

bool check(std::vector<double> a, double loss, double avg){

    double rem(0);
    for(long p = 0; p < a.size(); p++){rem += (a[p] - avg) * (1 - loss * (a[p] > avg));}
    return (rem >= 0);
}


int main(){

    const double EPS = 1e-8;
    long n, k; scanf("%ld %ld", &n, &k);
    std::vector<double> a(n, 0);

    double left(0), right(0);

    for(long p = 0; p < n; p++){
        scanf("%lf", &a[p]);
        right = (right > a[p]) ? right : a[p];
    }


    double res(0.0);
    while(right - left > EPS){
        double mid = (left + right) / 2;
        if(check(a, k / 100.0, mid)){res = mid; left = mid;}
        else{right = mid;}
    }

    printf("%.7lf\n", res);

    return 0;
}
Copy The Code & Try With Live Editor

Input

x
–
+
cmd
3 50
4 2 1

Output

x
–
+
cmd
2.000000000
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Demonstration

Codeforces Solution-B. Energy exchange-Solution in C, C++, Java, Python

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