2014 牡丹江预选赛 i题

I - Information Entropy

Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu

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Description

Information Theory is one of the most popular courses in Marjar University. In this course, there is an important chapter about information entropy.

Entropy is the average amount of information contained in each message received. Here, a message stands for an event, or a sample or a character drawn from a distribution or a data stream. Entropy thus characterizes our uncertainty about our source of information. The source is also characterized by the probability distribution of the samples drawn from it. The idea here is that the less likely an event is, the more information it provides when it occurs.

Generally, "entropy" stands for "disorder" or uncertainty. The entropy we talk about here was introduced by Claude E. Shannon in his 1948 paper "A Mathematical Theory of Communication". We also call it Shannon entropy or information entropy to distinguish from other occurrences of the term, which appears in various parts of physics in different forms.

Named after Boltzmann‘s H-theorem, Shannon defined the entropy Η (Greek letter Η, η) of a discrete random variable X with possible values {x₁, x₂, ..., x_n} and probability mass function P(X) as:

H(X)=E(−ln(P(x)))

Here E is the expected value operator. When taken from a finite sample, the entropy can explicitly be written as

H(X)=−∑i=1nP(xi)log b(P(xi))

Where b is the base of the logarithm used. Common values of b are 2, Euler‘s number e, and 10. The unit of entropy is bit for b = 2, nat for b = e, and dit (or digit) for b = 10 respectively.

In the case of P(x_i) = 0 for some i, the value of the corresponding summand 0 log_b(0) is taken to be a well-known limit:

0log b(0)=limp→0+plog b(p)

Your task is to calculate the entropy of a finite sample with N values.

Input

There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case:

The first line contains an integer N (1 <= N <= 100) and a string S. The string S is one of "bit", "nat" or "dit", indicating the unit of entropy.

In the next line, there are N non-negative integers P₁, P₂, .., P_N. P_i means the probability of the i-th value in percentage and the sum of P_iwill be 100.

Output

For each test case, output the entropy in the corresponding unit.

Any solution with a relative or absolute error of at most 10^-8 will be accepted.

Sample Input

3
3 bit
25 25 50
7 nat
1 2 4 8 16 32 37
10 dit
10 10 10 10 10 10 10 10 10 10

Sample Output

1.500000000000
1.480810832465
1.000000000000

按照题目中所给的第二个公式求出结果，当字符为“bit”时log的底数为2当字符为“nat”时底数为e字符为“dit”时底数为10注意所给数据中出现0时要把0 排除注：求log?X，log10X，lnx  直接调用math头文件中的log2（），log10（），log（）即可

#include <iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#define maxn 110
using namespace std;
int main()
{
    int t,n;
    double a[maxn];
    double sum;
    char s[4];
    scanf("%d",&t);
    while(t--)
    {
        sum=0;
        scanf("%d %s",&n,s);
        for(int i=1;i<=n;++i)
            scanf("%lf",&a[i]);
        if(strcmp(s,"bit")==0)
        {
            for(int i=1;i<=n;++i)
            {
                if(a[i]!=0)
                sum-=(a[i]*0.01*((log10(a[i]*0.01)/log10(2))));
			}
        }
        else if(strcmp(s,"nat")==0)
        {
            for(int i=1;i<=n;++i)
            {
            	if(a[i]!=0)
                sum-=(a[i]*0.01*log(a[i]*0.01));
            }
        }
        else if(strcmp(s,"dit")==0)
        {
            for(int i=1;i<=n;++i)
            {
            	if(a[i]!=0)
                sum-=(a[i]*0.01*log10(a[i]*0.01));
            }
        }
        printf("%.12lf\n",sum);
    }
    return 0;
}