2014ACM/ICPC亚洲区域赛牡丹江站现场赛-I （ ZOJ 3827 ） Information Entropy

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 functionP(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_i will 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

Source: The 2014 ACM-ICPC Asia Mudanjiang Regional Contest

题目链接：Information Entropy

解题思路：又是一道签到题！！！当时想多了，居然纠结在积分上，果真是想多了，根本不需要什么技术，就是按照那个数学公式算。又学到了个小方法，表示对数可以用log（m, n）= log(n) / log(m). 还有就是e = exp（1）。还有就是注意当p = 0时，函数值为0。

AC代码：

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
#define INF 0x7fffffff

#define e exp(1.0)

int main()
{
    #ifdef sxk
        freopen("in.txt","r",stdin);
    #endif
    int T, n, p;
    string s;
    scanf("%d",&T);
    while(T--)
    {
        double ans = 0;
        scanf("%d",&n);
        cin>>s;
        if(s == "bit"){
            for(int i=0; i<n; i++){
                scanf("%d", &p);
                if(p)
                    ans += -(p/100.0*log(p/100.0)/log(2));
            }
        }
        else if(s == "nat"){
            for(int i=0; i<n; i++){
                scanf("%d", &p);
                if(p)
                    ans += -(p/100.0*log(p/100.0)/log(e));
            }
        }
        else{
            for(int i=0; i<n; i++){
                scanf("%d", &p);
                if(p)
                    ans += -(p/100.0*log(p/100.0)/log(10));
            }
        }
        printf("%.14lf\n",ans);
    }
    return 0;
}