作业3：读入一个24bitRGB文件，输出该数据文件中R、G、B三个分量的概率分布示意图和熵。

c++代码

#include<iostream>
#include<math.h>
#include<stdio.h>
using namespace std;

int w=256;
int h=256;
unsigned char r[65535];
unsigned char g[65535];
unsigned char b[65535];
double p_r[256]={0};
double p_g[256]={0};
double p_b[256]={0};
double h_r,h_g,h_b;

int main()
{
	FILE* fp = fopen("D:\\x.rgb", "rb+");
	FILE* Red=fopen("Red.txt", "w");
	FILE* Green=fopen( "Green.txt", "w");
	FILE* Blue=fopen( "Blue.txt", "w");

	unsigned char a[196608];
   fread(a, sizeof(unsigned char), 196608, fp);
   for (int i=0,j=0;i<196608;i=i+3,j++)
   {
      b[j] = a[i];
      g[j] = a[i+1];
      r[j] = a[i+2];
   }

   for(int i=0;i<256;i++)
   {
      for(int j=0;j<65536;j++)
      { 
		  if(int(r[j])==i) {p_r[i]++;} 
	  }  
   }
   for(int i=0;i<256;i++)
   {
      for(int j=0;j<65536;j++)
      {
          if(int(g[j])==i) {p_g[i]++;}
      }
   }
   for(int i=0;i<256;i++)
   {
      for(int j=0;j<65536;j++)
      {
          if(int(b[j])==i) {p_b[i]++;}
      }
   }

   for(int i=0;i<256;i++)
   {
      p_r[i]=p_r[i]/double(w*h);
      p_g[i]=p_g[i]/double(w*h);
      p_b[i]=p_b[i]/double(w*h);
   }

   for(int i=0;i<256;i++)
	{
		if(p_r[i]!=0)
		{
			h_r+=p_r[i]*(log(1/p_r[i])/log(2));
		}
	}
	cout<<"r的熵为"<<h_r<<endl;
	for(int i=0;i<256;i++)
	{
		if(p_g[i]!=0)
		{
			h_g+=p_g[i]*(log(1/p_g[i])/log(2));
		}
	}
	cout<<"g的熵为"<<h_g<<endl;
	for(int i=0;i<256;i++)
	{
		if(p_b[i]!=0)
		{
			h_b+=p_b[i]*(log(1/p_b[i])/log(2));
		}
	}
	cout<<"b的熵为"<<h_b<<endl;

	fprintf(Red,"数值\t概率\n");
	for(int i=0;i<256;i++)
	{
		fprintf(Red,"%d\t%f\n",i,p_r[i]);
	}
	

	fprintf(Green,"数值\t概率\n");
	for(int i = 0; i < 256; i++)
	{
		fprintf(Green,"%d\t%f\n",i,p_g[i]);
	}

	fprintf(Blue,"数值\t概率\n");
	for (int i = 0; i < 256; i++)
	{
		fprintf(Blue,"%d\t%f\n",i,p_b[i]);
	}

	fclose(Red);
	fclose(Green);
	fclose(Blue);
    system("pause");
}

相关结果