作业3：读入一个24bitRGB文件（以down.rgb为例，其分辨率为256*256），输出该数据文件中R、G、B三个分量（各8bit表示）的概率分布示意图（类似下图）和熵。

思路：开辟3个width*height的unsigned char型数组；打开要读出的RGB文件（以“rb”方式打开），打开3个要输出的数据统计文件（以“w”方式打开，可命名为R_sat.txt等）；将RGB数据从RGB文件中读出，并分别保存到3个数组中，期间计算数据的概率分布和熵，并将这些数据写入3个数据统计txt文件中。

#include “pch.h”
#include<stdio.h>
#include<math.h>
#include
int main()
{
FILE fp;
fp = fopen(“down.rgb”, “rb”);
const int width = 256;
const int height = 256;
unsigned char cache[widthheight * 3];
fread(cache, sizeof(unsigned char), widthheight * 3, fp);
unsigned char red[widthheight] = { 0 };
unsigned char green[widthheight] = { 0 };
unsigned char blue[widthheight] = { 0 };
（这一部分是定义数组）

int ri = 0;
int gi = 0;
int bi = 0;
for (int i = 0; i < 3 * width*height; i++)
{
	if (i % 3 == 0)
	{
		blue[bi] = cache[i];
		bi++;
	}
	else if (i % 3 == 1)
	{
		green[gi] = cache[i];
		gi++;
	}
	else if (i % 3 == 2)
	{
		red[ri] = cache[i];
		ri++;
	}
}

（这一部分是计算rgb的数目）

int rtime[256] = { 0 };
int gtime[256] = { 0 };
int btime[256] = { 0 };
for (int i = 0; i < 256; i++)
{
	for (int j = 0; j < width*height; j++)
	{
		if (i == red[j])
			rtime[i]++;
		if (i == green[j])
			gtime[i]++;
		if (i == blue[j])
			btime[i]++;

	}
}

double rfreq[256] = { 0 };
double gfreq[256] = { 0 };
double bfreq[256] = { 0 };
for (int i = 0; i < 256; i++) 
{
	rfreq[i] = double(rtime[i]) / (width*height);
	gfreq[i] = double(gtime[i]) / (width*height);
	bfreq[i] = double(btime[i]) / (width*height);
}

（这一部分计算频率）

FILE *R;
FILE *G;
FILE *B;
R = fopen("red.txt", "w");
G = fopen("green.txt", "w");
B = fopen("blue.txt", "w");
fprintf(R, "symbol\tfreq\n");
for (int i = 0; i < 256; i++)
{
	fprintf(R, "%d\t%f\n", i, rfreq[i]);
}
fprintf(G, "symbol\tfreq\n");
for (int i = 0; i < 256; i++)
{
	fprintf(G, "%d\t%f\n", i, gfreq[i]);
}
fprintf(B, "symbol\tfreq\n");
for (int i = 0; i < 256; i++)
{
	fprintf(B, "%d\t%f\n", i, bfreq[i]);
}

（这一部分写入数据）

double hr = 0;
double hg = 0;
double hb = 0;
for (int i = 0; i < 256; i++) 
{
	if (rfreq[i] != 0)
		hr = hr - rfreq[i] * log(rfreq[i]) / log(2);
	if (gfreq[i] != 0)
		hg = hg - gfreq[i] * log(gfreq[i]) / log(2);
	if (bfreq[i] != 0)
		hb = hb - bfreq[i] * log(bfreq[i]) / log(2);
}
printf("H(红基色)=%f\n", hr);
printf("H(绿基色)=%f\n", hg);
printf("H(蓝基色)=%f\n", hb);
fclose(fp);
fclose(R);
fclose(G);
fclose(B);
return 0;
}

（这一部分计算熵）
结果如下：