程序代码：

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
//把字母统一对应26个数字
int charge(char c)
{
    int n;
    if (c>=65&&c<=90)
    {
    c+=32;
    }
    if (c>=97 && c<=122)
    {
    n=c-97;
    return n;
    }
    else return -1;
}
void main()
{
    int count[26][26]={0};
    char zifu1,zifu2;
    int i,n,m,j;
    int sum=0;
    FILE *fp;
    if ((fp = fopen("file.txt","rb")) == NULL)
    {
    printf("can't open file!\n");
    exit(0);
    }
    while(!feof(fp))
    {
    zifu1 = fgetc(fp);
    n =charge(zifu1);
    if (n!=-1)
    {
        zifu2 = fgetc(fp);
        m = charge(zifu2);
        if (m!=-1)
        {
            count[n][m]++;
            fseek(fp, -1, 1);
        }
    }
    }
    fclose(fp);

    for (i =0; i<26; i++)
    {
        for ( j=0; j<26; j++)
        {
            sum+=count[i][j];
        }
    }
    printf("the number of all the code is %d\n", sum);
    float    q = (float)sum;
    for (i =0;i<26;i++)
        for (j=0;j<26;j++)
        {
            if (j%3 == 0)
                printf("\n");
            printf("%c%c,%4d   %6.5f%%    ", i+97, j+97, count[i][j], count[i][j]*100/q);
        }
        float    sum1=0;
        printf("\n");
        for(i=0;i<26;i++)
            for(j=0;j<26;j++)
           
                if (count[i][j])
                    sum1 = sum1+(float)(
                        (count[i][j]/q) * log10(1/(double)(count[i][j]/q))/log10( (double)(2) )
                    );
                printf("\n信息熵为： H(X)=%f\n", sum1/2);
                float H0=sum1/2;
                float H2=3.32;
                printf("信源的剩余度为： R=%f",1-H2/H0);
}

程序代码：

#include <stdio.h>
#include <math.h>
#include <stdlib.h>
int charge(char c)
{
    int n;
    if (c>=65 && c<=90)
    c+=32;
    if (c>=97 && c<=122)
    {
        n=c-97;
        return n;
    }
    else
        return -1;
}
int main()
{
    int count[26] = {0};
    char ch;
    int sum=0;
    int i;
    int j;
    float fSum=0;
    float fH=0;
    FILE *fp;
    if ( (fp = fopen("file.txt", "rb")) == NULL)
    {
        printf("Can't open file to read!\n");
        exit(0);
    }
    while(!feof(fp))
    {
        ch = fgetc(fp);
        if (-1 != charge(ch))
            count[charge(ch)]++;
    }
    for (i = 0; i<26 ; i++)
        sum += count[i];
    printf("the number of all the code is %d\n", sum);
    fSum=(float)sum;

    for (i = 0; i<26; i++)
    {
        if (i%3==0)
            printf("\n");
        printf("%c, %5d    %6.5f%%  ", i+97, count[i], count[i]/fSum);
        fH += (float)((count[i]/fSum)* log10((double)1/(count[i]/fSum)) / log10((double)(2)));
    }
    printf("\n\n一维信源熵为：H(X) = %f\n", fH);
    printf("\n一维信源剩余度：R = %f\n", 1-4.03/fH);
    return 0;
} //香浓编码
#include<stdio.h>
#include<iostream.h>
#include<math.h>
#define MAX 7
double P[MAX]={0.01, 0.10, 0.15, 0.17,0.18, 0.19, 0.20},Pax[MAX],machang[MAX];
int main()
{
    double temp;
//递减排序(冒泡排序)
for(int a=1;a<MAX;a++)
{
    for(int i=0;i<MAX-a;i++)
    if(P[i]<P[i+1])
    {
        temp=P[i];
        P[i]=P[i+1];
        P[i+1]=temp;
    }
}
//输出消息概率
for(int i=0;i<MAX;i++)
    cout<<P[i]<<" ";
    cout<<endl;
    //累加概率
for(i=0;i<MAX;i++)
{
    Pax[0]=0.0;
    Pax[i+1] = Pax[i] + P[i];
}
    cout << "概率累加和为：" << endl;
    //输出对应的累加概率
for ( i = 0; i<MAX; i++)
{
    cout << Pax[i] << " ";
}
    cout << endl;
//求每个消息的码长
for (i=0; i<MAX; i++)
{
    double m = log(1/P[i])/log(2);
    if (m -int(m) == 0)
        machang[i] = log(1/P[i])/log(2);
    else
        machang[i] = int(m) +1;
    cout << P[i] << "的码长为：" << machang[i] << endl;

}
//小数点转化为二进制
for (i=0; i<MAX; i++)
{
    cout<<"消息概率" << P[i] <<"的二进制码组为: ";
    for (int j=0;j<machang[i];j++)
    {
        int n=int(Pax[i]*2);
        cout << n;
        //判断Pax[i]*2是否>1
        if ( (Pax[i]*2-1)>=0)
            Pax[i] = Pax[i]*2 -1;//大于1就减1
    else
        Pax[i] = Pax[i]*2;//否则就称2
    }
   
    cout << endl;
}
cout << endl;
return 0;
} Huffma编码
#include <stdio.h>
#include <stdlib.h>
#define n 5
#define m 2*n-1
//位域结构体
typedef struct Bit
{//只能放0和1
unsigned int a:1;//放0或1
}Mybit;
typedef struct tree
{
    float weight;
    int lChild;
    int rChild;
    int parent;
    Mybit bit;
    struct tree()//初始化树
    {
        weight = 0.0;
        lChild = -1;
        rChild = -1;
        parent = -1;
        bit.a = 1;
    }
}Huffman;

typedef int elemtype;//定义元素类型

typedef struct stack
{
    elemtype data[n-1];//
    int top;
}Stack;

void TowSmallNode(Huffman tree[],int k,int *lsmall, int *rsmall)
{
        int i = 0;
        int temp;
        *lsmall = *rsmall = 0;
       
    for ( i = 0; i<k; i++)//找到最小权的下标
    {
   
        if(tree[i].parent == -1)
        {
            if(tree[*lsmall].weight >= tree[i].weight || tree[*lsmall].parent != -1)
            {
               
               *lsmall = i;
            }   
        }
    }
   
    for (i=0; i<k; i++)//找到第二小的权的下标
    {
        if(tree[i].parent == -1 && i != *lsmall)
        {
           
            if(tree[*rsmall].weight >= tree[i].weight)
            {
               *rsmall = i;
            }
            else if ( (*lsmall == 0 && *rsmall ==0) || tree[*rsmall].parent != -1)
            {
                *rsmall = i;
            }
           
        }
    }

    //判断哪个做右边哪个做左边
    if (tree[*rsmall].lChild !=-1 && tree[*lsmall].lChild == -1 )//有孩子那边做为左边
    {
        temp = *rsmall;
        *rsmall = *lsmall;
        *lsmall = temp;
    }
   
}

void CreateTree(Huffman tree[])
{

    int i = 0;
    int lsmall = 0;
    int rsmall = 0;
    float w;
    printf("请输入消息的概率：\n");
    for (i = 0; i<n; i++)//输入叶子
    {
        scanf("%f", &w);
        tree[i].weight = w;
    }

    for (i=n; i<m; i++)//剩下的节点
    {
        TowSmallNode(tree,i, &lsmall, &rsmall);//找出最小权的两个标
        tree[i].lChild = lsmall;//左孩子
        tree[i].rChild = rsmall;//右孩子
        tree[i].weight = tree[lsmall].weight + tree[rsmall].weight;//两个最小权值相加
        tree[lsmall].parent = i;
        tree[rsmall].parent = i;
        tree[lsmall].bit.a = 0;//是左边的就放二进制的0
    }
}

//置空栈
void setNull(Stack *s)
{
    s ->top = -1;
return ;
}
//进栈
void push(Stack *s, elemtype x)
{
    s ->top++;
    s ->data[s->top] = x;
    return;
}
//出栈
elemtype pop(Stack *s)
{
    s ->top--;
    return (s ->data[s->top+1]);
}
//
int main()
{
    Huffman tree[m];
    CreateTree(tree);
    int i =0;
    int k;
    Stack *s = (Stack*)malloc(sizeof(Stack));//分配栈空间
    for (i =0 ; i<n; i++)
    {
        k = i;
        setNull(s);//置栈为空
        while(tree[k].parent !=-1)
        {
           
            push(s, tree[k].bit.a);//进栈
            k = tree[k].parent ;
        }
        //出栈
        printf("消息的概率为%.2f的二进制码为:",tree[i].weight);
        while(s->top != -1)
        {
            printf("%d", pop(s));//输出二进制码
        }
            printf("\n");
    }
    free(s);//释放空间
return 0;
}