DHT.rar_Chord-Source-Code_DHT

下面是我的快速ＤＣＴ变换的代码，调试通过了，肯能往这敲时会有点手误，你看看吧，兄弟我的毕业设

计就是做基于ＤＣＴ的图象压缩算法，你看看吧，分块的都考虑在主程序里了，后面那段是反变换的ＩＤＣ

Ｔ．希望对你有用

/* This program is for carrying out 2-D fast consine Transform of blocks subdividing a given

image .Block size is user specified .*/

#define PI 3.14159

#include <stdio.h>

#include <math.h>

#include <alloc.h>

void main()

{

int m,N,j,i,k1,k2,NB,NS,NT,k;

float *x,*C,**w;

unsigned int *L;

double nsq;

FILE *fptri,*fptro;

char file_name[14];

unsigned char *buffi;

float *buffo;

printf("No such file exists.[BS)]");

exit(1);

}

nsq=(double)filelength(fileno(fptri)); /*得到文件长度， 用 open()函数打开的文件句柄

返回文件长度，操作失败时返回 -1*/

/*assume image is square.*/

N=(int)sqrt(nsq);

m=(int)(log10((double)N)/log10((double)2.0));

k=1;

for(i=0;i<m;i++)

printf("Image must have dimensions which are multiples of 2.[BS)]");

exit(1);

printf("Enter name of output file -->");

scanf("%s",file_mane);

fptro=fopen(file_name,"wb");

printf("Enter block size(8X8,16X16,etc.)-->");

L=(unsigned int *)malloc(NB*sizeof(float));

C=(float *)malloc((NB-1)*sizeof(float));

w=(float* *)malloc(NB*sizeof(float*));

bit_reversal(L,m,NB);

WTS(C,m,NB);

NS=(N>>m);

for(i=0;i<NS;i++)

{

fread(buffi,NT,1,fptri);/*从文件的当前位置开始中读取 size*count 个字节的数据*/

/*输入：buf 存放读入数据的指针 ，size 每个数据单位的字节数，

count 读入的数据单位个数,

输出：buf 读出的数据

读入的字节数据，如果小于要求的字节数据，说明读到了文件结尾或出错*/

w[k1][k2]=buffi[k1*N+k2+j*NB];

for(k1=0;k1<NB;k1++)

{

for(k2=0;k2<NB;k2++)

x[k2]=w[k1][k2];

FCT(x,L,C,m,NB);

for(k2=0;k2<NB;k2++)

w[k1][k2]=x[k2];

}

for(k2=0;k2<NB;k2++)

FCT(x,L,C,m,NB);

for(k1=0;k1<NB;k1++)

}

for(k1=0;k1<NB;k1++)

for(k2=0;k2<NB;k2++)

buffo[k1*N+k2+j*NB]= w[k1][k2];

fcloseall();

int NK1,NK2,i,j,k,kk,ip,incr,L1,k1,k2,iter;

float T;

/* Rearranging the order of the input sequence.*/

Nk1=N>>1;

T=x[2];

x[2]=x[1];

x[1]=T;

k1=2;

k2=4;

for(i=1;i<(NK1-1);i++)

{

k1++;

k2+=2;

}

NK2=N-1;

for(i=0;i<(NK1>>1);i++)

{

T=x[NK2-i];

x[nk2-i]=x[NK1+i];

x[NK1+i]=T;