怎样用<em>excel</em>逐板计算法求精馏塔理论<em>塔板数</em

这个到底是什么问题........

我这里倒是有HUFFMAN法的文件压缩过程:
#include<stdio.h>
#include<stdlib.h>
#include<malloc.h>
#include<string.h>
#include<math.h>

struct htnode
{
char ch;
int weight,parent,lchild,rchild;
};
struct hc
{
char *code;
char word;
int size;
};

void select(struct htnode * ht,int n,int& s1, int& s2)
{
int small1=9999, small2=9999,i;
for(i=1;i<=n-1;i++)
{
if(ht[i].weight<small1&&ht[i].parent==0)
{
small1=ht[i].weight;
s1=i;
}
++i;
if(ht[i].weight<small2&&ht[i].parent==0)
{
small2=ht[i].weight;
s2=i;
}
}
}

void reduce()
{
FILE *fpin,*fpout;
int cut[32768];
int i,k;
char flag;
double j=0;
for(i=0;i<=32768;i++)
cut[i]=0;
fpin=fopen("code.txt","r");
fpout=fopen("reduce.txt","wb");
i=0;
while(fread(&flag,sizeof(char),1,fpin))
{

if(flag=='1')
{
cut[i]=(cut[i])|(int)(pow(2,(14-j)));
}
j=j+1;
if(j==15)
{
j=0;
i++;
}
}
for(k=0;k<=i;k++)
{
fwrite(&cut[k],sizeof(int),1,fpout);
}
printf("\n");
fclose(fpin);
fclose(fpout);
}

void unreduce()
{
FILE *fpin;
fpin=fopen("reduce.txt","r");
double j=0;
int flag,uncut[32768],k=0,i;
while(fread(&flag,sizeof(int),1,fpin))
{

for(j=0;j<=14;j=j+1)
if(flag&(int)(pow(2,(14-j))))
uncut[k++]=1;
else
uncut[k++]=0;
}
for(i=0;i<k-15;i++)
printf("%d",uncut[i]);
fclose(fpin);
}

void translate(struct hc *hcc,int n)
{
int i,j,k;
FILE *fpin,*fpout;
char text[32767];
for(k=0;k<32767;k++)
text[k]='\0';
k=0;
fpin=fopen("code.txt","r");
fpout=fopen("translate.txt","w");
while(!e69da5e6ba907a686964616f337feof(fpin))
{
text[k++]=fgetc(fpin);
for(i=1;i<=n;i++)
if(strcmp(text,hcc[i].code)==0)
{
fprintf(fpout,"%c",hcc[i].word);
for(j=0;j<k;j++)
text[j]='\0';
k=0;
}
}
fclose(fpin);
fclose(fpout);
}

void huffmancodeing(int w[32768],char word[32767],int n,char text[32767])
{
FILE *fp,*fpp;
fp=fopen("huffmancode.txt","w");
struct hc *hcc;
char *cd;
int i,c,f,start,k=0,s1,s2;
struct htnode *ht;
int m=2*n-1;
if(n<=1) return;
ht=(struct htnode *)malloc(sizeof(struct htnode)*(m+1));
for(i=1;i<=n;i++)
{
ht[i].ch=word[i-1];
ht[i].parent=0;
ht[i].lchild=0;
ht[i].rchild=0;
ht[i].weight=w[ht[i].ch];
}
for(i=1;i<=n;i++)
printf("%c,%d",ht[i].ch,ht[i].weight);
printf("\n");
for(i=n+1;i<=m;i++)
{
ht[i].ch=word[i-n];
ht[i].parent=0;
ht[i].lchild=0;
ht[i].rchild=0;
ht[i].weight=0;
}

for(i=n+1;i<=m;i++)
{
select(ht,i-1,s1,s2);
ht[s1].parent=i;
ht[s2].parent=i;
ht[i].lchild=s1;
ht[i].rchild=s2;
ht[i].weight=ht[s1].weight+ht[s2].weight;
}
hcc=(struct hc *)malloc((n+1)*sizeof(struct hc));
cd=(char *)malloc(n*sizeof(char));
cd[n-1]='\0';
for(i=1;i<=n;i++)
{
start=n-1;
for(c=i,f=ht[i].parent;f!=0;c=f,f=ht[f].parent)
if(ht[f].lchild==c)
cd[--start]='0';
else cd[--start]='1';
hcc[i].code=(char *)malloc((n-start)*sizeof(char));
strcpy(hcc[i].code,&cd[start]);
hcc[i].word=ht[i].ch;
hcc[i].size=n-start-1;
}
free(cd);
for(i=1;i<=n;i++)
fprintf(fp,"%c:%s",hcc[i].word,hcc[i].code);
fclose(fp);
fpp=fopen("code.txt","wb");
while(text[k]!='\0')
{
for(i=1;i<=n;i++)
if(text[k]==hcc[i].word)
{
fwrite(hcc[i].code,sizeof(char)*hcc[i].size,1,fpp);

}
k++;
}
fclose(fpp);
for(i=1;i<=n;i++)
printf("%c:%s\n",hcc[i].word,hcc[i].code);
translate(hcc,n);
reduce();
unreduce();
}

void main()
{
FILE *fpin;
char file[20];
printf("请输入要压缩的文件名\n");
gets(file);
fpin=fopen(file,"r");
int w[32768],i,j,k,m=0,flag,n=0;
for(i=0;i<=32767;i++) w[i]=0;
i=k=0;
char ch,word[32768],text[32768],code[32768];
while(!feof(fpin))
code[m++]=fgetc(fpin);
fclose(fpin);
code[--m]='\0';
m=0;
while(code[m])
{
ch=code[m];
m++;
text[k++]=ch;
flag=0;
w[ch]++;
for(j=0;j<=32767;j++)
if(word[j]==ch)
flag=1;
if(flag==0)
{
word[i]=ch;
n++;
i++;
}

}
word[i]='\0';
text[k]='\0';
huffmancodeing(w,word,n,text);
}

如果你指的是一股进料的普通精馏塔，那么精馏段塔板数是指从冷凝器以下那块顶板开始往下数到进料板上的那块为止。而提馏段指的则是进料板之下的塔板数
我有。。。。。。。。。。

给你I个现成的

本文标签：图解法 塔板数 excel(1)excel理论塔板数图解法(1)