TableOfContents

实验基本要求

集成开发环境指导

实验内容

1. 05瓯信算

先打开实验报告模板,查看需要完成哪些题目,然后打开判题系统地址,在判题系统上做对相应题目,最后将正确的程序复制到实验报告中,发到邮箱。

判题系统地址:[http://wzuacm.3322.org/]

完成如下实验报告:

在实验报告文件名后添加你的学号姓名,比如“面向对象实验1-0530215288-张三.doc”,然后email到: czk.cpp@gmail.com

===实验3===

  1. 拷贝5.4节中的程序:

       1    #define ALLOCSIZE 10000 /* size of available space */
       2 
       3    static char allocbuf[ALLOCSIZE]; /* storage for alloc */
       4    static char *allocp = allocbuf;  /* next free position */
       5 
       6    char *alloc(int n)    /* return pointer to n characters */
       7    {
       8        if (allocbuf + ALLOCSIZE - allocp >= n) {  /* it fits */
       9            allocp += n;
      10            return allocp - n; /* old p */
      11        } else      /* not enough room */
      12            return 0;
      13    }
      14 
      15    void afree(char *p)  /* free storage pointed to by p */
      16    {
      17        if (p >= allocbuf && p < allocbuf + ALLOCSIZE)
      18            allocp = p;
      19    }
    
  2. 1.9节getline

       1   /* getline:  read a line into s, return length  */
       2    int getline(char s[],int lim)
       3    {
       4        int c, i;
       5 
       6        for (i=0; i < lim-1 && (c=getchar())!=EOF && c!='\n'; ++i)
       7            s[i] = c;
       8        if (c == '\n') {
       9            s[i] = c;
      10            ++i;
      11        }
      12        s[i] = '\0';
      13        return i;
      14    }
    
  3. 5.6节

       #include <stdio.h>
       #include <string.h>
    
       #define MAXLINES 5000     /* max #lines to be sorted */
    
       char *lineptr[MAXLINES];  /* pointers to text lines */
    
       int readlines(char *lineptr[], int nlines);
       void writelines(char *lineptr[], int nlines);
    
       void qsort(char *lineptr[], int left, int right);
    
       /* sort input lines */
       main()
       {
           int nlines;     /* number of input lines read */
    
           if ((nlines = readlines(lineptr, MAXLINES)) >= 0) {
               qsort(lineptr, 0, nlines-1);
               writelines(lineptr, nlines);
               return 0;
           } else {
               printf("error: input too big to sort\n");
               return 1;
           }
       }
    
       #define MAXLEN 1000  /* max length of any input line */
       int getline(char *, int);
       char *alloc(int);
    
       /* readlines:  read input lines */
       int readlines(char *lineptr[], int maxlines)
       {
           int len, nlines;
           char *p, line[MAXLEN];
    
           nlines = 0;
           while ((len = getline(line, MAXLEN)) > 0)
               if (nlines >= maxlines || (p = alloc(len)) == NULL)
                   return -1;
               else {
                   line[len-1] = '\0';  /* delete newline */
                   strcpy(p, line);
                   lineptr[nlines++] = p;
               }
           return nlines;
       }
    
       /* writelines:  write output lines */
       void writelines(char *lineptr[], int nlines)
       {
           int i;
    
           for (i = 0; i < nlines; i++)
               printf("%s\n", lineptr[i]);
       }

    及qsort和swap

    void qsort(char *v[], int left, int right)
    {
        int i, last;
        void swap(char *v[], int i, int j);
    
        if (left >= right)  /* do nothing if array contains */
            return;         /* fewer than two elements */
        swap(v, left, (left + right)/2);
        last = left;
        for (i = left+1; i <= right; i++)
            if (strcmp(v[i], v[left]) < 0)
                swap(v, ++last, i);
        swap(v, left, last);
        qsort(v, left, last-1);
        qsort(v, last+1, right);
    }
    /* swap:  interchange v[i] and v[j] */
    void swap(char *v[], int i, int j)
    {
        char *temp;
    
        temp = v[i];
        v[i] = v[j];
        v[j] = temp;
    }
  4. 所有的qsort换一个名字,比如qsort2

2. 05计本

在判题系统[http://czk.8866.org/oj/]上完成3题,并提交实验报告:

实验报告发送到: czk19790827@gmail.com

ch3n2k.com | Copyright (c) 2004-2020 czk.