cplus第六版第十一章第十一题稀有可行版本_c语言第六版11章答案-优快云博客

本文链接：https://blog.youkuaiyun.com/User_manxxx/article/details/120836125

本文介绍了一个C语言程序，它包含pro函数和四个子函数p1到p4，主要涉及字符串的读取、交换和排序。通过字符处理和输入验证，实现了一种简单的字符操作和字符串管理。

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#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<ctype.h>

#define SIZE 11
#define LIM 101

void pro(char (*a)[LIM]);
void p1(char (*a)[LIM]);
void p2(char (*a)[LIM]);
void p3(char (*a)[LIM]);
void p4(char (*a)[LIM]);

int main()
{
    char a[SIZE][LIM];
    for (int i = 0; i < 10; i++)
    fgets(a[i], 101, stdin);

   pro(a);

   return 0;
}

void pro(char (*a)[LIM])
{
   char ch;

   printf("1 2 3 4 5\n");
       while ((ch = getchar()) != '5')
   {
       if (ch == '\n')
       continue;
       switch(ch)
       {
           case '1':
               p1(a);
               break;
           case '2':
               p2(a);
               break;
           case '3':
               p3(a);
               break;
           case '4':
               p4(a);
               break;
           default:
               printf("1 2 3 4 5\n");
       }
       printf("1 2 3 4 5\n");
   }
}

void p1(char (*a)[LIM])
{
for (int i = 0; i < 10; i++)
fputs(a[i], stdout);
}

void p2(char (*a)[LIM])
{
   int top, seek;

   for (top = 0; top < 9; top++)
   for (seek = top + 1; seek < 10; seek++)
   if (strcmp(a[top], a[seek]) > 0)
   {
       strcpy(a[10], a[top]);
       strcpy(a[top], a[seek]);
       strcpy(a[seek], a[10]);
           }
   p1(a);
}

void p3(char (*a)[LIM])
{
   char * temp;
   for (int i = 0; i < 9; i++)
   for (int m = i + 1; m < 10; m++)
   if (strlen(a[i]) > strlen(a[m]))
   {
       strcpy(a[10], a[i]);
       strcpy(a[i], a[m]);
       strcpy(a[m], a[10]);
           }
   p1(a);
}

void p4(char (*a)[LIM])
{
   char * temp;
   int q, w, e, r;

   q = w = e = r = 0;
   for (int i = 0; i < 9; i++)
   for (int l = i + 1; l < 10; l++)
   {
       while (a[i][e] == ' ')
       e++;
       while (a[i][e] != ' ' && a[i][e] != '\0')
       {
           q++;
           e++;
           }
       while (a[l][r] == ' ')
       r++;
       while (a[l][r] != ' ' && a[l][r] != '\0')
       {
           w++;
           r++;
           }
   if (q > w)
   {
       strcpy(a[10], a[i]);
       strcpy(a[i], a[l]);
       strcpy(a[l], a[10]);
           }
           q = w = e = r = 0;
}
p1(a);
}