四则运算（未解决）

1086阅读　0评论2012-10-21　时间看来
分类：C/C++


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  1. //3、输入一个表达式，没有括号，数字在0-9之间，输出计算结果，所有的中间结果化为整形。
  2. //例如: 输入：3+8×2/9-2
  3. //输出：2
  4. //函数原型
  5. //public int getMyRet(String str)

  7. #include<stdio.h>
  8. #include<stdlib.h>

  10. //#include <iostream>
  11. //#include <string>
  12. //#include <cstring>
  13. //using namespace std;

  15. int get_my_ret(char *, int);

  17. int
  18. main( int argc, char *argv[] )
  19. {
  20.     //输入字符串
  21.     char my_chars[] = "3+8*2/9-2";
  22.     int my_ret, my_chars_cnt;
  23.     my_chars_cnt = sizeof(my_chars)/sizeof(char);

  25.     //显示需计算的字符串
  26.     int i;
  27.     char *one_char_p = my_chars;
  28.     printf("Need to calculate :\n");
  29.     for(i=0; i<my_chars_cnt; one_char_p++, i++){
  30.         printf("%c", *one_char_p);
  31.     }
  32.     printf("\n");

  34.     //调用计算函数
  35.     my_ret = get_my_ret(my_chars, my_chars_cnt);

  37.     //显示计算结果
  38.     printf("\n my_ret is %d\n", my_ret);
  39. }

  41. int
  42. get_my_ret(char *my_chars_p, int cnt)
  43. {
  44.     int my_ret;
  45.     //获取操作符号个数
  46.     int operations_cnt = 0;
  47.     int i;
  48.     char *cpy_my_chars_p = my_chars_p;
  49.     for(i=0; i<cnt; i++){
  50.         if(*cpy_my_chars_p == '+' || \
  51.                 *cpy_my_chars_p == '-' || \
  52.                 *cpy_my_chars_p == '*' || \
  53.                 *cpy_my_chars_p == '/'){
  54.             operations_cnt++;
  55.         }
  56.         cpy_my_chars_p++;
  57.     }

  59.     //申请数字个数
  60.     int *nums_p_first = (int *) calloc(operations_cnt+1, sizeof(int));
  61.     char *operations_p_first = (char *) calloc(operations_cnt, sizeof(char));

  63.     //将字符串转化为数字存在数组里
  64.     int *nums_p = nums_p_first;
  65.     char *operations_p = operations_p_first;
  66.     cpy_my_chars_p = my_chars_p;

  68.     for(i=0; i<cnt; i++){
  69.         if(*cpy_my_chars_p >= '0' && *cpy_my_chars_p <= '9'){
  70.             *nums_p = *cpy_my_chars_p - '0';
  71.             nums_p++;
  72.         }else{
  73.             *operations_p = *cpy_my_chars_p;
  74.             operations_p++;
  75.         }
  76.         cpy_my_chars_p++;
  77.     }

  79.     //显示获取的操作数
  80.     nums_p = nums_p_first;
  81.     printf("the nums are:\n");
  82.     for(i =0; i<(operations_cnt+1); i++){
  83.         printf("%d, ", *nums_p);
  84.         nums_p++;
  85.     }
  86.     printf("\n");

  88.     //显示获取的操作符号
  89.     operations_p = operations_p_first;
  90.     printf("the oprations are:\n");
  91.     for(i =0; i<(operations_cnt); i++){
  92.         printf("%c, ", *operations_p);
  93.         operations_p++;
  94.     }
  95.     printf("\n");

  97.     //根据两个操作符号，而进行计算
  98.     char *now_opration_p = operations_p_first;
  99.     char *next_opration_p = ++operations_p_first;
  100.     for(i=0; i<operations_cnt; i++){
  101.         //bad beaf
  102.         //bad beaf
  103.         //bad beaf
  104.         //bad beaf
  105.         //bad beaf
  106.         //bad beaf
  107.         //bad beaf
  108.         //bad beaf
  109.     }

  111.     free(nums_p_first);
  112.     free(operations_p_first);
  113.     return my_ret;
  114. }


//强大的网友的答案……难以理解的递归！！！

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  1. #include <stdio.h>
  2. #include <string.h>
  3. #include <stdlib.h>

  5. //
  6. // 封装常用的判断是否为数字的判断函数，如果是返回1否则0
  7. //
  8. int IsNum(char ch)
  9. {
  10.     return (ch >= '0' && ch <= '9');
  11. }

  13. //
  14. // 把指向Calc的INDEX放于全局变量的位置，因为在递归的过程中需要递增
  15. //
  16. int INDEX = 0;


  19. int Calc(char sub_str[],int first)
  20. {
  21.     //
  22.     // 1.记录下读取到的符号
  23.     //
  24.     char item = sub_str[INDEX];

  26.     char temp[10] = {0};
  27.     char temp_index = 0;

  29.     //
  30.     // 2.如果为终结符，返回first
  31.     //
  32.     if (item == '=' || item == '\0' || item == ')')
  33.     {
  34.         printf("2.return %d\n",first);
  35.         INDEX++;
  36.         return first;
  37.     }

  39.     //
  40.     // 3.1如果为前括号，
  41.     //
  42.     int sub_first = 0;
  43.     if (item == '(')
  44.     {
  45.         INDEX++;
  46.         printf("3.1sub_first = Calc(%s,%d)\n",sub_str,0);
  47.         sub_first = Calc(sub_str,0);
  48.         return Calc(sub_str,sub_first);
  49.     }

  51.     //
  52.     // 3.2如果为数字，读取数字作为first送入
  53.     //
  54.     else if (IsNum(item))
  55.     {
  56.         while (IsNum(sub_str[INDEX]))
  57.         {
  58.             temp[temp_index++] = sub_str[INDEX++];
  59.         }
  60.         sub_first = atoi(temp);

  62.         printf("3.2Calc(%s,%d)\n",sub_str,sub_first);
  63.         return Calc(sub_str,sub_first);
  64.     }

  66.     INDEX++;

  68.     //
  69.     // 4.如果为前括号，把后面的内容作为second
  70.     //
  71.     int second = 0;
  72.     if (sub_str[INDEX] == '(')
  73.     {
  74.         INDEX++;
  75.         printf("4.second = Calc(%s,0);\n",sub_str);
  76.         second = Calc(sub_str,0);
  77.     }

  79.     //
  80.     // 5.如果为数字，把后面的数字作为second
  81.     //
  82.     else
  83.     {
  84.         while (IsNum(sub_str[INDEX]))
  85.         {
  86.             temp[temp_index++] = sub_str[INDEX];
  87.             INDEX++;
  88.         }
  89.         second = atoi(temp);
  90.     }

  92.     //
  93.     // 判断加减乘除的函数，是程序的精华所在
  94.     //
  95.     printf("5.first = %d\titem = %c\tsecond = %d\n", first,item,second);
  96.     if (item == '+' )
  97.     {
  98.         return first + Calc(sub_str, second);
  99.     }
  100.     else if (item == '-')
  101.     {
  102.         return first - Calc(sub_str, second);
  103.     }
  104.     else if (item == '*')
  105.     {
  106.         return Calc(sub_str, first * second);
  107.     }
  108.     else if (item == '/')
  109.     {
  110.         return Calc(sub_str, first / second);
  111.     }
  112. }

  114. //
  115. // 使用递归的思路解决代括号的四则运算的问题
  116. //

  118. int mainn(int argc, char *argv[])
  119. {
  120.     char str[100] = {"((1+((12+50)/2+10))-20*10)+9*(10*(18-11)-20)"};
  121. // char str[100] = {"8+9*5-2/2+6"};
  122.     while (1)
  123.     {
  124.         //
  125.         // 循环输入函数，请输入规范的四则运算式子
  126.         //
  127.         printf("str = %s\n", str);
  128.         int answer = Calc(str,0);

  130.         if(INDEX = strlen(str))
  131.         {
  132.             printf("final_answer = %d\n", answer);
  133.         }
  134.         else
  135.         {
  136.             printf("算式错误，请检查语法\n");
  137.         }

  139.         scanf("%s", str);
  140.         INDEX = 0;
  141.     }
  142.     return 0;
  143. }


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