Chinaunix首页 | 论坛 | 博客

时间看来的ChinaUnix博客

首页　| 　博文目录　| 　关于我

时间看来

  • 博客访问： 444329
  • 博文数量： 67
  • 博客积分： 2468
  • 博客等级： 上尉
  • 技术积分： 1050
  • 用 户 组： 普通用户
  • 注册时间： 2011-01-05 01:21
文章分类

全部博文（67）

文章存档

2013年（1）

2012年（65）

2011年（1）

我的朋友
最近访客
推荐博文
相关博文
输入12，如何转换成壹拾贰;输入1020，转成壹千零贰拾

分类： C/C++

2012-11-30 20:19:09

点击(此处)折叠或打开

  1. /*
  2.  *输入12，如何转换成壹拾贰;输入1020，转成壹千零贰拾
  3.  */
  4. #include <stdio.h>
  5. #include <stdlib.h>
  6. #include <string.h>

  8. void get4num(char *dest, char *src);
  9. void output4num(char *src);
  10. int output_unit(int left_bits);

  12. char *my_array[10] = {"零","壹","贰","叁","肆",\
  13.      "伍","陆","柒","捌","玖"};
  14. char *my_unit[5] = {"拾","佰","仟","万","亿"};
  15. int FLAG_FIRST;
  16. int
  17. main(int argc, char *argv[])
  18. {
  19.     /*test my_array*/
  20.     int i;
  21.     for(i=0; i<10; i++){
  22.         printf("%s\n", *(my_array+i));
  23.     }

  25.     char input_num[100];
  26.     printf("input_num:\n");
  27.     scanf("%s", input_num);
  28.     printf("%s\n", input_num);
  29.     
  30.     int input_len = strlen(input_num);
  31.     char nums_4[6];

  33.     /*output first nums*/
  34.     FLAG_FIRST = 1;
  35.     int first_num = input_len % 4;
  36.     if(first_num != 0){
  37.         strncpy(nums_4, input_num, first_num);
  38.         output4num(nums_4);
  39.         output_unit(input_len);
  40.     }

  42.     /*output the left 4 nums*/
  43.     FLAG_FIRST = 1;
  44.     for(i=first_num; i<input_len; i+=4){
  45.         get4num(nums_4, input_num+i);
  46.         output4num(nums_4);
  47.         output_unit(input_len-i);
  48.         FLAG_FIRST = 0;
  49.     }

  51.     printf("\0");
  52.     return 0;
  53. }

  55. /*
  56.  *get 4 numbers
  57.  */
  58. void
  59. get4num(char *dest, char *src)
  60. {
  61.     int i = 4;
  62.     strncpy(dest, src, i);
  63. }

  65. /*
  66.  * output 4 numbers
  67.  */
  68. void
  69. output4num(char *src)
  70. {
  71.     int i, j;
  72.     int num, nums[4];
  73.     num = atoi(src);
  74.     j = strlen(src);

  76.     /*low number is at low place*/
  77.     for(i=0; i<4; i++){
  78.         nums[i] = (num % 10);
  79.         num /= 10;
  80.     }

  82.     /*output first nums*/
  83.     if(nums[j-1] == 0){
  84.         if((j == 4) && (FLAG_FIRST == 0)){
  85.             printf("%s", *(my_array+nums[j-1]));
  86.         }
  87.     }else{
  88.         printf("%s", *(my_array+nums[j-1]));
  89.         printf("%s", *(my_unit+j-2));
  90.     }

  92.     /*output number and unit*/
  93.     for(i=j-2; i>=1; i--){
  94.         if(nums[i] == 0){
  95.             if(nums[i+1] != 0){
  96.                 printf("%s", *(my_array+nums[i]));
  97.             }
  98.         }else{
  99.             printf("%s", *(my_array+nums[i]));
  100.             printf("%s", *(my_unit+i-1));
  101.         }
  102.     }
  103.     if(nums[0] != 0){
  104.         printf("%s", *(my_array+nums[i]));
  105.     }
  106. }

  108. /*
  109.  * output the unit of 4 nums
  110.  */
  111. int
  112. output_unit(int left_bits)
  113. {
  114.     int m, n;
  115.     m = (left_bits-1) / 4;
  116.     switch(m){
  117.         case 0:
  118.             return 0;
  119.         case 1:
  120.             printf("%s", my_unit[3]);
  121.             return 0;
  122.         default:
  123.             for(n=m; n>=2; n--){
  124.                 printf("%s", my_unit[4]);
  125.             }
  126.     }

  128.     printf(",");
  129. }
看到有用shell script写的，。我尝试用C语言写下，终于拼凑出来了。
input_num:
9283749817509843219827409812365908
9283749817509843219827409812365908
玖拾贰亿亿亿亿亿亿亿,捌仟叁佰柒拾肆亿亿亿亿亿亿,玖仟捌佰壹拾柒亿亿亿亿亿,伍仟零玖拾捌亿亿亿亿,肆仟叁佰贰拾壹亿亿亿,玖仟捌佰贰拾柒亿亿,肆仟零玖拾捌亿,壹仟贰佰叁拾陆万伍仟玖佰零捌

/////////////////////////////////////////////////////////////////////////////////

2012.12.1
非常感谢楼下的回复，陪我写玩具程序。

发现我上面的程序每4位输出的单位上万亿后都是错的。把重写了int output_unit(int left_bits)，还有修改了穿过来的left_bits是4的整数倍。
根据人的读数习惯（以4个位数为一个单元），第一感觉也就这么写了。相信还有跟好的方法。
2楼的就是根据每位数来判断单位，不过我编译运行你的程序后，输出是乱码，这个我没有搞明白。我可是很想抓你小辫子哦！哇咔咔~对你再次表示感谢。

点击(此处)折叠或打开

  1. /*
  2.  *输入12，如何转换成壹拾贰，输入1020，转成壹千零贰拾
  3.  */
  4. #include <stdio.h>
  5. #include <stdlib.h>
  6. #include <string.h>

  8. void get4num(char *dest, char *src);
  9. void output4num(char *src);
  10. int output_unit(int left_bits);

  12. char *my_array[10] = {"零","壹","贰","叁","肆",\
  13.                           "伍","陆","柒","捌","玖"};
  14. char *my_unit[5] = {"拾","佰","仟","万","亿"};
  15. int FLAG_FIRST;
  16. int
  17. main(int argc, char *argv[])
  18. {
  19.         /*test my_array*/
  20.         int i;
  21.         for(i=0; i<10; i++){
  22.                 printf("%s\n", *(my_array+i));
  23.         }

  25.         char input_num[100];
  26.         printf("input_num:\n");
  27.         scanf("%s", input_num);
  28.         printf("%s\n", input_num);
  29.         
  30.         int input_len = strlen(input_num);
  31.         char nums_4[6];

  33.         /*output first nums*/
  34.         FLAG_FIRST = 1;
  35.         int first_num = input_len % 4;
  36.         if(first_num != 0){
  37.                 strncpy(nums_4, input_num, first_num);
  38.                 output4num(nums_4);
  39.                 output_unit(input_len - first_num);
  40.         }

  42.         /*output the left 4 nums*/
  43.         FLAG_FIRST = 1;
  44.         for(i=first_num; i<input_len; i+=4){
  45.                 get4num(nums_4, input_num+i);
  46.                 output4num(nums_4);
  47.                 output_unit(input_len - i - 4);
  48.                 FLAG_FIRST = 0;
  49.         }

  51.         printf("\0\0");
  52.         return 0;
  53. }

  55. /*
  56.  *get 4 numbers
  57.  */
  58. void
  59. get4num(char *dest, char *src)
  60. {
  61.         int i = 4;
  62.         strncpy(dest, src, i);
  63. }

  65. /*
  66.  * output 4 numbers
  67.  */
  68. void
  69. output4num(char *src)
  70. {
  71.         int i, j;
  72.         int num, nums[4];
  73.         num = atoi(src);
  74.         j = strlen(src);

  76.         /*low number is at low place*/
  77.         for(i=0; i<4; i++){
  78.                 nums[i] = (num % 10);
  79.                 num /= 10;
  80.         }

  82.         /*output first nums*/
  83.         if(nums[j-1] == 0){
  84.                 if((j == 4) && (FLAG_FIRST == 0)){
  85.                         printf("%s", *(my_array+nums[j-1]));
  86.                 }
  87.         }else{
  88.                 printf("%s", *(my_array+nums[j-1]));
  89.                 printf("%s", *(my_unit+j-2));
  90.         }

  92.         /*output number and unit*/
  93.         for(i=j-2; i>=1; i--){
  94.                 if(nums[i] == 0){
  95.                         if(nums[i+1] != 0){
  96.                                 printf("%s", *(my_array+nums[i]));
  97.                         }
  98.                 }else{
  99.                         printf("%s", *(my_array+nums[i]));
  100.                         printf("%s", *(my_unit+i-1));
  101.                 }
  102.         }
  103.         if(nums[0] != 0){
  104.                 printf("%s", *(my_array+nums[i]));
  105.         }
  106. }

  108. /*
  109.  * output the unit of 4 nums
  110.  */
  111. int
  112. output_unit(int left_bits)
  113. {
  114.         int m, n;
  115.         /*the last one*/
  116.         if(left_bits < 4){
  117.                 return 0;
  118.         } else{
  119.                 m = left_bits % 8;
  120.                 switch(m){
  121.                         case 4:
  122.                                 printf("%s", my_unit[3]);
  123.                                 break;
  124.                         case 0:
  125.                                 n = left_bits / 8;
  126.                                 for(m=0; m<n; m++){
  127.                                         printf("%s", my_unit[4]);
  128.                                 }
  129.                 }
  130.         }

  132.         printf(",");
  133.         return 0;
  134. }
input_num:
123456789012345678901234567890
123456789012345678901234567890
壹拾贰万,叁仟肆佰伍拾陆亿亿亿,柒仟捌佰玖拾万,壹仟贰佰叁拾肆亿亿,伍仟陆佰柒拾捌万,玖仟零壹拾贰亿,叁仟肆佰伍拾陆万,柒仟捌佰玖拾

////////////////////////////////////////////////////////////////////
2012.12.2

点击(此处)折叠或打开

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

  5. char *my_array[10] = {"零","壹","贰","叁","肆",\
  6.      "伍","陆","柒","捌","玖"};
  7. char *my_unit[5] = {"拾","佰","仟","万","亿"};

  9. void
  10. output_unit(int position)
  11. {
  12.         if(position == 1){
  13.                 return;
  14.         }

  16.         int left_bits = position -1;
  17.         int i, many_yi;
  18.         int case_unit;
  19.         if(left_bits % 8 == 0){
  20.                 many_yi = left_bits / 8;
  21.                 for(i=0; i<many_yi; i++){
  22.                         printf("%s", my_unit[4]);
  23.                 }
  24.                 printf(";");
  25.         } else{
  26.                 case_unit = left_bits % 4;
  27.                 switch(case_unit){
  28.                     case 0:
  29.                                 printf("%s,", my_unit[3]);
  30.                                 break;
  31.                     case 1:
  32.                         printf("%s", my_unit[0]);
  33.                             break;
  34.                     case 2:
  35.                         printf("%s", my_unit[1]);
  36.                         break;
  37.                     case 3:
  38.                         printf("%s", my_unit[2]);
  39.                             break;
  40.                 }
  41.         }
  42. }

  44. int
  45. str_convert(char *src)
  46. {
  47.         int i;
  48.         /*check the input string*/
  49.         int src_len = strlen(src);
  50.         if(src_len < 1){
  51.                 printf("dont't have enough digital");
  52.         }

  54.         /*keep the string is legal*/
  55.         for(i=0; i<src_len; i++){
  56.                 if(*(src+i) >= '0' && *(src+i) <= '9'){
  57.                         continue;
  58.                 } else{
  59.                         printf("the number is illegal\n");
  60.                         return 0;
  61.                 }
  62.         }

  64.         /*output Chinese*/
  65.         int first_bit = 0;
  66.         for(i=0; i<src_len && *(src+i) == '0'; i++){
  67.                 ;
  68.         }
  69.         first_bit = i;

  71.         char *cur_bit;
  72.         int cur_dig;
  73.         for(i=first_bit; i<src_len; i++){
  74.                 cur_bit = src + i;
  75.                 cur_dig = *cur_bit - '0';
  76.                 printf("%s",*(my_array + cur_dig));
  77.                 output_unit(src_len - i);
  78.         }

  80.         printf("\n");
  81.         return 0;
  82. }

  84. int main(int argc, char *argv[])
  85. {
  86.         char tmp[100];
  87.         printf("input a number\n");
  88.         scanf("%s", tmp);
  89.         printf("src:%s\n", tmp);
  90.         str_convert(tmp);

  92.         fflush(stdout);
  93.         return 0;
  94. }

input a number
src:00123 4567 | 8901 2345 | 6789 0123 | 4567 8901
壹佰贰拾叁万,肆仟伍佰陆拾柒亿亿亿;捌仟玖佰零拾壹万,贰仟叁佰肆拾伍亿亿;陆仟柒佰捌拾玖万,零仟壹佰贰拾叁亿;肆仟伍佰陆拾柒万,捌仟玖佰零拾壹

阅读(2604) | 评论(1) | 转发(0) |
0

上一篇：文件 数据库 数据库管理系统

下一篇：算法的常识估算

给主人留下些什么吧！~~

时间看来2012-11-30 20:22:35

看到有用shell script写的，我尝试用C语言写下，终于拼凑出来了。
http://www.linuxsir.org/bbs/thread47558.html

回复 | 举报
关于我们 | 关于IT168 | 联系方式 | 广告合作 | 法律声明 | 免费注册

Copyright 2001-2010 ChinaUnix.net All Rights Reserved 北京皓辰网域网络信息技术有限公司. 版权所有

感谢所有关心和支持过ChinaUnix的朋友们

16024965号-6