|
发表于 2023-11-1 15:41:08
|
显示全部楼层
这个问题可以通过使用循环和条件语句来解决。首先,我们可以创建一个数组来存储数字对应的英文单词。然后,我们可以按照千位、百位、十位和个位的顺序将amount的每个数字转换成对应的英文单词,并将其存储到buffer中。
下面是一个可能的解决方案:
- void written_amount(unsigned int amount, char *buffer) {
- const char *digits[] = {
- "ZERO", "ONE", "TWO", "THREE", "FOUR", "FIVE", "SIX", "SEVEN", "EIGHT", "NINE"
- };
- const char *teens[] = {
- "TEN", "ELEVEN", "TWELVE", "THIRTEEN", "FOURTEEN", "FIFTEEN", "SIXTEEN", "SEVENTEEN", "EIGHTEEN", "NINETEEN"
- };
- const char *tens[] = {
- "", "", "TWENTY", "THIRTY", "FORTY", "FIFTY", "SIXTY", "SEVENTY", "EIGHTY", "NINETY"
- };
- const char *powers[] = {
- "", "THOUSAND", "MILLION", "BILLION"
- };
-
- if (amount == 0) {
- strcpy(buffer, "ZERO");
- return;
- }
-
- int index = 0;
- while (amount > 0) {
- if (amount % 1000 != 0) {
- int part = amount % 1000;
- int hundreds = part / 100;
- int tens_and_ones = part % 100;
-
- if (hundreds > 0) {
- strcat(buffer, digits[hundreds]);
- strcat(buffer, " HUNDRED ");
- }
-
- if (tens_and_ones >= 10 && tens_and_ones <= 19) {
- strcat(buffer, teens[tens_and_ones - 10]);
- } else {
- int tens = tens_and_ones / 10;
- int ones = tens_and_ones % 10;
-
- if (tens > 0) {
- strcat(buffer, tens[tens]);
- strcat(buffer, " ");
- }
-
- if (ones > 0) {
- strcat(buffer, digits[ones]);
- strcat(buffer, " ");
- }
- }
-
- strcat(buffer, powers[index]);
- strcat(buffer, " ");
- }
-
- amount /= 1000;
- index++;
- }
-
- // 移除最后一个空格
- buffer[strlen(buffer) - 1] = '\0';
- }
复制代码
这个函数使用了四个字符串数组来存储数字对应的英文单词。在循环中,我们从amount的个位开始,依次将每个部分转换成英文单词,并将其拼接到buffer中。最后,我们需要移除最后一个空格。
请注意,这个解决方案假设buffer参数足够大,以便存储转换后的字符串。如果buffer的长度有限,请确保足够大以容纳转换后的字符串。 |
|