1082 Read Number in Chinese (25point(s))
Given an integer with no more than 9 digits, you are supposed to read it in the traditional Chinese way. Output Fu first if it is negative. For example, -123456789 is read as Fu yi Yi er Qian san Bai si Shi wu Wan liu Qian qi Bai ba Shi jiu. Note: zero (ling) must be handled correctly according to the Chinese tradition. For example, 100800 is yi Shi Wan ling ba Bai.
Input Specification:
Each input file contains one test case, which gives an integer with no more than 9 digits.
Output Specification:
For each test case, print in a line the Chinese way of reading the number. The characters are separated by a space and there must be no extra space at the end of the line.
Sample Input:
-123456789
Sample Output:
Fu yi Yi er Qian san Bai si Shi wu Wan liu Qian qi Bai ba Shi jiu
Sample Input:
100800
Sample Output:
yi Shi Wan ling ba Bai
题目大意：
输入一个整数，输出这个整数的汉语读法
设计思路：
主要是 ling 的处理，和位的处理
- 一个字符数组记录数字，一个字符数组记录位
- 字符数组第 0 位为空是因为使其下标从 1 开始用
- 字符数组最后一位为空是因为个位不输出
- 把输入的数字，先处理正负号，然后用前导 0 填充满 10 位数，
- 每四位为一组，以输出特殊的亿位，万位，个位
- 输出位信息前一定会先输出数字信息，所以在位的字符数组中提前加入空格，输出时就不用再考虑空格了，而个位的位为空
- 输出过程中注意 ling 的判断，多个连续的 0，要符合汉字读法，省略成一个 ling
编译器：C (gcc)
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(void)
{
char *num[]={"ling", "yi", "er", "san", "si", "wu", "liu", "qi", "ba", "jiu"};
char *wei[]={"", " Yi", " Qian", " Bai", " Shi", " Wan", " Qian", " Bai", " Shi", ""};
char s[11] = {0};
int i, j, len, flag;
gets(s);
if (atoi(s) == 0) {
printf("%s", num[0]);
} else {
if (s[0] == '-') {
printf("Fu ");
s[0] = '0';
}
len = strlen(s);
for (i = len - 1, j = 9; i >= 0; i--, j--)
s[j] = s[i];
for (;j >= 0; j--)
s[j] = '0';
for (i = 0; s[i] == '0'; i++)
;
printf("%s%s", num[s[i] - '0'], wei[i]);
flag = 0;
for (i++; i <= 9; i++) {
if (i % 4 == 1) {
if (s[i] == '0') {
printf("%s", wei[i]);
} else {
if (flag == 1) {
printf(" ling %s%s", num[s[i] - '0'], wei[i]);
flag = 0;
} else {
printf(" %s%s", num[s[i] - '0'], wei[i]);
}
}
} else {
if (s[i] == '0') {
flag = 1;
} else {
if (flag == 1) {
printf(" ling %s%s", num[s[i] - '0'], wei[i]);
flag = 0;
} else {
printf(" %s%s", num[s[i] - '0'], wei[i]);
}
}
}
}
}
return 0;
}