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 1:
-123456789
Sample Output 1:
Fu yi Yi er Qian san Bai si Shi wu Wan liu Qian qi Bai ba Shi jiu
Sample Input 2:
100800
Sample Output 2:
yi Shi Wan ling ba Bai
Code:
using namespace std;
int main(int argc, char *argv[]) {
char num[11], buffer[5];
scanf("%s", num);
int len = strlen(num), i = 0, digit1, digit2;
if (num[0] == '-') {
printf("Fu ");
i = 1;
}
char str1[10][5] = { "ling", "yi", "er", "san", "si", "wu", "liu", "qi", "ba", "jiu" };
char str2[5][5] = { "Shi", "Bai", "Qian", "Wan", "Yi" };
vector<char*> result;//这里不能char[5]，要写char*，另外char*是类型，声明变量的时候要写char *val
if (len == 1) {//个位数直接进下面的代码容易出错，干脆单独拿出来了
printf("%s", str1[num[0] - '0']);
system("pause");
return 0;
}
else if (len == 2 && i == 1) {
printf(" %s", str1[num[0] - '0']);
system("pause");
return 0;
}
else;
for (; i < len; i++) {//最后再考虑空格，逐情况考虑
if (num[i] == '0')
if (i + 1 < len && num[i + 1] != '0')//零有很多种情况，这句是最关键的
result.push_back(str1[num[i] - '0']);//其实就是把printf全都改成了vector.push_back
else;
else
result.push_back(str1[num[i] - '0']);
digit1 = (len - i) % 4;
digit2 = (len - i) / 4;
//当然可以把以下判断单位的情况整合到一两个公式里面，但没必要
if (digit1 == 1) {//也就是到个位了
if (digit2 == 0);
else if (digit2 == 1) {
if (i > 3 && i < len - 3) {//万和亿中可能只说一个，需要判断一下
strncpy(buffer, num + i - 3, 4);
if (atoi(buffer) == 0)
continue;
}
result.push_back(str2[3]);//Wan
}
else if (digit2 == 2)
result.push_back(str2[4]);//Yi
}
else if (digit1 == 2 && num[i] != '0')
result.push_back(str2[0]);//Shi
else if (digit1 == 3 && num[i] != '0')
result.push_back(str2[1]);//Bai
else if (digit1 == 0 && num[i] != '0')
result.push_back(str2[2]);//Qian
}
printf("%s", result[0]);
for (int k = 1; k < result.size(); k++)
printf(" %s", result[k]);
system("pause");
return 0;
}
//刚开始就想着这个题如果改成用容器写可能会好一点，后来发现空格不好判断，还是得用容器
//写这个代码在第四个测试点第一次见段错误，说是越界或递归过深导致栈溢出之类的
//这种情况下，高中的做题思维就很好用了，首先考虑特殊点，比如0或者上下限~这里果然是0/个位数