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
吐槽一波,这就不能算是算法题,纯属是字符串硬处理
1 #include <iostream> 2 #include <string> 3 #include <vector> 4 using namespace std; 5 string num[10] = { "ling","yi", "er", "san", "si", "wu", "liu", "qi", "ba", "jiu" }; 6 string c[6] = { "Ge","Shi", "Bai", "Qian", "Yi", "Wan" }; 7 int J[] = { 1, 10, 100, 1000, 10000, 100000, 1000000, 10000000, 100000000 }; 8 vector<string> res; 9 int main() { 10 int n; 11 cin >> n; 12 if (n == 0) { 13 cout << "ling"; 14 return 0; 15 } 16 if (n < 0) { 17 cout << "Fu "; 18 n = -n; 19 } 20 int part[3]; 21 part[0] = n / 100000000; 22 part[1] = (n % 100000000) / 10000; 23 part[2] = n % 10000; 24 bool zero = false; //是否在非零数字前输出合适的ling 25 int printCnt = 0; //用于维护单词前没有空格,之后输入的单词都在前面加一个空格。 26 for (int i = 0; i < 3; i++) { 27 int temp = part[i]; //三个部分,每部分内部的命名规则都一样,都是X千X百X十X 28 for (int j = 3; j >= 0; j--) { 29 int curPos = 8 - i * 4 + j; //当前数字的位置 30 if (curPos >= 9) continue; //最多九位数 31 int cur = (temp / J[j]) % 10;//取出当前数字 32 if (cur != 0) { 33 if (zero) { 34 printCnt++ == 0 ? cout << "ling" : cout << " ling"; 35 zero = false; 36 } 37 if (j == 0) 38 printCnt++ == 0 ? cout << num[cur] : cout << ' ' << num[cur]; //在个位,直接输出 39 else 40 printCnt++ == 0 ? cout << num[cur] << ' ' << c[j] : cout << ' ' << num[cur] << ' ' << c[j]; //在其他位,还要输出十百千 41 } 42 else { 43 if (!zero&&j != 0 && n / J[curPos] >= 10) zero = true; //注意100020这样的情况 44 } 45 } 46 if (i != 2 && part[i] > 0) cout << ' ' << c[i + 4]; //处理完每部分之后,最后输出单位,Yi/Wan 47 } 48 return 0; 49 }