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".
输入描述:
Each input file contains one test case, which gives an integer with no more than 9 digits.
输出描述:
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.
输入例子:
-123456789
输出例子:
Fu yi Yi er Qian san Bai si Shi wu Wan liu Qian qi Bai ba Shi jiu将阿拉伯数字转换为汉语拼音的数字读法,数字最大到亿位。做这道题之前必须要把各种读数的情况给分清楚
#include <iostream> #include <string> #include <vector> using namespace std; string num[10] = { "ling","yi", "er", "san", "si", "wu", "liu", "qi", "ba", "jiu" }; string c[6] = { "Ge","Shi", "Bai", "Qian", "Yi", "Wan" }; int J[] = {1, 10, 100, 1000, 10000, 100000, 1000000, 10000000, 100000000}; int main() { freopen("in.txt","r",stdin); int n; cin >> n; if (n == 0) { cout << "ling"; return 0; } if (n < 0) { cout << "Fu "; n = -n; } int part[3]; part[0]= n / 100000000; part[1]= (n % 100000000) / 10000; part[2] = n % 10000; bool zero = false; //是否在非零数字前输出合适的ling 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。 for (int i = 0; i < 3; i++) { 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 for (int j = 3; j >= 0; j--) { int curPos = 8 - i * 4 + j; //当前数字的位置 if (curPos >= 9) continue; //最多九位数 int cur = (temp / J[j]) % 10;//取出当前数字 if (cur != 0) { if (zero) { printCnt++ == 0 ? cout<<"ling" : cout<<" ling"; zero = false; } if (j == 0) printCnt++ == 0 ? cout << num[cur] : cout << ‘ ‘ << num[cur]; //在个位,直接输出 else printCnt++ == 0 ? cout << num[cur] << ‘ ‘ << c[j] : cout << ‘ ‘ << num[cur] << ‘ ‘ << c[j]; //在其他位,还要输出十百千 } else { if (!zero&&j != 0 && n / J[curPos] >= 10) zero = true; //注意100020这样的情况,0最小也要从10位开始才读,10位之前如果有0要读出来,肯定n/j[curpos]>=10 } } 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 } return 0; }
原文地址:https://www.cnblogs.com/kuroko-ghh/p/10289890.html
时间: 2024-10-09 07:45:44