PAT

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

---

给定条件：
1.给定一个整数

要求：
1.“读出”这个数

求解思路：
1.先试着每个数字都读出来，找出错误的地方
2.发现ling和wan需要特殊处理
4.依次处理每一位，需要特殊处理的增加判断条件即可
5.具体看代码

---

#include <cstdio>#include <iostream>#include <vector>using namespace std;int n;vector<string> v;string num[10] = {"ling", "yi", "er", "san", "si", "wu", "liu", "qi", "ba", "jiu"};int main() {scanf("%d", &n);if(n == 0){printf("ling\n");return 0;}if(n < 0) {n *= -1;printf("Fu ");}int cnt = 0;while(n > 0) {switch(cnt){case 1 : if(n%10)v.push_back("Shi"); break;case 2 : if(n%10)v.push_back("Bai"); break;case 3 : if(n%10)v.push_back("Qian"); break;case 4 : if(n%10 || n/10%10 || n/10/10%10 || n/10/10/10%10)v.push_back("Wan"); break;case 5 : if(n%10)v.push_back("Shi"); break;case 6 : if(n%10)v.push_back("Bai"); break;case 7 : if(n%10)v.push_back("Qian"); break;case 8 : if(n%10)v.push_back("Yi"); break;}if(n%10 == 0) {if(!v.empty() && v[v.size()-1] != "ling" ) {if(v[v.size()-1] != "Shi" && v[v.size()-1] != "Bai" && v[v.size()-1] != "Qian" && v[v.size()-1] != "Wan")v.push_back(num[n%10]);}}else {v.push_back(num[n%10]);}cnt++;n /= 10;}for(int i = v.size()-1; i > 0; i--){printf("%s ", v[i].c_str());}printf("%s\n", v[0].c_str());return 0;}

---