PAT (Advanced Level) Practise 1082 Read Number in Chinese (25)

1082. Read Number in Chinese (25)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

题意：输出一个数字的中文读法

#include <iostream>    #include <cstdio>    #include <string>    #include <cstring>    #include <algorithm>    #include <cmath>    #include <vector>    #include <map>    #include <set>    #include <queue>    #include <stack>    #include <functional>    #include <climits>    using namespace std;#define LL long long    const int INF = 0x7FFFFFFF;string a[10] = { "", "Shi", "Bai", "Qian", "Wan", "Shi", "Bai", "Qian", "Yi" };string b[10] = { "ling", "yi", "er", "san", "si", "wu", "liu", "qi", "ba", "jiu" };stack<string>s;void solve(int x){for (int i = 0; x; i++, x /= 10){if (x % 10){ if (i) s.push(a[i]);s.push(b[x % 10]);}else{if (i == 4) { if (x % 10000) s.push(a[i]); }if (s.empty() || s.top() == b[0] || s.top() == a[4]) continue;s.push(b[0]);}}}int main(){int n;while (~scanf("%d", &n)){solve(abs(n));if (n < 0) s.push("Fu");while (!s.empty()){cout << s.top();s.pop();if (!s.empty()) printf(" ");}if (!n) printf("ling");printf("\n");}return 0;}