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

最新推荐文章于 2017-10-07 10:43:36 发布

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最新推荐文章于 2017-10-07 10:43:36 发布

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本文介绍了一个使用C++实现的程序，该程序能够将输入的整数（最多9位）转换为对应的中文读法，并正确处理零的情况。文章提供了一个完整的代码示例，包括如何处理负数和特殊情况下零的读法。

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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:

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;
}