1082. Read Number in Chinese

最新推荐文章于 2024-06-26 23:48:09 发布

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最新推荐文章于 2024-06-26 23:48:09 发布

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分类专栏： PAT 文章标签： PAT C++

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本文介绍了一个C++程序，用于将整数（最多9位）转换为对应的中文读法。包括处理负数和零的情况，例如-123456789被读作“负一亿二千三百四十五万六千七百八十九”。文章提供了完整的代码实现。

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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

因为位数有限而且不多，就直接处理好了，h万和l(1-999)有类似的地方，注意空格的输出和ling的输出。

要对0这个输入特殊处理一下。

#include<iostream>
#include<string>
#include<cmath>

using namespace std;

string num[10]={"ling","yi","er","san","si","wu","liu","qi","ba","jiu"};
string unit[5]={"","Shi","Bai","Qian","Wan"};
int main()
{
	int n;
	cin>>n;
	if(n==0)
	{
		cout<<"ling";
		return 0;
	}
	if(n<0)
		cout<<"Fu ";
	n=abs(n);
	if(n/100000000)
		cout<<num[n/100000000]<<" Yi";
	bool f=n/100000000;
	n=n%100000000;
	if(n&&f)
		cout<<" ";
	int h=n/10000;
	bool f0=!f;
	int op=1000;
	int cnt=3;
	while(h)
	{
		if(h/op)
		{
			cout<<num[h/op];
			f0=false;
			if(cnt)
				cout<<" "<<unit[cnt];		
			if(h%op)
				cout<<" ";
		}
		else if(!f0)
		{
			cout<<"ling ";
			f0=true;
		}
		h=h%op;
		op=op/10;
		cnt--;
	}
	if(n/10000)
		cout<<" Wan";
	int l=n%10000;
	if(l&&n/10000)
		cout<<" ";
	f0=!(f||n/10000);
	op=1000;
	cnt=3;
	while(l)
	{
		if(l/op)
		{
			cout<<num[l/op];
			f0=false;
			if(cnt)
				cout<<" "<<unit[cnt];		
			if(l%op)
				cout<<" ";
		}
		else if(!f0)
		{
			cout<<"ling ";
			f0=true;
		}
		l=l%op;
		op=op/10;
		cnt--;
	}
}