雲林SONG: 高中生程式解題系統：羅馬數字 (Roman Numerals)

若您覺得文章寫得不錯，請點選文章上的廣告，來支持小編，謝謝。

題目連結 https://zerojudge.tw/ShowProblem?problemid=a013。

這題用查表法，先將兩個羅馬數字(a, b)轉成一般數字(c, d)後，算出 c 與 d 差的絕對值 e。再將 e 換成羅馬數字。

程式碼：

	#include <iostream>

	using namespace std;

	short number(char c)
	{
	short num;

	switch(c)
	{
	case 'I':
	num = 1;
	break;

	case 'V':
	num = 5;
	break;

	case 'X':
	num = 10;
	break;

	case 'L':
	num = 50;
	break;

	case 'C':
	num = 100;
	break;

	case 'D':
	num = 500;
	break;

	case 'M':
	num = 1000;
	break;

	default:
	num = 0;
	break;
	}

	return num;
	}

	short getValue(const string& s )
	{
	short preValue = 1000, curValue, value = 0;
	short len = s.length();

	for( int i = 0; i < len; i++ )
	{
	curValue = number(s[i]);
	value += curValue;

	if( curValue > preValue )
	value -= 2 * preValue;

	preValue = curValue;
	}

	return value;
	}

	void attachMsg(short check, string &result, string msg)
	{
	if( check == 9 )
	{
	result.append(msg.substr(2,1));
	result.append(msg.substr(0,1));

	check = 0;
	}

	if( check == 4 )
	{
	result.append(msg.substr(2,1));
	result.append(msg.substr(1,1));

	check = 0;
	}

	if( check >= 5 )
	{
	result.append(msg.substr(1,1));
	check -= 5;
	}

	for(int i = 0; i < check; i++)
	{
	result.append(msg.substr(2,1));
	}
	}

	void prtNum( short n )
	{
	string result;
	n = (n < 0 ? -n: n);

	short check = n / 1000;
	for(int i = 0; i < check; i++)
	result.append("M");

	n = n % 1000;
	check = n / 100;
	attachMsg(check, result, "MDC");

	n = n % 100;
	check = n / 10;
	attachMsg(check, result, "CLX");

	n = n % 10;
	check = n;
	attachMsg(check, result, "XVI");

	cout << result;
	}

	int main()
	{
	string a, b;
	short numA, numB, numDif;



	while( cin >> a )
	{
	if( a[0] == '#' )
	break;

	cin >> b;

	numA = getValue(a);
	numB = getValue(b);

	numDif = numA - numB;

	if( numDif != 0 )
	prtNum(numDif);
	else
	cout << "ZERO";

	cout << endl;
	}

	return 0;
	}

view raw a013.cpp hosted with ❤ by GitHub