整数转换成罗马数字

给定一个整数num，( 1<=num<=3999)，将整数转换成罗马数字。

如1,2,3,4,5对应的罗马数字分别位I，II，III，IV，V等。

格式：

   第一行输入一个整数，接下来输出对应的罗马数字。

输入：123

输出：CXXIII

我的解法（较差，有六十多行代码）：

#include <stdio.h>

int main()
{
	int num=0; //[1,3999]
	int thousand=0;
	int hundred=0;
	int ten=0;
	int unit=0;
	char* String_thousand[3]={"M","MM","MMM"};
	char* String_hundred[9]={"C","CC","CCC","CD","D","DC","DCC","DCCC","CM"};
	char* String_ten[9]={"X","XX","XXX","XL","L","LX","LXX","LXXX","XC"};
	char* String_unit[9]={"I","II","III","IV","V","VI","VII","VIII","IX"};
	char* String_Temp=NULL;
	scanf("%d",&num);
	
	thousand=num/1000;
	hundred=num/100-10*(num/1000);
	ten=num/10-10*(num/100);
	unit=num%1000%100%10;

//	printf("%d,%d,%d,%d\n",thousand,hundred,ten,unit);
	if(thousand!=0)
	{
		for(int i=0;i<3;i++)
		{
			if(thousand-1==i)
				String_Temp=String_thousand[i];
		}
	}
	if(String_Temp!=NULL)
		printf("%s",String_Temp);
	String_Temp=NULL;
	if(hundred!=0)
	{
		for(int i=0;i<9;i++)
		{
			if(hundred-1==i)
				String_Temp=String_hundred[i];
		}
	}
	if(String_Temp!=NULL)
		printf("%s",String_Temp);
	String_Temp=NULL;
	if(ten!=0)
	{
		for(int i=0;i<9;i++)
		{
			if(ten-1==i)
				String_Temp=String_ten[i];
		}
	}
	if(String_Temp!=NULL)
		printf("%s",String_Temp);
	String_Temp=NULL;
	if(unit!=0)
	{
		for(int i=0;i<9;i++)
		{
			if(unit-1==i)
				String_Temp=String_unit[i];
		}
	}
	if(String_Temp!=NULL)
		printf("%s\n",String_Temp);
	return 0;
}

目前找到的最优解法（只有三十多行代码，比我的优化了一半）：

#include <stdio.h>

void toRoman(int num)
{
    static char *roman[][10]={"","I","II","III","IV","V","VI","VII","VIII","IX",
                  "","X","XX","XXX","XL","L","LX","LXX","LXXX","XC",
                  "","C","CC","CCC","CD","D","DC","DCC","DCCC","CM",
                  "","M","MM","MMM"};
    int i, j, n;
    printf("%d = ", num);
    for(j = 0, i = 10000; j < 4; ++j, i/=10)
    {
        n = (num%i) / (i/10);
        printf("%s",roman[3-j][n]);
    }
    printf("\n");
}
 
int main()
{
    int n;
    printf("Please enter a decimal number (1-3999):");
    scanf("%d", &n);
    if(n < 1 || n > 3999)
    {
        printf("Illegal Parameter!\n");
        return 1;
    }
    toRoman(n);
     
    return 0;
}

经过分析，可以找到几点学习的地方：

1.我的程序中所构建的三个字符串数组，在最优解法中，直接构建成了一个二维数组。

2.我对于千，百，十，个位的数字获取太过复杂，最优解法只使用一个for循环即达到效果。

for(j = 0, i = 10000; j < 4; ++j, i/=10)
    {
        n = (num%i) / (i/10);
    }

printf("%s",roman[3-j][n]);