第13题：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。

 

样例输入：

123

 

样例输出：

CXXIII

 

遇到这道题的时候，不了解罗马数字的规则的话，还真不知道怎么写。没什么好说的，度娘。。

研究了一下，知道这一题的本意是让我们将整数的每一位都拆开，分别用罗马数字替换，当然得符合规则。。

我写了一份对应表：

这样这一题就只要对应替换就行了，代码如下：

#include<stdio.h>

#include<string.h>

 

int main(void) {

         intnum;

         intx, i, len;

         charres[16];

         chara[5], b[5], c[5], d[5];

 

         scanf("%d",&num);

 

         if(num < 1 || num > 3999) {

                   return0;

         }

 

         //个位

         x= num % 10;

         switch(x) {

                   case0 : a[0] = '\0'; break;

                   case1 : a[0] = 'I'; a[1] = '\0'; break;

                   case2 : a[0] = 'I'; a[1] = 'I'; a[2] = '\0'; break;

                   case3 : a[0] = 'I'; a[1] = 'I'; a[2] = 'I'; a[3] = '\0'; break;

                   case4 : a[0] = 'I'; a[1] = 'V'; a[2] = '\0'; break;

                   case5 : a[0] = 'V'; a[1] = '\0'; break;

                   case6 : a[0] = 'V'; a[1] = 'I'; a[2] = '\0'; break;

                   case7 : a[0] = 'V'; a[1] = 'I'; a[2] = 'I'; a[3] = '\0'; break;

                   case8 : a[0] = 'V'; a[1] = 'I'; a[2] = 'I'; a[3] = 'I'; a[4] = '\0'; break;

                   case9 : a[0] = 'I'; a[1] = 'X'; a[2] = '\0'; break;

         }

 

         //十位

         x= ( num / 10 ) % 10;

         switch(x) {

                   case0 : b[0] = '\0'; break;

                   case1 : b[0] = 'X'; b[1] = '\0'; break;

                   case2 : b[0] = 'X'; b[1] = 'X'; b[2] = '\0'; break;

                   case3 : b[0] = 'X'; b[1] = 'X'; b[2] = 'X'; b[3] = '\0'; break;

                   case4 : b[0] = 'X'; b[1] = 'L'; b[2] = '\0'; break;

                   case5 : b[0] = 'L'; b[1] = '\0'; break;

                   case6 : b[0] = 'L'; b[1] = 'X'; b[2] = '\0'; break;

                   case7 : b[0] = 'L'; b[1] = 'X'; b[2] = 'X'; b[3] = '\0'; break;

                   case8 : b[0] = 'L'; b[1] = 'X'; b[2] = 'X'; b[3] = 'X'; b[4] = '\0'; break;

                   case9 : b[0] = 'X'; b[1] = 'C'; b[2] = '\0'; break;

         }

 

         //百位

         x= ( num / 100 ) % 10;

         switch(x) {

                   case0 : c[0] = '\0'; break;

                   case1 : c[0] = 'C'; c[1] = '\0'; break;

                   case2 : c[0] = 'C'; c[1] = 'C'; c[2] = '\0'; break;

                   case3 : c[0] = 'C'; c[1] = 'C'; c[2] = 'C'; c[3] = '\0'; break;

                   case4 : c[0] = 'C'; c[1] = 'D'; c[2] = '\0'; break;

                   case5 : c[0] = 'D'; c[1] = '\0'; break;

                   case6 : c[0] = 'D'; c[1] = 'C'; c[2] = '\0'; break;

                   case7 : c[0] = 'D'; c[1] = 'C'; c[2] = 'C'; c[3] = '\0'; break;

                   case8 : c[0] = 'D'; c[1] = 'C'; c[2] = 'C'; c[3] = 'C'; c[4] = '\0'; break;

                   case9 : c[0] = 'C'; c[1] = 'M'; c[2] = '\0'; break;

         }

 

         //千位

         x= ( num / 1000 ) % 10;

         switch(x) {

                   case0 : d[0] = '\0'; break;

                   case1 : d[0] = 'M'; d[1] = '\0'; break;

                   case2 : d[0] = 'M'; d[1] = 'M'; d[2] = '\0'; break;

                   case3 : d[0] = 'M'; d[1] = 'M'; d[2] = 'M'; d[3] = '\0'; break;

         }

 

         i= 0;

         len= 0;

         while(d[i] != '\0') {

                   res[len]= d[i];

                   i++;

                   len++;

         }

         i= 0;

         while(c[i] != '\0') {

                   res[len]= c[i];

                   i++;

                   len++;

         }

         i= 0;

         while(b[i] != '\0') {

                   res[len]= b[i];

                   i++;

                   len++;

         }

         i= 0;

         while(a[i] != '\0') {

                   res[len]= a[i];

                   i++;

                   len++;

         }

         res[len]= '\0';

 

         printf("%s",res);

 

         return0;

}