用c++写一个程序，它接收用4位阿拉伯数字表示的一个年份，输出用罗马数字表示的年份

Question

一些重要的罗那数字包括：V代表5，X代表10，L代表50，C代表100，D代表500，M代表1000.记住某些数字是通过“减去”一个罗马“数字”而生成的；例如，IV代表4，它是通过V减I而生成的，XL代表40，CM代表900，以此类推，一些示范行的年份转换结果是：MCM代表1900，MCML代表1950，MCMLX代表1960，MCMXL代表1940，MCMLXXXIX代表1989，假定年份在1000~3000之间。你的程序应该包括一个循环，允许用户重复这一计算，直到他们表示操作完毕。

Answer 1

用笨办法--查表法：
注解掉的行是表格对应整型值 //d_v[]，c_v[],b_v[],a_v[] 供你读程序时参考

#include <stdio.h>main(){char d_s[][5]={"","I","II","III","IV","V","VI","VII","VIII","IX"};//int d_v[]={0,1,2,3,4,5,6,7,8,9}; char c_s[][5]={"","X","XX","XXX","XL","L","LX","LXX","LXXX","XC"};//int c_v[]={0,10,20,30,40,50,60,70,80,90};char b_s[][5]={"","C","CC","CCC","CD","D","DC","DCC","DCCC","CM"};//int b_v[]={0,100,200,300,400,500,600,700,800,900};char a_s[][5]={"","M","MM","MMM"};//int a_v[]={0,1,2,3};int a=0,b=0,c=0,d=0;int y;printf("input year 1-2100\n");scanf("%d",&y);while(1){d = y %10;y = y / 10;if (y == 0) break;c = y %10;y = y / 10;if (y == 0) break;b = y %10;y = y / 10;if (y == 0) break;a = y %10;y = y / 10;break;}
//printf("%d%d%d%d\n",a,b,c,d);
printf("%s%s%s%s\n",a_s[a],b_s[b],c_s[c],d_s[d]); }