POJ 1588 Reverse Roman Notation

题目链接：http://acm.pku.edu.cn/JudgeOnline/problem?id=1588

题目大意:实现一个关于罗马数字的计算器，只包含+ - * /四种运算，每个操作数范围在1~4999。还有几种异常的处理

题目分析：这是一个简单的数据结构的模拟题，关键是怎样把阿拉伯数字转化我为罗马数字和怎么样把罗马数字转化为阿拉伯数字；

AC代码：

#include<cstdio> #include<string> #include<cmath> #include<algorithm> using namespace std; const int maxn = 10001; int value[13]={1000,900,500,400,100,90,50,40,10,9,5,4,1}; char rom[13][3]={"M","CM","D","CD","C","XC","L","XL","X","IX","V","IV","I"}; int stk[maxn]; char rd[20]; int GetValue(char c) { switch(c) { case 'I':return 1; case 'V':return 5; case 'X':return 10; case 'L':return 50; case 'C':return 100; case 'D':return 500; case 'M':return 1000; } } //把罗马数字转化为阿拉伯数字 int ConvertTonum(char *s) { int i; int len=strlen(s); if(len==1) return GetValue(s[0]); if(GetValue(s[0])<GetValue(s[1])) return ConvertTonum(s+1)-GetValue(s[0]); else return ConvertTonum(s+1)+GetValue(s[0]); } //把阿拉伯数字转化为罗马数字 void ConvertToRom( int num) { rd[0]='/0'; int i=0; while(num>0) { if(num>=value[i]) { strcat(rd,rom[i]); num-=value[i]; } else i++; } } int main() { //freopen("input.in","r",stdin); char in[20]; int top=-1; int a,b; while(scanf("%s",in)!=EOF) { if(in[0]=='+') { if(top>=1) { a=stk[top--]; b=stk[top--]; stk[++top]=a+b; } else printf("stack underflow/n"); } else if(in[0]=='-') { if(top>=1) { a=stk[top--]; b=stk[top--]; stk[++top]=b-a; } else printf("stack underflow/n"); } else if(in[0]=='*') { if(top>=1) { a=stk[top--]; b=stk[top--]; stk[++top]=a*b; } else printf("stack underflow/n"); } else if(in[0]=='/') { if(top>=1) { a=stk[top--]; b=stk[top--]; if(a==0) { printf("division by zero exception/n"); stk[++top]=b; continue; } else stk[++top]=b/a; } else printf("stack underflow/n"); } else if(in[0]=='=') { if(top<0) printf("stack underflow/n"); else { int tmp=stk[top]; if(tmp<=0||tmp>4999) printf("out of range exception/n"); else { ConvertToRom(tmp); printf("%s/n",rd); } } } else stk[++top]=ConvertTonum(in); } return 0; }