http://acm.sdibt.edu.cn/JudgeOnline/problem.php?id=2325
题目大意:(如题)
输入输出:(如题)
解题思路:
1.用打表法将每个数N(1<=N<3500)中间“I”“V”“X”“L”“C”“D”“M”的个数统计出来,用一个二维数组cnt[3500][7]保存起来。
2.枚举。从千位开始枚举,一直枚举到个位为止,每次判断减掉那个数之后剩下的数是否还不小于0。如果不小于则继续,反之结束。
3.减小代码的方法。
(1) 10进制数到罗马数字的转换表:

string rec[4][9]={"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"};

(2) 字符到数组下标的转换表:

char res[7]={'I','V','X','L','C','D','M'};

核心代码:

for(mrk=1;mrk<3500;mrk++)
 {
  dat=mrk;
  for(i=3;i>=0;i--)
  {
   for(j=9;j>=1;j--)
   {
    tmp=pow((double)10,(double)i)*j;
    while(dat-tmp>=0)
    {
     dat-=tmp;
     for(k=0;k<rec[i][j-1].length();k++)
     {
      switch(rec[i][j-1][k])
      {
      case 'I':
       cnt[mrk][0]++;
       break;
      case 'V':
       cnt[mrk][1]++;
       break;
      case 'X':
       cnt[mrk][2]++;
       break;
      case 'L':
       cnt[mrk][3]++;
       break;
      case 'C':
       cnt[mrk][4]++;
       break;
      case 'D':
       cnt[mrk][5]++;
       break;
      case 'M':
       cnt[mrk][6]++;
       break;
      default:
       break;
      }
     }
    }
   }
  }
 }

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