2.2.1 preface numbering

Http://acm.sdibt.edu.cn/JudgeOnline/problem.php? Id = 2325
Question: (such as question)
Input and Output: (such as the question)
Solution:
1. use the table-hitting method to convert each number N (1 <= n <3500) the number of "I", "V", "X", "L", "C", "D", and "M" counts, save it with a two-dimensional array CNT [3500] [7.
2. enumeration. Start from a thousand bits and start until a single bit. Check whether the remaining number is no less than 0 after the number is removed. If the value is not less than, continue, and vice versa.
3. Reduce the code.
(1) 10-digit conversion table to the number of Roman numerals:

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) character-to-array conversion table:

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

Core code:

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;      }     }    }   }  } }

Bad environment ...... Awesome ......

2.2.1 preface numbering