[Link to this article]
Http://www.cnblogs.com/hellogiser/p/romantointegerandintegertoroman.html
【Question]
Returns a roman number and converts it to an Arabic number. This question only considers the number less than 3999.
The Roman numerals have the following symbols:
I (1) V (5) x (10) L (50) C (100) D (500) M (1000)
Counting rules:
(1). Several numbers in the same number are the sum of these Roman numerals, for example, III = 3;
(2) A small number represents a number before a large number, that is, a large number minus a small number, for example, IV = 4;
(3). A small number after a large number represents a large number plus a small number, such as VI = 6;
Combination rules:
(1) No more than three of the basic numbers I, X, and C can be used as the number or placed on the right of the large number; you can only use one value to the left of a large number.
(2) do not place any of the basic numbers V, L, and D as decimal places on the left of a large number and subtract them to form the number. add them to the right of a large number to form the number, only one instance can be used.
(3) the small numbers on the left of V and X can only be I.
(4) numbers on the left of L and C can only be X.
(5) the small numbers on the left side of D and M can only use C.
【Analysis]
(1) Conversion of Roman numerals to Arabic numerals:
Traverse the roman numerals from the beginning to the end. If a number is smaller than the previous one, add the number to the result. Otherwise, subtract the previous number twice in the result and add the current number;
For example, how does one obtain XVIII = 18? The corresponding Arabic number is 10_5_1_1_1, so the result is 10 + 5 + 1 + 1 + 1 = 18;
How do I get XIX = 19? The corresponding Arabic number is 10_000010, so the result is 10 + 1 + 102*1 = 19.
(2) Arabic numerals to Roman numerals:
Use the combination of all small numbers as the basic number to create a corresponding value ing table.
For example, 4 = 15 = IV, 9 = 110 = IX, 40 = 1050 = XL, 90 = 10100 = XC, 400 = 100500 = CD, 900 = 1001000 = CM.
The corresponding ing is as follows:
Unsigned int val [] = {1000,900,500,400,100, 90, 50, 9, 5 };
String r [] = {"M", "CM", "D", "CD", "C", "XC", "L", "XL ", "X", "IX", "V", "IV", "I "};
For the Arabic number n, traverse the val array. If n> = val [I], the result is retained by r [I] and n = nval [I], until n = 0.
Test]
Given a number n, use the integer2raman function to convert it to a number r, and then use the roman2integer function to convert r to m. If n! = M, it indicates that the function is faulty. If they are equal, the function is correct.
[Code]
C ++ Code
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 

// Roman2Integer. cpp: Defines the entry point for the console application. // /* Version: 1.0 Author: hellogiser Blog: http://www.cnblogs.com/hellogiser Date: 2014/5/26 */
# Include "stdafx. h" # Include <string> # Include <iostream> # Include <map> # Include <assert. h> Using namespace std;
// Roman to integer Unsigned int roman2integer (string str) { // 99 > 10,100, 1, 10 // 66 > 50, 10, 5, 1 If (str = "") Return 0; Map <char, int> m; M ['I'] = 1; M ['V'] = 5; M ['X'] = 10; M ['l'] = 50; M ['C'] = 100; M ['d] = 500; M ['M'] = 1000;
Int sum = m [str [0]; Int len = str. length (); For (int I = 0; I <len1; I ++) { If (m [str [I]> = m [str [I + 1]) { // M [I]> = m [I + 1], then add m [I + 1] to sum Sum = sum + m [str [I + 1]; } Else { // M [I] <m [I + 1], then add m [I + 1] to sum, and remove 2 * m [I] Sum = sum + m [str [I + 1]2 * m [str [I]; } } Return sum; }
# Deprecision MAX 3999
// Integer to roman String integer2roman (unsigned int n) { // We shoshould consider 400,900, 40, 90 String result = ""; If (n <1  n> MAX) Return result;
Unsigned int val [] ={ 1000,900,500,400,100, 90, 50, 40, 10, 9, 5, 4, 1 }; Unsigned int length = sizeof (val)/sizeof (int ); String r [] = {"M", "CM", "D", "CD", "C", "XC", "L", "XL ", "X", "IX", "V", "IV", "I "};
For (int I = 0; I <length; I ++) { While (n> = val [I]) { Result + = r [I]; N= val [I]; } } Return result; }
// Test case for two functions Void test_case (int n) { For (int I = 1; I <= n; I ++) { String roman = integer2roman (I ); Int integer = roman2integer (roman ); Assert (I = integer ); } }
Void test_main () { // Test_case (20 ); Test_case (MAX ); }
Int _ tmain (int argc, _ TCHAR * argv []) { Test_main (); Return 0; } 
The implementation of roman2integer and integer2roman is given above, and the function is tested. For the number n between 1 and 3999, obtain the corresponding Roman number as r, and then convert r to the Arabic number m, then n should be equal to m. Therefore, the assert (I = integer); statement in test_case can run normally without throwing an exception.
[Reference]
Http://www.cnblogs.com/dosxp/archive/2008/08/13/1266781.html
Http://blog.csdn.net/wzy_1988/article/details/17057929
Http://blog.csdn.net/fightforyourdream/article/details/12934139
[Link to this article]
Http://www.cnblogs.com/hellogiser/p/romantointegerandintegertoroman.html