58. Analysis, testing and summary: Conversion of roman and Arabic numerals [roman to integer and integer to roman in c ++]

[Link to this article]

Http://www.cnblogs.com/hellogiser/p/roman-to-integer-and-integer-to-roman.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 + 10-2*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 = 1-5 = IV, 9 = 1-10 = IX, 40 = 10-50 = XL, 90 = 10-100 = XC, 400 = 100-500 = CD, 900 = 100-1000 = 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 = n-val [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

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// 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 <len-1; 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/roman-to-integer-and-integer-to-roman.html