给定一个数字，找出对应的罗马数字。 例如：
null
Input : 9Output : IXInput : 40Output : XLInput : 1904Output : MCMIV
以下是罗马符号列表，其中也包括减法格：
SYMBOL VALUEI 1IV 4V 5IX 9X 10XL 40L 50XC 90C 100CD 400D 500CM 900 M 1000
这个想法是将给定数字的单位、十位、百位和千位分别转换。如果数字为0，则没有相应的罗马数字符号。数字4和9的转换与其他数字略有不同，因为这些数字紧跟其后 减法记数法 .
十进制数到罗马数字的转换算法 按1000、900、500、400、100、90、50、40、10、9、5、4、1的顺序将给定的数字与基值进行比较。小于或等于给定数字的基值将是初始基值（最大基值）。将数字除以其最大的基值，相应的基符号将被重复商倍，剩余的将成为未来除法和重复的数字。该过程将重复，直到数字变为零。
演示上述算法的示例：
Convert 3549 to its Roman Numerals
输出：
MMMDXLIX
说明：
说明：
第一步
- 初始数量=3549
- 因为3549>=1000；最初最大基值为1000。
- 除以3549/1000。商=3，余数=549。对应的符号 M 将重复三次。
- 我们将结果值附加到第二个列表中。
- 现在余数不等于0，所以我们再次调用该函数。
第二步
- 现在，数字=549
- 1000 > 549 >= 500 ; 最大基值为500。
- 除以549/500。商=1，余数=49。对应的符号 D 将重复一次并停止循环。
- 我们将结果值附加到第二个列表中。
- 现在余数不等于0，所以我们再次调用该函数。
第三步
- 现在，数字=49
- 50 > 49 >= 40 ; 最大基值为40。
- 除以49/40。商=1，余数=9。对应的符号 特大号 将重复一次并停止循环。
- 我们将结果值附加到第二个列表中。
- 现在余数不等于0，所以我们再次调用该函数。
第四步
- 现在，数字=9
- 数字9出现在列表ls中，因此我们直接从dictionary dict中获取值，并将余数设置为0&停止循环。
- 余数=0。对应的符号 九 将重复一次，现在余数值为0，因此我们不会再次调用该函数。
第五步
- 最后，我们加入第二个列表值。
- 获得的输出 MMMDXLIX。
以下是上述算法的实现：
C++
// C++ Program to convert decimal number to // roman numerals #include <bits/stdc++.h> using namespace std; // Function to convert decimal to Roman Numerals int printRoman( int number) { int num[] = {1,4,5,9,10,40,50,90,100,400,500,900,1000}; string sym[] = { "I" , "IV" , "V" , "IX" , "X" , "XL" , "L" , "XC" , "C" , "CD" , "D" , "CM" , "M" }; int i=12; while (number>0) { int div = number/num[i]; number = number%num[i]; while ( div --) { cout<<sym[i]; } i--; } } //Driver program int main() { int number = 3549; printRoman(number); return 0; } |
JAVA
// Java Program to convert decimal number to // roman numerals class GFG { // To add corresponding base symbols in the array // to handle cases that follow subtractive notation. // Base symbols are added index 'i'. static int sub_digit( char num1, char num2, int i, char [] c) { c[i++] = num1; c[i++] = num2; return i; } // To add symbol 'ch' n times after index i in c[] static int digit( char ch, int n, int i, char [] c) { for ( int j = 0 ; j < n; j++) { c[i++] = ch; } return i; } // Function to convert decimal to Roman Numerals static void printRoman( int number) { char c[] = new char [ 10001 ]; int i = 0 ; // If number entered is not valid if (number <= 0 ) { System.out.printf( "Invalid number" ); return ; } // TO convert decimal number to roman numerals while (number != 0 ) { // If base value of number is greater than 1000 if (number >= 1000 ) { // Add 'M' number/1000 times after index i i = digit( 'M' , number / 1000 , i, c); number = number % 1000 ; } // If base value of number is greater than or // equal to 500 else if (number >= 500 ) { // To add base symbol to the character array if (number < 900 ) { // Add 'D' number/1000 times after index i i = digit( 'D' , number / 500 , i, c); number = number % 500 ; } // To handle subtractive notation in case of number // having digit as 9 and adding corresponding base // symbol else { // Add C and M after index i/. i = sub_digit( 'C' , 'M' , i, c); number = number % 100 ; } } // If base value of number is greater than or equal to 100 else if (number >= 100 ) { // To add base symbol to the character array if (number < 400 ) { i = digit( 'C' , number / 100 , i, c); number = number % 100 ; } // To handle subtractive notation in case of number // having digit as 4 and adding corresponding base // symbol else { i = sub_digit( 'C' , 'D' , i, c); number = number % 100 ; } } // If base value of number is greater than or equal to 50 else if (number >= 50 ) { // To add base symbol to the character array if (number < 90 ) { i = digit( 'L' , number / 50 , i, c); number = number % 50 ; } // To handle subtractive notation in case of number // having digit as 9 and adding corresponding base // symbol else { i = sub_digit( 'X' , 'C' , i, c); number = number % 10 ; } } // If base value of number is greater than or equal to 10 else if (number >= 10 ) { // To add base symbol to the character array if (number < 40 ) { i = digit( 'X' , number / 10 , i, c); number = number % 10 ; } // To handle subtractive notation in case of // number having digit as 4 and adding // corresponding base symbol else { i = sub_digit( 'X' , 'L' , i, c); number = number % 10 ; } } // If base value of number is greater than or equal to 5 else if (number >= 5 ) { if (number < 9 ) { i = digit( 'V' , number / 5 , i, c); number = number % 5 ; } // To handle subtractive notation in case of number // having digit as 9 and adding corresponding base // symbol else { i = sub_digit( 'I' , 'X' , i, c); number = 0 ; } } // If base value of number is greater than or equal to 1 else if (number >= 1 ) { if (number < 4 ) { i = digit( 'I' , number, i, c); number = 0 ; } // To handle subtractive notation in case of // number having digit as 4 and adding corresponding // base symbol else { i = sub_digit( 'I' , 'V' , i, c); number = 0 ; } } } // Printing equivalent Roman Numeral System.out.printf( "Roman numeral is: " ); for ( int j = 0 ; j < i; j++) { System.out.printf( "%c" , c[j]); } } //Driver program public static void main(String[] args) { int number = 3549 ; printRoman(number); } } // This code is contributed by PrinciRaj1992 |
Python3
# Python3 program to convert # decimal number to roman numerals ls = [ 1000 , 900 , 500 , 400 , 100 , 90 , 50 , 40 , 10 , 9 , 5 , 4 , 1 ] dict = { 1 : "I" , 4 : "IV" , 5 : "V" , 9 : "IX" , 10 : "X" , 40 : "XL" , 50 : "L" , 90 : "XC" , 100 : "C" , 400 : "CD" , 500 : "D" , 900 : "CM" , 1000 : "M" } ls2 = [] # Function to convert decimal to Roman Numerals def func(no,res): for i in range ( 0 , len (ls)): if no in ls: res = dict [no] rem = 0 break if ls[i]<no: quo = no / / ls[i] rem = no % ls[i] res = res + dict [ls[i]] * quo break ls2.append(res) if rem = = 0 : pass else : func(rem,"") # Driver code if __name__ = = "__main__" : func( 3549 , "") print ("".join(ls2)) # This code is contributed by # VIKAS CHOUDHARY(vikaschoudhary344) |
C#
// C# Program to convert decimal number // to roman numerals using System; class GFG { // To add corresponding base symbols in the // array to handle cases which follow subtractive // notation. Base symbols are added index 'i'. static int sub_digit( char num1, char num2, int i, char [] c) { c[i++] = num1; c[i++] = num2; return i; } // To add symbol 'ch' n times after index i in c[] static int digit( char ch, int n, int i, char [] c) { for ( int j = 0; j < n; j++) { c[i++] = ch; } return i; } // Function to convert decimal to Roman Numerals static void printRoman( int number) { char [] c = new char [10001]; int i = 0; // If number entered is not valid if (number <= 0) { Console.WriteLine( "Invalid number" ); return ; } // TO convert decimal number to // roman numerals while (number != 0) { // If base value of number is // greater than 1000 if (number >= 1000) { // Add 'M' number/1000 times after index i i = digit( 'M' , number / 1000, i, c); number = number % 1000; } // If base value of number is greater // than or equal to 500 else if (number >= 500) { // To add base symbol to the character array if (number < 900) { // Add 'D' number/1000 times after index i i = digit( 'D' , number / 500, i, c); number = number % 500; } // To handle subtractive notation in case // of number having digit as 9 and adding // corresponding base symbol else { // Add C and M after index i/. i = sub_digit( 'C' , 'M' , i, c); number = number % 100; } } // If base value of number is greater // than or equal to 100 else if (number >= 100) { // To add base symbol to the character array if (number < 400) { i = digit( 'C' , number / 100, i, c); number = number % 100; } // To handle subtractive notation in case // of number having digit as 4 and adding // corresponding base symbol else { i = sub_digit( 'C' , 'D' , i, c); number = number % 100; } } // If base value of number is greater // than or equal to 50 else if (number >= 50) { // To add base symbol to the character array if (number < 90) { i = digit( 'L' , number / 50, i, c); number = number % 50; } // To handle subtractive notation in case // of number having digit as 9 and adding // corresponding base symbol else { i = sub_digit( 'X' , 'C' , i, c); number = number % 10; } } // If base value of number is greater // than or equal to 10 else if (number >= 10) { // To add base symbol to the character array if (number < 40) { i = digit( 'X' , number / 10, i, c); number = number % 10; } // To handle subtractive notation in case of // number having digit as 4 and adding // corresponding base symbol else { i = sub_digit( 'X' , 'L' , i, c); number = number % 10; } } // If base value of number is greater // than or equal to 5 else if (number >= 5) { if (number < 9) { i = digit( 'V' , number / 5, i, c); number = number % 5; } // To handle subtractive notation in case // of number having digit as 9 and adding // corresponding base symbol else { i = sub_digit( 'I' , 'X' , i, c); number = 0; } } // If base value of number is greater // than or equal to 1 else if (number >= 1) { if (number < 4) { i = digit( 'I' , number, i, c); number = 0; } // To handle subtractive notation in // case of number having digit as 4 // and adding corresponding base symbol else { i = sub_digit( 'I' , 'V' , i, c); number = 0; } } } // Printing equivalent Roman Numeral Console.WriteLine( "Roman numeral is: " ); for ( int j = 0; j < i; j++) { Console.Write( "{0}" , c[j]); } } // Driver Code public static void Main() { int number = 3549; printRoman(number); } } // This code is contributed by Rajput-Ji |
Javascript
<script> // JavaScript Program to convert decimal number to // roman numerals // Function to convert decimal to Roman Numerals function printRoman(number) { let num = [1,4,5,9,10,40,50,90,100,400,500,900,1000]; let sym = [ "I" , "IV" , "V" , "IX" , "X" , "XL" , "L" , "XC" , "C" , "CD" , "D" , "CM" , "M" ]; let i=12; while (number>0) { let div = Math.floor(number/num[i]); number = number%num[i]; while (div--) { document.write(sym[i]); } i--; } } //Driver program let number = 3549; printRoman(number); //This code is contributed by Manoj </script> |
输出
MMMDXLIX
参考资料： http://blog.functionalfun.net/2009/01/project-euler-89-converting-to-and-from.html
另一种方法1： : 在这种方法中，我们必须首先观察问题。问题陈述中给出的数字最多可以是4位。解决这个问题的想法是：
- 将给定的数字在不同的位置分成数字，如1、2、100或1000。
- 从千位开始打印相应的罗马值。例如，如果千位的数字是3，那么打印3000的罗马对应数字。
- 重复第二步，直到到达目的地。
实例 : 假设输入号码是3549。所以，从千禧开始，我们将开始印刷罗马版。在这种情况下，我们将按以下顺序打印： 第一 ：相当于3000的罗马货币 第二 ：相当于500的罗马货币 第三 ：罗马相当于40 第四 ：罗马相当于9 因此，输出将是： MMMDXLIX
下面是上述想法的实现。
C++
// C++ Program for above approach #include <bits/stdc++.h> using namespace std; // Function to calculate roman equivalent string intToRoman( int num) { // storing roman values of digits from 0-9 // when placed at different places string m[] = { "" , "M" , "MM" , "MMM" }; string c[] = { "" , "C" , "CC" , "CCC" , "CD" , "D" , "DC" , "DCC" , "DCCC" , "CM" }; string x[] = { "" , "X" , "XX" , "XXX" , "XL" , "L" , "LX" , "LXX" , "LXXX" , "XC" }; string i[] = { "" , "I" , "II" , "III" , "IV" , "V" , "VI" , "VII" , "VIII" , "IX" }; // Converting to roman string thousands = m[num / 1000]; string hundreds = c[(num % 1000) / 100]; string tens = x[(num % 100) / 10]; string ones = i[num % 10]; string ans = thousands + hundreds + tens + ones; return ans; } // Driver program to test above function int main() { int number = 3549; cout << intToRoman(number); return 0; } |
JAVA
// Java Program for above approach class GFG { // Function to calculate roman equivalent static String intToRoman( int num) { // storing roman values of digits from 0-9 // when placed at different places String m[] = { "" , "M" , "MM" , "MMM" }; String c[] = { "" , "C" , "CC" , "CCC" , "CD" , "D" , "DC" , "DCC" , "DCCC" , "CM" }; String x[] = { "" , "X" , "XX" , "XXX" , "XL" , "L" , "LX" , "LXX" , "LXXX" , "XC" }; String i[] = { "" , "I" , "II" , "III" , "IV" , "V" , "VI" , "VII" , "VIII" , "IX" }; // Converting to roman String thousands = m[num / 1000 ]; String hundreds = c[(num % 1000 ) / 100 ]; String tens = x[(num % 100 ) / 10 ]; String ones = i[num % 10 ]; String ans = thousands + hundreds + tens + ones; return ans; } // Driver program to test above function public static void main(String[] args) { int number = 3549 ; System.out.println(intToRoman(number)); } } |
Python3
# Python3 program for above approach # Function to calculate roman equivalent def intToRoman(num): # Storing roman values of digits from 0-9 # when placed at different places m = [" ", " M ", " MM ", " MMM"] c = [" ", " C ", " CC ", " CCC ", " CD ", " D", "DC" , "DCC" , "DCCC" , "CM " ] x = [" ", " X ", " XX ", " XXX ", " XL ", " L", "LX" , "LXX" , "LXXX" , "XC" ] i = [" ", " I ", " II ", " III ", " IV ", " V", "VI" , "VII" , "VIII" , "IX" ] # Converting to roman thousands = m[num / / 1000 ] hundreds = c[(num % 1000 ) / / 100 ] tens = x[(num % 100 ) / / 10 ] ones = i[num % 10 ] ans = (thousands + hundreds + tens + ones) return ans # Driver code if __name__ = = "__main__" : number = 3549 print (intToRoman(number)) # This code is contributed by rutvik_56 |
C#
// C# Program for above approach using System; class GFG { // Function to calculate roman equivalent static String intToRoman( int num) { // storing roman values of digits from 0-9 // when placed at different places String[] m = { "" , "M" , "MM" , "MMM" }; String[] c = { "" , "C" , "CC" , "CCC" , "CD" , "D" , "DC" , "DCC" , "DCCC" , "CM" }; String[] x = { "" , "X" , "XX" , "XXX" , "XL" , "L" , "LX" , "LXX" , "LXXX" , "XC" }; String[] i = { "" , "I" , "II" , "III" , "IV" , "V" , "VI" , "VII" , "VIII" , "IX" }; // Converting to roman String thousands = m[num / 1000]; String hundreds = c[(num % 1000) / 100]; String tens = x[(num % 100) / 10]; String ones = i[num % 10]; String ans = thousands + hundreds + tens + ones; return ans; } // Driver program to test above function public static void Main() { int number = 3549; Console.WriteLine(intToRoman(number)); } } |
PHP
<?php // PHP Program for above approach // Function to calculate roman equivalent function intToRoman( $num ) { // storing roman values of digits from 0-9 // when placed at different places $m = array ( "" , "M" , "MM" , "MMM" ); $c = array ( "" , "C" , "CC" , "CCC" , "CD" , "D" , "DC" , "DCC" , "DCCC" , "CM" ); $x = array ( "" , "X" , "XX" , "XXX" , "XL" , "L" , "LX" , "LXX" , "LXXX" , "XC" ); $i = array ( "" , "I" , "II" , "III" , "IV" , "V" , "VI" , "VII" , "VIII" , "IX" ); // Converting to roman $thousands = $m [ $num / 1000]; $hundreds = $c [( $num % 1000) / 100]; $tens = $x [( $num % 100) / 10]; $ones = $i [ $num % 10]; $ans = $thousands . $hundreds . $tens . $ones ; return $ans ; } // Driver Code $number = 3549; echo intToRoman( $number ); // This code is contributed by Akanksha Rai |
Javascript
<script> // JavaScript Program for above approach // Function to calculate roman equivalent function intToRoman(num) { // storing roman values of digits from 0-9 // when placed at different places let m = [ "" , "M" , "MM" , "MMM" ]; let c = [ "" , "C" , "CC" , "CCC" , "CD" , "D" , "DC" , "DCC" , "DCCC" , "CM" ]; let x = [ "" , "X" , "XX" , "XXX" , "XL" , "L" , "LX" , "LXX" , "LXXX" , "XC" ]; let i = [ "" , "I" , "II" , "III" , "IV" , "V" , "VI" , "VII" , "VIII" , "IX" ]; // Converting to roman let thousands = m[Math.floor(num/1000)]; let hundreds = c[Math.floor((num%1000)/100)]; let tens = x[Math.floor((num%100)/10)]; let ones = i[num%10]; let ans = thousands + hundreds + tens + ones; return ans; } // Driver program to test above function let number = 3549; document.write(intToRoman(number)); //This code is contributed by Mayank Tyagi </script> |
输出
MMMDXLIX
幸亏 沙什瓦特·贾因 用于提供上述解决方案方法。
另一种方法2： : 在这种方法中，我们考虑数字中的主要有效数字。例：在1234中，主要有效数字是1。类似地，在345中是3。 为了提取出主有效位，我们需要保持一个除数（我们称之为div），比如1000代表1234（因为1234/1000=1），100代表345（345/100=3）。 另外，让我们维护一个名为RomanDimetric={1:’I’，5:’V’，10:’X’，50:’L’，100:’C’，500:’D’，1000:’M’}的字典
以下是算法：
如果主有效位<=3
- 罗马数字[div]*main有效数字
如果主有效位==4
- 罗马数字[div]+罗马数字[div*5]
如果5<=主有效位<=8
- 罗马数字[div*5]+（罗马数字[div]*（主要有效数字-5））
如果主有效位==3
- 罗马数字[div]+罗马数字[div*10]
实例 : 假设输入的数字是3649。
Iter 1
- 初始数量=3649
- 主要有效数字为3。Div=1000。
- 所以，罗马数字[1000]*3
- 给：嗯
Iter 2
- 现在，数字=649
- 主要有效数字是6。Div=100。
- 所以罗马数字[100*5]+（罗马数字[div]*（6-5））
- 给：DC
Iter 3
- 现在，数字=49
- 主要有效数字为4。Div=10。
- 所以罗马数字[10]+罗马数字[10*5]
- 给出：XL
Iter 4
- 现在，数字=9
- 主要有效数字是9。Div=1。
- 所以罗马数字[1]*罗马数字[1*10]
- 给出：IX
通过使用上述所有工具获得最终结果：MMMDCXLIX
下面是上述想法的Python实现。
Python3
# Python 3 program to convert Decimal # number to Roman numbers. import math def integerToRoman(A): romansDict = { 1 : "I" , 5 : "V" , 10 : "X" , 50 : "L" , 100 : "C" , 500 : "D" , 1000 : "M" , 5000 : "G" , 10000 : "H" } div = 1 while A > = div: div * = 10 div / / = 10 res = "" while A: # main significant digit extracted # into lastNum lastNum = (A / / div) if lastNum < = 3 : res + = (romansDict[div] * lastNum) elif lastNum = = 4 : res + = (romansDict[div] + romansDict[div * 5 ]) elif 5 < = lastNum < = 8 : res + = (romansDict[div * 5 ] + (romansDict[div] * (lastNum - 5 ))) elif lastNum = = 9 : res + = (romansDict[div] + romansDict[div * 10 ]) A = math.floor(A % div) div / / = 10 return res # Driver code print ( "Roman Numeral of Integer is:" + str (integerToRoman( 3549 ))) |
输出
Roman Numeral of Integer is:MMMDXLIX
幸亏 斯利哈沙·桑梅塔 用于提供上述解决方案方法。 本文由 拉胡尔·阿格拉瓦尔 .如果你喜欢GeekSforgek，并想贡献自己的力量，你也可以使用 写极客。组织 或者把你的文章寄去评论-team@geeksforgeeks.org.看到你的文章出现在Geeksforgeks主页上，并帮助其他极客。 如果您发现任何不正确的地方，或者您想分享有关上述主题的更多信息，请写下评论。
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