Implement a 10-to-2 or 16-to-16 conversion.

1 public class test {2 3 Public static void main (string [] ARGs) {4 system. out. println (tobinary (9); 5 system. out. println (tohex (559); 6 system. out. println (integer. tobinarystring (9); 7 system. out. println (integer. tohexstring (559); 8/* 1001 9 22f10 100111 22f */12} 13 14 public static string tobinary (int n) {15 stringbuilder temp = new stringbuilder (); 16 while (n/2> = 1 | n % 2 = 1) {// except for 2, (| n % 2 = 1) in order to add the last 0 or 117 temp. append (N % 2); 18 N = n/2; 19} 20 return temp. reverse (). tostring (); 21} 22/** 23 10 to hexadecimal: 24 divide the given decimal integer by the base number 16, and the remainder is the equivalent hexadecimal decimal digit. 25. Divide the quotient in the previous step by the base number 16, and the remainder is the low value of the equivalent hexadecimal number. 26. Repeat the previous step until the final operator equals 0. The remainder of each division is the number of users in hexadecimal notation. The remainder of the last Division is the highest bit 27 */28 public static string tohex (int n) {29: except 16 get remainder */30 stringbuilder temp = new stringbuilder (); 31 While (N/16> = 1) {32 int AA = N/16; 33 int BB = n % 16; 34 // 0123456789 10 11 12 13 14 1535 // 0123456789 a B c d e f36 string STR = ""; 37 If (BB = 10) {38 STR = "A"; 39} else if (BB = 11) {40 STR = "B "; 41} else if (BB = 12) {42 STR = "C"; 43} else if (BB = 13) {44 STR = "D "; 45} else if (BB = 14) {46 STR = "e"; 47} else if (BB = 15) {48 STR = "F "; 49} else {50 STR = BB + ""; 51} 52 temp. append (STR); 53 n = AA; 54 if (N/16 <1) {// Add the last digit 55 temp. append (n); 56} 57} 58 return temp. reverse (). tostring (); 59} 60}

 

Implement a 10-to-2 or 16-to-16 conversion.