POJ 1220 NUMBER BASE CONVERSION-优快云博客

本文介绍了一种实现不同进制间数字转换的算法，并通过一个具体的C++程序示例进行了解释。该算法能够处理从2到62进制之间的转换，支持包括0-9、A-Z及a-z在内的62种不同的数字表示方式。

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NUMBER BASE CONVERSION

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 2470		Accepted: 1055

Description

Write a program to convert numbers in one base to numbers in a second base. There are 62 different digits:
{ 0-9,A-Z,a-z }
HINT: If you make a sequence of base conversions using the output of one conversion as the input to the next, when you get back to the original base, you should get the original number.

Input

The first line of input contains a single positive integer. This is the number of lines that follow. Each of the following lines will have a (decimal) input base followed by a (decimal) output base followed by a number expressed in the input base. Both the input base and the output base will be in the range from 2 to 62. That is (in decimal) A = 10, B = 11, ..., Z = 35, a = 36, b = 37, ..., z = 61 (0-9 have their usual meanings).

Output

The output of the program should consist of three lines of output for each base conversion performed. The first line should be the input base in decimal followed by a space then the input number (as given expressed in the input base). The second output line should be the output base followed by a space then the input number (as expressed in the output base). The third output line is blank.

Sample Input

8
62 2 abcdefghiz
10 16 1234567890123456789012345678901234567890
16 35 3A0C92075C0DBF3B8ACBC5F96CE3F0AD2
35 23 333YMHOUE8JPLT7OX6K9FYCQ8A
23 49 946B9AA02MI37E3D3MMJ4G7BL2F05
49 61 1VbDkSIMJL3JjRgAdlUfcaWj
61 5 dl9MDSWqwHjDnToKcsWE1S
5 10 42104444441001414401221302402201233340311104212022133030

Sample Output

62 abcdefghiz
2 11011100000100010111110010010110011111001001100011010010001

10 1234567890123456789012345678901234567890
16 3A0C92075C0DBF3B8ACBC5F96CE3F0AD2

16 3A0C92075C0DBF3B8ACBC5F96CE3F0AD2
35 333YMHOUE8JPLT7OX6K9FYCQ8A

35 333YMHOUE8JPLT7OX6K9FYCQ8A
23 946B9AA02MI37E3D3MMJ4G7BL2F05

23 946B9AA02MI37E3D3MMJ4G7BL2F05
49 1VbDkSIMJL3JjRgAdlUfcaWj

49 1VbDkSIMJL3JjRgAdlUfcaWj
61 dl9MDSWqwHjDnToKcsWE1S

61 dl9MDSWqwHjDnToKcsWE1S
5 42104444441001414401221302402201233340311104212022133030

5 42104444441001414401221302402201233340311104212022133030
10 1234567890123456789012345678901234567890

Source

Greater New York 2002

/* 进制之间的任意转换其实就是循环做除法，每趟循环除法得到的余数作为结果的当前最高位 */ #include <iostream> #include <string> using namespace std; string input, ouput; char index[62] = { '0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z', 'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z'}; int getV(char c) { if(c >= '0' && c <= '9') return c - '0'; else if(c >= 'A' && c <= 'Z') return c - 'A' + 10; else return c - 'a' + 36; } string change(int f, int t, string input) { string res = ""; int last = 0, quo, cur, residual, pos = 0; string quotient; while(input.length() != 0) { last = 0; cur = 0; for(pos = 0; pos < input.length(); pos++) { if((cur = last * f + getV(input[pos])) < t) { quotient = quotient + '0'; last = cur; continue; } else { quo = cur / t; quotient = quotient + index[quo]; last = cur % t; } } res = index[last] + res; while(quotient.length() >= 1 && quotient[0] == '0') quotient = quotient.substr(1, quotient.length() - 1); input = quotient; quotient = ""; } return res; } int main() { int caseNum, c, from, to; cin>>caseNum; for(c = 1; c <= caseNum; c++) { cin>>from>>to; cin>>input; cout<<from<<" "<<input<<endl; cout<<to<<" "<<change(from, to, input)<<endl<<endl; } return 0; }