POJ-1220 NUMBER BASE CONVERSION（大整数进制转换）

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

---

#include <iostream>
#include <string>
#include <algorithm>

using namespace std;

int charToNum(char ch)
{
    if ('0' <= ch && ch <= '9') return ch - '0';
    if ('A' <= ch && ch <= 'Z') return ch - 'A' + 10;

    return ch - 'a' + 36;
}

char numToChar(int n)
{
    if (0 <= n && n <= 9) return '0' + n;
    if (10 <= n && n <= 35) return 'A' + (n - 10);

    return 'a' + (n - 36);
}

inline bool isAllZero(string & s)
{
    return s.size() == 0 ? true : false;
}

string conversion(string & sequence, int sourceBase, int targetBase)
{
    if (sequence == "0" || sequence == "1") return sequence;
    string res;
    string quotient = sequence; //商数

    while (!isAllZero(quotient)) {
        int extra = 0;
        string tmpStr;
        for (size_t i = 0; i < quotient.size(); ++i) {
            int tmp = extra * sourceBase + charToNum(quotient[i]);
            tmpStr.push_back(numToChar(tmp / targetBase));
            extra = tmp % targetBase;
        }
        res.push_back(numToChar(extra)); //进制转换中的余数

        size_t pos = 0;
        while (tmpStr[pos] == '0') ++pos;
        quotient = tmpStr.substr(pos);
    }
    reverse(res.begin(), res.end());

    return res;
}

int main()
{
    std::ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);

    int caseNum;
    cin >> caseNum;

    while (caseNum--) {
        int sourceBase, targetBase;
        string sequence;
        cin >> sourceBase >> targetBase >> sequence;

        cout << sourceBase << " " << sequence << endl;
        cout << targetBase << " " << conversion(sequence, sourceBase, targetBase) << endl;
        if (caseNum != 0) cout << endl; 
    }

    return 0;
}

思路就以上面图片为例子，假如将十进制的6转为2进制，利用的是短除法。和这张图略有区别的是，应该在最后的1再进行一次除法，这样被除数就变为了0。程序就是去模拟这个过程，用变量quotient代表每次的被除数，用extra记录每一次除法的余数，并放在res里保存，因为要进行无法预知的多轮除法，所以采用while循环，终止条件就是被除数为0的时候，因为输入的数据是大整数，用string容器来存储，所以被除数为零意味着其长度为0。