大数进制转换 poj1220

普通的做法，大数除小数。

复杂度o( log(n)*log(n) ),其实就是位数的平方。

NUMBER BASE CONVERSION

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 4913		Accepted: 2246

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

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <stdlib.h>
#include <math.h>
using namespace std;

char str[];
int num[];
int ans[];

int chg(char c)
{
    if( c>=''&&c<='' ) return c-'';
    if( c>='A'&&c<='Z' ) return c-'A'+;
    if( c>='a'&&c<='z' ) return c-'a'+;
    return -;
}

char unchg(int x)
{
    if(x>=&&x<=) return x+'';
    if(x>=&&x<=) return 'A'+x-;
    else return 'a'+x-;
}

int main()
{
    int T;
    cin>>T;
    while(T--)
    {
        int from,to;
        scanf("%d%d",&from,&to);
        scanf("%s",str);
        int len=strlen(str);
        for(int i=;i<len;i++)
        {
            num[i] = chg(str[i]);
        }
        int cnt=;
        int wei=;
        while( cnt < len )
        {
            //然后做一次除法
            for(int i=cnt;i<len;i++)
            {
                num[ i+ ] += (num[i]%to)*from;
                num[ i ] /= to;
            }
            ans[ wei++ ] = num[len]/from;
            num[len]=;
            for(int i=cnt;i<len;i++)
            {
                if(num[i]==) cnt++;
                else break;
            }
        }
        printf("%d %s\n",from,str);
        printf("%d ",to);
        for(int i=wei-;i>=;i--)
        {
            printf("%c",unchg(ans[i]));
        }
        printf("\n\n");
    }
    return ;
}