(高精度运算4.7.26)POJ 1220 NUMBER BASE CONVERSION(高精度数的任意进制的转换——方法:ba1----->10进制----->ba2)

本文链接：https://blog.youkuaiyun.com/hjd_love_zzt/article/details/13621851

本文介绍了一个Java程序，用于解决POJ 1220问题，涉及不同数制间的转换。该程序使用BigInteger类处理大整数运算，通过堆栈实现从一种数制到另一种数制的转换。

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package com.njupt.acm;

import java.math.BigInteger;
import java.util.Scanner;

public class POJ_1220_1 {

	public static void main(String[] args) {
		Scanner scanner = new Scanner(System.in);
		
		int t = scanner.nextInt();
		
		while(t > 0){
			BigInteger ba1 = scanner.nextBigInteger();
			BigInteger ba2 = scanner.nextBigInteger();
			BigInteger sum = new BigInteger("0");
			
			String str = scanner.next();
			
			int len = str.length();
			int i;
			int count = 0;
			int temp;
			for(i = len - 1; i >= 0 ; --i){//先将输入的ba1进制的数按照给定的规则转化成10进制的数
				char w = str.charAt(i);
				temp = 0;
				if(Character.isDigit(w)){
					temp = w - '0';
				}else if(Character.isLowerCase(w)){
					temp = w - 'a' + 36;
				}else if(Character.isUpperCase(w)){
					temp = w - 'A' + 10;
				}
				
				sum = sum.add(new BigInteger(temp + "").multiply(ba1.pow(count++)));
				
			}

			BigInteger zero = new BigInteger("0");
			int top = 0;
			int stack[] = new int[2000];
			while(sum.compareTo(zero) != 0){//转化成指定ba2进制的数
				stack[++top] = sum.mod(ba2).intValue();
				sum = sum.divide(ba2);
			}
			
			System.out.println(ba1+" "+str);
			System.out.print(ba2+" ");
			if(top == 0){
				System.out.print(0);
			}
			
			while(top != 0){
				char w = 0;
				temp = stack[top--];
				if(temp < 10){
					w = (char) (temp +'0');
				}else if(temp>= 10 && temp < 36){
					w = (char) (temp +'A' - 10);
				}else if(temp >= 36){
					w = (char) (temp + 'a' - 36);
				}
				
				System.out.print(w);
			}
			
			System.out.println();
			System.out.println();
			t--;
		}
		
	}
}