(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)...

系统管理员 2022-05-12 23:10 123阅读 0赞

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--;
    		}
    		
    	}
    }