(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)...

xiaoxiao2021-03-01  13

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--; } } }
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