[PAT] 1010 Radix （25 分）Java

Given a pair of positive integers, for example, 6 and 110, can this equation 6 = 110 be true? The answer is yes, if 6 is a decimal number and 110 is a binary number.

Now for any pair of positive integers N​1​​ and N​2​​, your task is to find the radix of one number while that of the other is given.

Input Specification:

Each input file contains one test case. Each case occupies a line which contains 4 positive integers:

N1 N2 tag radix

Here N1 and N2 each has no more than 10 digits. A digit is less than its radix and is chosen from the set { 0-9, a-z } where 0-9 represent the decimal numbers 0-9, and a-z represent the decimal numbers 10-35. The last number radix is the radix of N1 if tag is 1, or of N2 if tag is 2.

Output Specification:

For each test case, print in one line the radix of the other number so that the equation N1 = N2 is true. If the equation is impossible, print Impossible. If the solution is not unique, output the smallest possible radix.

Sample Input 1:

6 110 1 10

Sample Output 1:

2

Sample Input 2:

1 ab 1 2

Sample Output 2:

Impossible

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.math.BigInteger;

/**
 * @Auther: Xingzheng Wang
 * @Date: 2019/2/18 23:08
 * @Description: pattest
 * @Version: 1.0
 */
public class PAT1010 {
    public static void main(String[] args) throws IOException {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        String[] split = reader.readLine().split(" ");
        String v1 = split[0], v2 = split[1];
        String know_v = v1, unknow_v = v2;
        BigInteger know_radix = BigInteger.valueOf(Integer.valueOf(split[3]));
        if ("2".equals(split[2])) {
            know_v = v2;
            unknow_v = v1;
        }
        //期望值转化为十进制 相当于把输入转化为 N1 N2 1 10
        BigInteger expect = convert2ten(know_v, know_radix);
        String s = binarySearch(unknow_v, expect);
        System.out.println(s);
    }
    //转化为十进制
    private static BigInteger convert2ten(String value, BigInteger radix) {
        char[] chars = value.toCharArray();
        BigInteger sum = BigInteger.ZERO; //
        for (char aChar : chars) {
            int v = char2int(aChar);
            if (radix.compareTo(BigInteger.valueOf(v)) < 0) {
                return BigInteger.valueOf(Long.MAX_VALUE);
            }
            sum = sum.multiply(radix).add(BigInteger.valueOf(v));
        }
        return sum;
    }
    //char转换成int， ASCII中'0' = 48; 'A' = 65;'a' = 97;    'a' - '0' - 39= 97 - 48 - 39 = 10
    private static int char2int(char c) {
        int ret = c - '0';
        return ret < 10 ? ret : ret - 39;
    }

        //重点在于最大值和最小值的界定 下界大于 所有位中最大值 考虑2 2 1 10 这个case
        //上界不小于下界,不大于expect 考虑200 1 1 10 这个case
    private static String binarySearch(final String value, final BigInteger expect) {
        BigInteger start = BigInteger.valueOf(maxBitValue(value) + 1); //下界 value数字中最大位+1
        BigInteger end = expect.compareTo(start) < 0 ? start : expect; //上界 expect
        BigInteger result_radix = null;
        while (end.compareTo(start) >= 0) {
            BigInteger middle = start.add(end).divide(BigInteger.valueOf(2));
            BigInteger ten_value = convert2ten(value, middle);
            int compare = ten_value.compareTo(expect);
            if (compare >= 0) {
                if (compare == 0 && (result_radix == null || middle.compareTo(result_radix) < 0))
                    result_radix = middle;
                end = middle.subtract(BigInteger.valueOf(1));  //subtract相减
            } else {
                start = middle.add(BigInteger.valueOf(1));
            }
        }
        return result_radix == null ? "Impossible" : result_radix + "";
    }
    //求其中value各位数字中最大数字
    private static int maxBitValue(String value) {
        int max = 0;
        for (char c : value.toCharArray()) {
            int i = char2int(c);
            if (i > max)
                max = i;
        }
        return max;
    }
}