Numbers CCPC 大数除法

题意：把n分成m份，使得or值最小

首先，我们要找到他的最高位，如果(2 * k - 1 ) * m > n > (2 *(k-1) - 1) * m, 然后我们就必须把k位赋为1， 为什么呢？ 你可以想一下 (2 * k - 1 ) * m 了， 然后如果 再不分的话，就会超过m份。

import java.math.BigInteger;
import java.util.Scanner;
 
public class Main 
{
    public static BigInteger two = BigInteger.valueOf(2);
    public static BigInteger one = BigInteger.ONE;
    public static BigInteger zero = BigInteger.ZERO;
    public static BigInteger n, m;
    public static BigInteger a[] = new BigInteger[4005];
    public static void init()
    {
        a[0] = one;
        for(int i = 1; i <= 4002; i++)
            a[i] = a[i - 1].multiply(two);
    }
    public static void main(String[] args) 
    {
        Scanner cin = new Scanner(System.in);
        int t = cin.nextInt();
        init();
        while(t > 0)
        {
            t--;
            int up = 0;
            n = cin.nextBigInteger();
            m = cin.nextBigInteger();
            BigInteger sum = zero, temp = n, ans = zero;
            for(int i = 0; sum.compareTo(n) < 0; i++)
            {
                sum = sum.add(m.multiply(a[i]));
                up = i;
            }
            for(int i = up; i >= 0; i--)
            {
                BigInteger b = a[i].subtract(one);
                //if(T.multiply(m).compareTo(temp) >= 0) continue;
                if((b.multiply(m)).compareTo(temp) >= 0) continue;
                BigInteger k = temp.divide(a[i]); // 可以分为几个
                k = k.min(m);
                ans = ans.add(a[i]);
                temp = temp.subtract(a[i].multiply(k)); // 减去
            }
            System.out.println(ans);
        }
        cin.close();
    }
}