二分法

用一组绳子的长度分别除以最长绳子的一半，如果得到的绳子总和小于目标，那么就减小每个单位的长度right = mid，反之，增加每个单位的长度left = mid，当left和right的差值在接近0的某个范围的时候，停止循环。

import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int target = in.nextInt();
        int[] strings = new int[n];
        int max = 0;
        for (int i = 0; i < n; i++) {
            strings[i] = in.nextInt();
            if (strings[i] > max) max = strings[i];
        }
        double res = binarySearch(strings, 0.0, max, target);
        System.out.println(res);
    }

    public static double binarySearch(int[] strings, double left, double right, int target) {
        while (Math.abs(left - right) > 1e-6) {
            double mid = (left + right) / 2;
            int count = 0;
            for (int len : strings) {
                count += (int) (len / mid);
            }
            if (count < target) {
                right = mid;
            } else {
                left = mid;
            }
        }
        return left;
    }
}

import java.util.Scanner;

public class Changes {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int cost = in.nextInt();
        in.close();
        int remaining = 1024 - cost;
        int res = 0;
        int[] changes = {64, 16, 4, 1};
        for (int i = 0; i < changes.length; i++) {
            res += remaining / changes[i];
            remaining = remaining % changes[i];
        }
        System.out.println(res);
    }
}