题解 | #画匠问题#

import java.util.Scanner;

public class Main {

    public static int getNeedNum(int[] arr, long lim) {
        int res = 1;
        long stepSum = 0;
        for (int i = 0; i < arr.length; i++) {
            if (arr[i] > lim) {
                return Integer.MAX_VALUE;
            }
            stepSum += arr[i];
            if (stepSum > lim) {
                res++;
                stepSum = arr[i];
            }
        }
        return res;
    }

    public static long solution3(int[]arr,int num){
        if(arr==null||arr.length==0||num<1){
            throw new RuntimeException("err");
        }
        if(arr.length<num){
            int max=Integer.MIN_VALUE;
            for (int i = 0; i < arr.length ; i++) {
                max=Math.max(max,arr[i]);
            }
            return max;
        }else {
            long minSum=0;
            long maxSum=0;
            for (int i = 0; i < arr.length ; i++) {
                maxSum+=arr[i];
            }
            //最终目标，num>=需要的num
            while (minSum!=maxSum-1){
                long mid=(maxSum+minSum)/2;  //员工的限制时间
                if(getNeedNum(arr,mid)>num){
                    minSum=mid;
                }else {
                    maxSum=mid;
                }
            }
            return maxSum;
        }

    }

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int k=in.nextInt();
        int[]arr=new int[n];
        for (int i = 0; i <n ; i++) {
            arr[i]=in.nextInt();
        }
        System.out.println(solution3(arr,k));
    }
}