题目来源：补提交卡
解题思路：假设未提交程序的天数为：a1,a2,....,an，补交的张数为M。依次从a1,a2,....,an中去掉连续的 K 天（0<=K<=M），然后再来计算剩余数组中最长连续提交天数。
具体算法（Java版，直接AC）
1 import java.util.Scanner; 2 3 public class Main { 4 5 //计算数组中最长连续提交天数 6 public static int getMax(int[] array) { 7 int[] copy = new int[array.length + 2]; 8 copy[0] = 0; 9 copy[array.length + 1] = 101; 10 for (int i = 0; i < array.length; i++) { 11 copy[i + 1] = array[i]; 12 } 13 int max = Integer.MIN_VALUE; 14 for (int i = 1; i < copy.length; i++) { 15 int sum = copy[i] - copy[i - 1] - 1; 16 max = max > sum ? max : sum; 17 } 18 return max; 19 } 20 21 //从下标为start开始，删除长度为count个元素 22 public static int[] remove(int[] array, int count, int start) { 23 int[] copy = new int[array.length - count]; 24 int index = 0; 25 for (int i = 0; i < start; i++) { 26 copy[index++] = array[i]; 27 } 28 for (int i = start + count; i < array.length; i++) { 29 copy[index++] = array[i]; 30 } 31 return copy; 32 } 33 34 //计算使用k张补交卡后，最长连续提交天数 35 public static int solve(int[] array, int k) { 36 int max = Integer.MIN_VALUE; 37 for (int i = 0; i <= array.length - k; i++) { 38 int sum = getMax(remove(array, k, i)); 39 max = max > sum ? max : sum; 40 } 41 return max; 42 } 43 44 public static void main(String[] args) { 45 Scanner scanner = new Scanner(System.in); 46 int T = scanner.nextInt(); 47 while (T > 0) { 48 int N = scanner.nextInt(); 49 int M = scanner.nextInt(); 50 int[] array = new int[N]; 51 for (int i = 0; i < N; i++) { 52 array[i] = scanner.nextInt(); 53 } 54 int max = Integer.MIN_VALUE; 55 for (int i = 0; i <= M; i++) { 56 int sum = solve(array, i); 57 max = sum > max ? sum : max; 58 } 59 System.out.println(max); 60 --T; 61 } 62 } 63 }