用dp[i]记录以第i个元素作为序列的开始。遍历一遍原始序列nums，对于第K个值，将大于maxLen的序列dp[i]+nums[k],判断是否为K的倍数，更新maxLen的值。 
public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		in.nextLine();
		int[] nums = new int[n];
		for (int i = 0; i < n; i++)
			nums[i] = in.nextInt();
		in.nextLine();
		int k = Integer.valueOf(in.nextLine());
		int maxLen = 0;
		int[] dp = new int[n];
		for (int i = 0; i < n; i++) {
			for (int j = 0; j <= i - maxLen; j++) {
				dp[j] += nums[i];
				if (dp[j] % k == 0 && (i - j + 1) > maxLen) {
					maxLen = (i - j) + 1;
					break;
				}
			}
		}
		System.out.println(maxLen);
		in.close();
	}