美团编程题AC JAVA版本

第一题找是K的倍数的最长子序列
sum[i]记录序列从0到i的和
问题转化为sum[]数组中两个元素差是k的倍数，两元素下标的最大差值
public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        while (sc.hasNext()) {
            int n = sc.nextInt();
            int[] p = new int[n];
            int[] sum = new int[n + 1];
            sum[0] = 0;
            for (int i = 0; i < n; i++) {
                p[i] = sc.nextInt();
                sum[i + 1] = sum[i] + p[i];
            }
            int k = sc.nextInt();
            int res = 0;
            for (int i = 0; i < n; i++) {
                for (int j = n; j > i; j--) {
                    if (getLength(sum, i, j, k) > res) {
                        res = getLength(sum, i, j, k);
                        break;
                    }
                }
                if (res >= n - i) {
                    break;
                }
            }
            System.out.println(res);
        }
        sc.close();
    }
    static int getLength(int[] sum, int begin, int end, int k) {
        int res = 0;
        if ((sum[end] - sum[begin]) % k == 0) {
            res = end - begin;
        }
        return res;
    }

第二题改卷子
只要最大数小于等于和的二分之一就行
public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		while (sc.hasNext()) {
			int n = sc.nextInt();
			int[] s = new int[n];
			int sum = 0;
			for (int i = 0; i < n; i++) {
				s[i] = sc.nextInt();
				sum += s[i];
			}
			Arrays.sort(s);
			if (sum >= 2 * s[n - 1]) {
				System.out.println("Yes");
			} else {
				System.out.println("No");
			}
		}
		sc.close();
	}
PS:前面的选择题有点炸