蓝桥杯vip之基础练习

1.阶乘计算

题目描述

代码实现

public class JieCheng {
	public static void main(String[] args) {
		Scanner sc=new Scanner(System.in);
		int n=sc.nextInt();
		int i=1;
		BigInteger res=new BigInteger("1");
		while (i<=n) {
			String s=String.valueOf(i);
			res=res.multiply(new BigInteger(s));
			i++;
		}
		System.out.println(res);
	}
}

2.高精度加法

题目描述

代码实现

public class GaoJingDuJiSuan {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		String a = sc.next();
		String b = sc.next();
		BigInteger res = new BigInteger("0");
		res = new BigInteger(a).add(new BigInteger(b));
		System.out.println(res);
	}
}

3. Huffuman树

题目描述

代码实现

public class Huffuman {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int size = sc.nextInt();
		int arr[] = new int[size];
		int res = 0;
		for (int i = 0; i < size; i++) {
			int n = sc.nextInt();
			arr[i] = n;
		}
		Arrays.sort(arr);
		for (int i = 1; i < arr.length; i++) {
			arr[i] = arr[i] + arr[i - 1];
			res += arr[i];
			arr[i - 1] = 0;
			Arrays.sort(arr);
		}
		System.out.println(res);
	}
}

4.报时助手

题目描述
样例输入

0 15

样例输出

zero fifteen

代码实现

public class BaoShiJiShu {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int h = sc.nextInt();
		int m = sc.nextInt();
		String s[] = { "zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine", "ten", "eleven",
				"twelve", "thirteen", "fourteen", "fifteen", "sixteen", "seventeen", "eighteen", "nineteen", "twenty",
				"twenty one", "twenty two", "twenty three", "twenty four" };
		String case1[] = { "o'clock", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine", "ten",
				"eleven", "twelve", "thirteen", "fourteen", "fifteen", "sixteen", "seventeen", "eighteen", "nineteen" };
		String case2[] = { "", "ten", "twenty", "thirty", "forty", "fifty", "sixty" };

		String strRes = "";
		strRes = s[h];
		if (m < 20 && m >= 0) {
			strRes = strRes + " " + case1[m];
		}
		if (m >= 20 && m <= 60) {
			if (m % 10 == 0) {
				strRes = strRes + " " + case2[m / 10];
			} else {
				strRes = strRes + " " + case2[m / 10] + " " + case1[m % 10];
			}
		}
		System.out.println(strRes);
	}
}

5.回形取数

题目描述

问题描述
　　回形取数就是沿矩阵的边取数，若当前方向上无数可取或已经取过，则左转90度。一开始位于矩阵左上角，方向向下。
　　
输入格式
　　输入第一行是两个不超过200的正整数m, n，表示矩阵的行和列。接下来m行每行n个整数，表示这个矩阵。
　　
输出格式
　　输出只有一行，共mn个数，为输入矩阵回形取数得到的结果。数之间用一个空格分隔，行末不要有多余的空格。

样例输入
3 3
1 2 3
4 5 6
7 8 9

样例输出
1 4 7 8 9 6 3 2 5

样例输入
3 2
1 2
3 4
5 6

样例输出
1 3 5 6 4 2

代码实现

public class HuiXingQuShu {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int rows = sc.nextInt();
		int cols = sc.nextInt();
		int arr[][] = new int[rows][cols];
		for (int i = 0; i < rows; i++) {
			for (int j = 0; j < cols; j++) {
				arr[i][j] = sc.nextInt();
			}
		}

		int cur = 0, x = -1, y = 0;
		while (cur < rows * cols) {
			while (x + 1 < rows && arr[x + 1][y] != -1) {
				System.out.print(arr[++x][y] + " ");
				arr[x][y] = -1;
				++cur;
			}
			while (y + 1 < cols && arr[x][y + 1] != -1) {
				System.out.print(arr[x][++y] + " ");
				arr[x][y] = -1;
				++cur;
			}
			while (x - 1 >= 0 && arr[x - 1][y] != -1) {
				System.out.print(arr[--x][y] + " ");
				arr[x][y] = -1;
				++cur;
			}
			while (y - 1 >= 0 && arr[x][y - 1] != -1) {
				System.out.print(arr[x][--y] + " ");
				arr[x][y] = -1;
				++cur;
			}

		}

	}

}