public class Main {
	public static void main(String[] args) {
		int count = 0;
		for(int i = 1;i <= 2020;i++) {
			if(gcd(i,2020) == 1) {
				count++;
			}
		}
		System.out.println(count);
	}
	private static int gcd(int a, int b) {
		if(b == 0)
			return a;
		return gcd(b,a%b);
	}
}

2.ASCII码

A	B	C	D	E	F	G	H	I	J	K	L	M	N	O	P	Q
65	66	67	68	69	70	71	72	73	74	75	76	77	78	79	80	81

3.Pierce序列

解法一

public class Main {
	public static void main(String[] args) {
		int v = 1160;
		int p = 2021;
		int ans = 1;
		while(p % v != 0) {
			v = p % v;
			++ans;
		}
		System.out.println(ans);
	}
}

解法二

public class Main {
	private static int count = 0;
	public static void main(String[] args) {
		fun(1160,2021);
		System.out.println(count);
	}
	private static void fun(int v, int p) {
		int[] a = new int [100];
		a[1] = v;
		for(int i = 2;i <= 50;i++) {
			if(a[i-1] != 0) {
				a[i] = p % a[i-1];
			}
		}
		for(int i = 1;i <= 50;i++) {
			if(a[i-1] != 0) {
				count ++;
				//System.out.println(a[i]);
			}
		}
	}
}

4.二叉树的叶结点

5.Excel的列名

解法一

public class Main {
	public static void main(String[] args) {
		int n = 2021;
		StringBuilder s = new StringBuilder("");
		while(n != 0) {
			s.append((char)(((n - 1) % 26) + 'A'));
			n = (n - 1) / 26;
		}
		System.out.println(s.reverse().toString());
	}
}

解法二

import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		char[] ans = new char[1000000];
		int count = 0;
		while(n > 0) {
			int temp = n % 26;
			if(temp == 0) {
				ans[count++] = 'Z';
				n = n / 26 - 1;
			}else {
				ans[count++] = (char) ('A' + temp -1);
				n = n / 26;
			}
		}
		for(int i = count -1;i >= 0;i--) {
			System.out.println(ans[i]);
		}
	}
}

6.增加分割符

import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		long v = sc.nextLong();
		char[] ch = new String(v + "").toCharArray();
		if(ch.length <= 3) {
			System.out.println(v);
		}else if(ch.length % 3 == 0){
			System.out.print("" + ch[0] + ch[1] + ch[2]);
			for(int i = 3;i < ch.length;i += 3) {
				System.out.print("," + ch[i] + ch[i+1] + ch[i+2]);
			}
		}else {
			for(int i = 0;i < ch.length % 3;++i) {
				System.out.print(ch[i]);
			}
			for(int i = ch.length % 3;i < ch.length;i += 3) {
				System.out.print("," + ch[i] + ch[i+1] + ch[i+2]);
			}
		}
	}
}

7.斐波那契数列

import java.math.BigInteger;
import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		BigInteger two = new BigInteger("2");
		if((F(N).mod(two)).compareTo(BigInteger.ZERO) == 0) {
			System.out.println(0);
		}else {
			System.out.println(1);
		}
	}
	private static BigInteger F(int N) {
		BigInteger F1 = BigInteger.ONE;
		BigInteger F2 = BigInteger.ONE;
		BigInteger F3 = BigInteger.ZERO;
		for(int i = 3;i <= N;i++) {
			F3 = F1.add(F2);
			F1 = F2;
			F2 = F3;
		}
		return F3;
	}
}

8.矩阵

import java.util.Scanner;
public class Main {
	static int[][] v;
	static int total = 0,n,m;
	static int r = 0,c = 0,d = Integer.MAX_VALUE;
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int m = sc.nextInt();
		v = new int [n][m];
		for(int i = 0;i < n;++i) {
			for(int j = 0;j < m;++j) {
				v[i][j] = sc.nextInt();
				total += v[i][j];
			}
		}
		int[][] dp = new int[n + 1][m + 1];
		for(int i = 1;i <= n;++i) {
			for(int j = 1;j <= m;++j) {
				dp[i][j] = dp[i-1][j] + dp[i][j-1] -dp[i-1][j-1] + v[i-1][j-1];
				check(dp[i][j],i,j);
			}
		}
		System.out.println(r + " "+ c);
	}
	private static void check(int x, int i, int j) {
		if(Math.abs(2 * x -total) < d || Math.abs(2 * x - total) == d && i * j < r * c) {
			r = i;
			c = j;
			d = Math.abs(2 * x - total);
		}
	}
}

9.像素

10.购买物品

import java.util.Arrays;
import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
// 模拟二进制，三种属性C、B、A，让它们分别代表4、2、1，
//即1<<2、1<<1、1<<0（注1<<n是位运算），
//当i==5时，dp[i]表示拥有CA属性时代价最小
		int[] dp = new int[7+1];
//7(十进制)==111(二进制)==CBA，最终dp[7]就表示同时拥有ABC属性时代价最小
		Arrays.fill(dp, Integer.MAX_VALUE);
		for(int i = 0;i < N;++i) {
			int c = sc.nextInt(),v = 0;//v用来记录当前物品有哪几种属性
			char[] ch = sc.next().toCharArray();
			for(int j = 0;j < ch.length;++j) {
				v |= 1 << (ch[j] - 'A');
			}
			for(int j = 1;j < dp.length;++j) {//j从1到7，分别表示A、B、C、BA、CA、CB、CBA
				if((j & v) != v && dp[j] != Integer.MAX_VALUE && c + dp[j] < dp[v | j])
					//j==5,v==4,则v中拥有的属性，j也有，j没必要与v进行组合
					dp[v | j] = c + dp[j];//当dp[j] == Integer.MAX_VALUE时，代表还没有j这种组合
			}
			dp[v] = Math.min(dp[v],c);
		}
		System.out.println(dp[7] == Integer.MAX_VALUE ? -1 : dp[7]);
	}
}

11.上升子序列

import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		int MOD = 1000007;
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int k = sc.nextInt();
		int[] a = new int[n];
		for(int i = 0;i < n;i++) {
			a[i] = sc.nextInt();
		}
		int[][] dp = new int[n][k+1];
		for(int i = 0;i < n;i++) {
			dp[i][1] = 1;
		}
		for(int i = 1;i < n;i++) {
			for(int j = 0;j < i;j++) {
				if(a[i] > a[j]) {
					for(int t = 1;t <= k;t++) {
						if(dp[j][t-1] != 0) {
							dp[i][t] = (dp[i][t] + dp[j][t-1]) % MOD;
						}
					}
				}
			}
		}
		int ans = 0;
		for(int i = 0;i < n;i++) {
			ans = (ans + dp[i][k] % MOD);
		}
		System.out.println(ans);
	}
}