动态日历

简介

根据一个年份一个月份显示当前月份的日历布

代码

灵活运用switch的特点计算总天数,在计算中需考虑平年闰年,大月小月和二月的特殊性

import java.util.Scanner; public class T2 { public int n(int y, int m) { int day = 0; for (int i = 1901; i <= y; i++) { if ((i % 4 == 0 && i % 100 != 0) || i % 400 == 0) { day += 366; } else if (y > 0) { day += 365; } } switch (m - 1) { case 12: day += 31; case 11: day += 30; case 10: day += 31; case 9: day += 30; case 8: day += 31; case 7: day += 31; case 6: day += 30; case 5: day += 31; case 4: day += 30; case 3: day += 31; case 2: if ((y % 4 == 0 && y % 100 != 0) || y % 400 == 0) { day += 29; } else if (y > 0) { day += 28; } case 1: day += 31; } return day; } public static void main(String[] args) { T2 s = new T2(); Test2 e = new Test2(); Scanner in = new Scanner(System.in); int j = in.nextInt(); int q = in.nextInt(); int day = e.getDayOfMonth(j, q); int a = s.n(j, q); int b = a % 7; int c; int i; int sum = b; if (b > 0) { for (c = 1; c <= b; c++) { System.out.print("\t"); } for (i = 1; i <= day; i++) { System.out.print(i + "\t"); if ((sum + i) % 7 == 0) { System.out.println(); } } } else { for (i = 1; i <= day; i++) { System.out.print(i + "\t"); if ((sum + i) % 7 == 0) { System.out.println(); } } } } } public class Test2 { public int getDayOfMonth(int y, int m) { int day = 0; switch (m) { case 1: case 3: case 5: case 7: case 8: case 10: case 12: day = 31; break; case 2: if ((y % 4 == 0 && y % 100 != 0) || y % 400 == 0) { day = 29; } else if (y > 0) { day = 28; } else { day = -1; } break; case 4: case 6: case 9: case 11: day = 30; break; default: day = -1; } return day; } public static void main(String[] args) { Test2 d = new Test2(); int day = d.getDayOfMonth(400, 2); if (day == -1) { System.out.println("输入错误"); } else { System.out.println(day); } } }

运行结果

小结

第一次做这道题感觉很麻烦,逻辑很复杂,现在再看感觉自己的代码很杂乱,可以有更好的解决办法,不需要用复杂的switch语句也能解决.