编程逻辑题java版答案完美版.pdf

编程逻辑题编程逻辑题+java+java 版答案完美版版答案完美版 ? import java.util.Scanner; public class ForExercise public static void main(String args) / Test1(); / Test2(); / Test3(); / Test4(); / Test5(); / Test6(); / Test7(); / Test8(); / Test9(); / Test10(); / Test11(); / Test12(); / Test13(); / Test14(); / Test15(); / Test16(); / Test17(); / Test18(); Test19(); public static void Test1() System.out.println(1.打印九九乘法表:); for (int i = 1; i 10; i+) for (int j = 1; j = i; j+) int a = j * i; System.out.print(j + * + i + = + a + ); System.out.println(); System.out.println(); public static void Test2() System.out.println(2.输入一个数字判断他是奇数还是偶数:); Scanner er = new Scanner(System.in); System.out.println(请您输入一个整数：); int a = er.nextInt(); if (a % 2 = 0) System.out.println(您输入的 + a + 是一个偶数); else System.out.println(您输入的 + a + 是一个奇数); System.out.println(); public static void Test3() System.out.println(3.求解 1 到 100 之间的偶数); for (int a = 1; a = 100; a+) if (a % 2 = 0) System.out.print(1 到 100 之间的偶数: + a + ;); System.out.println(); public static void Test4() System.out.println(4.一次酒店宴席安排宾客就座吃饭，5 人一桌剩 4 人，7 人一桌剩 6 人，9 人一桌剩 8 人，11 人一桌正好。问宴席共最少有多少人); int s = 0; for (int i = 1; i+) if (i % 5 = 4 i % 7 = 6 i % 9 = 8 i % 11 = 0) s = s + 1; if (s = 1) System.out.println(宴会最少 + i + 人); break; System.out.println(); public static void Test5() System.out.println(5.求 1 到 100 之间的素数(能被 1 和他本身整除的数); for (int i = 1; i = 100; i+) int b = 0; for (int j = 1; j = i; j+) if (i % j = 0) b = b + 1; if (b = 2) System.out.print(素数: + i + ;); System.out.println(); public static void Test6() System.out.println(6.1 到 10 的乘积之和); int sum = 0; for (int i = 1; i = 10; i+) for (int j = 1; j = 10; j+) sum = sum + i * j; System.out.println(1 到 10 的乘积之和: + sum); System.out.println(); public static void Test7() System.out.println(7.笼子里一共有鸡和兔子 35 只， 一共有 94 条退， 笼子里一共有鸡 和兔子共多少只); for (int i = 0; i = 35; i+) for (int j = 0; j = 35 - j; j+) if (i + j = 35 (2 * i + 4 * j = 94) System.out.println(i + 只鸡 + j + 只兔子); System.out.println(); public static void Test8() System.out.println(8.键盘输入 3 个值便将这三个值从小到大打 印出来.); Scanner xin = new Scanner(System.in); System.out.println(请您输入三个整数：); int a = xin.nextInt(); int b = xin.nextInt(); int c = xin.nextInt(); if (a b) int d = a; a = b; b = d; if (b c) int d = b; b = c; c = d; if (a b) int d = a; a = b; b = d; System.out.println(您输入的三个数字从小到大依次为： + a + + b + + c); System.out.println(); public static void Test9() System.out.println(9.有一分数序列：2/1，3/2，5/3，8/5，13/8，21/13.求出这 个数列的前 20 项之和); int a = 1, b, dang, qian; double sum = 0, er, yi; dang = a; qian = dang; for (int i = 1; i = 20; i+) er = dang; b = dang; dang = dang + qian; yi = dang; qian = b; sum = sum + yi / er; System.out.println(前 20 项之和是： + sum); System.out.println(); public static void Test10() System.out.println(10.输入一个大于 6 的偶数,可以拆分为两个素素相见的和的形式 ); Scanner sr = new Scanner(System.in); System.out.println(请您输入一个大于 6 的偶数：); int a = sr.nextInt(); for (int i = 1; i = a; i+) int b = 0; for (int j = 1; j = i; j+) if (i % j = 0) b = b + 1; if (b = 2) for (int c = 1; c = a; c+) int d = 0; for (int j = 1; j = c; j+) if (c % j = 0) d = d + 1; if (d = 2) if (i + c = a i = c) System.out.println(您输入的： + a + 可以拆分为： + i + + + c); System.out.println(); public static void Test11() System.out.println(11.打印一个三角型:); System.out.println(*); System.out.println(*); System.out.println(*); for (int i = 1; i = 4; i+) for (int j = 1; j = 8 - 2 * i; j+) System.out.print( ); for (int a = 1; a = 2 * i - 1; a+) System.out.print(*); System.out.println(); System.out.println(); public static void Test12() System.out .println(12.判定公历闰年遵循的规律为: 四年一闰,百年不闰,四百年再闰.公历闰年的 简单计算方法: （符合以下条件之一的年份即为闰年）1.能被 4 整除而不能被 100 整除 （如 2004 年就是闰年,1900 年不是） 2.能被 400 整除. （如 2000 年是闰年）题目:从键盘输入一个年份 ， 判断该年份是不是闰年，只有三次输入机会); Scanner xin = new Scanner(System.in); int b = 3; while (b = 1) b-; System.out.println(请您输入一个年份：); int a = xin.nextInt(); if (a % 400 = 0 | (a % 4 = 0 a % 100 != 0) System.out.println(a + 是闰年！); else System.out.println(a + 不是闰年 ！); System.out.println(); public static void Test13() System.out.println(13.已经知道：100 个房间，每个房间里有一盏灯，现在有一百个 人排成一队第一个人拉所有能把 1 整除的房间里的灯,第二个人拉所有能把 2 整除的房间里的 灯第 100 个人拉所有能把 100 整除的房间 里的灯问：最后哪些灯是亮的); for (int i = 1; i = 100; i+) int a = 0; for (int j = 1; j = 100; j+) if (i % j = 0) a = a + 1; if (a % 2 = 1) System.out.print(最后第 + i + 盏灯是亮的;); System.out.println(); public static void Test14() System.out.println(14.分配： 公司有 100 员工有 100 块钱保证每一个员工都得有一张 票；中国馆 5 元一张；美国馆 3 元一张；日本馆 一元 3 张); int sum = 0; for (int i = 0; i 21; i+) for (int j = 0; j 34; j+) for (int e = 0; e 101; e+) int a = 5 * i + 3 * j + (int) (e / 3); int b = i + j + e; int w = e % 3; if (a = 100 b = 100 w = 0) System.out.println(买中国馆 + i + 张；买美国馆 + j + 张；买日本馆 + e + 张。); sum = sum + 1; System.out.println(一共有 + sum + 种买票方案！); System.out.println(); public static void Test15() System.out.println(15.团团圆圆 3 个月生一对小熊猫以后每个月都会生一对。小熊猫 生下来 3 个月之后也会生小猫熊，24 个月后有多少只熊猫); int sum = 1; int zhi = 0; int qian = 0; for (int i = 1; i 25; i+) if (i 3) zhi = sum; if (i = 3) sum = zhi; zhi = zhi + qian; qian = sum; System.out.println(24 个月之后有： + zhi + 只熊猫); System.out.println(); public static void Test16() System.out.println(16. 求 100-999 之 间 的 所 有 的 水 仙 花 数 ：153 = 1*1*1+5*5*5+3*3*3); int s = 0; for (int i = 100; i 1000; i+) int a = i % 10; int b = (i / 10) % 10; int c = i / 100; if (a * a * a + b * b * b + c * c * c = i) System.out.print(i + ); s = s + 1; System.out.println(); System.out.println(100-1000 之间一共有 + s + 个水仙花数); System.out.println(); public static void Test17() System.out.println(17.打印输出以下图形:); System.out.println(*); System.out.println( * ); System.out.println(*); System.out.println( * ); System.out.println(*); System.out.println(); for (int i = 1; i 6; i+) for (int j = 1; j = Math.abs(3 - i); j+) System.out.print( ); for (int a = 1; a = 5 - 2 * (Math.abs(3 - i); a+) System.out.print(*); System.out.println(); System.out.println(); public static void Test18() System.out.println(18.打印如下图形：); System.out.println(*); System.out.println( * * ); System.out.println(*); System.out.println( * * ); System.out.println(*); System.out.println(); for (int i = 1; i 6; i+) for (int j = 1; j = Math.abs(3 - i); j+) Syste