A、哪天返回
小明被不明势力劫持。后被扔到x星站再无问津。小明得知每天都有飞船飞往地球，但需要108元的船票，而他却身无分文。
他决定在x星战打工。好心的老板答应包食宿，第1天给他1元钱。
并且，以后的每一天都比前一天多2元钱，直到他有足够的钱买票。
请计算一下，小明在第几天就能凑够108元，返回地球。
题解：
package action; public class demo { public static void main(String[] args) { int an = 1; int sn = 0; int n=0; for (n = 0; sn < 108; n++) { sn += an; an += 2; } System.out.println(n); } }
B、猴子分香蕉
5只猴子是好朋友，在海边的椰子树上睡着了。这期间，有商船把一大堆香蕉忘记在沙滩上离去。
第1只猴子醒来，把香蕉均分成5堆，还剩下1个，就吃掉并把自己的一份藏起来继续睡觉。
第2只猴子醒来，重新把香蕉均分成5堆，还剩下2个，就吃掉并把自己的一份藏起来继续睡觉。
第3只猴子醒来，重新把香蕉均分成5堆，还剩下3个，就吃掉并把自己的一份藏起来继续睡觉。
第4只猴子醒来，重新把香蕉均分成5堆，还剩下4个，就吃掉并把自己的一份藏起来继续睡觉。
第5只猴子醒来，重新把香蕉均分成5堆，哈哈，正好不剩！
请计算一开始最少有多少个香蕉。
题解:
package action; public class demo { public static void main(String[] args) { for (int n = 20;; n++) { double y = n; // 桃子的个数 for (int x = 1; x <= 4; x++) { y = y - x - (y - x) / 5; } if (y % 5 == 0) { System.out.println(n); break; } } } }
C、字母阵列
仔细寻找，会发现：在下面的8x8的方阵中，隐藏着字母序列：“LANQIAO”。
SLANQIAO
ZOEXCCGB
MOAYWKHI
BCCIPLJQ
SLANQIAO
RSFWFNYA
XIFZVWAL
COAIQNAL
我们约定: 序列可以水平，垂直，或者是斜向；
并且走向不限（实际上就是有一共8种方向）。
上图中一共有4个满足要求的串。
下面有一个更大的（100x100）的字母方阵。
你能算出其中隐藏了多少个“LANQIAO”吗？
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题解：
package action; import java.util.Scanner; public class demo { static int n = 100; // 长度 static char[] ch = { 'L', 'A', 'N', 'Q', 'I', 'A', 'O' }; // 寻找数据 static char[][] data = new char[n][n]; // 存储字母方阵 // 从正北方向绕顺时针旋转一圈的八个方向对应的增减值 static int[] dx = { 0, 1, 1, 1, 0, -1, -1, -1 }; static int[] dy = { 1, 1, 0, -1, -1, -1, 0, 1 }; static int count = 0; // 计数 public static void main(String[] args) { Scanner sc = new Scanner(System.in); // 插入数据 for (int i = 0; i < data.length; i++) { data[i] = sc.nextLine().toCharArray(); } // 循环 for (int i = 0; i < data.length; i++) { for (int j = 0; j < data[i].length; j++) { // 找第一个字符L if (data[i][j] == ch[0]) { sheck(i, j); // 查找方法 } } } System.out.println(count); // 打印 sc.close(); } public static void sheck(int i, int j) { // 八个方向 for (int d = 0; d < 8; d++) { int x = i + dx[d]; int y = j + dy[d]; int num = 1; // 寻找字符数组下标 while (x < n && x >= 0 && y < n && y >= 0 && data[x][y] == ch[num]) { // 当测试到最后一个字母以后，计数+1，并跳出while循环，进行下一个方向的寻找 if (num == 6) { count++; break; } num++; // 下一个 // 对应坐标加上相应的dx,dy x += dx[d]; y += dy[d]; } } } }
D、第几个幸运数
到x星球旅行的游客都被发给一个整数，作为游客编号。x星的国王有个怪癖，他只喜欢数字3,5和7。
国王规定，游客的编号如果只含有因子：3,5,7,就可以获得一份奖品。
我们来看前10个幸运数字是：3 5 7 9 15 21 25 27 35 45
因而第11个幸运数字是：49
小明领到了一个幸运数字 59084709587505，他去领奖的时候，人家要求他准确地说出这是第几个幸运数字，否则领不到奖品。
请你帮小明计算一下，59084709587505是第几个幸运数字。
题解：
package action; public class demo { public static void main(String[] args) { long y = 59084709587505l; int count = 0; for (int i = 0; Math.pow(3, i) < y; i++) { for (int j = 0; Math.pow(5, j) < y; j++) { for (int k = 0; Math.pow(7, k) < y; k++) { if (Math.pow(3, i) * Math.pow(5, j) * Math.pow(7, k) < y) { count++; System.out.println(count + " " + Math.pow(3, i) * Math.pow(5, j) * Math.pow(7, k)); } } } } System.out.println(count); } }
