From ffd83ca8cc5197b423af1b530172813a929607c8 Mon Sep 17 00:00:00 2001
From: Leo <582717189@qq.com>
Date: Fri, 15 Jan 2021 11:09:38 +0800
Subject: [PATCH] Tower test

---
 src/leo/class22/KillMonster.java | 43 +++++++++++++++
 src/leo/class22/splitNumber.java | 92 ++++++++++++++++++++++++++++++++
 2 files changed, 135 insertions(+)
 create mode 100644 src/leo/class22/KillMonster.java
 create mode 100644 src/leo/class22/splitNumber.java

diff --git a/src/leo/class22/KillMonster.java b/src/leo/class22/KillMonster.java
new file mode 100644
index 0000000..c352d18
--- /dev/null
+++ b/src/leo/class22/KillMonster.java
@@ -0,0 +1,43 @@
+package leo.class22;
+
+/**
+ * @author Leo
+ * @ClassName KillMonster
+ * @DATE 2021/1/14 9:06 下午
+ * @Description
+ *
+ * 给定3个参数，N，M，K
+ * 怪兽有N滴血，等着英雄来砍自己
+ * 英雄每一次打击，都会让怪兽流失[0~M]的血量
+ * 到底流失多少？每一次在[0~M]上等概率的获得一个值
+ * 求K次打击之后，英雄把怪兽砍死的概率
+ */
+public class KillMonster {
+
+    static class Recursion{
+        public static double kill(int n, int m, int k) {
+            if (n < 1 || m < 1 || k < 1) {
+                return 0;
+            }
+            long ans = process(n, m, k);
+            return (double) ans / Math.pow(m + 1, k);
+        }
+
+        private static long process(int hp, int m, int rest) {
+            if (hp <= 0) {
+                return (long) Math.pow(m + 1, rest);
+            }
+            if (rest == 0) {
+                return hp == 0 ? 1 : 0;
+            }
+            int ans = 0;
+            for (int i = 0; i <= m; i++) {
+                ans += process(hp - i, m, rest - 1);
+            }
+            return ans;
+
+        }
+
+    }
+
+}
diff --git a/src/leo/class22/splitNumber.java b/src/leo/class22/splitNumber.java
new file mode 100644
index 0000000..38363cb
--- /dev/null
+++ b/src/leo/class22/splitNumber.java
@@ -0,0 +1,92 @@
+package leo.class22;
+
+/**
+ * @author Leo
+ * @ClassName splitNumber
+ * @DATE 2021/1/14 10:02 下午
+ * @Description
+ *
+ * 给定一个正数1，裂开的方法有一种，
+ * (1) 给定一个正数2，裂开的方法有两种，(1和1)、
+ * (2) 给定一个正数3，裂开的方法有三种，(1、1、1)、(1、2)、
+ * (3) 给定一个正数4，裂开的方法有五种，(1、1、1、1)、(1、1、2)、(1、3)、(2、2)、 (4)
+ * 给定一个正数n，求裂开的方法数。 动态规划优化状态依赖的技巧
+ */
+public class splitNumber {
+    static class Recursion {
+        public static int number(int n) {
+            if (n <= 0) {
+                return 0;
+            }
+            return process(n, 1);
+        }
+
+        private static int process(int rest, int pre) {
+            if (rest == 0) {
+                return 1;
+            }
+            if (rest <= pre) {
+                return rest - pre == 0 ? 1 : 0;
+            }
+            int ans = 0;
+            for (int j = pre; j <= rest; j++) {
+                ans += process(rest - j, j);
+            }
+            return ans;
+        }
+
+    }
+
+    static class Dp{
+        public static int number(int n) {
+            if (n <= 0) {
+                return 0;
+            }
+            int[][] dp = new int[n + 1][n + 1];
+            for (int i = 1; i <= n; i++) {
+                dp[i][i] = 1;
+                dp[i][0] = 1;
+            }
+            for (int pre = n - 1; pre > 0; pre--) {
+                for (int rest = pre + 1; rest <= n; rest++) {
+                    int ans = 0;
+                    for (int j = pre; j <= rest; j++) {
+                        ans+= dp[j][rest - j];
+                    }
+                    dp[pre][rest] = ans;
+                }
+            }
+            return dp[1][n];
+        }
+    }
+
+
+    static class OptDp{
+        public static int number(int n) {
+            if (n <= 0) {
+                return 0;
+            }
+            int[][] dp = new int[n + 1][n + 1];
+            for (int i = 1; i <= n; i++) {
+                dp[i][i] = 1;
+                dp[i][0] = 1;
+            }
+            for (int pre = n - 1; pre > 0; pre--) {
+                for (int rest = pre + 1; rest <= n; rest++) {
+                    //下面的格子
+                    dp[pre][rest] = dp[pre+1][rest];
+                    //左侧的格子
+                    dp[pre][rest] += dp[pre][rest - pre];
+
+                }
+            }
+            return dp[1][n];
+        }
+    }
+    public static void main(String[] args){
+        System.out.println( Recursion.number(13));
+        System.out.println(Dp.number(13));
+        System.out.println(OptDp.number(13));
+    }
+
+}