diff --git a/README.md b/README.md
index 495e0843..4b3ce7c5 100644
--- a/README.md
+++ b/README.md
@@ -232,6 +232,10 @@ Power by [logomakr](https://logomakr.com/).
+
+ 
+
+
#### License
diff --git a/notes/剑指 offer 题解.md b/notes/剑指 offer 题解.md
index fbd0f07a..cc82b048 100644
--- a/notes/剑指 offer 题解.md
+++ b/notes/剑指 offer 题解.md
@@ -587,6 +587,8 @@ public int RectCover(int n) {
## 解题思路
+### 动态规划
+
```java
public int JumpFloorII(int target) {
int[] dp = new int[target];
@@ -598,6 +600,34 @@ public int JumpFloorII(int target) {
}
```
+### 数学式子推导
+
+跳上 n-1 级台阶，可以从 n-2 级跳 1 级上去，也可以从 n-3 级跳 2 级上去...也可以从 0 级跳上去。那么
+```
+f(n-1) = f(n-2) + f(n-3) + ... + f(0) ①
+```
+
+同样，跳上 n 级台阶，可以从 n-1 级跳 1 级上去，也可以从 n-2 级跳 2 级上去...也可以从 0 级跳上去。那么
+```
+f(n) = f(n-1) + f(n-2) + ... + f(0) ②
+```
+
+②-①：
+```
+f(n) - f(n-1) = f(n-1)
+f(n) = 2*f(n-1)
+```
+
+所以 f(n) 是一个等比数列：
+```
+f(n) = 2^(n-1)
+```
+
+```java
+public int JumpFloorII(int target) {
+ return (int) Math.pow(2, target - 1);
+}
+```
# 11. 旋转数组的最小数字