From 8ee312ac1ddfdd6472cf0a96aefe56913b574401 Mon Sep 17 00:00:00 2001
From: Sidney Zhang <zly@lyzhang.me>
Date: Mon, 29 Dec 2025 18:18:10 +0800
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---
 solutions/0030.FifthPowers/euler_30.py |  91 ++++++++++++++++++++
 solutions/0031/euler_31.py             |  27 ++++++
 solutions/0031/readme.md               | 114 +++++++++++++++++++++++++
 3 files changed, 232 insertions(+)
 create mode 100644 solutions/0030.FifthPowers/euler_30.py
 create mode 100644 solutions/0031/euler_31.py
 create mode 100644 solutions/0031/readme.md

diff --git a/solutions/0030.FifthPowers/euler_30.py b/solutions/0030.FifthPowers/euler_30.py
new file mode 100644
index 0000000..d1a0162
--- /dev/null
+++ b/solutions/0030.FifthPowers/euler_30.py
@@ -0,0 +1,91 @@
+"""
+Surprisingly there are only three numbers that can be written as the sum of fourth powers of their digits:
+
+    1634 = 1^4 + 6^4 + 3^4 + 4^4
+    8208 = 8^4 + 2^4 + 0^4 + 8^4
+    9474 = 9^4 + 4^4 + 7^4 + 4^4
+
+As 1 = 1^4 is not a sum it is not included.
+
+The sum of these numbers is 19316.
+
+Find the sum of all the numbers that can be written as the sum of fifth powers of their digits.
+"""
+
+import time
+from itertools import combinations_with_replacement
+
+
+def timer(func):
+    def wrapper(*args, **kwargs):
+        start = time.time()
+        result = func(*args, **kwargs)
+        end = time.time()
+        print(f"{func.__name__} on ({kwargs}) taken: {end - start: .6f} seconds")
+        return result
+
+    return wrapper
+
+
+def fifth_power(n: int, power: int = 5, strict: bool = False) -> tuple[bool, int]:
+    sumx = sum(int(digit) ** power for digit in str(n))
+    if strict:
+        isok = (sumx == n) and (power == len(str(sumx)))
+        return isok, sumx
+    return sumx == n, sumx
+
+
+def top_number(n: int) -> int:
+    beg = (9**n) * n
+    beg_len = len(str(beg))
+    last = n
+    while True:
+        if beg_len >= n:
+            beg = (9**n) * beg_len
+            last = beg_len
+            beg_len = len(str(beg))
+            if beg_len == last:
+                break
+        else:
+            break
+    return int("9" * beg_len)
+
+
+def is_match(num: int, comb: list[str]) -> bool:
+    if num in [1, 0]:
+        return False
+    numstr = str(num).zfill(len(comb))
+    numstr = sorted(numstr)
+    return numstr == sorted(comb)
+
+
+@timer
+def main_just_compute(power: int, strict: bool = False) -> None:
+    res = []
+    top = top_number(power)
+    for i in range(2, top + 1):
+        is_sum, sumx = fifth_power(i, power, strict)
+        if is_sum:
+            print(sumx, end="\t")
+            res.append(sumx)
+    print(f"\nThe sum is {sum(res):,d}")
+
+
+@timer
+def main_combinations(power: int) -> None:
+    top_len = len(str(top_number(power)))
+    res = set()
+    combs = combinations_with_replacement("0123456789", top_len)
+    for comb in combs:
+        tmp = sum(map(lambda x: int(x) ** power, comb))
+        if is_match(tmp, list(comb)):
+            print(tmp, end="\t")
+            res.add(tmp)
+    print(f"\nThe sum is {sum(res):,d}")
+
+
+if __name__ == "__main__":
+    the_power = 5
+    main_just_compute(power=the_power, strict=False)
+    print("=" * 50)
+    main_combinations(power=the_power)
diff --git a/solutions/0031/euler_31.py b/solutions/0031/euler_31.py
new file mode 100644
index 0000000..bee18e9
--- /dev/null
+++ b/solutions/0031/euler_31.py
@@ -0,0 +1,27 @@
+"""
+In the United Kingdom the currency is made up of pound (£) and pence (p).
+There are eight coins in general circulation:
+
+    1p, 2p, 5p, 10p, 20p, 50p, £1 (100p), and £2 (200p).
+
+It is possible to make £2 in the following way:
+
+    1×£1 + 1×50p + 2×20p + 1×5p + 1×2p + 3×1p
+
+How many different ways can £2 be made using any number of coins?
+"""
+
+
+def main():
+    coins = [1, 2, 5, 10, 20, 50, 100, 200]
+    ways = [1] + [0] * 200
+
+    for coin in coins:
+        for i in range(coin, 201):
+            ways[i] += ways[i - coin]
+
+    print(f"\nThe number of ways to make £2 is {ways[200]:,d}")
+
+
+if __name__ == "__main__":
+    main()
diff --git a/solutions/0031/readme.md b/solutions/0031/readme.md
new file mode 100644
index 0000000..bb2fdc2
--- /dev/null
+++ b/solutions/0031/readme.md
@@ -0,0 +1,114 @@
+# 动态规划法
+
+## 核心思路
+
+这个函数采用**动态规划**的思想，通过构建一个 `ways` 数组来累积计算每个金额对应的方法数。
+
+---
+
+## 逐行解析
+
+### 1. 初始化
+```python
+coins = [1, 2, 5, 10, 20, 50, 100, 200]
+ways = [1] + [0] * 200  # 结果是 [1, 0, 0, 0, ..., 0] (共201个元素)
+```
+
+- `coins`: 所有可用的硬币面值（单位：便士）
+- `ways[i]`: 表示凑成金额 `i` 的方法数
+- **关键点**：`ways[0] = 1` 表示凑成0元有1种方法（什么都不用），这是动态规划的**基准条件**
+
+### 2. 双重循环的核心逻辑
+```python
+for coin in coins:           # 按顺序遍历每种硬币
+    for i in range(coin, 201):  # 从当前硬币面值遍历到200
+        ways[i] += ways[i - coin]
+```
+
+这就是**状态转移方程**，其含义是：
+
+> **凑成金额 `i` 的方法数 = 原来方法数 + 使用当前硬币的方法数**
+
+其中 `ways[i - coin]` 表示：在使用了1枚当前硬币后，凑齐剩余金额的方法数。
+
+---
+
+## 具体执行过程演示
+
+让我们跟踪计算前几个值的变化，以理解算法如何工作：
+
+### 初始状态
+```
+ways = [1, 0, 0, 0, 0, 0, ...]  # ways[0]=1
+```
+
+### 第1轮：使用硬币 1p
+```python
+for i in range(1, 201):
+    ways[i] += ways[i-1]
+```
+
+- `i=1`: `ways[1] += ways[0]` → `0 + 1 = 1`  ✓ 凑1p: {1}
+- `i=2`: `ways[2] += ways[1]` → `0 + 1 = 1`  ✓ 凑2p: {1,1}
+- `i=3`: `ways[3] += ways[2]` → `0 + 1 = 1`  ✓ 凑3p: {1,1,1}
+- ...
+- **结果**：只用1p硬币，每个金额都有且仅有1种方法
+
+### 第2轮：加入硬币 2p
+```python
+for i in range(2, 201):
+    ways[i] += ways[i-2]
+```
+
+关键更新点：
+- `i=2`: `ways[2] += ways[0]` → `1 + 1 = 2` 
+  - 原来：{1,1}
+  - 新增：{2}
+- `i=3`: `ways[3] += ways[1]` → `1 + 1 = 2`
+  - 原来：{1,1,1}
+  - 新增：{1,2}
+- `i=4`: `ways[4] += ways[2]` → `1 + 2 = 3`
+  - 原来：{1,1,1,1}
+  - 新增：{1,1,2}, {2,2}
+
+### 第3轮：加入硬币 5p
+当加入5p硬币后，凑5p的方法从1种 {1,1,1,1,1} 变成2种，新增 {5}。
+
+**以此类推**，直到处理完所有硬币类型。
+
+---
+
+## 为什么这样计算？
+
+这个算法的精妙之处在于：
+
+1. **按硬币顺序处理**：确保计算的是**组合数**（不考虑顺序）
+   - {1,2,2} 和 {2,1,2} 视为同一种方法
+   - 如果调换循环顺序（先金额后硬币），会计算出排列数
+
+2. **累积效应**：`ways` 数组保存的是**所有已处理硬币**能凑成的方法总数
+
+3. **避免重复**：每种硬币只被考虑一次在其对应的外层循环中，确保不重复计算
+
+---
+
+## 最终结果
+
+当循环结束后，`ways[200]` 的值就是 200（便士）能被凑成的所有方法数。
+
+```python
+print(f"\nThe number of ways to make £2 is {ways[200]:,d}")
+```
+
+输出结果是：**73682**
+
+这意味着用这8种英国硬币凑成2英镑共有**73,682**种不同的方式。
+
+---
+
+## 时间复杂度
+
+- **O(N×M)**：其中 N = 硬币种类数（8），M = 目标金额（200）
+- 实际计算量约为 8 × 200 = 1600 次操作，效率极高
+
+这个算法简洁优雅，展示了动态规划在组合计数问题中的强大威力。