From a5829c4e72171824d80efd5825ea921d0953a989 Mon Sep 17 00:00:00 2001
From: Shadow <15656193+shadowatomic@user.noreply.gitee.com>
Date: Sat, 25 Oct 2025 00:21:14 +0800
Subject: [PATCH 1/2] =?UTF-8?q?20251025=E7=AC=AC=E4=B8=80=E6=AC=A1?=
 =?UTF-8?q?=E6=8F=90=E4=BA=A4?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 ..._\351\203\221\346\235\203\345\263\260.cpp" | 132 ++++++++++++++++++
 1 file changed, 132 insertions(+)
 create mode 100644 "exercise/LC2048_\351\203\221\346\235\203\345\263\260.cpp"

diff --git "a/exercise/LC2048_\351\203\221\346\235\203\345\263\260.cpp" "b/exercise/LC2048_\351\203\221\346\235\203\345\263\260.cpp"
new file mode 100644
index 0000000..0196d5d
--- /dev/null
+++ "b/exercise/LC2048_\351\203\221\346\235\203\345\263\260.cpp"
@@ -0,0 +1,132 @@
+#include <bits/stdc++.h>
+using namespace std;
+
+// 【题目】LC2048. 下一个更大的数值平衡数
+// 【难度】中等 （难度分：1734） 
+// 【提交】2025.10.24 https://leetcode.cn/problems/next-greater-numerically-balanced-number/submissions/673273348
+// 【标签】哈希表、数学、回溯、计数、枚举、第264场周赛 
+
+// 方法1：枚举（执行时间：659ms | 击败：57.14%）
+class Solution {
+public:
+    bool is_balance(int x){
+        vector<int> count(10);
+        while(x > 0){
+            count[x % 10]++;
+            x /= 10;
+        }
+        for(int i = 0; i < 10; i++){
+            if(count[i] > 0 && count[i] != i) return false;
+        }
+        return true;
+    }
+
+    int nextBeautifulNumber(int n) {
+        for(int i = n + 1; i <= 1224444; i++){
+            if(is_balance(i)) return i;
+        }
+        return -1;
+    }
+};
+
+// 方法2：倒序贪心（执行时间：0ms | 击败：100.00%）
+class Solution {
+    pair<vector<int>, bool> zero_one_knapspace(vector<int> &a, int target){
+        int n = a.size();
+        vector f(n + 1, vector<int8_t>(target + 1));
+        f[n][0] = true;
+
+        for(int i = n - 1; i >= 0; i--){
+            int v = a[i];
+            for(int j = 0; j <= target; j++){
+                if(j < v) f[i][j] = f[i + 1][j];
+                else f[i][j] = f[i + 1][j] || f[i + 1][j - v];
+            }
+        }
+        if(!f[0][target]) return {};
+        vector<int> ans;
+        int j = target;
+        for(int i = 0; i < n; i++){
+            int v = a[i];
+            if(j >= v && f[i + 1][j - v]){
+                ans.emplace_back(v);
+                j -= v;
+            }
+        }
+        return {ans, true};
+    }
+
+public:
+    int nextBeautifulNumber(int n) {
+        string s = "0" + to_string(n);
+        int m = s.size();
+        constexpr int MX = 10;
+        int cnt[MX]{};
+        for(int i = 1; i < m; i++) cnt[s[i] - '0']++;
+
+        for(int i = m - 1; i >= 0; i--){
+            if(i > 0) cnt[s[i] - '0']--;
+            for(int j = s[i] - '0' + 1; j < MX; j++){
+                cnt[j]++;
+                int free = m - 1 - i;
+                for(int k = 0; k < MX; k++){
+                    int c = cnt[k];
+                    if(k < c){
+                        free = -1;
+                        break;
+                    }
+                    if(c > 0) free -= k - c;
+                }
+                if(free < 0){
+                    cnt[j]--;
+                    continue;
+                }
+                vector<int> a;
+                for(int k = 1; k < MX; k++){
+                    if(cnt[k] == 0) a.emplace_back(k);
+                }
+                auto [missing, ok] = zero_one_knapspace(a, free);
+                if(!ok){
+                    cnt[j]--;
+                    continue;
+                }
+                for(int v : missing) cnt[v] = -v;
+                s[i] = '0' + j;
+                s.resize(i + 1);
+                for(int k = 1; k < MX; k++){
+                    int c = cnt[k];
+                    c = c > 0 ? k - c : -c;
+                    s += string(c, '0' + k);
+                }
+                return stoi(s);
+            }
+        }
+        return -1;
+    }
+};
+
+// 方法3：打表 + 二分查找（执行时间：0ms | 击败：100.00%）
+class Solution {
+public:
+    const vector<int> balance {
+        1, 22, 122, 212, 221, 333, 1333, 3133, 3313, 3331, 4444,
+        14444, 22333, 23233, 23323, 23332, 32233, 32323, 32332,
+        33223, 33232, 33322, 41444, 44144, 44414, 44441, 55555,
+        122333, 123233, 123323, 123332, 132233, 132323, 132332,
+        133223, 133232, 133322, 155555, 212333, 213233, 213323,
+        213332, 221333, 223133, 223313, 223331, 224444, 231233,
+        231323, 231332, 232133, 232313, 232331, 233123, 233132,
+        233213, 233231, 233312, 233321, 242444, 244244, 244424,
+        244442, 312233, 312323, 312332, 313223, 313232, 313322,
+        321233, 321323, 321332, 322133, 322313, 322331, 323123,
+        323132, 323213, 323231, 323312, 323321, 331223, 331232,
+        331322, 332123, 332132, 332213, 332231, 332312, 332321,
+        333122, 333212, 333221, 422444, 424244, 424424, 424442,
+        442244, 442424, 442442, 444224, 444242, 444422, 515555,
+        551555, 555155, 555515, 555551, 666666, 1224444
+    };
+
+    int nextBeautifulNumber(int n) {
+        return *upper_bound(balance.begin(), balance.end(), n);
+    }
+};
-- 
Gitee


From 00e4f4a96841c4221a63ce89e7125a3ed7a3c3dd Mon Sep 17 00:00:00 2001
From: Shadow <15656193+shadowatomic@user.noreply.gitee.com>
Date: Sun, 26 Oct 2025 01:08:54 +0800
Subject: [PATCH 2/2] =?UTF-8?q?20251025=E7=AC=AC=E4=B8=80=E6=AC=A1?=
 =?UTF-8?q?=E6=8F=90=E4=BA=A4?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 ..._\351\203\221\346\235\203\345\263\260.cpp" | 17 ++++
 ..._\351\203\221\346\235\203\345\263\260.cpp" | 79 +++++++++++++++++++
 ..._\351\203\221\346\235\203\345\263\260.cpp" | 47 +++++++++++
 ..._\351\203\221\346\235\203\345\263\260.cpp" | 55 +++++++++++++
 ..._\351\203\221\346\235\203\345\263\260.cpp" | 58 ++++++++++++++
 ..._\351\203\221\346\235\203\345\263\260.cpp" | 52 ++++++++++++
 6 files changed, 308 insertions(+)
 create mode 100644 "exercise/LC1716_\351\203\221\346\235\203\345\263\260.cpp"
 create mode 100644 "topic05/submit/LC2266_\351\203\221\346\235\203\345\263\260.cpp"
 create mode 100644 "topic05/submit/LC2466_\351\203\221\346\235\203\345\263\260.cpp"
 create mode 100644 "topic05/submit/LC300_\351\203\221\346\235\203\345\263\260.cpp"
 create mode 100644 "topic05/submit/LC377_\351\203\221\346\235\203\345\263\260.cpp"
 create mode 100644 "topic05/submit/LC746_\351\203\221\346\235\203\345\263\260.cpp"

diff --git "a/exercise/LC1716_\351\203\221\346\235\203\345\263\260.cpp" "b/exercise/LC1716_\351\203\221\346\235\203\345\263\260.cpp"
new file mode 100644
index 0000000..a772610
--- /dev/null
+++ "b/exercise/LC1716_\351\203\221\346\235\203\345\263\260.cpp"
@@ -0,0 +1,17 @@
+#include <bits/stdc++.h>
+using namespace std;
+
+// 【题目】LC1716. 计算力扣银行的钱
+// 【难度】简单 
+// 【提交】2025.10.25 https://leetcode.cn/problems/calculate-money-in-leetcode-bank/submissions/673373028
+// 【标签】数学、463双周赛 
+
+// 方法1：（执行时间：0ms | 击败：100.00%） 
+class Solution {
+public:
+    int totalMoney(int n) {
+        int cnt = n / 7;
+        int remain = n % 7;
+        return (cnt * 21 + 7 * (1 + cnt) * cnt / 2) + (1 + remain) * remain / 2 + remain * cnt;
+    }
+};
diff --git "a/topic05/submit/LC2266_\351\203\221\346\235\203\345\263\260.cpp" "b/topic05/submit/LC2266_\351\203\221\346\235\203\345\263\260.cpp"
new file mode 100644
index 0000000..40e0ca4
--- /dev/null
+++ "b/topic05/submit/LC2266_\351\203\221\346\235\203\345\263\260.cpp"
@@ -0,0 +1,79 @@
+#include <bits/stdc++.h>
+using namespace std;
+
+// 【题目】LC2266. 统计打字方案数
+// 【难度】中等 （难度分：1857） 
+// 【提交】2025.10.25 
+// 【标签】哈希表、数学、字符串、动态规划、第292场双周赛 
+
+// 方法1：动态规划（执行时间：4ms | 击败：80.82%） 
+#define ll long long
+
+int m = 1'000'000'007;
+int MAX = 100001;
+vector<ll> dp3 = {1, 1, 2, 4};
+vector<ll> dp4 = {1, 1, 2, 4};
+
+int init = [](){
+    for(int i = 4; i <= MAX; i++){
+        dp3.emplace_back((dp3[i - 1] + dp3[i - 2] + dp3[i - 3]) % m);
+        dp4.emplace_back((dp4[i - 1] + dp4[i - 2] + dp4[i - 3] + dp4[i - 4]) % m);
+    }
+    return 0;
+}();
+
+class Solution {
+public:
+    int countTexts(string pressedKeys) {
+        int n = pressedKeys.size();
+        ll res = 1;
+        int cnt = 1;
+        for(int i = 1; i < n; i++){
+            if(pressedKeys[i] == pressedKeys[i - 1]) cnt++;
+            else {
+                if(pressedKeys[i - 1] == '7' || pressedKeys[i - 1] == '9') res *= dp4[cnt];
+                else res *= dp3[cnt];
+                res %= m;
+                cnt = 1;
+            }
+        }
+        if(pressedKeys[n - 1] == '7' || pressedKeys[n - 1] == '9') res *= dp4[cnt];
+        else res *= dp3[cnt];
+        res %= m;
+        return res;
+    }
+};
+
+// 方法2： （执行时间：0ms | 击败：100.00%） 
+#define ll long long
+
+int m = 1'000'000'007;
+int MAX = 100001;
+vector<ll> dp3 = {1, 1, 2, 4};
+vector<ll> dp4 = {1, 1, 2, 4};
+
+int init = [](){
+    for(int i = 4; i <= MAX; i++){
+        dp3.emplace_back((dp3[i - 1] + dp3[i - 2] + dp3[i - 3]) % m);
+        dp4.emplace_back((dp4[i - 1] + dp4[i - 2] + dp4[i - 3] + dp4[i - 4]) % m);
+    }
+    return 0;
+}();
+
+class Solution {
+public:
+    int countTexts(string s) {
+        int n = s.size();
+        ll res = 1;
+        int cnt = 0;
+        for(int i = 0; i < n; i++){
+            char c = s[i];
+            cnt++;
+            if(i == n - 1 || s[i] != s[i + 1]) {
+                res = res * (c != '7' && c != '9' ? dp3[cnt] : dp4[cnt]) % m;
+                cnt = 0;
+            }
+        }
+        return res;
+    }
+}; 
diff --git "a/topic05/submit/LC2466_\351\203\221\346\235\203\345\263\260.cpp" "b/topic05/submit/LC2466_\351\203\221\346\235\203\345\263\260.cpp"
new file mode 100644
index 0000000..f97fcd5
--- /dev/null
+++ "b/topic05/submit/LC2466_\351\203\221\346\235\203\345\263\260.cpp"
@@ -0,0 +1,47 @@
+#include <bits/stdc++.h>
+using namespace std;
+
+// 【题目】LC2466. 统计构造好字符串的方案数
+// 【难度】中等 （难度分：1694） 
+// 【提交】2025.10.25 
+// 【标签】动态规划、第91场双周赛 
+
+// 方法1：用动态规划递推（执行时间：3ms | 击败：98.32%） 
+class Solution {
+public:
+    int countGoodStrings(int low, int high, int zero, int one) {
+        const int MOD = 1'000'000'007;
+        int ans = 0;
+        vector<int> dp(high + 1);
+        dp[0] = 1;
+        for(int i = 1; i <= high; i++){
+            if(i >= zero) dp[i] = dp[i - zero];
+            if(i >= one) dp[i] = (dp[i] + dp[i - one]) % MOD;
+            if(i >= low) ans = (ans + dp[i]) % MOD;
+        }
+        return ans;
+    }
+};
+
+// 方法2：DFS + 记忆化搜索（执行时间：11ms | 击败：20.09%） 
+class Solution {
+public:
+    int countGoodStrings(int low, int high, int zero, int one) {
+        const int MOD = 1'000'000'007;
+        vector<int> memo(high + 1, -1);
+
+        auto dfs = [&](this auto&& dfs, int i){
+            if(i < 0) return 0;
+            if(i == 0) return 1;
+            int &res = memo[i];
+            if(res != -1) return res;
+            return res = (dfs(i - zero) + dfs(i - one)) % MOD;
+        };
+
+        int ans = 0;
+        for(int i = low; i <= high; i++){
+            ans = (ans + dfs(i)) % MOD;
+        }
+        return ans;
+    }
+};
diff --git "a/topic05/submit/LC300_\351\203\221\346\235\203\345\263\260.cpp" "b/topic05/submit/LC300_\351\203\221\346\235\203\345\263\260.cpp"
new file mode 100644
index 0000000..30d9628
--- /dev/null
+++ "b/topic05/submit/LC300_\351\203\221\346\235\203\345\263\260.cpp"
@@ -0,0 +1,55 @@
+#include <bits/stdc++.h>
+using namespace std;
+
+// 【题目】LC300. 最长递增子序列
+// 【难度】中等 
+// 【提交】2025.10.25 https://leetcode.cn/problems/longest-increasing-subsequence/submissions/673376067
+// 【标签】数组、二分查找、动态规划 
+
+// 方法1：（执行时间：79ms | 击败：49.22%） 
+class Solution {
+public:
+    int lengthOfLIS(vector<int>& nums) {
+        int n = nums.size();
+        vector<int> dp(n + 1);
+        dp[1] = 1;
+        for(int i = 2; i <= n; i++){
+            int max_num = 0;
+            for(int j = 1; j < i; j++){
+                if(nums[j - 1] < nums[i - 1] && dp[j] > max_num) max_num = dp[j];
+            }
+            dp[i] = max_num + 1;
+        }
+        return ranges::max(dp);
+    }
+};
+
+// 贪心 + 二分查找 （执行时间：0ms | 击败：100.00%）
+class Solution {
+public:
+    int lengthOfLIS(vector<int>& nums) {
+        vector<int> g;
+        for(int x : nums){
+            auto it = ranges::lower_bound(g, x);
+            if(it == g.end()) g.emplace_back(x);
+            else *it = x;
+        }
+        return g.size();
+    }
+};
+
+// 原地修改（空间复杂度: O(1)） （执行时间：0ms | 击败：100.00%）
+class Solution {
+public:
+    int lengthOfLIS(vector<int>& nums) {
+        auto end = nums.begin();
+        for(int x : nums){
+            auto it = lower_bound(nums.begin(), end, x);
+            if(it == end){
+                end++;
+            }
+            *it = x;
+        }
+        return end - nums.begin();
+    }
+};
diff --git "a/topic05/submit/LC377_\351\203\221\346\235\203\345\263\260.cpp" "b/topic05/submit/LC377_\351\203\221\346\235\203\345\263\260.cpp"
new file mode 100644
index 0000000..127e587
--- /dev/null
+++ "b/topic05/submit/LC377_\351\203\221\346\235\203\345\263\260.cpp"
@@ -0,0 +1,58 @@
+#include <bits/stdc++.h>
+using namespace std;
+
+// 【题目】LC377. 组合总和 Ⅳ
+// 【难度】中等 
+// 【提交】2025.10.25 https://leetcode.cn/problems/combination-sum-iv/submissions/673467029
+// 【标签】数组、动态规划 
+
+// 方法1：（执行时间：0ms | 击败：100.00%） 
+class Solution {
+public:
+    int combinationSum4(vector<int>& nums, int target) {
+        vector<int> f(target + 1);
+        f[0] = 1;
+        for(int i = 1; i <= target; i++){
+            for(int x : nums){
+                if(x <= i && f[i - x] <= INT_MAX - f[i]) f[i] += f[i - x];
+            }
+        }
+        return f[target];
+    }
+};
+
+// 方法2：（执行时间：0ms | 击败：100.00%）
+using ll = long long;
+class Solution {
+public:
+    int combinationSum4(vector<int>& nums, int target) {
+        vector<unsigned> f(target + 1);
+        f[0] = 1;
+        for(int i = 1; i <= target; i++){
+            for(int x : nums){
+                if(x <= i) f[i] += f[i - x];
+            }
+        }
+        return f[target];
+    }
+};
+
+// 方法3：DFS剪枝（执行时间：0ms | 击败：100.00%）
+class Solution {
+public:
+    int combinationSum4(vector<int>& nums, int target) {
+        vector<int> memo(target + 1, -1);
+        
+        auto dfs = [&](this auto&& dfs, int i){
+            if(i == 0) return 1;
+            int &res = memo[i];
+            if(res != -1) return res;
+
+            res = 0;
+            for(int x : nums) if(x <= i) res += dfs(i - x);
+            return res;
+        };
+
+        return dfs(target);
+    }
+};
diff --git "a/topic05/submit/LC746_\351\203\221\346\235\203\345\263\260.cpp" "b/topic05/submit/LC746_\351\203\221\346\235\203\345\263\260.cpp"
new file mode 100644
index 0000000..01565e0
--- /dev/null
+++ "b/topic05/submit/LC746_\351\203\221\346\235\203\345\263\260.cpp"
@@ -0,0 +1,52 @@
+#include <bits/stdc++.h>
+using namespace std;
+
+// 【题目】LC746. 使用最小花费爬楼梯
+// 【难度】简单 （难度分：1358） 
+// 【提交】2025.10.25 https://leetcode.cn/problems/min-cost-climbing-stairs/submissions/673422808
+// 【标签】数组、动态规划 
+
+// 方法1：（执行时间：0ms | 击败：100.00%） 
+class Solution {
+public:
+    int minCostClimbingStairs(vector<int>& cost) {
+        int n = cost.size();
+        if(n == 2) return min(cost[0], cost[1]);
+        vector<int> f(n + 1);
+        f[0] = cost[0];
+        f[1] = cost[1];
+        for(int i = 2; i < n; i++){
+            f[i] = min(f[i - 1], f[i - 2]) + cost[i];
+        }
+        return min(f[n - 1], f[n - 2]);
+    }
+};
+
+// 方法2： （执行时间：0ms | 击败：100.00%）
+class Solution {
+public:
+    int minCostClimbingStairs(vector<int>& cost) {
+        int n = cost.size();
+        vector<int> f(n + 1);
+        f[0] = f[1] = 0;
+        for(int i = 2; i <= n; i++){
+            f[i] = min(f[i - 1] + cost[i - 1], f[i - 2] + cost[i - 2]);
+        }
+        return f[n];
+    }
+};
+
+// 方法3：（执行时间：0ms | 击败：100.00%）
+class Solution {
+public:
+    int minCostClimbingStairs(vector<int>& cost) {
+        int n = cost.size();
+        vector<int> f(n + 1);
+        f[0] = 0;
+        f[1] = min(cost[0], cost[1]);
+        for(int i = 2; i < n; i++){
+            f[i] = min(f[i - 1] + cost[i], f[i - 2] + cost[i - 1]);
+        }
+        return f[n - 1];
+    }
+};
-- 
Gitee