update: 添加问题“2327.知道秘密的人数”的代码和题解 (#1122)

LetMeFly666 · web-flow · commit c7e5241daf35 · 2025-09-11T19:10:42.000+08:00
* 2327: AC.cpp (#1121) cpp - AC,36.69%,23.02% * word: 今早CV的昨天题的解法二模板到现在也还没来得及写 * 2327: WA.cpp (#1121) * 2327: WA.cpp (#1121) * 2327: WA.cpp (#1121) - 果然要+mod再模（至少最后） * 2327: AC.cpp (#1121) cpp - AC,100.00%, 64.29% * update: 添加问题“2327.知道秘密的人数”的代码和题解 (#1122) Signed-off-by: LetMeFly666 <814114971@qq.com> * docs: multi 2327 (#1121) #1122 (comment) --------- Signed-off-by: LetMeFly666 <814114971@qq.com>
diff --git a/Codes/2327-number-of-people-aware-of-a-secret.cpp b/Codes/2327-number-of-people-aware-of-a-secret.cpp
@@ -0,0 +1,29 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-09-09 23:42:14
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-09-09 23:52:39
+ */
+#if defined(_WIN32) || defined(__APPLE__)
+#include "_[1,2]toVector.h"
+#endif
+
+using ll = long long;
+const ll MOD = 1e9 + 7;
+class Solution {
+public:
+    int peopleAwareOfSecret(int n, int delay, int forget) {
+        vector<ll> dp(n);
+        dp[0] = 1;
+        for (int i = 0; i < n; i++) {
+            for (int j = i + delay; j < i + forget && j < n; j++) {
+                dp[j] = (dp[j] + dp[i]) % MOD;
+            }
+        }
+        ll ans = 0;
+        for (int i = 0; i < forget; i++) {
+            ans = (ans + dp[n - i - 1]) % MOD;
+        }
+        return ans;
+    }
+};
diff --git a/Codes/2327-number-of-people-aware-of-a-secret_ChaFen.cpp b/Codes/2327-number-of-people-aware-of-a-secret_ChaFen.cpp
@@ -0,0 +1,50 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-09-09 23:42:14
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-09-11 10:49:54
+ */
+#if defined(_WIN32) || defined(__APPLE__)
+#include "_[1,2]toVector.h"
+#endif
+
+using ll = long long;
+const ll MOD = 1e9 + 7;
+class Solution {
+public:
+    int peopleAwareOfSecret(int n, int delay, int forget) {
+        vector<ll> diff(n + 1);
+        diff[1] = 1;
+        diff[2] = -1;
+        ll now = 0, ans = 0;
+        for (int i = 1; i <= n; i++) {
+            now = (now + diff[i]) % MOD;
+            if (i + forget > n) {
+                ans = (ans + now) % MOD;
+            }
+            if (i + delay <= n) {
+                diff[i + delay] = (diff[i + delay] + now) % MOD;
+            }
+            if (i + forget <= n) {
+                diff[i + forget] = (diff[i + forget] + MOD - now) % MOD;
+            }
+        }
+        return ans;
+    }
+};
+
+#if defined(_WIN32) || defined(__APPLE__)
+/*
+6 2 4
+
+5
+*/
+int main() {
+    int a, b, c;
+    while (cin >> a >> b >> c) {
+        Solution sol;
+        cout << sol.peopleAwareOfSecret(a, b, c) << endl;
+    }
+    return 0;
+}
+#endif
diff --git a/README.md b/README.md
@@ -740,6 +740,7 @@
 |2311.小于等于K的最长二进制子序列|中等|<a href="https://leetcode.cn/problems/longest-binary-subsequence-less-than-or-equal-to-k/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2025/06/26/LeetCode%202311.%E5%B0%8F%E4%BA%8E%E7%AD%89%E4%BA%8EK%E7%9A%84%E6%9C%80%E9%95%BF%E4%BA%8C%E8%BF%9B%E5%88%B6%E5%AD%90%E5%BA%8F%E5%88%97/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/148935308" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/longest-binary-subsequence-less-than-or-equal-to-k/solutions/3709306/letmefly-2311xiao-yu-deng-yu-k-de-zui-ch-2n27/" target="_blank">LeetCode题解</a>|
 |2312.卖木头块|困难|<a href="https://leetcode.cn/problems/selling-pieces-of-wood/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2024/03/15/LeetCode%202312.%E5%8D%96%E6%9C%A8%E5%A4%B4%E5%9D%97/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/136747100" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/selling-pieces-of-wood/solutions/2689067/letmefly-2312mai-mu-tou-kuai-dong-tai-gu-vk5a/" target="_blank">LeetCode题解</a>|
 |2316.统计无向图中无法互相到达点对数|中等|<a href="https://leetcode.cn/problems/count-unreachable-pairs-of-nodes-in-an-undirected-graph/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2023/10/21/LeetCode%202316.%E7%BB%9F%E8%AE%A1%E6%97%A0%E5%90%91%E5%9B%BE%E4%B8%AD%E6%97%A0%E6%B3%95%E4%BA%92%E7%9B%B8%E5%88%B0%E8%BE%BE%E7%82%B9%E5%AF%B9%E6%95%B0/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/133962709" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/count-unreachable-pairs-of-nodes-in-an-undirected-graph/solutions/2492186/letmefly-2316tong-ji-wu-xiang-tu-zhong-w-79vz/" target="_blank">LeetCode题解</a>|
+|2327.知道秘密的人数|中等|<a href="https://leetcode.cn/problems/number-of-people-aware-of-a-secret/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2025/09/11/LeetCode%202327.%E7%9F%A5%E9%81%93%E7%A7%98%E5%AF%86%E7%9A%84%E4%BA%BA%E6%95%B0/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/151586337" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/number-of-people-aware-of-a-secret/solutions/3779067/letmefly-2327zhi-dao-mi-mi-de-ren-shu-do-y3ey/" target="_blank">LeetCode题解</a>|
 |2335.装满杯子需要的最短总时长|简单|<a href="https://leetcode.cn/problems/minimum-amount-of-time-to-fill-cups/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2023/02/11/LeetCode%202335.%E8%A3%85%E6%BB%A1%E6%9D%AF%E5%AD%90%E9%9C%80%E8%A6%81%E7%9A%84%E6%9C%80%E7%9F%AD%E6%80%BB%E6%97%B6%E9%95%BF/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/128980819" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/minimum-amount-of-time-to-fill-cups/solutions/2105156/letmefly-2335zhuang-man-bei-zi-xu-yao-de-kq65/" target="_blank">LeetCode题解</a>|
 |2336.无限集中的最小数字|中等|<a href="https://leetcode.cn/problems/smallest-number-in-infinite-set/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2023/11/29/LeetCode%202336.%E6%97%A0%E9%99%90%E9%9B%86%E4%B8%AD%E7%9A%84%E6%9C%80%E5%B0%8F%E6%95%B0%E5%AD%97/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/134687046" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/smallest-number-in-infinite-set/solutions/2546486/letmefly-2336wu-xian-ji-zhong-de-zui-xia-wzz1/" target="_blank">LeetCode题解</a>|
 |2337.移动片段得到字符串|中等|<a href="https://leetcode.cn/problems/move-pieces-to-obtain-a-string/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2023/08/22/LeetCode%202337.%E7%A7%BB%E5%8A%A8%E7%89%87%E6%AE%B5%E5%BE%97%E5%88%B0%E5%AD%97%E7%AC%A6%E4%B8%B2/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/132421605" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/move-pieces-to-obtain-a-string/solutions/2399221/letmefly-2337yi-dong-pian-duan-de-dao-zi-oi31/" target="_blank">LeetCode题解</a>|
diff --git a/Solutions/LeetCode 2327.知道秘密的人数.md b/Solutions/LeetCode 2327.知道秘密的人数.md
@@ -0,0 +1,155 @@
+---
+title: 2327.知道秘密的人数：动态规划/差分数组O(n)
+date: 2025-09-11 13:00:17
+tags: [题解, LeetCode, 中等, 队列, 动态规划, 模拟]
+categories: [题解, LeetCode]
+---
+
+# 【LetMeFly】2327.知道秘密的人数：动态规划/差分数组O(n)
+
+力扣题目链接：[https://leetcode.cn/problems/number-of-people-aware-of-a-secret/](https://leetcode.cn/problems/number-of-people-aware-of-a-secret/)
+
+<p>在第 <code>1</code>&nbsp;天，有一个人发现了一个秘密。</p>
+
+<p>给你一个整数&nbsp;<code>delay</code>&nbsp;，表示每个人会在发现秘密后的 <code>delay</code>&nbsp;天之后，<strong>每天</strong>&nbsp;给一个新的人&nbsp;<strong>分享</strong>&nbsp;秘密。同时给你一个整数&nbsp;<code>forget</code>&nbsp;，表示每个人在发现秘密&nbsp;<code>forget</code>&nbsp;天之后会&nbsp;<strong>忘记</strong>&nbsp;这个秘密。一个人&nbsp;<strong>不能</strong>&nbsp;在忘记秘密那一天及之后的日子里分享秘密。</p>
+
+<p>给你一个整数&nbsp;<code>n</code>&nbsp;，请你返回在第 <code>n</code>&nbsp;天结束时，知道秘密的人数。由于答案可能会很大，请你将结果对&nbsp;<code>10<sup>9</sup> + 7</code>&nbsp;<strong>取余</strong>&nbsp;后返回。</p>
+
+<p>&nbsp;</p>
+
+<p><strong>示例 1：</strong></p>
+
+<pre><b>输入：</b>n = 6, delay = 2, forget = 4
+<b>输出：</b>5
+<strong>解释：</strong>
+第 1 天：假设第一个人叫 A 。（一个人知道秘密）
+第 2 天：A 是唯一一个知道秘密的人。（一个人知道秘密）
+第 3 天：A 把秘密分享给 B 。（两个人知道秘密）
+第 4 天：A 把秘密分享给一个新的人 C 。（三个人知道秘密）
+第 5 天：A 忘记了秘密，B 把秘密分享给一个新的人 D 。（三个人知道秘密）
+第 6 天：B 把秘密分享给 E，C 把秘密分享给 F 。（五个人知道秘密）
+</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre><b>输入：</b>n = 4, delay = 1, forget = 3
+<b>输出：</b>6
+<strong>解释：</strong>
+第 1 天：第一个知道秘密的人为 A 。（一个人知道秘密）
+第 2 天：A 把秘密分享给 B 。（两个人知道秘密）
+第 3 天：A 和 B 把秘密分享给 2 个新的人 C 和 D 。（四个人知道秘密）
+第 4 天：A 忘记了秘密，B、C、D 分别分享给 3 个新的人。（六个人知道秘密）
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>2 &lt;= n &lt;= 1000</code></li>
+	<li><code>1 &lt;= delay &lt; forget &lt;= n</code></li>
+</ul>
+
+
+    
+## 解题方法一：动态规划
+
+> (为了方便描述)也可以把这道题看成细菌繁殖，细菌delay天成熟forget天死亡并且期间每天复制自身一份。
+
+令`dp[i]`代表第`i-1`天新生细菌的数量，初始值`dp[0]=1`，其余`dp[i]=0`。
+
+从第`1`天开始遍历到第`n`天，第`i`天出生的细菌可以在第`[i+delay, i+forget)`天分别产下一个新细菌，所以有`dp[i+j] += dp[i]`，其中`i+delay<=j<i+forget`。
+
++ 时间复杂度$O(n\times(forget-delay))$
++ 空间复杂度$O(n)$
+
+### AC代码
+
+#### C++
+
+```cpp
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-09-09 23:42:14
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-09-09 23:52:39
+ */
+#if defined(_WIN32) || defined(__APPLE__)
+#include "_[1,2]toVector.h"
+#endif
+
+using ll = long long;
+const ll MOD = 1e9 + 7;
+class Solution {
+public:
+    int peopleAwareOfSecret(int n, int delay, int forget) {
+        vector<ll> dp(n);
+        dp[0] = 1;
+        for (int i = 0; i < n; i++) {
+            for (int j = i + delay; j < i + forget && j < n; j++) {
+                dp[j] = (dp[j] + dp[i]) % MOD;
+            }
+        }
+        ll ans = 0;
+        for (int i = 0; i < forget; i++) {
+            ans = (ans + dp[n - i - 1]) % MOD;
+        }
+        return ans;
+    }
+};
+```
+
+## 解题方法二：查分数组
+
+方法一中有一个耗时操作：对于第`i`天出生的细菌，令`i+delay`到`i+forget-1`天的细菌分别加上`dp[i]`。
+
+这个耗时操作不正是差分数组可以优化的吗？
+
+令`diff[i]=dp[i]-dp[i-1]`，想让`dp[i+delay]`到`dp[i+forget-1]`每个加`dp[i]`只需要令`diff[i+delay]+=dp[i]`和`diff[i+forget]-=dp[i]`。
+
+相应的，`dp[i]`就等于`sum(diff[0..i+1])`。
+
++ 时间复杂度$O(n)$
++ 空间复杂度$O(n)$
+
+### AC代码
+
+#### C++
+
+```cpp
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-09-09 23:42:14
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-09-11 10:49:54
+ */
+using ll = long long;
+const ll MOD = 1e9 + 7;
+class Solution {
+public:
+    int peopleAwareOfSecret(int n, int delay, int forget) {
+        vector<ll> diff(n + 1);
+        diff[1] = 1;
+        diff[2] = -1;
+        ll now = 0, ans = 0;
+        for (int i = 1; i <= n; i++) {
+            now = (now + diff[i]) % MOD;
+            if (i + forget > n) {
+                ans = (ans + now) % MOD;
+            }
+            if (i + delay <= n) {
+                diff[i + delay] = (diff[i + delay] + now) % MOD;
+            }
+            if (i + forget <= n) {
+                diff[i + forget] = (diff[i + forget] + MOD - now) % MOD;
+            }
+        }
+        return ans;
+    }
+};
+```
+
+
+> 同步发文于[CSDN](https://letmefly.blog.csdn.net/article/details/151586337)和我的[个人博客](https://blog.letmefly.xyz/)，原创不易，转载经作者同意后请附上[原文链接](https://blog.letmefly.xyz/2025/09/11/LeetCode%202327.%E7%9F%A5%E9%81%93%E7%A7%98%E5%AF%86%E7%9A%84%E4%BA%BA%E6%95%B0/)哦~
+>
+> 千篇源码题解[已开源](https://github.com/LetMeFly666/LeetCode)
diff --git a/Solutions/Other-English-LearningNotes-SomeWords.md b/Solutions/Other-English-LearningNotes-SomeWords.md
@@ -1423,6 +1423,10 @@ categories: [自用]
 |induction|n. 就职(仪式)，入门，诱导，归纳法，电磁效应|
 |||
 |abreast|adv. 并排地<br/>adj.(少见) 并肩的|
+|||
+|diction|n. 用词，用语，措辞，吐字|
+|||
+|faction|n. (大团体中的)派系，小集团，派系斗争，内讧|
 
 + 这个web要是能设计得可以闭眼(完全不睁眼)键盘控制背单词就好了。
 + 也许可以加个AI用最近词编故事功能(返回接口中支持标注所使用单词高亮?)
