From b61f47947df00859a4a0128e84d84f43c5b53a95 Mon Sep 17 00:00:00 2001
From: wisdompeak <guan.huifeng@gmail.com>
Date: 2025年9月14日 00:53:17 -0700
Subject: [PATCH 1/3] Create 3685.Subsequence-Sum-After-Capping-Elements.cpp
---
 ...Subsequence-Sum-After-Capping-Elements.cpp &#124; 29 +++++++++++++++++++
 1 file changed, 29 insertions(+)
 create mode 100644 Dynamic_Programming/3685.Subsequence-Sum-After-Capping-Elements/3685.Subsequence-Sum-After-Capping-Elements.cpp
diff --git a/Dynamic_Programming/3685.Subsequence-Sum-After-Capping-Elements/3685.Subsequence-Sum-After-Capping-Elements.cpp b/Dynamic_Programming/3685.Subsequence-Sum-After-Capping-Elements/3685.Subsequence-Sum-After-Capping-Elements.cpp
new file mode 100644
index 000000000..3633e575c
--- /dev/null
+++ b/Dynamic_Programming/3685.Subsequence-Sum-After-Capping-Elements/3685.Subsequence-Sum-After-Capping-Elements.cpp
@@ -0,0 +1,29 @@
+class Solution {
+public:
+ vector<bool> subsequenceSumAfterCapping(vector<int>& nums, int k) {
+ int n = nums.size();
+ vector<int>dp(k+1,0);
+ dp[0] = 1;
+ vector<bool>rets(n, 0);
+ 
+ sort(nums.begin(), nums.end());
+ int i = 0;
+ for (int x=1; x<=n; x++) { + while (i<n && nums[i]<x) { + for (int c = k; c>=1; c--) {
+ if (c<nums[i]) break; + dp[c] &#124;= dp[c-nums[i]]; + } + i++; + } + int m = n-i; + for (int j=0; j<=m && j*x<=k; j++) { + if (dp[k-j*x]) { + rets[x-1] = 1; + break; + } + } + } + return rets; + } +}; From a9399cb88505ee4d0b2d0a03811e747c33b28355 Mon Sep 17 00:00:00 2001 From: wisdompeak <guan.huifeng@gmail.com>
Date: 2025年9月14日 00:53:51 -0700
Subject: [PATCH 2/3] Update Readme.md
---
 Readme.md &#124; 1 +
 1 file changed, 1 insertion(+)
diff --git a/Readme.md b/Readme.md
index 9ced2324e..62a1937dc 100644
--- a/Readme.md
+++ b/Readme.md
@@ -888,6 +888,7 @@
 [2902.Count-of-Sub-Multisets-With-Bounded-Sum](https://github.com/wisdompeak/LeetCode/tree/master/Dynamic_Programming/2902.Count-of-Sub-Multisets-With-Bounded-Sum) (H) 
 [3489.Zero-Array-Transformation-IV](https://github.com/wisdompeak/LeetCode/tree/master/Dynamic_Programming/3489.Zero-Array-Transformation-IV) (H-) 
 [3592.Inverse-Coin-Change](https://github.com/wisdompeak/LeetCode/tree/master/Dynamic_Programming/3592.Inverse-Coin-Change) (H) 
+[3685.Subsequence-Sum-After-Capping-Elements](https://github.com/wisdompeak/LeetCode/tree/master/Dynamic_Programming/3685.Subsequence-Sum-After-Capping-Elements) (H) 
 * ``键盘型`` 
 [650.2-Keys-Keyboard](https://github.com/wisdompeak/LeetCode/blob/master/Dynamic_Programming/650.2-Keys-Keyboard) (M+) 
 [651.4-Keys-Keyboard](https://github.com/wisdompeak/LeetCode/tree/master/Dynamic_Programming/651.4-Keys-Keyboard) (M+) 
From 34b73c7a86fb529a3eb01b0ccad8afd5373b3b0a Mon Sep 17 00:00:00 2001
From: wisdompeak <guan.huifeng@gmail.com>
Date: 2025年9月14日 01:26:38 -0700
Subject: [PATCH 3/3] Create Readme.md
---
 .../Readme.md &#124; 24 +++++++++++++++++++
 1 file changed, 24 insertions(+)
 create mode 100644 Dynamic_Programming/3685.Subsequence-Sum-After-Capping-Elements/Readme.md
diff --git a/Dynamic_Programming/3685.Subsequence-Sum-After-Capping-Elements/Readme.md b/Dynamic_Programming/3685.Subsequence-Sum-After-Capping-Elements/Readme.md
new file mode 100644
index 000000000..d40cb52db
--- /dev/null
+++ b/Dynamic_Programming/3685.Subsequence-Sum-After-Capping-Elements/Readme.md
@@ -0,0 +1,24 @@
+### 3685.Subsequence-Sum-After-Capping-Elements
+
+对于n个元素,问是否可以组合出和为k的方案,就是一个典型的有界背包问题,用o(n*k)的时间复杂度可解。大致算法是:
+```cpp
+for (int i=0; i<nums.size(); i++) { + for (int c=k; c>=1; c--) {
+ dp[c] &#124;= dp[c-nums[i];
+ }
+}
+```
+
+对于此题而言,对于每一个给定的x,我们要用nums里所有小于x的元素,以及剩余的元素当做x使用,问是否能组合出和为k的方案。我们暂时先不考虑第二种用法,即只用nums里小于x的元素。我们发现,随着x的增大,其实第一个for循环里可选的元素种类的上界也在单调增大,故总体这依然是一个o(n*k)可解的问题。
+
+然后更深入地思考,我们将x从小到大遍历的时候,可以不断提升i,从而加入所有不超过x的新nums[i],就可以不断更新dp。同时我们还需要考虑等于x的元素:因为capping的缘故,这样的元素有n-i个。此时我们需要考虑这额外的n-i个元素能否对于组成k有帮助。显然我们可以用同样的背包思想:
+```cpp
+for (int j=1; j<n-i; j++) {
+ if (dp[k-x*j]) {
+ return true;
+ break;
+ }
+}
+```
+注意,这个过程中我们只是判断是否能组合成k,不会去更新dp。我们始终保持dp[c]表示&quot;能否用所有小于x的元素组成c&quot;。这样,即使x变大,dp的定义依然适用。
+
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