From 3be07e8bb706e625bc7517456826a6eff50df3b3 Mon Sep 17 00:00:00 2001
From: hanyixuanten <hanyixuanten@outlook.com>
Date: Wed, 28 Oct 2026 08:11:16 +0000
Subject: [PATCH] new file:   solutions/AT_arc073_b.md

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 solutions/AT_arc073_b.md | 47 ++++++++++++++++++++++++++++++++++++++++
 1 file changed, 47 insertions(+)
 create mode 100644 solutions/AT_arc073_b.md

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+# AT_arc073_b [ABC060D] Simple Knapsack 题解
+
+注意到 $W$ 的值很大，若是直接计算会 TLE/MLE/RE。
+
+```cpp
+#include <bits/stdc++.h>
+using namespace std;
+typedef pair<int, int> pii;
+int n; // 100
+int w; // 1e9
+pair<int, int> obj[105];
+unordered_map<int, int> f[105];
+int sumw[105];
+int main()
+{
+    freopen("satchel.in", "r", stdin);
+    freopen("satchel.out", "w", stdout);
+    scanf("%d%d", &n, &w);
+    for (int i = 1; i <= n; ++i)
+    {
+        scanf("%d%d", &obj[i].first, &obj[i].second);
+    }
+    sort(obj + 1, obj + n + 1, [](pii x, pii y)
+         { return x.first > y.first; });
+    for (int i = 1; i <= n; ++i)
+        sumw[i] = sumw[i - 1] + obj[i].first;
+    for (int i = 1; i <= n; ++i)
+    {
+        for (int j = obj[i].first; j <= min(w, sumw[i]); ++j)
+        {
+            f[i][j] = max(f[i - 1][min(j, sumw[i - 1])], f[i - 1][j - obj[i].first] + obj[i].second);
+        }
+    }
+    printf("%d\n", f[n][min(w, sumw[n])]);
+    return 0;
+}
+```
+
+令 $base = w_1, d_i = w_i - base$。由题，$0 \le d_i \le 3$。若选择 $k$ 个物品，总重量为 $\sum d_i + k \times base$，其中 $0 \le \sum d_i \le 300$。
+
+考虑用 $f_{i, j}$ 记录选择 $i$ 个物品，$\sum d_i = j$ 的最大价值。对于每个物品：
+
+$$f_{i + 1, j + d_i} = \max(f_{i, j} + v_i)$$
+
+最后枚举 $i, j$，若 $i \times base + j \le W$ 则更新答案。
+
+[code](../AT_arc073_b.cpp)