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docs/ACM/专题训练/DP.md

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+!!! abstract
+ 和DP有关的题目练习合集
+
+## CF366C Dima and Salad
+
+[题目链接](https://codeforces.com/problemset/problem/366/C)
+
+**题目：**
+
+有两个长度为n($1 \leq n \leq 100$)的整数序列 $a = [a_1, a_2, .. a_n], b = [b_1, b_2, .. b_n]$, 其中 $1 \leq a_i, b_i \leq 100$, 还有一个整数k ($1 \leq k \leq 10$).
+
+从中选出来m个下标，使得满足
+
+$$
+\frac{\sum_{j=1}^m{a_j}}{\sum_{j=1}^m{b_j}} = k
+$$
+
+求最大的$\sum_{j=1}^m{a_j}$, 否则输出-1
+
+**思路：**
+
+$$
+\begin{align}
+& \frac{\sum_{j=1}^m{a_j}}{\sum_{j=1}^m{b_j}} = k \\
+\Rightarrow& \sum_{j=1}^m{a_j} = k * \sum_{j=1}^m{b_j} \\
+\Rightarrow& \sum_{j=1}^m{a_j} = \sum_{j=1}^m{k * b_j} \\
+\Rightarrow& \sum_{j=1}^m{(a_j - k * b_j)} = 0 \\
+\end{align}
+$$
+
+问题转化为: 从n个物体中选一些物品, 每个物品的体积是 $a_i - k * b_i$, 每个物品的价值是 $a_i$. 求总体积为0的最大价值为多少
+
+即01背包
+
+其中通过增加偏移量来解决体积为负的问题, 大小为 $B = k * max(b_i) * n$
+
+
+**code**
+
+
+```cpp
+#include <bits/stdc++.h>
+
+#define LL long long
+#define ULL unsigned long long
+#define x first
+#define y second
+#define endl "\n"
+
+using namespace std;
+
+typedef pair<int, int> PII;
+typedef pair<LL, LL> PLL;
+
+const int INF = 0x3f3f3f3f;
+
+int main()
+{
+ ios::sync_with_stdio(false);
+ cin.tie(0), cout.tie(0);
+
+ int n, k;
+ cin >> n >> k;
+ vector<int> a(n + 1), b(n + 1);
+
+ for (int i = 1; i <= n; i++)
+ {
+  cin >> a[i];
+ }
+
+ for (int i = 1; i <= n; i++)
+ {
+  cin >> b[i];
+ }
+
+ int B = 10 * 100 * 100;
+ vector<int> v(n + 1), w(n + 1), f(2 * B + 10, -INF);
+ for (int i = 1; i <= n; i++)
+ {
+  v[i] = a[i] - k * b[i];
+  w[i] = a[i];
+ }
+
+ f[0 + B] = 0;
+ for (int i = 1; i <= n; i++)
+ {
+  auto tmp = f;
+  for (int j = -B; j <= B; j++)
+  if (0 <= j + v[i] + B && j + v[i] + B <= 2 * B)
+  {
+  tmp[j + B] = max(tmp[j + B], f[j + v[i] + B] + w[i]);
+  }
+  f = tmp;
+ }
+
+ cout << (f[0 + B] <= 0 ? -1 : f[0 + B]) << endl;
+ return 0;
+}
+```

mkdocs.yml

@@ -119,6 +119,7 @@ nav:
  - Math:
  - CF数学: ACM/专题训练/Math/CF数学.md
  - LG数学: ACM/专题训练/Math/LG数学.md
+ - DP: ACM/专题训练/DP.md
  - ML:
  - Pytorch:
  - Pytorch的基本用法: ML/Pytorch/Pytorch.md