1 /**\
2 https://codeforces.com/problemset/problem/1633/D
3 注意到b[i]的范围最大是 1000 ,从1-1000 变化最多大约在13次左右
4 13n复杂度不高，直接预处理每个数从1-1000变化花费
5 然后经典01背包
6 \**/
7 #include <bits/stdc++.h>
8 using namespace std;
9 #define int long long
10 #define sf(x) scanf("%lld",&x)
11 #define ytz int _; sf(_); while(_--)
12 #define go continue
13 #define fory(i,a,b) for(int i = a; i <= b; ++i)
14 #define forl(i,a,b) for(int i = a; i >= b; --i)
15 #define debug(a) cout<<#a<<" = "<<a<<endl;
16 const double eps = 1e-7, pi = acos(-1.0);
17
18 const int N = 2e3 + 10;
19 const int M = 1e6 + 10;
20
21 int n, k;
22 int b[N], c[N], f[M], cost[N];
23
24 void solve()
25 {
26 sf(n), sf(k);
27 k = min(13 * n, k);
28 fory(i, 1, n) sf(b[i]);
29 fory(i, 1, n) sf(c[i]);
30
31 memset(f, 0, sizeof f);
32
33 fory(i, 1, n)
34 {
35 int cc = cost[b[i]];
36 for(int j = k; j >= cc; --j)
37 {
38 f[j] = max(f[j], f[j - cc] + c[i]);
39 }
40 }
41 printf("%lld\n", f[k]);
42 }
43
44 signed main()
45 {
46 memset(cost, 0x3f, sizeof cost);
47 cost[1] = 0;
48 fory(i, 1, 1000)
49 {
50 for(int j = 1; j <= i; ++j)
51 {
52 int tmp = i / j;
53 if(i + tmp <= 1000) cost[i + tmp] = min(cost[i + tmp], cost[i] + 1);
54 }
55 }
56 ytz
57 {
58 solve();
59 }
60 return 0;
61 }
