[POJ1276 Cash Machine]

[题目来源]：

[关键字]：背包

[题目大意]：给出n个面额和每个面额的数量，问在不超过给定限制的情况下最大可达到多少。

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[分析]：多重背包。把每件物品价值看成面额，体积也看成面额，背包容量限制为给定限制，多重背包求最大价值。方程：f[k] = f[k] or f[k-w[i]]，f[k]指容量为k可不可行。由于此题直接把多重背包转换成01背包会导致空间过大，所以要用到进制优化（类似ST算法）。此处有点复杂，建议仔细研究《背包9讲》。

[代码]：

View Code

 1 program project1;
 2 var
 3   n, m: longint;
 4   v: array[0..2000] of int64;
 5   f: array[0..200000] of boolean;
 6   s: array[0..300] of longint;
 7 
 8 procedure init;
 9 var
10   i, j, tot, x, y, temp: longint;
11 begin
12   read(n,m);
13   tot := 0;
14   for i := 1 to m do
15     begin
16       read(x,y);
17       temp := 0;
18       while x >= s[temp] do
19         begin
20           inc(tot);
21           v[tot] := s[temp]*y;
22           dec(x,s[temp]);
23           inc(temp);
24         end;
25       if x >= 0 then
26         begin
27           inc(tot);
28           v[tot] := y*x;
29         end;
30     end;
31   readln;
32   m := tot;
33   //writeln(m);
34   //for i := 1 to m do write(v[i],' ');
35   //writeln;
36 end;
37 
38 procedure work;
39 var
40   i, j, ans: longint;
41 begin
42   fillchar(f,sizeof(f),false);
43   f[0] := true;
44   for i := 1 to m do
45     for j := n downto v[i] do
46       f[j] := f[j] or f[j-v[i]];
47   for i := n downto 0 do
48     if f[i] then
49       begin
50         writeln(i);
51         break;
52       end;
53 end;
54 
55 procedure done;
56 var
57   i: longint;
58 begin
59   s[0] := 1;
60   for i := 1 to 30 do
61     s[i] := s[i-1]*2;
62 end;
63 
64 begin
65   done;
66   while not seekeof() do
67     begin
68       init;
69       work;
70     end;
71 end.