Pku1252 euro Efficiency

There are 6 currencies with a nominal value. The smallest value is 1, and the minimum number of currencies is calculated as 1 .. 100. The maximum value and their average value are output. Subtraction can be used.

For example, 6 = 1 + 2 + 3, 6 = 7-1, then the minimum number of currencies for 6 is 2.

It is a backpack with no lower limit. You can open the volume array to 3000 + (after testing ).

View code

 1 program pku1252(input,output);
 2 var
 3    f       : array[-3000..3000] of longint;
 4    v       : array[1..6] of longint;
 5    cases,e : longint;
 6 procedure init;
 7 var
 8    i : longint;
 9 begin
10    for i:=1 to 6 do
11       read(v[i]);
12    readln;
13    fillchar(f,sizeof(f),63);
14    f[0]:=0;
15 end; { init }
16 function min(aa,bb : longint ):longint;
17 begin
18    if aa>bb then
19       exit(bb);
20    exit(aa);
21 end; { min }
22 procedure main;
23 var
24    i,j : longint;
25 begin
26    for i:=1 to 6 do
27    begin
28       for j:=v[i]-3000 to 3000 do
29      f[j]:=min(f[j],f[j-v[i]]+1);
30       for j:=3000-v[i] downto -3000 do
31      f[j]:=min(f[j],f[j+v[i]]+1);
32    end;
33 end; { main }
34 procedure print;
35 var
36    i,max,sum : longint;
37 begin
38    sum:=0;
39    max:=0;
40    for i:=1 to 100 do
41    begin
42       inc(sum,f[i]);
43       if f[i]>max then
44      max:=f[i];
45    end;
46    writeln((sum/100):0:2,' ',max);
47 end; { print }
48 begin
49    readln(cases);
50    for e:=1 to cases do
51    begin
52       init;
53       main;
54       print;
55    end;
56 end.