Pku1946 cow indexing ing

There are n cows running around the bicycle. If the cows need to run around X in one minute, the leader consumes x physical strength, and the latter consumes x physical strength. The number n of native cows is given now. The energy E of each native cow needs to be round D. As long as a cow reaches the D circle first, even if the loop is completed, the cow can complete the loop. If not, the output is 0. Otherwise, the shortest time of the output loop.

The optimal solution is obviously the nheaded cattle take turns to take the lead, and finally the nheaded bull line.

F [I, j, k] indicates that the I-lead has taken the J lap, and it has consumed the shortest time of K points of physical strength.

Write a forward equation f [I, j + X, K + x * x] = min {f [I, j, k] + 1} // leads the runner-up to continue running.

F [I + 1, J, J] = min {f [I, j, k]} // get started

View code

 1 program pku1946(input,output);
 2 const
 3    MAXW    = 10000000;
 4 var
 5    f      : array[0..25,0..110,0..110] of longint;
 6    n,e,d  : longint;
 7    answer : longint;
 8 procedure init;
 9 begin
10    readln(n,e,d);
11    fillchar(f,sizeof(f),63);
12 end; { init }
13 function min(aa,bb :longint ):longint;
14 begin
15    if aa<bb then
16       exit(aa);
17    exit(bb);
18 end; { min }
19 procedure main;
20 var
21    i,j,k,l : longint;
22 begin
23    f[1,0,0]:=0;
24    for i:=1 to n do
25       for j:=0 to d do
26      for k:=0 to e do
27         if f[i,j,k]<19950714 then
28            for l:=1 to e do
29            begin
30           if (j+l<=d)and(k+l*l<=e) then
31              f[i,j+l,k+l*l]:=min(f[i,j,k]+1,f[i,j+l,k+l*l]);
32           f[i+1,j,j]:=min(f[i+1,j,j],f[i,j,k]);
33            end;
34    answer:=MAXW;
35    for i:=0 to e do
36       if f[n,d,i]<answer then
37      answer:=f[n,d,i];
38 end; { main }
39 procedure print;
40 begin
41    writeln(answer);
42 end; { print }
43 begin
44    init;
45    main;
46    print;
47 end.