SKI

题面：

 

 

思路：

 

代码：

uses math;
var a:array[1..5000] of longint;
 f:array[1..5000] of longint;
  max:double;
   i,j,n,m:longint;


begin

assign(input,'ski.in');
reset(input);
assign(output,'ski.out');
rewrite(output);

read(n,m);
for i:=1 to n do read(a[i]);
f[1]:=1;
for j:=1 to n-1 do
 begin
  max:=-maxlongint;
  for i:=1 to min(m,n-j) do
  if max<=(a[i+j]-a[j])/i then
   begin
    if (f[i+j]=0) or (f[i+j]>f[j]+1) then f[i+j]:=f[j]+1;
    max:=(a[i+j]-a[j])/i;
   end;
 end;

writeln(f[n]);

close(input);close(output);

end.

 

转载于:https://www.cnblogs.com/godrose/p/9742142.html