【codevs】2918

细节理解题,注意体感有点问题,p只加一次

var
  codevs,poj:array[0..20] of longint;
  ans,ci,pi,n,k,p,i:longint;
begin
  readln(n,k,p);
  for i:=1 to n do read(poj[i]);
  readln;
  for i:=1 to n do read(codevs[i]);
  for i:=2 to n do poj[i]:=poj[i]+poj[i-1];
  for i:=2 to n do codevs[i]:=codevs[i]+codevs[i-1];
  ans:=-maxlongint;
  for i:=0 to n-k do
    begin
      ci:=codevs[i+k]-codevs[i];
      pi:=poj[i+k]-poj[i];
      if ci-pi>ans then ans:=ci-pi;
    end;
  ans:=ans+1-p;
  if ans<0 then writeln(-1)
    else writeln(ans);
end.

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posted @ 2015-09-20 18:33  ROLL-THE-FIRST  阅读(114)  评论(0编辑  收藏  举报