Daily Archives: 29 9 月, 2012

POJ 2271(HTML)

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这题字符串处理

注意Seekeof会自动把后面的空格吃掉（有时遇到回车，后面会漏一个空格……貌似有时没空格也会读到，字符串处理时慎用）

另外 空格的Ascii码是32，31及以下都是不可输入字符

program P2271;
var
   s:string;
procedure cin;
var
   c:char;
   i,len:longint;
begin
   s:='';
   c:=' ';
   i:=0;
   while not(seekeof) do
   begin
      read(c);
      while (ord(c)<=32) and not(eof) do read(c);
      if ord(c)<=32 then break;
      repeat
         s:=s+c;
         if eof then break;
         read(c);
      until (ord(c)<=32);

      len:=length(s);
      if s='<br>' then begin writeln; i:=0; end
      else
      if s='<hr>' then
      begin
         if i>0 then writeln;
         for i:=1 to 80 do write('-');
         writeln;
         i:=0;
      end
      else
      begin
         if (i=0) then
         begin
            if len<80 then
            begin
               write(s);
               i:=len;
            end;
         end
         else
         begin
            if (i+1+len<=80) then
            begin
               write(' ',s);
               i:=i+1+len;
            end
            else
            begin
               writeln;
               write(s);
               i:=len;

            end;
         end;



      end;



      s:='';
   end;






end;
begin
{   assign(input,'p2271.in');
   assign(output,'p2271.out');
   reset(input);
   rewrite(output);
 }  cin;
   writeln;
  { close(input);
   close(output);}
end.

POJ 1700(过河问题）

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玩过《雷顿》就知道这题可以贪心

小等于2人：1,2->

3人时：

1,3->

1<-

1,2->

1<-

否则：

1，2->

2<-

max1,max2->

1<-

OR：

1,max1->

1<-

2,max2->

2<-

于是数据规模-2

Program P1700;
var
   t,n,i,j:longint;
   l,r:longint;
   a:array[1..1000] of longint;
procedure qsort(l,r:longint);
var
   i,j,m,p:longint;
begin
   i:=l;
   j:=r;
   m:=a[(l+r) shr 1];
   repeat
      while (a[i]<m) do inc(i);
      while (a[j]>m) do dec(j);
      if i<=j then
      begin
         p:=a[i];a[i]:=a[j];a[j]:=p;
         inc(i);
         dec(j);
      end;
   until i>j;
   if l<j then qsort(l,j);
   if i<r then qsort(i,r);

end;
function max(a,b:longint):longint;
begin
   if a<b then exit(b) else exit(a);
end;
function min(a,b:longint):longint;
begin
   if a>b then exit(b) else exit(a);
end;

function ww(x:longint):longint;
var
   i,j:longint;
begin
   if x=1 then exit(a[1]);
   if x=2 then exit(a[2]);
   if x=3 then exit(a[1]+a[2]+a[3]);
   i:=2*a[1]+a[x]+a[x-1];
   j:=2*a[2]+a[1]+a[x];
   exit(ww(x-2)+min(i,j));
                           //1,2-> 1< r1,r2-> 2<=
end;
begin
   read(t);
   while (t>0) do
   begin
      read(n);
      for i:=1 to n do read(a[i]);
      qsort(1,n);
      l:=1;r:=n;

      writeln(ww(n));

      dec(t);
   end;
end.