Codeforces round #259 (Div. 2)

A. Little Pony and crystal mine

Question: Output something similar to a cross

Question: Print it directly...

Code:

 1 var n,i,j:longint; 2 begin 3   readln(n); 4   for i:=1 to n>>1 do 5    begin 6      for j:=1 to (n-i<<1+1)>>1 do write(‘*‘); 7      for j:=1 to i<<1-1 do write(‘D‘); 8      for j:=1 to (n-i<<1+1)>>1 do write(‘*‘); 9      writeln;10    end;11   for i:=1 to n do write(‘D‘);writeln;12   for i:=n>>1 downto 1 do13    begin14      for j:=1 to (n-i<<1+1)>>1 do write(‘*‘);15      for j:=1 to i<<1-1 do write(‘D‘);16      for j:=1 to (n-i<<1+1)>>1 do write(‘*‘);17      writeln;18    end;19 end.20        

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B. Little Pony and sort by shift

Question: Give You A series. You can adjust the last number to the first number each time, and find the minimum number of times required to change this series to monotonic increase.

Problem: copy the original series to the back and use a temporary variable to record the answer from the beginning.

Code:

 1 var n,i,cnt:longint; 2     a:array[0..250000] of longint; 3 begin 4   readln(n); 5   a[0]:=maxlongint; 6   for i:=1 to n do read(a[i]); 7   for i:=1 to n do a[n+i]:=a[i]; 8   cnt:=0; 9   for i:=1 to n<<1 do10    if a[i]>=a[i-1] then11     begin12       inc(cnt);13       if cnt=n-1 then break;14     end15    else cnt:=0;16   if cnt<>n-1 then writeln(‘-1‘)17   else if i=n then writeln(‘0‘)18   else writeln(n<<1-i);19 end.     

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