讨论 / 这样可以AC
Dendi 2015-11-04 01:45:33
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program tree;

var

i,j,k,n,m:longint;

x:array[1..3000] of char;

a:array[1..3000] of char;

procedure print(root:longint);

begin

if root<n then

begin

print(2*root);

print(2*root+1);

end;

write(a[root]);

end;

begin

assign(input,'fbi.in');

assign(output,'fbi.out');

reset(input);

rewrite(output);

readln(k);

n:=1;

for i:=1 to k do

n:=n*2;

for i:=1 to n do

read(x[i]);

for i:=n to n*2-1 do

if x[i-n+1]='1' then a[i]:='I' else a[i]:='B';

for i:=n-1 downto 1 do

if (a[i*2]='I') and (a[i*2+1]='I') then a[i]:='I' else

if (a[i*2]='B') and (a[i*2+1]='B') then a[i]:='B' else a[i]:='F';

print(1);

close(input);

close(output);

end.

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