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вернуться в форумПоказать все сообщения Спрятать все сообщенияNote: the rule stated in the prob indicates a palindromic string.
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program ural1007;
const
maxn=1000;
var
bit:array[1..maxn+1]of boolean;
n,l,i,j:integer;
c:char;
function pal(s,t:integer):boolean;
var
i:integer;
begin
for i:=s to (s+t) div 2 do
if bit[i]<>bit[s+t-i] then begin
pal:=false;
exit;
end;
pal:=true;
end;
begin
readln(n);
repeat
l:=0;
while not eoln(input) do begin
read(c);
if c='1' then begin
inc(l);bit[l]:=true;
end
else if c='0' then begin
inc(l);bit[l]:=false;
end;
end;
readln;
if l=0 then continue;
j:=0;
if l=n then begin
if not pal(1,l) then
for i:=1 to l div 2 do
if bit[i]<>bit[l+1-i] then begin
bit[i]:=false;
bit[l+1-i]:=false;
break;
end;
for i:=1 to l do
write(ord(bit[i]));
writeln;
end
else if l=n-1 then begin
for i:=1 to l div 2 do
if bit[i]<>bit[l+1-i] then begin
if pal(i,l-i) then j:=i else j:=l+2-i;
break;
end;
if j=0 then j:=l div 2+1;
for i:=1 to j-1 do
write(ord(bit[i]));
write(ord(bit[l+1-j]));
for i:=j to l do
write(ord(bit[i]));
writeln;
end
else if l=n+1 then begin
for i:=1 to l div 2 do
if bit[i]<>bit[l+1-i] then begin
if pal(i,l-i) then j:=l+1-i else j:=i;
break;
end;
if j=0 then j:=l div 2+1;
for i:=1 to j-1 do
write(ord(bit[i]));
for i:=j+1 to l do
write(ord(bit[i]));
writeln;
end;
until eof(input);
end. Note: the rule stated in the prob indicates a palindromic string. No, it is not necessarily. I had analogous illusion when wrote first version of solution. And I was really surprised when got WA at 1st test :)) Example of asymmetric string for N=5 is "11100" (1+2+3 = 6 = N+1) There are many cases similar to this for other values of N. program ural1007;
const
maxn=1001;
var
bit:array[1..maxn]of boolean;
n,i,j,l,m:integer;
c:char;
procedure nochange;
var
i:integer;
begin
for i:=1 to l do write(ord(bit[i]));
writeln;
end;
procedure clear(p:integer);
var
i:integer;
begin
for i:=1 to p-1 do write(ord(bit[i]));
write(0);
for i:=p+1 to l do write(ord(bit[i]));
writeln;
end;
procedure insert(p,b:integer);
var
i:integer;
begin
for i:=1 to p-1 do write(ord(bit[i]));
write(b);
for i:=p to l do write(ord(bit[i]));
writeln;
end;
procedure del(p:integer);
var
i:integer;
begin
for i:=1 to p-1 do write(ord(bit[i]));
for i:=p+1 to l do write(ord(bit[i]));
writeln;
end;
procedure tooshort;
var
i,c:integer;
begin
bit[n]:=false;c:=0;
for i:=n downto 1 do begin
if bit[i] then inc(c);
if (m+c) mod (n+1)=0 then begin insert(i,0);exit;end;
if (m+c+i) mod (n+1)=0 then begin insert(i,1);exit;end;
end;
end;
procedure toolong;
var
i,c:integer;
begin
c:=0;
for i:=n+1 downto 1 do begin
if (m-c-ord(bit[i])*i+(n+1)*2) mod (n+1)=0 then begin del(i);exit;end;
if bit[i] then inc(c);
end;
end;
begin
readln(n);
for i:=1 to n do begin
repeat
l:=0;
while not eoln(input) do begin
read(c);
if c='1' then begin
inc(l);bit[l]:=true;
end
else if c='0' then begin
inc(l);bit[l]:=false;
end;
end;
readln;
until (l>n-2) and (l<n+2);
m:=0;
for j:=1 to n do
if bit[j] then m:=(m+j) mod (n+1);
if l=n then
if m=0 then nochange else clear(m)
else if l=n-1 then
tooshort
else if l=n+1 then
toolong;
end;
end. your program is wrong ;
4
101
your answer is 0101
it should be 1001 ,isn't it?
Chicken QQ:372879887 |
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