POJ 1832(9连环）

格雷码做

0要特判

答案 就是读入的两个数充格雷码转为2进制的差的绝对值的十进制

Program P1832;Type   arr=record       a:array[1..500] of longint;       len:longint;       end;var   n,m,i:longint;   a,b,c,d:arr;   node2:array[1..1000] of arr;function max(a,b:longint):longint;begin   if a<b then exit(b) else exit(a);end;function compare:boolean;var   i,j:longint;begin   while (a.len>1) and (a.a[a.len]=0) do dec(a.len);   while (b.len>1) and (b.a[b.len]=0) do dec(b.len);   if a.len<b.len then exit(true)   else if a.len>b.len then exit(false);   i:=a.len;   while (i>1) and (a.a[i]=b.a[i]) do   begin      dec(i);   end;   if (a.a[i]<=b.a[i]) then exit(true) else exit(false);end;Procedure Subtract(a,b:arr;var c:arr);var   i,j:longint;   pa:arr;begin   if (compare) then   begin      pa:=a;      a:=b;      b:=pa;   end;   fillchar(c,sizeof(c),0);   for i:=1 to a.len do   begin      inc(c.a[i],a.a[i]-b.a[i]);      if c.a[i]<0 then      begin         inc(c.a[i],2);         dec(c.a[i+1]);      end;   end;   c.len:=a.len;   while (c.len>1) and (c.a[c.len]=0) do dec(c.len);end;procedure add(a,b:arr;var c:arr);var   i,j:longint;begin   fillchar(c,sizeof(c),0);   for i:=1 to max(a.len,b.len) do   begin      inc(c.a[i],a.a[i]+b.a[i]);      inc(c.a[i+1],c.a[i] div 10);      c.a[i]:=c.a[i] mod 10;   end;   c.len:=max(a.len,b.len)+1;   while (c.len>1) and (c.a[c.len]=0) do dec(c.len);end;begin   fillchar(node2,sizeof(node2),0);   node2[1].len:=1;   node2[1].a[1]:=1;   for i:=2 to 500 do   begin      add(node2[i-1],node2[i-1],node2[i]);   end; {  assign(input,'P1832.in');   assign(output,'P1832.out');   reset(input);   rewrite(output);  }   read(m);   while m>0 do   begin      fillchar(a,sizeof(a),0);      fillchar(b,sizeof(b),0);      read(n);      if n=0 then      begin         writeln('0');         dec(m);         continue;      end;      for i:=n downto 1 do read(a.a[i]);      for i:=n downto 1 do read(b.a[i]);      for i:=n-1 downto 1 do      begin         a.a[i]:=(a.a[i] xor a.a[i+1]);      end;      for i:=n-1 downto 1 do         b.a[i]:=b.a[i] xor b.a[i+1];      a.len:=n;      b.len:=n;      subtract(a,b,c);//      for i:=n downto 1 do write(c.a[i]);//      writeln;      fillchar(d,sizeof(d),0);      d.len:=1;      for i:=n downto 1 do         if c.a[i]>0 then add(node2[i],d,d);      for i:=d.len downto 1 do write(d.a[i]);      writeln;      dec(m);   end;{   close(input);   close(output);}end.