1.当田忌最慢的马比齐王最慢的马快,赢一场先
2.当田忌最慢的马比齐王最慢的马慢,和齐王最快的马比,输一场
3.当田忌最快的马比齐王最快的马快时,赢一场先。
4.当田忌最快的马比齐王最快的马慢时,拿最慢的马和齐王最快的马比,输一场。
5.当田忌最快的马和齐王最快的马相等时,拿最慢的马来和齐王最快的马比.”
这倒是正解……
var a,b:array[1..100000]of longint;
i,j,n,s,ax,an,bx,bn:longint;
procedure sort1(l,r:integer);
var i,j,x,y:integer;
begin
i:=l;
j:=r;
x:=a[(l+r) shr 1];
repeat
while a[i]<x do inc(i);
while x<a[j] do dec(j);
if i<=j then
begin
y:=a[i];
a[i]:=a[j];
a[j]:=y;
inc(i);
dec(j);
end;
until i>j;
if i<r then sort1(i,r);
if l<j then sort1(l,j);
end;
procedure sort2(l,r:integer);
var i,j,x,y:integer;
begin
i:=l;
j:=r;
x:=b[(l+r) shr 1];
repeat
while b[i]<x do inc(i);
while x<b[j] do dec(j);
if i<=j then
begin
y:=b[i];
b[i]:=b[j];
b[j]:=y;
inc(i);
dec(j);
end;
until i>j;
if i<r then sort2(i,r);
if l<j then sort2(l,j);
end;
begin
repeat
readln(n);
if n=0 then exit;
for i:=1 to n do read(a[i]);
sort1(1,n);
readln;
for i:=1 to n do read(b[i]);
sort2(1,n);
readln;
s:=0;
ax:=n;
an:=1;
bx:=n;
bn:=1;
for i:=1 to n do
begin
if a[ax]>b[bx] then
begin
inc(s,200);
dec(ax);
dec(bx);
continue;
end;
if a[an]>b[bn] then
begin
inc(s,200);
inc(an);
inc(bn);
continue;
end;
if a[an]>b[bx] then inc(s,200);
if a[an]<b[bx] then dec(s,200);
inc(an);
dec(bx);
end;
writeln(s);
until n=0;
end.