POJ1465 Multiple——Bfs+余数判重——Pku1465

这道题精在判重优化。因为如果存在两个数A、B，且满足A mod n=B mod n=C，那么，这两种情况可以看做是重复的，保留较小数即可。具体优化，开一个布尔数组即可。

特别需要注意一点：如果n=0，那么应当输出0！！！{本弱就因为这个WA了半天，看Discuss才明白的……}

CODE

Program Multiple;//By_Thispoet
Const
maxn=20000;
Var
h,t,j,k,m,n,i			:Longint;
seq,fa,mol				:Array[1..maxn]of Integer;
a						:Array[1..maxn]of Longint;
fg						:Array[0..maxn]of Boolean;
flag					:Boolean;
ans						:Array[1..maxn*2]of Longint;

Procedure Qsort(l,r:Longint);
var i,j,k,temp:Longint;
begin
i:=l;j:=r;
k:=a[(i+j)>>1];
repeat
while a[i]<k do inc(i);
while a[j]>k do dec(j);
if i<=j then
begin
temp:=a[i];a[i]:=a[j];a[j]:=temp;
inc(i);dec(j);
end;
until i>j;
if i<r then Qsort(i,r);
if l<j then QSort(l,j);
end;

Procedure Deal(i:Longint);
begin
k:=0;
while i<>0 do
begin
inc(k);
ans[k]:=seq[i];
i:=fa[i];
end;
for i:=k downto 1 do write(ans[i]);
writeln;
end;

BEGIN
readln(n);
readln(m);
while not eof do
begin
fillchar(fg,sizeof(fg),0);
fillchar(fa,sizeof(fa),0);
for i:=1 to m do readln(a[i]);
flag:=true;
Qsort(1,m);
h:=0;
if n=0 then
begin
writeln(0);
readln(n);readln(m);
continue;
end;
if a[1]=0 then
begin
for i:=2 to m do
begin
seq[i-1]:=a[i];
fg[a[i] mod n]:=true;
mol[i-1]:=a[i] mod n;
if a[i] mod n=0 then
begin
flag:=false;
writeln(a[i]);
end;
if not flag then break;
end;
t:=m-1;
end else
begin
for i:=1 to m do
begin
seq[i]:=a[i];
fg[a[i] mod n]:=true;
mol[i]:=a[i] mod n;
if a[i] mod n=0 then
begin
writeln(a[i]);
flag:=false;
end;
if not flag then break;
end;
t:=m;
end;
if not flag then
begin
readln(n);readln(m);continue;
end;
while h<t do
begin
inc(h);
for j:=1 to m do
begin
i:=(mol[h]*10+a[j])mod n;
if fg[i] then continue else
begin
fg[i]:=true;
inc(t);
seq[t]:=a[j];
fa[t]:=h;
mol[t]:=i;
end;
if i=0 then
begin
Deal(t);
flag:=false;
end;
end;
end;
if flag then writeln(0);
readln;
readln(n);
readln(m);
end;
END.