昨天梦到这道题了,所以一定要A掉(其实梦到了3道,有两道记不清了)
暴力枚举等的是哪张牌,将是哪张牌,然后贪心的判断就行了。
对于一个状态判断是否为胡牌,1-n扫一遍,然后对于每个牌,先mod 3,
如果还有剩余,就需要和i+1,i+2凑成顺子,减掉i+1,i+2的值就行了,然后
只要枚举到的i为负就是不可行,还有要枚举到n+2位,因为第n,n-1为可能
减了n+1,n+2但是没扫到。
/**************************************************************
Problem:
User: BLADEVIL
Language: Pascal
Result: Accepted
Time: ms
Memory: kb
****************************************************************/
//By BLADEVIL
var
n, m :longint;
count, jud :array[..] of longint;
i, j, k :longint;
x :longint;
ans :array[..] of longint;
function judge:boolean;
var
i :longint;
begin
for i:= to n+ do
begin
if jud[i]< then exit(false);
jud[i]:=jud[i] mod ;
if jud[i]> then
begin
dec(jud[i+],jud[i]);
dec(jud[i+],jud[i]);
jud[i]:=;
end;
end;
exit(true);
end;
begin
read(n,m);
for i:= to *m+ do
begin
read(x);
inc(count[x]);
end;
for i:= to n do
begin
for j:= to n do jud[j]:=count[j];
inc(jud[i]);
for j:= to n do if jud[j]>= then
begin
dec(jud[j],);
if judge then
begin
inc(ans[]);
ans[ans[]]:=i;
end;
if ans[ans[]]=i then break;
for k:= to n+ do jud[k]:=count[k];
inc(jud[i]);
end;
end;
if ans[]= then
begin
writeln('NO');
halt;
end;
write(ans[]);
for i:= to ans[] do write(' ',ans[i]); writeln;
end.