最小源代码擂台

谁 · 发表于 2008-5-26 09:34

CREATE OR REPLACE function SYSTEM.my_add_months(p_date_string varchar2, p_months number) return varchar2 IS
t int:=p_date_string;
l int:=100;
y int:= t/l/l; m int:= mod(t/l,l); d int:= mod(t,l);
BEGIN
t := case when m=2 and d>=28 then case when not ((mod(y,4)=0 and mod(y,100)!=0 or mod(y,400)=0) and d=28) then 1 end
         when m in(4,6,9,11) and d=30 then 1 when d=31 then 1 end;
m := m+p_months;
y := trunc(y+(m-1)/12);
m := mod(m,12);
if m<=0 then m:=m+12; end if;
l:=28;
if mod(y,4)=0 and (mod(y,100)!=0 or mod(y,400)=0) then l:=29; end if;
d := case when m in(4,6,9,11) and (d>30 or t=1) then 30
      when m=2 and (d>l or t=1) then l
      when t=1 then 31
      else d end;
RETURN y||lpad(m,2,'0')||d;
END;
/

SQL> call t();
Congratulation ... Code Length: 555 Bytes. Times: 00:00:03

走hotiice的思路，少4个字节，不过这已经到26号了
CREATE OR REPLACE function SYSTEM.my_add_months(p_date_string varchar2, p_months number) return varchar2 IS
t int:=p_date_string;
l int:=100;
y int:= t/l/l; m int:= mod(t/l,l); d int:= mod(t,l);
BEGIN
t := case when m=2 and d>=28 then case when not ((mod(y,4)=0 and mod(y,25)!=0 or mod(y,16)=0) and d=28) then 1 end
         when m in(4,6,9,11) and d=30 then 1 when d=31 then 1 end;
m := m+p_months;
y := trunc(y+(m-1)/12);
m := mod(m,12);
if m<=0 then m:=m+12; end if;
l:=28;
if mod(y,4)=0 and (mod(y,25)!=0 or mod(y,16)=0) then l:=29; end if;
d := case when m in(4,6,9,11) and (d>30 or t=1) then 30
      when m=2 and (d>l or t=1) then l
      when t=1 then 31
      else d end;
RETURN y||lpad(m,2,'0')||d;
END;
/

SQL> call t();
Congratulation ... Code Length: 551 Bytes. Times: 00:00:03

DragonBill · 发表于 2008-5-26 09:44

那个INSTR方法有些漏洞(当年份不为四位数时有问题), 不过在这里合理利用了你的规则(年份规定为四位数)

谁 · 发表于 2008-5-26 09:44

我的算法思路和hotiice差不多
不过字节比他的多多了

CREATE OR REPLACE function SYSTEM.my_add_months(p_date_string varchar2, p_months number) return varchar2 IS
t int:=p_date_string;
l int:=100;
y int:= t/l/l; m int:= mod(t/l,l); d int:= mod(t,l);
BEGIN
t := case when m=2 and d>=28 then case when not ((mod(y,4)=0 and mod(y,25)>0 or mod(y,16)=0) and d=28) then 1 end
         when m in(4,6,9,11) and d=30 then 1 when d=31 then 1 end;
m := m+p_months;
y := trunc(y+(m-1)/12);
m := mod(mod(m,12)+11,12)+1;
l:=28;
if mod(y,4)=0 and (mod(y,25)>0 or mod(y,16)=0) then l:=29; end if;
d := case when m in(4,6,9,11) and (d>30 or t=1) then 30
      when m=2 and (d>l or t=1) then l
      when t=1 then 31
      else d end;
RETURN y||lpad(m,2,'0')||d;
END;
/

SQL> /
Congratulation ... Code Length: 538 Bytes. Times: 00:00:04

这是按我的算法思路走，最少的字节数了吧

当然我的最终成绩，还是按照555字节来定
这三个数，也好听

谁 · 发表于 2008-5-26 09:45

原帖由 DragonBill 于 2008-5-26 09:44 发表
那个INSTR方法有些漏洞(当年份不为四位数时有问题), 不过在这里合理利用了你的规则(年份规定为四位数)

y>1582
至少是4位数
oracle的年不能超过9999

所以，没问题

DragonBill · 发表于 2008-5-26 09:48

但若避免这个漏洞仅仅只有一个Byte的差别

, 如下:

--Congratulation ... Code Length: 345 Bytes. Times: 00:00:08
create or replace function my_add_months(p_date_string varchar2,
                                       p_months    number)
  return varchar2
AS
C INT := p_date_string;
H INT := 100;
Y INT := C/H/H;
M INT := MOD(C/H, H);
D INT := MOD(C, H);
Z INT := Y * 12 + M + p_months;

PROCEDURE P
AS
BEGIN
   C := 27 + SUBSTR(43434434342 - SIGN(MOD(Y, 16 - 12 * SIGN(MOD(Y, H)))), 1 - M, 1);
END;

BEGIN

P;

M := MOD(Z - 1, 12) + 1;
Y := (Z - M) / 12;
Z := C;

P;

RETURN Y*H*H + M*H + LEAST(D + TRUNC(D/Z) * 3, C);
END;
/