SQL> edi 已写入 file afiedt.buf 1 declare 2 sum1 number:=1; 3 sum2 number:=0; 4 begin 5 for n in 1..10 loop 6 sum1:=sum1*n; 7 if mod(n,2)=0 then 8 sum2:=sum2+sum1; 9 end if; 10 end loop; 11 dbms_output.put_line('2!+4!+6!+8!+10! 的阶乘的合为:'||to_char(sum2)); 12* end; SQL> / 2!+4!+6!+8!+10! 的阶乘的合为:3669866
for n in 1..10 loop 6 sum1:=sum1*n; 7 if mod(n,2)=0 then 8 sum2:=sum2+sum1; 9 end if; 10 end loop;晕,这个编程思想真厉害...学习了.
SQL> edi
已写入 file afiedt.buf 1 declare
2 sum1 number:=1;
3 sum2 number:=0;
4 begin
5 for n in 1..10 loop
6 sum1:=sum1*n;
7 if mod(n,2)=0 then
8 sum2:=sum2+sum1;
9 end if;
10 end loop;
11 dbms_output.put_line('2!+4!+6!+8!+10! 的阶乘的合为:'||to_char(sum2));
12* end;
SQL> /
2!+4!+6!+8!+10! 的阶乘的合为:3669866
6 sum1:=sum1*n;
7 if mod(n,2)=0 then
8 sum2:=sum2+sum1;
9 end if;
10 end loop;晕,这个编程思想真厉害...学习了.