求出
10000以内，被
2除余
1，被
3除余
2，被
4除余
3，被
5除余
4，被
6除余
5，被
7除余
6，被
8除余
7，被
9除余
8的自然数。源程序
ex7_1.prg如下：
SET
TALK
OFF
CLEA
CLEA ALL
FOR
i
=
1
TO
10000
IF
MOD
(i,
2
)
=
1
AND
MOD
(i,
3
)
=
2
;
AND
MOD
(i,
4
)
=
3
AND
MOD
(i,
5
)
=
4
;
AND
MOD
(i,
6
)
=
5
AND
MOD
(i,
7
)
=
6
;
AND
MOD
(i,
8
)
=
7
AND
MOD
(i,
9
)
=
8
? i
ENDIF
ENDFOR
RETU
SET
TALK
OFF
CLEA
CLEA ALL
FOR
i
=
2520
TO
10000
+
1
STEP
2520
? i
-
1
ENDFOR
RETU
SET
TALK
OFF
CLEACLEA ALL
FOR
i
=
1
TO
10000
IF
MOD
(i,
2
)
=
1
AND
MOD
(i,
3
)
=
2
;
AND
MOD
(i,
4
)
=
3
AND
MOD
(i,
5
)
=
4
;
AND
MOD
(i,
6
)
=
5
AND
MOD
(i,
7
)
=
6
;
AND
MOD
(i,
8
)
=
7
AND
MOD
(i,
9
)
=
8
? i ENDIFENDFORRETU
按照这种方法，思维比较直观，也容易实现，但程序效率太低，有必要作一下优化。设x为满足条件的数，根据题意，则x+1是2、3、4、5、6、7、8、9的公倍数，即x+1是其最小公倍数2520的倍数，并且x+1在10001的范围内，所以第一个满足条件的数是2520-1。据此编写程序可以高速度2520倍。源程序ex7_2.prg如下：
SET
TALK
OFF
CLEACLEA ALL
FOR
i
=
2520
TO
10000
+
1
STEP
2520
? i
-
1
ENDFORRETU
