我在看一个算法,这个算法是pascal写的,可惜我没系统学过他,而且手头没书,大家给看看这个代表什么意思?
......
for i:=1 to len do l[i]=[1..n]-l[i]
......
前面是说做len次循环完成do后面的语句,不明白的就是后边的表达式
其中l为某集合的一维数组,这个集合的定义是 set of 1.....maxn 其中maxn为一常数=50
......
for i:=1 to len do l[i]=[1..n]-l[i]
......
前面是说做len次循环完成do后面的语句,不明白的就是后边的表达式
其中l为某集合的一维数组,这个集合的定义是 set of 1.....maxn 其中maxn为一常数=50
解决方案 »
- 请教:我怎么把 edit1 和 updown1 这两个控件联系起来呢?谢谢!
- .Dll中的一个问题
- 【¤¤¤¤】请问怎样判断一个路径是否存在?如不存在则创建【¤¤¤¤】
- 如何将clx组件下的程序变成vCL?
- 送书《MoreEffectiveC++(WQ版).doc》推荐,想学C++的朋友不要错过了呀
- 怎样取得硬盘序列号?
- 怎样用quickreprt打印treeview的内容?
- 请问怎样去掉windows2000 server的开机屏幕和关机屏幕!该死的金上毒霸,卸了之后还留下这么大一个尾巴让你难受!
- 5.1放假了,走时再替我解决一个问题好了!
- 程序中有两个form怎么实现form1执行后在执行form2?
- 关于窗口的设置问题(急!!!!)
- 急、急、急:怎样实现TreeView的开始拖拽?
Operator Operation Operand types Result type Example
+ union set set Set1 + Set2
- difference set set S - T
* intersection set set S * T
<= subset set Boolean Q <= MySet
>= superset set Boolean S1 >= S2
= equality set Boolean S2 = MySet
<> inequality set Boolean MySet <> S1
in membership ordinal, set Boolean A in Set1
The following rules apply to +, -, and *.An ordinal O is in X + Y if and only if O is in X or Y (or both). O is in X - Y if and only if O is in X but not in Y. O is in X * Y if and only if O is in both X and Y.
The result of a +, -, or * operation is of the type set of A..B, where A is the smallest ordinal value in the result set and B is the largest.The following rules apply to <=, >=, =, <>, and in.X <= Y is True just in case every member of X is a member of Y; Z >= W is equivalent to W <= Z. U = V is True just in case U and V contain exactly the same members; otherwise, U <> V is True.
For an ordinal O and a set S, O in S is True just in case O is a member of S.
是说将[1..n]这个集合中属于集合l[i]的元素去掉,然后再将余下的元素赋给l[i],这样l[i]中只有与[1..n]中不同的元素了,我这么理解对么?
http://www.cxsyzx.com/ReadNews.asp?NewsID=150
结贴了。