读n个和比较帕斯卡

Question

我要编写一个程序，要求用户输入“N”号，然后比较号码找到我使用二路最大数量，同时：

while i>n do read(number); i:=i+1; 

是真的？如何保存号码并进行比较？ 用于离如果用户有4号节省他们这样的：

一个：=数字1; b：= number2; ...

Answer 1

从集合中找出第二大数字的算法（假设这就是你之后的事情）相对简单。

get number into ultimate, on end-of-file exit 
get number into penult, on end-of-file exit 
if penult is greater than ultimate: 
    swap ultimate and penult 

while true: 
    get number into num, on end-of-file return penult 
    if num is greater than ultimate: 
     set penult to ultimate 
     set ultimate to num 
    else: 
     if num is greater than penult: 
      set penult to num 

换句话说，保持前两位数字[ultimate, penult]的“清单”和未来的每个号插槽成列表，如果它原来是属于那里。

一旦用完了数字，该列表的第二项就是整个集合中的第二高数字。

在你不想重复占用两个插槽（如1,2,1给你1为第二高，你真的想2）截止机会，你可以扔掉那些重复的，如果他们匹配：

get number into ultimate, on end-of-file exit 
set penult to ultimate 
while penult is equal to ultimate: 
    get number into penult, on end-of-file exit 
if penult is greater than ultimate: 
    swap ultimate and penult 

while true: 
    get number into num, on end-of-file return penult 
    if num is greater than ultimate: 
     set penult to ultimate 
     set ultimate to num 
    else: 
     if num is greater than penult and num is not equal to ultimate: 
      set penult to testnum 

重写在帕斯卡尔，我会离开你的练习，因为它可能是课堂作业。