CPP面试题笔试题 - SAP

Part 1

1 两个人轮流拿10个硬币，每次可拿1，2，4个，拿到最后一个的为输，问有无必胜条 件？

答：后拿者赢

2 有1000个表，每个表有若干个item，每个item形式为（x,y)，寻找这些表中overlapp ed的item

3 用一串节点存放N个数据，每个节点可放k个数据，其中包含额外的b个数据，问k为多

少最能节省存储空间（假设N/(k-b)mod1=1/2）

4 一个链表排序程序，补足其中一些丢失的语句（排序中不用额外的存储空间）

5 一篇英文，将打乱的各段排序并写个总结，大意是吹嘘SAP如何的好

6 两个仓库的进货，出货和仓库间的货物转移交易的流水帐，写出每次交易的货物数， 单价，交易后仓库中的货物数和amount(货物数*单价）

Part 2

1.Jeff and Diamond like playing game of coins,One day they designed a new set of rules: 1)Totally 10 coins

2)One can take away 1,2or 4 coins at one time by turns 3)Who takes the last loses.

Given these rules Whether the winning status is pre-determined or not

解答：

1：从后面开始考虑，最后肯定要留1个才能保证自己赢 2：所以要设法让对方留下2，3，5个

3：也就是要自己取后留下1，4，6，7，8，9。。。

4：如果自己取后留下6，对方取2个，与（3）矛盾，所以排除6

5：如果自己取后留下8，对方取4个，与（3）一样情况，所以也排除8

6：同样，9也不行，如果我抽后剩下9，对方抽2个，就反过来成对方抽剩成7个了，也与 （

3）矛盾，所以也排除

7：所以很显然，我只能抽剩1，4，7

8：因为只能抽后剩1，4，7才能赢，我先抽得话不可能达到这几个数，很显然，只能让 对

方先抽，也即是先抽的人输

2.the UI specialist Dafna found a problem that some of the

Items on the marketing document form overlapped with each other. because this form was implemented by differnt developers and

they didn't care the particular appearance of one item.Product manager Tidav decided to write one small checking tool to generate the overlapped items on all forms.He called in his guys to discuss about it.

Suppose the input is the integer coordinates (x,y)od the items (all rectangl es)

on one form.Construct an efficient methed to find out the overlapped items. Hint:The most direct way to do so is comparing each items with the others,Given 1000 forms.each with 100-1000items on average. the O(n2) algorithm is costly.Some guru suggested that one O(n) method could help only if 6.5 kilobytes extra storage is available.

One elite argued that he could cut down the number to 1%,It's now your turn to describe the idea.write out the pseudocodes,vertify his algorithm and propose more advanced optimization if possible.

3 in a file system ,data need not be sequentially located in physical blocks,We use a number of tables storing nodes imformation.Suppose now we use a fixed node size of variable-lenth n,it takes [n/(k-b)] nodes to store this item.(Here b is a constant, signifying that

b words of each node contain control information,such as a link to the next node).If the avarage length n of an Item is N,what choise of k minimizes the average amount of storage space required?

(Assume that the average value of (n/(k-b)) mod 1 is equal to 1/2 ,for any fixed k, as n varies)

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