※ 引述《bill790602 (REBORN)》之铭言： : 1.what is the Excess_128 representation of 5? : (a)00000101(b)10000100(c)10000101(d)10000001 超128就是加128... (5+128)10=(10000101)2 答案为c : 2.Change the decimal number-124 to an 8-bittwo's complement integer (-124)10用8bits，2的补数来表示 (124)10=(1111100)2 转2的补数为(0000100)2 答案为10000100 : 3.An alogorithm of complexity O(n log n)is not as efficient as one of order : (a)O(logˇ2n)(b)O(n^2 )(c)O(2^n )(d)O(n√n)(e)n^(log_(2^n ) ) 答案为d 这...是数学问题吗... : 4.Can you draw a binary tree from the following prefix and infix traversal : order?If yes,draw the tree.If not,state your reason : Prefix:D B A C I E H : Infix:A B H C D E I Prefix的D在B的左子树~~ Infix的D在A的右子树阶层中~~ 应该是画不出来吧.... : 5.We have an integer array foo[3][7]in a C program. Assume that each integer : is Stored in two bytes.The array is stored in row-major order .If foo[0][0] is : stored at 0x200,where is foo[2][3]? 0x200是(200)16... f00[2][3]=f[0][0]+[(2-0)*7+(3-0)]*2 =200+(34)10=200+22=222 答案为0x222 //解释到一半发现我写错了XD....以上更正 阵列的储存方式有两种~~row-major和column-major row-major的内部结构就是 xxxxxxx xxxxxxx ->一列一列存，存完一列再存下一列 xxx'x' column-major则是 xxx xxx xxx xxx xx ->一行一行存，存完一行再存下一行 xx xx 第一个x是f[0][0] 它的位址是0x200(16进制的200) 问你f[2][3]就是row-major的'x'的位址 那先算前面几个x-->2*7+3=17 一个x的大小是2byte-->17*2=34 转16进制=22 0x200+22=0x222 : 6x.What are the three main features of Object-Oriented : Programming Language? 封装、继承、多型 : 麻烦各位大大了~ : 小弟遇到瓶颈ˊˋ 请前辈多多指教!!!!^-^ --

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