coreBugZJ

此 blog 已弃。

Adding Reversed Numbers,spoj 42

42. Adding Reversed Numbers

Problem code: ADDREV

The Antique Comedians of Malidinesia prefer comedies to tragedies. Unfortunately, most of the ancient plays are tragedies. Therefore the dramatic advisor of ACM has decided to transfigure some tragedies into comedies. Obviously, this work is very hard because the basic sense of the play must be kept intact, although all the things change to their opposites. For example the numbers: if any number appears in the tragedy, it must be converted to its reversed form before being accepted into the comedy play.

Reversed number is a number written in arabic numerals but the order of digits is reversed. The first digit becomes last and vice versa. For example, if the main hero had 1245 strawberries in the tragedy, he has 5421 of them now. Note that all the leading zeros are omitted. That means if the number ends with a zero, the zero is lost by reversing (e.g. 1200 gives 21). Also note that the reversed number never has any trailing zeros.

ACM needs to calculate with reversed numbers. Your task is to add two reversed numbers and output their reversed sum. Of course, the result is not unique because any particular number is a reversed form of several numbers (e.g. 21 could be 12, 120 or 1200 before reversing). Thus we must assume that no zeros were lost by reversing (e.g. assume that the original number was 12).

Input

The input consists of N cases (equal to about 10000). The first line of the input contains only positive integer N. Then follow the cases. Each case consists of exactly one line with two positive integers separated by space. These are the reversed numbers you are to add.

Output

For each case, print exactly one line containing only one integer - the reversed sum of two reversed numbers. Omit any leading zeros in the output.

Example

Sample input: 

3
24 1
4358 754
305 794

Sample output:

34
1998
1


一段时间没写汇编了,使用输入缓冲提高效率。。。


  1 ; spoj42.s
  2 
  3 %define  L  1024
  4 
  5 section .bss
  6         pBuf  : resb  L
  7         pBufN : resd 1
  8         pBufI : resd 1
  9 
 10 section .text
 11         global _start
 12 
 13 _start : 
 14         mov dword[pBufN], 0x0
 15         mov dword[pBufI], 0x0
 16         call inInt
 17 CASE : 
 18         test eax, eax
 19         jz EXIT
 20         dec eax
 21         push eax
 22 
 23         call inIntR
 24         push eax
 25         call inIntR
 26         pop ebx
 27         add eax, ebx
 28         call outIntLnR
 29 
 30         pop eax
 31         jmp CASE
 32 EXIT : 
 33         mov eax, 0x1
 34         mov ebx, 0x0
 35         int 0x80
 36 
 37 in eax
 38 inChar : 
 39         mov eax, [pBufI]
 40         mov ebx, [pBufN]
 41         cmp eax, ebx
 42         jne NOEMPTY
 43         mov eax, 0x3
 44         mov ebx, 0
 45         mov ecx, pBuf
 46         mov edx, L
 47         int 0x80
 48         mov [pBufN], eax
 49         mov dword[pBufI], 0
 50         xor eax, eax
 51 NOEMPTY : 
 52         mov ebx, eax
 53         inc ebx
 54         mov [pBufI], ebx
 55         add eax, pBuf
 56         mov bl, byte[eax]
 57         xor eax, eax
 58         mov al, bl
 59         ret
 60 
 61 in eax
 62 inInt : 
 63         xor eax, eax
 64         push eax
 65 SKIPSPACE : 
 66         call inChar
 67         cmp al, '0'
 68         jb SKIPSPACE
 69         cmp al, '9'
 70         ja SKIPSPACE
 71 INDIGIT : 
 72         mov ebx, eax
 73         pop eax
 74         mov ecx, 0xA
 75         xor edx, edx
 76         mul ecx
 77         sub ebx, '0'
 78         add eax, ebx
 79         push eax
 80 
 81         call inChar
 82         cmp al, '0'
 83         jb INDIGITEND
 84         cmp al, '9'
 85         ja INDIGITEND
 86         jmp INDIGIT
 87 INDIGITEND : 
 88         pop eax
 89         ret
 90 
 91 in eax
 92 inIntR : 
 93         xor eax, eax
 94         push eax
 95         mov eax, 0x1
 96         push eax
 97 SKIPSPACER : 
 98         call inChar
 99         cmp al, '0'
100         jb SKIPSPACER
101         cmp al, '9'
102         ja SKIPSPACER
103 INDIGITR : 
104         sub eax, '0'
105         pop ebx
106         xor edx, edx
107         mul ebx
108         pop ecx
109         add eax, ecx
110         push eax
111         mov eax, ebx
112         mov ecx, 0xA
113         xor edx, edx
114         mul ecx
115         push eax
116         
117         call inChar
118         cmp al, '0'
119         jb inIntRend
120         cmp al, '9'
121         ja inIntRend
122         jmp INDIGITR
123 inIntRend : 
124         pop eax
125         pop eax
126         ret
127 
128 out eax
129 outIntLnR : 
130         push ebp
131         mov ebp, esp
132         sub esp, 100
133         mov ebx, esp
134 ZEROBEG : 
135         test eax, eax
136         jz ZEROEND
137         mov ecx, 0xA
138         xor edx, edx
139         div ecx
140         add edx, '0'
141         mov byte[ebx], dl
142         inc ebx
143         jmp ZEROBEG
144 ZEROEND : 
145         mov eax, esp
146 ZEROSKIP : 
147         mov cl, byte[eax]
148         cmp cl, '0'
149         jnz ZEROSKIPEND
150         inc eax
151         jmp ZEROSKIP
152 ZEROSKIPEND : 
153         mov byte[ebx], 0xA
154         inc ebx
155         mov edx, ebx
156         sub edx, eax
157         mov ecx, eax
158         mov eax, 0x4
159         mov ebx, 0x1
160         int 0x80
161 
162         mov esp, ebp
163         pop ebp
164         ret
165 

posted on 2011-05-18 15:27 coreBugZJ 阅读(649) 评论(0)  编辑 收藏 引用 所属分类: Assemble


只有注册用户登录后才能发表评论。
网站导航: 博客园   IT新闻   BlogJava   知识库   博问   管理