SPOJ 1112. Number Steps
1112. Number Steps
Starting from point (0,0) on a plane, we have written all non-negative integers 0, 1, 2,... as shown in the figure. For example, 1, 2, and 3 has been written at points (1,1), (2,0), and (3, 1) respectively and this pattern has continued.
You are to write a program that reads the coordinates of a point (x, y), and writes the number (if any) that has been written at that point. (x, y) coordinates in the input are in the range 0...10000.
Input
The first line of the input is N, the number of test cases for this problem. In each of the N following lines, there is x, and y representing the coordinates (x, y) of a point.
Output
For each point in the input, write the number written at that point or write No Number if there is none.
Example
Input: 3 4 2 6 6 3 4 Output: 6 12 No Number
我的代码：
; spoj 1112
section .data
noans : db 'No Number', 0x0a
noansL : equ $ - noans
section .text
global _start
_start :
push ebp
mov ebp, esp
sub esp, 12
mov eax, ebp
sub eax, 4
push eax
call inInt
L5 :
mov eax, [ebp-4]
test eax, eax
jz L6_A
dec eax
mov [ebp-4], eax
mov eax, ebp
sub eax, 8
push eax
call inInt
mov ebx, [ebp-8]
mov eax, ebp
sub eax, 12
push eax
call inInt
mov ecx, [ebp-12]
cmp ebx, 0
je X0
cmp ebx, 1
je X1
cmp ebx, ecx
je XY_A
add ecx, 2
cmp ebx, ecx
je XY2
;jmp NO
NO :
mov eax, 4
mov ebx, 1
mov ecx, noans
mov edx, noansL
int 0x80
jmp L5
L6_A :
jmp L6_B
X0 :
cmp ecx, 0
jne NO
;jmp X0Y0
X0Y0 :
push dword 0
call outInt
call outLn
jmp L5
XY_A :
jmp XY
X1 :
cmp ecx, 1
jne NO
;jmp X1Y1
X1Y1 :
push dword 1
call outInt
call outLn
jmp L5
L6_B :
jmp L6
XY2 :
mov eax, ebx
and eax, 0x1
jnz XoYo2
;jmp XeYe2
XeYe2 :
shl ebx, 1
sub ebx, 2
push ebx
call outInt
call outLn
jmp L5
XoYo2 :
shl ebx, 1
sub ebx, 3
push ebx
call outInt
call outLn
jmp L5
XY :
mov eax, ebx
and eax, 0x1
jnz XoYo
;jmp XeYe
XeYe :
shl ebx, 1
push ebx
call outInt
call outLn
jmp L5
XoYo :
shl ebx, 1
dec ebx
push ebx
call outInt
call outLn
jmp L5
L6 :
push 0
call exit
mov esp, ebp
pop ebp
ret
; func void exit( int )
exit :
mov eax, 1
mov ebx, [esp+4]
int 0x80
ret 4
; func void inInt( uint * )
inInt :
push ebp
mov ebp, esp
sub esp, 8
push eax
push ebx
push ecx
push edx
mov eax, [ebp+8]
mov dword [eax], 0
mov dword [ebp-8], 0
L4 :
mov eax, 3
mov ebx, 0
mov ecx, ebp
sub ecx, 4
mov edx, 1
int 0x80
xor ebx, ebx
mov bl, byte[ecx]
cmp ebx, 0x30
jb L3
cmp ebx, 0x39
ja L3
sub ebx, 0x30
mov ecx, [ebp+8]
mov eax, [ecx]
xor edx, edx
mov ecx, 10
mul ecx
add eax, ebx
mov ecx, [ebp+8]
mov [ecx], eax
mov dword [ebp-8], 1
jmp L4
L3 :
mov eax, [ebp-8]
test eax, eax
jz L4
L2 :
pop edx
pop ecx
pop ebx
pop eax
mov esp, ebp
pop ebp
ret 4
; func void outInt( uint )
outInt :
push ebp
mov ebp, esp
sub esp, 64
push edx
push eax
push ecx
mov eax, [ebp+8]
xor ebx, ebx
L1 :
xor edx, edx
mov ecx, 10
div ecx
inc ebx
add dl, 0x30
mov ecx, ebp
sub ecx, ebx
mov byte[ecx], dl
test eax, eax
jnz L1
mov edx, ebx
mov eax, 4
mov ebx, 1
mov ecx, ebp
sub ecx, edx
int 0x80
pop ecx
pop eax
pop edx
mov esp, ebp
pop ebp
ret 4
; func void outLn()
outLn :
push ebp
mov ebp, esp
sub esp, 4
push eax
mov eax, 4
push ebx
mov ebx, 1
push ecx
mov ecx, ebp
sub ecx, 4
mov byte[ecx], 0x0a
push edx
mov edx, 1
int 0x80
pop edx
pop ecx
pop ebx
pop eax
mov esp, ebp
pop ebp
ret