ArraySum

  1. Include Irvine32.inc
  2. .data
  3. array dword 11, 22, 44, 5
  4. theSum dword ?
  5. .code
  6. main proc
  7.     mov esi, offset array
  8.     mov ecx, lengthof array
  9.     call SumInt
  10.     mov theSum, eax
  11.     call writedec
  12. main endp
  13. ;------------------
  14. SumInt proc
  15. push esi
  16. push ecx
  17. mov eax, 0
  18. L1:
  19.     add eax, [esi]
  20.     add esi, type dword
  21.     loop L1
  22.     pop esi
  23.     pop ecx
  24.     ret
  25. SumInt endp
  26. end main
11 + 22 + 44 + 5 = 82

留言

張貼留言

這個網誌中的熱門文章

codeforces 271A --- beautiful year

BEAUTIFUL YEAR It seems like the year of 2013 came only yesterday. Do you know a curious fact? The year of 2013 is the first year after the old 1987 with only distinct digits. Now you are suggested to solve the following problem: given a year number, find the minimum year number which is strictly larger than the given one and has only distinct digits. Input The single line contains integer  y   (1000 ≤  y  ≤ 9000)  — the year number. Output Print a single integer — the minimum year number that is strictly larger than  y  and all it's digits are distinct. It is guaranteed that the answer exists. Examples input 1987 output 2013 input 2013 output 2014 #include <iostream> using   namespace  std ; int  iFind  (   int  iIn  ) { int  iFlag ;   int  iBucket [ 10 ] ; for   (   int  i = 0   ;  i < 10   ;  i ++ ...
閱讀完整內容

Uva 913

梯形公式 (a+b)*h / 2 amount                 number               1                           1 3                           3  5  7              (1+3)*2 -1                       5                           9 11 13 15 17  (1+3+5)*2-1 ... #include <iostream> using   namespace  std ; int  main ( ) { long   long   int  iSum, iNum ; long   long   int  iAmount, iHight ;           // (a+b)h/2 int  iLength, iCarry , iTemp ;     while   (   cin   >>  iAmount  )    ...
閱讀完整內容

Uva 591--- Box of Bricks

#include <iostream> using   namespace  std ; int  main ( ) { int  iStack ; int  iHight [ 101 ] ; int  iSum, iTime, iTemp ; int  i,j ; j  =   0 ;   while   ( cin   >>  iStack )      {      if   (  iStack  ==   0   )          break ;       iHight [ 101 ]   =   { 0 } ;     iSum  =  iTime  =   0 ;         for   (  i = 0   ;  i < iStack  ;  i ++   )                  {          cin   >>  iHight [ i ] ;            iSum  + =  iHight [ i ] ;            }     iSum  =  iSum  /  iStack...
閱讀完整內容