台灣數位學習數位教學平台 RSS feed provider zh-tw http://sites.xms.com.tw/board.php?courseID=151&f=doclist&folderID= http://sites.xms.com.tw/board.php?courseID=151&f=doc&cid=3907 #if 0#endif// --------------------------------------------------------#include <time.h>#include <stdlib.h>#include <stdio.h>// --------------------------------------------------------#define Num 10int A[10]={0};// --------------------------------------------------------void RandomNum()  //氣泡排序法之副程式{ int i; srand((unsigned)time(NULL)); printf("產生10個亂數值："); for (i = 1; i <= Num; i++)    { A[i] = rand() % 90+10;  //產生10~100的整數亂數值 printf("%4d",A[i]);    }}// --------------------------------------------------------void PrintBubSort(int A[], int n)  //氣泡排序法之副程式 {    int i;     printf("\n *** 排序10個亂數值： ");       for (i = 0; i <= Num; i++)    { printf("%4d", A[i]);    }   }// --------------------------------------------------------void BubSort(int A[], int n)  //氣泡排序法之副程式{    int i, j , t=1, Temp;    for (i=n-1; i>0; i--) { printf("\n *** i= %d\n", i); for (j =0; j <=i; j++) { if (A[j] > A[j+1]) {  //兩數交換位置 // show data printf("swap data of a[j]= %d, a[j+1]= %d\n", A[j], A[j+1]); Temp = A[j]; A[j] = A[j+1]; A[j+1] = Temp; }  } PrintBubSort(A, n);     //呼叫排序後的結果之副程式 printf("\n"); system("pause"); }}// --------------------------------------------------------// check_data(A, Num);void check_data(int a[], int n){ // int i, j; printf("\n in check_data(), n= %d\n", n); system("pause"); // 0 - (n-1) for (i=0;i<=(n-1);i++) { j= i + 1; // we hope, a[i] <= a[j] if (a[i] > a[j]) { printf("\n *** error in check_data(), \n"); printf("a[i]= %d, a[j]= %d\n", a[i], a[j]); system("pause"); exit(1); } } printf("\n *** check OK in check_data()\n"); system("pause"); }// end of check_data()// --------------------------------------------------------int main(){ //主程式 // // generate data     RandomNum();              //呼叫產生10個亂數值的副程式     printf("\n");     BubSort(A, Num);          //呼叫氣泡排序法的副程式 // check the data check_data(A, Num);     PrintBubSort(A, Num);     //呼叫排序後的結果之副程式    printf("\n");      system("PAUSE");     return(0);}// end of main() Fri, 24 Dec 2010 08:26:02 +0800 http://sites.xms.com.tw/board.php?courseID=151&f=doc&cid=3853 請參考 圖檔。 Mon, 20 Dec 2010 14:49:37 +0800 http://sites.xms.com.tw/board.php?courseID=151&f=doc&cid=3851 #if 0#endif// ----------------------------------------------#include <cstdlib>#include <iostream>#include <math.h>#include <stdlib.h>#include <time.h>// ----------------------------------------------using namespace std;// ----------------------------------------------int main(int argc, char *argv[]){   int i;   // srand((unsigned) time(NULL));   srand(327);   for(i= 0;i<10;i++) {      printf( "  %3d", (rand()%100));   }        printf("\n");    system("PAUSE");        return EXIT_SUCCESS;}// end of main() Mon, 20 Dec 2010 09:55:42 +0800 http://sites.xms.com.tw/board.php?courseID=151&f=doc&cid=3843 for 上課用。 Sat, 18 Dec 2010 10:06:24 +0800 http://sites.xms.com.tw/board.php?courseID=151&f=doc&cid=3787 #if 0hello, world!請按任意鍵繼續 . . .#endif// --------------------------------------------------------#include <cstdlib>#include <iostream>using namespace std;// --------------------------------------------------------int main(int argc, char *argv[]){    printf("hello, world!\n");    system("PAUSE");        return EXIT_SUCCESS;} Wed, 15 Dec 2010 11:22:11 +0800 http://sites.xms.com.tw/board.php?courseID=151&f=doc&cid=3768 #include "inc-01.h"// ----------------------------------------------//   rnd1(    &s1);void rnd1(int *s1){ double x1, x2; x1= (double) (*s1); // x1= (x1*a + b) mod c;// a= 16807, b= 0, c= 2147483647 x2= fmod((x1*16807.0), 2147483647.0); (*s1)= (int) (x2 + 0.5);}// end of rnd1()// ----------------------------------------------//   init_rnd(    &s1);void init_rnd(int *s1){ int t1, t2, i; time1(&t1);// get t1 t2= t1;// t2 == t1 while (t2 == t1) { time1(&t1);// get a new t1 } // t1 <> t2 for (i=0;i<300;i++) { rnd1(&t1); } (*s1)= t1;}// end of init_rnd()// ----------------------------------------------//   swap_int(&i1, &i2);void swap_int(int *i1, int *i2){ int temp; temp= *i1; *i1= *i2; *i2= temp; }// end of swap_int()// ----------------------------------------------//   rnd2(s1, &x);void rnd2(int *s1, double *x){ rnd1(s1); *x= ((double) *s1)/2147483648.0;}// end of rnd2()// ----------------------------------------------//   irnd(&s1, i1, i2, &ii);void irnd(int *s1, int i1, int i2, int *ii){ // must i1 <= i2 if (i1 > i2) { swap_int(&i1, &i2); } double x; rnd2(s1, &x); (*ii)= (int) (x*(i2 - i1 + 1) + i1);}// end of // ----------------------------------------------int main(){ int s1, t1, i, no; double dt; no= (int) (6000e4 + 0.5); init_rnd(&s1); time1(&t1); for (i=0;i<no;i++) { rnd1(&s1); } time2(t1, &dt); printf("no= %ld, dt= %.3lf\n", no, dt); pause(); return(0);}// ----------------------------------------------#if 0no= 60000000, dt= 10.125Press [Esc] for stop! other key for continue...#endif// ---------------------------------------------- Tue, 14 Dec 2010 12:21:23 +0800 http://sites.xms.com.tw/board.php?courseID=151&f=doc&cid=3764 假設，我要寫 計算時間 間隔的副程式，我必須知道，計算時間，使用 ms 做單位，如果 時間的間隔 超過 24天，就會 over- flow產生錯誤的結果。或是，計算的時間間隔，跨越 24天週期的 分界線，也是會產生錯誤的結果，而且不會有任何的compiler 給的錯誤，或是警告的訊息。跑程式的時候，即使發生錯誤，也不會有任何的錯誤，或是警告訊息。所以，一切都要靠自己。#include <stdio.h>#include <math.h>#include <stdlib.h>// ----------------------------------------------int main(){ int a= (int) ((pow(2, 31)) - 1 + 0.5); printf("a= %d\n", a); system("pause"); int b= a/1000/(1L*24*60*60); int c= b*(1L*24*60*60)*1000; printf("b= %d, c= %d\n", b, c); system("pause"); return(0);}// ----------------------------------------------#if 0a= 2147483647請按任意鍵繼續 . . .b= 24, c= 2073600000請按任意鍵繼續 . . .#endif// ---------------------------------------------- Tue, 14 Dec 2010 06:19:12 +0800 http://sites.xms.com.tw/board.php?courseID=151&f=doc&cid=3745 main(){}// ----------------------------------------------------------------------------#if 0--------------------Configuration: VC1214 - Win32 Debug--------------------Compiling...VC1214.CPPD:\VC1214\VC1214.CPP(3) : warning C4508: 'main' : function should return a value; 'void' return type assumedLinking...VC1214.exe - 0 error(s), 1 warning(s)#endif// ---------------------------------------------------------------------------- Mon, 13 Dec 2010 06:44:28 +0800 http://sites.xms.com.tw/board.php?courseID=151&f=doc&cid=3712 #if 0然後 產生 autocad script file 的範例ct1=   1, x= 0.000000, y= 0.000000ct1=   2, x= 0.196350, y= 0.195090ct1=   3, x= 0.392699, y= 0.382683ct1=   4, x= 0.589049, y= 0.555570ct1=   5, x= 0.785398, y= 0.707107ct1=   6, x= 0.981748, y= 0.831470ct1=   7, x= 1.178097, y= 0.923880ct1=   8, x= 1.374447, y= 0.980785ct1=   9, x= 1.570796, y= 1.000000ct1=  10, x= 1.767146, y= 0.980785ct1=  11, x= 1.963495, y= 0.923880ct1=  12, x= 2.159845, y= 0.831470ct1=  13, x= 2.356194, y= 0.707107ct1=  14, x= 2.552544, y= 0.555570ct1=  15, x= 2.748894, y= 0.382683ct1=  16, x= 2.945243, y= 0.195090ct1=  17, x= 3.141593, y= 0.000000 Press [Esc] for stop! other key for continue...#endif// ----------------------------------------------#include "inc-01.h"#include "sj-01.h"#include "sj-02.h"// ----------------------------------------------int main(){ FILE *f1, *f2; double x, y; f1= fopen("plot-sin.txt", "rt"); f2= fopen("plot-sin.scr", "wt"); int ct1= 0, ct2; fprintf(f2, "spline\n"); while (!feof(f1)) { ct2= fscanf(f1, "%lf %lf", &x, &y); if (ct2 != 2) break; ct1++; printf("ct1= %3d, x= %.6lf, y= %.6lf\n", ct1, x, y); fprintf(f2, "%.6lf,%.6lf\n", x, y); } fprintf(f2, "\n\n\n"); fclose(f1); fclose(f2); pause(); return 0;}// end of main() Wed, 08 Dec 2010 14:44:45 +0800 http://sites.xms.com.tw/board.php?courseID=151&f=doc&cid=3710 !dec$if(.false.)檔案的內容是y= sin(), x= 0 to 3.1415926 step dxdx= 16等分準備教   0.000000  0.000000   0.196350  0.195090   0.392699  0.382683   0.589049  0.555570   0.785398  0.707107   0.981748  0.831470   1.178097  0.923880   1.374447  0.980785   1.570796  1.000000   1.767146  0.980785   1.963495  0.923880   2.159845  0.831470   2.356194  0.707107   2.552544  0.555570   2.748894  0.382683   2.945243  0.195090   3.141593  0.000000!dec$endif! --------------------------------------------------------- program VF0904 implicit none integer no real x, y, x1, x2, dx ! plot y= sin(x), x= 0 to pi step dx no= 16 x1= 0.0 x2= 4.0*atan(1.0) dx= (x2 - x1)/no x2= x2 + 0.1*dx ! for x= x1 to x2 step dx do ... x= x1 open(unit= 1, file= 'plot-sin.txt', status= 'UNKNOWN'); do while (x .LE. x2)   y= sin(x)   write(unit= 1, fmt= '(1x, 2F10.6)')x, y   x= x + dx end do close(1); print *, 'Done!' pause end program VF0904 Wed, 08 Dec 2010 14:16:14 +0800 http://sites.xms.com.tw/board.php?courseID=151&f=doc&cid=3706 !dec$if(.false.)   11.0000000000000        10.0000000000000Fortran Pause - Enter command<CR> or <CR> to continue.      9007199450284736.000     9007199450284736.000Fortran Pause - Enter command<CR> or <CR> to continue.!dec$endif! --------------------------------------------------------- program VF0904 implicit none real*8 a, b a= 10.0D0 b= a + 1.0D0 ! b= 11, > 10, b > a print *, b, a pause do while (b .GT. a)   a= a*(1.0D0 + 1.0D-7)   b= a + 1.0D0 end do ! not for (b > a), --> b <= a write(*, '(1x, 2F25.3)')b, a pause end program VF0904 Wed, 08 Dec 2010 08:18:11 +0800 http://sites.xms.com.tw/board.php?courseID=151&f=doc&cid=3705 !dec$if(.false.)   11.00000       10.00000Fortran Pause - Enter command<CR> or <CR> to continue.    16777216.000   16777216.000Fortran Pause - Enter command<CR> or <CR> to continue.!dec$endif! --------------------------------------------------------- program VF0904 implicit none real a, b a= 10.0 b= a + 1.0 ! b= 11, > 10, b > a print *, b, a pause do while (b .GT. a)   a= a + 1.0   b= b + 1.0 end do ! not for (b > a), --> b <= a write(*, '(1x, 2F15.3)')b, a pause end program VF0904 Wed, 08 Dec 2010 07:23:00 +0800 http://sites.xms.com.tw/board.php?courseID=151&f=doc&cid=3704 !dec$if(.false.) b, a=           11          10Fortran Pause - Enter command<CR> or <CR> to continue. b, a=  -2147483648  2147483647Fortran Pause - Enter command<CR> or <CR> to continue.!dec$endif! --------------------------------------------------------- program VF0904 implicit none integer a, b a= 10 b= a + 1 ! 11 > 10, b > a print *, 'b, a= ', b, a pause do while (b .GT. a)   a= a + 1   b= a + 1 end do ! not of (b > a) --> b <= a, why? print *, 'b, a= ', b, a pause end program VF0904 Wed, 08 Dec 2010 06:49:58 +0800 http://sites.xms.com.tw/board.php?courseID=151&f=doc&cid=3696 請參考下面的 程式設計，double 是八個 Bytes 的變數型態，有 64個 bits, 第一個 bits 用來表示 浮點數的正負。使用 11個 bits 來表示 浮點數的 數量級。可以用來表示 (2^-1024) 到 (2^+1023) 範圍的 浮點數，大概是 十進位數字的(10^-308) 到 (10^308)剩下的 52個 bits 是用來表示 浮點數的 內容。所以，精確度可以到達 相對誤差：(1.0)/(2^52), 也就是 十進位的大概 16 - 17位的 有效位數。#if 0 b= 4.14159265358979310000, a= 3.14159265358979310000 Press [Esc] for stop! other key for continue... b= 9007199771293056.000, a= 9007199771293056.000 Press [Esc] for stop! other key for continue...#endif// ----------------------------------------------#include "inc-01.h"#include "sj-01.h"#include "sj-02.h"// ----------------------------------------------int main(){ double a, b; a= 4.0*atan(1.0); b= a + 1;// (a + 1) > a, so b > a printf("\n b= %.20lf, \n a= %.20lf\n", b, a); pause(); while (b > a) { a= a*(1.0 + 1e-7); b= a + 1.0; } // !(b > a), --> b <= a printf("\n b= %.3lf, \n a= %.3lf\n", b, a); pause(); return 0;}// end of main() Tue, 07 Dec 2010 16:33:29 +0800 http://sites.xms.com.tw/board.php?courseID=151&f=doc&cid=3695 以下的程式，可以證明 int 的範圍是-2147483648 到 +2147483647也就是 (-2^31) 到 (2^31) - 1int 是 四個 Bytes 的整數，是 32個 bits 的整數，有一個 bit 被用來表示 正，或是 負。剩下的 31 個 bits, 可以用來表示0 - +2147483647負的整數，使用 二的補數來表示。所以，40000*60000 --> ???其結果，是一個 負的整數。#if 0 b= 11, a= 10 Press [Esc] for stop! other key for continue... *** exit the loop, b= -2147483648, a= 2147483647 Press [Esc] for stop! other key for continue...#endif// ----------------------------------------------#include "inc-01.h"#include "sj-01.h"#include "sj-02.h"// ----------------------------------------------int main(){ int a, b; a= 10; b= a + 1;// b= 11, > 10, > a printf("\n b= %d, a= %d\n", b, a); pause(); while (b > a) { a= a + 1; b= a + 1; } // !(b > a), --> b <= a printf("\n *** exit the loop, \n b= %d, a= %d\n", b, a); pause(); return 0;}// end of main() Tue, 07 Dec 2010 16:15:29 +0800 http://sites.xms.com.tw/board.php?courseID=151&f=doc&cid=3679 ...RECTANG@200,100@DX,DY@R<THETALINEBPOLYALIGNDIM Mon, 06 Dec 2010 12:10:40 +0800 http://sites.xms.com.tw/board.php?courseID=151&f=doc&cid=3676 CIRCLEPDMODEDIVIDEOSNAPARCARRAYBPOLYAREA Mon, 06 Dec 2010 09:47:36 +0800 http://sites.xms.com.tw/board.php?courseID=151&f=doc&cid=3642 !dec$if(.false.)底下是 online help , 關於 date_and_time() 的說明DATE_AND_TIMEIntrinsic Subroutine: Returns character data on the real-time clock and date in a form compatible with the representations defined in Standard ISO 8601:1988. Syntax CALL DATE_AND_TIME ( [date] [, time] [, zone] [, values] )date (Optional; output) Must be scalar and of type default character; its length must be at least 8 to contain the complete value. Its leftmost 8 characters are set to a value of the form CCYYMMDD, where:CC  Is the century  YY  Is the year within the century  MM  Is the month within the year  DD  Is the day within the month time (Optional; output) Must be scalar and of type default character; its length must be at least 10 to contain the complete value. Its leftmost 10 characters are set to a value of the form hhmmss.sss, where:hh  Is the hour of the day  mm  Is the minutes of the hour ss.sss  Is the seconds and milliseconds of the minute  zone (Optional; output) Must be scalar and of type default character; its length must be at least 5 to contain the complete value. Its leftmost 5 characters are set to a value of the form hhmm, where hh and mm are the time difference with respect to Coordinated Universal Time (UTC) in hours and parts of an hour expressed in minutes, respectively. UTC (also known as Greenwich Mean Time) is defined by CCIR Recommendation 460-2.values (Optional; output) Must be of type default integer. One-dimensional array with size of at least 8. The values returned in values are as follows:values (1)  The 4-digit year  values (2)  The month of the year  values (3)  The day of the month  values (4)  The time difference with respect to Coordinated Universal Time (UTC) in minutes  values (5)  The hour of the day (range 0 to 23) - local time  values (6)  The minutes of the hour (range 0 to 59) - local time  values (7)  The seconds of the minute (range 0 to 59) - local time  values (8)  The milliseconds of the second (range 0 to 999) - local time  Compatibility CONSOLE STANDARD GRAPHICS QUICKWIN GRAPHICS WINDOWS DLL LIB See Also: GETDAT, GETTIM, IDATE, FDATE, TIME, ITIME, RTC, CLOCKExample Consider the following example executed on 2000 March 28 at 11:04:14.5:  INTEGER DATE_TIME (8)  CHARACTER (LEN = 12) REAL_CLOCK (3)  CALL DATE_AND_TIME (REAL_CLOCK (1), REAL_CLOCK (2), &                      REAL_CLOCK (3), DATE_TIME)This assigns the value "20000328" to REAL_CLOCK (1), the value "110414.500" to REAL_CLOCK (2), and the value "-0500" to REAL_CLOCK (3). The following values are assigned to DATE_TIME: 2000, 3, 28, -300, 11, 4, 14, and 500.The following shows another example:CHARACTER(10) tCHARACTER(5) zCALL DATE_AND_TIME(TIME = t, ZONE = z)!dec$endif! --------------------------------------------------------- program VF0904 implicit none !    INTEGER DATE_TIME(8), i, ms    CHARACTER(LEN= 12) REAL_CLOCK(3)    CALL DATE_AND_TIME (REAL_CLOCK (1), REAL_CLOCK (2), &                        REAL_CLOCK (3), DATE_TIME)    ! do i=1, 3   print *, 'i, data= ', i, real_clock(i) end do pause do i=1, 8   print *, 'i, data= ', i, date_time(i) end do pause ms= date_time(3) ! = day number ms= mod(ms, 21) ! max.= 21- days ms= ms*24 + date_time(5) ! --> hours ms= ms*60 + date_time(6) ! --> min ms= ms*60 + date_time(7) ! --> sec ms= ms*1000 + date_time(8) ! --> ms print *, 'ms= ', ms pause end program VF0904 ! -----------------------------------------------------!dec$if(.false.)!dec$endif Fri, 03 Dec 2010 09:27:50 +0800 http://sites.xms.com.tw/board.php?courseID=151&f=doc&cid=3638 模擬 你的需求寄件人gmail.com隱藏詳細資料 21:56 (8 分鐘前)1000 到 9999 取 200個亂數，然後，測試 是否為 質數，如果是 質數的話，寫入 檔案PRIME-N.DATN= 該質數的 千位數，你看 如何？為了 完成這個程式，需要 ㄧ些 副程式翰 函數。call skip(3)可以 空三行的 輸出call pause暫停call time1(t1)t1 能夠逮表目前的 時間刻度，單位是 mscall time2(t1, dt)計算 從 t1 到現在，所經過的 秒數，單位是 秒call rnd1(s1)把 s1 丟進去，產生 新的 s1s1 是亂數，1 <= s1 <= 2147483646call rnd2(s1, x1)s1 的說明，同上面。0.0 < x1 < 1.0call irnd(s1, i1, i2, ii)s1 的說明，同上面。i1, i2 是 input, ii 是 output,i1 <= ii <= i2call init_rnd(s1)呼叫 time1(), 產生 亂數的 種子數 s1is_prime(no)測試 no 是否為 質數 Thu, 02 Dec 2010 22:05:39 +0800 http://sites.xms.com.tw/board.php?courseID=151&f=doc&cid=3637 換成 real, 如何解釋那個 結果？寄件人gmail.com隱藏詳細資料 21:40 (22 分鐘前)       program VF0904       implicit none       real a, b       a= 4.0*atan(1.0)       b= a + 1.0       ! b= 4.1415926, > 3.1415926, b > a       print *, 'a, b= ', a, b       pause       ! -----------------------------------------------------       do while (b .GT. a)          a= a + 1.0          b= a + 1.0       end do       ! b <= a, 怎麼可能 (a + 1) <= a       print *, 'exit do while(), b <= a'       print *, 'a, b= ', a, b       pause       end program VF0904       ! -----------------------------------------------------!dec$if(.false.) a, b=    3.141593       4.141593Fortran Pause - Enter command<CR> or <CR> to continue. exit do while(), b <= a a, b=   1.6777216E+07  1.6777216E+07Fortran Pause - Enter command<CR> or <CR> to continue.!dec$endif Thu, 02 Dec 2010 22:04:26 +0800 http://sites.xms.com.tw/board.php?courseID=151&f=doc&cid=3636 阿，這個 程式呢？寄件人gmail.com隱藏詳細資料 21:16 (45 分鐘前)       program VF0904       implicit none       integer a, b       a= 10       b= a + 1       ! b= 11, > 10, b > a       print *, 'a, b= ', a, b       pause       ! -----------------------------------------------------       do while (b .GT. a)          a= a + 1          b= a + 1       end do       ! b <= a, 怎麼可能 (a + 1) <= a       print *, 'exit do while(), b <= a'       print *, 'a, b= ', a, b       pause       end program VF0904       ! -----------------------------------------------------!dec$if(.false.) a, b=           10          11Fortran Pause - Enter command<CR> or <CR> to continue. exit do while(), b <= a a, b=   2147483647 -2147483648Fortran Pause - Enter command<CR> or <CR> to continue.!dec$endif Thu, 02 Dec 2010 22:03:03 +0800 http://sites.xms.com.tw/board.php?courseID=151&f=doc&cid=3635 這個程式，有沒有哪裏 不了解的寄件人gmail.com隱藏詳細資料 21:10 (50 分鐘前)!  VF0904.f90!!  FUNCTIONS:!       VF0904      - Entry point of console application.!!       Example of displaying 'Hello World' at execution time.!!****************************************************************************!!  PROGRAM: VF0904!!  PURPOSE:  Entry point for 'Hello World' sample console application.!!****************************************************************************       program VF0904       implicit none       integer a, b, c       a= 40000       b= 60000       c= a*b       print *, 'Hello World, c= ', c       end program VF0904       ! -----------------------------------------------------!dec$if(.false.) Hello World, c=  -1894967296Press any key to continue!dec$endif Thu, 02 Dec 2010 22:01:44 +0800 http://sites.xms.com.tw/board.php?courseID=151&f=doc&cid=3566 ; FILE: PLOT-SIN.LSP; PLOT Y= SIN(X), X= X1 TO X2 STEP DX(SETQ NO 16)(SETQ X1 0.0      X2 (* 4.0 (ATAN 1.0))      DX (/ (- X2 X1) NO))(SETQ X2 (+ X2 (/ DX 10.0))); FOR X= X1 TO X2 STEP DX(SETQ X X1)(COMMAND "SPLINE")(WHILE (<= X X2)  (SETQ Y (SIN X))  (SETQ P1 (LIST X Y))  (COMMAND P1)  (SETQ X (+ X DX))  ); END OF WHILE()(COMMAND "" "" "")(PRINC); END OF FILE Mon, 29 Nov 2010 17:15:46 +0800 http://sites.xms.com.tw/board.php?courseID=151&f=doc&cid=3488 以下，20億個 加法運算，需要多少時間？只看 主程式的部份，是不是很簡潔啊？#if 0no=   2000000000, sum=  -1973237248, dt= 7.828 Press [Esc] for stop! other key for continue...#endif// ----------------------------------------------#include <stdio.h>#include <stdlib.h>#include <math.h>#include <conio.h>#include <sys/timeb.h>#include <time.h>// ----------------------------------------------#include "sj-01.h"// ----------------------------------------------void time1(int *t1){ struct _timeb timebuffer; int ms, sec; _ftime( &timebuffer ); ms= timebuffer.millitm; sec= timebuffer.time; sec%= (21L*86400L);// max. = 24.8 day, so ... (*t1)= sec*1000 + ms;}// end of time1()// ----------------------------------------------void time2(int t1, double *dt){ int t2; time1(&t2); (*dt)= ((double) (t2 - t1))/1000.0; // (*dt) must >= 0 if ((*dt) < 0) { (*dt)+= (21L*86400L); }}// end of time2()// ----------------------------------------------int main(){ // 20億個 加法運算，需要多少時間？ int no= (int) (20E8 + 0.5);// 四捨五入 int sum, i, t1; double dt; time1(&t1); sum= 0; for (i=1;i<=no;i++) { sum+= i; } time2(t1, &dt); printf("no= %12ld, sum= %12ld, dt= %.3lf\n", no, sum, dt); pause(); return 0;}// end of main() Fri, 19 Nov 2010 18:49:21 +0800 http://sites.xms.com.tw/board.php?courseID=151&f=doc&cid=3477 #if 0no=           10, sum=           55, dt= 0.000no=           20, sum=          210, dt= 0.000no=           40, sum=          820, dt= 0.000no=           80, sum=         3240, dt= 0.000no=          160, sum=        12880, dt= 0.000no=          320, sum=        51360, dt= 0.000no=          640, sum=       205120, dt= 0.000no=         1280, sum=       819840, dt= 0.000no=         2560, sum=      3278080, dt= 0.000no=         5120, sum=     13109760, dt= 0.000no=        10240, sum=     52433920, dt= 0.000no=        20480, sum=    209725440, dt= 0.000no=        40960, sum=    838881280, dt= 0.000no=        81920, sum=   -939483136, dt= 0.000no=       163840, sum=    536952832, dt= 0.016no=       327680, sum=  -2147319808, dt= 0.000no=       655360, sum=       327680, dt= 0.000no=      1310720, sum=       655360, dt= 0.015no=      2621440, sum=      1310720, dt= 0.032no=      5242880, sum=      2621440, dt= 0.046no=     10485760, sum=      5242880, dt= 0.079no=     20971520, sum=     10485760, dt= 0.093no=     41943040, sum=     20971520, dt= 0.219no=     83886080, sum=     41943040, dt= 0.313no=    167772160, sum=     83886080, dt= 0.625no=    335544320, sum=    167772160, dt= 1.453no=    671088640, sum=    335544320, dt= 2.718no=   1342177280, sum=    671088640, dt= 6.360Press any key to continue#endif// ----------------------------------------------#include <stdio.h>#include <stdlib.h>#include <math.h>#include <conio.h>#include <sys/timeb.h>#include <time.h>// ----------------------------------------------#include "sj-01.h"// ----------------------------------------------void time1(int *t1){ struct _timeb timebuffer; int ms, sec; _ftime( &timebuffer ); ms= timebuffer.millitm; sec= timebuffer.time; sec%= (21*86400L);// max. = 24.8 day, so ... (*t1)= sec*1000 + ms;}// end of time1()// ----------------------------------------------void time2(int t1, double *dt){ int t2; time1(&t2); (*dt)= ((double) (t2 - t1))/1000.0; // (*dt) must >= 0 if ((*dt) < 0) { (*dt)+= (21L*86400L); }}// end of time2()// ----------------------------------------------int main(){ int t1, no= 10, sum, i; double dt; while (no > 0) { time1(&t1); sum= 0; for (i=1;i<=no;i++) { sum+= i; } time2(t1, &dt); printf("no= %12ld, sum= %12ld, dt= %.3lf\n", no, sum, dt); no*= 2; } return 0;}// end of main() Fri, 19 Nov 2010 11:14:08 +0800 http://sites.xms.com.tw/board.php?courseID=151&f=doc&cid=3459 #if 0The time is Thu Nov 18 16:43:19.296 2010ms= 296 Press [Esc] for stop! other key for continue...The time is Thu Nov 18 16:43:20.46 2010ms= 046 Press [Esc] for stop! other key for continue... 上面，當 ms= 46的時候，必須印出來是 20.046, 卻發生錯誤，印成 20.46, 就是 20.460, 很明顯的，是一個 錯誤。微軟的 MSDN, 關於 ftime() 的範例程式，寫錯了。#endif// ----------------------------------------------#include <stdio.h>#include <stdlib.h>#include <math.h>#include <conio.h>#include <sys/timeb.h>#include <time.h>#include "sj-01.h"// ----------------------------------------------int main(){ struct _timeb timebuffer; char *timeline; int ms; for (;;) { skip(3); _ftime( &timebuffer ); ms= timebuffer.millitm; timeline = ctime( & ( timebuffer.time ) ); printf( "The time is %.19s.%hu %s", timeline, timebuffer.millitm, &timeline[20] ); printf("ms= %03d\n", ms); pause(); } return 0;}// end of main() Thu, 18 Nov 2010 16:48:09 +0800 http://sites.xms.com.tw/board.php?courseID=151&f=doc&cid=3441 關鍵是#include "file.h"// 這是 主程式的 檔案#if 0#endif// ----------------------------------------------#include <stdio.h>#include <stdlib.h>#include <math.h>#include <conio.h>#include "sj-01.h"// ----------------------------------------------int main(){ int i; for (i=1;i<=20;i++) { skip(i); printf("i= %5d, i*i= %5d, sqrt(i)= %10.6lf\n",  i,      i*i,      sqrt(i)); pause(); } return 0;}// end of main()// 以下，是副程式的檔案// file: sj-01.h//   skip(3);void skip(int no){ int i; // limits no in 1 .. 20 if ((no >= 1) && (no <= 20)) {// no is OK for (i=1;i<=no;i++) { // printf("*** in skip(), i= %d\n", i); printf("\n"); } } else {// no is invalid // printf("*** error in skip(), no= %d\n", no); printf("\n"); }}// end of skip()// ----------------------------------------------//   pause();void pause(void){ // remove type- ahead while (kbhit()) { getch(); } // ---------------------------------------------- printf(" Press [Esc] for stop! other key for continue..."); do { // wait for keyPressed! } while (!kbhit()); int ch1= getch(); printf("\n"); // ---------------------------------------------- if (ch1==0x1b) { exit(1); } // remove extra key- pressed while (kbhit()) { getch(); }}// end of pause() Tue, 16 Nov 2010 17:28:42 +0800 http://sites.xms.com.tw/board.php?courseID=151&f=doc&cid=3433 #if 0#endif// ----------------------------------------------#include <stdio.h>#include <stdlib.h>#include <math.h>#include <conio.h>// ----------------------------------------------//   skip(3);void skip(int no){ int i; // limits no in 1 .. 20 if ((no >= 1) && (no <= 20)) {// no is OK for (i=1;i<=no;i++) { // printf("*** in skip(), i= %d\n", i); printf("\n"); } } else {// no is invalid // printf("*** error in skip(), no= %d\n", no); printf("\n"); }}// end of skip()// ----------------------------------------------//   pause();void pause(void){ // remove type- ahead while (kbhit()) { getch(); } // ---------------------------------------------- printf(" Press [Esc] for stop! other key for continue..."); do { // wait for keyPressed! } while (!kbhit()); int ch1= getch(); printf("\n"); // ---------------------------------------------- if (ch1==0x1b) { exit(1); } // remove extra key- pressed while (kbhit()) { getch(); }}// end of pause()// ----------------------------------------------int main(){ int i; for (i=1;i<=20;i++) { skip(i); printf("i= %5d, i*i= %5d, sqrt(i)= %10.6lf\n",  i,      i*i,      sqrt(i)); pause(); } return 0;}// end of main() Tue, 16 Nov 2010 07:45:41 +0800 http://sites.xms.com.tw/board.php?courseID=151&f=doc&cid=3426 #if 0理想中的亂數是，一 要儘可能的產生比較長的週期，也就是說，在出現相同的數字之前，所產生的 所有數字的 個數二 每一個數字的出現頻率要相同，所以，週期要 固定三 產生的速度要 儘可能的快s1=  20826, s2=  20826, ct=   40354s1=  20829, s2=  20829, ct=   32390s1=  20833, s2=  20833, ct=    9846s1=  20836, s2=  20836, ct=   47585s1=  20839, s2=  20839, ct=    7268s1=  20842, s2=  20842, ct=   24388s1=  20846, s2=  20846, ct=    1273s1=  20849, s2=  20849, ct=   21765s1=  20852, s2=  20852, ct=   17792s1=  20856, s2=  20856, ct=   24732s1=  20859, s2=  20859, ct=   45344s1=  20862, s2=  20862, ct=   96963s1=  20865, s2=  20865, ct=   27246s1=  20869, s2=  20869, ct=   28311s1=  20872, s2=  20872, ct=   32888s1=  20875, s2=  20875, ct=    8121s1=  20878, s2=  20878, ct=   42342s1=  20882, s2=  20882, ct=   31186s1=  20885, s2=  20885, ct=   66311s1=  20888, s2=  20888, ct=   68819s1=  20891, s2=  20891, ct=    6028s1=  20895, s2=  20895, ct=   25150s1=  20898, s2=  20898, ct=   18894s1=  20901, s2=  20901, ct=   43998以上，一共跑了 180趟，平均值 接近 32768,符合預期。但是，週期的變化，min= 99, max= 18萬多，不合規定，而且也是顯得 週期太短。min=       99, max= 184309, ave.= 30662.267Press any key to continue#endif// ----------------------------------------------#include <stdlib.h>#include <stdio.h>#include <time.h>void main( void ){ int i, ct, s1, s2, t1, t2, sum= 0; int min, max; min= 2000000000; max= -min; for (i=1;i<=(60*3);i++) {// do the 10- times t1= time(NULL); t2= time(NULL); while (t2 == t1) { t2= time(NULL); } srand(t2); ct= 0; s1= rand(); s2= rand(); // s2 <> s1 while (s2 != s1) { s2= rand(); ct++; } printf("s1= %6d, s2= %6d, ct= %7d\n", s1, s2, ct); if (min > ct) min= ct; if (max < ct) max= ct; sum+= ct; } printf("\n\n min= %8d, max= %5d, ave.= %.3lf\n", min, max, sum/(60.0*3));} Mon, 15 Nov 2010 12:00:13 +0800 http://sites.xms.com.tw/board.php?courseID=151&f=doc&cid=3425 只有到 秒。以下的程式的 輸出，可以看出一些 奇怪的現象。#if 0no=           10, sum=           55, dt=     0no=           30, sum=          465, dt=     0no=           90, sum=         4095, dt=     0no=          270, sum=        36585, dt=     0no=          810, sum=       328455, dt=     0no=         2430, sum=      2953665, dt=     0no=         7290, sum=     26575695, dt=     0no=        21870, sum=    239159385, dt=     0no=        65610, sum=  -2142598441, dt=     0no=       196830, sum=  -2103713615, dt=     0no=       590490, sum=  -1754143841, dt=     0no=      1771470, sum=   1390803145, dt=     0no=      5314410, sum=   -372987993, dt=     0no=     15943230, sum=    922132129, dt=     0no=     47829690, sum=   -338575121, dt=     1上面的時間 1, 乘以 3, 為何是 0, ???no=    143489070, sum=   1104302137, dt=     0上面的時間 0, 乘以 3, 為何是 2, ???no=    430467210, sum=    918317431, dt=     2no=   1291401630, sum=  -1616479343, dt=     5Press any key to continue#endif// ----------------------------------------------#include <time.h>#include <math.h>#include <stdio.h>void main(){ int no, t1, t2, sum, i; no= 10; while (no > 0) { t1= time(NULL); sum= 0; for (i=1;i<=no;i++) { sum+= i; } t2= time(NULL); printf("no= %12d, sum= %12d, dt= %5d\n", no, sum, (t2 - t1)); no*= 3;// 為何不是 乘以2, ... }}// end of main()因為，電腦使用 二進位，累乘以 2, 會得到一些 奇怪的結果，有興趣的，可以自己 試試看，。。。 Mon, 15 Nov 2010 11:06:31 +0800 http://sites.xms.com.tw/board.php?courseID=151&f=doc&cid=3406 #if 0n= 2147483647, m= -2147483648請按任意鍵繼續 . . .目前的 int 所能夠表示的 最大整數為2147483647,這個數字加一，會變成 -2147483648.這是因為電腦的內部，使用二的補數來表示 負整數的關係。詳細，請繼續參考 二的補數#endif#include <stdlib.h>#include <stdio.h>int main(){ int n, m; n= 10; m= n + 1;// m > n while (m > n) { n++; m= n + 1; } printf("n= %d, m= %d\n", n, m); system("pause"); return 0;} Sat, 13 Nov 2010 08:21:30 +0800 http://sites.xms.com.tw/board.php?courseID=151&f=doc&cid=3396 底下這個程式，讀 excel 產生的 text file,使用 空格分離 不同欄位。產生 AutoCAD script file,即可 教AutoCAD 自動產生圖形物件#include "stdafx.h"#include <process.h>// --------------------------------------------------------int main(int argc, char* argv[]){       FILE *f1, *f2;       // open the input and output file       if ((f1= fopen("book2a.txt", "rt")) == NULL) {               printf("\n *** error of fopen() of f1\n");               system("pause");               exit(1);       }       if ((f2= fopen("T0435.scr", "wt")) == NULL) {               printf("\n *** error of fopen() of f2\n");               system("pause");               exit(1);       }       printf("\n OK for fopen() of f1 and f2, at 04:37\n");       system("pause");       // ------------------------------------------       // read the data from file       int ct= -1, no, i;       double x[100], y[100];       while (!feof(f1)) {               ct++;               no= fscanf(f1, "%lf %lf", &x[ct], &y[ct]);               if (no != 2) {                       ct--;                       break;               }       }// end of while       fprintf(f2, "spline\n");       for (i=0;i<=ct;i++) {               printf("%5d, %10.6lf, %10.6lf\n", i, x[i], y[i]);               fprintf(f2, "%.6lf,%.6lf\n", x[i], y[i]);       }       fprintf(f2, "\n\n\n");       fclose(f1);       fclose(f2);       return 0;} Thu, 11 Nov 2010 16:48:28 +0800 http://sites.xms.com.tw/board.php?courseID=151&f=doc&cid=3364 #include <stdio.h>#include <stdlib.h>int main(void){    printf("hello, world!\n"); system("pause");     return 0;} Wed, 10 Nov 2010 08:14:27 +0800 http://sites.xms.com.tw/board.php?courseID=151&f=doc&cid=3347 (defun dtor(d1 / )  (* (/ d1 180.0) pi)  ); end of dtor(); -----------------------------------------------; draw a arc by CDR, C:圓心，D:方向, R:半徑(defun c:arc-cdr( / pc p1 t1 t2 t3 p2 p3)  (setq pc (getpoint "\n 弧的圓心: "))  (setq p1 (getpoint "\n 大概的方向: "))  (setq t1 (angle pc p1))  (setq r1 (getdist "\n 輸入兩點 決定半徑: "))  (setq t2 (+ t1 (dtor 15.0)))  (setq t3 (- t1 (dtor 15.0)))  (setq p2 (polar pc t3 r1))  (setq p3 (polar pc t2 r1))  (setq p1 (polar pc t1 r1))  (command "arc" p2 p1 p3)  (princ)  ); end of c:arc-cdr Tue, 09 Nov 2010 11:06:43 +0800