C程序百例.docx

1、C程序百例语言编程经典100例【程序51】题目：学习使用按位与 & 。1.程序分析：0&0=0; 0&1=0; 1&0=0; 1&1=12.程序源代码：#include stdio.hmain()int a,b;a=077;b=a&3;printf(40: The a & b(decimal) is %d n,b);b&=7;printf(40: The a & b(decimal) is %d n,b);-【程序52】题目：学习使用按位或 | 。1.程序分析：0|0=0; 0|1=1; 1|0=1; 1|1=12.程序源代码：#include stdio.hmain()int a,b;a=0

2、77;b=a|3;printf(40: The a & b(decimal) is %d n,b);b|=7;printf(40: The a & b(decimal) is %d n,b);-【程序53】题目：学习使用按位异或 。1.程序分析：00=0; 01=1; 10=1; 11=02.程序源代码：#include stdio.hmain()int a,b;a=077;b=a3;printf(40: The a & b(decimal) is %d n,b);b=7;printf(40: The a & b(decimal) is %d n,b);-【程序54】题目：取一个整数a从右端开

3、始的47位。程序分析：可以这样考虑： (1)先使a右移4位。(2)设置一个低4位全为1,其余全为0的数。可用(04;c=(04);d=b&c;printf(%on%on,a,d);-【程序55】题目：学习使用按位取反。1.程序分析：0=1; 1=0;2.程序源代码：#include stdio.hmain()int a,b;a=234;b=a;printf(40: The as 1 complement(decimal) is %d n,b);a=a;printf(40: The as 1 complement(hexidecimal) is %x n,a); -【程序56】题目：画图，学用c

4、ircle画圆形。1.程序分析：2.程序源代码：/*circle*/#include graphics.hmain()int driver,mode,i;float j=1,k=1;driver=VGA;mode=VGAHI;initgraph(&driver,&mode,);setbkcolor(YELLOW);for(i=0;i=25;i+)setcolor(8);circle(310,250,k);k=k+j;j=j+0.3; -【程序57】题目：画图，学用line画直线。1.程序分析：2.程序源代码：#include graphics.hmain()int driver,mode,i;

5、float x0,y0,y1,x1;float j=12,k;driver=VGA;mode=VGAHI;initgraph(&driver,&mode,);setbkcolor(GREEN);x0=263;y0=263;y1=275;x1=275;for(i=0;i=18;i+)setcolor(5);line(x0,y0,x0,y1);x0=x0-5;y0=y0-5;x1=x1+5;y1=y1+5;j=j+10;x0=263;y1=275;y0=263;for(i=0;i=20;i+)setcolor(5);line(x0,y0,x0,y1);x0=x0+5;y0=y0+5;y1=y1-5

6、;-【程序58】题目：画图，学用rectangle画方形。1.程序分析：利用for循环控制100-999个数，每个数分解出个位，十位，百位。2.程序源代码：#include graphics.hmain()int x0,y0,y1,x1,driver,mode,i;driver=VGA;mode=VGAHI;initgraph(&driver,&mode,);setbkcolor(YELLOW);x0=263;y0=263;y1=275;x1=275;for(i=0;i=18;i+)setcolor(1);rectangle(x0,y0,x1,y1);x0=x0-5;y0=y0-5;x1=x1

7、+5;y1=y1+5;settextstyle(DEFAULT_FONT,HORIZ_DIR,2);outtextxy(150,40,How beautiful it is!);line(130,60,480,60);setcolor(2);circle(269,269,137);-【程序59】题目：画图，综合例子。1.程序分析：2.程序源代码：# define PAI 3.1415926# define B 0.809# include graphics.h#include math.hmain()int i,j,k,x0,y0,x,y,driver,mode;float a;driver=

8、CGA;mode=CGAC0;initgraph(&driver,&mode,);setcolor(3);setbkcolor(GREEN);x0=150;y0=100;circle(x0,y0,10);circle(x0,y0,20);circle(x0,y0,50);for(i=0;i16;i+) a=(2*PAI/16)*i; x=ceil(x0+48*cos(a); y=ceil(y0+48*sin(a)*B); setcolor(2); line(x0,y0,x,y);setcolor(3);circle(x0,y0,60);/* Make 0 time normal size le

9、tters */settextstyle(DEFAULT_FONT,HORIZ_DIR,0);outtextxy(10,170,press a key);getch();setfillstyle(HATCH_FILL,YELLOW);floodfill(202,100,WHITE);getch();for(k=0;k=500;k+) setcolor(3); for(i=0;i=16;i+) a=(2*PAI/16)*i+(2*PAI/180)*k; x=ceil(x0+48*cos(a); y=ceil(y0+48+sin(a)*B); setcolor(2); line(x0,y0,x,y

10、); for(j=1;j=50;j+) a=(2*PAI/16)*i+(2*PAI/180)*k-1; x=ceil(x0+48*cos(a); y=ceil(y0+48*sin(a)*B); line(x0,y0,x,y); restorecrtmode();-【程序60】题目：画图，综合例子。1.程序分析：2.程序源代码：#include graphics.h#define LEFT 0#define TOP 0#define RIGHT 639#define BOTTOM 479#define LINES 400#define MAXCOLOR 15main()int driver,mo

11、de,error;int x1,y1;int x2,y2;int dx1,dy1,dx2,dy2,i=1;int count=0;int color=0;driver=VGA;mode=VGAHI;initgraph(&driver,&mode,);x1=x2=y1=y2=10;dx1=dy1=2;dx2=dy2=3;while(!kbhit() line(x1,y1,x2,y2); x1+=dx1;y1+=dy1; x2+=dx2;y2+dy2; if(x1=RIGHT) dx1=-dx1; if(y1=BOTTOM) dy1=-dy1; if(x2=RIGHT) dx2=-dx2; if(

12、y2=BOTTOM) dy2=-dy2; if(+countLINES) setcolor(color); color=(color=MAXCOLOR)?0:+color; closegraph(); 【程序61】题目：打印出杨辉三角形（要求打印出10行如下图）1.程序分析： 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10105 1 2.程序源代码：main()int i,j;int a1010;printf(n);for(i=0;i10;i+) ai0=1; aii=1;for(i=2;i10;i+) for(j=1;ji;j+) aij=ai-1j-1+ai-1j

13、;for(i=0;i10;i+) for(j=0;j=i;j+) printf(%5d,aij); printf(n); -【程序62】题目：学习putpixel画点。1.程序分析：2.程序源代码：#include stdio.h#include graphics.hmain()int i,j,driver=VGA,mode=VGAHI;initgraph(&driver,&mode,);setbkcolor(YELLOW);for(i=50;i=230;i+=20) for(j=50;j=230;j+) putpixel(i,j,1);for(j=50;j=230;j+=20) for(i=

14、50;i=230;i+) putpixel(i,j,1);-【程序63】题目：画椭圆ellipse1.程序分析：2.程序源代码：#include stdio.h#include graphics.h#include conio.hmain()int x=360,y=160,driver=VGA,mode=VGAHI;int num=20,i;int top,bottom;initgraph(&driver,&mode,);top=y-30;bottom=y-30;for(i=0;inum;i+)ellipse(250,250,0,360,top,bottom);top-=5;bottom+=5

15、;getch();-【程序64】题目：利用ellipse and rectangle 画图。1.程序分析：2.程序源代码：#include stdio.h#include graphics.h#include conio.hmain()int driver=VGA,mode=VGAHI;int i,num=15,top=50;int left=20,right=50;initgraph(&driver,&mode,);for(i=0;inum;i+)ellipse(250,250,0,360,right,left);ellipse(250,250,0,360,20,top);rectangle

16、(20-2*i,20-2*i,10*(i+2),10*(i+2);right+=5;left+=5;top+=10;getch();-【程序65】题目：一个最优美的图案。1.程序分析：2.程序源代码：#include graphics.h#include math.h#include dos.h#include conio.h#include stdlib.h#include stdio.h#include stdarg.h#define MAXPTS 15#define PI 3.1415926struct PTS int x,y;double AspectRatio=0.85;void L

17、ineToDemo(void)struct viewporttype vp;struct PTS pointsMAXPTS;int i, j, h, w, xcenter, ycenter;int radius, angle, step;double rads;printf( MoveTo / LineTo Demonstration );getviewsettings( &vp );h = vp.bottom - vp.top;w = vp.right - vp.left;xcenter = w / 2; /* Determine the center of circle */ycenter

18、 = h / 2;radius = (h - 30) / (AspectRatio * 2);step = 360 / MAXPTS; /* Determine # of increments */angle = 0; /* Begin at zero degrees */for( i=0 ; iMAXPTS ; +i ) /* Determine circle intercepts */rads = (double)angle * PI / 180.0; /* Convert angle to radians */pointsi.x = xcenter + (int)( cos(rads)

19、* radius );pointsi.y = ycenter - (int)( sin(rads) * radius * AspectRatio );angle += step; /* Move to next increment */circle( xcenter, ycenter, radius ); /* Draw bounding circle */for( i=0 ; iMAXPTS ; +i ) /* Draw the cords to the circle */for( j=i ; jn2) swap(pointer1,pointer2);if(n1n3) swap(pointe

20、r1,pointer3);if(n2n3) swap(pointer2,pointer3);printf(the sorted numbers are:%d,%d,%dn,n1,n2,n3);swap(p1,p2)int *p1,*p2;int p;p=*p1;*p1=*p2;*p2=p;-【程序67】题目：输入数组，最大的与第一个元素交换，最小的与最后一个元素交换，输出数组。1.程序分析：谭浩强的书中答案有问题。2.程序源代码：main()int number10;input(number);max_min(number);output(number);input(number)int nu

21、mber10;int i;for(i=0;i9;i+) scanf(%d,&numberi); scanf(%d,&number9);max_min(array)int array10;int *max,*min,k,l;int *p,*arr_end;arr_end=array+10;max=min=array;for(p=array+1;p*max) max=p; else if(*p*min) min=p; k=*max; l=*min; *p=array0;array0=l;l=*p; *p=array9;array9=k;k=*p; return;output(array)int array10; int *p;for(p=array;parray+9;p+) printf(%d,*p);printf(%dn,array9);-【程序68】题目：有n个整数，使其前面各数顺序向后移m个位置，最后m个数变成最前面的m个数1.程序分析：2.程序源代码：main()int number20,n,m,i;printf(the total numbers is:);scanf(%d,&n);printf(back m:);scanf(%d,&m);for(i=0;in-1;i+) scanf(%d,&numberi);scanf(%d,&numbern-1);mov