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经典C语言程序设计100例61-70

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【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  10 10 5  1 
2.程序源代码:
main()
{int i,j;
int a[10][10];
printf("\n");
for(i=0;i<10;i++)
  {a[i][0]=1;
  a[i][i]=1;}
for(i=2;i<10;i++)
  for(j=1;j<i;j++)
  a[i][j]=a[i-1][j-1]+a[i-1][j];
for(i=0;i<10;i++)
  {for(j=0;j<=i;j++)
  printf("%5d",a[i][j]);
  printf("\n");
  }
}
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【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.h"
main()
{
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=50;i<=230;i++)
  putpixel(i,j,1);
}
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【程序63】
题目:画椭圆ellipse   
1.程序分析:
2.程序源代码:
#include "stdio.h"
#include "graphics.h"
#include "conio.h"
main()
{
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;i<num;i++)
{
ellipse(250,250,0,360,top,bottom);
top-=5;
bottom+=5;
}
getch();
}
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【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.h"
main()
{
int driver=VGA,mode=VGAHI;
int i,num=15,top=50;
int left=20,right=50;
initgraph(&driver,&mode,"");
for(i=0;i<num;i++)
{
ellipse(250,250,0,360,right,left);
ellipse(250,250,0,360,20,top);
rectangle(20-2*i,20-2*i,10*(i+2),10*(i+2));
right+=5;
left+=5;
top+=10;
}
getch();
}
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【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 3.1415926
struct PTS {
int x,y;
};
double AspectRatio=0.85;
void LineToDemo(void)
{
struct viewporttype vp;
struct PTS points[MAXPTS];
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 = h / 2;
radius = (h - 30) / (AspectRatio * 2);
step = 360 / MAXPTS; /* Determine # of increments */
angle = 0; /* Begin at zero degrees */
for( i=0 ; i<MAXPTS ; ++i ){ /* Determine circle intercepts */
rads = (double)angle * PI / 180.0; /* Convert angle to radians */
points[i].x = xcenter + (int)( cos(rads) * radius );
points[i].y = ycenter - (int)( sin(rads) * radius * AspectRatio );
angle += step; /* Move to next increment */
}
circle( xcenter, ycenter, radius ); /* Draw bounding circle */
for( i=0 ; i<MAXPTS ; ++i ){ /* Draw the cords to the circle */
for( j=i ; j<MAXPTS ; ++j ){ /* For each remaining intersect */
moveto(points[i].x, points[i].y); /* Move to beginning of cord */
lineto(points[j].x, points[j].y); /* Draw the cord */
} } }
main()
{int driver,mode;
driver=CGA;mode=CGAC0;

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