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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<10; 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"); } } 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(%26amp;driver, %26amp;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); } 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" /* 欢迎访问 c++builder研究 - www.ccrun.com */ main() { int x = 360, y = 160, driver = vga, mode = vgahi; int num=20, i; int top, bottom; initgraph(%26amp;driver, %26amp;mode, ""); top = y - 30; bottom = y - 30; for(i=0; i ellipse(250, 250, 0, 360, top, bottom); top -= 5; bottom += 5; } getch(); } 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(%26amp;driver, %26amp;mode, ""); for(i=0; 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(); } 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; }; /* 63 63 72 75 6e 2e 63 6f 6d */ 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( %26amp;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 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 for(j=i; j 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; initgraph(%26amp;driver, %26amp;mode, ""); setcolor(3); setbkcolor(green); linetodemo(); }  本文共2页: 第 [1] [2] 页
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