求“一箭穿心” C语言编程

其实这个程序也不难，只要用心形线悄者此方程（1-ρ×cosθ)来绘图就可以了。嫌悉程序如下：

#include<graphics.h>

#include<math.h>

#define FNX(x1) (int)(x+(x1)*sl)

#define FNY(y1) (int)(MAXY-(y+(y1)*sl))

#define R(theta) 1-pow(cos(1*theta),1)

int sl=50,MAXY

float x1,y1,xs,ys,r,theta

/*画心*/

void draw(int x,int y)

{

for(theta=0theta<2*3.14theta+=0.01)

{r=R(theta)

x1=r*sin(theta)y1=r*cos(theta)

xs=FNX(x1)ys=FNY(y1)

if(theta==0)moveto(xs,ys)else lineto(xs,ys)

}

}

/启迅*画箭头*/

void jian()

{

line(205,235,200,240)

line(200,240,205,245)

line(200,240,214,240)

line(400,240,500,240)

line(485,235,480,240)

line(480,240,485,245)

line(490,235,485,240)

line(485,240,490,245)

line(495,235,490,240)

line(490,240,495,245)

line(500,235,495,240)

line(495,240,500,245)

line(505,235,500,240)

line(500,240,505,245)

line(485,235,505,235)

line(485,245,505,245)

}

/*主程序*/

main()

{int driver=DETECT,mode

initgraph(&driver,&mode,"")

setbkcolor(1)

setcolor(4)

MAXY=getmaxy()

draw(280,280)/*左心*/

draw(370,280)/*右心*/

jian()

getch()

closegraph()

}

兄弟，你程序的for语句的变量递增是不是用得太不恰当了，应该说循环次数为纳樱慧整数类型，你把

for(theta=0theta<2*3.14theta+=0.01)

此句先定义theta为整型再改为

for(theta=0theta<洞答2*314theta++)

另外你的程序风格我不太喜欢，有个很好的建议：用PUTPIXCEL和GETPIXCEL函数将计算得出的座标数组动态输出。另外建议你有空的话多上网“道”点国外颂含名家的C程序看看。

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