发布时间 : 星期六 文章C璇█绋嬪簭璁捐100涓粡鍏镐緥瀛?- 鐧惧害鏂囧簱更新完毕开始阅读fe8ac3e242323968011ca300a6c30c225801f001
#include \ #define maxpts 15
#define pi struct pts { int x,y; };
double aspectratio=; void linetodemo(void)
{
struct viewporttype vp; struct pts points[maxpts]; int i, j, h, w, xcenter, ycenter;
int radius, angle, step;
double rads;
printf(\
getviewsettings( &vp );
h = - ; w = - ;
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 ; irads = (double)angle * pi / ; /* 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 ; ifor( j=i ; jmoveto(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(&driver,&mode,\
setcolor(3); setbkcolor(green); linetodemo();}
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【程序66】
题目：输入3个数a,b,c，按大小顺序输出。
1.程序分析：利用指针方法。
2.程序源代码： /*pointer*/ main() {
int n1,n2,n3;
int *pointer1,*pointer2,*pointer3; printf(\
scanf(\
pointer1=&n1; pointer2=&n2; pointer3=&n3;
if(n1>n2) swap(pointer1,pointer2); if(n1>n3) swap(pointer1,pointer3); if(n2>n3) swap(pointer2,pointer3);
printf(\
} swap(p1,p2) int *p1,*p2; {int p;
p=*p1;*p1=*p2;*p2=p;
}
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【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 number[10]; input(number); max_min(number); output(number);
} input(number) int number[10];
{int i; for(i=0;i<9;i++) scanf(\ scanf(\
}
max_min(array) int array[10]; {int *max,*min,k,l; int *p,*arr_end; arr_end=array+10; max=min=array;
for(p=array+1;p if(*p>*max) max=p;
else if(*p<*min) min=p;
k=*max; l=*min;
*p=array[0];array[0]=l;l=*p;