C语言编程 绘制曲线，会的来

Question

设计要求：绘制一个两维坐标，在坐标轴上标出刻度，并用阿拉伯数字标出。根据所给点的坐标，用线段将每个点连接起来，构成一条曲线。为简单起见，曲线的数据由一位数组给定，数组元素为8个，数组如下：int data[]={7,3,12,6,9,5,8,11};涉及知识：C的图形工作模式。相关函数：initgraph()；closegraph();

Answer 1

TC下调试通过

/*******************************************************
*Author :Wacs5
*Date :20090105(YYYY-MM-DD)
*Function :画简易的曲线图 *********************************************************/
#include <stdio.h>
#include <conio.h>
#include <stdlib.h>
#include <math.h>
#include <Graphics.h>

#define n 8

int main()
{
int i,j;
float data[]={7,3,12,6,9,5,8,11};
char str[40];

int gdrive=DETECT,gmode,errorcode;
int maxx,maxy;
int perx,pery;
int x0,x1,y0,y1;
int coloraxis=2,colorline=3;
float mindata,maxdata,ndata;

initgraph(&gdrive,&gmode,""); /*初始化设备*/
if ((errorcode=graphresult())!=grOk) /*查错*/
{
printf("Graphics error:%s\nPress any key to exit:",grapherrormsg(errorcode));
getch();
exit(1);
}
maxx=getmaxx();
maxy=getmaxy();
mindata=maxdata=data[0];
for (i=1;i<n;i++)
{
if (mindata>data[i])
mindata=data[i];
if (maxdata<data[i])
maxdata=data[i];
}
mindata=floor(mindata);
maxdata=ceil(maxdata);

perx=maxx/(n+4);
pery=maxy/(maxdata-mindata+4);
x0=2*perx;
y0=maxy-2*pery;

x1=maxx-2*perx;
y1=2*pery;
setcolor(coloraxis);
line(x0,y0,x1,y0);
line(x0,y0,x0,y1);

line(x1,y0,x1-4,y0+3);
line(x1,y0,x1-4,y0-3);
line(x0,y1,x0+3,y1+4);
line(x0,y1,x0-3,y1+4);

settextjustify(CENTER_TEXT,TOP_TEXT);
for (i=0;i<n;i+=n/3)
{
j=x0+i*perx;
line(j,y0,j,y0+2); /*刻度线*/
sprintf(str,"%d",i);
outtextxy(j,y0+4,str);
}

settextjustify(RIGHT_TEXT,CENTER_TEXT);
for (i=(maxdata-mindata)/3;i<=maxdata-mindata;i+=(maxdata-mindata)/3)
{
j=y0-i*pery;
line(x0,j,x0-2,j); /*刻度线*/
sprintf(str,"%d",(int)mindata+i);
outtextxy(x0-4,j,str);
}

setcolor(colorline);
x1=x0+perx;
y1=y0-(data[0]-mindata)*pery;
circle(x1,y1,2);
moveto(x1,y1);
i=1;
do
{
x1+=perx;
y1=y0-(data[i]-mindata)*pery;
lineto(x1,y1);
circle(x1,y1,2);
moveto(x1,y1);
i++;
}while(i<n);

getch();
closegraph();
return 0;
}