#include?windows.h
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LRESULT?CALLBACK?WndProc(HWND,?UINT,?WPARAM,?LPARAM);
int?WINAPI?WinMain(HINSTANCE?hInstance,?HINSTANCE?hPrevInstance,
PSTR?szCmdLine,?int?iCmdShow)
{
static?TCHAR?szAppName[]=TEXT("二次函數(shù)");
HWND?????????hwnd;
MSG??????????msg;
WNDCLASS?????wndclass;
wndclass.style=CS_HREDRAW|CS_VREDRAW;
wndclass.lpfnWndProc=WndProc;
wndclass.cbClsExtra=0;
wndclass.cbWndExtra=0;
wndclass.hInstance=hInstance;
wndclass.hIcon=LoadIcon(NULL,?IDI_APPLICATION);
wndclass.hCursor=LoadCursor(NULL,?IDC_ARROW);
wndclass.hbrBackground=(HBRUSH)GetStockObject(WHITE_BRUSH);
wndclass.lpszMenuName=NULL;
wndclass.lpszClassName=szAppName;
if?(!RegisterClass(wndclass))
{
MessageBox(NULL,?TEXT("Error"),
szAppName,?MB_ICONERROR);
return?0;
}
hwnd=CreateWindow(szAppName,?TEXT("二次函數(shù)"),
WS_OVERLAPPEDWINDOW,
CW_USEDEFAULT,?CW_USEDEFAULT,
CW_USEDEFAULT,?CW_USEDEFAULT,
NULL,?NULL,?hInstance,?NULL);
ShowWindow(hwnd,?iCmdShow);
UpdateWindow(hwnd);
while?(GetMessage(msg,?NULL,?0,?0))
{
TranslateMessage(msg);
DispatchMessage(msg);
}
return?msg.wParam;
}
LRESULT?CALLBACK?WndProc(HWND?hwnd,?UINT?message,?WPARAM?wParam,?LPARAM?lParam)
{
static?int??cxClient,?cyClient;
const?static?int?n=1000;
HDC?????????hdc;
int?????????i;
PAINTSTRUCT?ps;
POINT???????apt[n];
switch?(message)
{
case?WM_SIZE:
cxClient=LOWORD(lParam);
cyClient=HIWORD(lParam);
return?0;
case?WM_PAINT:
hdc=BeginPaint(hwnd,?ps);
MoveToEx(hdc,?0,?cyClient/2,?NULL);
LineTo(hdc,?cxClient,?cyClient/2);
MoveToEx(hdc,?cxClient/2,?0,?NULL);
LineTo(hdc,?cxClient/2,?cyClient);
for?(i=0;?i??n;++i)
{
apt[i].x=cxClient/4+i; apt[i].y=cyClient-(cyClient/2-i)*(cyClient/2-i)/300-cyClient/2+100;
}
Polyline(hdc,?apt,?n);
return?0;
case?WM_DESTROY:
PostQuitMessage(0);
return?0;
}
return?DefWindowProc(hwnd,?message,?wParam,?lParam);
}
用GDI繪圖吧,比較簡(jiǎn)單。繪圖的思想是讓x以固定的值在區(qū)間內(nèi)持續(xù)增長(zhǎng),比如x=0.1,0.2,0.3.....,以計(jì)算出的y值來確定y坐標(biāo)。用線連接所有的點(diǎn)就行了。MoveTo(),LineTo()函數(shù)你用得著,具體情況請(qǐng)自行查看MSDN。
和數(shù)學(xué)上一樣啊 來個(gè)坐標(biāo)x,y 分別表示列和行 。在函數(shù)上就打印一個(gè)* 給個(gè)范圍,雙層循環(huán)加條件就可以了
我畫了半個(gè)正弦函數(shù),其余的你可以照著畫,歡迎討論,
#include "stdio.h"
#include "conio.h"
#include "math.h"
#include "graphics.h"
void main()
{
int driver,mode,i;
driver=DETECT;
mode=0;
initgraph(driver,mode,"");
setcolor(15);
for(i=0;i360;i++)
line(i,200*sin(i*3.14/360),i+1,200*sin((i+1)*3.14/360));
getch();
restorecrtmode();
}
你要什么圖像 sin 還cos 要不我都給你把!
1、#includestdio.h
#includemath.h
void main()
{
double y;
int i,n;
for(y=1;y=0;y-=0.1)
{n=asin(y)*10;
for(i=1;i=n;i++)
printf(" ");
printf("*");
for(;i=31-n;i++)
printf(" ");
printf("*\n");}
for(y=0;y=1;y+=0.1)
{n=asin(y)*10;
for(i=-1;i=31+n;i++)
printf(" ");
printf("*");
for(;i=62-n;i++)
printf(" ");
printf("*\n");}
}
2、#includestdio.h
#includemath.h
void main()
{
double y;
int x,m;
for(y=1;y=-1;y-=0.1)
{m=acos(y)*10;
for(x=1;xm;x++)
printf(" ");
printf("*");
for(;x62-m;x++)
printf(" ");
printf("*\n");}
}
之后在給你個(gè)連個(gè)圖像相交的把
3、#includestdio.h
#includemath.h
void main()
{
double y;
int n,m,i,j,x,yy;
for(yy=0;yy=20;yy++)
{ y=0.1*yy;
m=acos(1-y)*10;
n=asin(1-y)*10;
i=32+asin(y-1)*10;
j=61-asin(y-1)*10;
for(x=0;x62;x++)
{if((x==n)(x==m)) printf("+");
else if((x==n)||(x==i)||(x==j)) printf("+");
else if((x==m)||(x==62-m)) printf("*");
else printf(" ");}
printf("\n");
}
}
需要Window圖像編程,Window編程就是基于C語言的,在Visual C++6.0可以編寫Win32程序
文章標(biāo)題:c語言函數(shù)圖像 c語言函數(shù)圖像繪制器代碼
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